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

## 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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