Operator matrix elements, opposite-parity functions

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SUMMARY

The discussion centers on the operator matrix elements for the operator H = f(r)γ, where γ is defined as the matrix \bigl(\begin{smallmatrix} 0 & I\\ I & 0 \end{smallmatrix}\bigr) and f(r) is an even function. It is established that the matrix element <\psi_{a}|H|\psi_{b}> is non-vanishing only if the wave functions ψ_a and ψ_b possess opposite parity, exemplified by the 2s and 2p1/2 functions. The confusion arises from the parity classification of the operator H, with the conclusion that H is indeed an even parity operator due to the even nature of f(r).

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Homework Statement



For operator
H=f(r)γ , where γ=\bigl(\begin{smallmatrix}<br /> 0 &amp; I\\ <br /> I&amp; 0<br /> \end{smallmatrix}\bigr) , f(r) some even function.

Show that matrix element of this operator

&lt;\psi_{a}|H|\psi_{b}&gt;

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 ?
 
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A simple answer like yes or no, on my conclusion that H is even would be
appreciated.
Anyone ?
 

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