Section 10: Many Particle Quantum Mechanics Solutions

Transcription

1 Physics 143a: Quantum Mechanics I Section 10: Many Particle Quantum Mechanics Solutions Spring 015, Harvard Here is a summary of the most important points from this week (with a few of my own tidbits), relevant for either solving homework problems, or for your general education. This material is covered in Sections 5.1 and 5. of [1]. Consider the Hamiltonian The change of variables to simplifies H to H = 1 + V (r 1 r ). (1) m 1 m R = m 1r 1 + m r m 1 + m, r = r 1 r () ] H = M R + [ µ r + V (r). (3) Because this H can be written as the sum H R + H r, the eigenfunctions are of the form ψ(r)e ik R. In general, if we have a two body problem of the form H = H 1 + H, and the form of H 1, are identical, then we can write down exact eigenstates that are products of the form ψ a (r 1 )ψ b (r ) each ψ a,b is a single-particle eigenstate. In quantum mechanics, if these wave functions describe two identical particles (such as two electrons), then the wave functions must be chosen to be symmetric (if the particles are bosons): ψ = ψ a (r 1 )ψ b (r ), (a = b), ψ = ψ a(r 1 )ψ b (r ) + ψ b (r 1 )ψ a (r ), (a b) (4) or antisymmetric (if the particles are fermions): ψ = ψ a(r 1 )ψ b (r ) ψ b (r 1 )ψ a (r ), (a b) (5) It is impossible to write down a fermionic state with a = b. This is referred to as the Pauli exclusion principle. In general, the two body problem contains an interacting potential between particles 1 and. In this case, the eigenstates above are no longer exact. However the fully interacting problem is almost always impossible to solve exactly, and if the interaction is small, these may provide good approximate eigenstates. The expected value of the interactions in terms of the separable wave functions is given by V (r 1, r ) = d 3 r 1 d 3 r ψ a (r 1 ) ψ b (r ) V (r 1, r ) ± d 3 r 1 d 3 r Re [ψa(r 1 )ψ a (r )ψ b (r )ψb (r 1)] V (r 1, r ) (6) 1

2 The first term refers to the expected value of the potential if we had identical, distinguishable particles; the second is inherently quantum and are called exchange interactions. It is simply related to the fact that the wave function must be a linear superposition of states which is symmetric (+) or antisymmetric ( ). If we include the spin of particles, and H = H pos + H spin, then the wave functions may be written as Ψ(r 1, s 1, r, s ) = φ(r 1, r )χ(s 1, s ). (7) The total product Ψ must be symmetric (bosons) or antisymmetric (fermions); φ and χ can be either symmetric or antisymmetric so long as the product has the right symmetry. If we have N bosons or N fermions, then the wave function is Ψ(r 1, s 1,..., r N, s N ) and must be symmetric (bosons) or antisymmetric (fermions) under the interchange of any two particles. Finding the electronic structure of a nucleus of charge Ze requires finding the ground state of H = Z i=1 [ p i m ] Ze + 4πɛ 0 r i i>j e 4πɛ 0 r i r j, (8) which cannot be done if Z > 1. We approximate that the wave function consists of orbitals (the wave functions of the hydrogen atom of charge Ze) along with an appropriate spin wave function. The rules are basically as follows: 1) states of higher l get a higher energy (due to the partial screening of the charge of the nucleus), ) if we have electrons with a partially unfilled orbital, then we choose the wave function to have maximal total spin. This is because an antisymmetric coordinate wave function minimizes the exchange interactions, which are repulsive (Hund s Rule 1). But these wave functions are not the real ground states and so there are exceptions to the rules above. A simple model for a metal is the free electron gas, which consists of N electrons in a cubical box of volume V with the Hamiltonian of N free particles. The wave functions are (if we choose periodic boundary conditions on the box) ψ nxn yn z = 1 V e πi(nxx+nyy+nzz)/l. (9) We often define k i = π L n i. (10) The filled energy levels will be (as N ) all states with k < k F ( 3π n ) 1/3 (11) where n N/V. The surface in k-space at k = k F is called the Fermi surface, and divides the occupied states from the empty states. There is a Fermi energy E F = k F m such that all states of E < E F are filled, and states of E > E F are empty. (1)

3 If we account for the fact that there are ions, then we have to account for a periodic potential from the ions: V (r) = V (r + a), though we still ignore electron-electron interactions: H = N i=1 [ ] p i m + V (r i). (13) Bloch s Theorem tells us that the wave functions can be written in the form ψ(r) = e ik r φ(r), (14) with φ(r) = φ(r + a), and k i < π/a i. The allowed energy levels E in such a system tend to form continuous bands with gaps in between them. If E F sits within a band, then we have a metal we can add a small amount of energy and move an electron from just below E F to just above. If E F sits in between two bands, then it takes a finite amount of energy to move an electron from a filled band to an empty band this is an insulator. Problem 1: Suppose that we have two spinless particles of the same mass m, and a Hamiltonian: H = p 1 m + p m + V (r 1, r ). (a) Define the exchange operator P, such that P Ψ(r 1, r ) = Ψ(r, r 1 ), for an arbitrary function Ψ. Show that P = 1. What are the eigenvalues of P? Solution: P Ψ(r 1, r ) = P P Ψ(r 1, r ) = P Ψ(r, r 1 ) = Ψ(r 1, r ), thus P = 1. The eigenvalues σ must obey P Ψ = σ Ψ = Ψ, so we get σ = 1 or σ = ±1. The eigenvalue σ = 1 corresponds to a bosonic wave function, and σ = 1 to a fermionic wave function. (b) Under what circumstances does [P, H] = 0? 1 Solution: One thing we need to be clear about is that 1Ψ(, ) always acts on the first slot of the wave function, and always acts on the second slot of the wave function. Let s evaluate P HΨ(r 1, r ) = P [ ( m 1 Ψ(r 1, r ) + Ψ(r 1, r ) ) ] + V (r 1, r )Ψ(r 1, r ) = ( m Ψ(r, r 1 ) + 1Ψ(r, r 1 ) ) + V (r, r 1 )Ψ(r, r 1 ) HP Ψ(r 1, r ) = HΨ(r, r 1 ) = [ ( m 1 Ψ(r, r 1 ) + Ψ(r, r 1 ) ) ] + V (r 1, r )Ψ(r, r 1 ). These equations are equivalent when V (r 1, r ) = V (r, r 1 ). (c) Suppose that [P, H] = 0, and that at time t = 0: Ψ(r 1, r, 0) = σψ(r, r 1, 0) where σ = ±1. Show that Ψ(r 1, r, t) = σψ(r, r 1, t). 1 It may help to multiply by a test function to evaluate this commutator! Be careful with the momentum terms in H! 3

4 Solution: Here is a simple way to do this: P Ψ(r 1, r, dt) = P (1 + Hi ) dt Ψ(r 1, r, 0) = = σ (1 + Hi dt ) P Ψ(r 1, r, 0) (1 + Hi dt ) Ψ(r 1, r, 0) = σψ(r 1, r, dt) In an infinitesimal time step, thus the wave function stays symmetric/antisymmetric. Thus add up many of these small time steps, and we conclude that P Ψ(r 1, r, t) = σψ(r 1, r, t). A bosonic/fermionic wave function will stay bosonic/fermionic for all times, and thus it makes sense to demand symmetry/antisymmetry of the wave function. (d) Does it make sense to talk about a pair of bosons or fermions if [P, H] 0? Solution: No, because from part (c) if [P, H] 0, then the wave function will not stay symmetric or antisymmetric. Alternatively, we can distinguish the particles by seeing which potential they feel, roughly speaking. Problem (Pigments): A typical pigment molecule has a structure sketched below []. The bonds highlighted in red are special: they are called conjugated π bonds. Suppose there are N conjugated π bonds: then each bond contributes one free spin-1/ electron, of mass m, which may move up and down the chain of bonds freely. If each bond has length a, when N is large, we may thus approximate these electrons as moving in a particle in a box of width L = a(n 1) Na. R R (a) Suppose N is an even number. Using the Pauli exclusion principle and the results for the particle in a box, describe which energy levels in the box are filled and which are empty. Ignore electron-electron interactions. Solution: The first N/ energy levels of the particle in the box will be filled with electrons; all the rest will be empty. (b) How much energy does it take to place all of these fermions in the ground state of this quantum system? Approximate your answer in the N 1 limit. Solution: The energy levels of the box are The energy stored in the lowest filled levels is N/ E gs = E j = π N/ ml E j = π j ml. j π 1 ml 3 ( ) N 3 = N π 4ma. 4

5 (c) Now, suppose we send a photon of wavelength λ at the pigment molecule. What is the largest value of λ such that the photon can be absorbed by an electron in the pigment molecule? When the photon is absorbed, the electron must be able to jump to an unoccupied state in the box. You should find that λ KN in the limit when N 1 what is the value of K? Solution: The smallest energy photon that can be absorbed excites an electron from the j = N/ energy level to the j = 1 + N/ energy level. This has energy [ ( E = π ml 1 + N ) ( ) ] N π mn a N. The wavelength of light is λ = π c E π c ma = π N = 4cma π N (d) Evaluate numerically the value of K, given that m kg and a m. A typical pigment molecule might have a chain with N 0. Does λ correspond to a photon in the visible spectrum? Solution: We find K = 4cma 30 nm. π So if N = 0, we get λ = 600 nm, which is indeed in the visible spectrum! (e) Suppose I give you pigment molecules, one of which is red, and one of which is blue. Which pigment molecule do we expect has a longer chain of conjugated π bonds? Solution: Red light has a longer wavelength than blue light. But also, a molecule which absorbs red light effectively will appear blue, since the blue light is the only light that gets reflected and vice versa. So since λ N, we conclude the red molecule absorbs blue light, and has a shorter chain of π bonds than the blue molecule. Problem 3 (Repulsive Bosons): Suppose that we have N identical bosons of mass m in a one dimensional box of length L = Na (0 < x < Na), with the Hamiltonian H = N i=1 p i m + λ δ(x i x j ). We are interested in the N 1 limit. This problem can be solved exactly [3, 4], but we will not attempt to do so here. (a) Suppose λ = 0. What is the ground state wave function? Call your answer Ψ 0 (x 1,..., x N ). What is the energy of this state, E 0 (0)? i j Solution: Since H is a sum of N single particle Hamiltonians when λ = 0, the ground states will be sums of products of particle in a box wave functions. The ground state is the symmetric wave function N Ψ 0 (x 1,..., x N ) = Na sin πx j Na. 5

6 The energy of this state is simply the sum of the energies of each particle: Note that E 0 (0) 0 as N. E 0 (0) = N π m(na) = π ma 1 N. (b) Now, let us suppose that λ > 0, so that these bosons tend to repel each other. Calculate E 0 (λ) = Ψ 0 H Ψ 0 and comment on your answer. What is happening physically? Solution: We ve already evaluated the kinetic energy in part (a), so all we have to do is evaluate Ψ 0 λ δ(x i x j ) Ψ 0 N λ ( ) dx 1 dx sin πx 1 πx Na Na sin Na δ(x 1 x ) i j = λ a dx sin 4 πx Na = 3λN 8a. The key thing to note above is that because each individual interaction term only depends on two coordinates, we can immediately integrate over N coordinates in the wave function, and we simply get factors of unity. So we find that as N, E 0 (λ) 3λ 8a N. The overall factor of N tells us that each boson feels a net energy of λ/a due to repulsive interactions with all other particles. (c) Let us temporarily consider that we had N fermions in the box. What would be the ground state Ψ 0 (x 1,..., x N ), if λ = 0? Solution: If we had fermions, then we would construct the Slater determinant: Ψ 0 (x 1,..., x N ) = 1 N! σ(i1,..., i N ) N Na sin i jπx j Na where σ(i 1,..., i N ) is 1 if it takes an even number of swaps to switch the integers to 1,..., N, and 1 if an odd number. If this seems confusing to you, we have just constructed explicitly the determinant of a very big matrix. (d) What is the energy of the state Ψ 0? Call it Ẽ, and verify that it is independent of λ. Solution: The first thing to notice is that Ψ 0 δ(x i x j ) Ψ 0 = 0 since a fermionic wave function necessarily vanishes if x i = x j! So all the energy is pure kinetic energy at arbitrary λ and we find Ẽ = N π m(na) π N 3 m(na) 3 = π 6ma N. 6

7 (e) Show that the function Ψ(x 1,..., x N ) = Ψ 0 (x 1,..., x N ) i>j sign(x i x j ) is a symmetric, normalized wave function. What is Ψ H Ψ? This is provably the exact ground state at λ =. Solution: Without loss of generality, let us suppose that x 1 < x < < x N (otherwise, just reorganize the xs in increasing order). Suppose we switch x i and x j with i < j. Then how many signs flip? Well, if k > j, then x k x i and x k x j are both positive, and so nothing happens to sign(x k x i )sign(x k x j ) a similar statement holds if k < i. But what if i < k < j. Well now we get sign(x k x i )sign(x j x k ) sign(x k x j )sign(x i x k ) = ( 1) sign(x k x i )sign(x j x k ), so this too is invariant. But the term sign(x j x i ) does flip sign, and so this product of sign functions flips sign if we switch x i and x j. Furthermore, the function Ψ is continuous, since the only place where the product of signs is ill-defined is at x i = x j, where Ψ 0 vanishes. Thus Ψ is a continuous wave function which is the product of two totally antisymmetric functions it therefore is symmetric. And it s normalized since Ψ = Ψ 0, as sign(x) = 1. Now since Ψ = 0 when x i = x j we conclude that all kinetic energy is again kinetic. And furthermore the x j derivatives acting on sign(x i x j ) lead to δ(x i x j ), but again these terms must vanish. We conclude that the energy of this state is the same as what we computed in part (d)! (f) Give a heuristic argument for what should happen to the ground state of H as we increase the parameter λ. Solution: For small λ, we expect Ψ 0 to be the approximate ground state, and for large λ we expect Ψ to be the ground state. What do we mean by small λ? Well, the energies of the two states are comparable when so we expect that at the scale 3λ 8a π 6ma λ 4 π 9ma there is a transition from free boson to free fermion like behavior of the ground state. [1] D. J. Grififths. Introduction to Quantum Mechanics (Prentice Hall, nd ed., 004). [] H. Kuhn. A quantum-mechanical theory of light absorption of organic dyes and similar compounds, Journal of Chemical Physics (1949). [3] M. Girardeau. Relationship between systems of impenetrable bosons and fermions in one dimension, Journal of Mathematical Physics (1960). [4] E. H. Lieb and W. Liniger. Exact analysis of an interacting Bose gas. I. The general solution and the ground state, Physical Review (1963). 7