(a) Equation 12.17: \hat{\rho}^{2}=|\Psi\rangle\langle\Psi \mid \Psi\rangle\langle\Psi|, \text { and since } \Psi \text { is normalized, } \hat{\rho}^{2}=|\Psi\rangle 1\langle\Psi|=\hat{\rho} .
\rho^{2}=\rho, \quad \text { (it is idempotent) } (12.17).
Equation 12.18: To find the Hermitian conjugate of an operator in Dirac notation, we reverse the order of the terms, switch bras and kets, place a † on every operator and a * on every number:
\rho^{\dagger}=\rho, \quad \text { (it is hermitian), } (12.18).
\hat{\rho}^{\dagger}=(|\Psi\rangle\langle\Psi|)^{\dagger}=|\Psi\rangle\langle\Psi|=\rho .
Equation 12.19: \operatorname{Tr}(\rho)=\sum_{i} \rho_{i i}=\sum_{i}\left\langle e_{i} \mid \Psi\right\rangle\left\langle\Psi \mid e_{i}\right\rangle=\left\langle\Psi\left|\left(\sum_{i}\left|e_{i}\right\rangle\left\langle e_{i}\right|\right)\right| \Psi\right\rangle=\langle\Psi \mid \Psi\rangle=1 .
\operatorname{Tr}(\rho)=\sum_{i} \rho_{i i}=1 \text { (its trace is } 1 \text { ) } (12.19).
Equation 12.20:
\langle A\rangle=\operatorname{Tr}(\rho A ) (12.20).
\operatorname{Tr}(\rho A )=\sum_{i}(\rho A )_{i i}=\sum_{i} \sum_{j} \rho_{i j} A_{j i}=\sum_{i} \sum_{j}\left\langle e_{i} \mid \Psi\right\rangle\left\langle\Psi \mid e_{j}\right\rangle\left\langle e_{j}|\hat{A}| e_{i}\right\rangle .
=\sum_{i} \sum_{j}\left\langle\Psi \mid e_{j}\right\rangle\left\langle e_{j}|\hat{A}| e_{i}\right\rangle\left\langle e_{i} \mid \Psi\right\rangle=\left\langle\Psi\left|\left(\sum_{j}\left|e_{j}\right\rangle\left\langle e_{j}\right|\right) \hat{A}\left(\sum_{i}\left|e_{i}\right\rangle\left\langle e_{i}\right|\right)\right| \Psi\right\rangle=\langle\Psi|\hat{A}| \Psi\rangle .
=\langle A\rangle .
(b) i \hbar \frac{d}{d t} \hat{\rho}=i \hbar(|\dot{\Psi}\rangle\langle\Psi|+| \Psi\rangle\langle\dot{\Psi}|) . From the Schrödinger equation: i \hbar|\dot{\Psi}\rangle=\hat{H}|\Psi\rangle , and, taking the adjoint: -i \hbar\langle\dot{\Psi}|=\langle\Psi| \hat{H} . Thus
i \hbar \frac{d}{d t} \hat{\rho}=\hat{H}|\Psi\rangle\langle\Psi|-| \Psi\rangle\langle\Psi| \hat{H}=\hat{H} \hat{\rho}-\hat{\rho} \hat{H}=[\hat{H}, \hat{\rho}] .