By the Born rule in Quantum Mechanics, a state's complex wave function $$\Psi(x,t) \in L^2$$ gives probabilities when we take its complex norm $$\overline\Psi(x,t)\cdot\Psi(x,t) = |\Psi(x,t)|^2$$. In this case, since I'm using the cartesian spatial wave function, the probability of finding the particle that is in the state $$|\Psi\rangle$$ in a small region around $$x$$ at time $$t$$ is: $$|\Psi(x,t)|^2\,dx$$ Now, say I have the spatial wave function for the ground state of hydrogen, $$\psi_{100}(r,\phi,\theta) = \frac{1}{\sqrt{\pi a^3}}e^{-r/a}$$, which is given in spherical coordinates (where $$\phi$$ is measured azimuthally). If I now want to know the probability of finding the particle in a small region around a certain radius, I turn out to have to consider a thin shell of radius $$r$$ as follows: $$P=|\psi_{100}(r,\phi,\theta)|^2 dV = |\psi_{100}(r,\phi,\theta)|^2d\left(\frac{4}{3}\pi r^3\right)=|\psi_{100}(r,\phi,\theta)|^24\pi r^2dr$$ ... and thus, if $$\frac{dP}{dr}=f(r)$$ is the radial probability density, we have $$f(r)=\frac{1}{\pi a^3}e^{-2r/a}4\pi r^2$$.

However, what if I want the probability densities $$g(\phi)$$, $$h(\theta)$$ or even $$k(\phi,\theta)$$? As an analogue to a thin spherical shell, I can only imagine constructing thin, outward-beaming rectangles with volume $$dV = r^2\sin(\theta)\,d\theta\,d\phi\,dr = d\Omega\,r^2\,dr$$ and thus, presumably, if I only want to vary $$\theta$$ and $$\phi$$, summing those volumes like you would for the full-solid-angle shells: $$dv = d\Omega \int^r_0r^2\,dr = d\Omega\,\frac{r^3}{3}$$ But now, what is $$g$$, $$h$$ or $$k$$? How, for one, do I sneak the last result into $$P=|\psi_{100}(r,\phi,\theta)|^2\,dV$$?

More generally, how does one go about this kind of converting to specific probability distribution variables for any wavefunction in spherical (and, while we're at it, cartesian) coordinates? I'm aware that in probability theory, doing a change of variables from a single to another single variable is always based on the equation $$f(x)dx = g(y)dy$$, but in this case, I'm stuck.

• I have corrected one point in your notation. The angular integration measure $d\Omega=\sin\theta\,d\theta\,d\phi$ already contains the $\sin\theta$ factor from the expression for $dV$ in spherical coordinates.
– Buzz
Jun 6, 2020 at 23:07
• Ah, yes, thanks for that.
– Mew
Jun 6, 2020 at 23:47

If you want to know the probably density $$k(\phi,\theta)$$ for the particle to be located in at an angular position $$(\phi,\theta)$$, you have to integrate out the radial coordinate: $$k(\phi,\theta)=\int_{0}^{\infty}r^{2}\, dr\,P(r,\phi,\theta)=\int_{0}^{\infty}r^{2}\, dr\,\left|\Psi(r,\phi,\theta)\right|^{2}.$$ For an $$S$$ state, this is not very interesting, because the wave function is independent of the angles, and thus $$k_{n00}(\phi,\theta)=\int_{0}^{\infty}r^{2}\,\left|\psi_{n00}\right|^{2}=\frac{1}{4\pi},$$ but for wave functions with nonzero angular momentum, you will get a nontrivial function $$k_{nlm}(\phi,\theta)=\left|Y_{lm}(\phi,\theta)\right|^{2},$$ where $$Y_{lm}(\phi,\theta)$$ is the spherical harmonic that represents the angular part of the wave function when it is in an angular momentum eigenstate.
• That's interesting. I do remember something along the lines of this from when I studied multivariate probability theory. I'm wondering, then: is it just a coincidence that the $d$-method I used for the radial probability distribution is equivalent to integrating over the entirety of angular space (giving $4\pi$, and leaving the $r^2$ from the Jacobian)? Also, I'm not quite sure what's happening between the second equality: isn't there some infinitesimal factor (like $d\Omega$) we're forgetting there, for $|\Psi|^2$ to count as a probability distribution?