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Calculating the probability of a certain measurement

  1. Mar 12, 2013 #1
    1. The problem statement, all variables and given/known data
    This really is not a homework problem but I am studying for the qualifying exam upcoming. I came across an objective that I am not familiar with. I'm given a wave function made of a linear combination of spherical harmonics with complex coefficients. I'm asked to calculate the possible values of measurement for [itex]L^{2}[/itex] and [itex]L_{z}[/itex] which is of course straight forward. What I am unsure of though is it asks me to calculate the probability of obtaining those values. Perhaps I missed this in my undergraduate coursework somehow, but I'll list the wavefunction.


    2. Relevant equations
    [itex]\psi=A[(1+2i)Y_{3}^{-3}+(2-i)Y_{3}^{2}+\sqrt{10}Y_{2}^{2}][/itex]

    3. The attempt at a solution

    I've already calculated the normalization constant A, and found it via integration and the orthogonality condition to be

    [itex]A=\frac{1}{4}\sqrt{\frac{7}{6\pi}}[/itex]

    I know the values of angular momentum [itex]L^{2}[/itex] are [itex]12\hbar^{2}[/itex] corresponding to [itex]\left|3,-3\right\rangle[/itex], and [itex]\left|3,2\right\rangle[/itex]. Also, [itex]6\hbar^{2}[/itex] corresponding to [itex]\left|2,2\right\rangle[/itex]. I am utterly lost on how to calculate this probability. I have tried this:

    [itex]|\left\langle 3, m\right|A[(1+2i)\left|3,-3\right\rangle+(2-i)\left|3, 2\right\rangle+\sqrt{10}\left|2,2\right\rangle]|^{2}[/itex] for m = -3, 2

    &

    [itex]|\left\langle 2, 2\right|A[(1+2i)\left|3,-3\right\rangle+(2-i)\left|3, 2\right\rangle+\sqrt{10}\left|2,2\right\rangle]|^{2}[/itex]


    This gives me values that do not add up to 1 as expected. Is this right, and if so perhaps I am doing it incorrectly? Any pointers would be appreciated.
     
  2. jcsd
  3. Mar 12, 2013 #2

    TSny

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    Usually the functions ##Y^m_l## are normalized so they form an orthonormal set. If so, I don't see how you got the value for ##A##.
     
  4. Mar 12, 2013 #3
    That is true but then how else would I calculate the value for A if not by the condition,

    [itex]\int\int Y_{l'}^{m'}\bar{Y_{l}^{m}}d\Omega = \frac{4\pi}{2l+1}\delta_{l,l'}\delta_{m,m'}[/itex]

    where [itex]d\Omega = sin(\theta)d\theta d\phi [/itex] for [itex]0≤\theta≤\pi[/itex], and [itex]0≤\phi≤2\pi[/itex] ?

    The condition, as always, is to calculate [itex]\int |\psi|^{2} dV[/itex]=1 over all space, is it not? According to this, and by orthogonality of the spherical harmonics the cross terms will cancel and the resulting integration yields the number I obtained. Perhaps I miss-calculated something.
     
  5. Mar 12, 2013 #4

    TSny

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    OK, you are using a normalization of the spherical harmonics such that they are orthogonal but not orthonormal. See here for some different choices of normalization. [EDIT: I agree with your result for A for your normalization.]

    The answers for the individual probabilities will depend on how the functions ##Y^m_l## are assumed to be normalized in the statement of the problem. But, nevertheless, the sum of the probabilities should add to 1.

    The probability for measuring ##L^2 = 6\hbar^2## would be ##\frac{|<2,2|\psi>|^2}{|<\psi|\psi>|^2|<2,2|2,2>|^2}## if ##|2,2>## denotes ##Y^2_2##.
     
  6. Mar 12, 2013 #5
    So it turns out I used the wrong normalization. In that case my normalization constant becomes

    [itex]A=\frac{1}{2\sqrt{5}}[/itex]

    Such that the probabilities become, 1/4 each for measuring the [itex]12\hbar^{2}[/itex] and 1/2 for measuring the [itex]6\hbar^{2}[/itex]. This of course adds up to 1 as expected. Thanks for your help!
     
  7. Mar 12, 2013 #6

    TSny

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    Looks good!
     
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