Quantum mechanics problem

[Nylex]

I have a question that I'm struggling with a bit.

The azimuthal part of the wavefunction of a particle is

\Psi(\phi) = Ae^{-iq\phi} where \phi 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 \Psi(\phi) = \Psi(\phi + 2\pi) because \phi and \phi + 2\pi are the same angle.

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

and Ae^{-iq\phi} = Ae^{-iq\phi}e^{-iq2\pi}

\Rightarrow e^{-iq2\pi} = 1

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 e^{ix} = \cos x + i\sin x, but I'm not sure.

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

\int \Psi^* \Psi d\phi = 1

\int_{0}^{2\pi} Ae^{iq\phi}Ae^{-iq\phi} = 1

A^2 \int_{0}^{2\pi} d\phi = 1

So A = \sqrt{ \frac{1}{2\pi} }

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

Thanks.

[dextercioby]

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

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

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

Daniel.

[Nylex]

That's where I was getting confused. How do you know sine is 0 there? I know \sin n\pi = 0 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?

[dextercioby]

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

\sin^2 x+\cos^2 x=1 \, \ x\in\mathbb{R}

Daniel.

P.S.As i said,don't worry about the sine.

[jtbell]

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

\cos (-q2 \pi) = 1

\sin(-q2 \pi) = 0

If we start with the first condition, that eliminates all values of q except the + and - integers, and zero. These remaining values of q 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 q 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.

[Nylex]

Ahh ok, thanks.