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Operator matrix elements, opposite-parity functions

  • Thread starter vst98
  • Start date
  • #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 ?
 

Answers and Replies

  • #2
51
0
A simple answer like yes or no, on my conclusion that H is even would be
appreciated.
Anyone ?
 

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