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Homework Help: Quantum Mechanics, Angular Momentum Measurement Problem

  1. May 2, 2010 #1
    The problem involves measurement in a quantum system with angular momentum. There are two parts I am struggling with. Apologies for the images, I tried to put it all in latex but failed!

    Part 1:

    [PLAIN]http://i2.sqi.sh/s_1/1xo/part1.jpg [Broken]

    I have found a long derivation in a book involving a function Ylm, but I am sure I must be missing a simple trick to derive the eigenvalues simply from this data given? I have found that the eigenvalue should be mh, but I am stumped as to how I can find this from L2. Any hints are welcome.

    Part 2:

    [PLAIN]http://i2.sqi.sh/s_1/1xp/part2.jpg [Broken]

    I have no problem with the first part of this question, finding the eigenvalues with respect to L2 and Lz, but the second part has me stumped.

    My eigenvalues are as follows, found using a form of L2 and Lz given previously in the question.

    Firstly for L2

    [tex]2\,{h}^{2}, 2\,{h}^{2}+{\frac {{h}^{2}}{{\sin}^{2}\theta}}, 2\,{h}^{2}-{\frac {{h}^{2}}{{\sin}^{2}\theta}[/tex]

    And for Lz

    [tex]0, -h, h[/tex]

    My problem is, I'm not sure what a simultaneous measurement is? I can see that the answers given are simply the eigenvalues for Lz, and one of the eigenvalues for L2, but I'm not quite sure how they are connected?

    I am also not sure about the final part of this question. Should I try to write [tex]\Psi[/tex] as a linear combination of u1, u2 and u3?

    Thanks in advance for any help, I think my main problem is interpreting what I should be finding, I don't think the maths will be that difficult.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 2, 2010 #2


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    For part one, assume that [itex]\psi(r,\theta,\phi)=R(r)\Theta(\theta)\Phi(\phi)[/itex] is an eigenfunction of [tex]\hat{L}_z[/itex] (i.e. Assume the eigenfunctions are separable). The eigenvalue equation for [tex]\hat{L}_z[/itex] will then give you a separable 1st order ODE to solve for [itex]\psi[/itex].
  4. May 2, 2010 #3


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    The second and third values are wrong. The eigenvalues are constants. They won't depend on θ.
    Yes, that's exactly what you want to do.

    When you have a state, how do you determine what the possible outcomes of a measurement are and their probabilities? Your textbook should cover this topic if you don't know the answer. It's one of the basic things you need to know in quantum mechanics.
  5. May 3, 2010 #4
    First part is so simple first you write the l^2 in the form of x,y and z-coordinates and then add it

    you see the book of neuredine zettlie of quantum mechanics
  6. May 3, 2010 #5
    Thanks for the responses.

    Thanks, I have had a go but not sure I am on the right lines.

    I start with;

    [tex]-i\hbar \frac{\partial}{\partial \theta} \psi = X \psi[/tex]

    where X is the eigenvalue? Is this right? If I proceed this way, I get the eigenfunction as an exponential.

    The form of [tex]\hat{L}^2[/tex] I am given gives me the [tex]sin^2(\theta)[/tex] term. I have redone my calculations and I still get this term. What should I get? I'm sorry I cannot see where I have gone wrong here.

    Am i correct in assuming the possible outcomes of a measurement are the eigenvalues? And the probability of these eigenvalues being obtained is the modulus squared of the constant in the linear combination?

    I still don't quite understand what a "simultaneous measurement" is?

    Sorry I don't quite understand what you mean. Can you be a bit more specific?
  7. May 3, 2010 #6


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    You should be differentiating with respect to ϕ, not θ, and to be a bit more precise, you found that the eigenfunction is proportional to an exponential. Did you assume the form of ψ that gabbagabbahey suggested?
    You should just get [itex]2\hbar^2[/itex] for all three. You might want to post your work here so we can see where you went astray.
    Yes and yes.
    When they say you do the measurements simultaneously, they mean you're doing both measurements on the same initial state. If you did one measurement followed by the other, the first measurement would cause the wavefunction to collapse to a new state, and the second measurement would then be performed on this new state.
  8. May 3, 2010 #7
    Sorry, just a mistype, should have been ϕ.

    So I get [tex] -i\hbar R(r)\Theta(\theta)\frac{\partial \Phi}{\partial \phi}=X R(r)\Theta(\theta)\Phi(\phi)[/tex]

    Cancelling the R and θ functions, I get

    [tex] -i\hbar \frac{\partial \Phi}{\partial \phi}=X \Phi(\phi)[/tex]

    Then solving for [tex]\Phi[/tex] I get

    [tex] \Phi = exp{\frac{-X\phi}{i\hbar}}[/tex]

    Is this correct? If so, where do I go from here?

    Finally sorted it and got the same answer for all 3, sorry, was just some bad calculus.

    Thanks for clearing this part up, I think I was just reading too much into it.

    Thanks for the continued support, I'm very grateful.
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