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I Possible measurements of z-component of angular momentum

  1. May 4, 2016 #1
    I'm looking through an old exam, and don't quite understand the solution given for one of the problems.

    We have given a wavefunction [itex] g(\phi,\theta) = \sqrt{\frac{3}{8\pi}}(-cos(\theta) + isin(\theta)sin(\phi)) [/itex]

    and are asked what possible measurements can be made of the z-component of the angular momentum.
    My instinct is to use the operator [itex] \hat{L}_z = \frac{\hbar}{i} \frac{\partial}{\partial \phi} [/itex] on the wavefunction, which shows that g is not an eigenfunction of the operator.

    In the solution however, they rewrote g as [itex] g(\phi,\theta) = \sqrt{\frac{3}{8\pi}}(-cos(\theta) + \frac{1}{2}sin(\theta)(e^{i\phi}-e^{-i\phi})) [/itex] and said that the first term is an eigenfunction of [itex] \hat{L}_z [/itex] with the eigenvalue m=0, the second term has the eigenvalue m=1 and the last m=-1. Therefore the possible measurements are [itex] m=0,\pm{1} [/itex].

    My question is why this is valid. Why does one of the components of the original wavefunction give you one of the measurement values? Don't you have to apply [itex] \hat{L}_z [/itex] to the entire wavefunction? Has this got something to do with your choice of [itex] \phi [/itex]?
  2. jcsd
  3. May 4, 2016 #2


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    Staff: Mentor

    They've just rewritten the wavefunction as a superposition of three eigenfunctions.
    If you apply ##L_z## to the entire wave function you will get the expectation value of ##L_z##. The possible results and their probabilities are given by the Born rule: The result will be an eigenvalue, and the probability of getting a particular eigenvalue is the square of the coefficient of that eigenfunction.
  4. May 5, 2016 #3


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    I think, it's just formulated in a somewhat strange way. I guess, what they wanted to know is, which possible values ##L_z## can take, if the system is prepared in the given state. Since the state belongs obviously to ##l=1##, you can just calculate
    $$\langle 1,m |g \rangle=\int_{\Omega} \mathrm{d}^2 \Omega [Y_1^m(\vartheta,\varphi)]^* g(\vartheta,\varphi) \quad \text{for} \quad m \in \{-1,1,0 \}.$$
    Then all ##m## are possible outcomes of an ##L_z## measurement, for which this product is different from 0. Of course this is equivalent to the answer given in #1, because indeed you can immediately read off the decomposition of ##g## in terms of ##Y_1^m##.
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