# Mean value of projection operator

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

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 $$|c_i|^2$$.

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