Solution to Density Matrix Pure State Problem

[LagrangeEuler]

Homework Statement

Find condition for which $\hat{\rho}$ will be pure state density operator?
$\hat{\rho} = \begin{bmatrix} 1+a_1 & a_2 \\[0.3em] a_2^* & 1-a_1 \end{bmatrix}$

Homework Equations

In case of pure state $Tr(\hat{\rho}^2)=Tr(\hat{\rho})=1$.

The Attempt at a Solution

Using that condition I got
$\hat{\rho}^2 = \begin{bmatrix} (1+a_1)^2+|a_2|^2 & (1+a_1)a_2+a_2(1-a_1) \\[0.3em] (1+a_1)a_2^*+a_2^*(1-a_1) & (1-a_1)^2+|a_2|^2 \end{bmatrix}$
and from that
$2|a|^2+2a_1^2=-1$ which can not be true. Because $a_1$ must be real, condition to $\hat{\rho}$ is hermitian.

 

[Robert_G]

I don't quite understand. That density operator you stated, it's not a pure state density operator, and that's why The trace of the square of rho will not be 1. It's correct. I afraid I did not understand your question.

 

[LagrangeEuler]

Question is find condition put on $a_1$ and $a_2$ for which $\hat{\rho}$ is pure state density operator.

 

[Fightfish]

The problem occurs because you have not normalized the original density operator yet.

 

[LagrangeEuler]

Tnx a lot Fightfish. I did not see that trace of given matrix is not $1$. Density matrix must be
$\hat{\rho} =\frac{1}{2} \begin{bmatrix} 1+a_1 & a_2 \\[0.3em] a_2^* & 1-a_1 \end{bmatrix}$