A composite system AB, its initial state is a product state of A and B given by $(I_A\otimes \rho_B)$ (A in a completely mixed state). The composite system undergoes a unitary operation $U_{AB}$. My calculation seems to show that we have

$$\mathrm{Tr}_{B}(U_{AB}(I_A\otimes \rho_B)U_{AB}^{+})=I_A$$

since if the dimensions of A and B are m and n respectively, then we have:

$$I_A\otimes \rho_B = \left[ {\begin{array}{cccc} \rho_B & & & \\ & \rho_B & & \\ & & \ddots& \\ & & & \rho_B \end{array} } \right] ,\quad U_{AB} = \left[ {\begin{array}{cccc} u_{11} & u_{12} & \cdots & u_{1m} \\ u_{21} & u_{22} & \cdots & u_{2m} \\ \vdots & \vdots & \ddots & \vdots \\ u_{m1} & u_{m2} & \cdots & u_{mm} \\ \end{array} } \right]$$

where $u_{ij}$ is a $n \times n$ matrix

Then $O_{AB}=U_{AB}(I_A\otimes \rho_B)U_{AB}^{+}$ is given by

$$O_{AB} = \left[ {\begin{array}{cccc} o_{11} & o_{12} & \cdots & o_{1m} \\ o_{21} & o_{22} & \cdots & o_{2m} \\ \vdots & \vdots & \ddots & \vdots \\ o_{m1} & o_{m2} & \cdots & o_{mm} \\ \end{array} } \right],$$

where $o_{ij}=\sum_{k}u_{ik}\rho_B u_{jk}^{+}$, and the $(i,j)$ item of $\mathrm{Tr}_{B}(U_{AB}(I_A\otimes \rho_B)U_{AB}^{+})$ is given by $\mathrm{Tr}(o_{ij})$.

And due to the fact that $U_{AB}$ is unitary, so $\sum_{k}u_{ik}u_{jk}^{+}=\sigma_{ij}I_{n\times n}$. This leads to

$Tr(o_{ij})=Tr(\sum_{k}u_{ik}\rho_B u_{jk}^{+})=Tr(\sum_{k}u_{jk}^{+}u_{ik}\rho_B)=Tr((\sum_{k}u_{jk}^{+}u_{ik})\rho_B)=\sigma_{ij}$

we can easily get

$$\mathrm{Tr}_{B}(U_{AB}(I_A\otimes \rho_B)U_{AB}^{+})=I_A$$

This means that starting from a completely mixed state $I_{A}$, subsystem A will stay in a completely mixed state no matter how it's jointly evolved with another system B if A and B are in an initial product state.

Is my calculation correct? Or I made a mistake somewhere?

PS: Finally I found I am wrong with my calculation. Please refer to my own comments on my question.

  • 2
    $\begingroup$ It seems to me that you are trying to use the property of the trace $\mathrm{Tr}(XY)=\mathrm{Tr}(YX)$, but you can't do that on the partial trace $\mathrm{Tr}_B(X_{AB}Y_{AB})$ since $Y_{AB}$ (and $X_{AB}$) are still operators with an $A$-dependence (and obviously do not commute) $\endgroup$ – yuggib Feb 25 '16 at 9:57
  • $\begingroup$ No, I am not using $Tr(XY)=Tr(YX)$ on partial trace. In fact I only used it in computing $Tr(o_{ij})$. $\endgroup$ – XXDD Feb 25 '16 at 11:10

This was the start of an answer but Martin got ahead of me, but the perspective might still be worth adding.

As was mentioned in the comments, the calculation looks ripe for application of the cyclicity of the trace, but this no longer holds for partial traces. There, instead, you have \begin{align} {}_A⟨i|\mathrm{Tr}_B(XY)|j⟩_A & = \sum_{klm}⟨ik|X|lm⟩⟨lm|Y|jk⟩ \\ & = \sum_{klm}⟨lm|Y|jk⟩⟨ik|X|lm⟩ \\ & = \mathrm{Tr}\left(Y|j⟩_A\!⟨i|X\right),\\ \text{so}\quad \mathrm{Tr}_B(XY) & =\sum_{ij}|i⟩\mathrm{Tr}\left(Y(|j⟩⟨i|\otimes I)X\right)⟨j| \end{align} which does not simplify.

In particular, applying this to $X=U$ and $Y=(I\otimes\rho) U^\dagger$, gets you the matrix element \begin{align} {}_A⟨i|\mathrm{Tr}_B(U(1\otimes\rho)U^\dagger) |j⟩_A & = \mathrm{Tr}\left((1\otimes\rho)U^\dagger(|j⟩⟨i|\otimes I)U\right), \end{align} and this makes it much easier to look for counterexamples, by concentrating on unitaries for which $$U^\dagger(|j⟩⟨i|\otimes I)U$$ will be nontrivial.

In particular, Martin's choice of swap is the cleanest, and it gives you $$U^\dagger(|j⟩⟨i|\otimes I)U=I\otimes |j⟩⟨i|,$$ in which case \begin{align} {}_A⟨i|\mathrm{Tr}_B(U(1\otimes\rho)U^\dagger) |j⟩_A & = \mathrm{Tr}\left(\rho|j⟩⟨i|\right)=⟨i|\rho|j⟩, \end{align} i.e. $\mathrm{Tr}_B(U(1\otimes\rho)U^\dagger)=\rho$ as in Martin's answer.


Yuggib already pointed out one possible fault in your reasoning: You cannot use the cyclicity of the trace, because this is the partial trace.

However, I'm not completely convinced that this is what you are doing as your explanation is not really clear. In particular, I believe your error lies (as ever so often) in the

And due to the fact that $U_{AB}$ is unitary, we can easily get

I just don't see where the unitarity of $U$ helps you unless you take some partial trace or believe that if $U$ is unitary, also the submatrices $u$ are unitary, which is false.

To give you an easy answer as to why your result cannot be correct:

Suppose the two systems A and B have the same (finite) dimension. Let $U$ be the swap-operation, i.e. $U=\sum_{ij} |ij\rangle \langle ji|$. This is clearly unitary since $U^{\dagger}U=\sum_{ij}|ij\rangle \langle ij|$ and what it does is it swaps the two systems or in other words $U(I_A\otimes \rho_B)U^{\dagger}=\rho_B\otimes I_A$. Clearly, $tr_B(\rho_B\otimes I_A)=\rho_B$ contradicting your result.

  • $\begingroup$ Nice and simple counterexample ;-) Also to me it was not so clear what the OP was inferring from unitarity (and so I assumed he was using the ciclicity in some sense). $\endgroup$ – yuggib Feb 25 '16 at 10:48
  • $\begingroup$ Yes, I thought that was what you were thinking. I didn't want to imply you were sure that was the answer (otherwise I guess you could have written an answer). The counterexample is one of my favourites and has helped me a lot. I truly believe everyone working in the field should have it in his/her toolbox - I basically wrote the answer to advertise it... $\endgroup$ – Martin Feb 25 '16 at 10:57
  • $\begingroup$ @Martin I mentioned that '$U_{AB}$ is unitary', since this is used in computing $Tr(o_{ij})=Tr(\sum(u_{ik}\rho u_{jk}^{+}))=Tr(\sum(u_{jk}^{+}u_{ik}\rho ))=\sigma_{ij}Tr(\rho)$. $\endgroup$ – XXDD Feb 25 '16 at 11:12
  • $\begingroup$ Again, I am not using $Tr(XY)=Tr(YX)$ on partial trace, I only used it in $Tr(o_{ij})$ as shown above. $\endgroup$ – XXDD Feb 25 '16 at 11:14
  • $\begingroup$ @Martin Of course $u_{ij}$ are not unitary matrix, the unitary matrix $U_{AB}$ means that $\sum_{k}u_{jk}^{+}u_{ik}=\sigma_{ij}I_{n\times n}$, this is used in the computation of $Tr(o_{ij})$ $\endgroup$ – XXDD Feb 25 '16 at 11:17

With all the helps above, I found my error.

The unitary matrix $U_{AB}$ will lead to $\sum_{k}u_{ik}u_{jk}^{+}=\sigma_{ij}I_{n\times n}$.

But the computation of $Tr(o_{ij})$ results in $Tr(o_{ij})=Tr((\sum_{k}u_{jk}^{+}u_{ik})\rho_{B})\neq Tr((\sum_{k}u_{ik}u_{jk}^{+})\rho_{B})$.

So I can not get the result as $Tr(o_{ij})=\sigma_{ij}Tr(\rho_B)$.


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