QUALIFYING EXAMINATION, Part 2. Solutions. Problem 1: Quantum Mechanics I

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1 QUALIFYING EXAMINATION, Part Solutions Problem 1: Quantum Mechanics I (a) We may decompose the Hamiltonian into two parts: H = H 1 + H, ( ) where H j = 1 m p j + 1 mω x j = ω a j a j + 1/ with eigenenergies ω ( n j + ) 1 for intergers n j 0 and j = 1, Since a j, a k [ ] [ ] = δ j,k, [a j, a k ] = 0, and a j, a k = 0, we have [H 1, H ] = ( ω) [ a 1a 1, a a ] = 0 Hence, we can find simultaneous eigenstates of the H j (b) For non-negative integers n j, the quantum number n = n 1 + n can take all possible non-negative integers n = 0, 1,, The eigenenergy is E n = ω (n 1 + n + 1) == ω (n + 1) We may have n + 1 possible solutions for n = n 1 + n, with (n 1, n ) = (0, n), (1, n 1),, or (n, 0) Hence the degeneracy of the level associated with quantum number n is n + 1 (c) The operator A jk = a j a k annihilates one quantum in oscillator i and creates another quantum in oscillator j Hence the total number of quanta n = n 1 + n is preserved and all A jk commute with H In particular, any linear combination of the A jk commutes with H The other two hermitian combinations are A 1 + A 1 and i(a 1 A 1 ) (d) Expressing x j and p j as linear combination of a j and a j, we find L x 1 p x p 1 = i(a a 1 a 1a ) = i(a 1 A 1 ) (e) We first use [A, BC] = [A, B] C + B [A, C] to compute the commutators of A jk : [ ] [ ] ] [A jk, A lm ] = a j a k, a l a m = a j a k, a l a m + a l [a j a k, a m = δ kl a j a m δ jm a l a k Defining J x = (A 11 A )/, J y = (A 1 + A 1 )/, J z = L/ = i(a 1 A 1 )/ 1

2 and using the above commutation relations, we find [J x, J y ] = 1 4 [A 11 A, A 1 + A 1 ] = 1 4 ([A 11, A 1 ] + [A 11, A 1 ] [A, A 1 ] [A, A 1 ]) = 1 (a 1a a a 1 + a 1a a a ) 1 = ij z 4 Similarly, we can verify [J y, J z ] = ij x and [J z, J x ] = ij y, which are the same as for a three-dimensional angular momentum (f) We can compute (no need to do that) J = ( ) Jx + Jy + Jz = 1 [ ( ) ( ) ( ) ] a 4 1a 1 a a + a 1a + a a 1 + ( i) a 1a a a 1 = 1 (a 1a 1 + a a ) ( ) a 4 1a 1 + a a + = j(j + 1), where in the last step we have restricted the operator J to eigenstates with quantum number n for which a 1a 1 + a a = n and define j = n/ The degeneracy associated with quantum number j is j + 1 = n + 1, which is consistent with the result from part (b) Since for a given j the allowed values of J z are j, j + 1,, j and L = J z, we find that the allowed values of L for level with quantum number n are L = n, n +,, n (g) The symmetry group responsible for the degeneracy is the one generated by J x, J y, J z, which is SO(3) [isomorphic to SU()] The SO() generated by the physical angular momentum L = J z is a subgroup of the above SO(3)

3 Problem : Quantum Mechanics II (a) For real Ω 1 = Ω = Ω, the Hamiltonian can be represented by a 3 3 matrix: 0 0 Ω H = 0 0 Ω Ω Ω 0 in the basis of g 1, g and e By solving det [H λi] = λ 3 + λω = 0, we have the eigenenegies The associated eigenstates are λ = 0, ± Ω φ 0 = g 1 g φ ± = g 1 + g ± e We can verify that H φ 0 = Ω ( e e ) = 0 and H φ ± = ± Ω φ ± (b) The initial state can be expressed in the eigenbasis of the Hamiltonian: With the Hamiltonian evolution, we have For t = T = ψ (0) = g 1 = 1 φ φ φ ψ (t) = e iht/ ψ (0) = 1 φ e i Ωt φ ei Ωt φ = 1 ( 1 + cos ) Ωt g ( 1 cos ) Ωt g i 1 sin Ωt e π Ω, we have ψ (T ) = g with all its population evolved into state g (c) Method I: Similar to part (a), we can solve the 3 3 matrix and find the zero energy eigenstate Method II: We expand the zero-energy eigenstate as φ 0 = c 1 g 1 + c g + c e e For zero-energy eigenstate, we have H φ 0 = (Ω 1 c 1 + Ω c ) e + Ω 1c e g 1 + Ω c e g = 0, 3

4 which implies c e = 0 and Ω 1 c 1 + Ω c = 0 Hence, we find 1 φ 0 = (Ω g 1 Ω 1 g ) Ω 1 + Ω for the zero-energy eigenstate (d) Method I: Expand the zero-energy eigenstate as It satisfies H φ = Hence, we have c e = 0 and n 1 φ = c j g j + c e e ( n 1 ) n 1 Ω j c j e + c e Ω j g j = 0 n 1 Ω j c j = 0 The last equation expresses the orthogonality of the vector c j to the vector Ω j in an (n-1)-dimensional space There are (n-) such orthogonal vectors, corresponding to the n zero-energy eigenstates More explicitly, we can find these eigenstates by having 1 φ l = (Ω l+ g 1 Ω 1 g l+ ) Ω 1 + Ω l+ for l = 0,, n 3 Note that part (c) is just the special case with n = 3 Method II: We may rewrite the Hamiltonian as ( n 1 ) H = Ω Ω j tot e Ω tot g j Ω tot with Ω tot = n = Ω tot ( e φ B + φ B e ) ( n 1 + Ω tot Ω 1 and normalized state φ B = n 1 Ω j Ω tot g j ) e Ω j Ω tot g j Any superposition of { g j },,n 1 orthogonal to φ B has zero energy For example, we can explicitly find n such linearly independent zero energy eigenstates: with φ l = H φ l = 0 1 Ω 1 + Ω l+ (Ω l+ g 1 Ω 1 g l+ ), for l = 0,, n 3, which span the n dimensional zero energy eigenspace 4

5 Statistical Mechanics I (a) In classical statistical mechanics the statistical distribution in phase space is given by ρ(p, x) e βh(p,x) where β = 1/kT and H is the Hamiltonian as a function of momenta p and cartesian coordinates x Since the kinetic energy is independent of the coordinates, we can integrate over the momenta and find that the probability density is given by a Boltzmann factor P (x) = dpρ(p, x) dpe βh(p,x) e βv (x) The normalized distribution is P (x) = e V (x) kt dx e V (x ) kt (b) In normal coordinates, the potential matrix is diagonal and the potential is given by V (y) = 1 λ m ym Using the result in (a), the distribution of the normal coordinates is then given by m P (y) e m λmy m kt The distribution of a normal coordinate y m is obtained by integrating all other normal coordinates and is a Gaussian ( ) 1/ λm P (y m ) = e λmy m kt πkt (c) According to the equipartition theorem, the thermal average of each separated quadratic term in the Hamiltonian is kt/ Thus 1 λ my m = kt We find y m = kt λ m = kt κ [ 1 cos ( )] pπ N+1 We can also calculate y m directly from the Gaussian distribution found in part (b), leading to the same result 5

6 (d) Since P (y) = Π m P (y m ), the variables y m are uncorrelated, ie, for m n y m y n = y m y n = 0 (e) The original coordinate are related to the normal coordinates by x = R 1 x Since R is orthogonal and symmetric R 1 = R T = R and x = Rx We then have x n = m R nmy m Using the results in (c), we find x n = m,k R nm R nk y m y k = m R nm y m Substituting the expression for y m from part (b) and R mn, we find x n = kt (N + 1)κ N m=1 ( ) πnm sin 1 N cos ( ) mπ N+1 Using the formula for the sum given in the problem, we have x n = kt ( κ n 1 n ) N + 1 A sketch of x n versus n is shown in the figure below for N = <x_n^> n 6

7 Problem 4: Statistical Mechanics II (a) The average occupation of each adsorption site is given by the Fermi-Dirac distribution 1/ [ e β( µ) + 1 ] Thus N 0 N a = z 1 e β + 1, where β = 1 and z = kt eβµ is the fugacity (b) Since the kinetic energy is p, we have m N f = V d 3 p (π ) 3 1 z 1 e βp /m + 1 = V (π ) 3 0 4πp dp z 1 e βp /m + 1 β For convenience in the following parts, it is useful to change variables x = p and m obtain N f = V ( m π ) x (kt ) dx π z 1 e x + 1 = c(kt ) 3 f3/ (z), where c = V ( m π ) 3 0 (c) At zero temperature, the state must be in the ground state consistent with the Pauli exclusion principle Thus, all N 0 adsorption sites are occupied at T = 0 and the remaining N N 0 atoms are free Thus N a = N 0, N f = N N 0 The Fermi energy is the highest occupied single-particle level at T = 0, ie, in the ground state There are N N 0 fermions occupying the levels in the box of volume V and N N 0 = V pf 4πp dp = V (π ) 3 3π 3 p3 F = V 3π (mɛ 3 F ) 3/ The Fermi energy is thus given by ( ) ( ) ɛ F = 3π /3 N N0 m V 0 (d) At sufficiently large T, z is small and we can approximate f 3/ (z) z Hence, N f c(kt ) 3 z Since N > N 0 some number of atoms must be free To get a finite number of free atoms, it must be that z (kt ) 3 Using this results in the probability for adsorption sites being occupied, we conclude that N a N 0 const(kt ) 3/ e β + 1 0, N f N 7

8 at large temperatures (e) The Fermi energy is the highest occupied level at T = 0 Since N = N 0 all atoms occupy the adsorption sites and the Fermi energy is 8