1. Checking the Consistency of Local Density Matrices
Yi-Kai Liu
Computer Science and Engineering
University of California, San Diego

2. The Consistency Problem
Consider an n-qubit system
We are given density matrices ρ1,…,ρm, where each ρi describes a subset of the qubits Ci
Is there a state σ (on all n qubits) whose reduced density matrices match ρ1,…,ρm? ρ2
C2 = {2,4,5} ρ1
C1 = {1,2,3}

3. An Example 3 qubits, ρA = ρB = |φ)(φ|, where |φ) = (|00) + |11)) / √2
ρB, B = {2,3}
ρA, A = {1,2}
3 qubits, ρA = ρB = |φ)(φ|, where |φ) = (|00) + |11)) / √2
There is no state σ s.t. tr3(σ) = ρA, tr1(σ) = ρB
Can see this using strong subadditivity: S(1,2,3) + S(2) ≤ S(1,2) + S(2,3)

4. A More General Problem Consider a finite quantum system
We are given a set of observables T1,…,Tr, together with expectation values t1,…,tr
Is there a state σ with these expectation values, that is, tr(Ti σ) = ti for i = 1,…,r ?
Consistency of local density matrices is a special case of this problem
For each subset of qubits C, knowing the density matrix for C is equivalent to knowing the expectation values of all Pauli matrices on C

5. Our Results
I. For the consistency problem: If ρ1,…,ρm are consistent with some state σ > 0, then they are also consistent with a state σ’ of the following form: σ’ = (1/Z) exp(M1+…+Mm), where each Mi is a Hermitian matrix that acts on the same qubits as ρi, and Z is a normal-izing factor.
So the existence of σ’ is a necessary and sufficient condition for consistency

6. Our Results
II. For the general problem: If there exists a state σ > 0 with expectation values t1,…,tr, then there exists a state σ’ which has the same expectation values, and is of the form: σ’ = (1/Z) exp(θ1T1+…+θrTr), where θ1,…,θr are real.
This holds under a technical assumption that T1,…,Tr and I are linearly independent over the reals

7. Related Work
These results were previously derived by Jaynes (1957), as part of the maximum-entropy principle for statistical inference
Jaynes’ proof uses the Lagrange dual of the entropy-maximization problem
We give a somewhat different proof, using the convexity of the partition function

8. Facts about the Partition Function
Given observables T1,…,Tr
Let Z(θ) = tr(exp(θ1T1+…+θrTr))
Let ψ(θ) = log Z(θ)
Consider the family of states ρ(θ) = exp(θ1T1+...+θrTr) / Z(θ)
Replacing Ti with Ti+αI does not change ρ(θ)
The function ψ is convex and differentiable, and ∂ψ/∂θi = tr(Ti ρ(θ))

9. Expectation Values and the Partition Function
Given expectation values t1,…,tr
Can assume ti = 0 (by translating Ti appropriately)
We want to find θ s.t. gradient(ψ(θ)) = 0
Example: a single qubit, want <σz> = –1 (this happens when θ → –infty)
ψ(θ) = log tr(exp(θ(σz+1))) = log(e2θ + 1)
ψ(θ) θ

10. Expectation Values and the Partition Function
We want to find θ s.t. gradient(ψ(θ)) = 0
Another example: a single qubit, want <σz> = –1 and <σx> = –1 (not possible)
ψ(θ) = log tr(exp(θ1(σz+1) +
θ2(σx+1)))
ψ(θ1,θ2) θ1 θ2

11. Proof Sketch
We prove claim II, which implies claim I as a special case
We know there is a state ρ > 0 s.t. tr(Ti ρ) = ti. We can write ρ in the form: ρ(θ,φ) = exp(θ1T1+…+θrTr φ1U1+…+φsUs) / Z(θ,φ) where T1,…,Tr,U1,…,Us are a complete set of observables.
Let ui = tr(Ui ρ) be the expectation values of the Ui. We can assume ti = 0, ui = 0, for all i.

12. Proof Sketch
Since the Ti and Ui are a complete set of observables, there is a unique (θ,φ) s.t. ρ(θ,φ) has the expectation values ti and ui.
So gradient(ψ(θ,φ)) vanishes at exactly one point.
Since ψ is a convex function, it follows that: ψ(θ,φ) → infty as ||θ,φ|| → infty, where ||θ,φ|| is the norm of the vector (θ,φ).

13. Proof Sketch
Now consider states ρ’ of the form: ρ’(θ) = exp(θ1T1+…+θrTr) / Z’(θ)
The partition functions of ρ’ and ρ are related: ψ’(θ) = ψ(θ,0)
Hence ψ’(θ) → infty as ||θ|| → infty.
Since ψ’ is convex, it follows that ψ’ attains its minimum at some point θmin.
Hence gradient(ψ’(θmin)) = 0; and so the state ρ’(θmin) has the desired expectation values ti.

14. Proof Sketch
Example: we want <σz> = –.6 σx plays the role of the extra observables Ui, with <σx> = –.3
(1): gradient(ψ(θ,φ)) = 0
ψ’(θ) = ψ(θ,0)
(2): gradient(ψ’(θ)) = 0
ψ(θ,φ)
(2)
(1) φ θ