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

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


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 $O_{AF}=|1><1$.

  • $\begingroup$ 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. $\endgroup$ – Norbert Schuch Feb 27 '16 at 13:44
  • $\begingroup$ @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. $\endgroup$ – XXDD Feb 27 '16 at 16:51
  • 1
    $\begingroup$ 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 $\mathcal E(\rho) = |0\rangle\langle0|\,(\langle0|\rho|0\rangle +\langle1|\rho|1\rangle) + \tfrac12(|2\rangle\langle2|+|3\rangle\langle3|)(\langle2|\rho|2\rangle+\langle3|\rho| 3 \rangle)$. $\endgroup$ – Norbert Schuch Feb 27 '16 at 21:47
  • $\begingroup$ @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. $\endgroup$ – XXDD Feb 28 '16 at 1:49
  • $\begingroup$ @NorbertSchuch Thanks again for the beautiful example. $\endgroup$ – 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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.