- #1

- 51

- 0

## Homework Statement

For operator

**H**=f(

**r**)γ , where γ=[itex]\bigl(\begin{smallmatrix}

0 & I\\

I& 0

\end{smallmatrix}\bigr)[/itex] , f(

**r**) some even function.

Show that matrix element of this operator

[itex]<\psi_{a}|H|\psi_{b}>[/itex]

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[itex]_{nm}[/itex]|psi_n><psi_m|

so the nm matrix element of H is then given as

**H**[itex]_{nm}[/itex] = <psi_a|

**H**|psi_b> = <psi_a|

**I H I**|psi_b> =<psi_a|ƩH[itex]_{nm}[/itex]|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 ?