3
$\begingroup$

Consider a bipartite state $\rho$, and let $\Pi^A\equiv \{\Pi^A_a\}_a$ and $\Pi^A\equiv\{\Pi^B_b\}_b$ be local projective measurements. Suppose that measuring $\rho$ in these bases gives uncorrelated outcomes. More precisely, this means that the probability distribution $$p(a,b)\equiv \operatorname{Tr}[(\Pi^A_a\otimes \Pi^B_b)\rho],\tag A$$ is uncorrelated: $p(a,b)=p_A(a)p_B(b)$ for some $p_A,p_B$.

I should stress that I'm considering this property for a fixed basis here, so this can easily hold regardless of the separability of $\rho$.

Given $\Pi^A,\Pi^B$, is there a way to characterise the set of $\rho$ producing uncorrelated measurement outcomes?

For example, if the measurements are rank-1 and diagonal in the computational basis, $\Pi^A_a=\Pi^B_a=|a\rangle\!\langle a|$, then any state of the form $$\sum_{ab} \sqrt{p_A(a) p_B(b)} e^{i\varphi_{ab}}|a,b\rangle\tag B$$ will give uncorrelated outcomes (assuming of course $p_A,p_B$ are probability distributions). These are product states for $\varphi_{ij}=0$, but not in general.

Nonpure examples include the maximally mixed state (for any measurement basis) and the completely dephased version of any pure state of the form (B).

Is there a way to characterise these states more generally? Ideally, I'm looking for a way to write the general state satisfying the constraints. If this is not possible, some other way to characterise the class more or less directly.

$\endgroup$
3
  • 1
    $\begingroup$ could mutually unbiased bases help you here? $\endgroup$
    – DaftWullie
    May 17 at 7:00
  • $\begingroup$ @DaftWullie mh... I'm not sure, can they? do you have something specific in mind? $\endgroup$
    – glS
    May 17 at 14:16
  • $\begingroup$ Well, for example, if I have Bell state and the two parties measure in mutually unbiased bases, their measurement results are uncorrelated. Now I guess you could generalise this. Imagine any state and Alice measures in any basis. The outcomes are some states $\{|\psi_i\rangle\}$ depending on Alice's outcome $i$. So, if Bob picked any basis that is mutually unbiased with respect to the $|\psi_i\rangle$, I guess you'd get what you want? So from that you might reverse engineer a valid set of conditions. $\endgroup$
    – DaftWullie
    May 17 at 14:42
1
$\begingroup$

Here's another example that works, essentially extending the rank-$1$ argument to arbitrary rank. Suppose all of the projectors within a set are orthogonal to each other. We can write each projector as $$\Pi^A_0=\sum_{i=0}^{n_0(A)} |i\rangle\langle i|,\quad \Pi^A_1=\sum_{i=n_0(A)+1}^{n_0(A)} |i\rangle\langle i|,\quad\cdots$$ for some integers $n_0<n_1<\cdots$ that depend on the fixed measurement basis and can differ between the two systems $n_i(A)\neq n_i(B)$. We can define a general eigenstate of each projection operator as $$|\phi_a^A\rangle=\sum_{i=n_a(A)}^{n_{a+1}(A)}\phi_i|i\rangle$$ for any amplitudes $\phi_i$ that could depend on $a$ and $A$, and similarly for system $B$. Then any state of the form $$\rho=\sum_k \rho_k|\Phi_k\rangle\langle\Phi_k|$$ with each component pure state taking the form $$|\Phi_k\rangle=\sum_{a,b}\Phi_{a,b} |\phi_a^A\rangle |\phi_b^B\rangle$$ will generate the measurement results $$p(a,b)=\sum_k \rho_k \langle\phi_a^A|\Pi^A_a|\phi_a^A\rangle\langle\phi_b^B|\Pi^B_b|\phi_b^B\rangle.$$ Each of $\rho_k$, $\langle\phi_a^A|\Pi^A_a|\phi_a^A\rangle$, and $\langle\phi_b^B|\Pi^B_b|\phi_b^B\rangle$ is positive, so I suspect $p(a,b)$ to only be uncorrelated when there is a single nonzero $\rho_k$. This means that one could only consider pure states that satisfy some strange condition that looks reminiscient of mixed-state separability. Thoughts?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.