Is this correct? $Tr_{B}(U_{AB}(I_A\otimes \rho_B)U_{AB}^{+})=I_A$

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $Tr(o_{ij})=\sigma_{ij}Tr(\rho_B)$.