Quantum Theory PH3210

Transcription

1 Quantum Theory PH3210 Dr Stephen West and Dr Andrew Ho Department of Physics, Royal Holloway, University of London, Egham, Surrey, TW20 0EX. 1

2 Contents 1 Books and Further reading Books Other Resources Introduction to Wavefunctions Schrödinger formulation ψ(x, t) and Probability Energy eigenstates and superposition Practice Exercises Operators and Observables Hermitian Operators Practice Exercises Dirac (Bra-Ket) notation Recovering Wavefunctions Manipulation with the momentum Operator Rules for Amplitudes Conserved Quantities Complete sets of Quantum Numbers The Uncertainty Principle Simple Harmonic Oscillator (SHO) Operator Method The Periodic Potential 54 2

3 7.1 Bloch Waves Angular Momentum Angular Momentum Eigenvalues and Eigenfunctions Angular Momentum: A better method Meaning of Angular Momentum Operator Results A close look at Spin Stern-Gerlach experiment Addition of Angular Momentum Approximate Methods I Time-independent perturbation theory Ex. 1: perturbed particle in a box Ex. 2: Quadratic Stark effect Approximate Methods II Degenerate perturbation theory (time independent) Linear Stark effect Approximate Methods III Effect of external magnetic field on the Hydrogen-like atom Orbital angular momentum Spin angular momentum Relativistic effect: spin-orbit coupling Strong field Zeeman effect Weak field Zeeman effect Approximate Methods IV 85 3

4 12.1 Variational Method Variational principle Ex.1: 1D simple harmonic oscillator Ex.2: Ground state energy of Helium Approximate Methods V Time-dependent Perturbation Theory Application to a special case: sinusoidal or constant in time perturbation Revision on Hydrogen-like atom Hydrogen-like atoms Angular (θ, φ) equation Radial (r) equation

5 1 Books and Further reading 1.1 Books Bransden and Joachain: Quantum Mechanics, Longman, 2nd edition ( BRA) C Cohen-Tannoudji, B Diu, F Laloe, Quantum Mechanics, vols 1 and 2, John Wiley ( COH) S Gasiorowicz, Quantum Mechanics, John Wiley, ( Gas) F Mandl, Quantum Mechanics, John Wiley, ( Man) Many other good books Other Resources Advanced but excellent notes/book can be found at This is the main book for the course. 5

6 2 Introduction to Wavefunctions 2.1 Schrödinger formulation De Broglie postulated that we associate a wave with every particle. λ = h p or λ = p where = h/2π and λ = λ/2π. Schrödinger introduced the wavefunction Wavefunction : ψ(x, t) and took ψ(x, t) to satisfy partial differential equation ψ(x, t) Ĥψ(x, t) = i t Schrödinger equation Ĥ is the Hamiltonian - partial differential operator for the total energy For a particle moving in 1D Ĥ = ˆp2 2m + V (x) where m is the mass of the particle and V is the potential. In QM the momentum operator ˆp = i x 1DSE then reads 2 2 ψ + V (x)ψ = i ψ 2m x2 t 6

7 Solution using separation of variables ψ(x, t) = φ(x)t (t) so divide both sides by ψ = φt ] [ ] [ 2 2 φ(x) T (t) + V (x)φ(x) T (t) = i φ(x) 2m x 2 t 1 φ [ 2 2 φ(x) + V (x)φ(x) 2m x 2 ] = i T T (t) t LHS is f(x) and RHS is a g(t). Only way they can be equal is if both equal to a constant (call it E). We get two equations 2 2m 2 φ + V (x)φ = Eφ (2.1) x2 T (t) i t = ET (2.2) Eq.(2.2) easy to solve T (t) = e iet/ constant. Eq.(2.1) is the Time Independent Schrödinger Equation or TISE, can be hard to solve, depends on V. Choose easy example - free particle, V (x) = 0. TISE becomes with solutions 2 2 φ 2m x = Eφ 2 φ e ikx, e ikx where 2 k 2 2m = E 7

8 Putting it all together ψ(x, t) e i(kx Et/ ) Wave moving in +ve x direction ψ(x, t) e i(kx+et/ ) Wave moving in ve x direction So we find the angular frequency ω = E/ E = ω = hν where ν is frequency and we see that the solutions automatically incorporate Einstein relation. Also for classical free particle E = p 2 /2m, so we get E = 2 k 2 2m = p2 2m p = k = 2π λ = h λ automatically incorporates De Broglie matter wave relation. But what does ψ(x, t) mean? ψ(x, t) cannot be a physical wave like an oscillating string, or EM wave of Maxwell theory TDSE is complex eqn, solutions are inherently complex (real and imag parts of ψ do not separately solve eqn) Something that is complex cannot be directly measured. So what is this wavefunction? 2.2 ψ(x, t) and Probability To understand ψ(x, t) must introduce Probablility Density: P (x, t) Basic idea is that can no longer be certain of exact position of particle 8

9 Only that Probability of measuring it between x and x + δx is Prob(in x to x + δx) = P (x, t)δx and a postulate of QM is that P (x, t) = ψ(x, t) ψ(x, t) = ψ(x, t) 2 We can use this to find averages or expectation values x xp (x, t)dx = x ψ(x, t) 2 dx similarly x 2 x 2 P (x, t)dx = x 2 ψ(x, t) 2 dx Total probability of finding particle anywhere must be 1, so 1 = P (x, t)dx = ψ(x, t) 2 dx which the normalisation condition on ψ(x, t). This must be imposed on ψ(x, t) as SE is linear in ψ. If ψ is a solution then so is ψ constant Constant is then fixed by the normalisation condition Complication for plane waves ψ(x, t) = Ge i (px Et) so ψ(x, t) 2 dx = G 2 dx Plane wave is not normalisable! 9

10 Reason: Plane waves are not physically realisable. They exist for all time and are spread over all space More physical situation is a plane wave confined to some finite region, (e.g. apparatus), then experimental L L G 2 dx = 1 G = 1 2L Look at an old example, the infinite square well Solving the Schrödinger equation for a number of examples should be familiar from PH2210 Quick look at the infinite square well, with potential as depicted in Fig 2.1 V = V = V =0 0 a x I II Figure 2.1: Infinite Square Well III Since V v(t) solutions of form ψ(x, t) = φ(x)e ite/ where 2 2 φ + V (x)φ = Eφ 2m x2 10

11 In regions where V =, φ must be zero for finite E solution (as K.E 0) Inside well so 2 2 φ = Eφ as V=0 2m x 2 φ = Ae ikx + Be ikx with 2 k 2 2m = E Now have to match inside to outside solutions via boundary conditions φ must be continuous At x = 0, 0 = A + B At x = a, 0 = Ae ika + Be ika = 2iA sin ka Either A = 0 (reject this) or sin ka = 0, k = nπ a, n = 1, 2, 3... Substitute back in to expression for energy E = 2 π 2 2m a 2 n2, n = 1, 2, 3... Possible energies of particle are quantised Typical for particles bound in a potential well (atoms,...) Also have to normalise solution so we have 1 = φ n (x) = A sin nπx a φ n (x) 2 dx = A 2 a A = 0 2 a sin 2 nπx a dx = A2 a 2 11

12 Putting it all together the spacial part of the wavefunction reads φ n (x) = 2 nπx sin a a Note that φ n are orthogonal a sin nπx a φ n(x)φ m (x)dx = 2 a 0 { 1 : n = m = 0 : n m mπx sin a dx n =5 E = 25E 1 n =4 E = 16E 1 n =3 E =9E 1 n =2 E =4E 1 n =1 0 a E = 2 π 2 2ma 2 E 1 Figure 2.2: Wavefunctions for Infinite Square Well From Fig. 2.2 we see that the ground state, n = 1 has no nodes 12

13 First excited state, n = 2 has one node It should be clear that the physics of low n states is very different from the classical expectation Classically: consider a ball bouncing back and forward in the well just a uniform distribution. So P (x) = 1 a x = x 2 = a 0 a 0 xp (x) dx = 1 a x 2 P (x) dx = a2 3 a 0 x dx = a 2 But quantum x 1 = a 2 same x 2 1 = a2 3 a2 2π 2 where the subscript 1 means we are using the n = 1 state. different from classical As n find that x 2 classical result The probability distribution becomes more spread out This is an example of the general behaviour 2.3 Energy eigenstates and superposition Wavefunction of the form ψ n (x, t) = φ n (x) e ient/ is special, it is an Energy Eigenstate or Stationary State (same thing) Equivalently: Ĥψ n (x, t) = ] [ 2 2 φ(x) + V (x)φ(x) ψ 2m x 2 n (x, t) = i t ψ n(x, t) = E n ψ n (x, t) by TDSE 13

14 So ψ n (x, t) is an Eigenfunction of Ĥ with Eigenvalue E n Physically: Energy Eigenstate Particle is in state of definite energy (here E n ) If energy is measured, guaranteed result E n Superpositions is not the most general solution to TDSE. ψ n (x, t) = φ n (x)e ient/ Since TDSE is linear equation, general solution is sum of above ψ(x, t) = n a n φ n (x)e ient/ where a n are arbitrary complex coefficients. Example: Superposition of 2 energy eigenstates act with energy operator ψ(x, t) = a 1 φ 1 (x)e ie 1t/ + a 2 φ 2 (x)e ie 2t/ Ĥψ(x, t) = a 1 E 1 φ 1 (x)e ie 1t/ + a 2 E 2 φ 2 (x)e ie 2t/ const ψ(x, t) But ψ is a solution of the TDSE, just not an energy eigenstate as E 1 E 2 What is the physical significance of this? For ψ to be valid wavefunction it must be normalised 1 = = a 0 ψ 2 dx = a 0 dx [ a 1φ 1 e ie 1t/ + a 2φ 2 e ie 2t/ ] [ a 1 φ 1 e ie 1t/ + a 2 φ 2 e ie 2t/ ] dx [ a 1 2 φ a 2 2 φ a 1 a 2φ 1 φ 2 e i(e 2 E 1 )t/ + a 1a 2 φ 1 φ 2 e i(e 1 E 2 )t/ ] 14

15 but φ 1 and φ 2 are orthogonal and are normalised so find 1 = a a 2 2 looks like a sum of probabilities, thus: interpretation of superposition If measurement made of energy of particle with this wavefunction will get the following result: E 1 with amplitude a 1, probability a 1 2 E 2 with amplitude a 2, probability a 2 2 The expectation value of energy should be Ĥ = E 1 a E 2 a 2 2 we can check this by doing the calculation directly. Ĥ = Ĥ which is the definition of in state ψ a 0 dx ψ Ĥ ψ Ĥ = a 0 dx [ a 1φ 1 e ie 1t/ + a 2φ 2 e ie 2t/ ] [ a 1 E 1 φ 1 e ie 1t/ + a 2 E 2 φ 2 e ie 2t/ ] = E 1 a E 2 a 2 2 using orthogonality of φs. This is the result we expected. Ĥ Note that is time independent Not all expectation values are time independent for energy superpositions, e.g. ˆp = a 0 = 2i a dx ψ ˆp ψ = a 0 dx a ( dx ψ i ) 0 x ψ [ a 1 sin πx a eie 1t/ + a 2 sin 2πx a ] [ π πx eie 2t/ a 1 cos a a e ie 1t/ 2π 2πx + a 2 cos a a where we have used the spatial wavefunctions for the infinite square well as our example. ] e ie 2t/ 15

16 Using the following useful identities a 0 dx sin πx a cos πx a = 0 = a 0 dx sin 2πx a 2πx cos a and for m n a 0 dx sin mπx a nπx cos a = am π(m 2 n 2 ) [1 ( 1)m+n ] = m + n even 2am π(m 2 n 2 ) (2.3) we arrive at the result ˆp = i 8 3a [ a1 a 2e i(e 2 E 1 )t/ c.c. ] where c.c. is the complex conjugate expression. Taking a simplified case where the coefficients are both real (a 1, a 2 R) ˆp = 16 3a a 1a 2 sin (E 2 E 1 )t Momentum oscillates in time!!!! 2.4 Practice Exercises (1) Show that in the case of the infinite square well presented in the notes x 1 = a 2 x 2 = a2 1 3 a2 2π. 2 What are these values for the nth state? That is x n and x 2 n. Do these values get closer to the classical values as we increase n? Is this what we expect? (2) Using Eq. (2.3) show that ˆp = 16 3a a 1a 2 sin (E 2 E 1 )t for the wavefunction ψ(x, t) = a 1 φ 1 (x)e ie 1t/ + a 2 φ 2 (x)e ie 2t/ where φ n (x) = and prove Eq. (2.3)) 2 a nπx sin. (Try a 16

17 3 Operators and Observables Numerical values of classical quantities (e.g. Momentum and energy) become differential operators in QM x- momentum : ˆp = i x energy : Ĥ = ˆp2 2 + V (x) = 2m 2m 2 x 2 + V (x) This is a general principle of QM Every physical observable corresponds to an operator, ˆQ Measured numbers in experiments are related to operators via eigenvalue equations In general we have Q χ n (x) = q n χ n (x) Eigenfunction χ n (x) labelled by n Eigenvalue q n (x) labelled by n The set of all q n is called the spectrum Example of energy Ĥ φ n = E n φ n Eigenvalue E n, Eigenfunction φ n General principle of QM is Measurements of a physical observable always give a result which is one of the eigenvalues of corresponding operator ˆQ 17

18 3.1 Hermitian Operators We know that experiments only return real measurements, this means that the class of QM operators must have real eigenvalues This class of operator are called Hermitian operators So Every operator representing a physical observable must be a Hermitian operator Definition of Hermitian operator; for 1d QM an operator ˆQ is Hermitian if χ ˆQ ψ dx = ( ˆQ χ) ψ dx for any functions χ(x), ψ(x) which are normalisable and vanish at x = ± What kinds of operators are Hermitian? i) The Position operator x so since x is a real number χ x ψ dx = (x χ) ψ dx = (xχ) ψ dx x = x where the dagger is notation for Hermitian conjugate. ii) The potential energy V (x) χ V (x) ψ dx = so as long as coefficients in V (x) are real we have V (x) = V (x) (V (x) χ) ψ dx which is what we expect as a complex potential does not make sense. 18

19 iii) Momentum operator ( χ i ) x ψ dx = i χ ψ x dx we need to get the operator acting on χ, so integrate by parts RHS = i {[χ ψ] } χ ψ dx x Since x is real, so x is also real and χ so RHS = = x ( ) χ x ( i χ ) ψ dx x Thus ˆp is hermitian. Note that the i in the definition of ˆp is crucial iv) Kinetic energy operator 2 2 2m x 2 is hermitian. Proof by integration by parts twice. v) Hamiltonian operator Since H = T + V and both T and V are Hermitian, H = H also. Sums of Hermitian operators are Hermitian. Proof of the reality of eigenvalues Let ˆQχ n = q n χ n, thus but ˆQ is hermitian so χ m ˆQχ n dx = χ mq n χ n dx LHS = ( ˆQχ m ) χ n dx = (q m χ m ) χ n dx = q mχ mχ n dx Putting the LHS side and the RHS of Eq. (3.1) together 0 = (q n qm) χ mχ n dx 19

20 now chose m = n 0 = (q n qn) χ n 2 dx = (q n qn) q n = q n q n is real But there is more...choose n m and use reality of qs 0 = (q n q m ) χ mχ n dx Thus χ m and χ n are orthogonal functions. This should be familiar from our infinite square well example where the wavefunctions had the forms 2 πnx φ n = sin a a We know from direct calculation that But yet more.. φ nφ m dx = 0 For φ m s that have fourier series any function can be expanded as f(x) = n=1 a n φ n (x) This is true in general using eigenfunctions of Hermitian operators. Theorem: Let χ n (x) be the eigenfunction of any hermitian operator Then any normalisable function f(x) can be written as f(x) = a n χ n (x) n=1 20

21 We say that the χ n (x)s form a complete set of functions (or states). It is this expansion theorem that enables a connection between operators and probabilities Let s see how... Suppose at a given time (choose t = 0) we have a wavefunction ψ(x, 0) and we measure Q Can expand ψ(x, 0) in eigenfunctions of Q ψ(x, 0) = n a n χ n (x) finding a n s is simple: multiply Eq. (3.1) by χ m(x) and integrate RHS = n χ m a n χ n dx = n a n χ m χ n dx = n a n δ nm = a m so LHS = a m = χ mψ(x, 0) dx n χ mψ(x, 0) dx Now substitute expanded ψ(x, 0) in equation for ˆQ ˆQ ψ n ( ( ) = (ψ(x, 0)) ˆQ ψ(x, 0) dx = a n χ n (x)) ˆQ a m χ m (x) = n ( ) a n χ n (x) a m q m χ m (x) n m = a na m q m χ n χ m dx n,m }{{} δ mn m = n a n 2 q n Remember that a n 2 is the probability that result of a single measurement of Q gives q n. q n s are the possible results of single measurements of Q. 21

22 Example: Back to the infinite square well. We know Hφ n = E n φ n with 2 πnx φ n = sin E n = n2 π 2 2 a a 2ma 2 Note that these are not momentum eigenfunctions ˆpφ n = i φ n 2 x = i nπ a a cos πnx a const φ n Easy to find the momentum eigenfunctions though φ pn = 1 a e ipnx/ satisfies i φ pn x = p n φ pn Now write the energy eigenfunctions in terms of φ pn 2 1 ( φ n = e inπx /a e inπx /a ) a 2i = 1 i φ pn 1 2 i φ pn 2 where p n = nπ a. So if a particle is in energy eigenstate φ n and we measure its momentum we find +p n with probability = p n with probability= 1 i 2 1 i 2 2 = = Practice Exercises (1) Prove that the kinematic operator is Hermitian m x 2 22

23 4 Dirac (Bra-Ket) notation More powerful and more general than wavefunctions Wavefunctions can be linearly superposed with complex coefficients Mathematical structure is a complex linear vector space Basic object is state ket ψ which represents the state of the system A particular set of states are energy eigenstates Ĥ n = E n n We also have complex conjugates of wavefunctions and eigenstates. Represent these by state Bras ψ or n The overlap of wavefunctions dx φ(x) ψ(x) generalises to the unitary inner product for bras and kets inner product = φ ψ It is clear from this generalisation that φ ψ = ψ φ A Ket is normalised if ψ ψ = 1 and ψ and φ are orthogonal if ψ φ = φ ψ = 0 23

24 The complex vector space possessing the structure of an inner product, which describes the possible states of the system is called the Hilbert Space of the system. Sometimes denoted by H. Like all linear spaces H has a dimension determined by maximal number of linearly independent vectors c 1 u 1 + c 2 u c N u N = 0 satisfied only if c 1 = c 2 =... = c N = 0. A difference with vector spaces is that Dimension of H is typically E.g. we know particle in 1D infinite square well has number of energy eigenstates φ n (x) n which are linearly independent. However, later we will meet very important physical systems where H is finite dimensional (spin,...) and cannot be described by wavefunctions Existence of inner product allows us to construct complete sets of orthonormal kets (a basis for H) via Schmidt procedure. Aside on Schmidt procedure Given a set of α linearly independent, normalisable functions (call them ψ α ) we can always construct a new set of α mutually orthogonal functions (call them φ α ) via the Schmidt procedure. Start with the function ψ 1 we write the following series φ 1 = ψ 1 φ 2 = a 21 φ 1 + ψ 2 φ 3 = a 31 φ 1 + α 32 φ 2 + ψ 3. φ α = a α1 φ 1 + a α2 φ a α,α 1 φ α 1 + ψ α 24

25 We need to find the forms of the a coefficients. Starting with a 21, we multiply φ 2 by φ 1 and integrate over all space φ 1φ 2 dv = a 21 φ 1φ 1 dv + φ 1ψ 2 dv but by definition the φ states are orthogonal and as φ 1 = ψ 1 and the ψ are normalised we are left with 0 = a 21 + φ 1ψ 2 dv giving a 21 = φ 1ψ 2 dv. This leads to the full expression for φ 2 as [ φ 2 = φ 1ψ 2 dv ] φ 1 + ψ 2 which is easily shown to be orthogonal to φ 1 as desired. Next, we wish to find the form for φ 3. To do so we first multiply the expression for φ 3 above by φ 1 and integrate over all space to get 0 = a 31 φ 1φ 1 dv + a 32 φ 1φ 2 dv + φ 1ψ 3 dv. Again the φ are orthogonal and φ 1 is normalised (the other φs are not) and we find a 31 = φ 1ψ 3 dv. To find a 32 we follow a similar procedure but multiply by φ 2. In this case we have φ a 32 = 2 ψ 3 dv φ 2 φ 2 dv. We can now write [ φ 3 = φ 1ψ 3 dv ] φ φ 1 2 ψ 3 dv φ 2 φ 2 dv φ 2 + ψ 3 We repeat this procedure until we reach the final state φ α which has the form α 1 φ α = i=1 φ i ψ α dv φ i φ i dv + ψ α. 25

26 We can redo this exercise using Dirac notation, write the series out again φ 1 = ψ 1 φ 2 = a 21 φ 1 + ψ 2 φ 3 = a 31 φ 1 + a 32 φ 2 + ψ 3. φ α = a α1 φ 1 + a α2 φ ψ α Following the same procedure as above to find a 21 we find in Dirac notation φ 1 φ 2 = a 21 φ 1 φ 1 + φ 1 ψ 2. Due to orthogonality φ 1 φ 2 = 0 and φ 1 is already normalised we can rearrange and then Similarly we find The final state is then given by a 21 = φ 1 ψ 2. φ 2 = φ 1 φ 1 ψ 2 + ψ 2. φ 3 = φ 1 φ 1 ψ 3 φ 2 φ 2 ψ 3 φ 2 φ 2 + ψ 3. α 1 φ α = φ i φ i ψ α φ i φ i + ψ α i=1 To have an orthonormal set we need to normalise these states. This is simply done and the result for the normalised states U m as U m = φ m φm ψ m Now we have a complete set of orthonormal kets (a basis for H) with U m with U n U m = δ nm If we have such a basis then any state ψ of system can be expanded ψ = dim H n=1 c n U n. 26

27 The c n s determined by taking inner product of this equation with U m U m ψ = dim H n=1 c n U m U n = c m. Thus the expansion for ψ is ψ = dim H n=1 U n U n ψ The bras and kets are related by taking the adjoint (complex conjugate transpose for usual vectors) so ψ = dim H n=1 c n U n dim H ( ψ ) = c n U n n=1 ψ = dim H n=1 c n U n The normalisation implies 1 = ψ ψ = dim H n=1 c n U n dim H m=1 U m c m = n,m c nc m U n U m = n,m c nc m δ mn = n c n 2 In QM also have (hermitian) operators which represent observables Key principle of QM is that all operators are linear, i.e. if ψ = c 1 U 1 + c 2 U 2 then Ô ψ = c 1 (Ô U 1 ) + c 2 (Ô U 2 ) 27

28 One exception to this, the time-reversal operator ˆT which is anti-linear - this is an advanced (and interesting) topic... Operators also act on vector space of bras linearly (Ô can act left or right ) ψ Ô = (c 1 U 1 + c 2 U 2 )Ô = c 1( U 1 Ô) + c 2( U 2 Ô) Since bras and kets are exchanged by taking adjoint, useful to define action of adjoint on operators Adjoint Ô of operator Ô defined by ψ Ô φ φ Ô ψ Hermitian (or self-adjoint ) operators satisfy Ô = Ô For hermitian ops (only) ψ Ô φ φ Ô ψ The quantity φ Ô ψ is called the matrix element of Ô between states φ and ψ If U n is an orthonormal basis then U m Ô U n is a matrix element in the conventional sense. A surprisingly useful operator secretly appears in the expansion of general ψ ψ = Since this holds for any ket ψ we find dim H n=1 U n U n ψ 28

29 Î = dim H n=1 U n U n where Î is the identity operator which leaves all states unchanged by expansion theorem for φ. Î φ = dim H n=1 The matrix element of Î is (in some a basis) U p Î U q = dim H n=1 U n U n φ = φ U p U n U n U q = dim H Thus written explicitly as a matrix Î is represented as Î dimh } {{ } dimh The unit matrix. n=1 δ pn δ nq = δ pq Another important example is the matrix element of a product of operators U p  ˆB U q = U p ÂÎ ˆB U q = }{{} can always insert Î n U p  U n U n ˆB U q = n A pn B nq just usual matrix multiplication of matrices representing  and ˆB. Check that our earlier definition of Ô agrees with adjoint of matrix (Ô ) mn U m Ô U n = U n Ô U m = (Ô) nm = (Ômn) T 29

30 Summary so far...fundamental Postulates of QM I) States of a system are represented by normalised kets ψ or bras φ in a Hilbert space (which varies from system to system) II) Observables are represented by linear hermitian operators acting on kets and bras III) Such hermitian operators  are assumed to possess a complete set of orthonormal eigenstates, e.g. 1 with eigenvalue a 1 2 with eigenvalue a 2 3 with eigenvalue a 3... Complete set means that 1, 2,... forms a basis for the space, so any ψ can be expanded in eigenstates of  (and for any Â) ψ = n n n ψ IV) The fundamental probability postulate for measurement is 1) Possible results of measurement of A are eigenvalues of  only. 2) Prob(A = a n ) = n ψ 2 3) After measurement of  with result a n the state ψ is reduced to ( collapsed to ) ψafter = n coefficient in front of n is changed to 1 where n is an eigenket of â with  n = a n n. Subsequent measurements of same  on ψafter = n return an with probability =1. As long as no measurements of other operators ˆB are made at intermediate times...more later. (Note: if a n is a degenerate eigenvalue (i.e. if more than one linearly independent ket has same a n eigenvalues, the procedure of reduction is slightly more involved - advanced topic) V) In the absence of a measurement the time evolution of the ψ(t) describing the state of the system at time t changes smoothly in time according to the TDSE i ψ(t) t = Ĥ ψ(t) 30

31 Notes 1) The TDSE is a linear eqn, which is also deterministic, i.e. given the state at t = 0, ψ(0). The state at a later time t is uniquely determined as long as no measurements are performed. 2) Thus the probabilistic, non-deterministic aspects of QM are purely due to the collapse of the state upon measurement. 3) Ĥ is the Hamiltonian - the operator corresponding to the energy of the system 4) We can formally integrate the TDSE (Ĥ ψ = i ψ t ) from time = t 0 to t f ψ(t f ) = e iĥ(t f t 0 )/ ψ(t 0 ) where the exponential is defined by its power series, e.g. eô = 1 + Ô + Ô2 2! +... since Ĥ = Ĥ (hermitian) the operator U e iĥ(t f t 0 )/ is unitary, U U = e i Ĥ (t f t 0 )/ e i Ĥ(t f t 0 )/ = 1 so the time evolution is unitary evolution. 4.1 Recovering Wavefunctions How do we recover wavefunctions? Consider position operator ˆx. This has a continuous spectrum of eigenvalues ˆx x = x x The eigenkets x are normalised as x x = δ(x x ) δ(x x ) is the analogue of n m = δ nm in the discrete case. Some useful identities with delta functions are e ikx dk = 2πδ(x) 31

32 This should be used inside integral as In 3-d this becomes f(x)δ(x a)dx = f(a) e ik.x d 3 k = (2π) 3 δ 3 (x) The expansion of a normalised state ψ of the particle in terms of position eigenkets reads ψ = dx x x ψ which is the analogue of ψ = n n n ψ General rules of the Dirac formalism tell us the interpretation of x ψ x ψ = probability amplitude that a particle in state ψ is located at x I.e. x ψ is precisely what we previously called the wavefunction The description of sates by wavefunctions is called the x-representation (or coordinate representation) Schrödinger s wave mechanics is the form QM takes if the coordinates of a particle are all one cares about. (E.g. if no spin, no antiparticle creation,...) All aspects of wave mechanics can be derived from Dirac description, e.g. overlap ψ φ = ψ dx x x φ } {{ } inserting Î in x-rep = dx ψ x x φ = dx x ψ x φ = dx ψ (x) φ(x) 32

33 4.1.1 Manipulation with the momentum Operator Momentum operator ˆp acting on some eigenstate of momentum (look at one dimension example ˆp p = p p We also have that r p = 1 2π e ipr/ Can check the normalisation by using the fact that p p = δ(p p ) and p p = dr p r r p = dr 1 2π eipr/ e ip r/ = δ(p p ) Also we can rewrite the following r ˆp ψ = dp r p p ˆp ψ = dp r p p p ψ = 1 2π = i dr r p p ψ = i r r r ψ dre ipr/ p p ψ Therefore r ˆp ψ = i r ψ(r) Similarly r ˆp ψ(r) ψ = 2m 2m r 2 33

34 4.2 Rules for Amplitudes In the x-representation (i.e. usual wavefunctions) the amplitudes are found by solving the TDSE with appropriate boundary conditions Now we come to very general rules for probability amplitudes that can be derived from the Dirac formulation Define an event in an experiment to be a situation in which all the initial and final conditions of the experiment are completely specified. I.e. all positions, angular momenta,... if all particles specified. Rules Rule 1: When an event can occur several alternative ways the amplitude is the sum of the amplitudes for each way considered separately (so we get interference) Rule 2: The amplitude for each separate way an event can occur can be written as the product of the amplitude for part of the event occurring that way with the amplitude of the remaining part e.g. Amp(particle x y) y x z Amp(x y) = Amp(x z).amp(z y) or y x = y z z x Rule 3: If an experiment is performed which is capable in principle of determining which of the alternative ways is actually taken (so in fact not all final conditions are the same) Then total probability (not amplitude) is sum of probabilities for each alternative P tot = P 1 + P

35 interference is lost. NOTE: capable in principle doesn t mean that its necessary for a human (or other scientific being) to check that all final conditions are the same. (its enough for the state of one atom to be different whether we are aware or not) Useful to manipulate amplitudes even when we don t know (yet) exactly their value E.g. two slit interference Source, S Slit 1 Observing Screen x Slit 2 1 S = Amplitude for particle to go from S to slit 1 x 1 = Amplitude for particle to go from 1 to slit x similar for 2 S and x 2. So the total amplitude from source S to x Amp 1+2 = x 1 1 S + x 2 2 S where we multiply amplitudes along the route and add the different routes. This leads to Prob(S x) = Amp1+2 2 and because of the interference term this is not simply equal to x 1 1 S 2 + x 1 1 S 2. 35

36 5 Conserved Quantities In classical physics many conserved quantities (e.g. Energy, Momentum, Ang momentum) What about in QM? Consider expectation value of some operator Q which does not have any explicit time-dependence (e.g. x or i / x) Q ψ = ψ (x, t) Q ψ(x, t) Due to the fact that ψ is a function of t we will find that, in general, Q ψ is also a function of t... d Q ψ dt = { ψ t Q ψ + ψ Q ψ } t dx Use TDSE Hψ = i ψ t d Q ψ dt = = 1 i using the fact that H is hermitian. {( ) Hψ Q ψ + ψ Q i ( )} Hψ i { ψ HQ ψ + ψ QHψ} dx dx d Q ψ dt = i ψ (HQ QH) ψ dx = i (HQ QH) ψ q is a conserved quantity if d Q ψ dt = 0 no matter what state ψ the particle is in. Only can happen if (HQ QH)ψ [H, Q] ψ = 0 for any normalisable function, ψ. The object [H, Q] is the commutator of H and Q. Example 36

37 Simplest case is Q = 1. as [H, 1] = 0 for any operator A. d dt ψ ψ = i ψ [H, 1] ψ = 0 Namely, the probability is conserved independent of time, i.e. ψ ψ = 1 Aside on Probability Current ρ = ψ ψ is the probability density but TDSE says for 1D QM ρ t = ψ ψ ψ + ψ t t (5.1) i ψ t = 2 2m 2 ψ + V ψ Using this in Eq. (5.1) we have ρ = i ) ( )} {( 2 t 2m 2 ψ + V ψ ψ ψ 2 2m 2 ψ + V ψ = i 2m. (ψ ψ ψ ψ) Defining j = i 2m (ψ ψ ψ ψ) and we finally have the conservation equation ρ t +.j = 0 37

38 Return to basic conservation equation d dt ψ Q ψ = i ψ [H, Q] ψ So far looked at case of Q = 1 Now explore more generally, with H = p2 2m + V (x) i) Total energy is conserved because [H, H] = HH HH = 0 so d ψ H ψ = 0 dt ψ ii) Now consider momentum p [H, p] = [ ] [ ] p 2 p 2 2m + V (x), p = 2m, p + [V (x), p] = 1 2m (p2 p pp 2 ) +[V (x), p] = [V (x), p] }{{} =0 but [V (x), p] = [V (x), i ]. To work this out we must put arbitrary function on RHS and we x find So... [V (x), p] f(x) = i [V (x), p] = i V (x) x f(x) V (x) x Not Zero! Therefore [H, p] = i V (x) x and p t V (x) = x Note the similarity of this result to classical equation of motion. 38

39 This shows that d p /dt cannot be zero unless V (x)/ x = 0. But this means that V =Const and we find linear momentum conservation only whenh is independent of position (system is independent of position) iii) Now take Q = X A useful identity is [ ] [ ] p 2 p 2 [H, x] = 2m + V (x), x = 2m, x + [V (x), x] }{{} =0 [A 2, B] = A[A, B] + [A, B]A Applying this we have [H, x] = 1 {p[p, x] + [p, x]p} 2m Finally we need [p, x] ( [p, x]f(x) = i x x + xi ) { } x f(x) = i f(x) + x f x x x x f = i f(x) x This is a fundamental property of x and p operators. Hence, Thus [H, x] = 1 2m {p( i ) + ( i )p} = i p m x t = p m And position is not concerved unless in special case with p = 0. Note: the above equation is analogue to classical ẋ = p/m From both the above equation and p = V (x) we see that the classical equations of motion t x are obeyed on the average in QM - - EHRENFEST THEOREM More on commutators We have seen that [H, Q] = 0 implies d Q dt = 0 So far we have been working with expectation values 39

40 We now would like to ask, does [H, Q] = 0 have special consequences for individual measurements...answer is yes... We know eigenstates of h satisfy Hφ n = E n φ n Recall that this means that if ψ(x, t) = e ient/ φ n (x) }{{} not a superposition then when we measure energy we will get E n (with probability =1). Now suppose the same functions φ n are are also eigenstates of another operator Q Qφ n = q n φ n This means that if the particle is described by φ n then it also has a definite value of Q given byq n Prob(Q = q n ) = 1 What is the condition that this is possible? Act on Hφ n = E n φ n with Q QHφ n = E n Qφ n = E n q n φ n Alternatively act on Qφ n = q n φ n with H HQφ n = q n Hφ n = q n E n φ n This is the same result. Subtracting these two equations (HQ QH)φ n = 0 Since, by assumption, this is true for all φ n [H, Q] = 0 Thus 40

41 If [H, Q] = 0 it is possible for the particle to be in state of definite energy a stationary state in which the value of Q is also definite What happens if ψ(x, t) is not an eigenstate? E.g. a superposition of 2 eigenstates with so Let us compute Q This is just Q = = ψ(x, t) = a 1 φ 1 (x)e ie 1t/ + a 2 φ 2 (x)e ie 2t/ 1 = a a 2 2 Prob(E = E 1 ) = a 1 2 Prob(E = E 2 ) = a 2 2 ψ Qψ dx (a 1φ 1 e ie1t/ + a 2φ 2 e ) ( ie 1=2t/ a 1 φ 1 e ie1t/ + a 2 φ 2 e ) ie 1=2t/ = q 1 a q 2 a 2 2 Q = q 1 (Prob in state 1) + q 2 (Prob in state 2) So if we make a measurement of both H and Q we find (E 1, q 1 ) with prob = a 1 2 (E 2, q 2 ) with prob = a 2 2 Fundamental point is the following: If [H, Q] = 0 then can measure both quantities and simultaneously find precise values of both H and Q 41

42 V = V = V =0 0 a x I II Figure 5.1: Infinite Square Well III What happens if [H, Q] 0? We have already studied this when we looked at Q = P in the case of the infinite square well. Since [H, P ] = i V x 0 at x = 0 V x andat x = a V x In this case we saw that If particle has definite energy then ψ(x, t) = e ient/ φ n (x) ψ is a superposition of states with different momenta Equivalent: We cannot know both E and P for sure since [H, P ] 0 General result in QM 42

43 If A and B are physical (hermitian) ops. with [A, B] 0 then cannot in general simultaneously know values A and B. 5.1 Complete sets of Quantum Numbers Suppose we have two operators Q 1, Q 2 satisfying [H, Q 1 ] = 0, and [H, Q 2 ] = 0 then d Q 1 dt and d Q 2 dt and both observables are conserved. However if [Q 1, Q 2 ] 0 then will not be able to make a definite measurement of both If [Q 1, Q 2 ] = 0 then always possible to find simultaneous eigenstates of Q 1 and Q 2 and H. In this case the wavefunction ψ(x, t) = e ient/ φ n (x) which has Definite energy E n Definite Q 1 q 1 Definite Q 2 q 2 These are the quantum numbers of the state ψ. In this case [Q 1, Q 2 ] = 0, Q 1 and Q 2 are compatible. In the case [Q 1, Q 2 ] 0, Q 1 and Q 2 are incompatible and it is possible for us to have state ψ(x, t) = e iet/ φ(x) with either a Definite energy E n Definite Q 1 q 1 43

44 or Definite energy E n Definite Q 2 q 1 But not both E.g. Since [p, x] = i we can never have a state in which both p and x have definite values We say that E,... form a complete set of Quantum numbers if maximal number of simultaneous eigenvalues are specified 5.2 The Uncertainty Principle We now want to know what happens when we make measurements on some general superposition Consider, e.g. energy superposition ψ(x, t) = a 1 φ 1 e ie1t/ + a 2 φ 2 e ie 2t/ Suppose measurement of energy at t = t 0 gives result E 1 Means we know for sure that energy is E 1 But this implies that now (for t t 0 ) system must be in energy eigenstate Thus ψ(x, t > t 0 ) = φ 1 e ie 1t/ (1) Part of original ψ which has different energy to the one measured disappears (2) Remaining part has its coeff changed so that new wavefunction is correctly normalised Procedure is the so-called collapse of the wavefunction Applies when any quantity (not just energy) is measured Collapse of the wavefunction also know as Projection Postulate Radical proposals such as many worlds which tries to explain why we experience collapse. In any case important to stress 44

45 Every experiment ever performed leads to results consistent with the simple collapse postulate and so we don t need anything more fancy Let s see what happens in an example of a series of measurements (1) Start with infinite square well ψ(x, t) = φ n (x)e ient/, E n = 2 π 2 n 2 2ma 2, φ n = (2) measure energy: Result: E 1, so wavefunction collapses to ψ(x, t) a = φ 1 (x)e ie 1t/, 2 nπx sin a a (3) Now measure momentum. To work out possible results must rewrite ψ a as a sum of momentum eigenstates ψ a = e ie 1t/ 2 1 ( e iπx/a e iπx/a). a 2i This is a superposition of two momentum states one with π a π and. a Suppose we now measure momentum π /a Wavefunction collapses to ψ b = e ie 1t/ eiπx/a a (4) Now measure energy again. To work out possible results must rewrite ψ a as a sum of energy eigenstates sin πx a e iπx/a a = 1 a ( cos πx a πx ) + i sin a πx is already an energy eigenstate but we must expand cos a 1 cos πx ( ) a a = 2 nπx a n sin a a n=1 }{{} remember this is an energy eigenfunction in terms of a Fourier series 45

46 where Thus a n = 1 a a 0 2 nπx sin a a e iπx/a a = i a (φ 1 (x)) + So when we measure energy we get { nπx 0 if n = odd cos a dx = if n = even n even 2 2n π(n 2 1) 2 2n π(n 2 1) (φ n(x)) E 1 with prob. P (E 1 ) = 1 2 E 2 with prob. P (E 2 ) = 32 9π 2 E 4 with prob. P (E 4 ) = π 2.. E n with prob. P (E n ) =. 8n 2 (n 2 1) 2 π 2 Thus even though we started with an energy eigenstate, by making a measurement of momentum (an operator incompatible with H) we have left a particle in a state of very indefinite energy! These considerations lead Heisenberg to his famous uncertainty principle Uncertainty principle not separate axiom of QM it is a consequence of rules already stated. When there is uncertainty in the outcome of measurement useful to quantify: Define Uncertainty as ( q) 2 q 2 q 2 q = 0 if q takes a single value. Heisenberg s Uncertainty principle considers x and p (remember [x, p] = i ) and states x p /2 But what does this mean? It says that is we prepare a particle in a state whereby its location x is know to within x, then the uncertainty in its momentum is at least /2 x Example: 46

47 Suppose we have a Gaussian wavefunction clearly also so that φ(x) = x = 1 /2a2 a 1/2 e x2 π1/4 x φ(x) 2 dx = 0 x 2 = 1 x 2 e x2 /a 2 dx = a2 aπ 1/2 2 x = [ x 2 x 2] 1/2 = a 2 Now for the momentum p and p 2 = 2 a π using the useful integral p = i a π e x2 /2a 2 2 /2a2 e x2 x2 e x2 /2a 2 x e x2 /2a2 dx = 0 ( dx = 2 a e x2 /a x 2 2 π a 1 ) dx = 4 a 2 2a 2 Finally we get Putting this all together e x2 /a 2 = a π p = a 2 p x = a a = Exactly the saturation of Heisenbergs s uncertainty principle bound. In fact, a Gaussian wavefunction gives the least possible value for p x. Since p 0 interesting to ask what the momentum distribution is. 47

48 We know x-space wavefunction φ(x) = 1 /2a2 a 1/2 e x2 π1/4 In Dirac notation this is x φ - Amplitude for particle in state φ to be found at x. We want p φ - Amplitude for particle in state φ to be measured with momentum p. This is just the Fourier transform of φ(x) φ(p) = 1 2π e ipx/ φ(x) Factor in front of the integral is there so dp φ 2 = 1 Or in Dirac notation where p φ = p x x φ all values of x p x = x p = ( e ipx/ 2π ) = eipx/ 2π Integral for φ(p) can be done by completing the square Another Gaussian! φ(p) = 1 2π dx e 1 2a 2 {(x ipa 2 / ) 2 +p 2 a 4 / 2 } = a a2 /2 2 e p2 π1/2 48

49 6 Simple Harmonic Oscillator (SHO) Many systems to leading approximation are the SHO (or many weakly coupled SHOs), so this is an important case Potential energy is V (x) = 1 2 kx2 so TISE convenient to rescale the x variable by x = αy 2 d 2 φ 2m dx kx2 = Eφ 2 d 2 φ 2mα 2 dy α2 ky 2 = Eφ and insist that coefficients of K.E. and P.E. are the same 2 2mα 2 = 1 2 kα2 α 2 = mk Then we have k 1 m 2 ( ) d2 φ dy + 2 y2 φ = Eφ Now set k/m = ω the classical angular frequency, so let ɛ = ( E ω ) 2 i.e., ɛ is the energy in units of ω/2. Thus our TISE becomes d2 φ dy 2 + y2 φ = ɛφ It is easy to check that φ 0 = e y2 /2 49

50 is a solution (and is normalisable), let plug it in... So TISE reads φ 0 = ye y2 /2 Solution if ɛ = 1 The energy eigenvalue. We will soon see that this is the ground state φ 0 = y 2 e y2 /2 e y2 /2 (y 2 e y2 /2 e y2 /2 ) + y 2 e y2 /2 = ɛe y2 /2 E 0 = ω 2 > 0 What about the other energy eigenstates? - There are two ways of getting rest... (I) Try φ = H(y)e y2 /2 Putting this into the TISE we find H 2yH + H(ɛ 1) = 0 This is called Hermite s equation, and can be solved by Frobeinius series method. See second year notes for details. There are normalisable solutions for ɛ = 2n + 1; H n (y) is a Hermite polynomial 6.1 Operator Method. (II): A much more instructive method which generalises to many other problems In y coords our Hamiltonian operator is Let s try to factorise this Define Operators H = 2 y 2 + y2 a + = y + y a = y + y 50

51 Acting on arbitrary f(y) a + a f(y) = ( y + y ) ( y + y ) f(y) Similarly So we learn = d2 f dy 2 + y2 f f a a + f(y) = d2 f dy 2 + y2 f + f i) [a +, a ] = 2 (6.1) ii) We can write TISE as (a + a + 1)φ = ɛφ (6.2) iii) Apply a to ground state wavefunction: ( ) a φ 0 = y + y e y2 /2 = 0 Thus and again we have found ɛ 0 = 1. (a + a + 1)φ 0 = 1.φ 0 iv) Now act on TISE (the energy eigenvalue eqn, 6.2) with a + and use commutator [a +, a ] So we have learnt a + (a + a + 1)φ = ɛa + φ a + (a a )φ = ɛa + φ [a + a a + + ( 2 + 1)a + ]φ = ɛa + φ (a + a + 1)(a + φ) = (ɛ + 2)(a + φ) If φ has eigenvalue ɛ a + φ has eigenvalue ɛ

52 v) Similarly easy to show that If φ has eigenvalue ɛ a φ has eigenvalue ɛ 2 Since a φ 0 = 0, we say that a annihilates φ 0 There is no state with lower energy than φ 0 φ 0 is indeed ground state, with ɛ = 1 vi) By repeated application of a + on φ 0 we generate entire spectrum a + and a as raising and lowering or ladder or creation and annihilation operators.. a. a + a + φ 0 a + Thus spectrum of SHO is ɛ 0 = 1 E 0 = ω φ 2 0 = e y2 /2 ɛ 1 = 3 E 1 = (1 + 1) ω φ 2 1 = ( y ) + y φ 0 = 2ye y2 /2 ɛ 2 = 5 E 2 = (2 + 1) ω φ 2 2 = ( y ) + y φ 1 = 2(2y 2 1)e y2 /2... leading to, for general n ɛ n = 2n + 1 E n = (n ) ω φ n = ( y + y ) n φ 0 = H n (y)e y2 /2 This is a complete solution of the problem! 52

53 So far we have not normalised the wavefunctions, e.g. φ 0 (x) = ce x2 /2α 2 (note: must convert back to x) 1 = dx φ 0 2 = c 2 dx e x2 /α 2 = c 2 π 1/2 α C = Similarly can normalise higher φ n (x)s... 1 α 1/2 π 1/4 recall α 2 = mk Just as important our wavefunctions are also orthogonal φ n (x)φ m (x)dx = 0 n m This must be true: they are eigenfunctions of hermitian operator H with different eigenvalues ɛ n and ɛ m. Note that the φ n (x) go even-odd-even and just like the infinite square well, the nth excited state has n nodes. Features of the ground-state φ(x) 0 = 1 α 1/2 π x 2 e 1/4 2α 2, E 0 = ω 2, α2 = mk A classical particle of total energy ω/2 would be confined to a region where V (x) E 0, 1 2 kx2 ω 2 x 2 ω k = k k m = k mk = α 2 i.e. α x α If we go back to TISE 2 d 2 φ 2m dx kx2 φ = Eφ at the point where E = V (x) (The limit of the classical motion) we see that d 2 φ dx 2 = 0 53

54 this is a point of inflexion for φ. We can calculate the probability of being outside of the classical region. Quantum effects most significant for low E states. Prob outside classical region n Below is a plot of the n = 0 case. 1.0 φ 0 / απ 1/ Classically forbidden region 0.4 Classically forbidden region 0.2 α α α α α α x

55 7 The Periodic Potential Final example of 1-d problems, consider the motion of a particle in a periodic potential of period, l, so that V (x + l) = V (x) (7.1) An example of such a potential is shown in Figure 7.1. This type of potential can be used to model the interactions to which an electron is subjected in a crystal lattice consisting of a regular array of single atoms separated by the distance l. V (x) b l 0 b x V 0 l Figure 7.1: Periodic potential with rectangle sections, called the Konig-Penny potential. 7.1 Bloch Waves. Although a real crystal is of finite length, assume here that Eq. (7.1) is true for all x. Thus, if ψ(x) is a solution of the Schrödinger equation corresponding to energy E, then so is ψ(x + l). 55

56 As SE is a linear equation, any solution ψ(x) can be represented as a linear combination of two linearly independent solutions ψ 1 (x) and ψ 2 (x) ψ(x) = c 1 ψ 1 (x) + c 2 ψ 2 (x). Now ψ 1 (x + l) and ψ 2 (x + l) are also solutions and can be represented as linear combinations of ψ 1 (x) and ψ 2 (x) ψ 1 (x + l) = a 11 ψ 1 (x) + a 12 ψ 2 (x) ψ 2 (x + l) = a 21 ψ 1 (x) + a 22 ψ 2 (x) Thus we can write ψ(x + l) as ψ 1 (x + l) = (c 1 a 11 + c 2 a 21 )ψ 1 (x) + (c 1 a 12 + c 2 a 22 )ψ 2 (x) = d 1 ψ 1 (x) + d 2 ψ 2 (x) The relationship between the coefficients (c 1, c 2 ) and (d 1, d 2 ) clearly involves the matrix multiplication ( d 1 d 2 ) ( ) ( a 11 a 21 = a 12 a 22 c 1 c 2 ) Let us see what happens if we diagonalise the 2 2 matrix in this equation. We must then solve the equation a 11 λ a 21 a 12 a 22 λ = 0 This generates a quadratic equation for λ, with two solutions λ 1 and λ 2. If (c 1, c 2 ) is a eigenvector corresponding to one of the eigenvalues λ, we have d 1 = λc 1 and d 2 = λc 2. Thus among the solutions there are two having the property ψ(x + l) = λψ(x) where λ is a constant factor. This result is known as Floquet s Theorem. We see immediately that ψ(x + nl) = λ n ψ(x), n = 0, ±1, ±2,... 56

57 Now let ψ λ1 and ψ λ2 be two solutions of the SE corresponding to the energy E, which satisfy ψ(x + l) = λψ(x), and correspond respectively to the eigenvalues λ 1 and λ 2 of Eq. (7.2). Let ψ λ 1 and ψ λ 2 are their derivatives ψ λ1 ψ λ2 W = ψ λ 1 ψ λ 2 denotes something called the Wronskian determinant of ψ λ1 and ψ λ2, we have, using ψ(x + l) = λψ(x), W (x + l) = λ 1 λ 2 W (x). Turns out that Wronskian determinant of two solutions of the SE equation corresponding to the same energy eigenvalue E is a constant, from this we deduce λ 1 λ 2 = 1. Returning to Eq. (7.2). If λ > 1, it is clear that for x ψ grows and grows, if λ < 1 for x ψ also grows and grows. Therefore, must have λ = 1. Given this, λ 1 = e ikl, λ 2 = e ikl. where K is a real number. We get a full range of values for the λs by restricting the value of K to the interval π l K π l. Therefore, all physically admissible solutions must satisfy the relation ψ(x + nl) = e inkl ψ(x), n = 0, ±1, ±2... which is the Bloch Condition. If we rewrite this by letting ψ(x + nl) = e ikx u K (x) then it follows that u K (x + l) = u K (x) which is called Bloch s Theorem. Then we see that the Bloch wave function, ψ(x+nl) = e ikx u K (x) represents a travelling wave of wavelength 2π/K, whose amplitude u K (x) is periodic with the same period l as the crystal lattice. 57

58 8 Angular Momentum Throughout this section we will work in 3-D. Classically angular momentum is given by L = r p so in QM we have the operator L = i r where = i + j + k. From this we find the individual components as x y z L x = yp z zp y L y = zp x xp z L z = xp y yp x and using p i = i x i [ L x = i y z z ] y [ L y = i z x x ] z [ L z = i x y y ] x with total angular momentum L 2 = L 2 x + L 2 y + L 2 z How many angular momentum quantum numbers are there? In classical case all L i and L 2 are all defined, and specifying any 3 completely determines the angular momentum This is not the case in QM, indeed L x, L y, L z and L 2 do not all commute with each other. In fact we have [L x, L y ] = i L z [L y, L z ] = i L x [L z, L x ] = i L y 58

59 so no 2 operators out of L x, L y, L z commute. We can right this compactly as [L i, L j ] = i ɛ ijk L k where +1 if (i, j, k) is (1, 2, 3), (3, 1, 2) or (2, 3,1) ɛ ijk = 1 if (i, j, k) is (1, 3, 2), (3, 2, 1) or (2, 1,3) 0 otherwise: i = j or j = k or k = i So out of L x, L y, L z we can at most have 1 operator which leads to a quantum number. Conventional to choose L z, however, this is not yet the maximal set. Easy to show that [L x, L 2 ] = 0 Thus, maximal set of commuting ops is L 2 and L z Can easily show that both L x and L y commute with L 2 and so we could use an alternative set of commuting operators L 2, L x or L 2, L y. 8.1 Angular Momentum Eigenvalues and Eigenfunctions We will not derive the eigenfunctions and eigenvalues of angular momentum here as they were covered in detail in 2nd year course but you should be familiar with the derivation - see second year notes The eigenfunctions and eigenvalues are L 2 Y l,m = 2 l(l + 1)Y l,m, L x Y l,m = my l,m where l = 0, 1, 2, 3,... and for a fixed l, m = l, l + 1,..., l 1, l. The eigenfunctions Y l,m are called the spherical harmonics and are functions of (θ, φ) and normalised via Y l,m(θ, φ)y l,m (θ, φ)dω = δ ll δ mm. where θ and φ are defined in terms of spherical polar coordinates as x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ. 59

60 The first few spherical harmonics are Y 0,0 (θ, φ) = Y 1,0 (θ, φ) = Y 1,±1 (θ, φ) = 1 4π 3 4π cos θ 3 sin θ e±iφ 8π 8.2 Angular Momentum: A better method. Recall for the SHO we had two ways of getting eigenvalues/eigenvectors Differential Equations Algebraic using a +, a ops We can do similar thing for angular momentum. We can introduce step or ladder operators and solve for eigenvalues of L 2 and L z algebraically Big difference though: algebraic method shows there are more possible eigenvalues of L 2 and L z than what is found using differential ops. How can this happen? Start from the beginning and propose eigenvalue equations L 2 ψ = 2 K 2 ψ, L z ψ = kψ To find out what K 2 and k are consider L + = L x + il y, L = L x il y these are the angular momentum step operators. Note that [L ±, L z ] = [L x ± il y, L z ] = L ± Now consider a ψ k which satisfies L z ψ k = kψ k and define a new function ψ = L + ψ k (if it exists - it might be zero - come back to this). 60

61 What L z value does ψ have, i.e. L z (L + ψ k ) =? Use commutation relation [L z, L + ] = L + L z L + = L + + L + L z Thus L z (L + ψ k ) = ( L + + L + L z )ψ k = ψ + L + kψ k = (k + 1)ψ So the state ψ must be proportional to a state with L z eigenvalue (k + 1) similarly easy to show that L ψ k satisfies so similarly we have L + ψ k = c + (k) ψ }{{} k+1 Normalisation L z (L ψ k ) = (k 1)(L ψ k ) L ψ k = c (k)ψ k 1 L + and L step up and down tower of L z eigenvalues by one unit What do L +, L do to L 2 eigenvalues Since, L 2 commutes with L x and L y and L ± = L x ± L y obvious that [L 2, L ± ] = 0 This means that if ψ k satisfies L 2 ψ k = 2 K 2 ψ k (so label as ψ k,k 2) then L 2 (L + ψ k,k 2) = L + L 2 ψ k,k 2 = L + 2 K 2 ψ k,k 2 = 2 K 2 (L + ψ k,k 2) and similarly for L ψ k,k 2. Thus L +, L do not change L 2 eigenvalue 61