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Quantum mechanics problem

  1. May 9, 2005 #1
    I have a question that I'm struggling with a bit.

    The azimuthal part of the wavefunction of a particle is

    [tex]\Psi(\phi) = Ae^{-iq\phi}[/tex] where [tex]\phi[/tex] is the azimuthal angle. Show that q must be an integer. By normalising the wavefunction, find the value of A. What is the value of L_z for this particle?

    Ok, I know that [tex]\Psi(\phi) = \Psi(\phi + 2\pi)[/tex] because [tex]\phi[/tex] and [tex]\phi + 2\pi[/tex] are the same angle.

    So, [tex]Ae^{-iq\phi} = Ae^{-iq(\phi + 2\pi)}[/tex]

    and [tex]Ae^{-iq\phi} = Ae^{-iq\phi}e^{-iq2\pi}[/tex]

    [tex]\Rightarrow e^{-iq2\pi} = 1[/tex]

    How does this imply that q is an integer? This was the way it was done in lectures, but we were just told that this shows q is an integer. I thought it was something to do with [tex]e^{ix} = \cos x + i\sin x[/tex], but I'm not sure.

    For the normalising bit, I know I need to use [tex]\int \Psi^* \Psi d\phi = 1[/tex] but I'm not sure about the limits. This is what I've done:

    [tex]\int \Psi^* \Psi d\phi = 1[/tex]

    [tex]\int_{0}^{2\pi} Ae^{iq\phi}Ae^{-iq\phi} = 1[/tex]

    [tex]A^2 \int_{0}^{2\pi} d\phi = 1[/tex]

    So [tex]A = \sqrt{ \frac{1}{2\pi} }[/tex]

    Is this correct? As for the angular momentum component, I'm working on it.

    Thanks.
     
  2. jcsd
  3. May 9, 2005 #2

    dextercioby

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    Sure,that's the Condon-Shortley convention.Actually the wave function is a phase factor;so other one would be superfluous.

    [tex] e^{-iq2\pi}=\cos\left(-q2\pi\right)+i\sin\left(-q2\pi\right)=1 [/tex]

    So when is the cosine =1 ...?(Don't worry,the sine in those points is automatically 0)

    Daniel.
     
  4. May 9, 2005 #3
    That's where I was getting confused. How do you know sine is 0 there? I know [tex]\sin n\pi = 0[/tex] where n is an integer, but if you don't know n is an integer in the first place, how can you assume that those sine terms are 0?
     
  5. May 9, 2005 #4

    dextercioby

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    If the cosine is "+1" (as it should be),then automatically the sine is 0,because we know that

    [tex] \sin^2 x+\cos^2 x=1 \, \ x\in\mathbb{R} [/tex]

    Daniel.

    P.S.As i said,don't worry about the sine.
     
  6. May 9, 2005 #5

    jtbell

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    Here's another way to look at it: [itex]q[/itex] must satisfy both of the following conditions:

    [tex]\cos (-q2 \pi) = 1[/tex]

    [tex]\sin(-q2 \pi) = 0[/tex]

    If we start with the first condition, that eliminates all values of [itex]q[/itex] except the + and - integers, and zero. These remaining values of [itex]q[/itex] all satisfy the second condition, so we're done.

    Alternatively, we can start with the second condition. In this case, we eliminate all values of [itex]q[/itex] except the + and - integers and half-integers, and zero. Now we apply the first condition to those remaining values, which eliminates the half-integers, and gives us the same final result as before.
     
  7. May 10, 2005 #6
    Ahh ok, thanks.
     
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