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I know that one of the requirements for a density matrix is that it is positive-semidefinite. This means that the eigenvalues are non-negative (and sum to 1, so we can assign them the meaning of probabilities).

My question is: Does this forbid the eigenvalues from being larger than $1$? I would think that yes. So is it true that the eigenvalues of a density matrix should be $0 \leq \lambda_{j}\leq 1$ and sum to 1 overall?

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    $\begingroup$ If the largest eigenvalue is $> 1$ and all the others are $\ge 0$, the sum of all of them must also be $> 1$. So each eigenvalue must be $\le 1$ (and it can only equal $1$ if all the others are $0$). $\endgroup$ – alephzero Aug 21 '19 at 13:32
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The diagonal entries are the relative populations of your various states, so must sum to 1 and be non-negative. Going to a basis where the density matrix is diagonal with the eigenvalues as diagonal entries shows you cannot have eigenvalues greater than one else this would imply at least one relative population greater than 1.

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