# Orbital angular momentum projection

1. Nov 18, 2014

### 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?

2. Nov 18, 2014

### The_Duck

Almost.
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

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

3. Nov 18, 2014

### 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?

4. Nov 18, 2014

### 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.

5. Nov 18, 2014

### VantagePoint72

Very helpful, thank you.

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