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Z-Component of Angular Momentum.

  1. Nov 7, 2013 #1
    1. The problem statement, all variables and given/known data

    A particle is in a state described by the wave function:

    [tex]\Psi = \frac{1}{\sqrt{4}}(e^{i\phi} sin \theta + cos \theta) g(r)[/tex],


    [tex]\int\limits_0^\infty dr r^{2} |g(r)|^{2} = 1[/tex]

    and [itex]\phi[/itex] and [itex]\theta[/itex] are the azimuth and polar angle, respectively.

    OBS: The first spherical harmonics are:

    [itex]Y_{0,0} = \frac{1}{sqrt{4 \pi}[/itex], [itex]Y_{1,0} = \frac{sqrt{3}}{sqrt{4 \pi} cos \theta[/itex] and [itex]Y_{1,\pm1} = \mp \frac{sqrt{3}}{sqrt{8 \pi}} e^{\pm i \phi} sin \theta[/itex]

    a - What are the possibles results of a measurement of the z-componet [itex]L_{z}[/itex] of the angular momentum of the particle in this state?
    b - What is the probability of obtaining each of the possible results of part (a)?
    c - What is the expectation value of [itex]L_{z}[/itex]?

    2. Relevant equations

    [tex]L_{z} = -i\hbar \frac{\partial}{\partial \phi}[/tex]

    [tex]L_{z} \Psi = m_{l} \hbar \Psi[/tex]

    3. The attempt at a solution

    I start by writing the wavefunction in terms of the spherical harmonics.

    [tex]\Psi = (- \frac{\sqrt{2}{\sqrt{3}}} Y_{1,1} + \frac{1}{sqrt{3} Y_{1,0}) g(r)[/tex]

    So, the possibles values for [itex]m_{l}[/itex] are [itex]0[/itex] and [itex]1[/itex]; and [itex]l_{z} = 0\hbar[/itex] and [itex]l_{z} = 1 \hbar[/itex]

    Thus, the probabilities should be [itex]P = \frac{2}{3}[/itex] for [itex]l_{z} = 1 \hbar[/itex] and [itex]P = 1/3[/itex] for [itex]l_{z} = 0 \hbar[/itex].

    For the expectation value, I should evaluate the integral:

    [tex]<L_{z}> = \int \Psi L_{z} \Psi \,d^{3}r = \int \Psi \frac{\partial \Psi}{\partial \phi} \,d^{3}r[/tex]

    [tex]<L_{z}> = \frac{2}{3} \int Y_{1,1} (-i \hbar \frac{\partial}{\partial \phi}) Y_{1,1} \,d^{3}r + \frac{1}{3} \int Y_{1,0} (-i \hbar\frac{\partial}{\partial \phi}) Y_{1,0} \,d^{3}r[/tex]

    The Y_{1,0} does not depend on the azimuth angle, so:

    [tex]<L_{z}> = \frac{2}{3} \int Y_{1,1} (-i \hbar \frac{\partial}{\partial \phi}) Y_{1,1} \,d^{3}r[/tex]

    [tex]<L_{z}> = \frac{2}{3} \hbar \int_0^\pi sin^{2} \theta \,d\theta \int_0^{2 \pi} e^{2 i \phi} \,d\phi[/tex]

    However, the result of this integral is zero.

    Is that right or am I doing something wrong?
  2. jcsd
  3. Nov 7, 2013 #2


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    Did you forget to take the complex conjugate somewhere?

    You can get the answer for ##<L_{z}>## without doing these integrations by just using the probabilities that you have found.
  4. Nov 7, 2013 #3
    Yes. The integral is actually:

    [tex]\int \Psi^{*} L_{z} \Psi \,d^{3}r = -i \hbar \int \Psi^{*} \frac{\partial \Psi}{\partial \phi} \,d^{3}r[/tex]


    [tex]Y_{l,m}^{*} = (-1)^{m}Y_{l,-m}[/tex]

    [tex]<L_{z}> = P_{m_{l} = 1} (m_{l} \hbar) + P_{m_{l} = -1} (m_{l} \hbar)[/tex]
    [tex]<L_{z}> = \frac{2}{3}(1 \hbar) + \frac{1}{3} (0 \hbar)[/tex]
    [tex]<L_{z}> = \frac{2}{3} \hbar[/tex]

    Is that it?

    By solvind the integral, I get [itex]<L_{z}> = \frac{2}{3} \pi^{2} \hbar[/itex].
    Last edited: Nov 7, 2013
  5. Nov 7, 2013 #4


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    Yes. (Maybe a typographical error in your first line.)

    It seems to come out ok for me. I don't think your original wave function ##\Psi## is normalized.
    (Your integrals over the spherical harmonics should not have ##d^3r## since you have presumably already integrated over ##r##.)
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