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Angular Momentum(Q&M)

  1. May 15, 2009 #1
    1. The problem statement, all variables and given/known data
    What are the possible measurements for Lz



    2. Relevant equations
    [tex] \psi(\theta,\phi) = \sqrt{\frac{3}{4 \pi}} sin(\phi) sin(\theta) [/tex]
    probability Lz quantum


    3. The attempt at a solution
    Well I'm sure I can expand [tex] sin(\phi)= \frac{e^{i \phi}-e^{-i \phi}}{2 i}[/tex]
    Getting m=1,-1.
    [tex] \psi(\theta,\phi) = \sqrt{\frac{3}{4 \pi}} sin(\theta) \frac{e^{i \phi}-e^{-i \phi}}{2 i}[/tex]
    Should the the probability be the coefficents mod squared?
     
  2. jcsd
  3. May 15, 2009 #2
    Yes, but you have to keep track of normalization; the coefficients should be those of normalized eigenfunctions of Lz. I suggest writing psi in terms of spherical harmonics.
     
  4. May 15, 2009 #3
    you are working with two quantum numbers, If I remember correctly you have add up the normalized coefficients squared by increasing their orbital quantum number l up to m.
     
  5. May 15, 2009 #4
    ok. So:
    [tex]\psi=\frac{\sqrt{2}i}{2}\left(Y_{1,1}+Y_{1,-1}\right)[/tex] with everything sorted
    I have equal probability of measuring [tex]\pm \hbar[/tex] for [tex]L_z[/tex], right?
     
  6. May 15, 2009 #5
    Wait, what I typed can't be right. m has to be -1,0,1 since l=1. so in terms of spherical harmonics I should have one more term [tex]Y_{1,0}[/tex] and then normalize that. That makes more sense.
     
  7. May 16, 2009 #6

    malawi_glenn

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    Homework Helper

    You don't need a Y(1,0) A priori.

    If your psi in post 4 equals the psi in Relevant equations in your first post, then it is ok.

    You also check the normalizability of your result in post #4 by simply integrating it over [itex]r^2 d\Omega[/itex] and see if you indeed get 1.

    Y(1,0) is proportional to cos(theta), and you only have sin(theta), so you should not expect a Y(1,0) term.
     
  8. May 16, 2009 #7
    ok.
    In post four the coefficents work out since[tex]\Sigma |c_k|^2=1[/tex]
     
  9. May 16, 2009 #8

    diazona

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    You could consider the coefficient of [tex]Y^1_0[/tex] to be 0 ;-)
     
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