I'm trying to find a way to verify that the following expansion is valid for any potential, including noncentral ones,

$$ \langle \textbf{k}' |V|\textbf{k}\rangle = \frac2\pi\sum_{lm} V_l (k', k) Y_{lm}^*(\hat{\textbf{k}}')Y_{lm}(\hat{\textbf{k}})), $$

and seeking the expression for $V_l (k', k)$.

My attempt is to insert the completeness relation $$ \frac2\pi\sum_{lm} \int_0 k^2 dk |klm\rangle\langle klm| $$

to get $$ \langle \textbf{k}' |V|\textbf{k}\rangle = \frac 4 {\pi^2} \sum_{l'm'lm}\int_0 p'^2p^2dp'dp \langle \textbf{k}'|p'l'm'\rangle\langle p'l'm'|V|plm\rangle \langle plm|\textbf{k}\rangle \\ =\frac 2\pi \sum_{l'm'lm} \langle k'l'm'|V|klm\rangle Y_{l'm'}(\hat{\textbf{k}'})Y_{lm}^*(\hat{\textbf{k}}) $$

and I seem to be stuck here.

  • $\begingroup$ What happened to the p integrals? $\endgroup$ – jacob1729 May 30 at 17:14
  • $\begingroup$ The overlap of momentum state and angular momentum state has a Dirac delta: < k |plm>=sqrt(pi/2) (1/k^2) delta(p-k) Ylm(k) $\endgroup$ – fieryslug May 30 at 17:24
  • $\begingroup$ Should the $p$'s in your last formula (in the $\langle p' | V | p \rangle$ part) be $k$'s then? $\endgroup$ – jacob1729 May 30 at 17:44
  • $\begingroup$ yes, thanks for pointing out $\endgroup$ – fieryslug May 30 at 17:52
  • $\begingroup$ Well now the only thing you can use is that $\langle l' m' | V (k',k) |lm \rangle = V_l (k',k) \delta_{ll'} \delta_{mm'}$, which means your potential is indeed central. The fact that you potential as no $m$ index and one total angular momentum index feels like it has to be central. I don't think such expression works for noncentral potentials? $\endgroup$ – gingras.ol May 30 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.