MODERN PHYSICS BUT MOSTLY QUANTUM MECHANICS

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1 MODERN PHYSICS BUT MOSTLY QUANTUM MECHANICS Seeing the invisible, Proving the impossible, Thinking the unthinkable Modern physics stretches your imagination beyond recognition, pushes the intellectual envelope to the limit, and takes you to the forefront of human knowledge. Masayasu AOTANI ( 青谷正妥 ) Kyoto University Kyoto, Japan Anax parthenope ( ギンヤンマ ) 013 No insects, no life!

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3 Modern Physics Masayasu AOTANI( 青谷正妥 ) Spring 013

4 Copyright c Masayasu AOTANI Permission is granted for personal use only.

5 Contents 1 Classical Physics vs. Modern Physics 13 Mathematical Preliminaries 15.1 Linear Vector Spaces Inner Product Spaces L -Space The Braket Notation Vector Subspace Linear Operators Matrix Representation of Linear Operators Matrix Representations of Operator Products The Adjoint of an Operator Eigenvalues and Eigenvectors Special Types of Operators Hermitian Operators Simultaneous Diagonalization Active and Passive Transformations The Postulates of Quantum Mechanics The Fundamental Postulates Unitary Time Evolution Time-Independent Hamiltonian Propagator as a Power Series of H Eigenstate Expansion of the Propagator The Uncertainty Principle Uncertainty and Non-Commutation Position and Momentum Energy and Time

6 Time rate of change of expectation values Time-energy uncertainty Classical Mechanics and Quantum Mechanics Ehrenfest s Theorem Exact Measurements and Expectation Values Spaces of Infinite Dimensionality Two Types of Infinity Countably Infinite Dimensions Uncountably Infinite Dimensions Delta Function as a Limit of Gaussian Distribution Delta Function and Fourier Transform f as a Vector with Uncountably Many Components Finite Dimensions Countably Infinite Dimensions Uncountably Infinite Dimensions Hermiticity of the Momentum Operator Checking P ij = Pji Hermiticity Condition in Uncountably Infinite Dimensions The Eigenvalue Problem of the Momentum Operator P Relations Between X and P The Fourier Transform Connecting X and P X and P in the X Basis Basis-Independent Descriptions in Reference to the X Basis X and P in the P Basis The Commutator X, P Schrödinger s Theory of Quantum Mechanics How was it derived? Born s Interpretation of Wavefunctions Expectation Values The Time-Independent Schrödinger Equation 11

7 7 Solutions of Time-Independent Schrödinger Equations in One Dimension The Zero Potential The Step Potential (E < V 0 ) The Step Potential (E > V 0 ) The Barrier Potential (E < V 0 ) The Infinite Square Well Potential The Simple Harmonic Oscillator Potential Classic Solution Raising and Lowering Operators Actions of a and a Higher Spatial Dimensions Degeneracy Angular Momentum Angular Momentum Angular Momentum Operators Quantum Mechanical Rotation Operator U R Rotationally Invariant Hamiltonian Raising and Lowering Operators: L + and L Generalized Angular Momentum J The Hydrogen Atom Particles Instead of Three-Dimensional System The Solutions for Φ The Solutions for Θ Associated Legendre Polynomials and Spherical Harmonics L, L z, and the Spherical Harmonics The Radial Function R Hydrogen-Like Atoms Simultaneous Diagonalization of H, L, and L z Revisiting the Fundamental Postulates Electron Spin What is the Electron Spin? Stern-Gerlach Experiment

8 11.1. Fine Structure of the Hydrogen Spectrum: Spin-Orbit Interaction Spin and Pauli Matrices Sequential Measurements Molecular Rotation Rotational Kinetic Energy Moment of Inertia for a Diatomic Molecule Two-Dimensional Rotation Confined to the x, y-plane Three-Dimensional Rotation Appendix A Matrix Algebra 57 A.1 Invertible Matrices A. Rank of a Matrix A.3 Change of Basis A.4 Diagonalizability and Simultaneous Diagonalization Appendix B Holomorphic Functional Calculus 77 Appendix C Wronskian and Linear Independence 79 Appendix D Newtonian, Lagrangian, and Hamiltonian Mechanics 83 D.1 An Overview D. Newtonian Mechanics D.3 Lagrangian Mechanics D.4 Hamiltonian Mechanics Appendix E Normalization Schemes for a Free Particle 87 E.1 The Born Normalization E. The Dirac Normalization E..1 Fourier Transform and the Delta Function E.. Wavefunctions as Momentum or Position Eigenfunctions88 E.3 The Unit-Flux Normalization Appendix F Symmetries and Conserved Dynamical Variables 93 F.1 Poisson Brackets and Constants of Motion F. l z As a Generator F.3 Rotation Around The Origin

9 F.4 Infinitesimal Rotations Around the z-axis Appendix G Commutators 99 G.1 Commutator Identities G. Commutators involving X and P G.3 Commutators involving X, P, L, and r Appendix H Probability Current 303 Appendix I Chain Rules in Partial Differentiation 309 I.1 One Independent Variable I. Three Independent Variables I.3 Reciprocal Relation is False I.4 Inverse Function Theorems Appendix J Laplacian Operator in Spherical Coordinates 313 Appendix K Legendre and Associated Legendre Polynomials 35 K.1 Legendre Polynomials K. Associated Legendre Polynomials Appendix L Laguerre and Associated Laguerre Polynomials 333 Bibliography 336 Answers to Exercises 337 Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Subject Index 359

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11 List of Tables 3.1 Physical observables and corresponding quantum operators Quantum operators in three dimensions Spherical Harmonics for l = 0, 1, and

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13 List of Figures 10.1 Spherical Coordinate System (Source: Wikimedia Commons; Author: Andeggs) Spherical Coordinate System Used in Mathematics (Source: Wikimedia Commons; Author: Dmcq) Volume Element in Spherical Coordinates (Source: Victor J. Montemayor, Middle Tennessee State University) Stern-Gerlach Experiment (Diagram drawn by en wikipedia Theresa Knott.) Results of Stern-Gerlach Experiment with and without a nonuniform external magnetic field (Source: Stern and Gerlach s original paper Gerlach and Stern, 19, p.350)

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15 Chapter 1 Classical Physics vs. Modern Physics Classical Physics Newtonian Mechanics Maxwell s Electricity and Magnetism Modern Physics Relativity Quantum Mechanics Newtonian mechanics no longer held under extreme conditions. Two Types of Extreme Circumstances 1. Microscopic Phenomena (atomic and subatomic) Tunneling, Wave-particle duality. At High Speeds (near the speed of light denoted by c) Fast moving particles live longer. Gallilean transformation does not work. Historical Events Discoveries and Theories 1. Black Body Radiation: (Quantization, Planck s Postulate) 13

16 14 CHAPTER 1. CLASSICAL PHYSICS VS. MODERN PHYSICS. Particle-Like Properties of Radiation: (Wave-Particle Duality) 3. Einstein s Quantum Theory: A bundle of energy is localized in a small volume of space. 4. Atomic spectra indicating discrete energy levels E 1, E, E 3,... as opposed to continuous distribution of energy levels E 0, ) 1 1 We will later see that E = 0 is impossible in some quantum mechanical systems such as a simple harmonic oscillator.

17 Chapter Mathematical Preliminaries This chapter is a summary of the mathematical tools used in quantum mechanics. It covers various important aspects of basic linear algebra, defines linear operators, and sets convenient notations for the rest of the book..1 Linear Vector Spaces Consider the three dimensional Cartesian space R 3 equipped with the x-, y-, and z-axes. We have three unit vectors, perpendicular/orthogonal to one another, denoted by (i, j, k), (î, ˆȷ, ˆk), or (ˆx, ŷ, ẑ). In the component notation, we have i = (1, 0, 0), j = (0, 1, 0) and k = (0, 0, 1). Any point p in this space identified by three coordinates x, y, and z such that p = (x, y, z) can also be regarded as a vector xi + yj + zk or (x, y, z) in the component notation. Note that we use (x, y, z) both for the coordinates and components by abuse of notation. You must be familiar with such arithmetic operations as addition, subtraction, and scalar multiplication for vectors; which are basically componentwise operations a. This is a canonical example of a linear vector space, and you should keep this example in mind as you study the following definition of a linear vector space. If you are not so inclined, you can safely skip this definition and still can understand the rest of this book 15

18 16 CHAPTER. MATHEMATICAL PRELIMINARIES fully. a To be more precise, these operations can be done component-wise though there are other equivalent definitions/formulations for these arithmetic operations. Definition.1 A linear vector space V is a set {v 1, v, v 3,...} of objects called vectors for which addition and scalar multiplication are defined, such that 1. Both addition and scalar multiplication generate another member of V. This property is referred to as closure under addition and scalar multiplication.. Addition and scalar multiplication obey the following axioms. Axioms for Addition: Consider v i, v j, and v k taken from V. (i) v i + v j = v j + v i (commutativity) (ii) v i + (v j + v k ) = (v i + v j ) + v k (associativity) (iii) There exists a unique null vector, denoted by 0, in V such that 0 + v i = v i + 0 = v i. (existence of the identity element) (iv) For each v i, there exists a unique inverse ( v i ) in V such that v i + ( v i ) = 0. (existence of the inverse) 1 Axioms for Scalar Multiplication: Consider arbitrary vectors v i, v j and arbitrary scalars α, β. (v) α(v i + v j ) = αv i + αv j (vi) (α + β)v i = αv i + βv i (vii) α(βv i ) = (αβ)v i Fact.1 The following facts can be proved using the axioms. 1. 0v = 0. α0 = 0 1 The axioms (i) through (iv) means that a linear vector space form an abelian group under addition.

19 .1. LINEAR VECTOR SPACES ( 1)v = ( v) Definition. If the allowed values for the scalars {α, β, γ,...} come from some field F, we say the linear vector space V is defined over the field F. In particular, if F is the field of real numbers R, V is a real vector space. Likewise, if F is the field of complex numbers C, V is a complex vector space. Clearly, R 3 is not a complex vector space, but a real vector space. Complex vector spaces are the important vector spaces in quantum mechanics. Definition.3 A set of vectors {v 1, v,..., v n } is linearly independent ( LI) if n α i v i = 0 = α 1 = α =... = α n = 0. i=1 Definition.3 is equivalent to saying that no vector in {v 1, v,..., v n } can be expressed as a linear combination of the other vectors in the set. Vectors i, j, and k are linearly independent because ai + bj + ck = 0 or (a, b, c) = (0, 0, 0) = a = b = c = 0. Generally speaking, vectors in R 3 that are perpendicular to one another are linearly independent. However, these are not the only examples of a linearly independent set of vectors. To give a trivial example, vectors i and i + j are linearly independent, though they are not perpendicular to each other. This is because ai + b(i + j) = 0 = (a + b, b) = (0, 0) = a = b = 0. Definition.4 A vector space is n-dimensional if it has at most n vectors that are linearly independent.

20 18 CHAPTER. MATHEMATICAL PRELIMINARIES Notation: An n-dimensional vector space over a field F is denoted by V n (F). The ones we encounter in this course are usually members of V n (C), including the cases where n =. It is often clear what the field F is. In this case, we simply write V n. Theorem.1 Suppose {v 1, v,..., v n } are linearly independent vectors in an n-dimensional vector space V n. Then, any vector v in V n can be written as a linear combination of {v 1, v,..., v n }. Proof As V n is n-dimensional, you can find a set of n+1 scalars {α 1, α,..., α n, α n+1 }, not all zero, such that ( n ) α i v i + α n+1 v = 0. i=1 Else, {v 1, v,..., v n, v} are linearly independent, and V n is at least n + 1- dimensional, which is a clear contradiction. Furthermore, α n+1 0. If not, at least one of {α 1, α,..., α n } is not 0, and yet n i=1 α i v i = 0, contradicting the assumption that {v 1, v,..., v n } are linearly independent. We now have v = n i=1 α i α n+1 v i. (.1) We will next show that there is only one way to express v as a linear combination of {v 1, v,..., v n } in Theorem.1. Theorem. The coefficients of Equation.1 are unique. Proof Consider any linear expression of v in terms of {v 1, v,..., v n } Subtracting Equation. from Equation.1, 0 = n v = β i v i. (.) i=1 n i=1 ( ) αi β i v i. α n+1

21 .. INNER PRODUCT SPACES 19 However, we know {v 1, v,..., v n } are linearly independent. Hence, n i=1 ( ) αi β i v i = 0 = α i β i = 0 or β i = α i for all i = 1,,..., n. α n+1 α n+1 α n+1 In order to understand Definition.5 below, just imagine how any vector in R 3 can be expressed as a unique linear combination of i, j, and k. Definition.5 A set of linearly independent vectors {v 1, v,..., v n }, which can express any vector v in V n as a linear combination (Equation.), is called a basis that spans V n. The coefficients of the linear combination β i are called the components of v in the basis {v 1, v,..., v n }. Turning the expression inside out, we also say that V n is the vector space spanned by the basis {v 1, v,..., v n } a. Once we pick a basis {v 1, v,..., v n }, and express v as a linear combination of the basis vectors v = n i=1 β i v i, we get a component notation of v as follows. n v = β i v i = (β 1, β,..., β n ). i=1 With this notation, vector additions and scalar multiplications can be done component by component. For vectors v = (β 1, β,..., β n ), w = (γ 1, γ,..., γ n ) and a scalar α, we simply have v +w = (β 1 +γ 1, β +γ,..., β n +γ n ) and αv = (αβ 1, αβ,..., αβ n ). a This means that we can define V n as the collection of all objects of the form n i=1 α iv i.. Inner Product Spaces This is nothing but a generalization of the familiar scalar product for V 3 (R) a. Namely, it is an extension of the inner product defined on R 3 to complex vectors. Readers should be familiar with the component-wise definition of the usual inner product on R 3, also called dot product, such that a b = (a 1, a, a 3 ) (b 1, b, b 3 ) = a 1 b 1 + a b + a 3 b 3. We can express

22 0 CHAPTER. MATHEMATICAL PRELIMINARIES this as a matrix product of a row vector a and a column vector b. a1 a a 3 b 1 b b 3 = a 1 b 1 + a b + a 3 b 3 ; where the one-by-one matrix on the right-hand side is a scalar by definition. This definition of inner product is extended to include complex components by taking the complex conjugate of the components of a. So, the extended inner product, which we simply call an inner product in the rest of the book, is given by a1 a a 3 b 1 b b 3 = a 1 a a 3 b 1 b b 3 = a 1b 1 + a b + a 3b 3. Definition.6 below is an abstract version of the familiar inner product defined on R 3. This definition contains the inner product defined on R 3, but not all inner products are of that type. Nevertheless, you should keep the inner product on R 3 in mind as you read the following general definition of inner product. In Definition.6, we are writing a as a and b as b so that a b = a b b in anticipation of Dirac s braket notation to be introduced on p.7. a This is the notation suggested on p.18, which is the same as R 3. b Naively, a b = a b, but we use only one vertical line and write a b. Definition.6 (Inner Product) An inner product of v i and v j, denoted by v i v j, is a function of two vectors mapping onto C ( : V n V n C) satisfying the axioms below. (i) v i v i 0 with the equality holding if and only if v i = 0 (ii) v i v j = v j v i, where denotes the complex conjugate (iii) v i αv j + βv k = α v i v j + β v i v k argument) (linearity in the second It is convenient to add the fourth property to this list, though it is not an axiom but a consequence of Axioms (i) and (iii).

23 .. INNER PRODUCT SPACES 1 (iv) αv i + βv j v k = α v i v k + β v j v k argument) (anti-linearity in the first We say the inner product is linear in the second vector and antilinear in the first vector. In order to see how (iv) follows from (ii) and (iii), see the derivation below. αv i + βv j v k = v k αv i + βv j by (ii) = (α v k v i + β v k v j ) by (iii) A moment s = α v i v k + β v j v k by (ii) thought would tell us that (iii) and (iv) can be extended to the following property. n n n α i v i β j v j = αi β j v i v j i=1 j=1 i,j=1 Definition.7 (Inner Product Space) A vector space on which an inner product is defined is called an inner product space. Our canonical example R 3 is naturally an inner product space. Definition.8 (Norm) The norm of a vector v, denoted by either v or v, is given by v v 1/. A vector is normalized if its norm is unity. We call such a vector a unit vector. Definition.9 (Metric Space) A metric space is a set where a notion of distance (called a metric ) between elements of the set is defined. In particular, a norm is a metric. There is a precise notion of a metric which is a real-valued function with two elements of a set as the arguments. Let us use the canonical notation d(x, y). Then, a metric d should satisfy the following conditions. 1. d(x, y) 0. d(x, y) = 0 if and only if x = y 3. d(x, y) = d(y, x) 4. d(x, z) d(x, y) + d(y, z) The first condition is redundant as d(x, y) = 1 d(x, y) + d(y, x) 1 d(x, x) = 0.

24 CHAPTER. MATHEMATICAL PRELIMINARIES Definition.10 (Complete Metric Space) A metric space M is called complete if every Cauchy sequence of points in M converges in M; that is, if every Cauchy sequence in M has a limit that is also in M. Definition.11 (Hilbert Space) An inner product space which is complete with respect to the norm induced by the inner product is called a Hilbert space. R 3 is also a Hilbert space. Definition.1 (Orthogonality) Two vectors u and w are orthogonal to each other if u w = 0. Note here that in the familiar -dimensional and 3-dimensional cases, namely V (R) and V 3 (R), the norm is nothing but the length of the vector, and orthogonality means the vectors are geometrically perpendicular to each other. Definition.13 (Orthonormal Set) A set of vectors {v 1, v,..., v n } form an orthonormal set if v i v j = δ ij ; where the symbol δ ij is known as Kronecker s delta with the following straightforward definition. δ ij = { 1 if i = j 0 if i j Definition.14 (Completeness) a {v 1, v,..., v n } is complete if An orthonormal set of vectors n v i v i = I. (.3) i=1 Alternatively, {v 1, v,..., v n } is complete if any vector v in the Hilbert space can be expressed as a linear combination of v 1, v,..., v n. That is, any v = n i=1 a i v i if and only if the set {v 1, v,..., v n } satisfies ni=1 v i v i = I.

25 .. INNER PRODUCT SPACES 3 Proof Suppose n i=1 v i v i = I. Then, ( n ) n v = Iv = v i v i v = v i v v i. (.4) i=1 i=1 Now, suppose v = n j=1 a j v j for any v in the Hilbert space. Then, ( n ) ( n v i v i v = v i v i ) n n a j v j = v i v i a j v j i=1 i=1 j=1 i,j=1 n n n = a j v i v i v j = a j v i δ i,j = a j v j = v. (.5) i,j=1 i,j=1 j=1 Because (.5) holds for an arbitrary vector v, we have n i=1 v i v i = I by definition. Definition.15 (Orthonormal Basis) If a basis of a vector space V n forms an orthonormal set, it is called an orthonormal basis. b The unit vectors along the x-, y-, and z-axes, denoted by i, j, and k, form an orthonormal basis of R 3. a This is also known as the closure condition. b Note that a set of vectors which is orthonormal and complete forms an orthonormal basis. Here is a caveat. Note that the representation of a vector v as an n 1 matrix (column vector) or a 1 n matrix (row vector) can be in terms of any basis. However, an orthonormal basis simplifies computations of inner products significantly. Let u = (u 1, u,..., u n ) and w = (w 1, w,..., w n ) in the component notation with respect to some orthonormal basis {e 1, e,..., e n }. Then, n n n n n u w = u i e i w j e j = u i w j e i e j = u i w j δ ij = u i w i i=1 j=1 i,j=1 i,j=1 i=1

26 4 CHAPTER. MATHEMATICAL PRELIMINARIES = u 1w 1 +u w u n 1w n 1 +u nw n. 3 Theorem.3 (Schwarz Inequality) Any inner product satisfies the following inequality known as Schwarz Inequality. Proof Consider the vector v i v j v i v j Then, as v v 0, we get v j v i v j v i v j v i v j v i v j v j v = v i v j v i v j v j. = v i v i v j v i v i v j v j v i v j v i v j v j + v j v i v j v i v j v j = v v j 4 i v j v i v i v j v j v i v j v i + v j v i v j v i v j v j v j v j 4 = v i v j v i v i v j v j v j v i v j v i v j + v j v i v j v i v j = v i v j v i v i v j v j 0 = v i v j v i v j. We can also see that the equality holds only when v = v i v j v i v j v j = 0 or v i = v j v i v j v j. Now, let v i = αv j for some scalar α. Then, v j v i v j v j = v j αv j v j v j = v j v j αv j v j = v j αv j v j = αv j = v i. Therefore, the equality holds if and only if one vector is a scalar multiple of the other. 4 3 Don t make a beginner s mistake of applying this to vectors expressed in a basis that is not orthonormal. For example, consider the space spanned by{ e 1 and e} ; i.e. the x, y-plane if you may. While {e 1, e } form an orthonormal basis, e 1, e1+e is a basis composed of normalized vectors which are not mutually orthogonal. Let u = e 1 and w = e 1+e. Then, u = (1, 0) and w = (0, 1), and u 1w 1 + u w = = 0. But, ( ) u w = e 1 e1 +e = e 1 e 1 +e 1 e = This is not surprising as v j v i vj v j is the projection of v i along v j. Needless to say, if v i is parallel to v j, its projection in the direction of v j is v i itself.

27 .3. L -SPACE 5.3 L -Space It was mentioned on p.0 that there are other inner products than the familiar one on R 3. And, there is an associated Hilbert space for each of those inner products. One important class of Hilbert spaces for us is the L -space. It is a collection of functions with a common domain, for which an inner product is defined as an integral. Most of the Hilbert spaces we encounter in this course are L -spaces. In the following, we will specialize to R as the domain of our functions, but the same definitions and descriptions apply to a general domain a. In particular, + if the domain is R or R 3, we have a double integral,, or triple + integral,, respectively, rather than a single integral shown here. Definition.16 (L -Function) An L -function f is a square integrable function; that is, f satisfies + f(x) dx <. b The next step is to define an inner product on this set of square integrable functions. Definition.17 (L -Inner Product) Given two square integrable functions f and g, we define an inner product by + f(x) g(x) dx c ; where f(x) signifies the complex conjugate. With this inner product defined on a set of L -functions, we have a Hilbert space. As stated in the beginning of this section, the common domain of the functions can be any other measurable set such as R, R 3, 0, 1, and ( a, + a ) for a > 0. Using the notation given in Footnote b, we will denote these L -spaces by L (R ), L (R 3 ), L 0, 1, and L ( a, + a).

28 6 CHAPTER. MATHEMATICAL PRELIMINARIES The central equation of quantum mechanics, called Schrödinger Equation, is a differential equation, and its solutions are complex valued functions on a common domain. Therefore, the L -space is our canonical Hilbert space. a To be precise, the domain should be a measure space so that an integral can exist. However, such purely mathematical details can be omitted as all the domains we consider are measure spaces. b We often have an interval of finite length as the domain of f. The closed interval 0, 1 is a canonical example. In this case, we require 1 0 f(x) dx <. We use notations such as L (, + ) and L 0, 1 to make this distinction. c We can easily check that this integral satisfies Definition.6. Some examples of orthonormal bases for L -spaces are as follows. 1. The trigonometric system {e iπnx } + n= is an orthonormal basis for L 0, 1. The expansion of a function in this basis is called the Fourier series of that function.. The Legendre polynomials, which are obtained by taking the sequence of monomials {1, x, x, x 3,...} and applying the Gram-Schmidt orthogonalization process to it, form an orthonormal basis for L 1, 1. The first few normalized polynomials are as follows. p 1 (x) = 1/ p (x) = x 3/ p 3 (x) = ( 3x 1 ) 5/8 p 4 (x) = ( 5x 3 3x ) 7/ p 5 (x) = ( 35x 4 30x + 3 ) 9/ 8 p 6 (x) = ( 63x 5 70x x ) 11/ 8

29 .4. THE BRAKET NOTATION 7.4 The Braket Notation 5 This is a notation first used by Dirac. In fact, we have already been using it. You may have noticed that the inner product between v and w was denoted by v w, and not by the more familiar v, w. We will see why shortly. Without much ado, let us define a bra and a ket. On p.0, we have already encountered a = a and b = b such that the inner product between a and b is given by a b = a 1b 1 +a b +a 3b 3. Here, a is called bra a, and b is called ket b. More generally, recall the component notation of a vector v = (v 1, v,..., v n ). This usually means that we have picked a particular orthonormal basis 6 {e 1, e,..., e n } and expressed v as a linear combination of e 1, e,..., e n such that v = v 1 e 1 + v e v n e n. We can arrange the components either horizontally or vertically, but it is more convenient for us to arrange it vertically as an n by 1 matrix, often called a column vector. In quantum mechanics, we denote this column vector by v called ket v. v = v 1. v n Now, what is bra v denoted by v? It is defined as the row vector whose components are the complex conjugates of the components of v. v = v 1,..., v n In this notation, a basis vector e i has the following bra and ket. We sometimes denote e i and e i by i and i, respectively. 5 Note that this is bra + ket = braket, and NOT bracket. 6 Actually, it does not have to be an orthonormal basis, but any basis would do. However, we will specialize to an orthonormal basis here in order to explain why we used the notation u v for the inner product. Besides, orthonormal bases are by far the most common and useful bases.

30 8 CHAPTER. MATHEMATICAL PRELIMINARIES 0 e i = i = 0, 0,..., 1,..., 0 and e i = i = The only nonzero element 1 is in the i-th place. We can express v as a linear combination of i s. v = Similarly, n v i i i=1 n v = vi i. i=1 This suggests that we should define αv and αv by αv = α v and αv = α v = v α. As shown above, it is customary, for conserving symmetry, to write v α rather than α v. We say α v and v α are adjoints of each other. More generally, the adjoint of n α i v i is i=1 n v i αi, i=1 and vice versa. In order to take the adjoint, we can simply replace each scalar by its complex conjugate and each bra/ket by the corresponding ket/bra.

31 .4. THE BRAKET NOTATION 9 Let us now try some manipulations with bras and kets in order to get used to their properties. First, consider two vectors u = u 1 e 1 + u e u n e n and w = w 1 e 1 + w e w n e n in an n-dimensional space. We have u = u 1,..., u n and w = w 1. w n. Hence, u w = u 1,..., u n w 1. w n = u 1w 1 + u w u nw n. We drop one vertical bar form u w and write u w, so that u w = u 1,..., u n w 1. w n = u 1w 1 + u w u nw n. As you can see, the right-hand side is the inner product. Hence, the braket notation is consistent with the notation we have been using for the inner product. In passing, we will make a note of the obvious fact that e i e j = i j = δ ij. Next, given an orthonormal basis { 1,,..., n }, any vector v can be expanded in this basis so that n v = v i i, (.6) i=1 where v i s are unique. Taking the inner product of both sides with j, we get j v = j n i n n v i = v i j i = v i δ ij = v j. i=1 i=1 i=1

32 30 CHAPTER. MATHEMATICAL PRELIMINARIES Plugging this back into v = n i=1 v i i, v = n i=1 i v i = n i=1 i i v. (.7) The second equality in Equation.7 may not be clear. So, we will show it below explicitly by writing out each vector. The term on the left-hand side is as follows. i v i = 0, 0,..., 1,..., 0 v 1. v n = v i = 0 0. v i. 0 Note that the only nonzero element 1 is in the i-th place. On the other hand, the term on the right-hand side is i i v = , 0,..., 1,..., 0 v 1. v n = v i = 0 0. v i. 0. (.8) Hence, the second equality in Equation.7 indeed holds term by term. In the case of (.8), we have a product, from the left, of an n-by-1 matrix, a 1-by-n matrix, and an n-by-1 matrix. By the associative law of matrix

33 .4. THE BRAKET NOTATION 31 multiplication, it is possible to compute i i 7 first. i i v = , 0,..., 1,..., 0 v 1. v n = 1 i n i n v 1. v n = δ ii v 1. v n = 0 0. v i. 0 (.9) The symbol δ ij in (.9) means an n by n matrix whose only nonzero entry is the 1 at the intersection of the i-th row and the j-th column. δ ij := 1 j n i n It is easy to see that 7 As (.9) shows, i i extracts the i-th component of vector v. So, i i is called the projection operator for the ket i.

34 3 CHAPTER. MATHEMATICAL PRELIMINARIES And, it is also easy to see that i >< j = δ ij. n n i >< i = δ ii = I n ; (.10) i=1 i=1 With this identity, Equa- where I n is the n-dimensional identity operator. 8 tion.7 reduces to triviality. ( n ) ( n ) n v = I v = i i v = i i v = i i v i=1 i=1 i=1 Because the identity matrix can be inserted anywhere without changing the outcome of the computation, we can insert n i=1 i i anywhere during our computation. This seemingly trivial observation combined with the associativity of matrix multiplication will prove useful later in this book. Example.1 ( i, i, and i i in Two Dimensions) In two dimensions, we have 1 = Then, 1 0, = 0 1, 1 = 1 0, and = 0 1 (.11) 1 0 i i = i= = + =. (.1) When there is no possibility of confusion, we will simply use I instead of I n.

35 .4. THE BRAKET NOTATION 33 Now, consider any vector v = (v 1, v ) and compute 1 1 v and v. 1 v 1 1 v = v1 v1 = = 0 v 0 0 v 0 0 v v = v1 0 = = 1 v 0 1 v v (.13) (.14) Let us consider the adjoint of (.6). n v = i vi (.15) i=1 From our previous computation and due to a property of the inner product, we get n n n v = i ( i v ) = i v i = v i i. i=1 Again, note the identity n i=1 i >< i = n i=1 δ ii = I. So, ( n n ) v i i = v i i = v I = v. i=1 i=1 i=1 It is convenient at this point to have a rule by which we can mechanically arrive at the adjoint of an expression. i=1 Taking the Adjoint: 1. Reverse the order of all the factors in each term, including numerical coefficients, bras, and kets.. Take complex conjugate of the numerical coefficients. 3. Turn bras into kets and kets into bras. The following example demonstrates what one should do. Consider α v = a u + b w s t What is the adjoint?

36 34 CHAPTER. MATHEMATICAL PRELIMINARIES After Step 1 v α = u a + t s w b +... After Step v α = u a + t s w b +... After Step 3 v α = u a + t s w b +... Therefore, the adjoint of α v = a u + b w s t +... is v α = u a + t s w b Theorem.4 (Gram-Schmidt Theorem) Given n linearly independent vectors { v 1, v,..., v n }, we can construct n orthonormal vectors { 1,,..., n }, each of which is a linear combination of v 1, v,..., v n. In particular, we can construct { 1,,..., n } such that 1 = v 1 v. 1 v 1 Proof We will construct n mutually orthogonal vectors { 1,,..., n } first. Once it is done, we will only need to normalize each vector to make them orthonormal. First let 1 = v 1. Next, let where = v 1 1 v, (.16) v 1 1

37 .4. THE BRAKET NOTATION 35 is nothing but the projection of v on 1 9. So, in (.16), the component of v in the direction of 1 is subtracted from v, making it orthogonal to 1. Indeed, 1 = 1 v v = The third ket is defined proceeding in the same manner. Then, 3 = v v = 1 v v 3 1 v and 3 = v v 3 v v 3. = 1 v 3 1 v 3 0 = 0 = v 3 0 v 3 0 = 0. If n 3, we are done at this point. So, assume n 4, and suppose we have found i mutually orthogonal vectors 1,,..., i for i n 1. Now, define (i + 1) as follows. (i + 1) = v i v i For any 1 j i, j (i + 1) = j v i+1 i k=1 v i+1 j k k v i+1 k k... i i v i+1 i i = j v i+1 j j j v i+1 j j Hence, by mathematical induction, we can find n mutually orthogonal vectors 1,,..., n. Finally define i for 1 i n by i = i i i 1/, 9 In the familiar x, y-plane, 1 1 v ( 1 1 ) ( 1 v cos θ) 1 1 = 1 = v cos θ 1, where θ is the angel between 1 and v, and 1 is the unit vector in the direction of 1. = 0.

38 36 CHAPTER. MATHEMATICAL PRELIMINARIES so that i i = i i = 1. i i 1/ i i 1/ We have now constructed n orthonormal vectors 1,,..., n. Question: Linear Independence If you are a careful reader, you may have noticed that we did not, at least explicitly, use the fact that the vectors { v 1, v,..., v n } are linearly independent. So, why is this condition necessary? In order to see this, suppose (i + 1) = v i v i+1 v i+1... i i v i+1 = i i (.17) for some i n 1. Because our construction of i requires v i, this means we have at most n 1 orthonormal vectors, and our construction fails. However, we know that j is a linear combination of v 1, v,..., and v j for any j by construction. Therefore, 1 1 v i v i+1... i i v i+1 i i in (.17) is a linear combination of v 1, v,..., and v i. So, (i + 1) is a linear combination of v 1, v,..., and v i+1 where the coefficient of v i+1 is 1. But, this means that there is a linear combination of v 1, v,..., and v i+1 which equals 0 though not all the coefficients are zero, violating the linear independence assumption for { v 1, v,..., v n }. Turning this inside out, we can conclude that (i + 1) 0 for all i n 1 if { v 1, v,..., v n } are linearly independent. This is why we needed the linear independence of { v 1, v,..., v n } in Theorem.4. Theorem.5 The maximum number of mutually orthogonal vectors in an inner product space equals the maximum number of linearly independent vectors. Proof Consider an n-dimensional space V n, which admits a maximum of n linearly independent vectors by definition. Suppose we have k mutually orthogonal vectors v 1, v,..., v k, and assume k α i v i = 0. i=1

39 .5. VECTOR SUBSPACE 37 Then, k k v j α i v i = α i v j v i = α j v j v j = 0 = α j = 0 for j = 1,,..., k. i=1 i=1 So, the vectors 1,,..., and k are linearly independent. This implies that k n as V n can only admit up to n linearly independent vectors. On the other hand, Theorem.4 (Gram-Schmidt Theorem) assures that we can explicitly construct n mutually orthogonal vectors. This proves the theorem..5 Vector Subspace Definition.18 (Vector Subspaces) A subset of a vector space V that form a vector space is called a subspace. Our notation for a subspace i of dimensionality n i is V n i i. Definition.19 (Orthogonal Complement) The orthogonal complement of a subspace W of a vector space V equipped with a bilinear form B, of which an inner product is the most frequently encountered example, is the set W of all vectors in V that are orthogonal to every vector in W. It is a subspace of V. Definition.0 (Direct Sum) The direct sum of two vector spaces V and W is the set V W of pairs of vectors (v, w) in V and W, with the operations: (v, w) + (v, w ) = (v + v, w + w ) c(v, w) = (cv, cw) With these operations, the direct sum of two vector spaces is a vector space..6 Linear Operators An operator Ω operates on a vector u and transforms it to another vector v. We denote this as below. Ω u = v The same operator Ω can also act on u to generate v. In this case, we will reverse the order and write

40 38 CHAPTER. MATHEMATICAL PRELIMINARIES u Ω = v. For an operator Ω to be a linear operator, it should have the following set of properties. Ω (α u ) = α (Ω u ) Ω{α u + β v } = α (Ω u ) + β (Ωv) ( u α) Ω = ( u Ω) α ( u α + v β) Ω = ( u Ω) α + ( v Ω) β Recall that u is a column vector and u is a row vector. So, the best way to understand a linear operator is to regard it as an n n square matrix. Then, its action on u and u as well as linearity follows naturally. The significance of linearity of an operator Ω is that its action on any vector is completely specified by how it operates on the basis vectors 1,,..., n. Namely, if then, n v = α i i, i=1 ( n ) n Ω v = Ω α i i = α i Ω i. i=1 i=1 The action of the product of two operators ΛΩ on a ket u is defined in an obvious way. (ΛΩ) u = Λ (Ω u ) Ω acts on u first followed by Λ. This is an obvious associative law if the operators Λ and Ω are regarded as matrices as suggested already. With this view, ΛΩ is nothing but a matrix multiplication, and as such, ΛΩ is not the same as ΩΛ generally speaking.

41 .7. MATRIX REPRESENTATION OF LINEAR OPERATORS 39 Definition.1 (Commutator) The commutator of operators Λ and Ω, denoted by Λ, Ω is defined as follows. Λ, Ω := ΛΩ ΩΛ When the commutator Λ, Ω is zero, we say Λ and Ω commute, and the order of operation does not affect the outcome. We have the following commutator identities. Ω, ΛΓ = Λ Ω, Γ + Ω, Λ Γ ΛΩ, Γ = Λ Ω, Γ + Λ, Γ Ω Definition. (Inverse) The inverse Ω 1 of an operator Ω satisfies ΩΩ 1 = Ω 1 Ω = I; where I is the identity operator such that I u = u and u I = u. Needless to say, I corresponds to the identity matrix. Note that not all operators have an inverse. Note that the inverse satisfies (ΩΛ) 1 = Λ 1 Ω 1, which is consistent with the matrix representations of Ω and Λ..7 Matrix Representation of Linear Operators Recall that the action of a linear operator Ω is completely specified once its action on a basis is specified. Pick a basis { 1,,..., n } of normalized vectors which are not necessarily mutually orthogonal; that is, it is not necessarily an orthonormal basis. The action of Ω on any ket v is completely specified if we know the coefficients Ω ji for 1 i, j n such that

42 40 CHAPTER. MATHEMATICAL PRELIMINARIES n Ω i = Ω ji j. (.18) j=1 The coefficients in (.18) can be computed as follows. n n n k Ω i = k Ω ji j = Ω ji k j = Ω ji δ kj = Ω ki j=1 j=1 j=1 So, Ω ji = j Ω i in (.18). Note that k j was used instead of k j to make it clear that this matrix product 0 1 k n is not necessarily an inner product unless our basis happens to be orthonormal. 10 There is another way to find Ω ji. From (.10), we have n i=1 i i = I, and so, Now, let Then, n n Ω i = IΩ i = j j Ω i = j Ω i j = Ω ji = j Ω i. j=1 j=1 n n v = v i i and v = Ω v = v j j. i=1 j=1 ( n ) v n n n ( n n ) = Ω v = Ω v i i = v i Ω i = v i Ω ji j = Ω ji v i j i=1 i=1 i=1 j=1 j=1 i=1 10 See the footnote on p.4 for more detail. 1 j n

43 .7. MATRIX REPRESENTATION OF LINEAR OPERATORS 41 ( n n ) = Ω ji v i j j=1 i=1 implies n v j = Ω ji v i. (.19) i=1 Consider the matrix Ω ji = j Ω i, where Ω ji is the j-th row and the i-th column entry. Then, (.19) implies or Because v 1 1 Ω 1 1 Ω n v 1. Ω 1 Ω n. = v n n Ω 1 n Ω n v n v 1 Ω 11 Ω 1n v 1. Ω = 1 Ω n v n Ω n1 Ω nn v n Ω 11 Ω 1k Ω 1n Ω k1 Ω kk Ω kn Ω n1 Ω nk Ω nn 0 the k-th column of Ω ij is nothing but Ω k. k Ω 1k. =., Ω kk. Ω nk 1 Example. (An Operator on C ) The column vectors and form an orthonormal basis for C 1. If Ω = and 0 1

44 4 CHAPTER. MATHEMATICAL PRELIMINARIES Ω 0 1 = 1 0, we should have Ω = because Ω is the first column of the matrix representation of Ω, 0 0 and Ω is the second column. Now consider an arbitrary vector v 1 given by v = v1 v = v v 0 1. Using the matrix representation Ω v = v1 v v1 + v = v 1. On the other hand, This checks. ( ) Ω v = Ω v 1 + v 0 = v 1 1 Ω + v 0 Ω = v 1 + v v1 + v = v 1 + v 1 =. 0 v 1 Recall that we denoted α v by αv, and so, it is natural to denote Ω v by Ωv. Then, it is also natural to consider its adjoint Ωv. For a scalar α, αv = v α while αv = α v. Therefore, it is not surprising if Ωv v Ω ij though v Ω ij makes sense as a matrix multiplication. We will next compute the matrix elements of Ω when it comes out of the Ωv and operate on v from the right. Looking ahead, let us denote this by Ω

45 .7. MATRIX REPRESENTATION OF LINEAR OPERATORS 43 such that Ωv = v Ω. We have n n v = v i i and v = Ω v = Ωv = v j j. i=1 j=1 So, n n v = i vi and v = adj (Ω v ) = Ωv = j v j, i=1 j=1 where adj means the adjoint. On the other hand, as n Ω i = Ω ji j, j=1 ( ) ( n n ) v n n = adj (Ω v ) = adj Ω v i i = adj v i Ω i = adj v i Ω ji j i=1 i=1 i=1 j=1 n n n n = adj v i Ω ji j = j vi Ω ji. j=1 i=1 j=1 i=1 Hence, v j = n vi Ω ji for each 1 j n. i=1 This translates to the following matrix relation. Ω 11 Ω 1 Ω n1 v 1... v n = v 1... vn Ω 1 Ω Ω n Ω 1n Ω n Ω nn or 1 Ω 1 Ω 1 n Ω 1 v 1... v n = v 1... vn 1 Ω Ω n Ω Ω n Ω n n Ω n

46 44 CHAPTER. MATHEMATICAL PRELIMINARIES Because Ω 11 Ω k1 Ω n k Ω 1k Ω kk Ω nk Ω 1n Ω kn Ω nn = Ω 1k Ω kk Ω nk, the k-th row of Ωji is nothing but kω. In summary, if the matrix representation of Ω in Ωv = Ω v is Ω 11 Ω 1n Ω Ω = Ω ij = 1 Ω n.,.... Ω n1 Ω nn then, the matrix representation of Ω such that Ωv = v Ω is Ω 11 Ω n1 Ω = Ω Ω 1 Ω n ji Ω 1n Ω nn That is, Ω is the transpose conjugate of Ω..7.1 Matrix Representations of Operator Products If we have matrix representations Ω ij and Γ kl for two operators Ω and Γ, it would be convenient if the matrix representation for the operator product ΩΓ is the product of the corresponding matrices. Luckily for us, this is indeed the case as shown below. First, recall that n k=1 k k = I. So, n n (ΩΓ) ij = i ΩΓ j = i Ω k k Γ j = Ω ik Γ kj. k=1 Hence, the matrix representation of a product of two operators is the product of the matrix representations for the operators in the same order. Notation: We often denote Ω ij by Ω for simplicity. k=1

47 .8. THE ADJOINT OF AN OPERATOR 45.8 The Adjoint of an Operator In Section.7, we have already encountered this, but let us formally define the adjoint of an operator and investigate its properties. Definition.3 (The Adjoint) Consider an operator Ω that operates in a vector space V. Ω v = Ωv If there exists an operator Ω such that Then, Ω is called the adjoint of Ω. Ωv = v Ω. We can show that the adjoint indeed exists by explicitly computing the matrix elements in a basis { 1,,..., n }. ( Ω ) ij = i Ω j = Ωi j = j Ωi = j Ω i = (Ω ji ) As shown in Section.7, the matrix representing Ω is the transpose conjugate of the matrix for Ω. Example.3 (Adjoint of an Operator) Consider Γ = i Then, Now, for v = Γv = i v1 v, v1 v = Γ = iv1 + v i v. = Γv = (iv 1 + v ) v

48 46 CHAPTER. MATHEMATICAL PRELIMINARIES = iv 1 + v v. On the other hand, This checks. v Γ = v 1 v i = iv 1 + v v Fact. The adjoint of the product of two operators (ΩΓ) is the product of their adjoints with the order switched; that is, (ΩΓ) = Γ Ω. Proof First, regard ΩΓ as one operator. Then, ΩΓv = (ΩΓ)v = v (ΩΓ). Next, regard ΩΓv as Ω acting on the vector (Γv) to get ΩΓv = Ω(Γv) = Γv Ω. But, means Therefore, Γv = v Γ ΩΓv = v Γ Ω. (ΩΓ) = Γ Ω. Note.1 (Taking the Adjoint) In order to take the adjoint: 1. Reverse the order of each element of each term.. Take the complex conjugate of the scalars. 3. Switch bras to kets and kets to bras. 4. Take the adjoint of each operator.

49 .9. EIGENVALUES AND EIGENVECTORS 47 For example, consider a linear expression After Step 1 After Step After Step 3 After Step 4 α v = β u + γ w q r + δω s + ϵγλ t. v α = u β + r w q γ + s Ωδ + t ΛΓϵ v α = u β + r w q γ + s Ωδ + t ΛΓϵ = v α = u β + r q w γ + s Ωδ + t ΛΓϵ v α = u β + r q w γ + s Ωδ + t ΛΓϵ v α = u β + r q w γ + s Ω δ + t Λ Γ ϵ However, it will be easier to take the adjoint of each term in one step and add up the terms once you get used to this operation..9 Eigenvalues and Eigenvectors Definition.4 (Eigenvalues, Eigenvectors, and Eigenkets) Consider an operator Ω operating on a vector space V. When a nonzero vector v or a nonzero ket v satisfies Ωv = λv or Ω v = λ v, v is called an eigenvector, v is called an eigenket, and λ is called an eigenvalue. Note that if v (v) is an eigenket (eigenvector) with the associated eigenvalue λ, any scalar multiple of the ket (vector) α v (αv), where α 0, is an eigenket (eigenvector) with the same eigenvalue λ.

50 48 CHAPTER. MATHEMATICAL PRELIMINARIES Example.4 (Eigenvalues and Eigenvectors) Consider Ω = Then, Ω Ω = = and 1 = 1 = = Hence, the eigenvalues of Ω are 1 and 1 with corresponding eigenvectors which are scalar multiples of the vectors given above. There is a standard way to find the eigenvalues of an operator Ω. Here is a sketch. Suppose λ is an eigenvalue for an operator Ω with a corresponding eigenket v. Then, Ω v = λ v. This means that (Ω λi) v = 0, which in turn implies that Ω λi is not invertible. suppose Ω λi has the inverse (Ω λi) 1. Then, In order to see this, (Ω λi) 1 (Ω λi) v = (Ω λi) 1 0 = v = 0. But, this is a contradiction because v 0 if v is an eigenvector. Now consider a matrix representation of the operator Ω in some basis. Then, from basic matrix theory, Ω λi is invertible if and only if det(ω λi) 0. If Ω is an n n matrix, det(ω λi), called a characteristic polynomial, is an n-th degree polynomial with n roots 11. We can find all the eigenvalues 11 We do not always have n distinct roots. The total number of the roots is n if multiplicity is taken into account. For example, if the polynomial has (x 1) when factored, 1 is a root with a multiplicity of, and it counts as two roots.

51 .10. SPECIAL TYPES OF OPERATORS 49 by solving the characteristic equation det(ω λi) = 0. In order to compute det(ω λi) we need to pick a particular basis. However, the resulting eigenvalues are basis-independent because the characteristic polynomial is basis-independent. For a proof, see Theorem A.4 in Appendix A. Notation: An eigenket associated with the eigenvalue λ is denoted by λ..10 Special Types of Operators.10.1 Hermitian Operators 1 Definition.5 (Hermitian Operators) A Hermitian operator is an operator which is the adjoint of itself; that is, an operator Ω is Hermitian if and only if Ω = Ω. In particular, this means that Ωv w = v Ωw as both sides are equal to v Ω w. In the linear algebra of real matrices, the matrix representation of a Hermitian operator is nothing but a symmetric matrix. But, our matrices are complex, and the matrix representing a Hermitian operator should equal its transpose conjugate. Definition.6 (Anti-Hermitian Operators) 13 An operator Ω is anti- 1 These are also called self-adjoint operators. You may hear that self-adjoint is the term used by mathematicians and Hermitian is used by physicists. However, there is a subtle difference between self-adjoint and Hermitian. We will not want to get bogged down worrying too much about such details, but interested readers should check a book titled Functional Analysis by Michael Reed and Barry Simon Reed and Simon, 1980, p These are also known as anti-self-adjoint operators.

52 50 CHAPTER. MATHEMATICAL PRELIMINARIES Hermitian if Ω = Ω. It is possible to decompose every operator Ω into its Hermitian and anti- Hermitian components; namely if we decompose Ω as Ω = Ω + Ω + Ω Ω, then, the first term Ω+Ω is Hermitian, and the second term Ω Ω is anti- Hermitian. Hermitian operators play a central role in quantum mechanics due to its special characteristics. Fact.3 (Eigenvalues and Eigenkets of a Hermitian Operator) 1. The eigenvalues of a Hermitian operator are real. Measurable values in physics such as position, momentum, and energy are real numbers, and they are obtained as eigenvalues of Hermitian operators in quantum mechanics.. The eigenkets of a Hermitian operator can always be chosen so that they are mutually orthogonal. Because an eigenket can be multiplied by any nonzero number to generate another eigenket of any desired length, this means that we can always choose the eigenkets which form an orthonormal set. 3. The eigenkets of a Hermitian operator is a complete set in the sense that any vector can be expressed as a linear combination of the eigenkets. Therefore, if you know the action of a Hermitian operator Ω on the eigenkets, you completely understand the action of Ω on any vector in the space. Verification: Let v = α i λ i where λ i s are eigenvalues and λ i s are the corresponding eigenkets. Then, Ωv = Ω ( α i λ i ) = αi (Ω λ i )

53 .10. SPECIAL TYPES OF OPERATORS 51 Example.5 (A -by- Hermitian Operator/Matrix) Let Ω = 1 i i 1 As the adjoint is the transpose conjugate, we have Ω = 1 i i 1 and Ω is Hermitian. Observe that i 1 i i Ω = 1 i 1 1 Ω i 1 = 1 i i 1 = and i = 1. = Ω, 0 0 i = 0 = So, the eigenvalues are indeed real; namely, 0 and. We have i i 0 = and =. 1 1 i 1 i 1 Because 0 = i 1 i 1 = i 1 i 1 = 0, 0 and are mutually orthogonal. Next note that i = 1 0 = and = 1 i = i

54 5 CHAPTER. MATHEMATICAL PRELIMINARIES Therefore, any ket vector v = v1 can be expressed as a linear combination of 0 and as follows. v = v1 v = v = v ( ) ( ) v = v v i ( v1 v ) ( v1 0 + i + v ) i So, any vector v is a linear combination of 0 and. If we normalize 0 and, we obtain an orthonormal basis for C. Namely, 1 i 1 form an orthonormal basis. Let us prove 1. and 1 i 1 Theorem.6 (Real Eigenvalues) The eigenvalues of a Hermitian operator Ω are real. Proof Let λ be an eigenvalue of Ω and v be an associated eigenket. Then, λ v v = λv v = Ωv v = v Ω Ω v = v v = v λv = λ v v = (λ λ) v v = 0 = λ = λ because v v is nonzero if v is nonzero. Theorem.7 Eigenkets v and w of a Hermitian operator Ω corresponding to different eigenvalues λ v and λ w are orthogonal. Proof ω Ω v = w λ v v = λ v w v

55 .10. SPECIAL TYPES OF OPERATORS 53 On the other hand, as Ω is Hermitian, and λ v and λ w are real, w Ω v = w Ω v = Ωw v = v Ωw = v λ w w = λ w v w = λ w w v. So, λ v w v λ w w v = (λ v λ w ) w v = 0 = w v = 0 as λ v λ w. We will now prove and 3 and some more. Lemma.1 Let Ω be a linear operator on a vector space V n and Ω ij B be its matrix representation with respect to a basis B = {u 1, u,..., u k..., u n }. If the k-th vector in the basis B, denoted by u k is an eigenvector with the corresponding eigenvalue λ k, such that Ωu k = λ k u k, the k-th column of Ω ij B is given by Ω jk = λ k δ jk for 1 j n. That is, the entries of the k-th column are zero except for the k-th entry, which is λ k. 0. the k-th column of Ω ij B =. λ k. 0 k Proof From p.41, we know that the k-th column of Ω ij is nothing but Ω u k. But, 0. Ω u k = λ k u k = λ k k 0. =. λ k. 0. k

56 54 CHAPTER. MATHEMATICAL PRELIMINARIES In matrix form, Ω 11 Ω 1k Ω 1n Ω k1 Ω kk Ω kn Ω n1 Ω nk Ω nn k = Ω 1k.. Ω kk. Ω nk = λ k k = 0.. λ k. 0 k Ω jk = λ k δ jk. Corollary.1 Let Ω be a Hermitian operator on a vector space V n and Ω ij B be its matrix representation with respect to a basis B = {u 1, u,..., u k,..., u n }. If the k-th vector in the basis B, denoted by u k is an eigenvector with the corresponding eigenvalue λ k, such that Ωu k = λ k u k, the k-th column of Ω ij B is given by Ω jk = λ k δ jk for 1 j n. That is, the entries of the k-th column are zero except for the k-th entry, which is λ k. Furthermore, the k-th row of Ω ij B is given by Ω kj = λ k δ kj for 1 j n. Proof Ω jk = λ k δ jk follows from Lemma.1. Ω kj = λ k δ kj follows from the definition of a Hermitian operator and Theorem.6. The matrix representation of Ω, denoted here by Ω ij B, is the transpose conjugate of the matrix for Ω denoted by Ω ij B. As the k-th column of Ω ij B is 0.. λ k. 0 k, the k-th row of Ω ij B is k 0 λ k 0.

57 .10. SPECIAL TYPES OF OPERATORS 55 But, λ k = λ k because λ k is real. In matrix form, Ω ij looks like. 0 λ k Lemma. (Block Structure of Hermitian Matrices) Consider an ordered list of all the eigenvalues {λ 1, λ,..., λ n } of a Hermitian operator Ω, where an eigenvalue λ is repeated m λ times if its multiplicity as a root of the characteristic polynomial is m λ. Without loss of generality, we can assume that the same eigenvalues appear next to each other without a different eigenvalue in between. 14 Then, the matrix representation of Ω in the basis { λ 1, λ,..., λ n } Including a root λ with multiplicity m λ m λ times, Theorem.8 If an operator Ω on V n is Hermitian, there is a basis of its orthonormal eigenvectors. In this basis, sometimes called an eigenbasis, the matrix representation of Ω is diagonal, and the diagonal entries are the eigenvalues. Note that this theorem only claims that there is at least one such basis. Theres is no claim about uniqueness as it is not true. Proof Consider the characteristic polynomial which is of n-th degree. If we set it equal to zero, the equation, called the characteristic equation, has at least 14 For simplicity, suppose we have a list of four eigenvalues consisting of two λ s, one γ, and one ω. Then, the ordered list is {λ, λ, ω, γ}, {λ, λ, γ, ω}, {ω, λ, λ, γ}, {γ, λ, λ, ω} and not {λ, ω, λ}.

58 56 CHAPTER. MATHEMATICAL PRELIMINARIES one root λ 1 and a corresponding normalized eigenket λ 1. This can be extended to a full basis B according to Theorem A.5. Now, following the Gram-Schmidt procedure, Theorem.4, we can find an orthonormal basis O = {o 1, o, o 3,..., o n } such that o 1 = λ 1. In this basis, according to Corollary.1, Ω has the following matrix representation which we denote by M or Ω in keeping with the notation set on p.44. M = Ω = λ If we denote the shaded submatrix by N, it can be regarded as the matrix representation of an operator acting on the subspace O spanned by {o, o 3,..., o n }. The characteristic polynomial for M is given by det(m λi n ) = (λ 1 λ) det(n λi n 1 ); (.0) where I n is the n n identity matrix, I n 1 is the (n 1) (n 1) identity matrix, and det(n λi n 1 ) is the characteristic polynomial of N. Because det(n λi n 1 ) is P n 1 (λ), an (n 1)st degree polynomial in λ, it must have at least one root λ and a normalized eigenket λ unless n = 1. Because we are regarding N as operating on O, λ is a linear combination of o, o 3,..., and o n, and is orthogonal to λ 1. As before, we can find an orthonormal basis { λ, o 3, o 4,..., o n} of O. Note that λ is an eigenvalue of M, λ is an eigenvector of M, and O 1, = { λ 1, λ, o 3, o 4,..., o n} is an orthonormal basis of the original vector space V n. Let us take a detour here and see how the relations between M and N work explicitly. It may help you understand what is going on better. First, λ is clearly an eigenvalue of M as it is a root of the characteristic polynomial for M because of the relation (.0). Next, as λ is a linear combination of o, o 3,..., and o n, it is also a linear combination of λ 1, o, o 3,..., and o n. We can write n λ = 0 λ 1 + e i o i. i=

59 .10. SPECIAL TYPES OF OPERATORS 57 So, λ is the following column vector in the original basis O = { λ 1, o, o 3,..., o n }. 0 e λ =. e n Hence, M λ = λ e. e n = 0 λ e. λ e n where the second equality in (.1) follows from 0 e 0. = 0 = 0 N λ. e n = λ λ ; (.1) 0 λ λ. (.) Note that λ is regarded as a vector in V n in (.1) and as a vector in the subspace spanned by o, o 3,..., and o n in (.). There is a trade-off between avoiding abuse of notation and simplicity, and our choice here is simplicity. In the basis O 1,, Ω takes the following form. λ λ 0 0 M = Ω = Repeating the same procedure, we will finally arrive at λ λ 0 0 M = Ω = 0 0 λ λ n

60 58 CHAPTER. MATHEMATICAL PRELIMINARIES Note that some eigenvalues may occur more than once, but the above proof still works as it is. Example.6 (Diagonalization of a Hermitian Matrix) Consider Ω = (.3) Ω is clearly Hermitian as it equals its own transpose conjugate. Observe that Ω = = = Ω 1 1 = and 1 = = Define normalized eigenvectors e 1 and e as follows. e 1 = and e = (.4) Then, Ω 11 = e 1 Ω e 1 = = = 1 4 =. (.5) Similarly, we also get Ω 1 = e 1 Ω e = 0 Ω 1 = e Ω e 1 = 0

61 .10. SPECIAL TYPES OF OPERATORS 59 and Ω = e Ω e = 0. Note that the eigenvalues are on the diagonal and all the off-diagonal entries are zero; i.e. we have diagonalized Ω. Fact.4 If Ω is an anti-hermitian operator: 1. The eigenvalues of Ω are purely imaginary.. There is an orthonormal basis consisting of normalized eigenvectors of Ω. A n = a n n A = n n a n n the spectral decomposition of the Hermitian operator A.10. Simultaneous Diagonalization Theorem.9 (Commuting Hermitian Operators) If Ω and Γ are commuting Hermitian operators, there exists a basis consisting of common eigenvectors, such that both Ω and Γ are diagonal in that basis. This very useful theorem is actually a corollary of the following more general theorems from matrix algebra. For details, check Appendix A. Theorem.10 (Commuting Diagonalizable Matrices) Let A and B be diagonalizable n n matrices. Then, AB = BA if and only if A and B are simultaneously diagonalizable. Theorem.11 (A Family of Diagonalizable Matrices) Let F be a family of diagonalizable n n matrices. Then, it is a commuting family if and only if it is simultaneously diagonalizable.

62 60 CHAPTER. MATHEMATICAL PRELIMINARIES.11 Active and Passive Transformations We will mainly deal with active transformations in this book. So, you only need to familiarize yourself with the general concept. In Classical Mechanics In classical mechanics, it is about a point in space and the coordinate system. An active transformation moves the point while leaving the coordinate axes unchanged/fixed. For example, a rotation of the point (x 0, y 0 ) in the x, y-plane by angle θ is effected by the matrix R θ given by cos θ sin θ R θ = (.6) sin θ cos θ such that the new position (x 0, y 0) is obtained as follows. x 0 x0 cos θ sin θ x0 x0 cos θ y y 0 = R θ = = 0 sin θ y 0 sin θ cos θ y 0 x 0 sin θ + y 0 cos θ (.7) Under an active transformation, the value of a dynamical variable, often denoted by ω(q, p), will not necessarily remain the same. A passive transformation does not move the point (x 0, y 0 ) but only changes the coordinate system. To take a -dimensional rotation as an example, this corresponds to rotating the x-y coordinate system without moving the point itself. Because the only change is in the value of the point s x- and y-coordinates relative to the new rotated axes, and the position itself does not change in this case, the physical environment of the point remains the same, and a dynamical variable ω(q, p) will have the same value after the transformation. Note that the new coordinates (x 0, y 0) are numerically the same whether you rotate the axes by θ or the point by θ. Hence, the effect of passive transformation by θ can be computed as below. x 0 x0 cos( θ) sin( θ) x0 x y 0 = R θ = = 0 cos θ + y 0 sin θ y 0 sin( θ) cos( θ) y 0 x 0 sin θ + y 0 cos θ (.8)

63 .11. ACTIVE AND PASSIVE TRANSFORMATIONS 61 Again, despite this relation, the two transformations should not be confused with each other because the active transformation is the only physically meaningful transformation, in the sense that the values of dynamical variables may change. In Quantum Mechanics In quantum mechanics, it is about an operator Ω and a basis vector v. Consider a unitary transformation U which causes a basis change v U v := U v. Under this transformation, the matrix elements of an operator Ω transforms as follows. v Ω v Uv Ω Uv = v U ΩU v (.9) In (.9), the change is caused to the vector v in Uv Ω Uv, and to the operator Ω in v U ΩU v. Nevertheless, the matrix elements of Ω are the same between the two formulations or interpretations. An active transformation corresponds to where the vector v is transformed as below. Uv Ω Uv ; (.30) v Uv (.31) A passive transformation picture is painted by where the operator Ω is transformed as follows. v U ΩU v ; (.3) Ω U ΩU (.33) Because the essence of quantum mechanics is the matrix elements of operators, and all of the physics of quantum mechanics lies in the matrix elements, the equality of (.30) and (.3) given in (.9) indicates that these two views are completely equivalent. Hence, active and passive transformations in quantum mechanics provide two equivalent ways to describe the same physical transformation.

64 6 Exercises 1. Consider Ω = The eigenvalues are 1 and 1. (a) Explain why Ω is Hermitian (b) Find a real normalized eigenvector for each eigenvalue. We will denote them by 1 and 1. (c) Verify that the eigenvectors are orthogonal. (d) Express an arbitrary vector v = 1 and 1. v1 (e) Express Ω as a matrix in the basis { 1, 1 }. v. as a linear combination of. Answer the following questions about the matrix representation of a Hermitian operator. (a) Given that the matrix M = 6 4 α 0 is Hermitian, what is α? (b) The eigenvalues are 8 and. Find NORMALIZED and REAL eigenvectors 8 and, remembering that λ means an eigenket associated with the eigenvalue λ. (c) Find the matrix representation of M in the basis consisting of 8 (the first basis ket) and (the second basis ket). 3. Let A = and B = (a) Show A, B = 0.

65 63 (b) The eigenvalues for A are and 1. Find two real normalized eigenvectors, A and 1 A. (c) Show that A and 1 A are also eigenvectors of B, but the eigenvalues are not the same. (d) Verify that { A, 1 A } forms a basis for C. (e) Find the matrix representations of A and B relative to the basis { A, 1 A }. 4. Consider two Hermitian matrices 1 1 M = 1 1 and N = (a) Prove that M and N commute. (b) Find the eigenvalues and real normalized eigenvectors of M and N. (c) Show that M and N are simultaneously diagonalizable.

66 64

67 Chapter 3 The Postulates of Quantum Mechanics This is where quantum mechanics markedly differs from mathematics as you know it a. The postulates of quantum mechanics cannot be proved or deduced in a purely mathematical fashion. The postulates are hypotheses, that are given to you from the beginning. If no disagreement with experiments is found, the postulates are accepted as correct hypotheses. Sometimes, they are called axioms, which means the postulates are regarded true though not provable. You should not feel excessively uncomfortable with this. Much of what we know in physics cannot be proved. For example, consider the celebrated Newton s Second Law of Motion; F = ma. There is no purely mathematical derivation of this relationship. In physics, a theory is deemed correct if there is a good agreement between observations and theoretical predictions. While classical Newtonian mechanics suffices for macroscopic systems, a huge body of experimental evidence supports the view that the quantum postulates provide a consistent description of reality on the atomic scale. a I am saying mathematics as you know it because mathematics also relies on many postulates often known as axioms. However, most of us are used to encountering definitions and proofs when doing mathematics and not necessarily postulates. 65

68 66 CHAPTER 3. THE POSTULATES OF QUANTUM MECHANICS 3.1 The Fundamental Postulates 1. An isolated physical system is associated with a topologically separable 1 complex Hilbert space H with inner product ϕ ψ. Physical states can be identified with equivalence classes of vectors of length 1 in H, where two vectors represent the same state if they differ only by a phase factor. The above is a mathematically correct and rigorous statement of Postulate 1. A more accessible simplified statement would be as follows. Each physical system has its associated Hilbert space H, so that there is one-to-one correspondence between a physical state s of the system and a vector v s, or a ket s, of unit length in H.. To every observable q in classical mechanics there corresponds a Hermitian operator Q in quantum mechanics. Table 3.1 summarizes the correspondences. In Table 3.1, I am showing the correspondence between classical observables and quantum mechanical operators for a single variable x for simplicity. If we include y and z, parts of Table 3.1 should be modified as shown in Table A topological space is separable if it contains a countable dense subset; that is, there exists a sequence {x i } i=1 of elements of the space such that every nonempty open subset of the space contains at least one element of the sequence. Separability means that the Hilbert space H has a basis consisting of countably many vectors. Physically, this means that countably many observations suffice to uniquely determine the state. 3 H or Ĥ is called Hamiltonian.

69 3.1. THE FUNDAMENTAL POSTULATES 67 Observable q Q Operation Position x x or ˆx M x : multiplication Momentum p p or ˆp î ( ) i Kinetic Energy T T or ˆT x m x Potential Energy V (x) V (x) or ˆV (x) M V (x) : multiplication Total Energy E H or Ĥ3 + V (x) m x Angular Momentum l x L x or ˆL x i ( y z ) z y Table 3.1: Physical observables and corresponding quantum operators Observable q Q Operation Momentum p p or ˆp i ( î + ˆȷ + ˆk ) x y z Kinetic Energy T T or ˆT ( ) + + m x y z Potential Energy V (r) V (r) or ˆV (r) M V (r) : multiplication ( x + Total Energy E H or Ĥ m Angular Momentum l x L x or ˆL x i ( y z z y + y ) l y L y or ˆL y i ( z x ) x z l z L z or ˆL z i ( x y ) y x z ) + V (r) Table 3.: Quantum operators in three dimensions 3. When a measurement is made for an observable q associated with operator Q, the only possible outcomes are the eigenvalues of Q denoted by {λ i }.

70 68 CHAPTER 3. THE POSTULATES OF QUANTUM MECHANICS 4. The set of eigenstates/eigenvectors of operator Q, denoted by λ i, forms a complete a orthonormal set of the Hilbert space H. a A collection of vectors {v α } α A in a Hilbert space H is complete if w v α = 0 for all α A implies w = 0. Equivalently, {v α } α A is complete if the span of {v α } α A is dense in H, that is, given w H and ϵ > 0, there exists w span{v α } such that w w < ϵ. 5. Suppose a physical system is in state s. If the vector v s H associated with the physical state s is given by n v s = a i λ i ; i=1 where n may go to, the possible outcomes of a measurement of the physical quantity q are the eigenvalues λ 1, λ,..., λ k,..., and the probability that λ k is observed is λ k v s = a k. 6. As soon as a measurement is conducted on a physical state s, yielding λ k as the outcome, the state vector v s collapses to the corresponding eigenstate λ k. Therefore, measurement affects the physical state of the system. a a This fact is often used in elaborate experimental verifications of quantum mechanics. 7. The average value of the observable q after a large number of measurements is given by the expectation value q or q = v s Q v s.

71 3.1. THE FUNDAMENTAL POSTULATES The state vector s associated with a physical state s typically depends on time and the spatial coordinates; that is, s = s(r, t). The state evolves in time according to the time-dependent Schrödinger equation. i s(r, t) = Ĥ s(r, t) (3.1) t In a typical application, our state vector is a differentiable function, called a wavefunction, and the canonical symbol used for the wavefunction is Ψ. With this notation, we get Ψ(r, t) i t = ĤΨ(r, t). 9. TO BE REWRITTEN: We can use the Schrödinger Equation to show that the first derivative of the wave function should be continuous, unless the potential is infinite at the boundary. notes/node141.html The wavefunction Ψ(r, t) and its spatial derivatives ( / x)ψ(r, t), ( / y)ψ(r, t), ( / z)ψ(r, t) are continuous in an isotropic medium a. a A good example of a non-isotropic medium is a medium where the potential energy varies discontinuously with position. We will encounter a potential of this kind when we discuss the infinite square well potential. 10. You can skip this for now. The total wavefunction must be antisymmetric with respect to the interchange of all coordinates of one fermion a with those of another. Electron spin must be included

72 70 CHAPTER 3. THE POSTULATES OF QUANTUM MECHANICS in this set of coordinates. The Pauli exclusion principle is a direct consequence of this antisymmetry principle. Slater determinants provide a convenient means of enforcing this property on electronic wavefunctions. a Particles are classified into two categories, fermions and bosons. Fermions have half-integral intrinsic spins such as 1/, 3/, and so forth, while bosons have integral intrinsic spins such as 0, 1, and so forth. Quarks and leptons, as well as most composite particles, like protons and neutrons, are fermions. All the force carrier particles, such as photons, W and Z bosons, and gluons are bosons, as are those composite particles with an even number of fermion particles like mesons. Of the 10 postulates, Postulates 8 is distinctly different from the other nine. The nine postulates describe the system at a given time t, while Postulate 8 specifies how the system changes with time. 3. Unitary Time Evolution Time evolution of a quantum system is always given by a unitary transformation Û, such that s(t) = Û(t) s(0). (3.) The nature of Û(t) depends on the system and the external forces it experiences. However, Û(t) does not depend on the state s. Hence, Û(α s 1 + β s ) = αû s 1 + βû s, and the time evolution operator Û is linear. Again, the key here is that the time evolution operator Û is a function of the physical system and not individual states. Sometimes Û(t) is referred to as a propagator. The unitarity of Û follows from the time-dependent Schrödinger equation (3.1) as follows. We have i t s(t) = Ĥ(t) s(t) or t i s(t) = Ĥ(t) s(t) ; where Ĥ is the Hamiltonian of the system and is Hermitian. Suppose s(t) = Û(t) s(0)

73 3.. UNITARY TIME EVOLUTION 71 for some operator Û. equation, we obtain t Plugging this into the time-dependent Schrödinger ( ) (Û(t) ) ( ) i i s(0) = Ĥ(t)Û(t) s(0) = tû(t) s(0) = Ĥ(t)Û(t) s(0) for any physical state s. This means that the actions of tû(t) and i Ĥ(t)Û(t) on any set of basis vectors are the same. Therefore, i = tû(t) Ĥ(t)Û(t)4. (3.3) Taking the adjoint, and noting that Ĥ = Ĥ, ( ) ( ) i tû(t) = Ĥ(t)Û(t) = tû (t) = Û i (t)ĥ (t) = i Û (t)ĥ(t). Now, at t = 0, Û(0) = I by necessity, and this gives Û (0)U(0) = I. On the other hand, ( ) ( ) ( ) ( ) (Û t (t)û(t)) = tû (t) Û(t)+Û (t) tû(t) = tû (t) Û(t)+Û (t) tû(t) = i Û (t)ĥ(t)û(t)+û (t) i Ĥ(t)Û(t) = i Û (t) ( Ĥ(t) Ĥ(t)) Û(t) = 0. Hence, Û (t)û(t) = I at all times t, and Û(t) is unitary. Because Û(t) is unitary, s(t) = s(t) s(t) = s(0) Û (t)û(t) s(0) = s(0) I s(0) = s(0), and the magnitude of the state vector is preserved. One way to interpret time evolution is to regard it as a rotation of the state vector in Hilbert space. 4 One way to understand what this equality means is to think in terms of their matrix representations with respect to a basis.

74 7 CHAPTER 3. THE POSTULATES OF QUANTUM MECHANICS 3.3 Time-Independent Hamiltonian So far, we have considered a Hamiltonian with explicit time-dependence as is clear from the notation Ĥ(t). However, for many physical systems, the Hamiltonian is not time-dependent. Fro example, consider a typical Hamiltonian given by ( ) H = m x + y + + V (r) z in Table 3.. In the most general case, the potential energy term V can have explicit time dependence so that V = V (r, t). However, if V does not depend on t, and neither does H as shown above, there are simple relations among the time evolution operator U(t), the system Hamiltonian H, and the eigenstates of H Propagator as a Power Series of H We start with the differential relation (3.3) d i = H(t)Û(t); (3.4) dtû(t) where we switched to total time derivative as we will only focus on t here, holding all the other variables fixed. Let us now remove time-dependence from H in (3.4). d i = HÛ(t); (3.5) dtû(t) We will compare this with the familiar case of a real-valued function f : R R satisfying the differential equation d i f(t) = dt Hf(t); where H is some scalar. As the answer to this differential equation is iht f(t) = f(0) exp,

75 3.3. TIME-INDEPENDENT HAMILTONIAN 73 you may think the answer to the analogous operator differential equation (3.5) is something like iht Û(t) = Û(0) exp = exp iht ; (3.6) where the second equality holds as Û(0) is the identity operator by necessity. As it turns out, this is the right answer with the interpretation of the righthand side as a power series in operator H. This is explained in more detail in Appendix B. But, we will only check it formally here to convince ourselves that the solution really works. Maclaurin series of the exponential function gives n n ih ih iht exp = t n = I + t n ; (3.7) n! n! n=0 where 0 ih = I by definition. Then, ( ) d iht exp = d I + dt dt n=1 n ih n! t n n=1 = d dt n=1 n ih n! t n = n=1 n ih n(n 1)! ntn 1 So, = ih n=1 n 1 ih (n 1)! tn 1 = ih n=0 n ih t n = i iht n! H exp. iht Û(t) = exp satisfies the differential equation (3.5). In Section 3., we showed that the time evolution operator Û(t) is unitary for any Hamiltonian H. When H does not depend on time, we have an alternative proof that Û(t) is unitary. However, note that our discussion below is somewhat heuristic and not completely rigorous mathematically speaking.

76 74 CHAPTER 3. THE POSTULATES OF QUANTUM MECHANICS First, taking the term-by-term adjoint of the right-hand side of (3.7), we get ( ) n ih ( ) iht exp = n=0 n ih n! t n = n=0 n! t n = It suffices to show iht iht Û(t)Û (t) = exp exp = I. n=0 n +ih t n = exp n! Recall from Theorem.8 that a matrix representing a Hermitian operator is diagonalizable with its eigenvalues on the diagonal. Because H is Hermitian, there exits an invertible matrix J such that λ 1 λ 1 JHJ 1 = D =... or H = J 1 DJ = J 1... J; λ m where λ 1,..., λ m are the eigenvalues of H. We also have n H n = ( J 1 DJ ) λ 1 n = J 1 D n J = J 1... J = J 1 Therefore, Û(t) = e i Ht = = J 1 ( n=0 = J 1 = J 1 n=0 n=0 ( ) i n H n t n 1 n! = n=0 ( ) i n D n t n 1 n!) ( i ) n λ n 1 ( ) n=0 n i λ n 1 t n 1 n! J = J 1... λ n m n=0 1 tn n!... n=0 λ m λ n 1 ( ) i n J 1 D n Jt n 1 n! = n=0 n ( ) i n λ 1... J ( ) n i λ n m t n 1 n! J λ m λ m... λ n m iht J.. ( ) i n J 1 D n t n 1 n! J t n 1 n! J

77 3.3. TIME-INDEPENDENT HAMILTONIAN 75 = J 1 = J 1 Similarly, We now have n=0 ( i λ 1t ) n 1 n! exp ( i λ 1t )... Û(t)Û (t) = J 1 = J 1 1 = J 1 Û (t) = e i Ht = J 1 exp ( i λ 1t ) ( i n=0 λ mt ) n 1 n! exp ( i λ mt ) exp ( i λ 1t )... J = I. J. exp ( i λ 1t )... exp ( i λ mt ) exp ( i λ mt ) J exp ( i λ mt ) JJ 1 exp ( i λ 1t )... J. exp ( i λ 1t )... exp ( i λ mt ) J exp ( i λ mt ) J Therefore, Û(t) is indeed unitary for all t Eigenstate Expansion of the Propagator Here, we will discuss how the propagator Û(t) can be expressed using the normalized eigenkets of H. The eigenvalue problem for H is H E = E E ;

78 76 CHAPTER 3. THE POSTULATES OF QUANTUM MECHANICS where E is an eigenvalue representing the total energy, and E is an associated normalized eigenket. From (.10), E E = I. Hence, s(t) = E E s(t), and E s(t) gives the coefficients when s(t) is expressed as a linear combination of { E }. For simplicity of notation, let c E (t) = E s(t), so that From Postulate 8, Ψ(r, t) i t = = i c E(t) t ĤΨ(r, t) = i s(t) t As E s are linearly independent, s(t) = c E (t) E. = H s(t) = i c E (t) E t E = c E (t)e E = ( i c E(t) t i c E(t) t Ec E (t) = 0 for each value of E. This implies that c E (t) is an exponential function. Ec E (t) ) = H c E (t) E E = 0. i c E(t) t Ec E (t) = 0 = c E(t) t = ie c E(t) = c E (t) = c E (0)e iet/. Remembering that c E (t) = E s(t), and in particular c E (0) = E s(0), we have s(t) = E s(0) e iet/ E = E E s(0) e iet/. Comparing this with (3.), we get Û(t) = E E E e iet/.

79 3.4. THE UNCERTAINTY PRINCIPLE 77 This is the eigenstate expansion of the propagator Û(t). Let us double check and see if this expression of Û(t) is unitary. Following the steps described in Note.1, ( ) Û (t) = E E e iet/ = E E ( E E e iet/ ) Step 1 E e iet/ E E Step E e iet/ E E Step 3 Û (t) = E e iet/ E E. Hence, ( ) ( ) Û(t)Û (t) = E E e iet/ e ie t/ E E = E E E E e i(e E )t/ E E E,E This checks. = E=E E E e i(e E )t/ = E 3.4 The Uncertainty Principle E E = I. The uncertainty principle is one of the celebrated consequences of quantum mechanics, whose influence has gone beyond the realm of physics into humanities such as philosophy Uncertainty and Non-Commutation Theorem 3.1 (The Uncertainty Relation) Consider two physical observables represented by Hermitian operators A and B. We denote the expectation value of an operator Ω by Ω ; that is, if the state of the system is u after normalization, In addition, define A by Ω = u Ω u.

80 78 CHAPTER 3. THE POSTULATES OF QUANTUM MECHANICS A = (A A I) ab. a From the formula, you can see that A is the standard deviation of the observable A in the state u. b This definition means ( A) = (A A ) = (A A A + A ) = A A A + A = A A or A = A A. Then, A, B = iα implies A B α ; where α is a real scalar. Proof We will first define two Hermitian operators C A and C B as follows. Then, C A = A A I and C B = B B I C A, C B = A A I, B B I = A A I, B A A I, B I = A, B = iα. Next define two vectors by Then, v = C A u and w = C B u. v v = u C AC A u = u C A C A u = u C A u = u (A A I) u = (A A ) and w w = u C BC B u = u C B C B u = u C B u = u (B B I) u = (B B ). Therefore, v v = ( A) and w w = ( B). Now, by Schwarz Inequality (Theorem.3), ( A) ( B) = v v w w v w = u C A C B u = u (C A C B C B C A + C A C B + C B C A )/ u = 1 4 u C A, C B + {C A, C B } u ;

81 3.4. THE UNCERTAINTY PRINCIPLE 79 where {C A, C B } = C A C B + C B C A is called an anti-commutator. Because C A and C B are Hermitian, u {C A, C B } u = u C A C B + C B C A u = u C A C B u + u C B C A u = u C AC B u + u C BC A u = C A u C B u + C B u C A u = C A u C B u + C A u C B u = Re C A u C B u. Fro simplicity of notation, let β = Re C A u C B u. Then, we have ( A) ( B) 1 4 iα + β = 1 4 (α + β ) 1 4 α. Therefore, A B α. (3.8) 3.4. Position and Momentum From Table 3.1, ˆp = i d dx and ˆx = M x. In order to compute the commutator ˆx, ˆp, we will let ˆx, ˆp act on a function f(x). ˆx, ˆpf(x) = x( i d dx )f(x) ( i d d x)f(x) = i x dx dx f(x)+i d dx (xf(x)) = i x d dx f(x)+i f(x)+i x d f(x) = i f(x) = ˆx, ˆp = i dx Therefore, x p Energy and Time The momentum-position uncertainty principle x p has an energytime analog, E t. However, this must be a different kind of relation- ship because t is not a dynamical variable. This uncertainty cannot have anything to do with lack of commutation.

82 80 CHAPTER 3. THE POSTULATES OF QUANTUM MECHANICS Time rate of change of expectation values How do expectation values change in time? That is, how can we compute d dt A (t) = d dt u(t) A u(t) ; where u(t) is the state of the system at time t. From (3.1), So, d dt A (t) = d dt u(t) = 1 i H u(t) and d dt u(t) = 1 i u(t) H5. ( ) ( ) d d dt u(t) A u(t) + u(t) A dt u(t) = 1 1 u(t) HA u(t) + u(t) A i i H u(t) = 1 1 u(t) HA+AH u(t) = A, H (t). i i We have d dt A (t) = 1 A, H (t). (3.9) i The indication here is that dynamical evolution of an observable depends on its level of commutativity with H, at least on the average Time-energy uncertainty Our derivation of the general form of the uncertainty principle (3.8) concerns two observables. However, as mentioned already, time t is not a dynamical variable. We need a way to make a connection between t and an observable A if we are to use the general inequality (3.8) to formulate time-energy uncertainty. We will use an observable A to characterize the change in the system in time. As A and the energy E are observables, we have A E 1 A, H 5 One way to understand this is to regard u(t) as an n 1 matrix (a column vector) and u(t) as its transpose conjugate (a row vector). With this interpretation, d dt U(t) becomes entry by entry differentiation of a matrix, and ( d dt u(t) ) = d dt u(t) and ( 1 i H u(t) ) = u(t) H naturally. 1 i

83 3.5. CLASSICAL MECHANICS AND QUANTUM MECHANICS 81 from (3.8). Now, from (3.9), 1 A, H = 1 i d dt A = d dt A. Hence, A E d dt A. The ratio of the amount of uncertainty in A, denoted by A, to the rate of change in the expected value of A, d dt A, is one candidate for t. With this definition, we have t = A d dt A. Then, E t = E A d A dt d dt A d dt A = = E t. Again this inequality is completely different from that for two observables. It depends on how you devise your way to measure time which is different from the usual time t. This may be the reason why time-energy uncertainty is so notoriously controversial. 3.5 Classical Mechanics and Quantum Mechanics According to the correspondence principle advocated by Niels Bohr, every new physical theory must contain as a limiting case the old theory. So, quantum mechanics has to be a superset of classical mechanics. Indeed, classical mechanics is a limiting case of quantum mechanics as h or 0. It was also shown by Ehrenfest that the expectation values of quantum mechanical observables satisfy the same equations as the corresponding variables in classical mechanics.

84 8 CHAPTER 3. THE POSTULATES OF QUANTUM MECHANICS Ehrenfest s Theorem One embodiment of the correspondence principle are the two relations proved by Ehrenfest that connect classical mechanics and quantum mechanics. Theorem 3. (Ehrenfest Theorem) If Ω is a quantum mechanical operator and Ω is its expectation value, we have d dt Ω = 1 Ω Ω, H +. (3.10) i t Proof This is a generalized version of (3.9); where the operator is possibly timedependent. We will write out the integral for the expectation values explicitly in order to see the reasons for the total derivative d and the partial derivative clearly. We will specialize to a one-dimensional case without loss dt t of generality and let Ψ(x, t) be a normalized wavefunction representing the state. Then, d dt Ω = d dt ( ) Ψ(x, t) Ω(x, t)ψ(x, t) dx = Ψ(x, t) Ω(x, t)ψ(x, t) dx 6 t ) = (Ψ(x, t) Ω(x, t)ψ(x, t) dx t + + Ψ(x, t) t ( Ω(x, t) Ψ(x, t) Ω(x, t) t Now, from Postulate 8 on p.69, we have ( ) Ψ(x, t) dx Ψ(x, t) ) dx. (3.11) i s(r, t) = Ĥ s(r, t) t = 1 s(r, t) = t i Ĥ s(r, t) and 1 s(r, t) = t i s(r, t) Ĥ7. (3.1) In our case (3.1) becomes ( ) Ψ(x, t) = 1 HΨ(x, t) and ) (Ψ(x, t) = 1 t i t i Ψ(x, t) H. (3.13) 6 We need to switch from d dt to t because the independent variables are x and t before the integration with respect to x removes x as a variable. 7 Note that 1 t s(r, t) = i s(r, t) Ĥ follows from ( ( ) t s(r, t) ). = 1 i Ĥ s(r, t)

85 3.5. CLASSICAL MECHANICS AND QUANTUM MECHANICS 83 Substituting these into (3.11), d dt Ω = = 1 i 1 i Ψ(x, t) HΩ(x, t)ψ(x, t) dx + + Ψ(x, t) Ω(x, t) 1 HΨ(x, t) dx i Ψ(x, t) (ΩH HΩ)Ψ(x, t) dx + = 1 Ω, H + i Ω t Ω t Ω t (3.14) In particular, if the Hamiltonian is time-independent as in Section 3.3, Ehrenfest Theorem implies the following. Corollary 3.1 (Time-Independent Hamiltonian) If the Hamiltonian H does not have explicit time dependence, the time rate of change of the expectation value of a variable Ω represented by operator Ω, < Ω >, is proportional to the expectation value of the commutator between Ω and H, Ω, H, with a proportionality constant i. Therefore, < Ω > is conserved if Ω commutes with a time-independent Hamiltonian H Exact Measurements and Expectation Values We have seen in Tables 3.1 and 3. that each classical observable has its associated quantum mechanical operator. This is an exact one-to-one correspondence. However, in terms of actual measurements, a quantum mechanical operator does not generate one exact and fixed value under repeated measurements as in the classical measurement. Instead, each quantum mechanical operator specifies a probability distribution of possible outcomes of the measurement by way of a sate function or wavefunction. This necessitates a probabilistic interpretation of classical laws of physics.

86 84 CHAPTER 3. THE POSTULATES OF QUANTUM MECHANICS As we saw in Ehrenfest Theorem, the general strategy is to use expectation values rather than exact measured values. In the quantum mechanical formulation of classical laws of physics, the expectation values play the role of the classical variables.

87 Exercises 1. Let Q be a quantum mechanical operator representing a physical observable q. An example of Q is the momentum operator P = i in x one dimension, representing the classical momentum p. For simplicity, we will only consider the cases with distinct eigenvalues. (a) What can you say about Q? In other words, what is the name of the class of operators/matrices Q belongs to? (b) Now consider the set { λ i } of all normalized eigenkets 8 of Q. i. What is λ i λ j? This property makes {λ i } an orthonormal set. ii. The set {λ i } is complete. What does this mean?. Let λ i and λ j be two distinct eigenstates of Q, i.e. i j. Note that λ i and λ j are normalized as they are states. (a) If α λ i + β λ j is a state, not necessarily an eigenstate, where α and β are scalars, what is the relation between α and β? (b) With what probabilities will λ i and λ j be observed if a measurement is made? (c) A measurement has been made for q and the value obtained was λ i. What is the state the system is in after the measurement? You can only answer this up to an arbitrary phase e iθ. (d) A second measurement is made on the resulting state above. What value/values will be observed with what probability/probabilities? 85 8 Recall we use eigenket, eigenvector, and eigenfunction interchangeably.

88 86

89 Chapter 4 Spaces of Infinite Dimensionality So far, we have focused mostly on finite-dimensional spaces. This was because finite dimensions make it easy to explain mathematics and physics for the author and also easy to understand mathematics and physics for the readership. However, the attentive reader may have wondered why finite dimensions will suffice. The simplest example is the position x and its quantum mechanical representation (operator) X or M x. The position observable is continuously distributed over the interval (, + ), and hence, there are infinitely many distinct eigenvalues for the operator X, which in turn implies that there are infinitely many eigenvectors which are mutually orthogonal according to Theorem.7. Needless to say, the dimensionality of the Hilbert space in which the operator X operates has to be infinite. Another example is the discrete energy distribution as indicated in 4 on p.14. This states that the total energy E can only take discrete values E 1, E, E 3,... and not any value E 0, ) as in classical mechanics where E is continuously distributed. However, it does not place an upper limit on E, and infinitely many discrete values of E are possible. 87

90 88 CHAPTER 4. SPACES OF INFINITE DIMENSIONALITY 4.1 Two Types of Infinity Two kinds of infinite dimensional vector spaces are encountered in quantum mechanics. One infinity is known as countable or denumerable infinity, and the other infinity is called uncountable/non-denumerable infinity. Discrete energy level distribution is an example of countable infinity, and the position x is an example of uncountable infinity. A canonical example of countable infinity is the number of natural numbers N = {1,, 3,...}, and a good example of uncountable infinity is the number of points in the interval (0, 1). In fact, countable infinity is far smaller than uncountable infinity. Namely, countably infinite number of points, when put next to each other, fit in a line segment of length 0, while uncountably many points form a line of strictly positive length. In order to understand the difference, let us see how we can show N is of length 0. Proof Let 0 < ε < 1 be an arbitrary number. Then, we have ( ε(1 ε) 1 1, 1 + (..., n ) ( ε(1 ε), ) n εn (1 ε), n + εn (1 ε) ε (1 ε),.... ), + ε (1 ε), This means n is contained in an interval of length ε n (1 ε). Therefore, N is contained in an interval of length ε n (1 ε). n=1 Noting that a geometric series n=1 ar n for 0 < r < 1 converges to we have (a, r) = (ε(1 ε), ε), a and 1 r ε n (1 ε) = n=1 ε(1 ε) 1 ε = ε. Therefore, N can be contained in an interval of arbitrary length ε, which implies the length of an interval containing all natural numbers is 0.

91 4.. COUNTABLY INFINITE DIMENSIONS 89 You may wonder why the same argument cannot be used for uncountably many points, but the reason is in the name itself. If there are uncountably many points, you cannot include all of them even if you do n=1 ε n (1 ε) because this is nothing but counting, and you cannot count the uncountable. Here is an additional qualitative description of what is happening. Each point is, by definition, of length 0. If there are only countably many points, the total length is 0 even if there are infinitely many points. However, if there are uncountably many points, the total length is strictly positive as uncountable infinity overwhelms the length of 0 of each point. Think of the following two situations as analogues of the above example. 1. Let f(x) = 1/x and g(x) = x. Then, lim f(x) = 0 and lim g(x) = x x, but lim f(x)g(x) = lim 1/ x = 0. x x. Let f(x) = 1/x and g(x) = x. Then, x lim f(x) = 0 and x lim g(x) =, but lim f(x)g(x) = lim x =. x x In Case 1, g(x) behaves like countable infinity, while g(x) is like uncountable infinity in Case. It is now clear that countable infinity and uncountable infinity should be treated separately. 4. Countably Infinite Dimensions If the Hilbert space is of countably infinite dimensions, we have a complete orthonormal basis satisfying and { e 1, e,..., e n,...} or e k ; k = 1,,...,, (4.1) e i e j = δ ij (orthonormality) (4.) e i e i = I (completeness). (4.3) i=1

92 90 CHAPTER 4. SPACES OF INFINITE DIMENSIONALITY In this case, aside from the infinitely many components for vectors and infinitely many elements matrices carry, everything is the same as for finite dimensional Hilbert spaces. 4.3 Uncountably Infinite Dimensions Let us consider the set of position eigenkets { x }, each labeled by the position eigenvalue x. As x is an observable and X is a Hermitian operator, { x } forms a basis. In particular, it is an orthonormal basis if the set { x } satisfies and x x = δ(x x ) (orthonormality) (4.4) + x x dx = I (completeness); (4.5) where the right-hand side of (4.4) is a real-valued function called the Dirac delta function with the following definition and properties. Delta function is necessary whenever the basis kets are labeled by a continuous index including the position variable x. Heuristic Description Qualitatively, the delta function can be described by δ(x) = { x = 0 0 x 0 (4.6) and + δ(x) dx = 1. (4.7) Needless to say, this is an extended function, and no usual function has these properties. Rather than regarding the delta function as a function as we know them, we should think of it as a device to make quantum mechanics as well as other theories of physics work in a consistent fashion. Properties of the Dirac Delta Function Let us look at the properties of δ(x x ) found in (4.4).

93 4.3. UNCOUNTABLY INFINITE DIMENSIONS As before, we have δ(x x ) = 0 if x x + and (4.8) δ(x x ) dx = 1.. The delta function is sometimes called the sampling function as it samples the value of the function f(x ) at one point x; namely, + δ(x x )f(x ) dx = f(x). (4.9) 3. The delta function is even as shown below. δ(x x ) = x x = x x = δ(x x) = δ(x x) (4.10) The last equality holds because the delta function is a real-valued function as stated on p.90. Now, let f(x) be a differentiable function. Then, d dx f(x f(x x ) f((x h) x ) x ) = lim h 0 h and d dx f(x f(x (x + h)) f(x x ) x ) = lim h 0 h = lim h 0 f(x h x ) f(x x ) h = lim h 0 f(x x ) f((x h) x ) h (4.11) (4.1) = d dx f(x x ) (4.13) imply the following relation for the formal derivatives of the delta function. 1 δ (x x ) = d dx δ(x x ) = d dx δ(x x ) (4.14) 1 As the delta function is even, its derivative is an odd function. Generally speaking, let f(x) be a differentiable even function and y = x, then, for an arbitrary value x 0, d dx f(x) = d x=x0 dx f( x) = d x0 d( x) f( x) = d dy f(y) = d dx f(x). x= x0 So, we have f (x 0 ) = f ( x 0 ) for any x 0. x0 y= x0

94 9 CHAPTER 4. SPACES OF INFINITE DIMENSIONALITY 4. We have δ (x x ) = δ(x x ) d dx, and more generally, d n δ(x x ) dx n = δ(x x ) dn. (4.15) dx n Therefore, δ (x x )f(x ) dx = δ(x x ) d dx f(x ) dx = δ(x x )f (x ) dx = f (x), (4.16) and more generally, δ n (x x )f(x ) dx = δ(x x ) dn dx n f(x ) dx = δ(x x )f n (x ) dx = f n (x). (4.17) Next, consider δ(ax) for any real number a 0. If we let y = ax, then, dy = a = dx = dy, y : + if a > 0, and y : + if dx a a < 0. So, + + ( + ( ) y 1 f(x)δ(ax) dx = f δ(y) dy a) a = 1 ( ) 0 a f = 1 f(0) if a > 0 (4.18) a a and ( ( ) y 1 ( + ( ) y 1 f(x)δ(ax) dx = f δ(y) dy + a) a = f δ(y) dy a) a ) = 1 a f ( 0 a = 1 f(0) if a < 0. (4.19) a 5. The following rescaling property holds for the delta function. δ(ax) = δ(x) a for any a 0 (4.0)

95 4.4. DELTA FUNCTION AS A LIMIT OF GAUSSIAN DISTRIBUTION Delta Function as a Limit of Gaussian Distribution The Gaussian distribution g σ (x x ) labeled by the standard deviation σ is given by g σ (x x 1 ) = exp (x x ). (4.1) (πσ ) 1 σ The Gaussian function g σ is symmetric about x = x, has width σ, and its peak height is (πσ) 1 at x = x. Furthermore, the area under the curve is unity irrespective of the value of the standard deviation σ or variance σ. Hence, it is at least qualitatively obvious that g σ provides a better and better approximation of the delta function as σ approaches zero. 4.5 Delta Function and Fourier Transform For a function f(x), its Fourier transform is f(k) = 1 + e ikx f(x) dx, (4.) π while the inverse Fourier transform is f(x ) = 1 + π e ikx f(k) dk. (4.3) Substituting (4.) into (4.3), we obtain the following relation. + ( 1 + ) f(x ) = e ik(x x) dk f(x) dx (4.4) π Now, compare (4.4) with (4.9). Then, we can see + 1 π e ik(x x) dk = δ(x x) (4.5) or 1 + e ik(x x ) dk = δ(x x ). π (4.6) This is another characterization of the delta function. The width is defined as the distance between the two points of inflection of g σ.

96 94 CHAPTER 4. SPACES OF INFINITE DIMENSIONALITY 4.6 f as a Vector with Uncountably Many Components In order to understand a function f and its value at x, normally denoted by f(x), we will consider vectors in a finite dimensional space, a space of countably infinite dimension, and a space spanned by uncountably many basis vectors, in this order. Without loss of generality, we will consider orthonormal bases Finite Dimensions Consider a vector v in V n (an n-dimensional vector space). If a particular orthonormal basis B = { e 1, e,..., e n } is chosen, we can express v as a column vector with respect to B. We have 1 v = Countably Infinite Dimensions n B (4.7) n v = i e i (4.8) i=1 and i = e i v. (4.9) If the vector space has a countably infinite dimension, and an orthonormal basis B = { e 1, e,..., e n,...}, we can express v as an infinitely long column vector with respect to B. 1. v = ṇ.. B (4.30)

97 4.6. F AS A VECTOR WITH UNCOUNTABLY MANY COMPONENTS 95 We have v = i e i (4.31) i=1 and i = e i v. (4.3) Uncountably Infinite Dimensions When the vector space has an uncountably infinite dimension and an orthonormal basis B = { e x } x R or x (,+ ); where the kets in the basis B are indexed by a continuous variable x, each vector v has uncountably many components { x } x R with a continuous index x R, as opposed to a discrete index i N. We can still regard v as a column vector.. v = x (4.33). + By abuse of notation, we can write B v = x R x e x (4.34) and, as before, x = e x v. (4.35) Let us switch to a more familiar notation. Instead of e x and v, we write x and f. Then, we have f. f = f x, (4.36). f + B

98 96 CHAPTER 4. SPACES OF INFINITE DIMENSIONALITY f = x R f x x, (4.37) and Now let us write f(x) instead of f x to get f x = x f. (4.38) f = x R f(x) x, (4.39) and f(x) = x f. (4.40) Notationally more appropriate expression for f than (4.39) is obtained using the completeness relation (4.5). f = I f = + x x f dx = + f(x) x dx (4.41) In this view, a function f is a vector which is a linear combination of orthonormal basis vectors { x } x R, and its coefficients, components of the column vector with respect to the basis { x } x R, are our familiar f(x). We can rewrite (4.36) as follows. f = f( ). f(x). f(+ ) { x } x R (4.4) Finally, you can see that the inner product is the one defined for L -functions in Section.3. All we need to do is to use the completeness property (4.5). + + f g = f I g = f x x dx g = f x x g dx + + = x f x g dx = f(x) g(x) dx (4.43)

99 4.7. HERMITICITY OF THE MOMENTUM OPERATOR Hermiticity of the Momentum Operator Recall that the momentum operator P in one dimension is given by P = i x = i d dx. (4.44) This operator acts on differentiable L functions {f(x)} and has to be Hermitian as the momentum p is a physical observable. Let us verify this. We have the following in our notation. df P f = i (4.45) dx x P f = x i d dx f = x Checking P ij = P ji i df dx = i df (x) = i df(x) dx dx (4.46) In this section, we will check if P is represented by a matrix which is equal to its transpose conjugate. Because our basis consists of uncountably many orthonormal vectors { x } x R, it is actually more appropriate to write P xx = Px x; where P xx = x P x. Using the completeness property (4.5), we can rewrite (4.46). b x P f = x P I f = x P x x dx f = b a x P x x f dx = b a a x P x f(x ) dx = i df(x) dx (4.47) Note that we used a and b as our limits of integration because the domain of the wavefunction is not always (, + ). Now compare (4.16) and (4.47) to obtain x P x = i δ (x x ) = i δ(x x ) d dx. (4.48) As shown in the footnote on p.91, the first derivative of the real-valued δ- function δ (y) is an odd function. Hence, P x x = x P x = ( i δ (x x)) = ( i ) (δ (x x))

100 98 CHAPTER 4. SPACES OF INFINITE DIMENSIONALITY = i δ (x x) = i ( δ (x x )) = i δ (x x ) = P xx. (4.49) Therefore, the matrix representation of P = i d, denoted by P dx xx, is equal to its transpose conjugate. This was a necessary and sufficient condition for an operator to be Hermitian if we have a finite dimensional Hilbert space (See Definition.5). However, this condition alone is not sufficient when the dimension is uncountably infinite Hermiticity Condition in Uncountably Infinite Dimensions We will go back to the definition of Hermiticity. Definition.5 states that an operator Ω is Hermitian if and only if which is equivalent to Ω = Ω, (4.50) Ωv w = v Ωw (4.51) for all v and w. As we are interested in P = i d, we would like to see if dx i df dx g dg df = f i i dx dx g dg = i f dx df dx g dg = f. (4.5) dx Let us first work on the left-hand side. Completeness condition and integration by parts give us the following. df df dx g = dx I df g b = x x dx dx a g b df ( ) b df(x) b = a dx x df (x) x g dx = g(x) dx = g(x) dx a dx a dx = f (x)g(x) + b f (x)g (x) dx. (4.53) On the other hand, the right-hand side is dg b dg f = f x x dx dx a dx a b = f x a x dg dx dx

101 4.7. HERMITICITY OF THE MOMENTUM OPERATOR 99 b = f (x)g (x) dx. (4.54) a The expressions (4.53) and (4.54) are equal if and only if f (x)g(x) b a = 0. (4.55) Needless to say, this is not satisfied by all functions f and g. However, there are a couple of notable cases where (4.55) holds. Single-Valuedness : When the spherical coordinates (r, θ, ϕ) are used as in Chapter 10, (r, θ, 0) and (r, θ, π) are the same point in the three-dimensional space. Therefore, we require Ψ(r, θ, 0, t) = Ψ(r, θ, π, t) due to the single-valuedness condition, Condition, on p.14. In this case, a = 0 and b = π, and we indeed have f (x)g(x) b = f (ϕ)g(ϕ) a π 0 = Ψ (r, θ, ϕ, t)ψ(r, θ, ϕ, t) π 0 = 0 with respect to the ϕ-integral. 3 One class of functions that satisfy (4.55) consists of periodic functions. For example, if a = 0 and b = π, trigonometric functions such as sin x and cos x and their complex counterparts including e ix satisfy this condition. Functions Vanishing at Infinity : This is when a = and b = +, and we need f (x)g(x) + = 0. (4.56) In particular, this equality holds for L -functions, described on p.5 of Section.3, as f L (, + ) implies lim f(x) = 0. x The Eigenvalue Problem of the Momentum Operator P The eigenvalue problem for P in an arbitrary basis is P λ = λ λ. (4.57) 3 We will see in Chapter 10 that we can separate the four variables (r, θ, ϕ, t) and write Ψ as a product of functions of r, θ, ϕ, and t; namely Ψ(r, θ, ϕ, t) = R(r)Θ(θ)Φ(ϕ)T (t).

102 100 CHAPTER 4. SPACES OF INFINITE DIMENSIONALITY Taking the projection on x, x P λ = λ x λ. (4.58) Using completeness, ( x P λ = x P I λ = x P ) x x dx λ = x P x x λ dx = λ x λ. (4.59) Now, from (4.48), we have Hence, x P x = i δ(x x ) d dx. (4.60) x P λ = x P x x λ dx = If we denote x λ by f λ (x), we have i δ(x x ) d dx x λ dx = i d x λ (4.61) dx i d dx f λ(x) = λf λ (x). (4.6) Had we started with the X-basis made up of the eigenvectors of the position operator X, we would have obtained (4.6) immediately as P = i d in dx this basis. Solving the ordinary differential equation (4.6), we get f λ (x) = Ce iλx/ ; (4.63) where C is an arbitrary constant. It is now clear that P takes any real number λ as its eigenvalue with the corresponding eigenfunction Ce iλx in the X-basis. In order to normalize the eigenfunctions, we let C = 1 π. Then, x λ = f λ (x) = 1 π e iλx/, (4.64)

103 4.8. RELATIONS BETWEEN X AND P 101 and, due to (4.6) and (4.0), λ λ = + + = 1 π λ x x λ dx = 1 π 1 e i 1 (λ λ)x dx = 1 = 1 ( ) 1 δ (λ λ ) = 1 So, { λ } λ R forms an orthonormal basis. ( 1 π δ(λ λ ) 1/ + e iλx/ e iλ x/ dx π Relations Between X and P ) e i 1 (λ λ)x dx = δ(λ λ ). (4.65) The Fourier Transform Connecting X and P Given a physical system S, there is an associated Hilbert space H according to Postulate 1 on p.66. This Hilbert space is spanned either by the eigenkets of P (the P basis) or the eigenkets of X (the X basis) as both X and P are Hermitian operators. Given a ket f, it can be expanded either in the X basis { x } or in the P basis { p }. Note here that we switched the notation for the P basis from { λ } to { p } for clarity. Recall that we have f = x f x and f = p f p. (4.66) x R p R So, f(x) = x f is the x-th coefficient/component of f, and f(p) = p f is the p-th coefficient/component of f. They are coefficients in view of the expansion (4.66), but they are also components of the column vector f. Let us compare f(p) and f(x) using (4.64). f(p) = p f = + p x x f dx = + x p x f dx = f(p) = 1 + e ipx/ f(x) dx (4.67) π f(x) = x f = + x p p f dp

104 10 CHAPTER 4. SPACES OF INFINITE DIMENSIONALITY = f(x) = 1 + e ipx/ f(p) dp (4.68) π These are nothing but the Fourier transform and the Fourier inverse transform. Therefore, the expressions of a function f in this Hilbert space in terms of the complete X basis and complete P basis are related through the familiar Fourier transform. You may note that the more familiar form of the Fourier transform can be obtained if we set = 1. 4 Indeed, if we consider the operator K = P / and its orthonormal eigenbasis { k } k R, which satisfies x k = 1 π e ikx as opposed to x p = 1 π e ipx/, f(k) and f(x) are related as follows. f(k) = k f = f(x) = x f = + + k x x f dx = + x k x f dx = 1 + e ikx f(x) dx (4.69) π x k k f dk = 1 + e ikx f(k) dk (4.70) π As advertised, these are the Fourier transform and its inverse in their more familiar forms X and P in the X Basis We already know that the position operator X in the X basis is x or M x, which is a scalar multiplication. We also know that the momentum operator P in the X basis is i d. We can check the action of X in the framework dx of this chapter as follows. X x = x x = x X x = x x x = x x x = xδ(x x) (4.71) ( ) = x X f = x XI f = x X x x dx f = x X x x f dx = x δ(x x )f(x ) dx = xf(x) (4.7) In order to express this action in a basis-independent manner, we can adopt Shankar s notation Shankar, 1980, p This is the reason why we left π as it was even though π = π h π = h.

105 4.8. RELATIONS BETWEEN X AND P Basis-Independent Descriptions in Reference to the X Basis Recall that f is a basis-independent expression for a vector (function) in our Hilbert space. One way to specify f is by giving x f = f(x) for the orthonormal basis { x } x R. Differently put, we cannot really specify the nature of f unless we go to some basis such as the position basis {x} x R. With this specification, it is now possible to express f as a basis-independent ket in reference to the position orthonormal basis { x } x R. This potentially confusing strategy reduces to writing the basis-independent ket f as f(x). What does this mean? It means the following. 1. f itself is basis-independent.. However, we know f is fully characterized by the coefficients {f(x)} x R. 3. So, we write f(x) for an abstract basis-independent vector whose coefficients with respect to { x } x R are {f(x)} x R. For example, from (4.7), the coefficients of X f with respect to { x } x R, namely { x X f } x R, are {xf(x)} x R. Therefore, the action of X can be expressed in a basis-independent manner, but in reference to the orthonormal position basis { x } x R, as below. X f(x) = xf(x) (4.73) Similarly, a basis-independent description of the momentum operator P in reference to { x } x R is X and P in the P Basis From (4.57), we have P f(x) = i df(x). (4.74) dx P p = p p (4.75) in the P basis. Hence, the matrix element P pp = p P p is given by p P p = p p p = p p p = p δ(p p ). (4.76)

106 104 CHAPTER 4. SPACES OF INFINITE DIMENSIONALITY It remains to show how X operates in this basis. X pp = p X p can be obtained as follows. The matrix element First, note that i d dp Using (4.64) and (4.77), we get ( e ipx/ ) = i ( ix/ )e ipx/ = xe ipx/. (4.77) ( ) p X p = p XI p = p X x x dx p = p X x x p dx = p x x x p dx = x p x x p dx = x x p x p dx ( ) ( ) 1 1 = x e ipx/ e ip x/ dx = 1 xe ipx/ e ip x/ dx π π π = 1 i d ( e ipx/ e ) ip x/ dx = i d ( 1 ) e i(p p)x/ dx π dp dp π = i d dp δ(p p ) = i δ (p p ). (4.78) The in front of the delta function arises in the following manner. If we let y = x/, dx = dy, and y : + as x : +. So, 1 + e i(p p)x/ dx = 1 π π + ( 1 e i(p p)y dy = π + ) e i(p p)y dy = δ(p p) = δ(p p ). (4.79) You may realize that this can be modified to generate an alternative proof of Property 5 or (4.0) on p.9. Hence, the action of X in the K basis is described by Xf(k) = i df(k) dk. (4.80) As before, the basis independent form of this with our notational convention is X f(k) = i d dk f(k). (4.81)

107 4.8. RELATIONS BETWEEN X AND P 105 Let us summarize what we have shown in Sections 4.8. and In the X basis: In the P basis: X x or M x (4.8) P i d dx (4.83) X i d dp (4.84) The Commutator X, P P p or M p (4.85) Let X, P X and X, P P be commutators of X and P in the X and P bases, respectively. Then, ( X, P X f(x) = M x i d ) ( i d ) dx dx M x f(x) = i x df(x) dx + i d ( (xf(x)) = i xdf(x) dx dx + i f(x) + x df(x) ) dx = i f(x) = X, P X = i I (4.86) ( and ) ( M p M p X, P P f(p) = i d i d f(p) dp dp = i d ( (pf(p)) i pdf(p) dp dp = i f(p) + p df(p) ) i p df(p) dp dp ) = i f(p) = X, P p = i I. (4.87) Because the actual expression for the identity operator I is basis-independent, we do not write I X or I P. We conclude that in either basis. X, P = i I (4.88)

108 106 CHAPTER 4. SPACES OF INFINITE DIMENSIONALITY

109 Chapter 5 Schrödinger s Theory of Quantum Mechanics In 195 an Austrian physicist Erwin Rudolf Josef Alexander Schrödinger proposed the Schrödinger Equation which is a differential equation whose solutions are wavefunctions. He chose a differential equation to describe new physics since a differential equation was the most common type of equation known to the physicists with a function for a solution. In Chapter 3, the postulates of quantum mechanics were presented. However, these postulates were put together in a post hoc fashion. In this chapter, we will learn how quantum mechanical framework was initially constructed, drawing on the analogy to classical mechanics. 5.1 How was it derived? In his derivation, he was guided by the following. 1. The classical traveling wavetraveling wave: a simple sinusoidal traveling wave ( ) x Ψ(x, t) = sin π λ νt (5.1). The de Broglie-Einstein postulates λ = h p, ν = E h (5.) 107

110 108 CHAPTER 5. SCHRÖDINGER S THEORY OF QUANTUM MECHANICS 3. The Newtonian energy equation E = P m + V (5.3) 4. The equation should be linear in terms of its solutions. If Ψ 1 (x, t), Ψ (x, t) (5.4) are two solutions of the Schrödinger Equation, then any linear combination of the two Ψ(x, t) = c 1 Ψ 1 (x, t) + c Ψ (x, t) (5.5) is also a solution; where c 1 and c are arbitrary constants. = This makes superposition and interference possible. Here is what we want. A differential equation, later named the Schrödinger Equation, which looks like the energy equation. Wave functions, which are the solutions of the Schrödinger Equation and behave like classical waves, including interference and superposition. Let us first consider the energy equation and observe the following. Let k = π 1 and ω = πν. Then, we have λ On the other hand, the energy equation is Ψ(x, t) = sin(kx ωt). (5.6) E = P m + V. 1 This k is called the wavenumber or angular/circular wavenumber to make a clear distinction from another wavenumber 1/λ, which is also called the spectroscopic wavenumber. By definition, k is the number of wavelengths per π units of distance.

111 5.1. HOW WAS IT DERIVED? 109 Plug in P = h λ and E = hν to obtain h + V (x, t) = hν. (5.7) mλ Recall that k = π λ and ω = πν. So, we have the following relations. ( 1 λ ) = ( ) k ν = ω π π (5.8) We now substitute these back into the energy equation. E = P m + V = h mλ + V (x, t) = hν = h k = ( h π ) k Now we introduce a new notation. 1 + V (x, t) = m ( ) (π) 1 ω m + V (x, t) = h π ( ) h ω (5.9) π = h π (5.10) When you read it aloud, it is pronouced either h bar or h cross, with h bar seeming more dominant these days. With this notation, we finally arrive at k m + V (x, t) = ω. (5.11) We will next look at the classical sinusoidal traveling wave given by Ψ(x, t) = sin(kx ωt). Here are three of its partial derivatives which are of interest to us. Ψ = k cos(kx ωt) x (5.1) Ψ x = k sin(kx ωt) (5.13) Ψ = ω cos(kx ωt) t (5.14)

112 110 CHAPTER 5. SCHRÖDINGER S THEORY OF QUANTUM MECHANICS From these, we arrive at the implications below. = x gets us k. t gets us ω. (5.15) Note that this argument is all qualitative and of strory-telling type. So is the rest of the argument. By abuse of notation, let Ψ(x, t) be the desired wave function which acts like the traveling wave. Since the energy equation, whose solution is Ψ(x, t), contains k and ω, we consider the correspondences below. k m Ψ(x, t) (5.16) x ω Ψ(x, t) (5.17) t We have used guiding assumptions (hypotheses) 1 through 3 to arrive at α Ψ(x, t) Ψ(x, t) + V (x, t) = β x t Why α and β? Recall de Broglie. We were off there by a constant. (5.18) How about the fourth assumption; the linearity assumption? Let Ψ 1 and Ψ be two solutions of the above equation. Then we have αψ 1,xx + V = βψ 1,t and αψ,xx + V = βψ,t = α(ψ 1,xx + Ψ,xx ) + V = β(ψ 1,t + Ψ,t ). (5.19) We are unfortunately off by the factor in front of the potential V. And one way around this problem is to modify the equation in the following way. α Ψ(x, t) Ψ(x, t) + V (x, t)ψ(x, t) = β x t (5.0)

113 5.1. HOW WAS IT DERIVED? 111 With a little more work, we can determine α and β. The easiest way is to consider V = 0 which gives us a free particle or a free traveling wave Substituting V = 0 and (5.1) into (5.0), α Ψ(x, t) Ψ(x, t) + V (x, t)ψ(x, t) = β x t Ψ(x, t) = e i(kx ωt). (5.1) α x ei(kx ωt) = β t ei(kx ωt) = αk e i(kx ωt) = iωβe i(kx ωt) = αk = iωβ (5.) Comparing (5.) with (5.11) when V (x, t) = 0, we get Therefore, αk = iωβ k m = ω. (5.3) And our end product is α = m and β = i = i. (5.4) Ψ(x, t) Ψ(x, t) + V (x, t)ψ(x, t) = i. (5.5) m x t This is known as the Schrödinger Equation or the Time-Dependent Schrödinger Equation. We often write it as below. Ψ(x, t) + V (x, t) Ψ(x, t) = i (5.6) m x t The solution set {Ψ(x, t)} represents the wave functions which are to be associated with the motion of a particle of mass m under the influnece of the force F described by V (x, t); namely F (x, t) = V (x, t). (5.7) x Agreement with experiment? How do you check it? We first need a physical interpretation of Ψ to do so.

114 11 CHAPTER 5. SCHRÖDINGER S THEORY OF QUANTUM MECHANICS 5. Born s Interpretation of Wavefunctions We now have Ψ(x, t) m + V (x, t)ψ(x, t) = i x or + V (x, t) m x Ψ(x, t) = i Ψ(x, t) t Ψ(x, t). t (5.8) So, if V (x, t) is given, a wavefunction Ψ(x, t) can be obtained at least in principle. But, what is Ψ? Since Ψ is a complex function, Ψ itself can not be a real physical wave. How can we extract physical reality from this? Max Born realized that Ψ Ψ = Ψ is real while Ψ itself is not and postulated that P (x, t) = Ψ(x, t) Ψ(x, t) can be regarded as a probability. He postulated that the probability P (x, t)dx of finding the particle at a coordinate between x and x + dx is equal to Ψ(x, t) Ψ(x, t)dx. However, there is a problem with this view as it is. It is too simplistic. To understand the nature of the problem simply note that Ψ(x, t) is also a solution if Ψ(x, t) is a solution of Ψ(x, t) Ψ(x, t) + V (x, t)ψ(x, t) = i. m x t This means that the probability is not unique, which of course is not acceptable. Furthermore, the total probability is not necessarily one if we used an arbitrary solution. Our necessary condition is + In other words, we need to find Ψ such that + P (x, t)dx = P (x, t)dx = 1. (5.9) + Ψ (x, t)ψ(x, t)dx = 1. (5.30) This is the physically meaningful Ψ called a normalized wavefunction. This process is called normalization.

115 5.3. EXPECTATION VALUES Expectation Values Recall that for a die value probability = = ( ) 1 6 gives the expectation value. Question : How do we calaulate the average position of a particle? Answer : By computing the expectation value of the position x denoted by x or x. Analogously to the die above, with x as the value, we have The continuous version of the above is x = x = + x = x probability. (5.31) xψ (x, t)ψ(x, t)dx = + How about P and E? We are tempted to write and It turns out P = E = + + Ψ (x, t)xψ(x, t)dx. (5.3) Ψ (x, t)p Ψ(x, t)dx (5.33) Ψ (x, t)eψ(x, t)dx. (5.34) Recall we are always to be guided by the classical wave. Instead of a sine wave, let us look at a particular solution of the Schrödinger Equation called a free particle solution. The free-ness of a free partilce derives from the lack of any external force. In other words, V (x, t) = 0 for a free particle. + V (x, t) m x Ψ(x, t) = i Ψ(x, t) t

116 114 CHAPTER 5. SCHRÖDINGER S THEORY OF QUANTUM MECHANICS = t) Ψ(x, t) = i Ψ(x, m x t (5.35) We will solve this in detail in the next chapter, but the answer is the classical traveling wave. Previously, we used Now we will use its complex version. Ψ(x, t) = sin(kx ωt). Ψ(x, t) = cos(kx ωt) + i sin(kx ωt) = e i(kx ωt) (5.36) Taking the first partial derivative with respect to x, where the last equality follows from Therefore, Ψ x = ikψ(x, t) = ip Ψ(x, t); (5.37) k = π λ = π h/p = P h/π = P. Similarly, P Ψ(x, t) = i Ψ(x, t). (5.38) x Ψ(x, t) t where the last equality follows from Therefore, = iωψ(x, t) = i E Ψ(x, t); (5.39) ω = πν = π E h = E h/π = E. E Ψ(x, t) = i Ψ(x, t). (5.40) t

117 5.3. EXPECTATION VALUES 115 On the other hand, let us compare with term by term. The first terms give us Working formally, + V (x, t) m x P + V (x, t) = E m m Ψ(x, t) Ψ(x, t) = i t x = P m. x = P = ±i x = P We will choose the minus sign because of our previous work. How about E? Quite straightforwardly, we obtain E = i t. Hence, it is consistent, so to speak, with the Schrödinger Equation. We have P Ψ(x, t) = i Ψ(x, t) x E Ψ(x, t) = i Ψ(x, t) t. (5.41) So, and + ( + P = Ψ (x, t)p Ψ(x, t)dx = Ψ (x, t) i ) Ψ(x, t)dx x + = i Ψ Ψdx (5.4) x E = + + Ψ (x, t)eψ(x, t)dx = i Ψ Ψdx. (5.43) t

118 116 CHAPTER 5. SCHRÖDINGER S THEORY OF QUANTUM MECHANICS Incidentally, this is the same as E = + Ψ (x, t) ( An Infinite Square Well Potential + V (x, t) m x ) Ψ(x, t)dx. (5.44) Consider a particle with total energy E in the potential well given below. 0 a < x < a V (x, t) = x a (a > 0) (5.45) One normalized solution to the Schrödinger Equation is Ψ(x, t) = A cos πx a e iet/ a < x < a 0 x a (a > 0) ; (5.46) where A is known as a normalization constant. Let us conduct a consistency check with this function. Since we get Ψ t = ie e iet/ A cos πx a = ie Ψ(x, t), (5.47) + E = i Ψ (x, t) Ψ(x, t) t = E + + dx = i However, since we have a normalized wavefunction, + Ψ (x, t) ie Ψ(x, t) Ψ Ψdx. (5.48) Ψ Ψdx = 1 and E = E.

119 5.3. EXPECTATION VALUES 117 I said A was a normalization constant, but I never told you what its value was. Let s find out. + = + a a Ψ Ψdx = = A + a + a a ( A cos πx ) πx a e iet/ A cos a e iet/ dx A cos πx a e+iet/ A cos πx a e iet/ dx = a cos πx dx = A a + a 0 + a a A A cos πx a e+iet/ e iet/ dx cos πx dx = 1 (5.49) a If we let y = πx, we get dx = a dy and y : 0 π as x : 0 a. So, integration a π by this subsitution gives us + a 0 Therefore, cos πx + a dx = 0 π ( ) ( ) a a y cos ydy = π π sin y π/ = a π π 4 = a 4. (5.50) A a 4 = A a = 1 = A = a. (5.51) Note here that the normalization constant A can not be detrmined uniquely. Indeed, the above relation indicates that there are infinitely many values of A that serves the purpose. A = a = A = a eiθ ; (5.5) where θ is any real number. This fact that the normalization constant A can only be specified up to the argument is another reason why the wavefunction itself does not represent physical reality.

120 118 Exercises 1. Consider a classical sinusoidal traveling wave given by the equation below; where λ is the wavelength and ν is the frequency so that νλ = (the speed of its propagation). π Ψ(x, t) = A sin λ (x t) Show that Ψ(x, t) = Ψ(x + t 0, t + t 0 ), and briefly explain what this means.. Write down, but do not derive, the time-dependent Schrödinger equation for V (x, t). This is a one-dimensional case, and the only variables are the spatial variable x and the temporal variable t. 3. If Ψ 1 (x, t) and Ψ (x, t) are both solutions of the Schrödinger equation, show that any linear combination αψ 1 (x, t)+βψ (x, t) is also a solution; where α and β are scalars. 4. Define e iθ by e iθ = cos θ + i sin θ and prove the following relationships. (a) e iθ e iϕ = e i(θ+ϕ) (b) ( e iθ) n = e inθ 5. This problem is the same as Chapter 7 Problem 1. Consider a particle of mass m and total energy E which can move freely along the x-axis in the interval a, + a, but is strictly prohibited from going outside this region. This corresponds to what is called an infinite square well potential V (x) given by V (x) = 0 for a < x < + a and Note that νλ = = ω = πν = π λ. Therefore, we have π Ψ(x, t) = A sin λ x ωt.

121 119 V (x) = elsewhere. If we solve the Schrödinger equation for this V (x), one of the solutions is Ψ(x, t) = A cos ( ) πx a e iet/ a < x < a 0 x a (a > 0) (a) Find A so that the function Ψ(x, t) is properly normalized. (b) In the region where the potential V (x) = 0, the Schrödinger equation reduces to t) Ψ(x, t) = i Ψ(x,. m x t Plug Ψ(x, t) = A cos ( ) πx e iet/ into this relation to show E = π ma. (c) Show that P = 0. a (d) Evaluate x for this wavefunction. You will probably need an integral table for this. (Of course, you can always try contour integration or some such. But, that is beyond this course.) (e) Evaluate P for the same wavefunction. You will not need an integral table if you use the fact that the function is normalized. Of course, a brute force computation will yield the same result as well. 6. This problem is the same as Chapter 7 Problem. If you are a careful student who pays attention to details, you may have realized that A can only be determined up to the argument, or up to the sign if you assume A is a real number. (a) What is the implication of this fact as to the uniqueness of wavefunction? (b) What is the implication of this fact as to the probability density Ψ (x, t)ψ(x, t)? How about P, P, and x?.

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123 Chapter 6 The Time-Independent Schrödinger Equation In the Schrödinger Equation + V (x, t) m x Ψ(x, t) Ψ(x, t) = i t (6.1) the potential often does not depend on time; i.e. V (x, t) = V (x). An example of this would be the infinite square well potential. 0 a < x < a V (x, t) = x a (a > 0) (6.) When V (x, t) = V (x), we can express the multivariable function Ψ(x, t) as a product of a function of x and a function of t as follows. Ψ(x, t) = ψ(x)ϕ(t) (6.3) This technique is called separation of variables. Let us substitute the above Ψ into the Schrödinger Equation. m x + V (x) ψ(x)ϕ(t) = i ψ(x)ϕ(t) t = ψ(x)ϕ(t) + V (x)ψ(x)ϕ(t) = i ψ(x)ϕ(t) m x t 11 (6.4)

124 1 CHAPTER 6. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION = ϕ(t) ψ(x) + V (x)ψ(x)ϕ(t) = ψ(x)i ϕ(t) m x t = ϕ(t) ψ(x) + V (x)ψ(x) = ψ(x)i ϕ(t) m x t (6.5) Dividing through by ψ(x)ϕ(t), we get 1 d ψ(x) + V (x)ψ(x) = i 1 ψ(x) m dx ϕ(t) } {{ } H(x) dϕ(t) dt } {{ } K(t) ; (6.6) where we changed the partial derivatives to total derivatives as we now have one-variable functions. Generally speaking, H(x) = K(t) for all pairs (x, t) (6.7) means that both H(x) and K(t) are a constant. Let H(x) = K(t) = G (a constant). Then, K(t) = G i 1 dϕ(t) ϕ(t) dt = G dϕ(t) dt = 1 i Gϕ(t) = Gϕ(t). (6.8) i Therefore, ϕ(t) = Ae igt/. (6.9) Compare this with the time-dependent part of the travelling wave e i(kx ωt) = e ikx e iωt. (6.10) We realize that G = π G h should be ω. π G h We now have G = E, and ωh = ω = G = π = πνh π = hν = E. (6.11) ϕ(t) = e iet/. (6.1)

125 13 This also implies H(x) = 1 ψ(x) m where E is the total energy. d d ψ(x) + V (x)ψ(x) dx = E = ψ(x) + V (x)ψ(x) = Eψ(x); (6.13) m dx d ψ(x) + V (x)ψ(x) = Eψ(x) (6.14) m dx or d m dx + V (x) is the Time-Independent Schrödinger Equation. Note here that the full wavefunction is given by ψ(x) = Eψ(x) (6.15) Ψ(x, t) = ψ(x)e iet/. (6.16) In the rest of this course, we will solve the Time-Independent Schrödinger Equation for various systems, culminating in a solution for the hydrogen atom. Physical Sense Revisited: Previously, I mentioned that only normalized solutions are the physically meaningful solutions in order for the Born s interpretation to make sense. However, this restriction alone would not eliminate all the mathematically correct solutions which are not acceptable on physical grounds. Hence, we need to put further restrictions on the nature of the solutions. We will place three types of restrictions on both ψ(x) and dψ(x). dx To be precise, the wavefunction ψ will most likly depend on x, y, and z for actual physical systems, and we will have ψ(x, y, z) instead of ψ(x). These are ad-hoc and post-hoc conditions as usual albeit physically very reasonable. 1. Finiteness and Integrability: The solution and its first derivative have to take a finite value for all x. Furthermore, the wavefunction has to be normalizable; i.e. + ψ ψdx <.

126 14 CHAPTER 6. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION. Single-valuedness: The solution and the first derivative should have one value at each point in space. 3. Continuity: Both the solution and its first derivative have to be continuous everywhere. We will find solutions of this type. Incidentally, Condition 3 on the derivative can not be imposed if the potential V (x) itself has a discontinuity. We will see such an example in Chapter 7.

127 Exercises Write down, but do not derive, the time-independent Schrödinger equation for V (x), that is, V is now a function of x only. This is a onedimensional case, and the only spatial variable is x.. The time-dependent Schrödinger Equation is Ψ(x, t) Ψ(x, t) + V (x, t)ψ(x, t) = i. m x t Consider a case where the potential does not depend on time; i.e. V (x, t) = V (x). Assume Ψ(x, t) = ψ(x)e iet/ and derive the timeindependent Schrödinger Equation. 3. Discuss in concrete terms and sufficient detail why Φ(ϕ) = sin ϕ is an acceptable wavefunction while Φ(ϕ) = ϕ is not. We are using spherical polar coordinates here.

128 16

129 Chapter 7 Solutions of Time-Independent Schrödinger Equations in One Dimension Though you may, perhaps naively, think our world is three-dimensional, there are many physical systems which are of lower dimensions. For example, a thin film forms a two-dimensional system, and a particle moving freely without any external force is the simplest one-dimensional system. Furthermore, the Schrödinger equations for one-dimensional systems are easier to solve. Hence, it makes good mathematical and physical sense to start with one dimensional cases. 7.1 The Zero Potential This is the case where V (x) = 0 for all x. Our Time-Independent Schrödinger Equation is d ψ(x) = Eψ(x). (7.1) m dx The most general solution for this second order linear ordinary differential equation is of the form ψ(x) = A 1 sin me x + A cos me x = C 1 e ikx + C e ikx ; (7.) 17

130 18 CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRÖDINGER EQUATIONS IN ONE DIMENSION where A 1, A, C 1, and C are arbitrary constants, and k = me. On the other hand, we always have for time dependence. Then, ϕ(t) = e iet/ = e ihνt/(h/π) = e iπνt = e iωt (7.3) Therefore, the full time-dependent wavefunction is Let us take Ψ(x, t) = ψ(x)ϕ(t) = ( C 1 e ikx + C e ikx) e iωt + = C 1 e i(kx ωt) + C e i(kx+ωt). (7.4) Ψ(x, t) = C 1 e i(kx ωt) ( = C 1 e ikx e iet/ ). + P = P = Ψ P Ψdx = C1e ikx e iet/ ( i ) x C 1e ikx e iet/ dx + ( + ) = C1e ikx e iet/ ( i )(ik)c 1 e ikx e iet/ dx = Ψ Ψdx This makes sense as + = k ψ ψdx = k = h π me P m = E = P = me = me. (7.5) classically, and (7.5) serves as yet another consistency checking device. As simple as the concept and the wavefunctions of a free particle are, there is one serious complication regarding normalization. To see this, consider an eigenfunction ψ(x) = Ae ikx. In order for this function to be normalizable, it has to be integrable to begin with. However, + ( ) Ae ikx ( Ae ikx) + + dx = A e ikx e ikx dx = A 1 dx =,

131 7.. THE STEP POTENTIAL (E < V 0 ) 19 and ψ(x) is not normalizable in the usual way. In order to circumvent this problem, three normalization schemes have been proposed; namely, the Born normalization, Dirac normalization, and unit-flux normalization. As these normalization procedures are not crucial for the rest of this book, their discussions are relegated to Appendix??. 7. The Step Potential (E < V 0 ) We now consider a step potential given by V (x) = { V0 x > 0 0 x 0 (7.6) for E < V 0. In Newtonian (classical) mechanics, the particle can not enter the region (0, ). We have E = K.E. + P.E. = K.E. + V 0 in (0, ) = K.E. = E V 0 < 0. This does not make any sense of any kind classically. Is this also the case in quantum mechanics? Let us just plunge in and solve it! The time-independent Schrödinger Equation is In our case, we have the following. m d ψ(x) + V (x)ψ(x) = Eψ(x). m dx d ψ(x) m dx = Eψ(x) x 0 (7.7) d ψ(x) + V dx 0 ψ(x) = Eψ(x) x > 0 (7.8) Equation 7.7 is that for a free particle. Therefore, the general solution is ψ(x) = Ae ik 1x + Be ik 1x ; (7.9)

132 130 CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRÖDINGER EQUATIONS IN ONE DIMENSION where k 1 = me, and A and B are arbitrary constants. On the other hand, (7.8) can be rewritten as follows. d ψ(x) m dx The general solution for x > 0 is = (E V 0 )ψ(x) d ψ(x) = (V m dx 0 E)ψ(x) d ψ(x) = m(v 0 E) ψ(x) (7.10) dx ψ(x) = Ce kx + De kx ; (7.11) m(v 0 E) where k =, and C and D are arbitrary constants. Note that V 0 E is positive in this region. As stated above, these are indeed the solutions, or the (most) general solutions, since we have linear second order ordinary differential equations. We have ψ(x) = Ae ik1x + Be ik 1x (k 1 = me ) x 0 ψ(x) = Ce kx + De k x m(v (k = 0 E) ) x > 0. (7.1) Recall the conditions on (ψ, dψ dx ). 1. Finite. Single-valued 3. Continuous We will determine A, B, C, and D using these conditions. Let us begin with the region x > 0. In this region, m(v ψ(x) = Ce kx + De kx 0 E) ; where k =. (7.13) Therefore, ψ (x)ψ(x) = ( C e k x + D e k x ) ( Ce k x + De k x )

133 7.. THE STEP POTENTIAL (E < V 0 ) 131 = C e k x + ReC D { }} { D C + C D + D e k x. (7.14) Since the wavefunction has to be normalizable, it has to satisfy the finiteness condition + ψ (x)ψ(x)dx = + ψ(x) dx <. (7.15) Noting that this ψ is the wavefuntion only in the region x > 0, we actually need + 0 ψ(x) dx <. (7.16) Hence, we have to have C = 0. At this point, we have ψ(x) = Ae ik 1x + Be ik 1x ψ(x) = De k x (k = k 1 = me x 0 m(v 0 E) ) x > 0. (7.17) Next, we will use the continuity condition on ψ(x) and dψ(x) dx. Continuity of ψ(x) at x = 0 gives De k 0 = Ae ik Be ik 1 0 = D = A + B. (7.18) Since the first derivative is dψ(x) = A(ik dx 1 )e ik1x + B( ik 1 )e ik 1x k 1 = me x 0 dψ(x) = D( k dx )e k x m(v 0 E) (k = ) x > 0, (7.19) continuity of dψ(x) dx at x = 0 gives k De k 0 = ik 1 Ae ik 1 0 ik 1 Be ik 1 0 = k D = ik 1 (A B) = k ik 1 D = A B. (7.0)

134 13 CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRÖDINGER EQUATIONS IN ONE DIMENSION We now have a simultaneous equation { ik k 1 D = A B D = A + B (7.1) for A, B, and D with the solution This gives ( ) A = ik k 1 D ( ) B = 1 1 ik. (7.) k 1 D ψ(x) = { D (1 + ik /k 1 )e ik 1x + D (1 ik /k 1 )e ik 1x x 0 De k x x > 0. (7.3) So, the full solution is Ψ(x, t) = ψ(x)ϕ(t) = ψ(x)e iet/ { D = (1 + ik /k 1 )e i(k1x Et/ ) + D(1 ik /k 1 )e i(k 1x+Et/ ) x 0 De kx e iet/ x > 0. (7.4) Recall that This is because Therefore, in the region x 0, e iet/ = e iωt. Et/ = hνt/ = hνt/( h ) = πνt = ωt. π Ψ(x, t) = Ae i(k 1x ωt) + Be i( k 1x ωt). (7.5) Note that is traveling to the right while Ae i(k 1x ωt) Be i( k 1x ωt) is traveling to the left. In other words, Ae i(k 1x ωt)

135 7.. THE STEP POTENTIAL (E < V 0 ) 133 is the incident wave, incident on the step form the left, while is the wave reflected by the step. Be i( k 1x ωt) What is the reflection coefficient? Since the reflection coefficient is the ratio of the amplitude of the reflected wave and the amplitude of the incident wave, we have the reflection coefficient = = Bei( k 1x ωt) = B e i( k1x ωt) Ae i(k 1x ωt) A e i(k 1x ωt) ( ) D ( ) ( 1 i k D k = 1 1 i k k i k ( ) D ( ) = 1 + i k D k i k k 1 And, we have a total reflection. the amplitude of the reflected wave the amplitude of the incident wave = B A = B B A A ) ( ) k 1 1 i k k ( ) ( 1 ) = 1 (7.6) 1 i k k i k k 1 Indeed, we have a standing wave in the region x 0. To see this, plug e ik 1x = cos k 1 x + i sin k 1 x into to obtain ψ(x) = { D (1 + ik /k 1 )e ik1x + (1 ik /k 1 )e ik 1x x 0 De k x x > 0 (7.7) ψ(x) = { D cos k1 x D k k 1 sin k 1 x x 0 De k x x > 0. (7.8) So, for x 0, ( Ψ(x, t) = D cos k 1 x k ) sin k 1 x e iet/. (7.9) k 1 Remember the following from high school math, α cos kx β sin kx = ( α β cos kx α + β α + β sin kx ) α + β

136 134 CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRÖDINGER EQUATIONS IN ONE DIMENSION = cos(kx + θ); (7.30) where θ is such that cos θ = α β and sin θ = α + β α + β. (7.31) Therefore, the nodes of Ψ(x, t) Ψ(x, t) do not move, and we indeed have a standing wave. While Now, in the region x > 0, Ψ (x, t)ψ(x, t) = D De k x. lim x Ψ Ψ = 0, Ψ Ψ 0 for x > 0. This means that the probability of finding the particle to the right of the step is not zero even if the total energy E is smaller than the step height V 0. This phenomenon is definitely non-classical and is known as barrier penetration. 7.3 The Step Potential (E > V 0 ) We will now consider the case where the total energy E is greater than the step height. Classically, we have the following. 1. Not a total reflection. Has to be a total penetration into (0, ) But, quantum mechanically, the Schrödinger Equation predicts the following. 1. Not a total reflection (agreement with the classical theory). There is partial reflection at the boundary x = 0 (disagreement)

137 7.3. THE STEP POTENTIAL (E > V 0 ) 135 Let us see how it works. As usual, Ψ(x, t) = ψ(x)e iet/, and the Time-Independent Schrödinger Equation is { d ψ(x) m d ψ(x) m dx = Eψ(x) x < 0 dx = (E V 0 )ψ(x) x > 0 (7.3) As E V 0 > 0 this time, the two equations are basically the same. They are both free particles, and the solutions are { Ae ik 1 ψ(x) = x + Be ik 1x x < 0 Ce ikx + De ik x x > 0 ; (7.33) where k 1 = me and k = m(e V 0 ). Suppose the particle is in the region x < 0 or (, 0) at t = 0. In other words, the particle is incident on the step from the left. Then, since the wave is travelling to the right in the region (0, ), we should set D = 0. This is simply because e ik x e iet/ is a wave travelling to the left. Imposing the continuity condition on ψ and dψ dx the constants B and C in terms of A as follows. at x = 0, we can express ψ(x) x=0 = ψ(x) x=0+ = dψ(x) x=0 dψ(x) dx dx x=0+ = { A + B = C k 1 (A B) = k C = { B = A k 1 k k 1 +k C = A k 1 k 1 +k (7.34) Therefoere, ψ(x) = The reflection coefficient R is given by { Ae ik 1 x + A k 1 k k 1 +k e ik 1x A k 1 k 1 +k e ik x (7.35) R = B B A A = ( ) k1 k. (7.36) k 1 + k

138 136 CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRÖDINGER EQUATIONS IN ONE DIMENSION Since the transmission coefficient T is related to R via R + T = 1, T = 1 R = 1 ( ) k1 k = (k 1 + k + k 1 k )(k 1 + k k 1 + k ) k 1 + k (k 1 + k ) = 4k 1k (k 1 + k ). (7.37) As advertised, we do not have a total reflection, and we do not have a total penetration into (0, ), either. We now solve the same problem for a wave incident on the potential boundary at x = 0 from the right. In other words, the particle is in the region (0, ) at t = 0 and traveling to the left. Our general solution is ψ(x) = where k 1 = me and k = { Ae ik 1 x + Be ik 1x x < 0 Ce ik x + De ik x x > 0 ; (7.38) m(e V 0 ). Since there is no boundary to reflect the wave in the region (, 0), the wave should be traveling to the left in this region. We set A = 0. ψ(x) = { Be ik 1 x x < 0 Ce ik x + De ik x x > 0 (7.39) The first derivative with respect to x is ψ (x) = { ik1 Be ik 1x x < 0 ik Ce ik x + ( ik )De ik x x > 0. (7.40) Imposing the continuity condition on ψ(x) and dψ(x) at x = 0, we get dx { { B = C + D C + D = B = ik 1 B = ik (C D) C D = k 1 k B { C = k k 1 k = B D = k +k 1 k B. (7.41)

139 7.4. THE BARRIER POTENTIAL (E < V 0 ) 137 What are the reflection coefficient and the transmission coefficient in this case? R = C C D D = ( ) k k 1 k B B ( ) k +k 1 B B k ( ) k1 k = (7.4) k 1 + k and T = 1 R = 4k 1k (k 1 + k ) (7.43) Note that these are the same as before when we computed R and T for the particle moving in from the left. Incidentally, B 0 in (7.33) can be proved on purely mathematical gorunds as follows. If B = 0, ψ(x) = { { Ae ik 1 x x < 0 A = C Ce ik x x > 0 = k 1 A = k C. But, it is obvious that this can not be solved for A and C unless we accept A = C = 0 which is physically meaningless. Hence, B 0 on this ground. 7.4 The Barrier Potential (E < V 0 ) The potential V (x) is V 0 in the region (0, a) and 0 elsewhere. V (x) = { V0 0 < x < a 0 x < 0, a < x Our Time-Independent Schrödinger Equation is Ae ik 1x + Be ik 1x x < 0 ψ(x) = F e k x + Ge k x 0 < x < a Ce ik 1x + De ik 1x x > a (7.44) ; (7.45)

140 CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRÖDINGER 138 EQUATIONS IN ONE DIMENSION where k 1 = me m(v 0 E) and k =. Assuming x < 0 at t = 0, D = 0. On the other hand, our boundary conditions are ψ(x) x=0 = ψ(x) x=0+ ψ(x) x=a = ψ(x) x=a+ dψ(x) = dψ(x). (7.46) dx x=0 dx x=0+ = dψ(x) x=a x=a+ dψ(x) dx We have the above four equations and the five unknowns A, B, C, F, and G. But, we also have a normalization condition. So, we can solve for the five unknowns. As it turned out, C 0, and the probability density Ψ (x, t)ψ(x, t) = ψ (x)ψ(x) has a (nonzero) tail in the region x > a. Of course, it is nonzero also inside the barrier. Therefore, we have both barrier penetration and tunneling. dx 7.5 The Infinite Square Well Potential The potential is given by V (x) = { x > a 0 x < a, (7.47) and we get, as we have already seen, ψ(x) = { 0 x > a A sin kx + B cos kx x < a ; (7.48) where k = me. As usual, the full wavefunction is Ψ(x, t) = ψ(x)e iet/. (7.49) It turned out the continuity condition of the first derivative is too restrictive for this system as the potential blows up to unlike any known physical

141 7.5. THE INFINITE SQUARE WELL POTENTIAL 139 system. The continuity condition on ψ(x) at x = a gives Therefore, { A sin ka + B cos ka { = 0 B cos ka A sin ka + B cos ka = 0 = = = 0 A sin ka = 0 { A = 0 and cos ka = 0 B = 0 and sin ka = 0. (7.50) ψ(x) = B cos kx where cos ka = 0 or ψ(x) = A sin kx where sin ka = 0. (7.51) We have two families of solutions. { ψn (x) = B n cos k n x where k n = nπ n = 1, 3, 5,... a ψ n (x) = A n sin k n x where k n = nπ n =, 4, 6,... a. (7.5) Note that n is a positive integer in both cases because k n > 0. We can also note that negative n s give redundant solutions and n = 0 gives a physically meaningless trivial solution. Indeed, n = 0 = ψ 0 (x) = A 0 sin 0 = 0. Let us compute the normalization constants A n and B n. We will obtain real and positive constants. + a a (B n cos k n x) (B n cos k n x) dx = B n + a = B n + a = B n = B n = B n = 1 + cos k n x dx = B n a x + a + a nπ sin nπ a x a ( a + a ) ( nπ sin nπ a a a cos k n x dx x + 1 k n sin k n x + a a a ) nπ sin nπ = B n a = 1

142 140 + a a CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRÖDINGER EQUATIONS IN ONE DIMENSION (A n sin k n x) (A n sin k n x) dx = A n + a = A n + a = A n = A n = A n = 1 cos k n x dx = A n a x a + a nπ sin nπ a x a ( a a ) ( nπ sin nπ a + a a sin k n x dx x 1 k n sin k n x + a a a ) nπ sin nπ = A n a = 1 Hence, (7.5) becomes ψ n (x) = a cos k nx where k n = nπ a n = 1, 3, 5,... ψ n (x) = a sin k nx where k n = nπ a n =, 4, 6,.... (7.53) We can also write (7.53) as follows. ψ n (x) = a sin nπ ( ) x + a = a ( x a sin nπ a + 1 ) (7.54) Furthermore, if we shift the x-axis so that V (x) = 0 for 0 < x < a, we will get ψ n (x) = ( ) nπx a sin. (7.55) a Because k n = me n /, the total energy E n can only take discrete values. men k n = = me n = kn = E n = kn m = π n n = 1,, 3, 4, 5,... (7.56) ma Note that E n n and the spacing between neighboring energy levels E n+1 E n = π (n + 1) ma π n ma = π (n + 1) ma (7.57)

143 7.6. THE SIMPLE HARMONIC OSCILLATOR POTENTIAL 141 is proportional to n + 1. This is the reason for the widening of the gap as n. More significant implication of this is that the particle can not have zero total energy. Indeed, the lowest energy level E 1 = π > 0. (7.58) ka This is one reason why absolute zero degree temperature can not be achieved. While the infinite square well potential is a mathematical idealization, it approximates the potential well experienced by atoms or nuclei in a crystal for example. Another type of bound system frequently encountered in nature is a diatomic molecule. 7.6 The Simple Harmonic Oscillator Potential Classic Solution Consider a typical spring encountered in high school physics with a spring constant k such that the restoring force F is given by F = kx for the displacement x from the equilibrium position. For no other reason than convention, we will use c instead of k for the spring constant. Because the Time-Independent Schrödinger Equation is V (x) = 1 cx, (7.59) d ψ(x) + c m dx x ψ(x) = Eψ(x). (7.60) We will need some fancy footwork to solve this equation. As it turns out, it is convenient to introduce a new variable ν analogously to the classical theory. Hence,

144 14 CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRÖDINGER EQUATIONS IN ONE DIMENSION ν = 1 π ω = 1 c a. π m After substituting c = (π) ν m into the equation and dividing through, we get d by m ψ me dx + ( ) πmν x ψ = 0. (7.61) Now, let α = πmν/ and β = me/ to simplify the equation to d ψ dx + (β α x )ψ = 0. (7.6) a To see this, we start with the classical equation F = mα. For us, cx = m d x dt where cos θ = = d x dt = c ( ) ( ) c c x = x(t) = A sin m m t + B cos m t = ( ) c A + B sin m t + θ ; A and sin θ = B A +B A. Therefore, we have +B ω = c m. Here are further manipulations. u = αx = πm π ( c 1/ m )1/ x = (cm)1/4 x (7.63) 1/ dψ dx = dψ du du dx = α dψ du (7.64)

145 7.6. THE SIMPLE HARMONIC OSCILLATOR POTENTIAL 143 d ψ dx = d dx ( ) dψ = dx dψ d( ) dx dx = d( α dψ ) du dx = α α d du ( dψ du ) = du d dx du You may be more comfortable with the following. ( ) α dψ du = α d ψ du (7.65) du = αdx = dx = 1 α du (7.66) d ψ dx = d 1 α du dψ 1 α du = α d ψ du (7.67) Either way is fine, and we now have d ψ dx + (β α x )ψ = α d ψ = d ψ du + Here comes a very inexact argument. du + (β αu )ψ = 0 ) ( β α u ψ = 0. (7.68) As u, the differential equation behaves like or d ψ du u ψ = 0 (7.69) d ψ du = u ψ (7.70) as β/α is a constant. The (mathematical) general solution of this differential equation is ap-

146 144 proximately a CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRÖDINGER EQUATIONS IN ONE DIMENSION ψ = Ae u / + Be u / as u. (7.71) But, we have to set B = 0 to satisfy the finiteness condition. So, ψ(u) = Ae u / as u. (7.7) We will look for a solution of the form ψ(u) = Ae u / H(u); (7.73) where H(u) must be slowly varying in comparison to e u /. Compute d ψ(u), then plug it back into the equation du d ( ) ψ(u) β + du α u ψ(u) = 0 to obtain as follows. a d H du udh du + ( ) β α 1 H = 0 (7.74) d ( ) du e u / = d ( du d ( ) du e u / = d ( du Hence, (7.71) holds. ue u / ) = e u / u u u e u / ue u / ) = e u / + u u u e u / ( ue u / ) = (u 1)e u / ( ue u / ) = (u + 1)e u /

147 7.6. THE SIMPLE HARMONIC OSCILLATOR POTENTIAL 145 Without loss of generality, set A = 1, and consider Then, we have Therefore, ψ(u) = e u / H(u). (7.75) dψ du = ue u / H(U) + e u / H (u) and d ψ du = e u / H(u) u ( ue u / ) H(u) ue u / H (u) ue u / H (u) + e u / H (u) = e u / H(u) + u e u / H(u) ue u / H (u) + e u / H (u). d ( ) ψ(u) β + du α u ψ(u) = e u / H(u) + u e u / H(u) ue u / H (u) +e u / H (u) + ( ) β α u e u / H(u) = e u / H(u) ue u / H (u) + e u / H (u) + β α e u / H(u) = 0. (7.76) Dividing through by e u /, we get which is (7.74). H(u) uh (u) + H (u) + β α H(u) ( ) β = H (u) uh (u) + α 1 H(u) = 0, We will try a power series solution for H(u). Let H(u) = a l u l = a 0 + a 1 u + a u + a 3 u 3 +. (7.77) l=0

148 146 Then, u dh(u) du and CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRÖDINGER EQUATIONS IN ONE DIMENSION ( ) = (a 1 + a u + 3a 3 u + 4a 4 u 3 + )u = (l + 1)a l+1 u l u l=0 = (l + 1)a l+1 u l+1 = la l u l (7.78) l=0 l=1 d H(u) du = a + 3a 3 u + 3 4a 4 u + = Plugging the above into gives us d H du udh du + ( ) β α 1 H = 0 (l + 1)(l + )a l+ u l. l=0 (7.79) (l + 1)(l + )a l+ u l la l u l + l=0 l=0 or (l + 1)(l + )a l+ la l + l=0 ( ) β α 1 a l u l = 0 (7.80) l=0 ( ) β α 1 a l u l = 0. (7.81) For a power series to be zero, each term has to be zero; that is to say all the coefficients have to be zero. We now have the following recursion relation. (β/α 1 l) a l+ = (l + 1)(l + ) a l (7.8) The general solution is a sum of even and odd series; namely, ( H(u) = a a u + a 4 u 4 + a ) 6 u 6 + a 0 a 0 a 0

149 7.6. THE SIMPLE HARMONIC OSCILLATOR POTENTIAL 147 u 5 + a ) 7 u 7 + (7.83) a 1 a 1 = a 0 (1 + b 1 u + b u 4 + b 3 u ) + a 1 u(1 + c 1 u + c u 4 + c 3 u ) = a 0 where b l ( +a 1 u + a 3 u 3 + a 5 a 1 l even = a l a 0 and c l b l u l + a 1 u l even c l u l ; (7.84) = a l+1 a 1 for l =, 4, 6,..., (7.85) and the reason for the awkward notations b l and c l for the coefficients will become clear when compared with (7.88). What (7.83) means is that one can get all even coefficients using (7.8) if a 0 is given 1, and all odd coefficients can be computed if a 1 is chosen. However, there is no relation between the even and odd coefficients, and the even series and odd series of (7.83) or (7.84) are completely independent. As l, So, we can see immediately that a l+ (β/α 1 l) = a l (l + 1)(l + ) l l l. (7.86) b l +1 b l = a l+ a l l l and c l +1 c l = a l+3 a l+1 l l. (7.87) Now compare this with the Taylor series expansion (about 0) of e u where 1 we change the running index from i to l = i and denote by d ( )! l l. e u = i=0 ((u ) i ) i! = i=0 u i i! = even l=0 u l even ( l = )! l=0 d l u l (7.88) Observe that d l +1 d l = 1 ( l +1 )! 1 ( l )! = ( ) ( ) l! l ( l + 1)! =! ( l + ( ) 1) l! 1 For example, a6 a 0 = a6 a 4 a a 4 a a 0.

150 148 CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRÖDINGER EQUATIONS IN ONE DIMENSION = 1 l l l/ = l. (7.89) Therefore, both the even series and the odd series of (7.84) behave like the Taylor expansion of e u for higher order terms, or as l. Guided by this, we assume H(u) behaves like H(u) = a 0 Ke u + a 1 K ue u. (7.90) So, e u / H(u) of (7.75) and (7.73) behaves like a 0 Ke u / + a 1 K ue u / as u a. (7.91) In order to satisfy the integrability condition, Condition 1 on p.13, H(u) has to terminate after some n to prevent the exponential terms in (7.91) from blowing up. This means we have to set β/α = n + 1 for n = 0, 1,, 3, 4, 5,... due to (7.8). When β/α = n+1 we get a Hermite polynomial denoted by H n (u). This gives us a series of eigenfunctions ψ n (u) = A n e u / H n (u). The first six functions look like this. n = 0 ψ 0 = A 0 e u / 1 ψ 1 = A 1 ue u / ψ = A (1 u )e u / 3 ψ 3 = A 3 (3u u 3 )e u / 4 ψ 4 = A 4 (3 1u + 4u 4 )e u / 5 ψ 5 = A 5 (15u 0u 3 + 4u 5 )e u / (7.9) a As u, higher order terms with large values of l become dominant, and this is where the series in (7.91) begin to behave like the Taylor expansion (7.88). Now recall that u = αx, α = πmν/, and β = me/.

151 7.6. THE SIMPLE HARMONIC OSCILLATOR POTENTIAL 149 β/α = me/ πmν/ = Setting this equal to n + 1, E πν = h π E E = πν hν β/α = n + 1 = E ( n hν = n + 1 = E n = n + 1 ) hν n = 0, 1,, 3, 4, 5,.... (7.93) Recall that ν = 1 c π m. Therefore, if the force constant c and the mass m are known, we know the energy levels of this harmonic oscillator. These are the two important facts about the simple harmonic oscillator. 1. The energy levels are equally spaced. This is markedly different from the infinite square well. This means many higher energy levles can be achieved more easily.. The lowest energy posssible is not zero but E 0 = 1 hν for n = 0. Since vibrations of diatomic molecules can be closely approximated by the simple harmonic oscillator, this is another reason why the absolute zero degree temperature can not be achieved. When a molecule drops from one energy level to a lower level, a photon carrying that much energy is released. On the other hand, a molecule moves to a higher energy level if it absorbs a photon carrying the energy corresponding to the difference between the two levels. These phenomena are called emission and absorption respectively.

152 150 CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRÖDINGER EQUATIONS IN ONE DIMENSION 7.6. Raising and Lowering Operators Our primary goal in this section is to simplify the computation of the energy levels for a simple harmonic oscillator. The beauty of this alternative approach is that we will be able to obtain the energy levels without explicitly computing the wavefunctions. The concept of raising and lowering operators has applications to angular momentum discussed in Chapter 9 and electron spin, a special type of angular momentum, described in Chapter 11. From (7.60), we have d ψ(x) + c m dx x ψ(x) = Eψ(x). (7.94) This equation is often expressed in terms of ω = πν. ν = 1 c c π m = ω = πν = m Substituting mω for c, we get m = c = mω d dx + 1 mω x ψ(x) = Eψ(x). (7.95) Our first step is to modify and rescale the variables. Let ˆX = mω x (7.96) and ˆP = 1 m ω p; (7.97) where x is the usual multiplication operator M x and p = i as given in x Table 3.1. We will not use capital letters X and P in this section in order to

153 7.6. THE SIMPLE HARMONIC OSCILLATOR POTENTIAL 151 make a clear distinction between the derived ( ˆX, ˆP ) and the original (X, P ). ˆX and ˆP are dimensionless variables satisfying a simpler commutation relation. Indeed, ˆX ˆP ˆP ˆX = mω x 1 m ω p 1 mω p m ω x = 1 (xp px), (7.98) and we have ˆX, ˆP = i. (7.99) We also introduce a rescaled Hamiltonian Ĥ defined by Now note that ˆX + ˆP = mω x + 1 m ω Therefore, = ω Ĥ = 1 H. (7.100) ω ( i d ) ( i d ) dx dx ( d m dx + 1 ) mω x = mω ω x m ω d dx = 1 H. (7.101) ω Ĥ = 1 ( ˆX + ˆP ). (7.10) We will now define a new operator a and its adjoint a, which will turn out to be raising and lowering operators in the title of this section. a = 1 ( ˆX + i ˆP ) (7.103)

154 15 CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRÖDINGER EQUATIONS IN ONE DIMENSION a = 1 ( ˆX i ˆP ) (7.104) Going backwards, we get the following if the above relations are solved for ˆX and ˆP. ˆX = 1 ( a + a ) (7.105) ˆP = 1 ( a a ) (7.106) Next, we will compute the commutator of a and a. 1 ( ) 1 ( ) ˆX + i ˆP, ˆX i ˆP = 1 ˆX + i ˆP, ˆX i ˆP = i { } i ˆP, ˆX ˆX, ˆP = ( i i) = 1 (7.107) Therefore, a, a = 1. (7.108) Now, a a = 1 ( ) ( ) 1 ( ˆX i ˆP ˆX + i ˆP = ˆX + ˆP + i ˆX ˆP ) i ˆP ˆX = 1 ( ˆX + ˆP 1 ) = Ĥ 1. (7.109) So, Ĥ = a a + 1. (7.110) On the other hand, And, aa = a, a + a a = 1 + Ĥ 1 = Ĥ + 1. (7.111)

155 7.6. THE SIMPLE HARMONIC OSCILLATOR POTENTIAL 153 Ĥ = aa 1. (7.11) At this point, let us introduce another self-adjoint operator N defined by N = a a. (7.113) Note that ( a a ) = a ( a ) = a a, which proves that N is self-adjoint. The operator N satisfies the following commutation relations. N, a = a a, a = a aa + (a aa a aa) aa a = a, aa = a (7.114) N, a = a a, a = a aa + (a a a a a a) a a a = a a, a = a (7.115) We have now replaced the initial eigenvalue problem for H, which is a function of x and p, with that of N, which is a product of a and a. Because we have H = ωĥ, Ĥ = N + 1 ( = H = ω N + 1 ), (7.116) ( N n = n n = H n = ω n + 1 ) n ; (7.117) where we are looking ahead and already using suggestive notations n and n, as opposed to the usual λ and λ. However, it is only on p.155 that we actually show that the eigenvalues of N are nonnegative integers. Relations (7.116) and (7.117) show that the eigenvalue problem for N is equivalent to the eigenvalue problem for H. When the eigenvalue problem for N is solved, we will get E n = ( n + 1 ) ω as the total energy for the simple harmonic oscillator.

156 154 CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRÖDINGER EQUATIONS IN ONE DIMENSION We will need the following three facts before solving the eigenvalue problem for N. Fact 7.1 Each eigenvalue n of the operator N is nonnegative. Proof Suppose N n = n n. Then, On the other hand, Therefore, n N n = n n n. (7.118) n N n = n a a n = a n. (7.119) n n n = a n = n = a n n n 0. (7.10) Fact 7. If the eigenvalue n = 0 for N, a 0 = 0. Conversely, if a v = 0, N v = 0 v. For other values of n, i.e. n > 0 from Fact 7.1, a n is an eigenvector of N with the eigenvalue n 1. Proof Suppose N 0 = 0 0. Then, a 0 = 0 a a 0 = 0 N 0 = = 0 = a 0 = 0. (7.11) Now suppose a v = 0, then, a v = 0 = a a v = N v = 0 v. (7.1) So, any nonzero vector v with a v = 0 is an eigenvector of N with the eigenvalue n = 0.

157 7.6. THE SIMPLE HARMONIC OSCILLATOR POTENTIAL 155 Next, consider a strictly positive eigenvalue n of the operator N. Then, from (7.114), N, a n = Na n an n = Na n an n = Na n na n = a n = N (a n ) = na n a n = (n 1) (a n ). (7.13) Therefore, a n is an eigenvector of N with the eigenvalue n 1. Fact 7.3 If n is an eigenvector of N with the eigenvalue n. Then, a n is an eigenvector of N with the eigenvalue n + 1. Proof We will first show a n is nonzero. From (7.108), a n = n aa n = n a, a + a a n = n 1 + N n = n n + n n n = (n + 1) n. (7.14) As n is an eigenvector of N, it is nonzero, and a n is nonzero. Now from (7.108), N, a n = Na n a N n = Na n a n n = Na n na n = a n = N ( a n ) = (n + 1) ( a n ). (7.15) Therefore, a n is an eigenvector of N with the eigenvalue n + 1. We are now ready to show that the eigenvalues of N are non-negative integers. Suppose an eigenvalue λ of N is not an integer, and let λ denote an associated eigenvector. From Fact 7.1, we already know that λ is strictly positive. From Fact 7., we know that a λ is an eigenvector of N with the eigenvalue λ 1. After repeated applications of a, on λ, k times, we get a k λ which is an eigenvector of N with the associated eigenvalue λ k. As k can be as large as we wish it to be, this implies that N has strictly negative eigenvalues, contradicting Fact 7.1. Therefore, the eigenvalues of N are all nonnegative integers.

158 156 CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRÖDINGER EQUATIONS IN ONE DIMENSION Consider an eigenvalue n, which is a positive integer, and an associated eigenvector denoted by n. Then, due to Fact 7., a k n > is an eigenvector for k = 1,,..., n. When k = n, the eigenvalue associated with a n n is 0. Then, from Fact 7., a (a n n ) = 0, terminating the chain of eigenvectors n, a n, a n,... at k = n. So, an eigenvalue n of N can only be a nonnegative integer. On the other hand, given an eigenvector 0 of N, we can apply a to the vector any number of times, with each application generating another eigenvector whose associated eigenvalue is greater than the previous eigenvalue by 1. This is guaranteed by Fact 7.3. Therefore, the eigenvalues of N, Ĥ, and H are all nonnegative integers. We have E n = ( n + 1 ) ω for n = 0, 1,,... (7.16) as before Actions of a and a Let { 0, 1,,...} be an orthonormal basis 3 for the Hilbert space for H, Ĥ, or N. Recall that the eigenvectors are the same for these three operators. We would like to find exactly what a and a do to these basis vectors. We already know for some constant c n. Taking the adjoint, So, a n = c n n 1 (7.17) n a = n 1 c n. (7.18) n a a n = n 1 n 1 c nc n = c nc n = n N n = c nc n 3 Because these are eigenvectors of a Hermitian operator H, they form a basis automatically. The only additional condition imposed here is normalization of each ket.

159 7.6. THE SIMPLE HARMONIC OSCILLATOR POTENTIAL 157 = n n n = c nc n = n = c n. (7.19) We have c n = ne iϕ. (7.130) By convention, we normally set ϕ = 0. Then, c n = n and a n = n n 1. (7.131) Similarly, we also know a n = d n n + 1 (7.13) for some constant d n. Taking the adjoint, n a = n + 1 d n. (7.133) This gives us n aa n = n + 1 d nd n n + 1. (7.134) The left-hand side can be transformed as below using the relation (7.108). n aa n = n a, a + a a n = n 1 + N n = 1 + n (7.135) Therefore, d n = n + 1. (7.136) Setting the phase equal to 0 as before, we get d n = n + 1 and a n = n + 1 n + 1. (7.137) We can represent the actions of a and a as matrices with infinitely many entries a (7.138)

160 158 CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRÖDINGER EQUATIONS IN ONE DIMENSION a (7.139) I hope it is now clear to you why a is the lowering operator and a is the raising operator 4. 4 Note that a and a are also called destruction and creation operators, respectively.

161 Exercises This problem is the same as Chapter 5 Problem 5. Consider a particle of mass m and total energy E which can move freely along the x-axis in the interval a, + a, but is strictly prohibited from going outside this region. This corresponds to what is called an infinite square well potential V (x) given by V (x) = 0 for a < x < + a and V (x) = elsewhere. If we solve the Schrödinger equation for this V (x), one of the solutions is A cos ( ) πx a e iet/ a < x < a Ψ(x, t) = (a > 0). 0 x a (a) Find A so that the function Ψ(x, t) is properly normalized. (b) In the region where the potential V (x) = 0, the Schrödinger equation reduces to t) Ψ(x, t) = i Ψ(x,. m x t Plug Ψ(x, t) = A cos ( ) πx e iet/ into this relation to show E = π ma. (c) Show that P = 0. a (d) Evaluate x for this wavefunction. You will probably need an integral table for this. (Of course, you can always try contour integration or some such. But, that is beyond this course.) (e) Evaluate P for the same wavefunction. You will not need an integral table if you use the fact that the function is normalized. Of course, a brute force computation will yield the same result as well.. This problem is the same as Chapter 5 Problem 6. If you are a careful student who pays attention to details, you may have realized that A in Problem 1 can only be determined up to the argument, or up to the sign if you assume A is a real number. (a) What is the implication of this fact as to the uniqueness of wavefunction?

162 160 (b) What is the implication of this fact as to the probability density Ψ (x, t)ψ(x, t)? How about P, P, and x? 3. Solve the time-independent Schrödinger equation for a free particle. (Note that a free particle is a particle with no force acting on it. In terms of the potential V (x), this means that it is a constant. For simplicity, assume that V (x) = 0 in this problem.) 4. Consider a particle of mass m which can move freely along the x- axis anywhere from x = a/ to x = +a/ for some a > 0, but is strictly confined to the region a/, +a/. This corresponds to what is called an infinite square well potential V (x) given by V (x) = 0 for a/ < x < +a/ and V (x) = elsewhere. Where the potential is infinite, the wavefunction is zero. Use the continuity condition on ψ(x) at x = a/, but not on ψ (x), to find at least two of the solutions. 5. Give a rigorous mathematical proof that H(x) = K(t) for all pairs of real numbers (x, t) implies that H(x) and K(t) are a constant. 6. Given a potential V (x) with the following profile V (x) = { 0 x < 0 V 0 x > 0 solve the Time-Independent Schrödinger Equation for a particle with the total energy E < V 0 traveling from x = to x = +. In fact, we have already solved this problem in this chapter. I just want you to review the procedure. 7. Show that we have total reflection in 6 above by direct computation using probability current described in Appendix H. 8. Consider the potential V (x) given by V (x) = 0 for x < 0 and V (x) = V 0 > E for x > 0, and show that we have a standing wave in the region x < 0 for a particle incident on the discontinuity at x = 0 from the left, that is, from the region characterized by negative values of x. 9. Consider the potential V (x) given by V (x) = 0 for x < 0 and V (x) = V 0 < E.

163 161 (a) Study the complete solution of the time-independent Schrödinger equation we solved in this chapter. (b) Sow that the transmission coefficient is given by 4k 1k (k 1 +k ). 10. Consider a step potential given by { 0 x < 0 V (x) = V 0 x > 0. (a) Write down, but do not derive or solve, the time-independent Schrödinger equation for a particle of mass m and total energy E > V 0 for each region. (b) For a particle moving toward the step from the left, the solutions are ψ(x) = Ae ik1x + Be ik 1x (x < 0) and ψ(x) = Ce ik x (x > 0), where k 1 = me m(e V 0 ) and k =. Use the two boundary conditions at x = 0 to express C in terms of A, k 1, and k. (c) As it turns out, we also get B = k 1 k k 1 +k A. Compute the reflection coefficient R; that is, express it in terms of k 1 and k. 11. In order to solve the Time-Independent Schrödinger Equation for a particle incident on the potential barrier given by { V0 > E 0 < x < a V (x) = 0 elsewhere from the left, that is, x < 0 at t = 0, you need to consider the wavefunction given by Ae ik 1x + Be ik 1x x < 0 ψ(x) = F e k x + Ge k x 0 < x < a ; (7.140) Ce ik 1x + De ik 1x x > a with the following set of boundary conditions. ψ(x) x=0 = ψ(x) x=0+ ψ(x) x=a = ψ(x) x=a+ dψ(x) = dψ(x) (7.141) dx x=0 dx x=0+ = dψ(x) x=a x=a+ dψ(x) dx dx

164 16 (a) What are k 1 and k in (7.140)? (b) Why should D in (7.140) be 0? (c) Write down the boundary conditions (7.141) after setting D = Consider a particle of energy E > V 0 moving from x = + to the left in the step potential: { V0 x > 0 V (x) = 0 x < 0. (a) Write the time-independent Schrödinger equations in the regions x < 0 and x > 0. (b) The general solutions are of the form { A (wave traveling to the right) + B (wave traveling to the left) x < 0 C (wave traveling to the right) + D (wave traveling to the left) x > 0. Complete the general solutions; i.e. substitute actual formulas for (wave traveling to the right) and (wave traveling to the left). For simplicity of notation, let k 1 = me m(e V 0 ) and k =. (c) Determine the value of one of the constants based on the fact that there is no wave coming back from x = and traveling to the right. (d) Use the boundary conditions at x = 0 to get the relations among the constants. (e) Compute the reflection coefficient R. 13. Consider a barrier potential given by 0 (x < a) V (x) = V 0 ( a < x < 0) 0 (x > 0). A stream of particles are originating at x = and traveling to the right. The total energy E is less than the barrier height V 0. (a) Write down, but do not derive or solve, the time-independent Schrödinger equation for a particle of mass m and total energy E < V 0 for each region.

165 163 (b) For a particle moving toward the step from the left, the solutions are Ae ik1x + Be ik 1x (x < a) ψ(x) = Ce kx + De k x ( a < x < 0) Ge ik 1x (x > 0) where k 1 = me m(v 0 E) and k =. i. Explain in words why we have ψ(x) = Ge ik1x and not ψ(x) = Ge ik1x + He ik1x in the region x > 0. ii. Give the two continuity coonditions at x = 0. iii. On physical grounds, we also need continuity of ψ ψ as well as its first derivative at the boundaries. Use this continuity condition on dψ ψ at x = 0 to show neither C nor D is 0. dx 14. Show the following for the infinite square well potential described in the class. { 0 a V (x) = < x < a elsewhere (a) Two general solutions ψ(x) = A sin(kx) + B cos(kx) and ψ(x) = Ce ikx + De ikx are equivalent. Here, k = me/, and A, B, C, and D are arbitrary constants. (b) The usual boundary conditions at x = 0 and a allow only discrete values of k, which in turn implies that the total energy is discretized. 15. The potential for the infinite square well is given by V (x) = { x > a 0 x < a, and the solutions for the time-independent Schrödinger equation given in the class were { ψn (x) = B n cos k n x where k n = nπ n = 1, 3, 5,... a ψ n (x) = A n sin k n x where k n = nπ. n =, 4, 6,... a (a) Find a normalized solution for n = for which the normalization constant A is purely imaginary.

166 164 (b) What is the expectation value of the position x for this solution? Use the full wavefunction to solve this problem. (c) Plug the above solution into the lefthand side of the time-independent Schrödinger equation to determine E, the total energy associated with this state. 16. The potential for the infinite square well is given by V (x) = { x > a 0 x < a. (a) Write down, but do not derive or solve, the time-independent Schrödinger equation for a particle of mass m and total energy E in the region a < x < a. (b) Find a solution of the form ψ(x) = A sin(bx) for a < x < a, where A and B are constants, and compute the total energy following the steps below. i. Express B as a function of E so that ψ(x) = A sin(bx) is a solution of the Shcrödinger equation. (Choose the positive square root.) ii. Find the smallest value of B that satisfies the boundary conditions. iii. Compute the total energy corresponding to the B above. iv. Determine A so that the wavefunction is properly normalized. 17. Consider an infinite square well whose potential is given by x < 0 V (x) = 0 0 < x < L. L < x (a) Write down, but do not derive or solve, the time-independent Schrödinger equation for a particle of mass m and total energy E in the region 0 < x < L. (b) The most general solution of the time-independent Schrödinger equation in the region 0 < x < L is of the form ψ(x) = A sin kx + B cos kx; where A and B are arbitrary constants and k = me.

167 165 i. Using one of the boundary conditions on ψ(x), determine the value of B. ii. Find all possible positive values k can take by applying one of the boundary conditions on ψ(x). Express k in terms of π, L, and an arbitrary natural number n. We will refer to each of these values as k n. With this, we can write our wavefunction as ψ n (x) = A n sin k n x + B n cos k n x in order to make (potential) dependence on n more explicit. iii. Impose the normalization condition to find the positive real value of A n given the above k n. (c) Now, use the relation k n = men to determine all possible values the total energy can take. In other words, express E n in terms of m, h, L, and n. 18. Try a series solution for the differential equation f(x) = a j x j j=0 and show that df(x) dx + f(x) = 3x + 8x + 3, a j + (j + 1)a j+1 = 0 for j For a simple harmonic oscillator, show that the wavefunction corresponding to n = 1 is given by ψ 1 = A 1 ue u /, and the wavefunction for n = takes the form ψ = A (1 u )e u /.

168 A time-independent ground state wavefunction of the simple harmonic oscillator is given by Φ 0 = A 0 exp u, where u = (Cm) 1 4 / 1 x. Find the expectation value of x as a function of the normalization constant A 0. Use the full wavefunction with the total energy E This question concerns the simple harmonic oscillator. (a) The full ground state wavefunction Ψ 0 can be expressed as with ψ 0 e iet/ ψ 0 = A 0 e u / ; where A 0 is a normalization constant, and u = (Cm) x. Compute the expectation value < x > of the position for this state. (b) Show by explicit computation that the first two wavefunctions of a simple harmonic oscillator are orthogonal. Use the functions given below, where A 0 and A 1 are normalization constants. ψ 0 = A 0 e u / ψ 1 = A 1 ue u / (c) Why do you know the third and the fifth eigenfunctions, denoted by ψ and ψ 4 as the first eigenfunction is ψ 0, are orthogonal to each other without conducting explicit integration? ψ = A (1 u )e u / ψ 4 = A 4 (3 1u + 4u 4 )e u / (d) Explain in fewer than 50 words why a solid made of diatomic molecules can not be cooled to absolute zero. Here, I am only asking you to reproduce the crude argument I gave you in the class as incomplete as it was.. Suppose n is an eigenket of a a with the eigenvalue n. Show that a n is an eigenket of a a with the eigenvalue n + 1.

169 Describe, in a sentence or two, one of the differences between classical mechanics and quantum mechanics. Your answer should be about an actual physical phenomenon or its interpretation. Hence, for example, the difference in form between the Schrödinger equation and the Newton s equation is not an acceptable answer.

170 168

171 Chapter 8 Higher Spatial Dimensions So far, we have only dealt with one dimensional cases partly for simplicity. However, most physical systems have multiple spatial dimensions, and there are a few new phenomena which you can observe only if you have more than one dimension. We will discuss two of such new features; degeneracy and angular momentum. 8.1 Degeneracy In a typical dictionary, degeneracy is defined as a state of being degenerate, and the primary or ordinary meanings of the adjective degenerate are Having declined, as in function or nature, from a former or original state and Having fallen to an inferior or undesirable state, especially in mental or moral qualities. But, degeneracy in quantum mechanics does not have much to do with declining or becoming inferior. According to Wikipedia Wikipedia contributors, nd, degeneracy in quantum mechanics has the following definition. In quantum mechanics, a branch of physics, two or more different states of a system are said to be degenerate if they are all at the same energy level. It is represented mathematically by the Hamiltonian for the system having more than one linearly independent eigenstate with the same eigenvalue. Conversely, an energy level is said to be degenerate if it contains two or more different states. The number of different states at a particular energy level is called the level s degeneracy, and this phenomenon is generally known as a quantum degeneracy. 169

172 170 CHAPTER 8. HIGHER SPATIAL DIMENSIONS From the perspective of quantum statistical mechanics, several degenerate states at the same level are all equally probable of being filled. One example of this is the hydrogen atom discussed in Chapter 10. In this chapter, we will give a general proof that there is no degeneracy for bound states in one dimension, leaving discussions of degeneracy for specific cases to other chapters. Definition 8.1 (Bound States) Bound states occur whenever the particle cannot move to infinity. That is, the particle is confined or bound at all energies to move within a finite and limited region of space which is delimited by two classical turning points. As the particle cannot move to infinity, we need to have for any bound state. ψ(x ± ) 0 (8.1) Theorem 8.1 There is no degeneracy for one-dimensional bound states. Proof Suppose there is degeneracy for total energy E. This means there are two unit vectors u and w, representing distinct physical states, such that H u = E u and H w = E w. Let x be the only position variable in this one-dimensional space. writing ψ u (x) and ψ w (x) for u and w respectively, we get Then, m d dx ψ u(x) + V ψ u (x) = Eψ u (x) () and d m dx ψ w(x) + V ψ w (x) = Eψ w (x). (3) Multiplying () by ψ w (x) and (3) by ψ u (x) gives us ( d m dx ψ u(x) ) ψ w (x) + V ψ u (x)ψ w (x) = Eψ u (x)ψ w (x) (4)

173 8.1. DEGENERACY 171 ( d m dx ψ w(x) ) and Now, subtracting (5) from (4) leads to Noting that ( ) ( d m dx ψ u(x) ψ w (x) ( ) d = dx ψ u(x) ψ w (x) = d dx = d dx ψ u (x) + V ψ w (x)ψ u (x) = Eψ w (x)ψ u (x). (5) ) ( d m ( d dx ψ w(x) dx ψ w(x) ) ) ψ u (x) = 0 ψ u (x) = 0. (6) ( ) ( ) d d dx ψ u(x) ψ w (x) dx ψ w(x) ψ u (x) ( ) d = dx ψ u(x) ψ w (x) + d dx ψ u(x) d dx ψ w(x) d dx ψ u(x) d ( ) d dx ψ w(x) dx ψ w(x) ψ u (x) ( ) d = dx ψ u(x) ψ w (x) + d dx ψ u(x) d dx ψ w(x) ( d dx ψ u(x) d ( ) ) d dx ψ w(x) + dx ψ w(x) ψ u (x) ( ) d dx ψ u(x) ψ w (x) d ( ) d dx dx ψ w(x) ψ u (x) ( ) ( ) d d dx ψ u(x) ψ w (x) dx ψ w(x) ψ u (x), (7) (6) implies ( ) ( ) d d dx ψ u(x) ψ w (x) dx ψ w(x) ψ u (x) = c (8) where c is some constant. Because ψ u and ψ w are bound states, ψ u (x) x ± x ± 0 and ψ w (x) 0. (9)

174 17 CHAPTER 8. HIGHER SPATIAL DIMENSIONS Therefore, the constant c in (8) is 0. 1 positive for now, we have Naively, i.e. assuming ψ u and ψ w are 1 d ψ u (x) dx ψ u(x) = 1 d ψ w (x) dx ψ w(x) = d ln ψ u(x) = d ln ψ w(x) dx dx d ln ψu (x) d ln ψw (x) = dx = dx = ln ψ u (x) = ln ψ w (x) + k dx dx = ln ψ u (x) = ln e k ψ w (x) = ψ u (x) = e k ψ w (x). (10) So, in this case, ψ u (x) is a scalar multiple of ψ w (x), and they represent the same physical state. This in turn implies that there is no degeneracy. If you think there is too much hand-waving in this proof, you are extremely right! I gave this naive argument first in order to make this proof more accessible for, say, freshmen. Having said that, a more rigorous proof can be given resorting to the Wronskian argument described in Appendix C. According to Appendix C, two solutions, ψ u and ψ w, of a linear second-order ordinary differential equation of the form ψ (x) + p(x)ψ (x) + q(x)ψ(x) = 0 (11) where p(x) and q(x) are continuous, are linearly dependent if and only if the Wronskian, W (ψ u, ψ w ), given by W (ψ u, ψ w )(x) = ψ u (x)ψ w(x) ψ u(x)ψ w (x) (1) is 0 for all x in the domain. As this is the case for us, we have ψ u (x) = Kψ w (x) for some constant K, and this proves the theorem. We have now shown that degeneracy, which requires two linearly independent eigenfunctions with the same energy eigenvalue E, is a phenomenon specific to more than one dimension. In passing, let us make a note of the following theorem found on p.17 of Quantum Mechanics: Concepts and Applications (second edition) by 1 d A more rigorous proof also requires an examination of the behavior of dx ψ u(x) and d dx ψ w(x) as x tends to ±. For now, just think of exponential functions e kx with k > 0 and their first derivatives as x ±.

175 8.. ANGULAR MOMENTUM 173 Nouredine Zettili Zettili, 009, p.17, which includes discreteness of energy levels for bound states. Theorem 8. (Discrete and Nondegenerate) In a one-dimensional problem the energy levels of a bound state system are discrete and not degenerate. The condition (8.1) together with Condition 3 (continuity conditions) on p.14 can be satisfied only for certain special values of energy, making allowed total energy values discrete. We saw how this works repeatedly in Chapter 7. Zettili also provides the following theorem. Theorem 8.3 The wave function ψ n (x) of a one-dimensional bound state system has n nodes (i.e., ψ n (x) vanishes n times) if n = 0 corresponds to the ground state and (n 1) nodes if n = 1 corresponds to the ground state. Remark 8.1 (Unbound One-Dimensional States) Note that we do have degeneracy if we have unbound states. A good example is a free particle traveling wave where we have according to (7.4). e i(kx ωt) and e i(kx+ωt) (8.13) These are linearly independent states corresponding to particles/waves traveling to the right e i(kx ωt) and to the left e i(kx+ωt). Because k = me for both waves, they have the same energy, and we have two-fold degeneracy. 8. Angular Momentum The only momentum we encounter in a one-dimensional system is the linear momentum p = mv as we only have a linear motion in one-dimension. However, in higher dimensions, rotational motions are also possible, and we have another variety of momentum called the angular momentum. The classical definition of the angular momentum L of a particle of mass m with respect to a chosen origin is given by

176 174 CHAPTER 8. HIGHER SPATIAL DIMENSIONS L = r p; (8.14) where r is the position vector connecting the origin and the particle, p = mv is the linear momentum, and is the usual vector cross product. If we denote the angle formed by r and p by θ, we have L = p r sin θ = m v r sin θ or L = mvr sin θ. a (8.15) a We can choose either angle for this purpose as sin θ = sin (π θ). The angular momentum is subject to the fundamental constraints of the conservation of angular momentum principle if there is no external torque, rotation-causing force, on the object. Using the usual substitutions x i M xi and p i i x i ; (8.16) where x 1 = x, x = y, x 3 = z, and likewise for p i s, we can derive quantum mechanical angular momentum operators L x, L y, and L z drawing on the classical relation (8.14). Because the study of the angular momentum is a large and thriving industry, we will devote the entire Chapter 9 to a detailed discussion of quantum mechanical angular momentum.

177 Chapter 9 Angular Momentum As we move from one-dimensional systems to higher dimensionality, more and more interesting physics emerges. In this chapter, we will discuss how angular momentum is dealt with in quantum mechanics and the consequences of quantum mechanical computations of angular momenta. The angular momentum relations are very important in the study of atoms and nuclei. Like the total energy E, the magnitude of angular momentum, usually denoted by L, as well as its z-component, L z are conserved. Furthermore, angular momentum is quantized like energy levels. 9.1 Angular Momentum Operators Quantum mechanical operators representing the x-, y-, and z-component of angular momentum were already presented in Table 3. without derivation. Now, we will see how we can derive quantum mechanical counterpart drawing on the classical theory of angular momentum. The angular momentum of a particle about the origin is given by L = r p = l x î+l yˆȷ+l zˆk = î ˆȷ ˆk x y z p x p y p z = l x l y l z = yp z zp y zp x xp z xp y yp x (9.1) ; 175

178 176 CHAPTER 9. ANGULAR MOMENTUM where the expression L = r p is coordinate-independent, but the componentby-component expressions on the right are in Cartesian coordinates. The quantum mechanical operator L = L x î + L yˆȷ + L zˆk is obtained as follows. First, noting that p x i, p x y i, and p y z i, we get: z ( ) ( L x = y i z i = i y z y z z y ( L y = z i ) ( x i ) ( = i z x z x x ) z ( L z = x i ) ( y i ) ( = i x y x y y ) x ) ( ) (9.) This book culminates in a full solution of the Schrödinger equation for the hydrogen atom; where we only have a central force, and the spherical coordinate system is employed to simplify the computation. In spherical coordinates, we get: L x = i L y = i ( ( sin ϕ θ ) + cot θ cos ϕ ϕ cos ϕ + cot θ sin ϕ θ ϕ L z = i ϕ. ) (9.3) Note that L z is particularly simple. As for the magnitude of angular momentum, we will work with the squared magnitude, denoted by L. L = L x + L y + L z = 1 sin θ ( sin θ ) + 1 θ θ sin θ ϕ (9.4) We have the following commutation relations.

179 9.1. ANGULAR MOMENTUM OPERATORS 177 L x, L y = i L z (9.5) L y, L z = i L x (9.6) L z, L x = i L y (9.7) L, L x = L, L y = L, L z = 0 (9.8) The relations (9.5), (9.6), and (9.7) can be summarized as L L = i L. (9.9) Let us verify (9.5) only, as (9.6) and (9.7) can be verified in the same manner. We will verify (9.5) in three different ways for the sake of comparison. The first verification uses (9.). For simplicity of notation, we will denote,, x y and by z x, y, z, respectively. You might as well familiarize yourself with this notational scheme at this point, unless you are already comfortable with it, because it is widely used in physics and engineering. L x, L y = ( i ) (y z z y ) (z x x z ) (z x x z ) (y z z y ) = ( i ) y z z x y z x z z y z x + z y x z (z x y z z x z y x z y z + x z z y ) = ( i ) ( y x + yz z x yx z z y x + zx y z zy x z +z x y + xy z x y xz z y ) = i ( i ) (x y y x ) = i ( i ) Next, we will verify (9.5) in spherical coordinates. ( x y y ) x = i L z (9.10) L x, L y = L x L y L y L x = i (sin ϕ θ + cot θ cos ϕ ϕ ) i ( cos ϕ θ + cot θ sin ϕ ϕ ) i ( cos ϕ θ + cot θ sin ϕ ϕ ) i (sin ϕ θ + cot θ cos ϕ ϕ ) = (i ) sin ϕ θ ( cos ϕ θ ) + sin ϕ θ (cot θ sin ϕ ϕ ) + cot θ cos ϕ ϕ ( cos ϕ θ ) + cot θ cos ϕ ϕ (cot θ sin ϕ ϕ ) (i ) cos ϕ θ (sin ϕ θ ) cos ϕ θ (cot θ cos ϕ ϕ ) + cot θ sin ϕ ϕ (sin ϕ θ ) + cot θ sin ϕ ϕ (cot θ cos ϕ ϕ )

180 178 CHAPTER 9. ANGULAR MOMENTUM ( = (i ) sin ϕ cos ϕ θ + sin ϕ 1 sin θ ϕ + sin ϕ cot θ θ ϕ + cot θ cos ϕ sin ϕ θ cot θ cos ϕ ϕ θ + cot θ cos ϕ ϕ ) + cot θ sin ϕ cos ϕ ϕ ( (i ) sin ϕ cos ϕ θ cos ϕ 1 sin θ ϕ cos ϕ cot θ θ ϕ + cot θ sin ϕ cos ϕ θ + sin ϕ cot θ ϕ θ cot θ sin ϕ ϕ ) + cot θ sin ϕ cos ϕ ϕ ( = (i ) sin ϕ 1 sin θ ϕ + cot θ cos ϕ ϕ + cos ϕ 1 sin θ ϕ ) + cot θ sin ϕ ϕ = (i ) ( 1 sin θ ϕ + cot θ ϕ ) ( ) 1 = i ( i ) sin θ cot θ ϕ = i ( i ) 1 cos θ sin ϕ = i ( i ) θ ϕ = i L z (9.11) Finally, we will use the commutator identity (G.5). L x, L y = yp z zp y, zp x xp z Now, we will verify (9.8). = yp z, zp x yp z, xp z zp y, zp x + zp y, xp z = yp z, zp x + zy, p x p z + zyp z, p x + y, zp z p x (yp z, xp z + xy, p z p z + xyp z, p z + y, xp z p z ) (zp y, zp x + zz, p x p y + zzp y, p x + z, zp y p x ) + zp y, xp z + xz, p z p y + xzp y, p z + z, xp y p z = y( i I)p x + z 0 p z + zy p z p x (y 0 p z + x 0 p z + xy p z p z ) (z 0 p x + z 0 p y + zz p y p x ) + z 0 p z + x(i I)p y + xz p y p z = i (xp y yp x ) = i L z (9.1) L, L x = L x + L y + L z, L x = L x, L x + L y, L x + L z, L x = 0 + L y, L x L y + L y L y, L x + L z, L x L z + LzL z, Lx = i L z L y i L y L z + i L y L z + i L z L y = 0 (9.13)

181 9.. QUANTUM MECHANICAL ROTATION OPERATOR U R 179 Likewise for L y and L z. 9. Quantum Mechanical Rotation Operator U R Let R(ϕk) be a physical rotation, a counterclockwise rotation around the z-axis by ϕ, and U R be the quantum mechanical operator associated with the physical rotation R. Then, the action of U R is characterized by U R x, y, z = x cos ϕ y sin ϕ, x sin ϕ + y cos ϕ, z (9.14) In order to derive U R, we will first consider an infinitesimal physical rotation R(εk) and its quantum mechanical counterpart U Rε given to first order in ε by As it turns out, the correct form is U Rε = I + εω. (9.15) U Rε = I iε L z; (9.16) where L z is the quantum mechanical angular momentum operator around the z-axis (9.). We can convince ourselves that (9.16) is indeed the desired quantum mechanical operator as follows. Note that this is not a very rigorous proof, but a rough sketch of how it should be shown. By definition, we have It follows from (9.17) that U Rε x, y, z = x yε, y + xε, z. (9.17) U Rε ψ = U Rε = + + x, y, z x, y, z ψ dxdydz U Rε x, y, z x, y, z ψ dxdydz

182 180 CHAPTER 9. ANGULAR MOMENTUM = + x yε, y + xε, z x, y, z ψ dxdydz. (9.18) Now, let x = x yε, y = y + xε, and z = z. Then, the Jacobian is x x x y x z y x y y y z z x z y z z to first order in ε. So, 1 = 1 ε 0 ε = ε 1 ε = 1 U Rε ψ = = = to first order in ε. Hence, x, y, z x, y, z ψ ε dx dy dz x, y, z x, y, z ψ dx dy dz x, y, z x + yε, y xε, z ψ dx dy dz (9.19) + x, y, z U Rε ψ = x, y, z x, y, z x + yε, y xε, z ψ dx dy dz = = + So, we have + x, y, z x, y, z x + yε, y xε, z ψ dx dy dz δ(x x )δ(y y )δ(z z ) x + yε, y xε, z ψ dx dy dz = x + yε, y xε, z ψ. = x, y, z U Rε ψ = (U Rε ψ ) (x, y, z) = ψ(x + yε, y xϵ, z). (9.0)

183 9.. QUANTUM MECHANICAL ROTATION OPERATOR U R 181 Expanding ψ(x + yε, y xε, z) in a Taylor series to order ε, we obtain ψ(x + yε, y xε, z) = ψ(x, y, z) + ψ ψ (yε) + ( xε). (9.1) x y On the other hand, Therefore, x, y, z U Rε ψ = x, y, z I + εω ψ = x, y, z ψ + ε x, y, z Ω ψ = ψ(x, y, z) + ε x, y, z Ω ψ. (9.) x, y, z U Rε ψ = ψ(x + yε, y xϵ, z) = ψ(x, y, z) + ε x, y, z Ω ψ = ψ(x, y, z) + ψ ψ (yε) + x y ( xε) = ε x, y, z Ω ψ = ε y ψ x + ( x) ψ y ( = (Ω ψ ) (x, y, z) = y x x ) ψ (x, y, z) y = Ω = y x x ( = i Ω = i y y x x ) y ( = i x y y ) = L z = Ω = L z x i = i L z (9.3) This verifies (9.16). U Rε = I iε L z The quantum mechanical operator U R for a finite rotation R(ϕ 0 k) can be constructed from (9.16) as follows. U R = lim N ( I i ) N ϕ 0 N L z = exp ( iϕ 0 L z / ) (9.4) The exponential here is to be regarded as a convergent power series. ( iϕ 0 / ) n exp ( iϕ 0 L z / ) = L n z (9.5) n=0 n!

184 18 CHAPTER 9. ANGULAR MOMENTUM Now, from (9.3), L z = i ϕ. So, in spherical coordinates, ( U R = exp ( iϕ 0 L z / ) = exp ( iϕ 0 i ) ) ( ) / = exp ϕ 0. (9.6) ϕ ϕ 9.3 Rotationally Invariant Hamiltonian Suppose the only force in the system is a central force F (r), which derives from a central potential V (r) where r is the distance from the origin, so that Then, the Hamiltonian F = V (r) = dv (r). dr H = p m + V (r) is invariant under arbitrary rotations, and it commutes with L x, L y, L z, and L. From (G.19) and (G.0), we have the following. H, L x = H, L y = H, L z = 0 and H, L = 0 Therefore, L, L x, L y, and L z are constants of motion and conserved. We know from Theorem.9 that commuting Hermitian operators Ω and Γ have common eigenvectors which diagonalize Ω and Γ simultaneously. Because L x, L y, and L z do not commute with each other, we cannot diagonalize H, L, L x, L y, and L z simultaneously. The best we can do is to diagonalize H, L, and only one of the components of L. It is customary to choose L z. So, we will diagonalize H, L, and L z simultaneously 1, which amounts to finding all eigenfunctions {ψ(r, θ, ϕ)} which are, by definition, 1 We can draw a contradiction if we assume there is a common eigenstate l for L z and L x such that This implies L x l = l x l and L z l = l z l. L z, L x l = i L y l = L z L x l L x L z l = l z l x l l x l z l = 0.

185 9.4. RAISING AND LOWERING OPERATORS: L + AND L 183 stationary under H, L, and L z. The explicit forms of the eigenfunctions will be found in Chapter 10. But, there is a way to solve the eigenvalue problem of L and L z using raising and lowering operators as for the simple harmonic oscillator discussed in Section Raising and Lowering Operators: L + and L In this section, we will find the eigenvalues of L and L z without finding the eigenfunctions explicitly. Looking ahead, we will denote the eigenvalues of L by l(l + 1) and those of L z by m l. We will write { l, m l } for the orthonormal eigenstates with associated eigenvalues l(l + 1) and m l. L l, m l = l(l + 1) l, m l L z l, m l = m l l, m l (9.7) l, m l l m l = δ ll δ ml,m l We will show that l is a nonnegative integer and m l = l, l+1,..., 0,..., l 1, l. The number l is referred to as an orbital quantum number, and m l is called either a magnetic quantum number or an azimuthal 3 quantum number. We first define the raising and lowering operators as follows. The reason for the naming will become clear shortly. This in turn implies We now have L x, L y l = i L z l = L x L y l L y L x l = 0 L y (l x l ) = 0 0 = 0. L y l = L z l = 0 = L x l = 0. Therefore, only possible common states are those for zero angular momentum. Needless to say, we cannot find a set of orthonormal eigenstates that simultaneously diagonalizes L x, L y, and L z. As we will see later, the energy levels among the states with different m l s are the same in the absence of an external magnetic field. 3 It is so called as the angle θ in spherical coordinate system is called an azimuth.

186 184 CHAPTER 9. ANGULAR MOMENTUM L + = L x + il y (9.8) L = L x il y (9.9) Because L x and L y are Hermitian, it follows that L + = (L x + il y ) = L x il y = L x il y = L (9.30) and L = (L x il y ) = L x + il y = L x + il y = L +. (9.31) Incidentally, using (9.3), we can express L + and L in spherical coordinates. ( L ± = e ±iϕ ± θ + i cot θ ) ϕ As we already know L commutes with L x and L y, we have (9.3) L, L + = L, L = 0. (9.33) Let us also compute other commutators. L +, L = L x + il y, L x il y = L x, L x + L x, il y + il y, L x + il y, il y = 0 + ( i)i L z + i( i L z ) + 0 = L z (9.34) L +, L z = L x + il y, L z = L x, L z + il y, L z = i L y + i(i L x ) = L x L x = (L x + il y ) = L + (9.35) L, L z = L x il y, L z = L x, L z il y, L z = i L y i(i L x ) We also have the following. = L x i L y = (L x il y ) = L (9.36) L + L = (L x + il y )(L x il y ) = L x il x L y + il y L x + L y

187 9.4. RAISING AND LOWERING OPERATORS: L + AND L 185 = L L z il x, L y = L L z i(i L z ) = L L z + L z (9.37) L L + = (L x il y )(L x + il y ) = L x + il x L y il y L x + L y = L L z + il x, L y = L L z + i(i L z ) = L L z L z (9.38) Because the operators L x, L y, and L z are all Hermitian, l, m l L l, m l = l, m L x + L y + L z l, m l On the other hand, Therefore, = l, m L x l, m l + l, m L y l, m l + l, m L z l, m l = l, m L xl x l, m l + l, m L yl y l, m l + l, m L zl z l, m l = L x (l, m l ) L x (l, m l ) + L y (l, m l ) L y (l, m l ) (9.39) + L z (l, m l ) L z (l, m l ) = L x (l, m l ) + L y (l, m l ) + L z (l, m l ) 0 l, m l L l, m l = l(l + 1) l, m l l, m l = l(l + 1). (9.40) l(l + 1) 0 = l 0 or l 1. (9.41) Because the parabola y = x(x + 1) is symmetric about the axis x = 1, we can choose l 0 and still get all possible nonnegative values l(l + 1) can take. Now, let + := L + l, m l ; (9.4) where l, m l is a normalized eigenstate of L with the eigenvalue l(l + 1) and also a normalized eigenstate of L z with the eigenvalue m l from (9.7). Then, L + = L L + l, m l = L + L l, m l = L + ( l(l + 1) l, m l ) = l(l + 1)L + l, m l = l(l + 1) +. (9.43)

188 186 CHAPTER 9. ANGULAR MOMENTUM Hence, + is still an eigenstate of L with the same eigenvalue l(l + 1). On the other hand, due to (9.7) and (9.35), L z + = L z L + l, m l = (L z, L + + L + L z ) l, m l = ( L + + m l L + ) l, m l = (m l + 1)L + l, m l = (m l + 1) +. (9.44) Hence, + is still an eigenstate of L z but with an eigenvalue which is greater than the original m l. Likewise, if we let := L l, m l, (9.45) we get ( L = L L l, m l = L L l, m l = L l(l + 1) l, m l ) = l(l + 1)L l, m l = l(l + 1) (9.46) and L z = L z L l, m l = (L z, L + L L z ) l, m l = ( L + m l L ) l, m l = (m l 1)L l, m l = (m l 1). (9.47) This shows that is an eigenstate of L with an eigenvalue l(l + 1) and is an eigenstate of L z with an eigenvalue that is smaller than the original m l. In sum, we have shown that L + /L raises/lowers the magnetic quantum number m l by one, but preserves the total angular momentum l. In other words, L + l, m l is proportional to l, m l + 1, and L l, m l is proportional to l, m l 1. Let us now compute the square of the norm for + and, respectively. + = + + = L + (l, m l ) L + (l, m l )

189 9.4. RAISING AND LOWERING OPERATORS: L + AND L 187 = l, m l L L + l, m l = l, m l L L z L z l, m l = ( ) l(l + 1) m l m l l, ml l, m l = (l(l + 1) m l (m l + 1)) = (l + l m l m l ) = ((l + m l )(l m l ) + (l m l )) = (l m l )(l + m l + 1) (9.48) = = L (l, m l ) L (l, m l ) = l, m l L + L l, m l = l, m l L L z + L z l, m l = ( l(l + 1) m l + m l ) l, ml l, m l = (l(l + 1) m l (m l 1)) = (l + l m l + m l ) = ((l + m l )(l m l ) + (l + m l )) = (l + m l )(l m l + 1) (9.49) Because we have to have + 0 and 0, from (9.48) and (9.49), we can see that The inequality (9.50) requires either or I. l m l and l m l 1 II. l m l and l m l 1; (l m l )(l + m l + 1) 0 (9.50) and while the other inequality (9.51)implies either or III. l m l and l m l 1 IV. l m l and l m l 1. (l + m l )(l m l + 1) 0. (9.51) Therefore, we should have either condition I or condition II holding true, and at the same time, one of conditions III and IV to be true. First, suppose m l 0. As specified on p.185, we chose l 0. If m l is nonnegative, l m l 1 is impossible. Hence, condition I should be

190 188 CHAPTER 9. ANGULAR MOMENTUM satisfied. In particular, we need l m l as l m l 1 is automatically satisfied in this case. Consider conditions III and IV next. Condition III reduces to l m l 1, and condition IV is impossible because l m l and l m l 1 implies l 1 which contradicts l 0. So, we have to satisfy l m l and l m l 1 simultaneously. Therefore, m l l if m l 0. (9.5) Next, suppose m l < 0. Condition I reduces to l m l 1, and condition II is impossible as l > 0 > m l. Now, condition III reduces to l m l as l m l 1 is automatically satisfied, and condition IV is impossible because l 0 > m l > m l 1. Hence, we require l m l 1 and l m l. Therefore, m l l if m l < 0. (9.53) The inequalities (9.5) and (9.53) together with the fact that L + /L raises/lowers the magnetic quantum number m l by one imply m l = l, l + 1,..., l 1, +l ; where l 0. (9.54) As the spacing for each pair of neighboring m l values is 1, it is necessary that l ( l) = l is a nonnegative integer. But, this is possible only for an integral or half-integral l. It turns out l is an integer for the orbital angular momentum 4. Before closing this section, we will briefly discuss a simple implication of (9.43), (9.44), and (9.48), which will be useful later. A similar result follows from (9.46), (9.47), and (9.49). From (9.43), we can see that L + l, m l is an eigenvector of L with the eigenvalue l(l+1). Furthermore, (9.44) indicates that L + l, m l is an eigenvector of L z with the eigenvalue (m l + 1). Hence, we have L + l, m l = C l, m l + 1 (9.55) for some scalar C. However, we already know from (9.48) that 4 Half-integral l s correspond to an internal or intrinsic angular momentum called the spin of a particle. We will discuss electron spin in Chapter 11.

191 9.5. GENERALIZED ANGULAR MOMENTUM J 189 L + l, m l = l, m l + 1 C C l, m l + 1 = C = (l m l )(l + m l + 1), (9.56) and so, C = e iθ (l m l )(l + m l + 1). (9.57) According to standard convention, we choose θ = 0 or e iθ = 1 Shankar, 1980, p.336 to get C = (l m l )(l + m l + 1) = Similarly, we can also show L + l, m l = (l m l )(l + m l + 1) l, m l + 1. (9.58) L l, m l = (l + m l )(l m l + 1) l, m l 1. (9.59) Finally, noting that l, l, l, l + 1,..., l, l 1, and l, l are the only eigenvectors of L z from (9.54), we have to have L + l, l = 0 (9.60) and L l, l = 0 (9.61) as in Section Generalized Angular Momentum J As shown on p.177, the three operators associated with the components of an arbitrary classical angular momentum, L x, L y, and L z, satisfy the commutation relations (9.5), (9.6), and (9.7). These relations originate from

192 190 CHAPTER 9. ANGULAR MOMENTUM the geometric properties of rotations in three-dimensional space. However, in quantum mechanics it is sometimes necessary to consider more abstract angular momentum with no counterpart in classical mechanics. What we do is to go backwards, from the commutation relations to rotations, and define quantum mechanical angular momentum J as any set of three observables/hermitian operators J x, J y, and J z which satisfies the same commutation relations as L x, L y, and L z. Namely, J x, J y = i J z, (9.6) J y, J z = i J x, (9.63) J z, J x = i J y, (9.64) and J, J x = J, J y = J, J z = 0; (9.65) where (9.65) follows from (9.6), (9.63), and (9.64) as in (9.13) on p.178. Note that J = J x + J y + J z (a scalar) by definition, and (9.65) implies J, J = 0. With this theoretical framework, quantum mechanical angular momentum is based entirely on the commutation relations (9.6), (9.63), and (9.64). If you examine the computations we conducted for L, L, L x, L y, and L z, you can see that we only used the commutation relations (9.5), (9.6), and (9.7), which are equivalent to (9.6), (9.63), and (9.64). Therefore, all the results we obtained for L apply to J without modification. In particular, we can define raising and lowering operators by which satisfy J + = J x + ij y (9.66) and J = J x ij y, (9.67) J + j, m j = (j m j )(j + m j + 1) j, m j + 1 (9.68)

193 9.5. GENERALIZED ANGULAR MOMENTUM J 191 as well as and J j, m j = (j + m j )(j m j + 1) j, m j 1 (9.69) J + j, j = 0 (9.70) and J j, j = 0. (9.71) So, we have j m j +j for j 0, (9.7) J j, m j = j(j + 1) j, m j, (9.73) and J z j, m j = m j j, m j. (9.74) Finally, recall that one immediate implication of this theory for experimental physics is that simultaneous measurements of two components of J is impossible while J and one component of J, typically chosen to be J z, can be determined at the same time. We will revisit the concept of generalized angular momentum in Chapter 11, where we discuss the electron spin.

194 19 Exercises 1. Show that L z, L x = i L y.

195 Chapter 10 The Hydrogen Atom So far 1. TDSE (the Time-Dependent Schrödinger Equation) + V (x, t) m x Ψ(x, t) Ψ(x, t) = i t (10.1). TISE (the Time-Independent Schrödinger Equation) d m dx + V (x) ψ(x) = Eψ(x) (10.) 3. The Relation between Ψ and ψ Ψ(x, t) = ψ(x)e iet/ (10.3) Terminology In above, ψ(x) is an eigenfunction, and E is an eigenvalue. Here is a more familiar -by- matrix example = = 1

196 194 CHAPTER 10. THE HYDROGEN ATOM 1 is an eigenvector for 1 0 with an eigenvalue. 0 We have not really tried any real system yet. But, we need to carry out a real computation for a real system for an experimental confirmation of the correctness of the Schrödinger Equation One of the simplest systems is a hydrogen atom. However, this poses a couple of difficulties since (1) two particles are involved now and also since () it is no longer one-dimensional, but threedimensional Particles Instead of 1 Let M be the mass of the nucleus and m the mass of the electron. The nucleus and the electron are moving about their fixed center of mass. But, introduction of the reduced mass µ given by µ = Mm M + m (10.4) solves this problem. In particular, for a hydrogen atom, µ = and we can simply use m instead of µ. ( ) M m m (10.5) m + M 10. Three-Dimensional System The classical energy equation is p µ + V = E (10.6)

197 10.. THREE-DIMENSIONAL SYSTEM 195 or 1 ( ) p µ x + p y + p z + V (x, y, z) = E. (10.7) We want to transform this into a three-dimensional Schrödinger Equation. Our solution, the wavefunction, now depends on four variables x, y, z, and t. Ψ(x, t) = Ψ(x, y, z, t) (10.8) Here are the operator correspondences we are going to use. p x i x p y i y p z i z E i t (10.9) Of course, this means we have the following. p x = x p y = y p z = z (10.10) Therefore, p µ = p x + p y + p z µ = 1 µ ( = µ x + ( x + y + z y ) ) + ( ) z. (10.11) On the other hand, the potential energy is now a function of x, y, and z as below.

198 196 CHAPTER 10. THE HYDROGEN ATOM V = V (x, y, z) = e 4πε 0 x + y + z (10.1) We now have the following Time-Dependent Schrödinger Equation in three dimensions. µ ( ) x + y + z Ψ(x, y, z, t) + V (x, y, z)ψ(x, y, z, t) Ψ(x, y, z, t) = i t Let us introduce a Laplacian operator or del squared defined by (10.13) = x + y + z. (10.14) This simplifies the equation to the following form. where Now let µ Ψ + V Ψ = i Ψ t ; (10.15) V = V (x, y, z) and Ψ = Ψ(x, y, z, t). (10.16) Ψ(x, y, z, t) = ψ(x, y, z)e iet/. (10.17) Then, we get the Time-Independent Schrödinger Equation.

199 10.. THREE-DIMENSIONAL SYSTEM 197 µ ψ(x, y, z) + V (x, y, z)ψ(x, y, z) = Eψ(x, y, z) (10.18) As it turns out, it is easier to solve this equation in spherical (polar) coordinates, where (x, y, z) are replaced by (r, θ, ϕ). 1 Since a hydrogen atom possesses spherical symmetry, this is a better coordinate system. Figure 10.1 shows the spherical coordinates used in physics, while Figure 10. is the convention used in mathematics. It is unfortunate that the roles of θ and ϕ are switched between the two systems, but there is not much chance of confusion once you become used to the spherical coordinate system. In this class, we will use the physicists convention. z (r, θ, φ) θ r φ y x Figure 10.1: Spherical Coordinate System (Source: Wikimedia Commons; Author: Andeggs) Right away, V (x, y, z) simplifies as follows. 1 Note that in pure math books θ and ϕ are usually switched.

200 198 CHAPTER 10. THE HYDROGEN ATOM z (r, θ, φ) φ r θ y x Figure 10.: Spherical Coordinate System Used in Mathematics (Source: Wikimedia Commons; Author: Dmcq) V (x, y, z) = e 4πε 0 x + y + z = e 4πε 0 r = V (r, θ, φ) (10.19) In fact, the potential energy only depends on the distance from the origin. Of course, V (r, θ, φ) = e 4πε 0 r = V (r) (10.0) ψ(x, y, z) = ψ(r, θ, φ). (10.1) This means that we have to convert = + + to another equivalent x y z operator expressed as derivatives with respect to r, θ, and φ. Then, we will have

201 10.. THREE-DIMENSIONAL SYSTEM 199 The answer is µ ψ(r, θ, φ) + V (r)ψ(r, θ, φ) = Eψ(r, θ, φ). (10.) = 1 ( r ) + 1 r r r r sin θ ( sin θ ) + θ θ 1 r sin θ φ, (10.3) and we can see that appears far more complicated in the spherical coordinates. Here is how it is done. We begin with the three relations between (x, y, z) and (r, θ, φ). x = r sin θ cos φ y = r sin θ sin φ z = r cos θ (10.4) We can see how etc. can be converted. What we should do is to keep x using chain rules and relations like x = sin θ cos φ. But, the actual operation r is quite tedious. Let us prove a part of this conversion. As Consider a one-variable function ψ(r) for r = x + y + z. ψ x = r ψ x r = x ψ x + y + z r = x ψ r r (10.5) and r x = 1 ( x + y + z ) 1/ x = r, (10.6)

202 00 CHAPTER 10. THE HYDROGEN ATOM we have ψ x = x ( ) x ψ = ( x 1 ) ψ = x ( ) 1 ψ + x ( 1 r r x r r x r r x r = 1 ψ r r + x r ( ) 1 ψ = 1 ψ x r r r r r + x x ( ) 1 ψ r r r r = 1 ( ) ψ r r + x 1 ψ r r r r ) ψ r. (10.7) Similarly, ψ y = 1 ψ r r + y r r ( 1 r ) ψ r (10.8) and ψ z = 1 r ψ r + z r r ( 1 r ) ψ. (10.9) r Therefore, = 3 r ψ r + x + y + z r ) ψ = 3 r ( ψ r + r 1 ψ r r + 1 r ψ r r = r ( 1 r ) ψ = 3 ψ r r ψ r + ψ r r + r r = 1 ( r r In order to get other parts, try ψ = ψ(φ) and ψ = ψ(θ). ( ) 1 ψ r r ) r ψ r. (10.30) In preparation for the rest of the course as well as your career as a physicist or just as someone who needs to understand and use physics, please do get used to the spherical coordinates. For example, the volume element becomes dxdydz r dr sin θdθdφ or r drd(cos θ)dφ

203 10.. THREE-DIMENSIONAL SYSTEM 01 Figure 10.3: Volume Element in Spherical Coordinates (Source: Victor J. Montemayor, Middle Tennessee State University) in spherical coordinates as shown in Figure Let us pause for a minute here, take a deep breath, and summarize what we have achieved so far to clearly understand where we are now, as well as, hopefully, where we are heading. 1. Due to the spherical symmetry inherent in the hydrogen atom, we made a decision to use the spherical coordinates as opposed to the familiar Cartesian coordinates. In particular, this simplifies the expression for the potential energy greatly. It is now a function only of the radial distance between the electron and the proton nucleus. a V (x, y, z) = V (r) = e 4πε 0 r (10.31). However, a rather heavy trade-off was the conversion of to the

204 0 CHAPTER 10. THE HYDROGEN ATOM equivalent expression in the spherical coordinates. The conversion process is cumbersome, and the resulting expression is far less palatable. = 1 ( r ) + 1 r r r r sin θ ( sin θ ) + θ θ 1 r sin θ φ (10.3) a If the charge on the nucleus is Ze rather than just e, we have V (x, y, z) = V (r) = Ze 4πε 0r instead. This makes extending the results obtained for the hydrogen atom to hydrogen-like atoms quite straightforward as we will see in Section We are now ready to solve the Time-Independent Schrödinger Equation in spherical coordinates given by µ ψ(r, θ, φ) + V (r)ψ(r, θ, φ) = Eψ(r, θ, φ). (10.33) We will resort to the separation of variables technique yet one more time. So, consider ψ as a product of three single variable functions of r, θ, and φ respectively. ψ(r, θ, φ) = R(r)Θ(θ)Φ(φ) (10.34) Then, the Time-Independent Schrödinger Equation is µ (r,θ,φ) + V (r) R(r)Θ(θ)Φ(φ) = ER(r)Θ(θ)Φ(φ). (10.35) Let us write it out in its full glory and start computing! µ ( 1 r (RΘΦ) ) + 1 r r r r sin θ ( sin θ (RΘΦ) ) + θ θ +V (r)r(r)θ(θ)φ(φ) = ER(r)Θ(θ)Φ(φ) 1 r sin θ (RΘΦ) φ

205 10.. THREE-DIMENSIONAL SYSTEM 03 = µ = µ = µ ( 1 r R ) r r r ΘΦ + 1 r sin θ ΘΦ 1 r r ΘΦ 1 r d dr ( ( r R r r dr dr ) ) Now, multiply through by +V (r)rθφ ( sin θ Θ ) θ θ RΦ + 1 Φ r sin θ φ RΘ + 1 r sin θ RΦ ( sin θ Θ ) 1 Φ + θ θ r sin θ RΘ φ +V (r)rθφ + 1 r sin θ RΦ d ( sin θ dθ ) 1 Φ + dθ dθ r sin θ RΘd dφ +V (r)rθφ = ER(r)Θ(θ)Φ(φ) (10.36) µ r sin 1 θ RΘΦ. µ µ r sin 1 θ ΘΦ 1 ( d r dr ) + 1 RΘΦ r dr dr r sin θ RΦ d ( sin θ dθ ) dθ dθ + 1 Φ r sin θ RΘd + V (r)rθφ µ dφ r sin 1 θ RΘΦ = ERΘΦ µ r sin 1 θ RΘΦ (10.37) = sin θ R ( d r dr ) + sin θ dr dr Θ = 1 d Φ Φ dφ = θ sin R ( d sin θ dθ ) + 1 d Φ dθ dθ Φ = µ r sin θe ( d r dr ) sin θ dr dr Θ dφ + µ r sin θv (r) d dθ ( sin θ dθ dθ ) µ r sin θ E V (r) (10.38)

206 04 CHAPTER 10. THE HYDROGEN ATOM The LHS (lefthand side) is a function of φ, and the RHS (righthand side) is a function of r and θ. And the equality holds for all values of r, θ, and φ. Therefore, both sides have to equal a constant. For later convenience, we denote this by m l. So, we have or 1 d Φ Φ dφ = m l (10.39) d Φ dφ = m l Φ. (10.40) According to (10.38), this also means sin θ R ( d r dr ) sin θ dr dr Θ Dividing through by sin θ, 1 ( d r dr ) R dr dr = 1 ( d r dr ) R dr dr 1 Θ sin θ ( d sin θ dθ ) µ dθ dθ r sin θ E V (r) = m l. (10.41) ( d sin θ dθ ) µ dθ dθ r E V (r) = m l sin θ + µ r E V (r) = m l sin θ 1 Θ sin θ ( d sin θ dθ dθ dθ ). (10.4) The LHS of (10.4) depends on r, and the RHS is a function of θ. As the equality holds for any values of r and θ, both sides have to be a constant. Denote that constant by l(l+1), for the reason that will become clear shortly, to get We are placing any restriction on the values l can take. Because l can be a complex number, any scalar can be represented as l(l + 1). However, we will later show that l is in fact a nonnegative integer.

207 10.3. THE SOLUTIONS FOR Φ 05 and ( 1 d sin θ dθ ) + m l Θ sin θ dθ dθ sin θ = l(l + 1)Θ (10.43) ( 1 d r dr ) + µ r dr dr E V (r) R = l(l + 1) R r. (10.44) We have now reduced the problem to that of solving the following three equations, each in one variable. 1 r d dr ( 1 d sin θ dθ ) sin θ dθ dθ ( r dr ) dr d Φ dφ = m l Φ (10.45) + m l Θ sin θ = l(l + 1)Θ (10.46) + µ E V (r) R = l(l + 1) R r (10.47) 10.3 The Solutions for Φ The easiest to solve is equation (10.45). So, d Φ dφ = m l Φ Φ(φ) = e im lφ. (10.48) To be precise, we have Ae im lφ (10.49)

208 06 CHAPTER 10. THE HYDROGEN ATOM for an arbitrary scalar A. However, we can adjust the constant when we normalize the total wavefunction Ψ(r, φ, θ, t). So, we have set A = 1 for now. Consider single-valuedness. Φ(0) = Φ(π) = e i m l 0 = e i m l π = 1 = e im l π = cos(m l π) + i sin(m l π) (10.50) Now, let m l = a + bi for (a, b R). 1 = e im l π = e i(a+bi)π = e iaπ e bπ (10.51) It is now obvious that m l is a real number and e bπ = e bπ = 1 (10.5) πb = 0 = b = 0 (10.53) m l = 0, 1,, 3, 4, (10.54) For each value of m l, there is a corresponding solution Φ ml (φ) = e im lφ. (10.55) The numbers m l are known as magnetic quantum numbers. Now we need to solve equations (10.46) and (10.47) The Solutions for Θ Equation (10.46) is known as the angular equation. 1 ( d sin θ dθ ) + m l Θ sin θ dθ dθ sin = l(l + 1)Θ θ Note that d cos θ = sin θdθ.

209 10.4. THE SOLUTIONS FOR Θ 07 So, and sin θ d dθ = sin θ d d cos θ sin θ 1 sin θ d dθ = d d cos θ, = sin d θ d cos θ = (1 d cos θ) d cos θ. This suggests that (10.46) can be simplified if we let z = cos θ. We have ( ) d (1 cos dθ θ) d cos θ d cos θ which leads to ( d (1 z ) dθ ) dz dz + m l Θ 1 cos θ = l(l + 1)Θ, + m l Θ = l(l + 1)Θ = 1 z ( d (1 z ) dθ ) ( ) + l(l + 1) m l Θ = 0 (10.56) dz dz 1 z At the time Schrödinger was solving his equation for the hydrogen atom, this was already a well-known differential equation, and the solutions are called the associated Legendre functions denoted by Θ lml (z). This is related to a better known set of functions called the Legendre polynomials denoted by P l (z). The relation between Θ lml (z) and P l (z) is as follows. Our first job is to prove this. Now P l (z) is a solution to Θ lml (z) = (1 z ) m l / d m l P l (z) dz m l (10.57)

210 08 CHAPTER 10. THE HYDROGEN ATOM (1 z ) d P l dz z dp l dz + l(l + 1)P l = 0. (10.58) Differentiate both sides with respect to z to obtain Try d dz again. z d P l dz = (1 z ) d3 P l dz 3 d P l dz Suppose we get + (1 z ) d3 P l dz 3 z d3 P l dz 3 = (1 z ) d4 P l dz 4 4z d P l dz + z d3 P l dz 3 z d3 P l dz 3 6z d3 P l dz 3 dp l dz z d P l dz + l(l + 1)dP l dz ( (l(l + 1) ) dp ) l dz + (1 z ) d4 P l dz 4 + l(l + 1)d P l dz d P l dz = 0. (10.59) d P l dz + (l(l + 1) 6)d P l dz = 0 (10.60) (1 z ) dk+ P l dz k+ (k + 1)z dk+1 P l dz k+1 dk + (l(l + 1) k(k + 1)) dz P k l = 0 (10.61) after differentiating k times. Relation (10.61) holds for k = 0, 1, and according to (10.58), (10.59), and (10.60). Now apply d one more time. dz z dk+ P l dz k+ + (1 z ) dk+3 P l dz k+3 (k + 1)dk+1 P l dz k+1 +(l(l + 1) k(k + 1)) dk+1 P l dz k+1 (k + 1)z dk+ P l dz k+ = (1 z ) d(k+1)+ P l dz (k+1)+ ((k + 1) + 1)z d(k+)+1 P l dz (k+)+1 +(l(l + 1) (k + 1)((k + 1) + 1) dk+1 P l = 0 (10.6) dzk+1

211 10.4. THE SOLUTIONS FOR Θ 09 So, by mathematical induction, (1 z ) dk+ dz P k+ l (k + 1)z dk+1 P l dz + (l(l + 1) k(k + P l k+1 1))dk dz = 0 k (10.63) when we apply dk dz k to both sides of (10.58). On the other hand, consider and substitute this into Θ lml = (1 z ) m l / Γ(z) (10.64) ( d (1 z ) dθ ) ( ) + l(l + 1) m l Θ = 0. (10.65) dz dz 1 z We have the following lengthy computation. ( d (1 z ) d ) ( dz dz (1 z ) k/ Γ + l(l + 1) k 1 z ) (1 z ) k/ Γ = z d dz ((1 z ) k/ Γ) + (1 z ) d dz ((1 z ) k/ Γ) ( ) k + l(l + 1) 1 z (1 z ) k/ Γ ( ) k = z (1 z ) (k/ 1) ( z)γ + (1 z k/ dγ ) dz +(1 z ) d ( ) k dz (1 z ) (k/ 1) ( z)γ + (1 z k/ dγ ) dz ( ) = z k + l(l + 1) 1 z ( (1 z ) k/ Γ kz(1 z ) (k/ 1) Γ + (1 z ) ) k/ dγ dz

212 10 CHAPTER 10. THE HYDROGEN ATOM +(1 z ) k (k/ 1)(1 z ) (k/ ) ( z)( z)γ + k (1 z ) (k/ 1) ( )Γ + k (1 z ) (k/ 1) ( z) dγ dz + k (1 z ) (k/ 1) ( z) dγ dz + (1 z ) k/ d Γ dz ( ) + k l(l + 1) 1 z (1 z ) k/ Γ +(1 z ) = kz (1 z ) (k/ 1) Γ z(1 z k/ dγ ) ( ) dz k kz 1 kz(1 z (k/ 1) dγ ) dz kz(1 z ) ( ) + l(l + 1) 1 z (1 z ) (k/ ) Γ k(1 z ) (k/ 1) Γ k (k/ 1) dγ dz + (1 z ) k/ d Γ dz (1 z ) k/ Γ = 0 Multiplying through by (1 z ) k/, kz ( ) dγ k Γ z 1 z dz + (1 z ) kz 1 (1 z ) Γ k(1 z ) 1 Γ ( ) kz(1 z 1 dγ ) dz kz(1 z 1 dγ ) dz + d Γ + l(l + 1) k Γ dz 1 z ( ) = kz dγ k Γ z 1 z dz + kz 1 (1 z ) 1 Γ kγ kz dγ dz ( ) +(1 z ) d Γ dz + l(l + 1) k Γ 1 z kz dγ dz = (1 z ) d Γ + ( kz z)dγ dz dz + l(l + 1) + kz + kz ( k 1) k k Γ 1 z = (1 z ) d Γ dγ (k + 1)z dz dz + l(l + 1) + k z k k Γ 1 z

213 10.4. THE SOLUTIONS FOR Θ 11 = (1 z ) d Γ dγ (k + 1)z dz dz + l(l + 1) k k Γ = 0. (10.66) So, we have shown or (1 z ) d Γ dγ (k + 1)z + l(l + 1) k(k + 1) Γ = 0 (10.67) dz dz (1 z ) d Γ dz ( m l + 1)z dγ dz + l(l + 1) m l ( m l + 1) Γ = 0. (10.68) Comparing this with (10.61), which is given below again for reference, (1 z ) dk+ dz P k+ l (k + 1)z dk+1 P l dz + (l(l + 1) k(k + P l k+1 1))dk dz = 0, k we conclude It remains to solve (10.58) Θ lml = (1 z ) m l / d m l P l dz m l. (10.69) (1 z ) d P l dz z dp l dz + l(l + 1)P l = 0 for P l. Try P l (z) = a k z k. (10.70) k=0 We get a recursion relation as we did before on p.146 in Section for the simple harmonic oscillator.

214 1 CHAPTER 10. THE HYDROGEN ATOM a j+ = j(j + 1) l(l + 1) a j (10.71) (j + )(j + 1) Also as before, we do not want an infinite series; that is, we want the series to terminate after a finite number of terms, so that the function is integrable/ normalizable. We need j(j + 1) l(l + 1) = j + j l l = (j + l)(j l) + j l = (j l)(j + l + 1) = 0 = l = j or j 1. (10.7) As j is a nonnegative integer, l can be any integer, positive, 0, or negative. However, the only meaningful quantity is l(l + 1). Noting that the parabola y(l) = l(l + 1) = ( l + 1 ( 1 ) ) is symmetric about l = 1, it is sufficient to consider l = 0, 1,, 3,.... So, l = 0, 1,, 3,... give all acceptable solutions. In fact, y( l 1) = ( l 1)( l 1+1) = ( l 1)( l) = l(l +1) proves that the pairs (0, 1), (1, ), (, 3), (3, 4)... give the same value for l(l + 1), and negative l values are clearly redundant. We get P 0 = 1, P 1 = z, P = 1 3z, P 3 = 3z 5z 3,.... So, the corresponding Θ lml s are Θ 00 = 1, Θ 10 = z, Θ 1±1 = (1 z ) 1/, Θ 0 = 1 3z, Θ ±1 = (1 z ) 1/ z, Θ ± = 1 z,... Note that P l (z) is an lth degree polynomial. Therefore, for each l, we have m l = l, l + 1,..., 0,..., l 1, + l. (10.73)

215 10.5. ASSOCIATED LEGENDRE POLYNOMIALS AND SPHERICAL HARMONICS 10.5 Associated Legendre Polynomials and Spherical Harmonics Recall that P l (z) s = P l (cos θ) s are called the Legendre polynomials, and Θ lml s are called associated Legendre functions/polynomials as explained on p.07. The associated Legendre polynomials are also denoted by P m l l (cos θ) or simply by Pl m (cos θ). So, Θ lml (cos θ) = P m l l (cos θ). (10.74) We now introduce a new set of functions known as the spherical harmonics, which are the products of Φ and Θ up to the normalization constant. It is customary to denote each spherical harmonic by Y m l l or simply by Yl m. So, we have Y m l l (θ, ϕ) Φ ml (ϕ)θ lml (cos θ) = Φ ml (ϕ)p m l l (cos θ) = e imlϕ P m l l (cos θ). (10.75) With the appropriate normalization constant 3, Y m l l (θ, ϕ) = ( 1) m l where the normalization is chosen such that In fact, Y m l l = π π 0 0 π l + 1 (l m l )! 4π (l + m l )! P m l l (cos θ)e imlϕ ; (10.76) (Y m l l (θ, ϕ)) Y m l l (θ, ϕ) sin θdθdϕ (Y m l l (θ, ϕ)) Y m l l (θ, ϕ)d(cos θ)dϕ = 1. (10.77) s are orthonormal such that π π 0 0 (Y m l l (θ, ϕ)) Y m l l (θ, ϕ) sin θdθdϕ 3 Different definitions for spherical harmonics are used in different fields such as geodesy, magnetics, quantum mechanics, and seismology. The definition given here is commonly adopted in the quantum mechanics community.

216 14 CHAPTER 10. THE HYDROGEN ATOM = π (Y m l l (θ, ϕ)) Y m l l (θ, ϕ)d(cos θ)dϕ = δ ll δ m l m. (10.78) l We will see how this comes about in Section In passing let us make a note of the fact that the normalization constant from the ϕ-dependent portion of the integral is 1, and the normalization constant from the θ- π (l+1)(l m)! dependent portion of the integral is (l+m l. )! Spherical harmonics for m l < 0 has the following alternative formula Y m l l (θ, ϕ) = l + 1 (l m l )! 4π (l + m l )! P m l l (cos θ)e imlϕ. (10.79) This is a direct consequence of the following identity satisfied by the associated Legendre polynomials. 4 We will write m instead of m l for simplicity. Indeed, for m = m < 0, we have Y m l Pl m m (l m)! (x) = ( 1) (l + m)! P l m (x) (10.80) (θ, ϕ) = Y m l (θ, ϕ) = ( 1) m l + 1 (l + m )! 4π (l m )! P m l (cos θ)e i m ϕ = ( 1) m l + 1 (l + m )! m (l m )! ( 1) 4π (l m )! (l + m )! P m l (cos θ) e i( m )ϕ = l + 1 (l m )! 4π (l + m )! P m l (cos θ)e imϕ. (10.81) We can combine (10.76) and (10.81) in the following manner. l (θ, ϕ) = ( 1) (m l+ m l )/ l + 1 (l m l )! 4π (l + m l )! P m l l (cos θ)e im lϕ Y m l (10.8) Note that the phase factor ( 1) m l or ( 1) (m l+ m l )/ does not change the physical observables, and it is more or less a mathematical book keeping device. 4 See Fact K.1 of Appendix K on p.36 for a proof.

217 10.6. L, L Z, AND THE SPHERICAL HARMONICS 10.6 L, L z, and the Spherical Harmonics The orthonormality relation (10.78) is a consequence of the following fact. Fact 10.1 The spherical harmonics are simultaneous eigenstates of L and L z corresponding to the eigenvalues l(l + 1) and m l. L Y m l l (θ, ϕ) = l(l + 1) Y m l l (θ, ϕ) (10.83) and L z Y m l l (θ, ϕ) = m l Y m l l (θ, ϕ) (10.84) Because L and L z are Hermitian operators, and also because Y m l l s are normalized, the set {Y m l l } l m l +l l=0,1,,... forms an orthonormal basis. Hence, Y m l l s are not only orthonormal, but also complete, and any function of θ and ϕ can be represented as a superposition (linear combination) of spherical harmonics. Recall that for each of l = 0, 1,, 3,... we have m l = l, l + 1, l +,... 0,..., l, l 1, l. All the spherical harmonics for l = 0, 1, and are listed in Table Next on our agenda is the radial function R The Radial Function R The general radial equation for a one-electron atom is ( 1 d r dr ) + µ E + Ze R = l(l + 1) R r dr dr 4πε 0 r r ; (10.85) where Ze is the charge on the nucleus. We will only solve this for Z = 1 or for the hydrogen atom.

218 16 CHAPTER 10. THE HYDROGEN ATOM l = 0 m l = 0 Y 0 0 (θ, ϕ) = 1 l = 1 l = m l = 1 Y 1 Spherical Harmonics 1 (θ, ϕ) = 1 m l = 0 Y 0 1 (θ, ϕ) = 1 m l = +1 Y 1 1 (θ, ϕ) = 1 m l = m l = 1 Y (θ, ϕ) = 1 4 Y 1 (θ, ϕ) = 1 m l = 0 Y 0 (θ, ϕ) = 1 4 m l = +1 Y 1 (θ, ϕ) = 1 m l = + Y (θ, ϕ) = π 3 3 π e iϕ sin θ = 1 π cos θ = z π r π eiϕ sin θ = π e iϕ sin θ = x iy π r π e iϕ sin θ cos θ = 1 π (3 cos θ 1) = π eiϕ sin θ cos θ = 1 15 π eiϕ sin θ = x+iy π r 15 (x iy) π r 15 (x iy)z π r 5 z x y π r 15 (x+iy)z π r 15 (x+iy) π r Table 10.1: Spherical Harmonics for l = 0, 1, and Let us first invoke some substitutions to put the radial equation in a more manageable form. β = µe (β > 0) (10.86) ρ = βr (10.87) With these substitutions, we get γ = µe 4πε 0 β (10.88) As ρ ( 1 d ρ dr ) + 1 l(l + 1) + γ R = 0. (10.89) ρ dρ dρ 4 ρ ρ R(ρ) e ρ/. (10.90)

219 10.7. THE RADIAL FUNCTION R 17 So, let us write and substitute this into the radial equation to get d ( ) F df γ 1 dρ + ρ 1 dρ + ρ Once again, we will try a series solution with R(ρ) = e ρ/ F (ρ), (10.91) l(l + 1) F = 0. (10.9) ρ F (ρ) = ρ s k=0 a k ρ k (a 0 0, s 0); (10.93) where the condition on s is in place to prevent F from blowing up at the origin, while a 0 0 assures that our solution is not trivial as we will see in (10.96). After substituting this trial series solution into the differential equation (10.9), we get s(s + 1) l(l + 1)a 0 ρ s + {(s + j + 1)(s + j + ) l(l + 1)a j+1 j=0 (s + j + 1 γ)a j }ρ s+j 1 = 0. (10.94) For the left-hand side to be zero for any value of ρ, the following two relations should hold. s(s + 1) l(l + 1) = 0 s + j + 1 γ a j+1 = (s + j + 1)(s + j + ) l(l + 1) a j (10.95) (10.96) The first equality (10.95) gives s = l and s = (l + 1). But, s = (l + 1) should be rejected as s 0. The second relation (10.96)

220 18 CHAPTER 10. THE HYDROGEN ATOM indicates that a j+1 = j + l + 1 γ (j + l + 1)(j + l + ) l(l + 1) a j γ = j + l + 1 = n (Call this n.) (10.97) assures termination of the series after a finite number of terms; namely, after the jth term. Hence, n = l + 1, l +, l + 3, as j goes from 0 to with l = 0, 1,, 3,.... Now recall from (10.86) on p.16 and from (10.88) on p.16 So, γ = n β = µe = µe 4πε 0 β µe 4 ( ) ( ) µe E n = (4πε 0 ) n = 3π ε 0 n = µa 0 n for n = 1,, 3,. (10.98) Here,

221 10.7. THE RADIAL FUNCTION R 19 is analogous to the Bohr radius a B defined by a 0 = 4πε 0 µe (10.99) a B = 4πε 0 m e e, where m e is the rest mass of an electron. Putting it all together, we get where ψ nlml (r, θ, φ) = R nl (r)y m l l (θ, φ) = R nl (r)θ lml (θ)φ ml (φ); (10.100) Φ ml (φ) = e im lφ m l = 0, 1,, 3,, (10.101) Θ lml (θ) = Θ lml (cos θ) = P m l l (cos θ) = sin m l θf l ml (cos θ), (10.10) and R nl (r) = e r/na 0 ( r a0 ) Gnl ( r a0 ). (10.103) The functions F and G are both polynomials; is a polynomial in cos θ, and F l ml (cos θ)

222 0 CHAPTER 10. THE HYDROGEN ATOM G nl ( r a 0 ) is a polynomial in r/a 0. More specifically, the normalized radial wavefunctions are 5 R nl (r) = ) ( 3 ( ) (n l 1)! na 0 n(n + l)! 3 e r/na 0 r l ( ) r L l+1 n l 1 na 0 na 0 (10.104) and the normalized full wavefunctions are ψ nlml = R nl (r)y m l l (θ, ϕ) ) = ( 3 ( ) (n l 1)! na 0 n(n + l)! 3 e r/na 0 r l ( ) r L l+1 n l 1 na 0 na 0 l + 1 (l m l )! 4π (l + m l )! P m l l (cos θ)e im lϕ. (10.105) Let us now summarize the relations among n, l, and m l. We have obtained m l = 0, 1,, 3,, (10.106) l = m l, m l + 1,, m l + 3,, (10.107) and n = l + 1, l +, l + 3, (10.108) in this order, but this can be reorganized as follows. n = 1,, 3, 4, (10.109) l = 0, 1,,, n 1 (10.110) m l = l, l + 1,, 0,, +l (10.111) 5 L in (10.104) is an associated Laguerre polynomial described in Appendix L.

223 10.8. HYDROGEN-LIKE ATOMS 1 The first number n is associated with R(r) and is called the principal quantum number. The second number l is associated with Θ(θ) and is called the orbital quantum number or the angular momentum quantum number. Finally, the number m l is called the magnetic quantum number which is associated with Φ(φ). In addition to these, we also have a spin quantum number m s, which we will discuss in Chapter Hydrogen-Like Atoms A hydrogen-like atom, or hydrogen-like ion, is an atomic nucleus with one electron. Some examples other than the hydrogen atom itself are He +, Li +, Be 3+, and B 4+. If we denote the atomic number by Z, the charge carried by hydrogen-like ions are e(z 1). With this notation, we can derive all the wavefunctions for the hydrogen-like ions from the wavefunctions for the hydrogen atom by replacing the charge e on proton by Ze. You can see this from the time-independent Hamiltonian. { µ ( 1 r ) r r r + 1 r sin θ + 1 r sin θ φ ( θ Ze 4πε 0 r sin θ θ } ) ψ(r, θ, φ) = Eψ(r, θ, φ) (10.11) The only difference is Z appearing once in the potential term. The resulting wavefunctions are ψ nlml,z = R nl,z (r)y m l l (θ, ϕ) ) = ( Z 3 ( ) (n l 1)! na 0 n(n + l)! 3 e Zr/na 0 Zr l ( ) Zr L l+1 n l 1 na 0 na 0 l + 1 (l m l )! 4π (l + m l )! P m l l (cos θ)e im lϕ. (10.113) And, the total energy E Z is given by

224 CHAPTER 10. THE HYDROGEN ATOM E Z,n = Z µe 4 ( Z (4πε 0 ) n = µe 4 ) ( 1 Z 3π ε 0 n = ) 1 µa 0 n for n = 1,, 3,. (10.114) 10.9 Simultaneous Diagonalization of H, L, and L z Let us examine the big picture now. In Section 10.6, we saw that the spherical harmonics {Y m l l } l m l +l l=0,1,,... form an orthonormal basis that simultaneously diagonalizes L and L z. Likewise, the set of normalized eigenfunctions or eigenkets {ψ nlml } or { n, l, m l } (10.115) for n = 1,, 3,..., l = 0, 1,..., n 1, and m l = l, l+1, , l 1, +l form an orthonormal basis that simultaneously diagonalizes the Hamiltonian H, the magnitude of the orbital angular momentum squared L, and the z-component of the orbital angular momentum L z. We have the following set of simultaneous eigenvalue problems with a common solution { n, l, m l }. H ψ = E n ψ L ψ = l(l + 1) ψ (10.116) L z ψ = m l ψ The reason why such common eigenkets can be found is that the commutators H, L and H, L z are both zero in addition to L, L z = 0 which we know from (9.8). As you can see in Appendix J, there are alternative forms of. The form given in (10.3) is = 1 ( r ) + 1 r r r r sin θ ( sin θ ) + θ θ 1 r sin θ φ,

225 10.9. SIMULTANEOUS DIAGONALIZATION OF H, L, AND L Z 3 and one alternative form is = r + r r + 1 r θ + = r + r r r sin θ From (9.4) and (10.118), we have = r + r r r sin θ = r + r = r + r θ cos θ r sin θ or ( sin θ θ θ + 1 r sin θ ) ( sin θ ) + 1 θ θ sin θ ) ϕ (10.117) + 1 sin. (10.118) θ ϕ ϕ r 1 ( 1 r ( ) sin θ + 1 sin θ θ θ sin θ ϕ r L r. (10.119) We will use this form of in order to show H, L = 0 and H, L z = 0. 6 Fact 10. (H, L = 0) The full Hamiltonian H = µ + V (r) = µ r + r r L + V (r) (10.10) r commutes with the squared magnitude of the angular momentum L = 1 sin θ ( sin θ ) + 1 θ θ sin θ. (10.11) ϕ Proof The important point here is that the expression of H contains only the radial variable r and L, while the expression of L contains the azimuthal and polar variables ϕ and θ. With this in mind, the commutator H, L can be computed as follows. H, L = µ r + r r L + V (r), L r 6 Both H, L = 0 and H, L z = 0 are proved in a different manner in Appendix G.

226 4 CHAPTER 10. THE HYDROGEN ATOM = µ r + r r L, L + V (r), L r = µ r + L, L + r r µr, L + V (r), L. (10.1) It is now clear that each term of (10.1) is zero, and we have verified H, L = 0. (10.13) Fact 10.3 (H, L z = 0) The full Hamiltonian H = µ + V (r) = µ r + r r L + V (r) (10.14) r commutes with the z-component of the angular momentum Proof L z = i ϕ. (10.15) H, L z = µ r + r r L + V (r), Lz r = µ r + L, Lz + r r µr, Lz + V (r), Lz = Revisiting the Fundamental Postulates Let us recall the Fundamental Postulates presented in Section 3.1. The following is the set of guiding principles that connects the physical reality with mathematical abstraction.

227 REVISITING THE FUNDAMENTAL POSTULATES 5 Each physical system S has an associated Hilbert space H. Each physical state s of the physical system S has an associated normalized ket s, called a state ket, in H. Every physical observable q is represented by a Hermitian Operator Q whose domain is H. Measurement of q in the physical system S is mirrored in the abstract quantum mechanical system by the action of the corresponding operator Q on the state ket s. The only possible outcomes of quantum mechanical measurements are the eigenvalues of the corresponding operator Q. A state ket is initially a linear combination of normalized eigenkets of Q. However, after the measurement, the state ket becomes the eigen ket associated with the measured eigenvalue. You can note here that the quantum mechanical system is composed of the threesome (H, s, Q), which are purely mathematical and abstract constructs. In particular, no reference is made to any coordinate system or a function defined relative to such coordinates. Therefore, the generic simultaneous eigenvalue problems (10.116) and their solutions (10.115) are actually coordinate free. In other words, the eigenkets { n, l, m l } live in some abstract unspecified Hilbert space H. From Section 4.6, we can expand n, l, m l in the coordinate basis { r, θ, ϕ } to get n, l, m l = I n, l, m l = r, θ, ϕ r, θ, ϕ n, l, m l = or more appropriately 0 r<+ 0 θ π 0 ϕ<π = 0 r<+ 0 θ π 0 ϕ<π 0 r<+ 0 θ π 0 ϕ<π r, θ, ϕ n, l, m l r, θ, ϕ (10.16) ψ nlml (r, θ, ϕ) r, θ, ϕ (10.17) π π n, l, m l = r, θ, ϕ r, θ, ϕ r dr sin θdθdϕ n, l, m l 0 0 0

228 6 CHAPTER 10. THE HYDROGEN ATOM = = π π π π where we used the identities r, θ, ϕ n, l, m l r, θ, ϕ r dr sin θdθdϕ (10.18) ψ nlml (r, θ, ϕ) r, θ, ϕ r dr sin θdθdϕ; (10.19) 0 r<+ 0 θ π 0 ϕ<π π π 0 When we write 0 0 and r, θ, ϕ r, θ, ϕ = I r, θ, ϕ r, θ, ϕ r dr sin θdθdϕ = I. ψ nlml (r, θ, ϕ) = r, θ, ϕ n, l, m l (10.130) for the coefficients in (10.16) and (10.18), we are taking the projection of n, l, m l onto the infinite-dimensional coordinate system whose axes are defined by { r, θ, ϕ }, to use a geometric language. In a more common language, this amounts to solving the set of eigenvalue problems (10.116) using the spherical coordinate system (r, θ, ϕ) in order to obtain the eigenvectors as functions of r, θ, and ϕ. The operators H, L, and L z are also abstract and coordinate-independent generic objects whose representations in the spherical coordinate system are H = µ ( 1 r ) + 1 r r r r sin θ L = 1 sin θ ( sin θ ) + θ θ ( sin θ ) + 1 θ θ sin θ and 1 r sin θ + V (r), φ (10.131), (10.13) ϕ L z = i ϕ. (10.133)

229 REVISITING THE FUNDAMENTAL POSTULATES 7 Similarly, we can project n, l, m l onto the Cartesian system defined by the familiar { x, y, z }. 7 In this case, we already know H = µ ψ nlml (x, y, z) = x, y, z n, l, m l (10.134) x + y + z + V (x, y, z) (10.135) and ( L z = i x y y ). 8 (10.136) x 7 This is not to be confused with the usual projection onto the x-, y-, and z-axes of a vector v, which allows us to write v = (v x, v y, v z ). Our coordinate system here is not the usual three-dimensional system, but it is of infinite dimensions, each axis of which is defined by one of { x, y, z }; where ( < ) x, y, z < +. 8 As we know L x = i y z z y and L y = i ( z x x ) z as well as L = L x + L y + L z, we can also express L in terms of x, y, z and x, y, z. However, the resulting messy formula is not of much instructional value or other use to us. That is the reason why L is not included in this list.

230 8 Exercises 1. Consider a one-variable function ψ(r) for r = x + y + z, and derive the relation below. ψ x = 1 ( ) ψ r r + x 1 ψ r r r r. Without showing detailed computations, simply sketch the proof that Θ lml = (1 z ) m l / d m l P l dz m l. 3. Hydrogen, deuterium, and singly ionized helium are all examples of one-electron atoms. The deuterium nucleus has the same charge as the hydrogen nucleus, and almost exactly twice the mass. The helium nucleus has twice the charge of the hydrogen nucleus, and almost exactly four times the mass. Make an approximate prediction of the ratios of the ground state energies of these atoms. 4. Consider n = 3 state of a hydrogen atom. (a) What l values are possible? (b) For each value of l, what m l values are possible? (c) Finally, how many degenerate states do we have for n = 3? 5. What is the energy of a photon emitted when the electron drops from the 3rd highest energy level (n = 3) to the ground state (n = 1)? Leave µ, e, π, ε 0, and as they are. 6. Answer the following questions about the hydrogen atom. (a) The differential equation of the ϕ-dependent function Φ(ϕ) is given by d Φ = k Φ with a solution Φ(ϕ) = e ikϕ. Find all values dϕ of k = a + bi (a, b real) consistent with the single-valuedness condition. Show all your work. (b) For each principal quantum number n, there are n degenerate energy levels. Prove this. 7. Answer the following questions about the hydrogen atom.

231 9 (a) Consider n = 4. i. What is the largest allowed value of l? ii. What is the magnitude of the corresponding angular momentum? Leave as it is. iii. How many different z-components may this angular momentum vector have? iv. What is the magnitude of the largest z component? Leave as it is. (b) Consider an electron in the ground state of the hydrogen atom characterized by n = 0. The normalized ground state wavefunction is given by ψ 000 = 1 π ( 1 a 0 ) 3/ e r/a 0. You may use ( x x e bx dx = e bx b x b + ) + C b 3 x 3 e bx dx = e bx ( x 3 and b 3x b + 6x b 6 ) + C. 3 b 4 i. What is the probability P (0 r < a 0 ) that the electron lies inside a sphere of radius a 0 centered at the origin? Leave e as it is. ii. Find the average distance < r > of the electron from the nucleus. (c) For a hydrogen-like atom with Z = 3, what is the energy of a photon emitted when the electron drops from the 3rd highest energy level (n = 3) to the ground state (n = 1)? Leave µ, e, π, ε 0, and as they are. Do not use a 0 in your answer.

232 30

233 Chapter 11 Electron Spin There are quantum mechanical phenomena that do not fit in the framework based on the postulates given in Section 3.1 in a straightforward manner. While the postulates draw heavily on the analogy based on the corresponding classical mechanical system, these phenomena have no classical analogues. They require an additional variable called intrinsic spin and a special Hilbert space on which the spin operators act. The intrinsic spin S is a very important signature for many particles. We will devote this chapter to the discussion of electron spin as it is the most representative example. Intrinsic spin can be characterized and understood best if we regard it as another kind of angular momentum that requires a separate treatment different from that for the familiar orbital angular momentum represented by L x, L y, and L z What is the Electron Spin? The electron was discovered in 1897 by Joseph John Thompson. It was the first subatomic and fundamental particle discovered. Physicists knew from the beginning that an electron had charge and mass. But, the concept of electron spin was proposed by Samuel Abraham Goudsmit and George Uhlenbeck much later in 195. Because the electron was usually regarded as a point particle with no internal structure and no physical dimension, it was actually impossible for the electron to spin like a top. Therefore, it was necessary to assume that the electron was born with intrinsic angular momentum, not associated with its orbital motion, called electron spin. The first sign of electron spin came from an experiment conducted by Otto Stern 31

234 3 CHAPTER 11. ELECTRON SPIN and Walther Gerlach in Stern-Gerlach Experiment In this experiment, electrically neutral silver atoms passed through a nonuniform magnetic field B = Bẑ, oriented along the z-axis, which deflected the atoms. The atoms, after passing through the magnetic field, hit a photographic plate generating small visible dots. The deflection is along the z-axis and the magnitude of the force on the atoms is proportional to db. In the dz classical picture, the orbiting electron produces magnetic dipole moments whose magnitudes are continuously distributed. If this is the case, the extent of vertical deflection should also be continuously distributed forming a blob on the plate. On the other hand, in the quantum mechanical picture, there are l + 1 z-components, m l s, for l = 1,, 3,.... In particular, note that m l = 0 is always possible. When m l = 0, there is no force on the atom and the atoms with m l = 0 will not be deflected. Hence, if l = 0, no atom will be deflected. Figure 11.1: Stern-Gerlach Experiment (Diagram drawn by en wikipedia Theresa Knott.) The results of Stern-Gerlach experiment presented in their original paper Gerlach and Stern, 19, p.350 are reproduced in Figure 11.. The picture on the left shows only one line as there was no external magnetic field. When an external non-uniform magnetic field was turned on, there were two lines

235 11.1. WHAT IS THE ELECTRON SPIN? 33 shown on the right due to the force proportional to db. However, this was dz inconsistent with their expectations in two ways. 1. There was no line at the center corresponding to m l = 0.. There should be an odd number of lines; namely l + 1. But, they observed two lines. Figure 11.: Results of Stern-Gerlach Experiment with and without a nonuniform external magnetic field (Source: Stern and Gerlach s original paper Gerlach and Stern, 19, p.350) Later in 197, T. E. Phipps and J. B. Taylor reproduced this effect using hydrogen atoms in the ground state (l = 0 and m l = 0). This was a convincing piece of evidence that there was something other than the orbital angular momentum because l = 0 implies the orbital angular momentum is Fine Structure of the Hydrogen Spectrum: Spin- Orbit Interaction Generally speaking, the term fine structure refers to the splitting of spectral lines due to first order relativistic effects. These relativistic effects add three