# Motivation

$$\newcommand{\expect}[1]{\langle \hat{#1}\rangle}$$ 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.

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

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.