University of California at San Diego – Department of Physics – Prof. John McGreevy
Physics 239 Spring 2016
Assignment 2 –
Solutions
Due 11am Tuesday, April 19, 2016
1. Entanglement entropy in a quantum not-so-many-body system. [from
Tarun Grover]
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, 2. The dynamics is governed
by the following (Hubbard) Hamiltonian:
X †
X
c1σ c2σ + c†2σ c1σ + U
H = −t
ni↑ ni↓ .
i
σ=↑,↓
σ =↑, ↓ labels the electron spin. c and c† are fermion creation and annihilation
operators,
{ciσ , c†i0 σ0 } = δii0 δσσ0
and niσ ≡ c†iσ ciσ is the number operator. The condition that there is a total of
two electrons means we only consider states |ψi with
!
X
niσ − 2 |ψi = 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).
(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†2↓ |0i where |0i is the state with no
electrons, ciσ |0i = 0. Find a basis for this subspace, {φa }, a = 1..N .
(c) Find the matrix elements of the Hamiltonian in this basis,
hij ≡ hφa |H|φb i , a, b = 1..N.
1
(d) Find the eigenstate and eigenvalue of the matrix h with the lowest eigenvalue. Write the groundstate as
|Ψi =
N
X
αa |φa i .
a=1
(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 = H1 ⊗ H2 , where Hi = span{|0i , c†i↑ |0i , c†i↓ |0i , c†i↑ c†i↓ |0i}. Using this bipartition, construct the reduced density matrix for the first site
in the groundstate:
ρ1 ≡ trH2 |Ψi hΨ| .
(f) Find the eigenvalues λα of ρ1 . Calculate the von Neumann entropy of ρ1 ,
P
S(ρ1 ) = − α λα log λα as a function of U/t. What is the numerical value
when U/t → ∞?
(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 = 2. 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
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
~Si ≡ c† ~σσσ0 c 0 .
iσ
iσ
The sign is important!
The effective Hamiltonian at second order acting on the degenerate subspace
is
X H |ii hi| H
H=P
P
E0 − Ei
i
where P is a projector onto the degenerate subspace with energy E0 and i
P
runs over ‘virtual’ states outside the subspace (note that i |ii hi| = 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
Ei − E0 = U above the groundstate space. The hopping matrix elements are
2
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(2) spin rotation symmetry generated
P
P
by i Si ≡ i c†i ~σ ci , we must be able to write the effective hamiltonian in
terms of SU(2) invariant combinations of these operators Therefore
H = −c
t2 X ~ ~
Si · Sj
U
hiji
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)
and means that the induced interaction is antiferromagnetic.
Therefore the groundstate for two sites at U t is a singlet, i.e. a Bell pair
√1 (|↑↓i − |↓↑i) which is maximally entangled, in the sense that tr1 |ψi hψ| =
2
1
and hence the entanglement entropy is S = 1.
1,
2
(h) Redo all the previous parts for the case where the two particles are spin-half
bosons,
ciσ
biσ , [biσ , b†i0 σ0 ] = δii0 δσσ0 .
Nearly everything is the same, except the wavefunctions are symmetric, and
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.
2. Chain rules.
Show that for a joint distribution of n random variables p(X1 · · · Xn ), the joint
and conditional entropies satisfy the following chain rule:
H(X1 · · · Xn ) =
n
X
H(Xi |Xi−1 · · · X1 ).
i=1
Show that the n = 2 case is the expectation of the log of the BHS of Bayes rule.
Then repeatedly apply the n = 2 case to increasing values of n.
3. Learning decreases ignorance only on average.
0a
Consider the joint distribution pyx =
, 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
P
implies xy pxy = a + 2b = 1, so we have a one-parameter family of distributions,
labelled by b.
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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 < 21 .
Since a = 1 − 2b, we need 0 ≤ b ≤ 12 to keep all the probabilities positive. The
marginals are px = (b, a + b) and py = (a, 2b). The conditional probabilities are
!
0 1
01
p(x|y) = 1 1
, p(y|x) =
.
a
b
2 2
a+b a+b
yx
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
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
H(X|Y ) ≡ hH(X|Y )iXY = py=↓ H(X|y =↓) + py=↑ H(X|y =↑) = 2b.
And indeed and H(b, 1 − b) > 2b for b < 21 , 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 − 2b
H(Y ) = H(1 − 2b) ≥ H(Y |X) = (1 − b)H
1−b
but H(Y |X =↓) = H 1−2b
can be either larger or smaller than H(Y ) (as in the
1−b
leftmost figure below).
4