# Dirac Hydrogen atom

1. Jun 4, 2015

### ChrisVer

Hey I was reading through a text and came across:
I can understand the second statement from the Pauli matrices... However I think that I don't understand the 1st statement as it is... why would the diagonal elements of an odd-operator be zero if parity is definite?

Last edited: Jun 4, 2015
2. Jun 4, 2015

### blue_leaf77

I think what they mean as "fixed" is actually "definite".

3. Jun 4, 2015

### ChrisVer

Yup understand that... (a flaw in my translation )...I don't understand why the diagonal elements are zero.

4. Jun 4, 2015

### blue_leaf77

If $A$ is an odd operator and $\pi$ is a parity operator, then we have $\pi^\dagger A \pi = -A$. If furthermore a state $|\psi\rangle$ which has a definite parity sandwhiches the expression $\pi^\dagger A \pi = -A$ from left and right in each side of the equation, you will get $\langle \psi| A |\psi \rangle = -\langle \psi|A|\psi \rangle$ because $\langle \psi|\pi^\dagger A \pi|\psi \rangle = \langle \psi| A |\psi \rangle$. You see then we must require $\langle \psi| A |\psi \rangle = 0$ so that the previous relation can hold.