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Expectation value of z component of angular momentum for a particle on a ring

  1. Oct 14, 2012 #1
    I have to find the expectation value of the z component of the angular momentum for a particle on a ring and the expectation value of the z component of the angular momentum squared for a particle on a ring.

    The wavefunction is e^((± imx))

    I've determined that the expectation value for the z component is -[itex]\hbar[/itex]/m and that the expectation value for the square of the z component is [itex]\hbar[/itex] squared over m squared.

    This would mean that the uncertainty in the z component of the angular momentum for a particle on a ring is 0.

    Is this correct?
     
  2. jcsd
  3. Oct 15, 2012 #2

    Jano L.

    User Avatar
    Gold Member

    No, the angular momentum should come out as proportional to m. You have to calculate the average value of L_z in this way:

    [tex]
    \langle L_z \rangle = \int \psi^* \hat{L}_z \psi~d\tau
    [/tex]

    where [itex]\tau[/itex] stands for the coordinates of the particle. On a circle, the position of particle is given by just one coordinate, usually angle. Let us denote it by [itex]\varphi[/itex]. It takes values from 0 to 2[itex]\pi[/itex]. In this coordinate, the operator of angular momentum is given by

    [itex]
    \hat{L}_z = i\hbar \frac{\partial }{\partial \varphi}
    [/itex]

    Tha last thing you need is the wave function. You gave [itex]e^{im\varphi}[/itex], but this is not correct wave function because it is not normalized.

    The correct function has to satisfy

    [tex]
    \int_0^{2\pi} \psi^* \psi ~d\varphi = 1,
    [/tex]

    so you will have to change your function little bit.
     
  4. Oct 15, 2012 #3
    I get

    [itex]\frac{1}{2π}[/itex][itex]\int[/itex](-i[itex]\hbar[/itex]/im) d[itex]\varphi[/itex] which, integrated over 0 to 2π yields -[itex]\hbar[/itex]/m

    following the same procedure, I find the expectation of L[itex]_{z}^{2}[/itex] is -[itex]\hbar^{2}[/itex]/m[itex]^{2}[/itex]

    Still the same result
     
  5. Oct 16, 2012 #4

    Jano L.

    User Avatar
    Gold Member

    How did you get m into denominator?
     
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