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.

  • 1
    $\begingroup$ could mutually unbiased bases help you here? $\endgroup$
    – DaftWullie
    May 17, 2021 at 7:00
  • $\begingroup$ @DaftWullie mh... I'm not sure, can they? do you have something specific in mind? $\endgroup$
    – glS
    May 17, 2021 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, 2021 at 14:42

1 Answer 1


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?


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