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Is there a unitary $U_{AB}$ such that, for any density operator $\rho$, we have $${\rm {Tr}}_A \left[U_{AB} \left(\frac{I_A}{2} \otimes \rho_B\right)U_{AB}^{\dagger}\right]= \frac{\rho_B}{2}+\frac{I_B}{4} \\ {\rm {Tr}}_B \left[U_{AB} \left(\frac{I_A}{2} \otimes \rho_B\right)U_{AB}^{\dagger}\right]= \frac{\rho_A}{2}+\frac{I_A}{4},$$ where $I_A=I_B=I$ is the identity matrix, $\rho_A=\rho_B=\rho$, A and B are both single qubit systems.

I have thought about decomposing $U_{AB}$ to express the partial trace but failed. I also considered about searching for such unitary, but $U_{AB}$ and $\rho$ are both unknown, which holds me back. Any ideas or comments, both in the analytic way or a computation way, would be appreciated.

It seems that symmetry can be used to derive the proof.


Cross-posted on math and physics

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  • $\begingroup$ How are you defining $\rho_{AB}$? $\endgroup$
    – Rammus
    Sep 18 at 19:36
  • $\begingroup$ It is not given. I have assumed that $\rho_{AB} = \rho_A \otimes \rho_B$. Then we can have $I_A/2=\rho_A=\rho_B=\rho=I/2$ according to the conditions. In this case, we can obtain the results above. But this is not what the original problem's intention. So the assumption $\rho_{AB} = \rho_A \otimes \rho_B$ is not right. $\endgroup$ Sep 19 at 6:52
  • $\begingroup$ @Michael.Andy is this a problem from somewhere? If so, can you link to the source? $\endgroup$
    – glS
    Sep 19 at 12:41
  • $\begingroup$ @glS: This problem is designed by a quantum computation researcher. So no links. $\endgroup$ Sep 19 at 12:45
  • $\begingroup$ @glS: What do you think about the definition of $\rho_{AB}$ in this problem? $\endgroup$ Sep 19 at 12:48

1 Answer 1

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I think the unitary is $\sqrt{\frac{1}{2}}I+\sqrt{\frac{1}{2}}iS$ where $S$ is swap operator such that $S|i\rangle\otimes|j\rangle=|j\rangle\otimes|i\rangle$, and have matrix form $S=\sum_{ij}{|ij\rangle \langle ji|}$. The original idea is to mix two unitary matrices, while generally not a unitary matrix, hence I add $i$ before $S$.

Mind that $$U_{AB}\frac{I}{2}\otimes {\rho U_{AB}}^{\dagger} \\ =\left( \sqrt{\frac{1}{2}}I+\sqrt{\frac{1}{2}}iS \right) \left( \frac{I}{2}\otimes \rho \right) \left( \sqrt{\frac{1}{2}}I+\sqrt{\frac{1}{2}}iS \right) ^{\dagger} \\ =\left( \sqrt{\frac{1}{2}}I+\sqrt{\frac{1}{2}}iS \right) \left( \frac{I}{2}\otimes \rho \right) \left( \sqrt{\frac{1}{2}}I-\sqrt{\frac{1}{2}}iS \right) \\ =\frac{1}{2}\frac{I}{2}\otimes \rho -\frac{i}{2}\frac{I}{2}\otimes \rho S+\frac{i}{2}S\frac{I}{2}\otimes \rho +\frac{1}{2}S\frac{I}{2}\otimes \rho S \\ =\frac{1}{2}\frac{I}{2}\otimes \rho +\frac{1}{2}\rho \otimes \frac{I}{2}+\frac{i}{2}S\frac{I}{2}\otimes \rho -\frac{i}{2}\frac{I}{2}\otimes \rho S$$ Then we only need to calculate $Tr_B$ of it, I only show that $Tr_B\left( \frac{i}{2}S\frac{I}{2}\otimes \rho -\frac{i}{2}\frac{I}{2}\otimes \rho S \right) =0$, the residual part is easy to show. This part may be calculated with tensor graph type method while I am not familiar with it, so I directly expand the index as follows: $$Tr_B\left( \frac{i}{2}S\frac{I}{2}\otimes \rho -\frac{i}{2}\frac{I}{2}\otimes \rho S \right) \\ =Tr_B\left( \frac{i}{2}\sum_{ij}{|ij\rangle \langle ji|}\frac{\sum_k{|k\rangle \langle k|}}{2}\otimes \sum_{mn}{\rho _{mn}|m\rangle \langle n|}-\frac{i}{2}\frac{\sum_k{|k\rangle \langle k|}}{2}\otimes \sum_{mn}{\rho _{mn}|m\rangle \langle n|}\sum_{ij}{|ij\rangle \langle ji|} \right) \\ =Tr_B\left( \frac{i}{4}\sum_{ijn}{\rho _{in}|ij\rangle \langle jn|}-\frac{i}{4}\sum_{ijm}{\rho _{mj}}|im\rangle \langle ji| \right) \\ =\frac{i}{4}\sum_{ij}{\rho _{ij}|i\rangle \langle j|}-\frac{i}{4}\sum_{ij}{\rho _{ij}}|i\rangle \langle j|=0.$$

As for $Tr_A$, I think the calculation should be the same.

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  • $\begingroup$ Great! How do you come up with the idea of constructing such a unitary matrix? $\endgroup$ Sep 20 at 16:00
  • $\begingroup$ @Michael.Andy Mix $I$ and $S$ with probability. While adding them up directly I didn't get a unitary, so I added an $i$ in front of $S$. $\endgroup$
    – narip
    Sep 20 at 23:57
  • $\begingroup$ @Michael.Andy By setting $U=I$ and $U=S$, we get partial trace is $\rho$ and $I/2$, hence a mix of them two will get the right answer, while setting $U$ in the answer form, there will be 4 terms so I hope the two terms that I don't want can cancel them out. Luckily they did. $\endgroup$
    – narip
    Sep 21 at 2:28
  • $\begingroup$ Great! Thank you narip! $\endgroup$ Sep 22 at 12:05
  • $\begingroup$ Could you recommend some related books to me? $\endgroup$ Sep 22 at 12:07

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