Lecture notes for Atomic and Molecular Physics, FYSC11, HT Joachim Schnadt

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1 Lecture notes for Atomic and Molecular Physics, FYSC11, HT 015 Joachim Schnadt August 31, 016

2 Chapter 1 Before we really start 1.1 What have you done previously? Already in FYSA1 you have seen nearly all the quantum mechanical concepts that we are going to talk about. These are: Particle-wave duality, de Broglie wavelengths: λ = h/p Heisenberg s uncertainty relationship for position and momentum: x p h/ Energy quantisation, for example in the radiation of a black body with radiation density S ν, S νdν dω da = hv3 c dν dω da e hν/kt 1 Wave functions: Ψ(x) or Ψ(x, t) (cf. Fig. 1.1) (NB: I often write the position vector as x with components x 1 = x, x = y, and x 3 = z, although I sometimes also might write it as r.) Wave functions are interpreted as the probability amplitude of finding the particle described by Ψ at a position x and at time t. Thus, the corresponding probability is Ψ(x, t), and it is normalised to 1: Ψ(x, t) d 3 x = 1 Orthogonality of wave functions: ϕ a and ϕ b are said to be orthogonal if ϕ a(x)ϕ b (x)d 3 x = δ ab

3 Quantum mechanical operators and measurements, eigenvalue equations: Â = Ψ (x)âψ(x)d3 x, and ÂΨ(x) = aψ(x), i.e., Â = a The superposition principle: if ϕ a and ϕ b are wave functions, then also Ψ(x, t) = ϕ a (x, t) + ϕ b (x, t) is a wave function Complete mutually orthogonal sets Particle in a potential well Compatible and non-compatible variables Commutators [A, B] The canonical commutation relations: [x i, x j ] = [p i, p j ] = 0 and [x i, p j ] = i hδ ij The parity operator P : P Ψ p (x, t) = ±Ψ p ( x, t) The one-dimensional oscillator and ladder operators â,â Vibrational spectra During the course we will re-encounter many of these concepts, although they sometimes might look somewhat different from what you are used to. 1. The Dirac notation Table 1.1 shows a couple of quantum mechanical concepts both in the way you have written them previously and in the notation introduced by Paul Dirac ( ), one of the founders of quantum mechanics. The notation is called Dirac notation or bra-ket notation. The main difference between wave function and Dirac notation, which indeed also is a conceptual difference, is that the wave function notation always specifies coordinates, while the Dirac notation is coordinate-free. If coordinates have to be specified, e.g. when you want to describe the outcome of a position measurement, you specify that you want to use the position basis written as { x } (the 3

4 Figure 1.1: Spectral distribution of the radiation density S (λ) of a blackbody (Figure taken from teledetection/report1/node16.html). curly brackets indicate that the position basis is made up from a set of vectors x ). Depending on your problem you also might specify other bases, such as the momentum basis { p }. The primary advantage of the Dirac notation is that it makes life easier: as you see from Table 1.1 instead of having to write out integrals for e.g. the scalar product and expectation value of an operator, one just can resort to the time and space saving ϕ a ϕ b and Ψ A Ψ. Beyond this, there exists indeed a conceptual difference between the concept of a wave function and that of a quantum mechanical state: wave functions are generally tied to their interpretation as finding a system within a volume d 3 x at position x with probability Ψ(x), i.e. they are tied to the position basis of the quantum mechanical state space. The quantum mechanical state, in contrast, is independent of the choice of basis of the quantum mechanical vector state and thus more general (note, though, that one frequently also uses e.g. wave functions which depend on linear momentum rather than position). As an example we may take the hydrogen atom. In Dirac notation, the quantum mechanical state of a hydrogen atom in the state with quantum numbers n, l, m l the knowledge of which fully characterises the state of the hydrogen atom is written nlm l. In wave function notation, the probably 4

5 Concept Notation that you have Dirac notation used before Wave function Ψ(x) = x Ψ Quantum mechanical state Scalar product Ψ ϕ a (x, t)ϕ b (x, t)d 3 x = ϕ a ϕ b Normalisation Orthogonality ψ(x, t) d 3 x = 1 ϕ a (x, t)ϕ b (x, t)d 3 x = δ ab = ϕ ϕ = ϕ a ϕ b = δ ab Operator  A Eigenvalue equation ÂΨ(x) = aψ(x) = x A Ψ = a x Ψ A Ψ = a Ψ Expectation value  = Ψ (x, t)âψ(x, t)d3 x = A = Ψ A Ψ Table 1.1: The Table shows some quantum mechanical concepts in both conventional and Dirac notation. 5

6 of finding the same hydrogen atom within a volume element d 3 x at x is written ψ n,l,ml (x) d 3 x. The connection between both notations is provided by x nlm l = ψ n,l,ml (x). 6

7 Chapter Quantum mechanics The primary goals of this chapter are to introduce a particular angular momentum, the spin angular momentum, and to introduce the concepts of quantum mechanical states and operators and the tools needed to use them. Beyond that, we will also take an excursion in translations and time evolution to show that it is possible to derive the canonical commutator relationships, the replacement prescription of p by the spatial derivative in the wave function formulation of quantum mechanics, and the Schrödinger equation..1 The Stern-Gerlach experiment The Stern-Gerlach experiment was conceived in 191 by Otto Stern and Walther Gerlach and carried out by Stern and Gerlach in Frankfurt in 19. A beam of (neutral) Ag atoms is transmitted through an inhomogenous magnetic field and further onto a photographic plate (see Fig..1). Ag has altogether 47 electrons, out of which 46 form a spherical shell with no resulting angular momentum. The total angular momentum of the neutral Ag atom is thus given by the spin of the 47 electron, or, in other words, the neutral Ag atom behaves like a single electrons when it comes to its interaction with a magnetic field 1. 1 The existence of an angular momentum intrisic to the electron has to be postulated in non-relativistic quantum mechanics it is an experimental fact, which finds one of its clearest evidences in the Stern-Gerlach experiment, although the concept was unknown at the time of the experiment and introduced first a couple of years later. 7

8 Figure.1: Stern-Gerlach experiment (Figure taken from 8

9 The magnetic moment of the atom is proportional to the spin, µ S, (.1) and the force onto the atom is then, entirely as in classical physics, given by the gradient of the scalar product of the magnetic moment and the magnetic field, F z = z (µb) = µ B z z z. (.) Classically, one would expect that, assuming that the spin angular momentum corresponds to a spinning motion of the electron, all values of µ z are assumed between µ and µ. Hence, the photographic plate should exhibit the recording of a beam of atoms elongated along z. What instead is observed are two distinct components that correspond to µ z S z = ± h, (.3) where h is the Planck constant h = ev s. The Stern-Gerlach experiment thus shows that if the magnetic isotropy of space is removed by application of a magnetic field in a particular direction the component of the spin in this direction is quantised, i.e., it can assume only particular values. The Stern-Gerlach experiment represents one of the most direct proofs of quantisation and has had a profound influence on the development of quantum mechanics..1.1 Sequential Stern-Gerlach experiments Of course it is possible to apply magnetic fields in other directions of space than in the (arbitrarily chosen) z-direction. Then one can also let the silver atom beam transmit sequential arrangements of Stern-Gerlach setups, such as shown schematically in Fig... In each of these sequential Stern-Gerlach experiments the beam traverses first a magnetic field in the z-direction, which leads to a split-up of the atomic beam into a beam with electrons with S z = + h/ and one with S z = h/. Only the former is allowed to enter the second Stern-Gerlach setup with a magnetic field in either the z- or x-direction. As expected, if the beam is transmitted through a second magnetic field in the z-direction only one beam with S z = + h/ exits the setup. Also as expected, the application of a magnetic field in the x-direction leads to a splitting-up of the beam into two beams with S x = + h/ and S x = h/, respectively. If one now again stops one of the beams, namely that with S x = h/, and lets the other go through yet another Stern-Gerlach setup with 9

10 z x Figure.: Sequential Stern-Gerlach experiments (Figure taken from J. J. Sakurai, Modern Quantum Mechanics). a magnetic field in the z-direction, something strange happens: classically, one would expect only one beam to exit this setup with S z = + h/. The reason is that the first beam stopper already removed all electrons with S z = h/. What is observed instead is two beams: one with S z = + h/ and one with S z = h/! It is very important to see that this behaviour is entirely unclassical. It was only the advent of the quantum mechanical theory which could render this result plausible. Quantum mechanically, S x and S z cannot be determined simultaneously, S x and S y are non-compatible observables. The determination of S x in the second Stern-Gerlach experiment destroys all information obtained in the first Stern-Gerlach setup with a magnetic field in the z-direction. A very peculiar behaviour indeed! We have to leave the classical description behind, and instead go over to an alternative formulation, which makes use of the concept of states. It is said that the Ag atoms are in spin states, which are represented by vectors in a two-dimensional vector space. The vector space is two-dimensional because they are two possible outcomes of a spin direction measurement, spin-up and spin-down. We write the spin states as S z, + = 1 S x, + 1 S x, (.4) and S z, = 1 S x, S x,, (.5) 10

11 or by re-arrangement S x, + = 1 S z, S z, (.6) and S x, = 1 S z, S z,. (.7) This implies that all states which can be produced in the measurement of the z-component of the spin can be written as superpositions of states produced in the measurement of the spin s x-component and vice versa. The outcome of the sequential Stern-Gerlach experiments can now be explained: after the first experiment only the beam with atoms in the S z, + is allowed to enter the second setup. This beam is a superposition of the S x, + and S x, states, and thus after the second setup with a magnetic field in the x-direction two beams are found. Only the S x, + beam is allowed to go on but the atoms of this beam with S x, + are from (.6) seen to be either in the S z, + or S z, state! Now, of course there is a third component of the spin, the S y component. It is incorporated in the above system by complex addition: S y, + = 1 S z, + + i S z,, (.8) S y, = 1 S z, + i S z,. (.9). The vector space of quantum mechanics..1 Requirements In the preceding section we have seen that the quantum mechanical states can be described as vectors in a vector space. So far we had restricted ourselves to a two-dimensional space with basis { S z, +, S z, } (alternatively: { S x, +, S x, } or { S y, +, S y, }, but it is convention to use the { S z, +, S z, } basis). The concept can be generalised to include all necessary physical quantities. What is needed is the following: a complex vector space with a norm (i.e., a length of the vectors, which is going to represent the probability of a measurement), which requires a scalar (inner) product, 11

12 with infinite dimensions (the number of dimensions defines the number of possible measurement outcomes, and, for example, a position measurements thus requires an infinite number of dimensions), which is complete (mathematically: every Cauchy sequence converges, which in turn implies that each superposition of vectors from that space also converges to a vector in that space, even though the difference between the vectors might be arbitrarily small)... States: kets The requirements are exactly the requirements which are fulfilled by a Hilbert space, named after the German mathematician David Hilbert ( ). Thus the quantum mechanical state space is a Hilbert space, with physical states represented by ket vectors α. Also a superposition of states is a vector on the quantum mechanical state space: and, likewise, α + β = γ, c α = α c, where c is a complex number. Actually, c α represents the same state as α alone...3 Operators Now that we have defined the physical states, we also have to tell how to arrive at a measurement of a physical quantity such as S z above. First of all, a physical quantity which can be observed is called an observable. Observables are represented by operators on the vector space. Operators act on states: A α. As in mathematics, operators may have eigenvectors or eigenstates, as they are called in quantum mechanics. If the eigenstates of an observable A are denoted by a, a, a,..., then A a = a a, A a = a a,..., 1

13 where the a, a,... are numbers. The introduction of operators in this way makes it possible to properly describe what happens in the sequential Stern-Gerlach experiments above. The z-component of the spin angular momentum is an observable and thus represented by an operator S z. It has eigenstates S z, + and S z, with eigenvalues h/ and h/ (you already see that the eigenvalues will represent the outcomes of a measurement!): S z S z, + = h S z, +, (.10) S z S z, = h S z,. (.11) Similar definitions apply to the operators S x and S y. In the sequential Stern-Gerlach experiments the first Stern-Gerlach setup is a measurement of the spin in the z-direction, which corresponds to an application of S z to the states of the silver atom beam, which can be in either state S z, + or S z, ; the outcome of the measurement is thus either h/ or h/ and leaves the atoms in the corresponding states S z, ± : S z S z, ± = ± h S z, ±. (.1) For the second and third experiment in Fig.. the second setup traversed by the S z, + beam corresponds to a measurement of the spin in the x- direction. According to (.4), an application of S x to S z, + yields ± h/ and leaves the atoms in states S x, ±. In the third experiment of Fig.. only S x, + is allowed to traverse the final magnetic field in z-direction. This corresponds again to a measurement of S z. An application of the operator S z to (.6) yields immediately the outcome of the experiment: the z-component of the spin can either be h/ or h/, and after measurement the atoms are either in state S z, + or S z,. Other examples for observables are the position and momentum operators x and p: x x = x x, (.13) p p = p p. (.14) Observe the difference between operators x and p and their eigenvalues x and p, which are numbers! Finally, two operators A and B are equal if their application to an arbitrary state yields the same results: A = B if (and only if) A α = B α α. (.15) 13

14 ..4 Dimensionality of the vector space As already indicated above, the dimensionality of the quantum mechanical vector space of interest i determined by the number of possible outcomes of a measurement (we will typically restrict ourselves to considering only a relevant subspace of the total Hilbert space of quantum mechanics, which comprises dimensions for all possible outcomes of measurements of all physical quantities). In the Stern-Gerlach experiment one really has only two different outcomes of the measurement, which in a bit more colloquial language are called spin-up and spin-down. You might argue that you could measure the z-component of the spin in the x-, y-, and z-directions, with a total of six dimensions. However, equations (.4) to (.9) show that the outcome of a measurement of S x and S y always can be described in terms of the z-component of the spin (of course the use of the z-component is a convention); hence two dimensions are enough. Any spin state can be written as superposition of the two possible eigenstates of S z : Ψ = c + S z, + + c S z,. More generally, if an observables has eigenstates { a } (observe the use of curly brackets, which again indicates a set) then a general state can be written as Ψ = a c a a. (.16)..5 States: bras Now from Mathematics it is known that each vector space has a dual space, which also is a vector space a mirror image, so to say. Of course also the quantum mechanical Hilbert space has a dual space. The vectors of this dual space are written as a, and they are called bra s. The bra which corresponds to a ket c a is c a, where c is the complex conjugate of c. The scalar product (inner product) of a bra a and a ket b is written as It has the following property: a b. (.17) 14

15 a b = b a. (.18) This implies that a a is a real number (how do you see that?). The scalar product of a bra with its corresponding ket is positive or zero: a a 0. (.19) This property is important since the scalar product defines the norm and thus the length of a vector. Since the square of the length of a quantum mechanical vector a a is coupled to its probability interpretation (cf. Table 1.1) this value has to be zero or larger and cannot be negative. It is also practical (but not necessary) to require that a a = 1. (.0) For an non-normalised state a this length normalisation is achieved by a = 1 a a a. (.1) Finally, two states a and b are said to be orthogonal if..6 More on operators a b = b a = 0. Operators in bra space and Hermitian operators Since we have operators on ket space, we also need operators on bra space. In the Dirac notation operators are defined on kets from the left and on bras from the right side (the arrows merely indicate, into which direction the operators act in the equations): b = X a, b = a Y. The operator in dual (bra) space that corresponds to the operator X in ket space is X, the Hermitian adjoint of X. Thus X a does not correspond to a X, but X a corresponds to a X. 15

16 Operators have matrix representations (we will come back to this point soon); the Hermitian adjoint of an operator is then obtained by taking the complex conjugate of the transpose of the operator s matrix in a matrix representation: Hermitian adjoint: a ij a ji. (.) The operator is then said to be Hermitian (or self-adjoint) if the Hermitian adjoint is the same as the matrix, which corresponds to the operator. An example of Hermitian operators is given by the Pauli matrices: a ij ( a ji ( ) ( 0 i i 0 ) ( 0 i i 0 ) ( ) ( ) ) The eigenvalues of a Hermitian operator A are real, and, furthermore the eigenkets of A corresponding to different eigenvalues are orthogonal. This is easily shown: Assume that A is Hermitian. Then but also A a = a a a A = a a, a A = a A, (.3) since A is Hermitian. Multiply now the first equation in (.3) by a from the left and the second by a from the right, taking into account that A is Hermitian. This gives Taking the difference yields a A a = a a a, a A a = a a a. 0 = (a a ) a a. If a = a, then a a = 1, and thus a = a = a and thus a is real. Hence the eigenvalues of a Hermitian operator A are real. 16

17 If, in contrast, a a, then a a = a a 0 and hence a a = 0 or, in other words, a and a are orthogonal. The eigenvectors, which corresponds to different eigenvalues of the same Hermitian operator A are orthogonal! Operator multiplication and outer product Two operators might act onto a state one after the other (this corresponds to two subsequent measurements). In general, operators do not commute, i.e., if X and Y are two operators, then in general XY a = Y X a. You can rationalise this fact by considering a real problem. You are familiar with the Heisenberg position-momentum uncertainty principle, which states that x p h/. This implies that first measuring the momentum and then the position of a quantum mechanical particle does not necessarily give the same measurement result as first measuring the position and then the momentum - x and p do not commute. You should also realise that the order of operators (in ket space), which corresponds to the order of measurements, is from right to left. The rightmost operator acts first onto the state. We have already considered operators in bra space and stated that X a corresponds to a X. From mathematics we know that the Hermitian adjoint of an operator product is given by (XY ) = Y X. This also makes sense physically, since operators in bra space act to the left. One can build operators from a product of states. The outer product of two states a and b is defined as X = b a ; forming such an outer product results in the formation of an operator, which here is called X. The Hermitian adjoint of this operator is X = a b. Finally, we have already encountered matrix elements (cf. Table 1.1) of the form b X a (you will soon see why these constructs are called matrix elements). For these the following relationship is valid: 17

18 b X a = a X b. Of course, if X is Hermitian, i.e., if X is an observable, then b X a = a X b...7 Bases of state space In section..4 we already discussed briefly that the state space of interest is spanned by the eigenvectors of the observable of interest. For example, in the Stern-Gerlach experiment the spin state space is two-dimensional and spanned by S z, + and S z,, which are the eigenvectors of the observable S z. After an S z measurement the silver atom is either in state S z, + or in state S z,. S z, + cannot be described in terms of S z, and vice versa; S z, + and S z, are linearly independent, which implies that they are orthogonal: S z, + S z, = 0. S z, + and S z, form a basis of the spin state space. More generally, if we call the basis vector of the state space of interest a, a,..., and if we normalise them, a N = a a a a N = a a a..., then a, a,... form a complete orthonormal basis of state space. Thus an arbitrary normalised state α in state space can be written as a sum of basis vectors, which here are assumed to be already normalised (i.e., we skip the index N): α = a c a a. (.4) The coefficients c a are retrieved by multiplying equation (.4) with a from the left: α can thus also be written as a α = c a. α = a a α a = a a a α. This in turn implies the very important relationship that 18

19 a a a = 1 (.5) (.5) is called the closure or completeness relation (SV: fullständighetsrelationen). The relation expresses the completeness of the basis { a } (what does this imply for an arbitrary vector on the quantum mechanical state space?). The meaning of the coefficients c a = a α is that they give the probability amplitude for finding the system described by α to be in state a, i.e., c a is the corresponding probability. We also can assign a meaning to the outer product a a : it projects the state it acts on (it is an operator!) onto the state a. a a is said to be a projector: a a α = a a a a a α = a a δ a a c a = c a a...8 Matrix representations Of course one always can choose different bases for the state space of interest. For example, one basis for the spin space of the Stern-Gerlach experiment is the { S z, +, S z, } basis. Other choices would be, e.g., { S x, +, S x, } or { S y, +, S z, }. For a particular basis { a (n) }, with n = 1... N, operators and vectors can be written in matrix notation : a (1) a X (1) a (1) a X ()... a (1) a X (N) a () X a (1) a X ˆ= () X a ()... a () X a (N), (.6) a (N) a X (1) a (N) a X (N) α ˆ= a (1) α a () α... a (N) α, (.7) α ˆ= ( a (1) α, a () α,..., a (N) α ). (.8) Equation (.6) shows clearly why the construct b X a is called a matrix element. You should read the sign ˆ= as corresponds to. 19

20 Observe that in equation (.6) to (.8) above there is no equality between what is to the left and the right of the ˆ= sign this is a correspondence. If you want to write an equation that corresponds to equation (.6) then you should apply the closure relation two times: X = a a a a X a a. (.9) Exercise: Matrix notation vs Dirac notation 1. Write the operator β α in matrix notation using the basis { a (n) }.. Consider, again, the electronic spin. If we define + S z, + and S z, and knowing that S z ± = ± h ± write the matrix/vector notations for +,, S z, S + h +, and S h The position operator and the position basis of state space The action of the position operator is defined by x x = x x, (.30) where x is an operator (actually, it is an observable) and x a scalar the outcome of the position measurement represented by x. Typically, the possible values of x are infinitesimally closely spaced. Just think of a free particle anywhere in space. Since x is an observable its eigenvectors must span a subspace of the quantum mechanical state space. This is the position subspace. Since the eigenvalues are infinitesimally closely spaced, there must exist an indefinite number of them, and hence the vectors x cannot form a discrete basis, but the basis must be continuous and the number of dimensions of the position subspace is infinite. Therefore, the sum in the closure relation must be replaced by an integral: a a = 1 a An arbitrary state α can then be expanded as dx x x = 1. (.31) or, in three dimensions: α = dx x x α, (.3) 0

21 α = d 3 x x x α. (.33) If you compare this to the discrete expansion α = a a a α = a c a a, then you see that x α must be a probability amplitude, as are the coefficients c a. From before you know that probability amplitudes related to a position measurement are given by the wave functions, and we can identify Scalar products of position states For discrete bases we found for the basis states that ψ(x ) = x α. (.34) a a = δ a a, where δ a a is the Kronecker delta. For a continuous basis such as the position basis things are bit more complicated: x x = δ(x x ), (.35) which looks very similar to the expression for the discrete basis. However, δ(x x ) is the delta function, a generalised function with the following two properties: dx δ(x x 0 )f(x) = f(x 0 ), (.36) dx dδ(x) dx f(x) = The latter property further implies that dδ( x) dx dx δ(x) df(x) dx = dδ(x) dx, = df(x) dx. (.37) x=0 x dδ( x) = δ(x). dx If the delta-function has a vector argument, this implies that it represents the product of the delta functions of the vector components: δ(x x 0 ) = δ(x x 0 )δ(y y 0 )δ(z z 0 ). (.38) 1

22 ..10 Comparison of relations for a discrete basis and for a continuous basis The following list compares the expressions for a discrete basis and a continuous basis (here the position basis has been chosen, but it could be any continuous basis, e.g., the momentum basis): Orthonormality: a a = δ a a x x = δ(x x ), Closure relation: a a a = 1 dx x x = 1, Expansion of an arbitrary state: α = a a a α α = dx x x α, Normalisation: a a α = 1 dx x α = 1, Scalar product of arbitrary states: β α = a β a a α β α = dx β x x α, Matrix elements of a matrix A with eigenstates { a (i) } : a (j) A a (i) = a (i) δ ij x x x = x δ(x x ). Question: Can you think of any observables (other than the position), which typically would have a continuous spectrum of eigenvalues? Task: Using equation (.34), rewrite the closure relation for a discrete basis in wave function notation. Give a physical interpretation of the outcome..3 The spin vectors and spin operators.3.1 Explicit construction Already in section.1.1 we wrote down expressions for the states S z, ±, S x, ±, and S y, ±. While we at that time postulated what they have to look like, we will now construct them explicitly. The same we will do for the spin operators S x, S y, and S z. We start by again considering the third of the sequential Stern-Gerlach experiments in Figure.. Since the outcome of this experiment clearly shows that the silver atoms coming out of the Stern-Gerlach analyser with the magnetic field in the x-direction can be in both the S z, + and S z,

23 states, and since the probability for both of these possibilities is 1/ we can write or, equivalently, S z, + S x, + = S z, S x, + = 1, S z, + S x, + = S z, S x, + = 1. (.39) From equation (.39) we get the normalisation factors for expressing the states S x, + and S x, in terms of the states S z, ±. By convention we furthermore set the complex phase factor of the S z, + state in the expressions for S x, + to 1, while the complex phase factor of the S z, is yet to be determined: S x, + = 1 S z, e iδ 1 S z,. (.40) Since we know that, necessarily, S x, S x, + =0 (why?) we can easily obtain an expression for S x,. Suppose that S x, = A S z, + + B S z,, where A and B are complex constants (how do we know that this a valid statement?). Again, by convention, A is set to 1, and only B has to be determined from equation (.40). Since and thus S x, = 1 S z, + + B S z,, S x, S x, + = 1 S z, + S z, eiδ 1 S z, + S z, + B S z, S z, + + B e iδ 1 S z, S z,, and since furthermore S z, + S z, + = 1, S z, S z, = 1, and S z, S z, + = S z, + S z, = 0, we find B = e iδ 1 and thus S x, = 1 S z, + 1 e iδ 1 S z,. (.41) In the same way expressions are obtained for S y, ± : 3

24 S y, ± = 1 S z, + ± 1 e iδ S z,. (.4) δ 1 and δ can now be determined from the scalar product of S y, ± with S x, ±, which by symmetry can be determined from equation (.39): S y, ± S x, ± = 1. Putting in the expressions for S x, ± and S y, ±, equations (.41) and (.4), one obtains 1 1 ± e i(δ 1 δ ) 1 =, i.e., δ δ 1 = ±π/. We choose δ 1 = 0 and δ = π/ and thus arrive at S x, ± = 1 S z, + ± 1 S z,, (.43) S y, ± = 1 S z, + ± i S z,. (.44) This means that the expressions for the components of the spin in the x and y directions now have been fully expressed in terms of the z-component of the spin. We still have to write the observables S x, S y, and S z in terms of the z- component of the spin, since such a formulation will tell us exactly how these observables act on the different spin states. This is straightforwardly done by using the fact that any observable can be written in terms of a sum of projectors onto the eigenstates of the observable. This is seen by applying the closure relation (.5) two times to an observable A: A = a a a a A a a = a a a a, (.45) since a A a = a δ a a. The eigenvalues of S z (and likewise of S x and S y ) are known from experiment to be ± h/. Since the eigenstates of S z are S z, + and S z, one readily obtains S z = h { S z, + S z, + S z, S z, }. (.46) The expressions for S x and S y are obtained in the same way starting from S x,y = h { S x,y, + S x,y, + S x,y, S x,y, }, 4

25 and then putting in the expressions for S x, ± and S y, ±, equations (.43) and (.44), respectively. This gives S x = h { S z, + S z, + S z, S z, + }, (.47) S y = h { i S z, + S z, + i S z, S z, + }. (.48).3. The commutators and anticommutators of the spin operators Definition of commutators and anticommutators We will see shortly that the commutators of observables play an immensely important role in quantum mechanics. The commutator [A, B] of two operators A and B is defined as Likewise, the anticommutator {A, B} is defined as [A, B] AB BA. (.49) {A, B} AB + BA. (.50) Exercise It is now your task to show that where [S i, S j ] = iε ijk hs k, (.51) +1 if the ijk are an even permutation ε ijk 1 if the ijk are an odd permutation 0 otherwise (.5) is the so-called Levi-Civita symbol. Observe that the ijk are ordered and that they can assume the values 1,, and 3, which stand for x, y, and z. Show also that Useful relations are {S i, S j } = 1 h δ ij. (.53) 5

26 {A, B} = [A, B] + BA, {A, B} = {B, A}. Relation (.5) is fundamental in the theory of angular momentum: it is the defining relation for an angular momentum. Moreover, it expresses that the components of the angular momentum fulfil an uncertainty relationship. The square of the spin angular momentum Another very important spin operator is the square of the spin S. It is the sum of the squares of the spin components: S S x + S y + S z. The squaring of the operators S x, S y, and S z expresses that they are applied twice: S x α = S x S x α. For example, with α = S z, + one gets S x S z, + = S x S x S z, + = S x h/ S z, + = h /4 S z, +. It is important that you realise that S = S S is a scalar product on cartesian space. It is not a scalar product on the quantum mechanical state space! Since {S i, S j } = 1 h δ ij we find that S = 3 4 h 1, (.54) where 1 is the unit operator. Often the unit operator is not written out so that it is written S = 3 4 h, but of course since the left-hand side of the equation is an operator, also the right-hand side must be an operator. Another important property of S is that it commutes with every component of the spin: [ S, S i ] = 0. (.55) This is easily shown when recognising that [A, B + C] = [A, B] + [A, C] for any operators A, B, and C, and then using the definition of the commutator. 6

27 .4 Commuting (Compatible) observables.4.1 The case of non-degenerate eigenstates We have just encountered an example of two observables which commute, namely S and, e.g., S z. The compatability of the two observables has a very important implication for their eigenstates, and that is what we will show now. Let us assume that we have two observables A and B which commute, i.e. [A, B] = 0. Furthermore, let us assume that the eigenstates of A are non-degenerate, i.e., to each eigenstate of A there exists a unique eigenvalue a and vice versa. If we now label the eigenvalues of A as previously by a, a,... then a [A, B] a = (a a ) a B a = 0. If a a this must mean that a B a = 0, or, in other words, the matrix elements of B in the basis { a } are diagonal. This implies that the states { a } are eigenstates of B, as well, as is shown by applying the closure relation to B two times: B = a a a a B a a = a a a B a a, which is multiplied by a from the right to give B a = a B a a. (.56) This is a very important result that shows that if [A, B] = 0 then any eigenstate a of A is also an eigenstate of B with eigenvalue a B a..4. Degeneracy Two (or more) eigenstates of an observable A are said to be degenerate if the corresponding eigenvalues are the same: A a (i) = a a (i), with i larger than 1. For example the eigenstates of the energy operator, the Hamiltonian, for the hydrogen atom are degenerate, as is illustrated in Figure.3. The problem is now that just specifying the energy of a particular state is not enough for exactly specifying which state the hydrogen atom is in. Further labels are needed (and shown in the Figure). 7

28 E n=3 3s 3p 3d n= s p n=1 1s Figure.3: Energy levels of the non-relativistic hydrogen atom. n is the principal quantum number which labels the energy; s, p, and d are the labels for the angular momentum quantum number. The degeneracy of the levels of a particular energy is n -fold, i.e., there exist n states of the same energy..4.3 Commuting observables with degenerate eigenstates These additional labels are provided by the eigenvalues of further observables. It can be shown that if A is an observable with degenerate eigenstates and if it commutes with an observable B, i.e., [A, B] = 0, then there exist simultaneous eigenstates of A and B, very much as in the non-degenerate case. Then A a, b = a a, b, B a, b = b a, b. (.57) Note that the common eigenstates of A and B now were labelled by the corresponding eigenvalues for both A and B. Example As we will see later during the course, the orbital angular momentum squared L is an observables with eigenvalues h l(l + 1), where l = 1... n is the angular momentum quantum number. Each eigenstate of L is (l + 1)-fold 8

29 degenerate. The necessary additional label is provided by the eigenvalues of the z-component of the angular momentum L z, which are m l h, with m l = l, l + 1,..., +l. Complete sets of compatible observables In the general case one needs to find a maximum set of commuting observables [A, B] = [B, C] = [A, C] =... = 0 to fully describe the system. The eigenstates are then labelled by the eigenvalues of the different observables: A a, b, c,... = a a, b, c,..., B a, b, c,... = b a, b, c,..., C a, b, c,... = c a, b, c, The scalar product of two such states is then a, b, c,... a, b, c,... = δ a a δ b b δ c c..., and the closure relation is written... a, b, c,... a, b, c,... = 1. (.58) a b c What happens now if one measures the properties corresponding to two commuting observables A and B? If all eigenstates of A and B are nondegenerate, then the measurement of an eigenvalue of A determines the corresponding eigenvalue of B, as well ( A depicts a measurement of A): α A a, b B a, b A a, b. This is different if A and B are degenerate. Then the determination of A does not specify B, i.e., the state after measurement of A is still in a superposition of all possible eigenstates of B: α A n i=1 c (i) a a (j), b (i) a (j) B, b (k) a (j) A, b (k), where j and k are fix labels, but i is a running label. From the last measurement you also see that A and B can be determined simultaneously if (and only if) A and B commute. 9

30 .5 Translation, momentum, and time evolution In principle, we now have the main tools for being able to proceed studying orbital angular momentum, with the hydrogen atom as the primary example. It is quite interesting, though, that the fundamental equation, the Schrödinger equation, as well as important relationships such as the canonical commutator relations can be derived from quite simple principles. Also the standard prescription of replacing momentum p by the operation when acting on a (position-dependent) wave function can be derived quite easily. Therefore, we here make a little detour to explore these things. What we will need to do to derive the relationships for the components of x and p is to consider translations. We will start with infinitesimal translations and then consider finite ones. Since the order of position measurement and translation matters (i.e., whether the system first is translated and the position measured then, or vice versa), we will see that the corresponding operators do not commute. Translation and momentum are connected, in that momentum generates translation. To derive the Schrödinger equation we will have to consider the time evolution of quantum systems. This will leads us to the Schrödinger equation in a straightforward manner, but we will still have to postulate that it is the Hamiltonian which governs time evolution. Question: What is a translation?.5.1 Infinitesimal translations We will start out with infinitesimal translations, i.e., by translations by an infinitesimal distance dx. If the initial state of the system before the translation is characterised by a position state x, then after the infinitesimal translation it is T (dx ) x = x + dx. The operator T (dx ) x is called the infinitesimal translation operator. If T (dx ) x is applied to a general state α, then one finds the following: T (dx ) α = T (dx ) d 3 x x x α = = d 3 x x + dx x α = d 3 x x x dx α = (.59) = d 3 x x ψ α (x dx ). 30

31 In the first step we used the closure relation. The last step is due to a change of variable from x to x dx. This does not affect the limits of integration, which are ±, and neither does it affect the volume element. You see here, that the consideration of the translation implies that we have to consider the wave function of the translated state ψ α (x dx ). It seems that we do not know very much about the translation operator T (dx ). However, we can require a number of properties: We require that if α is normalised, then also T (dx ) α needs to be normalised (why?), i.e. 1 = α α = α T (dx ) T (dx ) α. This implies that T (dx ) is unitary: T (dx )T (dx ) = 1 T (dx ) = T 1 (dx ). It should be possible to combine subsequent translations to a single translation: T (dx ) T (dx ) = T (dx + dx ) (observe that the first translation carried out is that to the right by dx ) Reverse translations are the same as translations in the negative direction: T ( dx ) = T 1 (dx ) Zero translations amount to the unity operator (no change): These four requirements are fulfilled by lim dx 0 T (dx ) = 1. T (dx ) = 1 i K dx, (.60) where K stands for the triple of Hermitian operators (K x, K y, K z ). Note that the marks the scalar product in 3d space (not in quantum mechanical state space!). Task: Show that the infinitesimal translation operator (.60) fulfills all requirements! Question: What do you think is the physical meaning of K? 31

32 Relationship between x and K We now consider the order of position measurement and translation. If the translation is carried out first, one finds x T (dx ) x = (x + dx ) x + dx. (.61) If one, instead, first carries out the position measurement then one arrives at T (dx )x x = x x + dx. (.6) Obviously, and not very surprising, the operators x and T (dx ) do not commute the order matters. For the commutator we find [x, T (dx )] x = dx x + dx dx x. (.63) Since the relation is valid for an arbitrary state x, we can identify [x, T (dx )] = dx. (.64) Putting in the definition of T (dx ) (equation (.60)) leads to ix(k dx ) + i(k dx )x = dx. You can now choose dx to be oriented along a vector x j, i.e., dx dx j = dx ˆx j, where ˆx j is the unit vector along that direction. Multiplying the equation by a vector dx i = dx i ˆx i, which is chosen to be normal to ˆx j (i.e., dx j dx i = δ ij ), and then dividing by dx i dx j leads to The canonical commutator relations [x i, K j ] = iδ ij. (.65) We have already said that momentum generates translation. Since we have seen that K generates an infinitesimal translation, there must exist a linear relationship between both quantities: From a comparison to de Broglie s relation K = const p. (.66) π λ = p h and by identifying k with the wave vector, k = π λ, 3

33 we see that the constant in (.66) must be given by 1/ h. For the infinitesimal translation operator (.60) this results in and relation (.65) turns into T (dx ) = 1 i p dx h, (.67) [x i, p j ] = i hδ ij. (.68) This is one of the canonical commutator relations for position and momentum, that you have talked about in FYSA1! Consider once more how we have arrived at it: it follows from the fact that it matters in which order one executes an infinitesimal translation and a position measurement! Since (.68) is another formulation of the uncertainty relation for position and momentum ( x) ( p) h 4, this uncertainty relation is related to the fact that the order of translation and position measurement matters!! As you know there are two other canonical commutator relations, namely that for the components of the position and that for the components of the momentum: [x i, x j ] = 0, (.69) [p i, p j ] = 0. The former of the these two relations derives from the fact that all three components of the position vector can be determined simultaneously to an arbitrary degree of accuracy; the latter from the fact that the order of subsequent translations does not matter for the result of the total translation. We do not show this here formally, but you should remember all three canonical commutator relations (.68) and (.69)..5. Finite translations In the preceding section we considered infinitesimal translations. We now direct our interest at finite translations of length x. Such a finite translations is achieved by combining N infinitesimal translation of length dx = x /N and letting N go to infinity. Without any loss of generality, we also assume that x is parallel to ˆx. The action of the finite translation operator 33

34 T ( x ) is then given by T ( x ) x = lim N T ( x N = lim N ( 1 ip x x N h ) ( ) ( ) T x... T x x = N N (.70) ) N ( ) = exp i p x x. In this equation p x is an operator! To understand what the meaning of a function of an operator is, consider the Taylor expansion of the operator. In our case, if A is an operator, then exp A corresponds to exp A = 1 + A + A , i.e., it is the identity operator plus A applied once plus A applied twice, etc..5.3 Momentum operator in the position basis The aim of this section is to derive the standard prescription of replacing the momentum operator p by the operator i h when dealing with wave functions. We will do this here in one dimension, but the result is readily generalised to three dimensions. Consider the action of the finite translation operator onto a general state α : T ( x ) α = (1 i p x x ) α. (.71) h The left-hand side can be rewritten using the closure relation for the position basis: T ( x ) α = dx T ( x ) x x α = dx x + x x α = = dx x x x α = dx x ( x α x x x α ) = α dx x x x x α. In this derivation use has been made of x ψ α(x ) = ψ α x = ψ α(x x ) ψ α (x ) x. h (.7) Combining equations (.71) and (.7) provides an expression for the action of the operator p x onto a state α : ( p x α = dx x i h ) x x α. (.73) 34

35 Now we consider the scalar product of another position state x with p x x and the scalar product of an arbitrary state β with p x x : Using x x = δ(x x ) and x α = ψ α (x ) we find and β p x α = x p x α = i h x x α (.74) dx ψ β(x )( i h x )ψ α(x ). (.75) These two equations together correspond to the prescription p x i h x, (.76) for the action of the momentum operator onto wave functions. You have encountered this prescription already in FYSA1. Here this prescription has been derived from the translation properties of momentum. In three dimensions, the prescription is.5.4 Dynamics in quantum systems p i h. (.77) The term dynamics always refers to a time evolution. The time evolution of a quantum mechanical state can be put into an operator, very much in the same way as we put the translation of a position state in a translation operator. So we ask ourselves how a state α α t 0 at time t 0 will evolve in time. At later times that same state is denoted by α t t 0, where the specification of t 0 represents the boundary condition that α t 0 and α t t 0 are identical at t 0. We introduce the time evolution operator U(t, t 0 ) which takes the state α t 0 to α t t 0 : α t t 0 = U(t, t 0 ) α t 0. (.78) As for the infinitesimal translation operator we require the following: the conversation of probability, expressed by that both the state at t 0 and the time-evolved state have to be normalised: α t t 0 α t t 0 = α t 0 U (t, t 0 )U(t, t 0 ) α t0 = 1, from which follows U (t, t 0 )U(t, t 0 ) = 1, i.e., U(t, t 0 ) is unitary, 35

36 that subsequent time evolutions can be combined to a single time evolution (t > t 1 > t 0 ) (observe that U(t 1, t 0 ) acts prior to U(t, t 1 )): U(t, t 0 ) = U(t, t 1 )U(t 1, t 0 ), that infinitesimal time evolutions tend toward the unity operator: lim dt 0 U(t 0 + dt, t 0 ) = 1. A solution for the infinitesimal time evolution operator is readily found (remember the solution for the infinitesimal translation operator): U(t 0 + dt, t 0 ) = 1 i H dt h, (.79) where H is an Hermitian operator (which will turn out to be the Hamiltonian). Consider now two subsequent evolutions from t 0 to t and then further to t + dt: or in differential form: U(t + dt, t 0 ) = U(t + dt, t)u(t, t 0 ) = (1 i H dt h )U(t, t 0) U(t + dt, t 0 ) U(t, t 0 ) = i H h dtu(t, t 0), i h t U(t, t 0) = HU(t, t 0 ). (.80) This is the Schrödinger equation, although it might look oddly unfamiliar to you. The normal formula is obtained by multiplying by α t 0 : i h t U(t, t 0) α t 0 = HU(t, t 0 ) α t 0 i h t α t t 0 = H α t t 0. (.81) We have derived the Schrödinger equation by just considering the time evolution of a physical state. Unfortunately, we a priori do not know very much about the operator H. Actually, that H is the classical Hamiltonian needs to be postulated. But it makes sense: the Hamiltonian has the right dimension, namely [energy], which is required by the form of the infinitesimal time evolution operator (.79), since the dimension of h is [energy time]. Furthermore, also in classical mechanics the Hamiltonian is the generator of time evolution. So it makes sense to postulate that the same is true in quantum mechanics. 36

37 .6 Where do we stand now? At this point it seems appropriate to take a step back and look at what has been achieved. We have introduced the quantum mechanical state space and vectors and operators on that space. We have seen the connection to wave functions, and the connection between observables, which are Hermitian operators, and measurements has been discussed. Spin has been introduced, together with the spin states and spin operators. We have discussed compatible (commuting) observables and seen that the eigenvalues to the common eigenvectors of observables with degenerate spectra can be used to label the eigenvectors. Finally, we have considered translation, which led to the canonical commutator relations for position and momentum, and time evolution, which resulted in the time-dependent Schrödinger equation. In the following the aim is to discuss another angular momentum, namely orbital angular momentum. Orbital angular momentum can be cast into the same picture as spin angular momentum and follows the same set of rules as spin. Its handling in wave function notation is, however, a bit more complicated than spin angular momentum. Since it is advantageous to work with a real problem, we will start out with the hydrogen atom, which is a twobody system and therefore can be described in terms of a central potential. Orbital angular momentum plays a central role in the discussion. 37

38 Chapter 3 The hydrogen atom 3.1 The hydrogen atom in classical and semiclassical physics The complete description of the hydrogen atom is the most basic problem of Atomic Physics. Here, the starting point is the classical description of the hydrogen atom. Already in the 19 th century it became clear that the experimental observations did not agree with any classical solution, however, and therefore Niels Bohr introduced a new, phenomenological model, now known as the Bohr Model, which could explain some of the observations. The Bohr Model was one of the highlights of early quantum mechanics, but was soon replaced by a better, truly quantum mechanical model, which we will treat in detail Statement of the problem The hydrogen atom is the smallest of all atoms. It contains one electron and a proton, which are held together by Coulomb attraction. Hence, we are dealing with a two-body problem characterised by a force acting along a line between the proton and electron and a potential, which depends on the distance r e r p only: V = V ( r e r p ). Classical physics show that such a two-body problem can be reduced to a one-body central field problem. 38

39 e - r ep r e p + r p R O Figure 3.1: Illustration of the vectors involved in the description of the hydrogen atom. R is the vector to the centre-of-mass of the hydrogen atom The classical approach Figure 3.1 shows the notation that we are going to use in the following treatment. Newton s second law gives us the following: where r ep = r e r p. Likewise, m e r e = e V ( r e r p ) = dv dr ep r e r p r ep, m p r p = e V (r ep ) = dv dr ep r p r e r ep. The vectors r e and r p can be exchanged against the vectors R and r ep the centre-of-mass and between the proton and electron, respectively: to r e = me+mp r M e = m e r M e + mp r M e + mp r M p mp r M p = = m er e +m p r p M + m p M r ep = R + m p M r ep, (3.1) and, in the same way, r p = R m e M r ep. (3.) 39

40 M is the mass of the total system of the proton and electron. Adding the two equations gives that is m e r e + m p r p = m e R + m e m p r M ep + m p R m e m p r M ep = M R = = dv dr ep r ep r ep + dv dr ep r ep r ep = 0, M R = 0, (3.3) which means that the centre-of-mass moves without acceleration, quite as expected for a hydrogen atom in no outer field. Taking the difference of equations (3.1) and (3.) one sees that 1 (m { e r e m p r p ) = 1 me R + m e m p r M ep m p R + m e m p r } M ep = m e m p r ( ) M ep µ r ep = 1 dv r e r p dr ep r ep dv dr ep ˆr ep, where ˆr ep denotes the unit vector along r ep and µ is the reduced mass of the system. In other words, we have shown that the separation of proton and electron, r ep, obeys a Newtonian law µ r ep = dv dr ep ˆr ep, (3.4) and that the two-body problem can be reduced to a one-body problem. V is the central potential, which in the case of the hydrogen atom is the Coulomb potential e V =. (3.5) 4πε 0 r ep Since the force is purely radial, i.e. F (r ep ) = f(r ep )ˆr ep = dv dr ep ˆr ep, we see that f(r ep ) = dv dr ep. (3.6) Orbital angular momentum The torque r ep F is zero. This means that the angular momentum l = r ep p is a constant of the motion. It follows that the motion is strictly planar: r ep l = 0, and we may choose a coordinate system such that x = r ep cos ϕ, y = r ep sin ϕ. (3.7) 40

41 Using these cylindrical coordinates one finds for the length of the angular momentum l = µr ep ϕ, (3.8) which, like l, is a constant of the motion. It turns out that also the z- component of the angular momentum is a constant of the motion. l z = µ (xẏ yẋ) (3.9) Remark: Here we find both parallels and differences to the quantum mechanical treatment, as we will see later during the course. Classically, l, l, and l z are constants of the motion. Quantum mechanically, in contrast, l is no constant of the motion, while both l and l z are. The reason is that quantum mechanics does not allow the full determination of the angular momentum. Task: Show that the torque r ep F vanishes. Show also that the angular momentum l = r ep p is a constant of the motion. Show that the motion of the electron around the nucleus (or rather the motion of the nucleus and electron around the centre-of-mass) is planar. Finally, show also that l and l z are constants of the motion. Phyiscally, why must l z be a constant of the motion? Centrifugal barrier The goal of a classical treatment would be to obtain the (classical) trajectory of the electron around the nucleus r ep (t). The starting point for finding r ep (t) would be the energy E = E kin + V = 1 µ (ẍ + ÿ ) + V (r ep ) = 1 µ ( r ep + r ep ϕ ) + V (r ep ) = = 1 µ r ep + l µr ep + V (r ep ) = 1 µ r ep + V eff (r ep ), (3.10) 41

42 l µ r ep Energy V eff - A / r ep r ep Figure 3.: Illustration of the shape of the effective potential. l where V eff (r ep ) contains the centrifugal barrier. For a 1/r-potential the µr ep potential curves are drawn in Figure 3.. Remark: We will also encounter the centrifugal barrier in the full quantum mechanical treatment of the hydrogen atom. Remark: The energy, which we consider above, is not a constant of the motion, i.e., we have neglected an important contribution. Which one? If we had the full energy, we could, in principle, determine the classical trajectory by calculating the time dependence t(r ep ) and then inverting it to give r ep (t). Even in the classical theory this becomes very demanding, due to the occurring integrals. One gets quite a bit further by instead explicitly considering the forces which act on the electron and neglecting radiation. With the potential given by Eq. (3.5), the relevant forces are the Coulomb force and the balancing centrifugal force: e 4πε 0 r ep = µv r ep = µω r ep, 4

43 where ω is the angular frequency ω = v/r ep. Solving for ω gives ω = e /4πε 0 µr 3 ep = 1 T, (3.11) with T the time period for one revolution. This corresponds exactly to Kepler s third law, with the gravitational potential replaced by a Coulomb potential: Kepler s third law states that the orbital period T is proportional to the cube a 3 (= r 3 ep) of the semi-major axis of its orbit. From the balance of forces we also find an expression for the energy: E = 1 µv + V (r ep ) = 1 µv e /4πε 0 r ep = e /4πε 0 r ep. (3.1) This is this far classical physics takes us under the neglect of radiation. In order to get a step further we now need to introduce the quantum hypothesis of Planck. It was Niels Bohr in Copenhagen ( ), who did this first in his semi-classical model of the atom. If you are interested you can find his original work on the internet (N. Bohr, Philosoph. Mag. Ser. 6 (1913) 6, 1, accessible from db=all) The Bohr Model (1913) In contradiction to classical theory, Bohr assumed that there are only certain allowed orbits for the electron, and that radiation due to electron acceleration is not allowed. The only way radiation can be emitted or absorbed is by electrons passing from one orbit to another. The lost or gained energy is assumed to be a multiple of Planck s energy quantum hω = hν: E = k hω, i.e., ω = E k h. (3.13) Without any loss of generality we here assume that E > 0. From Eq. (3.1) an expression for the difference of the energies of an electron in different orbits is found ( E = e /4πε 0 1 r ep 1 r ep ), i.e., ( ) ω = 1 e /4πε 0 1 k h r ep 1 e /4πε 0 r ep k h 43 r ep. r ep

44 Here it has been assumed that r ep r ep, which is true only for two orbits with a large distance of the electron from the nucleus. This is no further limitation, since the physics for large orbits carry over to smaller orbits. We can now equate the classical result in Eq. (3.11) with the present value: ω = e /4πε 0 = (e /4πε 0) ( r ep ) µr 3 ep 4k h, i.e., r ep ( r ep ) = 4 e /4πε 0 h m e m e µ k r ep = 4a 0 m e µ k r ep, = r ep = a 0 r ep me µ k. a 0 is the Bohr radius (a 0 = 0.59 Å= m). Let k = 1 and assume that there exists a quantum number n (a quantum number is always an integer number) so that r ep = an x m x e µ, r ep = a(n + 1) x m x e µ. This ansatz comes about from requiring that the orbit radius should scale with a label (the quantum number n) and that it needs to have the dimension of a length therefore the length parameter a. For n 1 and using a Taylor expansion of (n + 1) x one finds { ( ) x a me µ r ep = (a(n + 1) x an x ) m x e µ n x + x ( m e µ ) x 1 = ax(n + 1) ( ) x 1 x 1 m e µ (n + 1) x 1 n x ( m e µ = a 0 r me = x ep a µ 0 an = a = n1 x a0 x axn ( ) x 1 x 1 m e µ ( ) x me 4 µ ( ) 1 x me. µ ) x } = By construction, a should be independent of n, and this implies x = and a = a 0, which gives r ep = a 0 n m e µ, (3.14) 44 =

45 where n, as said before, is an integer number. Approximating m e µ (or defining a 0 using the reduced mass of the hydrogen atom rather than the electron rest mass) one gets the more commonly seen For the energy one finds r ep = a 0 n. (3.15) E = e /4πε 0 a 0 1 n µ m e m e µ : E = e /4πε 0 a 0 1 n. (3.16) This result for the energy of the hydrogen atom turns out to be correct even in a proper quantum mechanical treatment! The result applies, approximately, also for atoms composed of a heavier nuclei and an electron. The Rydberg constant and Rydberg formula One can express the energy in wave numbers instead of the here used energy. Wave numbers ( ν = 1/λ = ν/c = E/hc) are a commonly used quantity in some kinds of spectroscopy. In wave numbers the difference of energies becomes: ν = e /4πε 0 a 0 hc µ m e ( 1 n 1 n ) R µ m e ( 1 n 1 n ) R ( 1 n 1 n ). (3.17) The Rydberg constant of indefinite mass is R = m 1. The formula is called Rydberg formula. Good to remember: With n = 1 and n = you find hcr = 13.6 ev, which is the ground state ionisation energy for an atom with one electron and an indefinitely heavy nucleus, but of course the value also applies to the hydrogen atom due to the small difference between m e and µ. For real nuclei with atomic number Z and mass m z the Rydberg constant becomes ( µ m Z R Z = R = R R 1 m ) e. (3.18) m e m Z + m e m Z In addition, we need to take into account that real nuclei have a charge +Ze, so that the Coulomb interaction with the electron becomes Ze /4πε 0 r ep. 45

46 The energy of a hydrogen-like atom (i.e., nucleus plus one electron) is then (remember that a 0 contains the factor e /4πϵ 0!) Standing waves E = e /4πε 0 Z a 0 n = hcr Z Z n. (3.19) The derivation of the distance between electron and nucleus in the Bohr model is more straightforward if one applies the wave/matter duality (which was not known to Bohr at the time he developed his model). Any stationary state of the electron must be described by a standing wave. This condition can be written as πr ep = nλ electron, (3.0) with n an integer number. From de Broglies wave/matter duality we find the wavelength associated with the electron as which gives for the speed of the electron λ electron = h p = h µv, (3.1) v = n h. πµr ep From the balance of centrifugal and central forces we have and hence r ep = e /4πε 0 µv = e 4πε 0 (πµr ep ) µn h, r ep = n h 1 1 e /4πε 0 µ 4π = a 0n m e µ, which is the same result as we found before in Equation (3.14) Experimental observations One of the primary driving forces for the introduction of new atomic models in the 19th and 0th century and among these models in particular the Bohr Model was the observation that atoms emit and absorb light only at certain, discrete energies. This was first shown by Gustav Kirchhoff and Robert Bunsen in 1859, and they probably used setups similar to those shown in Figure 3.3 and 3.4 (well, without the computer and diode). Also the 46

47 Figure 3.3: Typical optical emission setup. The source is hot, so that the source atoms are heated up and emit light (Figure from Demtröder, Atoms, Molecules and Photons, Springer, Berlin, Heidelberg, 006). Figure 3.4: Typical optical absorption setup. The light source needs to be continuous (Figure from Demtröder, Atoms, Molecules and Photons, Springer, Berlin, Heidelberg, 006). emission series of the hydrogen atom, for which an empirical formula was developed by Johann Jakob Balmer in the 1880s, will have been measured by similar setups. Both the emission and absorption series of the hydrogen atom are shown in Figure 3.5. Balmer s formula was later generalised by Janne Rydberg to formula (3.17), which also describes the other emission and absorption series of the hydrogen atom. The Balmer series corresponds to transitions to/from the Bohr orbit with label n =. Its main part is found in the visible regime. Other series are the Lyman (from/to n = 1), Paschen (from/to n = 3), Brackett (from/to n = 4), and Pfund series (from/to n = 5). The wavelength, energy, and wavenumber regimes of these series are specified in Table

48 Figure 3.5: The Balmer series in emission and absorption (Figure from 48

49 Energy (ev) Wavelength (nm) Wavenumber (cm 1 ) Lyman (n=1) n = n = Balmer (n=) n = n = Paschen (n=3) n = n = Table 3.1: Energy, wavelength, and wavenumber regimes of the Lyman, Balmer, and Paschen series Moseley s Law In the 1910s Henry Gwyn Jeffreys Moseley measured the x-ray spectra of many elements by bombarding the materials by high-voltage electrons. This leads to the excitation of inner electrons. The excited atoms will eventually decay under the emission of x-rays with characteristic energies. Moseley found a relationship between the frequency of the emitted x-rays and the atomic number: ν Z. (3.) This relationship is today called Moseley s Law and is in qualitative agreement with Rydbergs formula { } Z ν = cr n Z. n The formula and measured energies do not comply too well, since the formula is formulated for atoms with one electron. The agreement can be improved by the introduction of shell-dependent screening factors σ n : { (Z σn ) ν = cr (Z σ n ) }. n n 49

50 Figure 3.6: Moseley s recording of the frequencies of the characteristic x-ray emitted by atoms with atomic number Z (from Phil. Mag. 7, 703 (1914)). 50