How can you show that for any pure state, the purity = 1?

Pure state: $\rho^2 = \rho$ and $Tr(\rho^2)=1$

Mixed state: $\rho^2 \neq \rho$ and $Tr(\rho^2)<1$ .


1 Answer 1


For a pure state, by definition, $$\rho = |\psi\rangle\langle \psi| $$ So it is a projection operator onto the pure state $|\psi\rangle$. Note that ${\rm Tr}(\rho L)=\langle\psi|L|\psi\rangle$ for this density matrix. So it follows that $$\rho^2 = |\psi\rangle\langle \psi|\psi\rangle\langle \psi|=|\psi\rangle\langle \psi|=\rho $$ and ${\rm Tr}(\rho^2)=1$ follows from the usual normalization conditions for the overall probability ${\rm Tr}(\rho)=\langle\psi|\psi\rangle=1$.

  • $\begingroup$ Brilliant! This method is what I used, I was concerned that there is a more rigorous way of proving this. $\endgroup$ Commented Nov 12, 2012 at 12:52
  • $\begingroup$ Why don't you consider the above method rigorous enough ? $\endgroup$ Commented Nov 12, 2012 at 13:40
  • $\begingroup$ It was more that I wasn't sure if there was another way. Using this technique, I also proved that $Tra(\rho)<1$ $\endgroup$ Commented Nov 12, 2012 at 14:20

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