0
$\begingroup$

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 $Tr(\rho^2)<1$ .

$\endgroup$

1 Answer 1

1
$\begingroup$

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 ${\rm Tr}(\rho)=\langle\psi|\psi\rangle=1$.

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.