Spherical Harmonics (QM)

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1. Homework Statement



upload_2015-9-13_12-23-43.png




Homework 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.
 

Answers and Replies

  • #2
stevendaryl
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First of all, since [itex]L_z = -i \hbar \frac{\partial}{\partial \phi}[/itex], the [itex]\theta[/itex]-dependence is irrelevant. So the real question is: how do you write [itex]sin 3\phi[/itex] as a sum of terms of the form [itex]e^{i m \phi}[/itex]?
 
  • #3
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## \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
stevendaryl
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## \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 ..

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

[itex]f(\theta, \phi) = \sum_{m l}C_{lm} Y_{lm} (\theta, \phi)[/itex]

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

[itex]sin(\theta) sin(3 \phi) = \sum_l (C_{l, +3} Y_{l, +3} + C_{l, -3} Y_{l, -3})[/itex]

You don't actually need to solve for the coefficients [itex]C_{l, \pm 3}[/itex]
 
  • #5
vela
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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.
 

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