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

  • $\begingroup$ 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


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.

  • $\begingroup$ 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

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.