Orbital angular momentum projection

[VantagePoint72]

Suppose I have particle in three dimensional space whose position space wavefunction in spherical coordinates is $\psi(r,\theta,\phi)$. The spherical harmonics $Y_{\ell,m}$ are a complete set of functions on the 2-sphere and so any function $f(\theta,\phi)$ can be expanded as $f(\theta,\phi) = \sum_{\ell=0}^\infty \sum_{m=-\ell}^{\ell} f_{\ell,m} Y_{\ell,m}(\theta,\phi)$. Since $\psi$ may, in general, have radial dependence, the coefficients $f_{\ell,m}$ will too, i.e., $f_{\ell,m}(r)$.

I'm curious, then, what the probability is that the particle will be measured have total orbital angular momentum $\sqrt{\ell_0(\ell_0+1)}\hbar$. I'm used to seeing expansions of quantum states in which the coefficients are just constants and so I'm a bit thrown by the radial dependence in this case. Intuitively, I'd expect that a total orbital angular momentum measurement would yield $\sqrt{\ell_0(\ell_0+1)}\hbar$ with probability
$P(\ell_0) = \sum_{m=-\ell_0}^{\ell_0} \int_0^\infty dr |f_{\ell_0,m}(r)|^2$

Is that right?

 

[The_Duck]

Almost.

Intuitively, I'd expect that a total orbital angular momentum measurement would yield $\sqrt{\ell_0(\ell_0+1)}\hbar$ with probability
$P(\ell_0) = \sum_{m=-\ell_0}^{\ell_0} \int_0^\infty dr |f_{\ell_0,m}(r)|^2$

Is that right?

Almost. It should be the norm of the part of the wave function with $\ell = \ell_0$, so

$$P(\ell_0) = \int_0^\infty dr r^2 \int_0^{2 \pi} d\phi \int_0^\pi d\theta \cos \theta \left| \sum_{m = -\ell_0}^{\ell_0} f_{\ell_0, m}(r) Y_{\ell_0, m}(\theta, \phi) \right|^2$$

You can use the orthogonality of the spherical harmonics to rewrite this as

$$P(\ell_0) = \sum_{m = -\ell_0}^{\ell_0} \int_0^\infty dr r^2 |f_{\ell_0, m}(r) |^2 \int_0^{2 \pi} d\phi \int_0^\pi d\theta \cos \theta \left|Y_{\ell_0, m}(\theta, \phi) \right|^2$$

If we assume that the spherical harmonics are normalized so that

$$\int_0^{2 \pi} d\phi \int_0^\pi d\theta \cos \theta |Y_{\ell, m}(\theta, \phi)|^2 = 1$$

Then we get

$$P(\ell_0) = \sum_{m=-\ell_0}^{\ell_0} \int_0^\infty dr r^2 |f_{\ell_0,m}(r)|^2$$

 

[VantagePoint72]

Interesting, I wouldn't have thought you'd need to do the whole spatial integral, particularly the angular part. I suppose that's just to take the inner product between spherical harmonics?

 

[The_Duck]

Well, you can see that you don't actually have to do any angular integrals in the end. But the norm is defined in terms of the integral over all space, so I started from that.

 

[VantagePoint72]

Very helpful, thank you.