Unitary conjugation preserves product forms

Question

I would like to know if, given a finite Hilbert space $H$, and two (unitary) linear operators $A,B:H\to H$, $$ W^* (A\otimes B)W:H\otimes H\to H \otimes H $$ is also a product form operator, given any (unitary) linear $W:H\otimes H\to H\otimes H$. I would be very pleased for any hint. Thanks in advance

Answer 1

androidAssuming that by "product form operator" you mean "elementary tensor", the answer is no. Even in finite dimension. Projections of the same rank are unitarily equivalent, so the result would imply that all projections are given by elementary tensors.

Consider $H=\mathbb C^2$, $A=B=\begin{bmatrix} 1&0\\0&0\end{bmatrix}$. Then $A\otimes B$ is the rank-one projection $$ A\otimes B=\begin{bmatrix} 1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}. $$ Another rank-one projection in $M_2(\mathbb C)\otimes M_2(\mathbb C)$ is $$ P=\begin{bmatrix} 1/2&0&0&1/2\\0&0&0&0\\0&0&0&0\\ 1/2&0&0&1/2\end{bmatrix}, $$ which is well-known not to be an elementary tensor. Being of the same rank, there exists a unitary $W$ such that $$W(A\otimes B)W^*=P.$$

The answer is the same when $A,B$ are unitary. For instance, again with $H=\mathbb C^2$ and using the usual representation of $A\otimes B$ as $\{A_{kj}B_{st}\}$ the unitary $$ U=\begin{bmatrix} 1/\sqrt2&0&0&1/\sqrt2\\ 0&1&0&0\\ 0&0&-1&0\\ 1/\sqrt2&0&0&-1/\sqrt2\end{bmatrix} $$ is not an elementary tensor. If it were, we would have $A_{11}B_{11}=1/\sqrt2$, $A_{11}B_{12}=0$, $A_{12}B_{12}=1/\sqrt2$, which is impossible.

On the other hand, if $A=B=\begin{bmatrix} 0&i\\-i&0\end{bmatrix}$, then the eigenvalues of $A\otimes B$ are $1,1,-1,-1$. As $A\otimes B$ and $U$ have the same eigenvalues with the same multiplicities, there exists a unitary $W$ with $W(A\otimes B)W^*=U$.