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It is obvious that $B = OAO^\dagger$ have equal trace if $O$ is Hermian. But is is also so that:

$Tr(A) = Tr(B) \rightarrow A = OBO^\dagger$

For some Hermitian $O$ ?

I know this should be trivial to prove/disprove but Is just can't get it sorted out...

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  • $\begingroup$ Ps: sorry for the bad Tex. Stackexchange doen't deel to like me phone's dollar sign :/ $\endgroup$ – gertian Jun 28 '17 at 11:48
  • $\begingroup$ What do you mean by $O^{\dagger}?$ $\endgroup$ – RideTheWavelet Jun 28 '17 at 12:09
  • $\begingroup$ The hermitian conjugate $\endgroup$ – gertian Jun 28 '17 at 12:35
  • $\begingroup$ I think you mean that $O$ is unitary (and maybe that $A$ and $B$ are Hermitian) $\endgroup$ – Omnomnomnom Jun 28 '17 at 12:40
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I think that you meant that $O$ is unitary, and that $A$ and $B$ are Hermitian. If that's the case, your statement is not true.

Note that $OAO^\dagger$ has the same eigenvalues as $A$. If we take $$ A = \pmatrix{1&0\\0&1}, \quad B = \pmatrix{2&0\\0&0} $$ Then we see that $\operatorname{tr}(A) = \operatorname{tr}(B)$. Since they have different eigenvalues, however, there can be no unitary $O$ such that $B = OAO^\dagger$.

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I am that sure that I understood your question, but I suppose that $A=\left(\begin{smallmatrix}0&0\\0&0\end{smallmatrix}\right)$ and $B=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ is a counter-example.

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