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Many properties of entropic quantities are shown by resorting to related properties of the relative entropy of suitable quantities. For instance, subadditivity of entropy may follow from non negativity of relative entropy by observing that

$$H(A)_{\rho^{A}}+H(B)_{\rho^{B}}-H(A, B)_{\rho^{AB}}=D\left(\rho^{AB} \vert\vert \rho^A\otimes \rho^{B}\right)\,,$$

or concavity of conditional entropy from convexity of relative entropy by observing that

$$H(A\vert B)_{\rho^{AB}}=-D\left(\rho^{AB} \vert\vert I_A\otimes \rho^{B}\right)\,.$$

To obtain such identities, the following formula

$$\log\left( P\otimes Q\right)= \log P\otimes I_B+I_A\otimes \log Q$$

is often invoked. Here I assume the logarithm of a positive definite operator to be defined by means of the spectral decomposition; the eignevalues of a positive definite operator are all positive, so their logarithms (as positive real numbers) make sense. Essentially, if $P=\sum_p p \vert p\rangle$, then $\log P=\sum_p \log p \vert p\rangle$.

However, what happens when $P$ and $Q$ are positive semidefinite but not necessarily definite? This is especially relevant in the context of entropic quantities, where a density matrix may have the zero eigenvalue. Each book I have consulted makes [possibly – see this comment for some "refs"] use of $\log \left( P\otimes Q\right)= \log P\otimes I+I\otimes \log Q$ . How do we justify the use of such formula for $P$ and $Q$ semidefinite?

PS: In the semidefinite case I'd expect the identity operator to be replaced by a projection on the support, i.e., $\log \left( P\otimes Q\right)= \log P\otimes \Pi_Q+\Pi_P\otimes \log Q$. On the other hand, I can't see different well reputed textbooks to be wrong on the same point. Should, by any chance, my guess make sense, there would be the problem of deriving expressions such as $H(A: B)=D\left(\rho^{AB} \vert\vert \rho^A\otimes \rho^{B}\right)$, or $H(A\vert B)=-D\left(\rho^{AB} \vert\vert I_A\otimes \rho^{B}\right)$. Is there some trick when we take partial traces?

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    $\begingroup$ The key is that for those entropic quantities you end up multiplying operators of the form $\rho\log \sigma$ for some density operators $\rho,\sigma$ and then use the same logic as why $x\log x\to 0$. For relative entropy you actually allow for divergent answer if the supports of the second entry is smaller than the first. $\endgroup$ Commented Apr 10 at 16:03
  • $\begingroup$ Thanks a lot. I see your point on the support of the entries of the relative entropy. Unfortunately, I still cannot justify all the steps in the calculations. To be concrete, let us consider $D\left(\rho^{AB}\vert\vert \rho^A\otimes \rho^B\right)$. First: is the support of $\rho^{AB} $ always no larger than that of $\rho^A\otimes\rho^B$? Then, if/when it is, how do we go from ${\rm tr}_{AB} \left(\rho^{AB}\log( \rho^A\otimes \rho^B)\right)$ to ${\rm tr}_A \left( \rho^{A}\log\rho^A\right)+ {\rm tr}_B\left( \rho^{B}\log\rho^B\right)$? $\endgroup$
    – atlantropa
    Commented Apr 10 at 16:59
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    $\begingroup$ where are you saying they use $\log(P\otimes Q)=\log P\otimes I+I\otimes \log Q$? Because I'm not seeing it eg in Nielsen&Chuang, nor Nielsen's paper about strong subadditivity (arxiv.org/abs/quant-ph/0408130), nor Watrous' book, nor Preskill's notes (though admittedly, I might be missing the statement you're looking at). Anyway, yes, it's clearly wrong in general, you see it easily taking eg $P=|0\rangle\!\langle 0|$, $Q=I/2$. $\endgroup$
    – glS
    Commented Apr 10 at 17:08
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    $\begingroup$ @atlantropa One way to justify the steps I guess is to "perturb" the density operators by some small parameter, which will make both full rank, use the known result for positive-definite case, and then take the limit as the perturbation vanishes. For reasons I mentioned it will work out because of the "$x\log x$"-type intuition. However, $\log(P\otimes Q)$ alone will fail as gIS mentioned because log of zero eigenvalues is ill-defined. $\endgroup$ Commented Apr 10 at 17:14
  • $\begingroup$ Thanks a lot to both. Refs for the question: Wilde asks to prove that statement in ex 11.8.1, asking semi-definiteness. Nielsen and Chuang maybe use it in corollary 11.13. As to Watrous, see (5.108); however, that formula is stated in the introduction for definite operators. Watrous takes into account the problem of images/supports, so I suspect he is doing things in the right way. It seems to me that everything would be solved if we had that the image of a density matrix is within the image of the tensor product of its reduced density matrices. Any suggestion/hint in this direction? $\endgroup$
    – atlantropa
    Commented Apr 10 at 17:57

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