# Is a density matrix still positive semidefinite after applying a projection operator?

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)

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 $$\vert\phi'\rangle:=X^\dagger \vert\phi\rangle$$.
• 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? Commented Aug 6 at 11:54
• @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 Commented Aug 6 at 14:49
• @QuantumMechanic Normalization should not matter: Nowhere any property of $X$ is needed. Commented Aug 6 at 15:00