# How to obtain the unitary operator to get specific partial trace?

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

• How are you defining $\rho_{AB}$? Sep 18, 2022 at 19:36
• 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. Sep 19, 2022 at 6:52
• @Michael.Andy is this a problem from somewhere? If so, can you link to the source?
– glS
Sep 19, 2022 at 12:41
• @glS: This problem is designed by a quantum computation researcher. So no links. Sep 19, 2022 at 12:45
• @glS: What do you think about the definition of $\rho_{AB}$ in this problem? Sep 19, 2022 at 12:48

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.
• @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 $S$. Sep 20, 2022 at 23:57
• @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. Sep 21, 2022 at 2:28