# Compute hatP^2 and is hatP idempotent?

## Let $| {\psi}_{n} >$ be the eigenstates of some Hermitian operator, $\hat{O}$ Consider the operator hatP = 1/2 (|ψ_m><ψ_m| + |ψ_n><ψ_n| )

Feb 17, 2018

Assuming you really mean

$\hat{P} = \frac{1}{2} \left(| {\psi}_{m}\right\rangle \left\langle{\psi}_{m} | + | {\psi}_{n}\right\rangle \left\langle{\psi}_{n} |\right)$,

then since $\hat{P} \ne \hat{O}$, the |psi_m >> and |psi_n >> need not be eigenstates of $\hat{P}$. We have:

${\hat{P}}^{2} = \frac{1}{4} \left(| {\psi}_{m}\right\rangle \left\langle{\psi}_{m} | + | {\psi}_{n}\right\rangle {\left\langle{\psi}_{n} |\right)}^{2}$

$= \frac{1}{4} \left(| {\psi}_{m}\right\rangle \left\langle{\psi}_{m} | + | {\psi}_{n}\right\rangle \left\langle{\psi}_{n} |\right) \left(| {\psi}_{m}\right\rangle \left\langle{\psi}_{m} | + | {\psi}_{n}\right\rangle \left\langle{\psi}_{n} |\right)$

$= \frac{1}{4} \left(| {\psi}_{m}\right\rangle \left\langle{\psi}_{m} | {\psi}_{m}\right\rangle \left\langle{\psi}_{m} | + | {\psi}_{m}\right\rangle \left\langle{\psi}_{m} | {\psi}_{n}\right\rangle \left\langle{\psi}_{n} | + | {\psi}_{n}\right\rangle \left\langle{\psi}_{n} | {\psi}_{m}\right\rangle \left\langle{\psi}_{m} | + | {\psi}_{n}\right\rangle \left\langle{\psi}_{n} | {\psi}_{n}\right\rangle \left\langle{\psi}_{n} |\right)$

By default, proper eigenstates of Hermitian operators are conventionally such that the | psi_m >> and | psi_n >> are orthonormal, so

= 1/4 [color(green)(|psi_m >> color(black)(cancel(color(green)(<< psi_m | psi_m >>))^(1)) color(green)(<< psi_m| + color(purple)(|psi_m >> color(black)(cancel(color(purple)(<< psi_m | psi_n >>))^(0)) color(purple)(<< psi_n| + color(orange)(|psi_n >> color(black)(cancel(color(orange)(<< psi_n | psi_m >>))^(0)) color(orange)(<< psi_n| + color(red)(|psi_n >> color(black)(cancel(color(red)(<< psi_n | psi_n >>))^(1)) color(red)(<< psi_n | ]

$= \frac{1}{4} \left(| {\psi}_{m}\right\rangle \left\langle{\psi}_{m} | + | {\psi}_{n}\right\rangle \left\langle{\psi}_{n} |\right)$

$= \frac{1}{2} \hat{P} \ne \hat{P}$

Even if the eigenstates are not normalized, each eigenvector must be orthogonal for distinct eigenvalues. Orthonormal states are the best-case scenario for simplifying this, so no matter how you spin it, $\hat{P}$ is not idempotent.