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Angular momentum values and probabilities

  1. May 2, 2015 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    upload_2015-5-2_16-43-17.png

    2. Relevant equations
    upload_2015-5-2_16-46-45.png
    upload_2015-5-2_16-47-33.png
    upload_2015-5-2_16-48-1.png
    3. The attempt at a solution
    For a angular momentum ##J##, there exists the eigenvalue problems
    $$J^{2} \mid j \hspace{0.02 in} m \rangle = j(j +1)\hbar^{2} \mid j \hspace{0.02 in} m \rangle $$
    $$ J_{z} \mid j \hspace{0.02 in} m \rangle = m \hbar \mid j \hspace{0.02 in} m \rangle $$

    (a)
    So I will plug in the values to get me started with this position and spin state
    $$ \frac {1}{\sqrt{24}} a^{-3/2} \frac {r}{a} e^{-r/2a} \Big [ \sqrt {\frac {1}{4 \pi}} cos \hspace{0.02 in} (\theta) \begin{pmatrix} 1 \\ 0 \end{pmatrix} - \sqrt {\frac {1}{4 \pi}} sin \hspace{0.02 in} (\theta) \hspace{0.02 in} e^{i \phi} \begin{pmatrix} 0 \\ 1 \end{pmatrix} \Big ] $$

    But I have no clue what theta, phi, or r are supposed to be, or what I should do from here if this is even the correct first step.
     
    Last edited: May 2, 2015
  2. jcsd
  3. May 2, 2015 #2

    BruceW

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    You Have written it out correctly. But that is not necessary yet. Look at what the question is asking, it wants to know about eigenvalues. And the state you are given contains terms of spherical harmonics. What can you say about the spherical harmonics? (It makes the first question easier to deal with, rather than having to write out the entire state as a function of position coordinates).
     
  4. May 2, 2015 #3

    Maylis

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    Okay, so ##L^{2} \mid l m \rangle = l(l+1) \mid l m \rangle##, but how do I know what ##l## is? I didn't really understand your response, particularly
    I have no idea what you are getting at.

    I suppose based on the coefficients that one probability is ##\sqrt{1/3}## and the other is ##\sqrt{2/3}##, but I don't know the values.
     
  5. May 2, 2015 #4

    BruceW

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    If you look up some information on the spherical harmonics in quantum mechanics, then you can find what ##l## is.
     
  6. May 2, 2015 #5

    Maylis

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    Oh yes, so it is ##l = 1##, so then the value is ##1(1+1)\hbar^{2} = 2 \hbar^{2}##. So then you have ##\sqrt{1/3}## chance of getting ##2 \hbar^{2}##? But the azimuthal number is the same for both spherical harmonics, so then it seems like you just have 100% chance of getting ##L^{2} = 2 \hbar^{2}##?
     
  7. May 2, 2015 #6

    BruceW

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    Yep. That looks right.
     
  8. May 2, 2015 #7

    Maylis

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    Okay, so for (b) then it would be ##L_{z} = 0## and ##P(0) = \sqrt{1/3}##, and ##L_{z} = \hbar## with ##P(\hbar) = \sqrt{2/3}##?
     
  9. May 2, 2015 #8

    BruceW

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    We need to take the square of the coefficients to get the Probabilities. And as a sanity check, the probabilities of all possible outcomes should add up to 1.
     
  10. May 2, 2015 #9

    Maylis

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    Okay great. Now I am on part (e), knowing ##J = L + S##

    I have

    $$J^{2} \mid j \hspace{0.02 in} m \rangle = j(j+1) \hbar^{2} \mid j \hspace{0.02 in} m \rangle$$

    Does it mean ##j = l + s##? If so, I know ##l = 1## in both cases. But also, isn't ##s = 1/2## in both cases? Since ##S^{2} \mid s \hspace {0.02 in} m \rangle = s(s+1) \mid s \hspace {0.02 in} m \rangle##

    For ##\chi_{+}##
    $$S^{2} \mid \frac {1}{2} \hspace {0.02 in} \frac {1}{2} \rangle = \frac {1}{2}(\frac {1}{2}+1)\hbar^{2} \mid \frac {1}{2} \hspace {0.02 in} \frac {1}{2} \rangle$$

    And for ##\chi_{-}##
    $$S^{2} \mid \frac {1}{2} \hspace {0.02 in} -\frac {1}{2} \rangle = \frac {1}{2}(\frac {1}{2}+1) \hbar^{2} \mid s \hspace {0.02 in} - \frac {1}{2} \rangle$$

    So in both cases ##j = 1 + \frac {1}{2} = \frac {3}{2}##

    Therefore,
    $$J^{2} \mid \frac {3}{2} m \rangle = \frac {3}{2}(\frac {3}{2} + 1) \hbar^{2} \mid \frac {3}{2} m \rangle $$

    What should ##m## be?
    ##J^{2} = \frac {15}{4} \hbar^{2}## with ##P = 1##

    P.S. Since we didn't learn about Clebsch-Gordon coefficients, we do not need to actually find the probability
     
    Last edited: May 2, 2015
  11. May 6, 2015 #10

    BruceW

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    unfortunately, it is not possible to add up ##j = l + s##. It is slightly more complicated. Thinking about it intuitively, the orbital and spin angular momenta may be somewhat misaligned, therefore the measurement of the magnitude of angular momentum will have more than one possible value. On the other hand, ##m## is more simple. The operator for the z component of the sum of momenta is simply the addition of the operators for the z components of the two individual momenta.
     
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