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Show that the set of density operators is invariant under the induced action of $U(H)$ on $End(H)$.

I know that a density operator must be positive and have a trace equal to one. But I don't know how to prove the invariance under the unitary group.

I assume I need to show that the set of density operators consists of the same elements even after applying $U(H)$ to $End(H)$, or am I misunderstanding the question entirely?

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I'm not sure what $End(H)$ is but here is a proof that unitaries take density operators to density operators. Let $X$ be a positive semidefinite matrix with unit trace. We have to show that for a unitary $U$, $U^\dagger XU$ is also positive semidefinite with unit trace. Using the cyclicity of trace and $UU^\dagger = I$

$$Tr(U^\dagger X U) = Tr(XUU^\dagger) = Tr(XI) = Tr(X) = 1$$

As for positive semidefiniteness, this means that for any $\vert \psi\rangle$, you have $\langle \psi\vert X\vert\psi\rangle \geq 0$. Now we have

$$\langle \psi\vert U^\dagger XU\vert\psi\rangle = \langle \phi\vert X\vert\phi\rangle \geq 0,$$

where I have set $\vert\phi\rangle = U\vert\psi\rangle$ and the inequality holds since $X$ is positive semidefinite.

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    $\begingroup$ A quicker way to get both properties at once is just to note that conjugation by a unitary operator preserves the spectrum. $\endgroup$
    – tparker
    Apr 12, 2020 at 19:32

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