Let a quantum system be given which has two subsystems $A$ and $B$ so that the Hilbert space decomposes $\mathscr{H}\simeq \mathscr{H}_A\otimes \mathscr{H}_B$.

If the state of the system is $\rho$, one defines the classical correlation with respect to measurements on $B$ as

$$J_{\overleftarrow{AB}}(\rho)=\max_{\{1\otimes \Pi_i\}}\{S(\rho_A)-\sum_i p_i S(\rho_A^i)\}$$

where the maximum is taken over all measuremenst on $B$ with probabilities $p_i$ and with post-measurement states $\rho^i$, and where $\rho_A = \operatorname{Tr}_B(\rho)/\operatorname{Tr}(\rho)$ and $\rho_A^i=\operatorname{Tr}_B(\rho^i)/\operatorname{Tr}(\rho^i)$.

The idea here is simple:

  1. Prior to measurement the state of $A$ is $\rho_A$ and the uncertainty on its specification is $S(\rho_A)$.

  2. If upon a specific measurement $\{1\otimes \Pi_i\}$ is conducted on $B$ one has obtained result $i$ the post-measurement state of $A$ is $\rho_A^i$ and the uncertainty on its specification is $S(\rho_A^i)$. Therfore, the uncertainty on the state of $A$ has reduced by $$S(\rho_A)-S(\rho_A^i).$$ As such, the mean information about the state of $A$ gained in the process of measurement is $$\sum_i p_i (S(\rho_A)-S(\rho_A^i))=\sum p_i S(\rho_A)-\sum p_i S(\rho_A^i)=S(\rho_A)-\sum p_i S(\rho_A^i).$$

  3. Finally, we may say that the maximum information we can obtain about $A$ by measuring $B$ is the maximum of such quantity over all measurements on $B$, which is $J_{\overleftarrow{AB}}(\rho)$.

That's fine, but why this is called "classical correlation" and not the total correlation?

I mean, for me correlation means the following

If whenever we observe $B$ we gain knowledge of $A$ then they are correlated.

The only way to observe $B$ is to measure something.

So why this quantity is classical correlation and not full correlation? In what sense quantum correlation differs from this and how then quantum correlations agree with the usual meaning of correlation?

  • $\begingroup$ can you add a reference for this usage of the term? $\endgroup$
    – glS
    Commented Feb 12, 2019 at 9:37
  • $\begingroup$ Sure, one such reference is arxiv.org/abs/1006.2460. Many of the cited papers there also use it. $\endgroup$
    – Gold
    Commented Feb 12, 2019 at 10:22
  • $\begingroup$ In a sense this adds to my doubt. The point is that it seems that in the context of quantum information this seems to be well established as "classical correlation" - as I said there are quite a few references with such terminology. I just don't get the idea behind this terminology. That's the reason for the question really. $\endgroup$
    – Gold
    Commented Feb 13, 2019 at 16:57
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    $\begingroup$ en.wikipedia.org/wiki/Quantum_discord - for an introduction to the distinction between classical and quantum correlations $\endgroup$
    – rnva
    Commented Feb 15, 2019 at 11:07

1 Answer 1


By "measure of correlation", in this context, we mean the mutual information $$I(X:Y)=H(X)+H(Y)-H(X,Y)=H(X)-H(X|Y).$$

Let us consider a few different classical scenarios and try to work out how this quantity should be computed for quantum states.

Full correlation

In the case in which observing $Y$ gives full information about $X$, we have $H(X|Y)=0$ and therefore the mutual information is maximal and equal to the total amount of information contained in $X$: $I(X:Y)=H(X)$. You can understand this as saying that, in this scenario, knowing $Y$ is the same as knowing $X$.

What is a quantum state which gives you the same effect? An easy example could be something like $$\newcommand{\ket}[1]{\lvert #1\rangle}\newcommand{\proj}[1]{\mathbb P(\,#1\,)}\newcommand{ketbra}[1]{\lvert #1\rangle\!\langle #1\rvert} \rho=\sum_i p_i \ketbra{i,i}.$$ As is readily verified, for such a state you have $\rho^A=\sum_i p_i\ketbra i$ and thus $S(\rho^A)=H(\boldsymbol p)$, and moreover, if $B$ measures in the computational basis, knowing the result of a measurement of $B$ fully determines the upcoming results of measuring $A$, so that $S(\rho^A|\rho^B)=0$ (again, in this choice of measurement basis).

Is this consistent with your definition of classical information via $J_{AB}(\rho)$? Yes it is, because with our choice of measurement we minimised the second term $\sum_i p_i S(\rho_i^A)$, by trivially having $S(\rho_i^A)=0$.

Finally, you might notice how taking the corresponding pure state $$\ket\psi=\sum_i \sqrt{p_i}\ket{i,i},$$ we get the same exact results. Now this might seem strange, as the correlations given by $\rho$ and $\ket\psi$ are clearly very different, but it is consistent with the fact that $J_{AB}$ is only measuring the classical correlations that can be given by these states.

More general scenario

Let us now consider a more general scenario in which $H(X|Y)\neq 0$. This means that knowing $Y$ is not enough to fully determine $X$, and thus $I(X:Y)<H(X)$.

A class of states that reproduces this type of correlation is for example $$\rho=\sum_i p_i \rho_i^A\otimes \ketbra i.$$ Again, $S(\rho_A)=H(\boldsymbol p)$, and $S(\rho_A|\rho_B)=\sum_i p_i S(\rho_A^i)$ can be any value depending on the choice of $\rho_A^i$.

Now, however, another problem arises: does this choice of measurement of $B$ maximise the mutual information between the observations? This is not obvious, as there could be another choice of measurement which makes $A$'s state collapse to a pure state, thus achieving larger correlations. Because we are looking for the maximum amount of classical correlation that can be obtained using the given state, it makes sense to define $J$ via the maximisation.

Sure, but why classical correlations?

Because there is nothing quantum about the mutual information measured this way.

What $J$ quantifies is the amount of correlation between the measurements results of $A$ and $B$, for a fixed measurement choice of $B$. This is the amount of correlation that can be used to implement a channel between $A$ and $B$ and transmit (classical) information.

If one restricts to this sort of scenario, there is no quantumness to be observed. Sure, we talked about "collapse" when discussing the results of measurements, but this sort of collapse is no different than the "collapse" of the probability distribution over $X$ induced by knowing that $Y=y$.

Saying it in yet another way, all correlations observable in such a scenario can be explained via local hidden variable models, by simply assuming the two parties to share some appropriate amount of (classical) correlation beforehand.

Quantum correlations, one the other hand, arise from the inability to explain the observation by simply assuming pre-shared correlations between the parties, and this can only be observed when different measurement bases are used, as otherwise it is not possible to observe the failure of classical models. The parameter $J_{AB}$ does not take this into account, and therefore cannot detect the "quantumness" of a given state.

  • $\begingroup$ I would add that it's a bit weird though that $J(\rho) = 1$ for both the maximally mixed state and the maximally entangled state. Surely, a measure of correlations ought to give you 0, at least for the maximally mixed state? $\endgroup$ Commented Feb 22, 2019 at 11:52
  • $\begingroup$ @user1936752 well, not really. For the max mixed state, you have a purely classical situation, with $X$ and $Y$ being fully correlated. The correlation is maximal because knowing $Y$ gives you full information about $X$, and thus $J(\rho)=I(X:Y)=H(\rho_A)=H(\vec p)$. The fact that you get the same value for the maximally entangled state is because you are not actually exploiting the entanglement of the state, and therefore you cannot see the "quantumness" at all. This is like $|0\rangle+|1\rangle\sim |0\rangle\!\langle0|+|1\rangle\langle1|$ if you only measure in the computational basis $\endgroup$
    – glS
    Commented Feb 22, 2019 at 12:20
  • $\begingroup$ Sorry, I was wrong. $J(\rho) = 0$ for max mixed state (the POVM is on B!). And this is correct since there are no correlations of any kind in the maximally mixed state. $\endgroup$ Commented Feb 22, 2019 at 12:58
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    $\begingroup$ According to the defintion of $J$, what you do is take a bipartite state. So here we have $I_{AB} = I_A\otimes I_B$. $S(\rho_A) = 1$ since the reduced state is also maximally mixed. According to the defintion, what you do next is measure system $B$ in the computational basis assuming the outcome is $i$, check $S(\rho_A^i)$. This is still maximally mixed since measuring in $B$ can't change the $A$ part. So for the maximally mixed state, $J(\rho) = 0$. If Alice and Bob hold a maximally mixed state, Bob can do whatever he wants with his state and there can be no effect on Alice's side. $\endgroup$ Commented Feb 22, 2019 at 13:52
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ Commented Feb 22, 2019 at 13:56

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