Physics 239/139 Spring 2018 Assignment 2 Solutions

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1 University of California at San Diego Department of Physics Prof. John McGreevy Physics 39/139 Spring 018 Assignment Solutions Due 1:30pm Monday, April 16, Classical circuits brain-warmer. (a) Show that this circuit adds the input bits (at left) mod two: XOR = XOR XOR XOR We can proceed by induction on the number of input bits. The initial step with n = input bits is true since XOR(x, y) = (x + y). For induction step, the input to the n + 1st XOR gate is (( n i=1 x i), x n+1 ), so the output is indeed ( n+1 i=1 x i). (b) [Optional] Construct a circuit with n input bits and one output bit which gives zero unless exactly one of the bits is one. The ingredients available are any gates that take two bits to at most two bits. 1 Here s a perhaps-overly-complicated solution I found, with some help from Hans Singh. First, let s grant ourselves a gate T n which takes n bits and outputs 1 if greater than one bit is 1 (we ll make it later). Then the following gives a recursive solution for the gate O n which outputs 1 if exactly one bit is 1: O n = O n 1 T n 1 XOR The gate with two labels O n 1, T n 1 has two outputs the top output is O n 1 and the bottom is T n 1. The gate indicated by 1 Thanks to Hans Singh and Brian Shotwell for pointing out that this function cannot be computed with only two-to-one gates! I had an error in my solution, and the problem is a little trickier than I thought

2 is a control-0 gate if the top bit is 0, the output is the bottom bit, if the top bit is 1, the output is zero (and nothing happens to the top bit, which we can throw away). In quantum language, it is The recursion starts by O = XOR. C Now we construct T n. The construction is also recursive and very similar to the construction of O n. The recursion starts by T = AND. T n = NOR T n 1 OR 0 where OR is 1 if either or both of the inputs are 1 and NOR is OR followed by NOT (i.e. gives 1 only if all inputs are 0). The NOR on n bits can be accomplished by NOR = OR OR OR X where X is the NOT gate. Here s a more explicit representation. Define a two bit adder a b + c d by: d = XOR(a, b), c = AND(a, b). Then the following will work: + OR + OR + X AND The top part of the input to the final AND gate makes sure there is not more than one one being input, and the bottom input makes sure the number of ones being input is one mod two.

3 . Entanglement entropy in a quantum not-so-many-body system made from spins. Consider the transverse-field Ising model on a lattice with only two (L = ) sites, i = 1,, so that the Hilbert space is H = H 1 H where each of H 1, is a two-state system, and the Hamiltonian is H = J (Z 1 Z + gx 1 + gx ). (a) Find the matrix elements of the Hamiltonian in the eigenbasis of Z 1, Z h ab = s a H s b where a, b = 1..N. What is N in terms of the system size L? Check that your matrix is hermitian. N = L = 4 here g g 0 01 g 1 0 g h ab = H ( 00, 01, 10, 11 ) = 10 g 0 1 g g g 1 (b) Find the eigenvalues of h and plot them as a function of g. (You may wish to use a computer for this and other parts of this problem.) (c) Find the eigenvector (the groundstate) and eigenvalue of the matrix h with the lowest eigenvalue. Write the groundstate as Ψ = N α a φ a. a=1 For the -site case, we can do this by hand, but in preparation for the L > case, let s just use this mathematica notebook. 3

4 (d) The Hilbert space is of the form H 1 H where H 1, are the Hilbert spaces of a single spin. Construct the reduced density matrix for the first site in the groundstate ρ 1 tr H Ψ Ψ. (e) Find the eigenvalues λ α of ρ 1. Calculate the von Neumann entropy of ρ 1, S(ρ 1 ) = α λ α log λ α as a function of g. What is the numerical value when g? What about g 0? Do they agree with your expectations? (f) [Bonus] Redo this problem with L = 3 sites (or more): H = J (Z 1 Z + Z Z 3 + Z 3 Z 1 + gx 1 + gx + gx ). Consider L = 3. There should be three gs in each row and column, since the X i operators move the bit string s 1 s s 3 by Hamming distance one, i.e. it changes exactly one of the bits. The diagonal entries are 3 (if the spins are all the same) and 1 for the other six entries. The spectrum looks like: 3. Entanglement entropy in a quantum not-so-many-body system made from electrons. [This problem is optional for students in physics 139.] Consider a system consisting of two electrons, each with spin one-half, and each of which can occupy either of two sites labelled i = 1,. The dynamics is governed by the following (Hubbard) Hamiltonian: H = t (c 1σc σ + c σc ) 1σ + U n i n i. i σ=, σ =, labels the electron spin. c and c are fermion creation and annihilation operators, {c iσ, c i σ } = δ ii δ σσ I got this problem from Tarun Grover. 4

5 and n iσ c iσ c iσ is the number operator. The condition that there is a total of two electrons means we only consider states ψ with ( ) n iσ ψ = 0. i,σ The first term is a kinetic energy which allows the electrons to hop between the two sites. The second term is a potential energy which penalizes the states where two electrons sit at the same site, by an energy U > 0. (a) Enumerate a basis of two-electron states (make sure they satisfy the Pauli exclusion principle). States with two electrons on two sites are: {, 0, 0,,,,,,,,, } where I ll use the convention that the fermion creation operators are ordered as written, e.g.:, 0 = c 1, c 1, 0. (b) The Hamiltonian above has some symmetries. In particular, the total electron spin in the ẑ direction is conserved. For simplicity, let s focus on the states where it is zero, such as c 1 c 0 where 0 is the state with no electrons, c iσ 0 = 0. Find a basis for this subspace, {φ a }, a = 1..N. States with S1 z + S z = 0 are {, 0, 0,,,,, } (c) Find the matrix elements of the Hamiltonian in this basis, h ab φ a H φ b, a, b = 1..N. There are some signs we need to worry about in the hopping terms from the fermionic nature of the electron 3 In particular, writing H t t σ=, (c 1σc σ + c σc ) 1σ, we have e.g. H t, 0 = t (c σc ) 1σ ) = t ( c c 1 c c 1 3 Thanks to Hans Singh for an important correction here. c 1 c 1 0 (1) 0 () = t (, + ) = t (, ). (3) 5

6 Since all the basis states have norm 1 ( e.g.,, = 1), this gives the matrix elements, H t, 0 = +t,, H t, 0 = t. Similarly, H t 0, = t (c 1σc ) σ c c 0 = t (, ). (4) So in the given basis, we find: U 0 t t 0 U t t h = t t 0 0 t t 0 0 (d) Find the eigenstate and eigenvalue of the matrix h with the lowest eigenvalue. Write the groundstate as Ψ = N α a φ a. a=1 For U > 0, the lowest eigenvalue is E 0 = U ( ) 4t U and the associated normalized eigenvector has ( ) 1 4t α a = 3t + ( U + (4t) + U, 4t U + (4t) + U, 1, 1. ) U+ (4t) +U (e) Before imposing the global constraints on particle number and S z, the Hilbert space can be factored (up to some signs because fermions are weird) by site: H = H 1 H, where H i = span{ 0, c i 0, c i 0, c i c i 0 }. Using this bipartition, construct the reduced density matrix for the first site in the groundstate: ρ 1 tr H Ψ Ψ. 6

7 ρ 1. = ( 1, 0 1, 1, 1 ) (α 1, 0 + α 0, + α 3, + α 4, ) 1 (α1, 0 + α 0, + α3, + α4 0, ) α = ( 0 ) α 0 1, α 1 1, α 4 1, α 3 α 1 1 α4 (5) 1 α3 1 = α α 1 + α 4 + α 3. (6) (f) Find the eigenvalues λ α of ρ 1. Calculate the von Neumann entropy of ρ 1, S(ρ 1 ) = α λ α log λ α as a function of U/t. What is the numerical value when U/t? Using the form of α a, the density matrix is of the form with a = 1. trρ = 1 then determines N =. So the eigenval- +a ues are 4t U+ (4t) +U a and 1 +a +a S 1 = trρ 1 log ρ 1 = ρ 1 = N diag (a, a, 1, 1),. The von Neumann entropy is ( ( ) a a + a log + 1 ( )) 1 + a + a log. + a Since a 0 as U/t, this approaches S 1 U t log. This entropy reflects the fact that in this limit, the groundstate is degenerate between the two states with one particle per site:, and,. (g) Super-Exchange. Go back to the beginning and consider the limit U t. What are the groundstates when U/t, so that we may completely ignore the hopping term? Let us consider the case where we have N sites and N electrons, instead of just N =. In the limit U/t, we must have exactly one electron per site, since if we had no electrons at any site, we would have to have more 7

8 than one at some other site, which would cost large energy U. So there is a single spin-half at each site. At second order in degenerate perturbation theory, find the effective Hamiltonian which splits the degeneracy for small but nonzero t/u. Write the answer in terms of the spin operator The sign is important! Si c iσ σ σσ c iσ. The effective Hamiltonian at second order acting on the degenerate subspace is H = P H i i H P E i 0 E i where P is a projector onto the degenerate subspace with energy E 0 and i runs over virtual states outside the subspace (note that i i i = 1 P ). The most important (lowest energy) virtual intermediate state comes from hopping a single electron to a neighboring site. This virtual state has energy E i E 0 = U above the groundstate space. The hopping matrix elements are each ( t). The tricky bit is that if the neighboring electrons are in the same spin state then the hopping matrix element vanishes by the Pauli principle, since the hopping term doesn t change the spin. Since the perturbation by finite hopping term respects the SU() spin rotation symmetry generated by i S i i c i σc i, we must be able to write the effective hamiltonian in terms of SU() invariant combinations of these operators Therefore H = c t U Si S j (7) for some constant c. The minus sign is ubiquitous in second-order perturbation theory (since the excited state has higher energy than the ground state). But actually c < 0! To see this we can compute just a few of the terms in the case of L =. H eff = 1 U H t (, 0, 0 + 0, 0, ) H t (8) ij (??),(??) = t [(,, ) (,, ) + (,, ) (,, )] U (9) = t U (,, + h.c.,,,, ). (10) 8

9 Now let s compare to (??). S z 1S z =,, +,,,,,, = 1 (,, +,, ). And the off-diagonal terms are S x 1 S x + S y 1S y = S + 1 S + h.c. = (,, +,, ). Comparing these expressions, we find c = 4. Therefore the groundstate for two sites at U t is a singlet, i.e. a Bell pair 1 ( ) which is maximally entangled, in the sense that tr 1 ψ ψ = 11, and hence the entanglement entropy is S = 1. (h) Redo all the previous parts for the case where the two particles are spin-half bosons, c iσ b iσ, [b iσ, b i σ ] = δ ii δ σσ. Nearly everything is the same, except the wavefunctions are symmetric, the signs in the hopping matrix elements are missing. The missing relative sign in the hopping matrix elements means that the spin-spin interaction will be ferromagnetic. If we had L >, it would matter that it is no longer true that hopping is blocked when the neighboring spins are in the same state; this adds an extra term to the effective hamiltonian proportional to the identity operator. 4. Chain rules. Show that for a joint distribution of n random variables p(x 1 X n ), the joint and conditional entropies satisfy the following chain rule: H(X 1 X n ) = n H(X i X i 1 X 1 ). i=1 Show that the n = case is the expectation of the log of the BHS of Bayes rule. Then repeatedly apply the n = case to increasing values of n. 5. Learning decreases ignorance only on average. ( ) 0 a Consider the joint distribution p yx =, where y =, is the row index and b b yx x =, is the column index (so yx are like the indices on a matrix). Normalization implies xy p xy = a+b = 1, so we have a one-parameter family of distributions, labelled by b. 9

10 What is the allowed range of b? Find the marginals for x and y. Find the conditional probabilities p(x y) and p(y x). Check that H(X Y ) H(X) and H(Y X) H(Y ) for any choice of b. Show, however, that H(X Y = ) > H(X) for any b < 1. Since a = 1 b, we need 0 b 1 to keep all the probabilities positive. The marginals are p x = (b, a + b) and p y = (a, b). The conditional probabilities are ( ) ( ) p(x y) =, p(y x) =. 1 1 yx So we have H(X) = b log b (a + b) log(a + b) = b log b (1 b) log(1 b) = H(b, 1 b). So if we measure Y and get, we know nothing about X. This means that for this particular outcome a a+b H(X y = ) = 1 > H(X) But if we get y = then we know for sure X = : H(X y = ) = 0, no uncertainty. The average conditional entropy for X is then b a+b H(X Y ) H(X Y ) XY = p y= H(X y = ) + p y= H(X y = ) = b. And indeed and H(b, 1 b) > b for b < 1, as you can see in the rightmost figure below. Similarly, H(Y x = ) = 0 means that the uncertainty for x = must be larger. Specifically, ( ) 1 b H(Y ) = H(1 b) H(Y X) = (1 b)h 1 b but H(Y X = ) = H ( ) 1 b 1 b can be either larger or smaller than H(Y ) (as in the leftmost figure below). yx 10