Partial trace of local operators applied to maximally entangled states I was looking at a problem where two invertible local operators were applied to a maximally entangled state, and didn't quite understand how some of it works out. We have local operators $A \otimes \mathbb{1}$ and $\mathbb{1} \otimes B$ with $Tr(A^{\dagger}A) = 1$ and $Tr(B^{\dagger}B) = 1$. We also have the maximally entangled state $\rho$. I do not really undersand this step
$$ Tr_B (A \otimes B \rho A^{\dagger} \otimes B^{\dagger}) = Tr_B (A^{\dagger}A \otimes B^{\dagger}B \rho) = Tr_B ((A^{\dagger}A \otimes \mathbb{1})\rho) $$
I assume this is some combination of the fact that $\rho$ is maximally entangled (so its partial trace is equal to the identity) and the fact that $Tr(B^{\dagger}B) = 1$, but I cannot really understand how to argue that.
It might also be the case that I'm just completely misunderstanding something.
 A: I tried to prove it elegantly below but hit a road block. Trying to do things by brute force, it looks as though this result cannot be true:
Start with a maximally entangled state
$$|\psi\rangle=\frac{1}{\sqrt{N}}\sum_{i=0}^{N-1} |i\rangle\otimes |i\rangle$$ and decompose the operators as
$A=\sum_{ij}A_{ij}|i\rangle\langle j|$ and $B=\sum_{ij}B_{ij}|i\rangle\langle j|$. We have
$$A\otimes B|\psi\rangle=\sum_{ik}(\sum_m A_{im} B_{km})|i\rangle\otimes |k\rangle.$$ This lets us write $$A\otimes B \rho A^\dagger\otimes B^\dagger =
\sum_{ik}(\sum_m A_{im} B_{km})|i\rangle\otimes |k\rangle
\sum_{i^\prime k^\prime}(\sum_{m^\prime} A_{i^\prime m^\prime}^* B_{k^\prime m^\prime}^*)\langle i^\prime|\otimes \langle k^\prime|
$$ and, tracing out the second subsystem, we find
$$\text{Tr}_B\left(A\otimes B \rho A^\dagger\otimes B^\dagger \right)=
\sum_{ii^\prime }\sum_k\left((\sum_m A_{im} B_{km})
(\sum_{m^\prime} A_{i^\prime m^\prime}^* B_{k m^\prime}^*)\right)|i\rangle\langle i^\prime|.
$$ There is still clearly some dependence on the operator $B$, unless it is unitary. We can try to simplify the sum over $k$ but cannot proceed with only the information $\text{Tr}(B^\dagger B)=\sum_{ij}|B_{ij}|^2$
$$\sum_k B_{km}B_{km^\prime}^*=???$$

A maximally entangled state
$$|\psi\rangle=\frac{1}{\sqrt{N}}\sum_{i=0}^{N-1} |i\rangle\otimes |i\rangle$$ satisfies the identity
$$U\otimes\mathbb{I} |\psi\rangle=\mathbb{I}\otimes U^\top |\psi\rangle$$ for all unitary operators $U$, where $^\top$ denotes the transpose in the same basis in which we have maximal entanglement. Moreoever, it holds for all entangled states with full Schmidt rank, so this in turn means that it holds for maximally entangled states and for all operators $A$:
$$A\otimes\mathbb{I} |\psi\rangle=\mathbb{I}\otimes A^\top |\psi\rangle.$$ (To prove this, set $A=\sum_{ij}A_{ij}|i\rangle\langle j|$ and compare the actions of $A\otimes\mathbb{I}$ and $\mathbb{I}\otimes A^\top$ on the maximally entangled state.) We can thus perform the simplifications, with maximally entangled $\rho=|\psi\rangle\langle \psi|$,
\begin{align}
\text{Tr}_B\left(A\otimes B \rho A^\dagger\otimes B^\dagger \right)&=\text{Tr}_B\left[(\mathbb{I}\otimes B)(A\otimes \mathbb{I}) \rho (A\otimes \mathbb{I})^\dagger(\mathbb{I}\otimes B)^\dagger \right]
\\&=\text{Tr}_B\left[\mathbb{I}\otimes BA^\top \rho \mathbb{I}\otimes (BA^\top)^\dagger \right].
\end{align} Next, we are allowed to cyclically permute operators acting on the $B$ system within the trace over $B$:
\begin{align}
\text{Tr}_B\left[\mathbb{I}\otimes BA^\top \rho \mathbb{I}\otimes (BA^\top)^\dagger \right]&=\text{Tr}_B\left[\mathbb{I}\otimes (BA^\top)^\dagger BA^\top \rho  \right]\\&
=\text{Tr}_B\left[(\mathbb{I}\otimes A^*B^\dagger BA^\top) \rho  \right].
\end{align} Finally, we use the maximal entanglement property again to move operators back to the $A$ subsystem:
\begin{align}
\text{Tr}_B\left[(\mathbb{I}\otimes A^*B^\dagger BA^\top) \rho  \right]&=\text{Tr}_B\left[(AB^\top B^* A^\dagger\otimes \mathbb{I}) \rho  \right].
\end{align} This doesn't look like the desired result...
A: This is not true as stated. I'd say the statement is close to the fact that
$${\rm Tr}_B((\Phi_A\otimes\Phi_B)\rho)=\Phi_A(\operatorname{Tr}_B(\rho)),$$
for any trace preserving map $\Phi_B$, and any bipartite state $\rho$, and map $\Phi_A$.
This holds because
$$\operatorname{Tr}_B[(\Phi_A\otimes\Phi_B)\rho]= \sum_{\alpha\beta\gamma\delta ijk\ell}K(\Phi_A)_{\alpha\beta,ik} K(\Phi_B)_{\gamma\delta,j\ell}\rho_{ij,k\ell} \operatorname{Tr}_B[|\alpha\rangle\!\langle\beta|\otimes|\gamma\rangle\!\langle \delta|] \\
= \sum_{\alpha\beta ijk} K(\Phi_A)_{\alpha\beta,ik} \rho_{ij,kj} |\alpha\rangle\!\langle \beta|
= \Phi_A(\operatorname{Tr}_B(\rho)),$$
where $K(\Phi)_{ij,k\ell}\equiv \langle i|\Phi(|k\rangle\!\langle \ell|)|j\rangle$ is the natural representation of the channel, and we used the property that for trace-preserving maps we have $\sum_i K(\Phi)_{ii,k\ell}=\delta_{k\ell}$.
In your case, you have this structure with $\Phi_A(X)\equiv AXA^\dagger$ and $\Phi_B(X)\equiv BXB^\dagger$. But this $\Phi_B$ is not trace-preserving. For that, you need the stronger condition $B^\dagger B=I$.
As an easy counterexample, try the statement with $B=|i\rangle\!\langle j|$ for some $i\neq j$, and you should find it doesn't work.
A: This is not true. A simple counterexample is $A\propto I$ and $B=\lvert0\rangle\langle0\rvert$.
