# An augmented version of the optimization problem

In my last question, I proposed the following 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 $U_{AB}$)?

I was wondering if the maximally mixed state will always be a solution. Thank for the help from Martin and Norbert Schuch, now we know this is not the case.

Norbert Schuch's example: A and B are qubits, $U_{AB}=|(00+11)/\sqrt{2}><00|+|(01+10)/\sqrt{2}><01+|10><10|+|11><11|$, then we know that for $\rho_B=|1><1|$, optimal $\rho_A=|0><0|$ but not $I/2$.

Now I would like to augment the problem as follows:

If the operation is repeated, which means we take the output of subsystem A, $O_A$, as the new input and iterate the operation till the system converge to a final output $O_{AF}$, then what's the optimal initial input $\rho_A$ that will lead to a $O_{AF}$ with the maximal entropy? Will the maximally mixed state always be a solution?

Note: There are cases that the iterative operation will not converge, but I'd like to believe this is relatively rare.

PS: It can be verified that the maximally mixed state of A is a solution of my new augmented version with the configuration of Norbert's example since any input $\rho_A$ will finally converge to the same output $O_{AF}=|1><1$.

• As long as the channel has a unique fixed point (which is generically the case), any initial state (i.e., positive semi-definite operator) which has full rank will converge to it. – Norbert Schuch Feb 27 '16 at 13:44
• @NorbertSchuch Thanks. If the fixed point is unique, then my problem is trivial. In fact I am interested in the case that there are multiple fixed states, then I am curious if the initial state as a maximally mixed state will converge to the output with a maximal entropy. – XXDD Feb 27 '16 at 16:51
• I have to admit that I find your habit of changing your questions on the go somewhat annoying. -- As to your new question, the answer is no: Consider e.g. a $4$-level system and a channel $\mathcal E(\rho) = |0\rangle\langle0|\,(\langle0|\rho|0\rangle +\langle1|\rho|1\rangle) + \tfrac12(|2\rangle\langle2|+|3\rangle\langle3|)(\langle2|\rho|2\rangle+\langle3|\rho| 3 \rangle)$. – Norbert Schuch Feb 27 '16 at 21:47
• @NorbertSchuch First thanks for the example, I will study it. But this time I did not change my question, since that's exactly my question (more than 1 fixed point depending on different initial $\rho_A$). Otherwise if there is a unique fixed point, I will not ask which $\rho_A$ leads to the maximal entropy output. – XXDD Feb 28 '16 at 1:49
• @NorbertSchuch Thanks again for the beautiful example. – XXDD Feb 28 '16 at 2:41

The maximally entangled state will not always lead to the output with the maximum entropy. Consider e.g. a $4$-level system and a channel $$\mathcal E(\rho) = |0\rangle\langle0|\,(\langle0|\rho|0\rangle +\langle1|\rho|1\rangle) + \tfrac12(|2\rangle\langle2|+|3\rangle\langle3|)(\langle2|\rho|2\rangle+\langle3|\rho| 3 \rangle)$$

On the maximally mixed state, this will have give an output $$\tfrac12|0\rangle\langle0|+ \tfrac14(|2\rangle\langle2|+|3\rangle\langle3|)$$ while the fixed point with maximal entropy is $$\tfrac13(|0\rangle\langle0|+ |2\rangle\langle2|+|3\rangle\langle3|)$$ (which obviously can be reached with itself as an input).