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Quantum Mechanics - Spherical Harmonics

  1. Feb 11, 2009 #1
    1. The problem statement, all variables and given/known data

    The angular part of a system’s wavefunction is

    [tex]<\theta, \phi | \psi>\propto (\sqrt{2}\cos\theta + \sin{\theta}e^{−i\psi} - \sin{\theta}e^{i\psi} ). [/tex]

    What are the possible results of measurement of (a) [tex]L^2[/tex] , and (b) [tex]L_z[/tex] , and their probabilities? What is the expectation value of [tex]L_z[/tex]?

    2. Relevant equations

    [tex]L^2|E> = l(l+1)|E>[/tex]

    3. The attempt at a solution

    I can see that the wavefunction takes the form of the spherical harmonics for l = 1, and from that I think I can say that [tex]L^2[/tex] = 2. However, I'm unsure whether this is correct, or how to find the probabilities.

    I haven't had much luck at all with [tex]L_z[/tex].
  2. jcsd
  3. Feb 11, 2009 #2
    eigenvalues of [itex]L^2[/itex] are [itex]l(l+1) \hbar^2[/itex]
    and eigenvalues of [itex]L_z[/itex] are [itex]m \hbar[/itex] where m ranges from l to -l
  4. Feb 11, 2009 #3
    Thanks, so the eigenvalue of [tex]L^2[/tex] is [tex]2 \hbar^2[/tex], which must presumably have a probability of 1 as it's the only eigenvalue.

    The eigenvalues of [tex]L_z[/tex] are [tex]-\hbar, 0, \hbar[/tex], but how do I figure out their probabilities?

    Sorry if this is a really stupid/easy question - as you can probably tell, QM isn't my forte by a long shot.
  5. Feb 11, 2009 #4
    why is l=1? l runs from 0 to n-1
  6. Feb 11, 2009 #5
    Because the functions that form the systems wave function are the three spherical harmonics of l=1.
  7. Feb 11, 2009 #6


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    Try writing the wave function in the form
    and find the [itex]c_{1,m}[/itex] coefficients. Can you say what the probabilities are in terms of these?
  8. Feb 11, 2009 #7
    Through a combination of the textbook and lecture notes, I get the constants to be:

    [tex]c_1^0 = \sqrt{\frac{6}{8 \pi}}[/tex]
    [tex]c_1^{\pm 1} = \pm \sqrt{\frac{3}{8 \pi}}[/tex]

    However, the sum of the square of these doesn't equal 1 - am I supposed to include the Y function when squaring to find the probability?
  9. Feb 11, 2009 #8
    I don't think the sum of the squares needs to be one in this case since they tell you proportional to and not equals. (Essentially you will need to find the correct constant that normalizes the wave function)
  10. Feb 11, 2009 #9
    Right, I think I've got it. So since we know that the coefficient of m=0 is a factor of [tex]\sqrt{2}[/tex] larger than m=±1, we find that...

    P(m=0) = 0.5
    P(m=1) = 0.25
    P(m=-1) = 0.25

    Hopefully, that's right. Thanks for everyone's help, I think I understand it all slightly better now (cue everyone telling me that I'm wrong).
  11. Feb 11, 2009 #10
    seeing as your coefficients were root(6/8pi) and root(3/8pi) you can see that one is double the other two and you know they have to sum to be 1 so your answer above is correct. i reckon you just hadn't properly normalised them earlier and that's why they weren't summing to 1.
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