In the discussions about quantum correlations, particularly beyond entanglement (discord, dissonance e.t.c), one can often meet two definitions of mutual information of a quantum system $\rho^{AB}$: $$ I(\rho^{AB}) = S(\rho^A) + S(\rho^B) - S(\rho^{AB}) $$ and $$ J(\rho^{AB}) = S(\rho^A)-S_{\{\Pi^B_j\}}(\rho^{A|B}), $$ where $S$ is the Von-Neumann entropy, $\rho^A$ and $\rho^B$ are the reduced states of the individual subsystems of $\rho^{AB}$ and the second term in $J$ is the quantum analogue of the conditional entropy $$ S_{\{\Pi^B_j\}}(\rho^{A|B}) = \sum_j p_j S(\rho^{A|\Pi^B_j}). $$ In the expression for the conditional entropy $\rho^{A|\Pi^B_j} = \text{Tr}_B[\rho^{AB} (\mathbb{I}^A \otimes \Pi^B_j )]/p_j $ are the states of the subsystem $A$ after getting a particular projector $\Pi^B_j$ in $B$, which happens with a probability $p_j = \text{Tr}[\rho^{AB} (\mathbb{I}^A \otimes \Pi^B_j ) ]$. While $I$ characterizes the total correlations between $A$ and $B$ the second expression involves a measurement process, in which non-classical features of $\rho^{AB}$ are lost, and therefore $J$ characterizes classical correlations in $\rho^{AB}$.

While measuring $J$ is relatively straightforward, (for 2 qubits one can just measure 4 probabilities $p(\Pi^A_i \Pi^B_j), \, i,j = 1,2$ and calculate the mutual information of the resulting probability distribution) I can't think of an easy way of estimating $I$. So my question is: is it possible to measure $I$ without performing a full tomography of $\rho^{AB}$?

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    $\begingroup$ What do you mean "Measuring J is straightforward"? How do you measure it? (What do you precisely mean by "measuring" -- given N copies of the state, experimentally?) $\endgroup$ – Norbert Schuch Nov 19 '18 at 13:10
  • $\begingroup$ I can measure the frequencies of observing $\Pi_1^A \Pi_1^B$, $\Pi_1^A \Pi_2^B$, $\Pi_2^A \Pi_1^B$, $\Pi_2^A \Pi_2^B$ (for polarization qubits by counting coincidences between different polarizations) and form a classical 2x2 probability distribution out of them. Then, I can calculate the mutual information from this distribution. $\endgroup$ – Ilya Nov 19 '18 at 14:00

The mutual information can be written in terms of the relative entropy, please see Nielsen and Chuang (the entropy Venn diagram figure 11.2). I am writing the equation in the question's notation: $$I(\rho^{AB}) = S(\rho^{AB}|\rho^{A} \otimes \rho^{B})$$ The relative entropy can be estimated without full tomography. The procedure is described in Bengtsson and Życzkowski (equation 12.55-12.59) based on Lindblad's work:

The estimation procedure for the estimation of $S(\rho|\sigma)$ is performed as follows:

  1. Preparation of a composite system: $$\rho^N = \otimes^{N} \rho$$
  2. Measurement a set of POVMs $\{E\}$: $$p_i = \operatorname{Tr}(\rho^N E_i)$$ $$q_i = \operatorname{Tr}(\sigma^N E_i)$$
  3. Computation of the "Classical" relative entropy: $$S_N(\rho|\sigma) = \frac{1}{N} \sum_i p_i \log{\frac{p_i}{q_i}}$$

The relative entropy is estimated by optimization over a large set of POVMs and for a large number of copies $N$ due to the result: $$ S(\rho|\sigma) = \lim_{N\rightarrow \infty}\operatorname{Sup}_E S_N(\rho|\sigma) $$

Of course, as in any statistical estimation, there are estimation errors due to finite samples, however, I don't know how to obtain these error bounds.

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    $\begingroup$ how many copies $N$ are needed for this? Because if $N$ increases exponentially with $N$ then this is not so different than doing tomography on the state $\endgroup$ – glS Nov 19 '18 at 17:02
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    $\begingroup$ @glS "N increases exponentially with N" typo? $\endgroup$ – AHusain Nov 19 '18 at 20:53
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    $\begingroup$ yes sorry, of course. I meant increases exponentially with the number of qubits, that is, with the dimension of the system $\endgroup$ – glS Nov 19 '18 at 22:12
  • $\begingroup$ @glS I don't know of a general result comparing the computational complexities of the two schemes. Moreover, in the above scheme, one needs to select an optimal POVM which is not trivial either. However, in the full tomography option, one needs to estimate $2^M \times 2^M$ matrix elements of $\rho$ ($M$ – is the number of qubits), in addition there is a large $N$ hiding in the averaging required for each element estimation. There is a lot of literature on optimal POVM selection. $\endgroup$ – David Bar Moshe Nov 20 '18 at 9:09
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    $\begingroup$ @DavidBarMoshe "There is a lot of literature on optimal POVM selection" could you please point me in that direction? $\endgroup$ – Ilya Nov 20 '18 at 9:50

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