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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 $Tr(\rho^2)<1$

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

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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
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