$\newcommand{\expect}[1]{\langle \hat{#1}\rangle}$ Let me start with thoughts I had so far; there is a TLDR with my question at the bottom.

Imagine a pure system $\rho$ and with a bipartite subsystem $\rho^{AB}$. Now imagine you want to calculate the correlation function $\langle \hat{A} \hat{B}\rangle$, where $\hat{A}$ ($\hat{B}$) exclusively acts on system $A$ ($B$). Now comes the the fiddly part: You only have access to $\rho^A$ and $\rho^B$.

  1. If one of $\rho^A$ or $\rho^B$ is a pure state we may deduce that $\rho^{AB} = \rho^A \otimes \rho^B$. Consequently $\langle\hat{A}\hat{B}\rangle = \expect{A}\expect{B}$. Notice, that the mutual information in this case is $S(A:B) = 0$.

If, on the other hand, both $\rho^A$ and $\rho^B$ are mixed we can discuss multiple scenarios.

  1. In the simplest case $\rho^{AB} = \rho^A \otimes \rho^B$ and the correlator factors out as above. Again $S(A:B) = 0$.

  2. Next, if $\rho^{AB} \ne \rho^A \otimes \rho^B$ there is no direct deduction. As an extreme example consider the Bell state $(|\uparrow\uparrow\rangle + |\downarrow\downarrow\rangle)/\sqrt{2}$ and $\hat{A} = \sigma_z^A$, $\hat{B} = \sigma_z^B$. Then $\rho^A$ and $\rho^B$ are the completely mixed states and we find $1 = \langle\hat{A}\hat{B}\rangle \ne \expect{A}\expect{B} = 0$. Notice that here the mutual information reaches its maximal value $S(A:B) = S(A) + S(B)$ (generally $S(A:B) \le 2 \min(S(A),S(B))$, where $S(A)$ denotes the von Neumann entropy).

In the last example, we discussed the extrem case in which both $\rho^A$ and $\rho^B$ are the maximally mixed states. To give the final motivation to my question consider now the more well-behaved case of the separable state

$$ \rho^{AB} = p\rho_1^A\otimes\rho_1^B + (1-p)\rho_2^A\otimes\rho_2^B $$

with orthogonal (possibly still mixed) states $\rho_1^{A,B} \perp \rho_2^{A,B}$. After some calculations we arrive at

$$ |\langle\hat{A}\hat{B}\rangle - \expect{A}\expect{B}| \le S(A:B) \left|\left[\frac{\expect{A}_1 - \expect{A}_2}{2} \right] \left[\frac{\expect{B}_1 - \expect{B}_2}{2} \right]\right|, $$

where $\expect{A}_i = \mathrm{Tr}[\hat{A} \rho_i]$ and similar for $\expect{B}_i$.

Finally, observe that an inequality of the form holds $|\langle\hat{A}\hat{B}\rangle - \expect{A}\expect{B}| \le c S(A:B)$, $c \in \mathbb{R}$ holds for all examples (1. to 3.) from above.

Question - TLDR

Can we estimate / bound a correlation function $\langle\hat{A}\hat{B}\rangle$ of a bipartite system by $\expect{A}$, $\expect{B}$ and some measure of entanglement, like the mutual information.

More precisely, I hope to find a relation along the line

$$ S(A:B) f_1(\rho^A, \hat{A}, \rho^B, \hat{B}) \le |\langle\hat{A}\hat{B}\rangle - \expect{A}\expect{B}| \le S(A:B) f_2(\rho^A, \hat{A}, \rho^B, \hat{B}), $$ but maybe $S(A:B)$ is also part of $f_1$ and $f_2$.

Footnote: $S(A:B) = S(\rho^{AB} || \rho^A\otimes\rho^B)$, where $S(\rho^{AB} || \rho^A\otimes\rho^B)$ is the relative entropy between $\rho^{AB}$ and $\rho^A\otimes\rho^B$, to give another point of view.

  • 1
    $\begingroup$ Are you aware of arxiv.org/abs/1206.2947? $\endgroup$ Commented Feb 1, 2021 at 18:11
  • $\begingroup$ @NorbertSchuch No, I wasn't. But this looks definitely worth for investigation; thanks! $\endgroup$
    – manthano
    Commented Feb 1, 2021 at 19:57

1 Answer 1


For one direction, you can use the bound $$ S(\rho\|\sigma)\ge \tfrac12\|\rho-\sigma\|_1^2\ , $$ which implies \begin{align} S(A:B)&=S(\rho^{AB}\|\rho^A\otimes \rho^B)\\ &\ge \tfrac12\|\rho^{AB}-\rho^A\otimes \rho^B\|_1^2\\ &\ge\tfrac12\big\lvert \langle\hat A\hat B\rangle -\langle\hat A\rangle\langle\hat B\rangle\big\rvert^2 \end{align} for operators $\hat A$, $\hat B$ with operator norm bounded by one.

  • $\begingroup$ I should say that I don't think there will be a comparable bound the other way round, since the entropy function is just too steep around 0. (So any bound would have to reproduce this x log x scaling.) $\endgroup$ Commented Feb 1, 2021 at 21:10

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