1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spherical Harmonics (QM)

  1. Sep 13, 2015 #1


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



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



    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. jcsd
  3. Sep 13, 2015 #2

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    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]?
     
  4. Sep 13, 2015 #3
    ## \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 ..
     
  5. Sep 13, 2015 #4

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    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]
     
  6. Sep 13, 2015 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Spherical Harmonics (QM)
  1. Spherical harmonics (Replies: 0)

  2. Spherical Harmonics (Replies: 3)

  3. Spherical Harmonics (Replies: 3)

Loading...