# Operator matrix elements, opposite-parity functions

## Homework Statement

For operator
H=f(r)γ , where γ=$\bigl(\begin{smallmatrix} 0 & I\\ I& 0 \end{smallmatrix}\bigr)$ , f(r) some even function.

Show that matrix element of this operator

$<\psi_{a}|H|\psi_{b}>$

is non-vanishing only if ψ_a and ψ_b functions are of the opposite parity
(for example 2s and 2p1/2 functions)

## The Attempt at a Solution

I can write H as

H = I H I = Ʃ H$_{nm}$|psi_n><psi_m|

so the nm matrix element of H is then given as

H$_{nm}$ = <psi_a|H|psi_b> = <psi_a|I H I|psi_b> =<psi_a|ƩH$_{nm}$|psi_n><psi_m||psi_b>

But I'm not sure how to proceed from here.

Maybe I should take a different approach. For example first to prove that operator
H is an odd parity operator , then <psi_a|H|psi_b> is non-vanishing only if
psi_a and psi_b are of the opposite parity , right ?
But H is even parity operator since f(r) is even, that is
f(-r)γ=f(r)γ , so I'm stuck again.

Can someone please clarify things form me here, how to evaluate matrix elements,
or is my conclusion that H is even wrong ?