I came up with an intuitive guess that tensor product $\phi\otimes\psi$ of two pure states $\phi,~\psi$ (which are density operators) is again pure. However, I tried to use basic linear algebra and proved nothing. Should I use Schmidt composition? Or is there an easier way to prove it?
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2$\begingroup$ $|\phi \rangle \otimes |\psi\rangle$ is a vector so it is by definition pure. $\endgroup$– RammusCommented Jun 1, 2021 at 12:21
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1$\begingroup$ What are you taking as your definition of a pure state? (If you've been trying to prove something, you presumably know what it is, mathematically, that you're trying to detect) $\endgroup$– DaftWullieCommented Jun 1, 2021 at 12:33
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$\begingroup$ @Rammus, DaftWullie I'm pretty sorry that I made a mistake when typing the question. Now I have edited. I was asking for the density operator case indeed, but mistakenly written $|\phi\rangle$ rather than $\phi$, leading it looked like talking a closed quantum system. $\endgroup$– SharaCommented Jun 1, 2021 at 13:06
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By far the easiest way to prove this is to realise that if a density matrix is pure, it is a rank 1 projector. Hence, to prove that a Hermitian $\rho$ is pure, you simply have to verify that $\text{Tr}(\rho)=1$ (thus showing the rank is 1 if it's a projector, although this is automatic for a correctly constructed density matrix) and $\rho^2=\rho$.
Now, you already know that $\phi^2=\phi$ and $\psi^2=\psi$ (they are pure by assumption), so you just use properties of the tensor product to show that $$ (\phi\otimes\psi)^2=(\phi^2)\otimes(\psi^2)=\phi\otimes\psi, $$ and you're done!
answered Jun 1, 2021 at 13:09
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$\begingroup$ Oh I see. I didn't notice the useful fact that the trace of a projector equals to its rank. I was trying to prove it using something more complicated (e.g. Schmidt decomposition.) Thanks a lot!!! $\endgroup$– SharaCommented Jun 1, 2021 at 13:12