Suppose we have a density operator $\hat{\rho}$ and a projection operator $\hat{\Pi}$, are the matrices $$\hat{\rho}'=\hat{\Pi}\hat{\rho}\hat{\Pi}^{\dagger}$$ and $$\hat{\rho}''=(\hat{I}-\hat{\Pi})\hat{\rho}(\hat{I}-\hat{\Pi}^{\dagger})$$ also positive semidefinite (i.e they can be renormalised to give physical density matrices)
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Yes: For $P$ positive semi-definite, $XPX^\dagger$ is also positive, since $$ \langle \phi\vert XPX^\dagger \vert\phi\rangle = \langle \phi'\vert P\vert\phi'\rangle \ge 0 $$ with $\vert\phi'\rangle:=X^\dagger \vert\phi\rangle$.
answered Aug 6 at 11:49
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$\begingroup$ great thanks, just to clarify, $ \langle \phi\vert XPX^\dagger \vert\phi\rangle \ge 0 $ $\forall \vert\phi\rangle$ is equivalent to saying $P$ has only positive eigenvalues? $\endgroup$ Commented Aug 6 at 11:54
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$\begingroup$ My issue is that applying such a transformation in a simulation leaves me with a state that has negative eigenvalues, although im pretty certain this must be numeric. $\endgroup$ Commented Aug 6 at 11:55
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1$\begingroup$ @AdrienAmour yes (well it is equivalent to saying that $XPX^\dagger$ has non-negative eigenvalues!). if you have numerical issues it could also come from how you're defining the projection, because if you haven't normalized it properly it could be that $\hat{I}-\hat{Pi}$ goes negative (ie, make sure that $\hat{\Pi}^2=\hat{\Pi}$ and not $\hat{\Pi}^2\propto \hat{\Pi}$); or you can check that your state itself before the projection is indeed positive $\endgroup$ Commented Aug 6 at 14:49
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1$\begingroup$ @AdrienAmour In complex vector spaces, <x|A|x> >= 0 is equivalent to saying that A is hermitian and only has non-negative eigenvalues. $\endgroup$ Commented Aug 6 at 14:59
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1$\begingroup$ @QuantumMechanic Normalization should not matter: Nowhere any property of $X$ is needed. $\endgroup$ Commented Aug 6 at 15:00