Checking the Consistency of Local Density Matrices

1. Checking the Consistency of Local Density Matrices Yi-Kai Liu Computer Science and Engineering University of California, San Diego y9liu@cs.ucsd.edu

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 ρB, B = {2,3} • 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) ρA, A = {1,2}

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(Tiσ) = 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(Tiρ(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(Tiρ) = 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(Uiρ) 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> = –.6σx 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