# Spherical Harmonics (QM)

1. Sep 13, 2015

### Safinaz

1. The problem statement, all variables and given/known data

2. Relevant equations

Here we have to express $\psi(\theta,\phi)$ in terms of spherical harmonics $Y_{lm}$ to find the angular momentum.

If $\psi(\theta,\phi) = i \sqrt{\frac{3}{4\pi}} \sin{\theta} \sin{\phi}$, it can be written as:
$$\frac{i}{\sqrt{2}} (Y_{1,1}- Y_{1,-1})$$
since :
$Y_{1,\pm1} = \mp \sqrt{\frac{3}{8\pi}} \sin{\theta} e^{\pm i \phi}$,

But now as $\psi(\theta,\phi)$ has $\sin{3 \phi}$ instead of $\sin{ \phi}$, how will it be represented ?

S.

2. Sep 13, 2015

### stevendaryl

Staff Emeritus
First of all, since $L_z = -i \hbar \frac{\partial}{\partial \phi}$, the $\theta$-dependence is irrelevant. So the real question is: how do you write $sin 3\phi$ as a sum of terms of the form $e^{i m \phi}$?

3. Sep 13, 2015

### Safinaz

$\sin{3\phi} = \frac{e^{3i\phi } - e^{-3i\phi } }{2i}$ , but alternatively I want to express $\sin{3\phi}$ in terms of $\sin{\phi}$ (I guess ), in order to use $Y_{1,\pm 1}$. Also I tried to find any $Y_{lm}$ defined by $\sin{m\phi}$ or $e^{\pm3i\phi }$ but I didn't find, look for example at Table : 5.2 " [Nouredine_Zettili]_Quantum_Mechanics_Concepts ". So any ideas ..

4. Sep 13, 2015

### stevendaryl

Staff Emeritus
Why do you want to write it in terms of $Y_{1, \pm 1}$? In general, you write a function $f(\theta, \phi)$ in the form:

$f(\theta, \phi) = \sum_{m l}C_{lm} Y_{lm} (\theta, \phi)$

Since each $Y_{lm} \propto e^{im\phi}$, there are only two terms involved in the sum over $m$: $m=\pm 3$. So your case boils down to:

$sin(\theta) sin(3 \phi) = \sum_l (C_{l, +3} Y_{l, +3} + C_{l, -3} Y_{l, -3})$

You don't actually need to solve for the coefficients $C_{l, \pm 3}$

5. Sep 13, 2015

### vela

Staff Emeritus
To elaborate on what steven said, you could find the coefficients $C_{lm}$ by using the orthogonality of the spherical harmonics:
$$C_{lm} = \int Y^*_{lm}(\theta,\phi) \psi(\theta,\phi)\,d\Omega.$$ Try evaluating the phi integral, and you'll see why $m=\pm 3$ and why the $\theta$ dependence doesn't really matter in this problem.