1
$\begingroup$

I have an integral involving spherical harmonics and a cross product. It reads

$$ \int d^3k d^3Q \phi^{*}_{L'}(k+Q)Y^{*}_{L',M'_{L'}}(\widehat{k+Q})C^{J'M'_{J'}}_{L'S'}\mathbf{S}\cdot \mathbf{(k\times Q)} \phi_{L}(k)Y_{L,M_{L}}(\widehat{k})C^{JM_{J}}_{LS} $$

with

$$ \phi_{L}(k)=e^{-k^2} $$

and

$$ C^{JM_{J}}_{LS}=<sm,\bar{s}\bar{m}|S M_S><S M_S,L M_L|J M_J> $$

are the Clebsch-Gordan coefficients associated with the addition of angular momentum, with sums over magnetic quantum numbers are assumed. Also, $L=L'=1$ but the $M_L$ and $M'_{L'}$ are unspecified. I want to perform the angular integrals first, so that the radial integrals over Gaussians can be easily performed after. My first attempt was to write the triple product as

$$ \mathbf{S}\cdot \mathbf{(k\times Q)} =\epsilon_{ijk}S_{i}k_{j}Q_{k} $$

and then write the cartesian components of $k$ and $Q$ in terms of spherical harmonics via

$$ k_{i}=\sqrt{\frac{4\pi}{3}}k\sum_{\alpha}\epsilon^{i}_{\alpha}(\hat{z})Y_{1 \alpha}(\hat{k}) $$

where the $\epsilon$ are the usual polarization vectors. This gives me something like

$$ \int d^3k d^3Q kQ\phi^{*}_{L'}(k+Q)Y^{*}_{L',M'_{L'}}(\widehat{k+Q})C^{J'M'_{J'}}_{L'S'}\frac{4\pi}{3}\sum_{\alpha,\alpha'}\mathbf{S}\cdot \mathbf{(\epsilon_{\alpha}\times \epsilon_{\alpha'})} Y_{1 \alpha}(\hat{k})Y_{1 \alpha'}(\hat{Q})\phi_{L}(k)Y_{L,M_{L}}(\widehat{k})C^{JM_{J}}_{LS}. $$

From here, I was going to integrate the spherical harmonics of the same variable, like

$$ \int d\Omega_{Q} Y^{*}_{1,a}(\hat{Q})Y_{1,b}(\hat{Q})=\delta_{a,b} $$

but I have spherical harmonics of $k$, $Q$ and $k+Q$, so I am stumped. Any help would be appreciated. Mostly, I was wondering if my approach is completely off.

Thanks.

$\endgroup$

1 Answer 1

0
$\begingroup$

Why don't you change integration variables from $\vec Q$ to $\vec q = \vec Q+\vec k$? Your $\vec S \cdot (\vec k \times \vec Q) = \vec S\cdot (\vec k\times \vec q)$, and your expansion in components will now be with $Y(\hat q)$ and $Y(\hat k)$.

$\endgroup$
1
  • $\begingroup$ That is actually what I was looking for. Thank you for your help! $\endgroup$
    – Christian
    Commented Jun 13, 2022 at 13:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.