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Consider two finite dimensional Hilbert spaces $A$ and $B$. If I have an isometry $V:A\rightarrow A\otimes B$, under what condition can I find a unitary $U:A\otimes B\to A\otimes B$ such that $$U\rho_{A}\otimes |\psi\rangle\langle \psi|_BU^\dagger=V\rho_{A}V^\dagger$$

for some state $|\psi\rangle_B$? The motivation is I have an isometry like $V$ and I would like it very much for a state to satisfy $S(A)_\rho=S(AB)_{V\rho V^\dagger}$ where $S$ is quantum entropy. Is this possible?

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  • $\begingroup$ The second part of your question (the motivation) does not require the first! $\endgroup$ – Norbert Schuch May 2 at 19:31
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    $\begingroup$ @NorbertSchuch Yes, part of the reason I accepted the answer that I accepted is that it made me realize that with its final remark $\endgroup$ – user2723984 May 2 at 19:38
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You don't need any additional conditions beyond those already stated in the question. That is, for any isometry $V: A \rightarrow A\otimes B$ and any unit vector $|\psi\rangle_B$, there will always be a unitary $U$ satisfying the equation in the question (simultaneously for every choice of $\rho_A$).

One way to see this is to first pick any orthonormal basis $\{|1\rangle,\ldots,|n\rangle\}$ for $A$, and then consider the two sets $\{|1\rangle\otimes|\psi\rangle, \ldots, |n\rangle\otimes|\psi\rangle\}$ and $\{V|1\rangle,\ldots,V|n\rangle\}$. These are orthonormal sets (using the fact that $V$ is an isometry in the second case), so you can extend them both to be orthonormal bases of $A\otimes B$ and choose a unitary $U$ that maps the first completed basis to the second. Of course there are many ways to do this in general, but in particular you can choose $U$ so that it maps $|k\rangle\otimes|\psi\rangle$ to $V|k\rangle$ for each $k \in \{1,\ldots,n\}$. This implies that the equation in the question is satisfied.

Note that if what you really want is $S(A)_{\rho} = S(AB)_{V\rho V^{\dagger}}$, then it is simpler to conclude that this is always true from the observation that $\rho$ and $V \rho V^{\dagger}$ must agree on their nonzero eigenvalues.

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You can always do that.
Subspaces $\text{Im}(V)$ and $A\otimes |0\rangle$ have the same dimension, so there must be some unitary that translates one subspace to another. That is, $\exists W \in \text{Unitary}(A\otimes B), W(\text{Im}(V)) = A\otimes |0\rangle$. Now $WV$ translates $A$ to $A\otimes |0\rangle$. Since $V$ is isometry $WV$ is also isometry, hence there exists unitary $Y$ on $A$ such that $WV|\phi\rangle = Y|\phi\rangle\otimes |0\rangle$.
Now the unitary you are looking is $U = W^{-1}\cdot Y\otimes I$.

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