1 / 14

Checking the Consistency of Local Density Matrices

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

tammy
Download Presentation

Checking the Consistency of Local Density Matrices

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related