- #1

Nylex

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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.