Is this correct? $Tr_{B}(U_{AB}(I_A\otimes \rho_B)U_{AB}^{+})=I_A$
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 $U_{AB}$. 680
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. 677