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

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}

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)

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

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

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

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

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))

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

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

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.

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).

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.

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