## Trace and its square of mixed state density operator using integral

1. The problem statement, all variables and given/known data
I want to show that
$$tr\left(\hat{\rho}_{mixed}\right)=1$$
$$tr\left(\hat{\rho}_{mixed}^{2}\right)<1$$
when
$$\hat{\rho}_{mixed}=\frac{1}{2\pi}\int_{0}^{2\pi}d \alpha \hat{\rho}(\psi)$$

2. Relevant equations
$$tr\left(\psi\right)= \sum_{n}\langle n|\psi|n\rangle$$
$$\hat{\rho}=\sum_{a}\omega_{a}|\psi\rangle\langle \psi|$$

3. The attempt at a solution
$$\hat{\rho}_{mixed}=\frac{1}{2\pi}\int_{0}^{2\pi}d\alpha\hat{\rho}(\psi) =\frac{1}{2\pi}\left[\alpha\right]_{0}^{2\pi}\hat{\rho}(\psi)=\frac{1}{2\pi}\left[2\pi-0\right]\hat{\rho}(\psi)=\hat{\rho}(\psi)$$
$$tr\left(\hat{\rho}_{mixed}(\psi)\right)= tr\left( \hat{\rho}(\psi)\right)=\sum_{n}\sum_{a}\langle n| \underbrace{\psi_{a}\rangle\langle\psi_{a}}_{=1}|n\rangle=\sum_{n}\lang le n|n\rangle=1$$
$$tr \left(\hat{\rho}_{mixed}^{2}( \psi)\right)= tr\left( \hat{\rho}^{2}(\psi)\right)= tr\left(\hat{\rho}(\psi) \cdot \hat{\rho}(\psi)\right) = \sum_{n}\sum_{a,b} \langle n| \underbrace{\psi_{a}\rangle \langle\psi_{a}}_{=1}| \underbrace{\psi_{b} \rangle\langle\psi_{b}}_{=1}|n\rangle= \sum_{n}\langle n|n\rangle= 1$$

That`s not correct, at least not the square of the trace.

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 Where is my mistake?