I'm having some difficulty understanding the physical meaning of the mutual information that two subsystems share with each other. For example, if $\rho_{AB}$ defines the matrix of a bipartite state, where where A is for the Hilbert space $\mathcal{H}_{A}$, B for the Hilbert space $\mathcal{H}_{B}$ and $\mathcal{H} = \mathcal{H}_{A}\otimes\mathcal{H}_{B}$ is the tensor product space where the density $\rho_{AB}$ matrix acts, so the mutual information is defined as:

$$I(A,B) = S(B) - S(B|A)$$

$S(B)$ is for the Von Neumann entropy of the reduced density matrix $\rho_{B}$ and $S(B|A)$ is the entropy of reduced density matrix $\rho_{B}$ conditioned to a measure of some operator that acts in $\mathcal{H}_{A}$.

I already did some calculations using certain states, and and I understood the equation. But I realized I didn't understand the physics behind.

  • $\begingroup$ Do you understand the classical notion of mutual information? $\endgroup$ May 27 at 14:58

1 Answer 1


Classical mutual information

First of all, one oughts to keep in mind the "physical" intuition behind the classical mutual information. Roughly speaking, given a joint probability distribution $p(x,y)$, its mutual information $$I(X:Y)\equiv H(X)+H(Y)-H(X,Y)= H(X) - H(X|Y)$$ quantifies the amount of correlations between the parties. For example, $I(X:Y)=0$ for disjoint distributions with no correlations, and $I(X:Y)=H(X)=H(Y)$ for fully correlated distributions.

A more precise operative way to understand it is in terms of achievable transmission rates. A classical transmission channel is essentially a conditional probability distribution $p(y|x)$. Given such a channel and a probability distribution for the "inputs", write it as $p(x)$, the mutual information of the associated joint probability distribution, $p(x,y)=p(x)p(y|x)$, tells you how many different input sequences whose letters are sampled from $p(x)$ can be transmitted and decoded with vanishing error probability through the channel. Or better said, that using the channel $n$ times, and using as input sequences typical according to $p(x)$, then roughly $2^{n I(X:Y)}$ of these sequences can be sent through the channel and recovered with vanishing error probability for $n\to\infty$. This comes essentially from one possible argument used to prove Shannon's noisy channel coding theorem.

Quantum mutual information

Having said that, let's jump to the quantum case. Here things get considerably more subtle. The quantum mutual information of a bipartite state $\rho$ is defined as $$I(\rho) \equiv S(\rho_A) + S(\rho_B) - S(\rho),$$ where $S(\rho)$ is the von Neumann entropy, and $\rho_A\equiv\operatorname{Tr}_B(\rho),\rho_B\equiv \operatorname{Tr}_A(\rho)$ are the partial states obtained partial tracing from $\rho$. One might naively think that this quantifies correlations between the two sides of the system, but this is not as straightforward as it might seem.

For example, for any pure state $\rho\equiv \mathbb{P}_\psi$, where $\mathbb{P}_\psi\equiv|\psi\rangle\!\langle\psi|$, computing the above gives double the amount of mutual information one might have expected. If you consider say a maximally entangled two-qubit state, $|00\rangle+|11\rangle$, you could think that the two parties are fully correlated, and you should thus get $I=1$. But in fact, computing the QMI gives you $I(\rho)=2.$ Contrast this with what you get for the corresponding classically correlated state $\rho=\frac12(\mathbb{P}_0+\mathbb{P}_1)$, which gives $I(\rho)=1$. The difference can be traced back to the fact that a fully entangled state contains more correlations than the ones that can be observed measuring both sides in the computational basis.

As the above example suggest, the QMI does not just tell you about accessible correlations. In fact, arguably, the QMI cannot be straightforwardly interpreted as quantifying "correlations" between the two parties, at least not in the sense of correlations that are observables by simply performing local measurements in a specific basis and comparing results. It also contains a component that cannot be directly observed by local measurements on the two sides, the so-called quantum discord.

This issue is not strictly due to entanglement either. Separable states can have nonvanishing discord as well. For example, $$\rho = \frac12(\mathbb{P}_0\otimes\mathbb{P}_0 + \mathbb{P}_1\otimes\mathbb{P}_+)$$ is a separable two-qubit state which nonetheless has nonzero discord (from right to left), because there is no way to measure the second party without the measurement introducing some disturbance.


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.