Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

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

share|cite|improve this question
up vote 1 down vote accepted

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

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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