# Can an isometry leave entropy invariant?

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?

• The second part of your question (the motivation) does not require the first! – Norbert Schuch May 2 '19 at 19:31
• @NorbertSchuch Yes, part of the reason I accepted the answer that I accepted is that it made me realize that with its final remark – user2723984 May 2 '19 at 19:38

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.
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 $$U = W^{-1}\cdot Y\otimes I$$.