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please tell me where I am wrong in the following chain of reasoning
1. trace of a matrix isn't affected by unitary change of basis
2. therefore trace of hermitian matrix is the sum of eigenvalues
3. if $U=e^{iH}$ where $H$ is hermitian, then $U$ is unitary
4. $\det U=e^{i\times Tr{H}}$
5. But $\det U=1$.
6. therefore $Tr(H) = 2n\pi$ which I think is false

martini
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euler16
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  • Note that the trace of _any_ matrix is the sum of its eigenvalues. See here: https://math.stackexchange.com/questions/546155/proof-that-the-trace-of-a-matrix-is-the-sum-of-its-eigenvalues – Rio Alvarado Apr 28 '22 at 04:03

1 Answers1

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An unitary does not necessarialy have $\det U = 1$, we can only say that $|\det U| = 1$, which implies $\mathrm{tr}\, H \in \mathbf R$, which is correct, due to the fact that all diagonal entries of $H$ are real.

martini
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