Projection of a 3D spherical function to a carteasian axis

Question

I have a 3D function defined in a spherical coordinate system $(r,\theta,\phi)$, which is written as a product of a radial function $R_{nl}(r)$ and a spherical harmonic $Y_{lm}(\theta,\phi)$ I.e $$ \Psi_{nlm}(r,\theta,\phi) = R_{nl}(r)Y_{lm}(\theta,\phi). $$ Here $n,l$ and $m$ represent the quantum numbers and $\Psi_{nlm}(r,\theta,\phi)$ is a wavefunction, for some context and I only have numerical data for the radial functions. I want to plot the projection of the modulus squared of the wavefunction versus a specific carteasian axis. For example the projection in the z-axis can be defined as $$ f(z) = \int\int dxdy~ |\Psi_{nlm}(r,\theta,\phi)|^2. $$ I've been told this is equivalent to writing $$ f(z) = \int d\vec{r}~ \delta(\vec{r}\cdot\hat{k}-z)|\Psi_{nlm}(r,\theta,\phi)|^2 $$ where $\vec{r}$ is the vector in spherical coordinates and $\hat{k}$ is the unit vector in the z direction. Is there a way to calculate this function of z (also for x and y) using one of the two equivalent forms, without knowing the analytic form of the radial function?

Answer 1

I'll do it for the $z$ axis to give you the idea; it's also possible for the other axes, but a bit more complicated.

The $z$ axis case is simpler both because, as noted in the comments, the integration over $\phi$ yields $2\pi$, and because in this case we have simply $\vec r\cdot\hat k=r\cos\theta$. Thus after integrating over $\phi$, we're left with

$$ f(z)=2\pi\int_0^\infty\mathrm drr^2\int_0^\pi\mathrm d\theta\sin\theta\,\delta(r\cos\theta-z)\left|R_{nl}(r)\right|^2\left|Y_{lm}(\theta,0)\right|^2\;. $$

Since you don't have $R$ in functional form (I guess you have samples?), we want to perform the integral over the delta distribution in $\theta$, not in $r$:

\begin{eqnarray} f(z)&=&2\pi\int_0^\infty\mathrm drr^2\int_{-1}^1\mathrm d\cos\theta\,\delta(r\cos\theta-z)\left|R_{nl}(r)\right|^2\left|Y_{lm}(\theta,0)\right|^2\\ &=&2\pi\int_0^\infty\mathrm drr^2\int_{-1}^1\mathrm d\cos\theta\frac1r\delta(\cos\theta-z/r)\left|R_{nl}(r)\right|^2\left|Y_{lm}(\theta,0)\right|^2\\ &=&2\pi\int_0^\infty\mathrm drr\left|R_{nl}(r)\right|^2\left|Y_{lm}(\arccos (z/r),0)\right|^2\;. \end{eqnarray}

This you can evaluate numerically using your numerical representation of $R_{nl}(r)$.

The principle applied here that you'll need for the other cases is

$$ \delta(f(t)-z)=\sum_{f(t_i)=z}\frac{\delta\left(t-t_i\right)}{f'(t_i)}\;. $$