/quantum information – Can an isometry leave entropy invariant? (via Qpute.com)
quantum information - Can an isometry leave entropy invariant?

quantum information – Can an isometry leave entropy invariant? (via Qpute.com)


You don’t need any additional conditions beyond those already stated in the question. That is, for any isometry $V: A rightarrow Aotimes B$ and any unit vector $|psirangle_B$, 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 ${|1rangle,ldots,|nrangle}$ for $A$, and then consider the two sets ${|1rangleotimes|psirangle, ldots, |nrangleotimes|psirangle}$ and ${V|1rangle,ldots,V|nrangle}$. 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 $Aotimes B$ 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 $|krangleotimes|psirangle$ to $V|krangle$ 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 $S(A)_{rho} = S(AB)_{Vrho V^{dagger}}$, 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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