Homework Help: Quantum mechanics problem

1. May 9, 2005

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.

2. May 9, 2005

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.

3. May 9, 2005

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?

4. May 9, 2005

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.

5. May 9, 2005

Staff: Mentor

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.

6. May 10, 2005

Nylex

Ahh ok, thanks.