Dirac Hydrogen Atom: Parity and Odd-Operator

[ChrisVer]

Hey I was reading through a text and came across:

"[ Having extracted the Dirac version of Schrödinger's equation of the H atom...] Since the states | j j_z l > have definite parity, the odd-operator \vec{S} \cdot \hat{r} will have vanishing diagonal elements. Also since \big(\vec{S} \cdot \hat{r} \big)^2 =1 then its offdiagonal elements will be \frac{1}{2} e^{\pm i \phi} (we can choose the phase \phi=0)[...]"

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?

 

[blue_leaf77]

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

 

[ChrisVer]

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

 

[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.