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SolubleFish
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In this case, there are $2$ projection operators : \begin{align} P_1 = |\psi_1\rangle\langle\psi_1| = \frac{H - \lambda_2}{\lambda_1-\lambda_2} \\ P_2 = |\psi_2\rangle\langle\psi_1| = \frac{H-\lambda_1 }{\lambda_2-\lambda_1} \end{align}\begin{align} P_1 = |\psi_1\rangle\langle\psi_1| = \frac{H - \lambda_2}{\lambda_1-\lambda_2} \\ P_2 = |\psi_2\rangle\langle\psi_2| = \frac{H-\lambda_1 }{\lambda_2-\lambda_1} \end{align}

These to operators satisfy : $P_i P_j = \delta_{ij}P_j$, so if the product defining your $\mathbb P$ is over both $i$ and $j$, you have $\mathbb P = 0$.

The expected value of $P_i$ in the state $|\psi\rangle = c_1 |\psi_1 \rangle + c_2 |\psi_2 \rangle$ is $|c_i|^2$.

In this case, there are $2$ projection operators : \begin{align} P_1 = |\psi_1\rangle\langle\psi_1| = \frac{H - \lambda_2}{\lambda_1-\lambda_2} \\ P_2 = |\psi_2\rangle\langle\psi_1| = \frac{H-\lambda_1 }{\lambda_2-\lambda_1} \end{align}

These to operators satisfy : $P_i P_j = \delta_{ij}P_j$, so if the product defining your $\mathbb P$ is over both $i$ and $j$, you have $\mathbb P = 0$.

The expected value of $P_i$ in the state $|\psi\rangle = c_1 |\psi_1 \rangle + c_2 |\psi_2 \rangle$ is $|c_i|^2$.

In this case, there are $2$ projection operators : \begin{align} P_1 = |\psi_1\rangle\langle\psi_1| = \frac{H - \lambda_2}{\lambda_1-\lambda_2} \\ P_2 = |\psi_2\rangle\langle\psi_2| = \frac{H-\lambda_1 }{\lambda_2-\lambda_1} \end{align}

These to operators satisfy : $P_i P_j = \delta_{ij}P_j$, so if the product defining your $\mathbb P$ is over both $i$ and $j$, you have $\mathbb P = 0$.

The expected value of $P_i$ in the state $|\psi\rangle = c_1 |\psi_1 \rangle + c_2 |\psi_2 \rangle$ is $|c_i|^2$.

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SolubleFish
  • 6.1k
  • 7
  • 31

In this case, there are $2$ projection operators : \begin{align} P_1 = |\psi_1\rangle\langle\psi_1| = \frac{H - \lambda_2}{\lambda_1-\lambda_2} \\ P_2 = |\psi_2\rangle\langle\psi_1| = \frac{H-\lambda_1 }{\lambda_2-\lambda_1} \end{align}

These to operators satisfy : $P_i P_j = \delta_{ij}P_j$, so if the product defining your $\mathbb P$ is over both $i$ and $j$, you have $\mathbb P = 0$.

The expected value of $P_i$ in the state $|\psi\rangle = c_1 |\psi_1 \rangle + c_2 |\psi_2 \rangle$ is $|c_i|^2$.