In an orthonormal eigenbasis $\{ \left|\psi_1\right\rangle , \left|\psi_2\right\rangle\}$ of the Hamiltonian operator $\hat{H}$ we have the projection operator:

$$ \hat{\mathbb{P}} = \prod_{i \neq j} \left( \frac{\hat{H} - \lambda_i}{\lambda_j - \lambda_i} \right)$$

and :

$$ \hat{H} \left|\psi_1\right\rangle = \lambda_1 \left|\psi_1\right\rangle$$

$$ \hat{H} \left|\psi_2\right\rangle = \lambda_2 \left|\psi_2\right\rangle$$

Dependent upon our choice:

$\lambda_i = \lambda_1 \longrightarrow \lambda_j = \lambda_2$

$$ \hat{\mathbb{P}} = \left|\psi_1\right\rangle \left\langle\psi_1\right|$$

or $\lambda_i = \lambda_2 \longrightarrow \lambda_j = \lambda_1$:

$$ \hat{\mathbb{P}} = \left|\psi_2\right\rangle \left\langle\psi_2\right|$$

If we have:

$$ \left|\psi\right\rangle = c_1 \left|\psi_1\right\rangle + c_2 \left|\psi_2\right\rangle$$

we just conclude that the mean value of $\hat{\mathbb{P}}$ is $|c_1|^2$ or $|c_2|^2$ or do we unify it somehow? This is confusing me because of the product, I know we only consider when $i \neq j$ but that is dependent on our initial choice for $\lambda_i$, right?


1 Answer 1


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 $|c_i|^2$.

  • $\begingroup$ There is a typo in your projection operator $P_2=|\psi_2\rangle\langle \psi_\mathbf 1| $. $\endgroup$
    – Hans Wurst
    Commented May 19, 2021 at 19:01
  • $\begingroup$ @HansWurst Thanks ! Edited $\endgroup$ Commented May 19, 2021 at 19:01

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