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I am considering such a problem:

(1) Given a finite dimensional composite system AB whose initial state is a product state of A and B so that $\rho_{AB}=\rho_A\otimes \rho_B$.

(2) Assuming AB undergoes a joint unitary operation $U_{AB}$ on AB, and the output of system A is given by

$O_A=Tr_B (U_{AB}\rho_{AB}U_{AB}^{+})$

Question:

What's the initial state $\rho_A$ that will result in an $O_A$ with a maximal Von Neumann entropy (for a given $\rho_B$ and $U_{AB}$)?

It seems that $\rho_A=I/n$ (n is the dimension of system A) might be the answer (at least for some special cases I tried). But I have no idea if it's an universal answer.

This reminds me of the channel capacity problem of classical information theory (some similarities if $U_{AB}$ is regarded as a channel). I am not sure if this is a solved problem in quantum information field. Thanks.

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    $\begingroup$ Just a very small remark. In general QM systems (i.e. set in possibly infinite dimensional Hilbert spaces), the identity operator - I am assuming $I$ is the identity operator - is not an allowed state (it is not trace class). Since finite dimensional spaces are usually taken as a toy model, it is not a good idea to use states there that are not such in realistic infinite dimensional systems. $\endgroup$ – yuggib Feb 25 '16 at 15:24
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    $\begingroup$ Please be aware that whatever your choice of $\rho_A,$ it needs to fulfill $\text{Tr}{\rho_A}=1$. Furthermore note that under unitary evolution of subsystems, say for system $A,$ the quantum relative entropy does not change, i.e. given $\rho_A$ (before) and $\rho'_A$ (after): $D(\rho_A ||\rho'_A)=\text{constant},$ thus the answer is independent of both the choice of $U_{AB}$ and $\rho_B$ (as you initiate them in a product state). $\endgroup$ – Phonon Feb 25 '16 at 15:33
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    $\begingroup$ @X.Dong Please read my comments carefully, you cannot set $\rho_A$ to identity as it needs to satisfy $\text{Tr}\rho_A=1.$ $\endgroup$ – Phonon Feb 25 '16 at 15:55
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    $\begingroup$ To reopen this question please make the title question match the question in the main body. Note that res. recom. q + tag cannot be combined with an actual physics question. (Conversely, any actual physics question is always implicitly a plea for res. recom.) $\endgroup$ – Qmechanic Feb 25 '16 at 16:03
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    $\begingroup$ Note that the problem at hand -- maximizing the output entropy -- can be solved by convex optimization methods (you are trying to maximize a concave function). The converse problem -- minimizing the output entropy -- is generally hard and relevant in the context of channel capacities in quantum information. $\endgroup$ – Norbert Schuch Feb 26 '16 at 13:23
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The answer will obviously depend very much on $\rho_B$ and $U_{AB}$.

Let's first assume that the output state with maximal entropy is the maximally mixed state.

First case: This might not be possible. In particular, there are channels that will map every state to a pure state. Here is one that you could already have known from my answer to your previous question: As $U_{AB}$ you take the swap operator $U_{AB}:=\sum_{ij} |ij\rangle \langle ji|$ and as $\rho_B$ an arbitrary pure state. Then $O_A=\rho_B$ always.

Second case: However, large classes of channels are not of this form. In particular, there are a lot of interesting channels that are unital, meaning that the maximally mixed state will map to itself. This is true, because a lot of channels (such as many noise channels or all channels without an environment) are entropy-increasing and clearly, entropy increasing channels have to map the maximally mixed state to itself. It doesn't have to be the only state that gets mapped to the mixed state - it could be many of them or even all. For instance, consider the channel where $\rho_B$ is maximally mixed and $U_{AB}$ is the swap operation.

Third case: There are also channels that are not unital but still map some state to the maximally mixed state. Norbert Schuch in the comments gave a very simple construction: You let $|0\rangle\mapsto I/2$ the maximally mixed state and $|1\rangle\mapsto |1\rangle$.


However, I misunderstood your question and you just wanted the maximal entropy possible per channel. This quantity is known as the maximum output entropy. Still, you will not be able to say so easily which state will achieve the maximum output entropy, because the third example is supposed to tell you that even in the case where that entropy is maximal, it's not always clear which state achieves it.

You are right to think that this has something to do with channel capacities: If you take a channel, the maximum output entropy is a (trivial) upper bound to the classical channel capacity, i.e. the maximum rate at which a quantum channel can send usual classical information.

Given a quantum channel $T$, this capacity is given by

$$ C(T)= \limsup_{n\to \infty} \chi(T^{\otimes n})/n $$

where $\chi$ is the Holevo quantity:

$$ \chi(T)=\max_{p_i} S(T(\sum_{i}p_i\rho_i))-\sum_i p_i S(\rho_i) $$

where $\rho_i$ are the possible messages (see Peter Shor's article for an overview).

Clearly, this is upper bounded by the maximum entropy (just leave out the second part of the term and don't take the limit).

The bound I believe is mostly not very tight and sometimes referred to as "trivial upper bound". In particular, all unital channels will have maximal output entropy given to the maximum possible von Neumann entropy. In these cases, it will not be very useful to know the maximum output entropy in order to say something about the channel capacity without also knowing the minimum output entropy, which is a much more interesting object: it tells you something about how much noise the channel introduces. If this is high, no matter what you do, all output states will be close to maximally mixed, hence the capacity will probably not be very high.

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  • $\begingroup$ For the third case, you can simply take a channel which maps the $|0\rangle$ state to the maximally mixed state and e.g. all other basis states to themselves. $\endgroup$ – Norbert Schuch Feb 25 '16 at 22:51
  • $\begingroup$ @Martin Thanks a lot. If we change a little bit of the question, not to 'find' the initial state of A, but ask: Will the maximally mixed state always be one (but not necessarily the only one) solution? Then can this be true? $\endgroup$ – XXDD Feb 26 '16 at 1:00
  • $\begingroup$ @Martin Another thing about your answer, I think my question is not the same as 'give an arbitrary channel and ask for the state that gets mapped to the maximally mixed state', since it might be the case that for some $\rho_B$ and $U_{AB}$, no input of $\rho_A$ can be mapped to maximally mixed state. So what I mean by 'maximal entropy' of $O_A$ is only under the constraints of $\rho_B$ and $U_{AB}$, but $O_A$ achieving this goal is not necessarily to be maximally mixed state. So your first case, maximally mixed state of A as an input is a solution. $\endgroup$ – XXDD Feb 26 '16 at 1:06
  • $\begingroup$ @NorbertSchuch Do you mean such a case: A,B are qubits, $U_{AB}$ is given by $|00+11><00|+|01+10><01|+|10><10|+|11><11|$ so that on A, it maps $|0> $to maximally mixed state and leave $|1>$ untouched. Then in this model, if $\rho_B=|1><1|$, the answer of $\rho_A$ should be $|0><0|$ but not $I/2$? $\endgroup$ – XXDD Feb 26 '16 at 1:32
  • $\begingroup$ @NorbertSchuch I think you are right, thanks for the example. $\endgroup$ – XXDD Feb 26 '16 at 1:43

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