# QM question, angular momentum operator and eigen functions

1. Apr 10, 2010

### indie452

For the operator L(z) = -i[STRIKE]h[/STRIKE][d/d(phi)]

phi = azimuthal angle

1) write the general form of the eigenfunctions and the eigenvalues.
2) a particle has azimuthal wave function PHI = A*cos(phi)
what are the possible results of a measurement of the observable L(z) and what is the probability of each.

this is a past paper qu im doing for revision
i think 1) is = A*R(r)*sin(theta)exp[i*phi] and the eigenvalue is [STRIKE]h[/STRIKE]

2. Apr 10, 2010

### gabbagabbahey

That's certainly an eigenfunction, but it isn't the most general form. What is the eigenvalue equation (expanded in the position basis) for the operator $L_z$? What do you get if you assume that the eigenfunctions are separable?

3. Apr 10, 2010

### morphemera

The question only ask for the eigenfunction of the operator L(z)
So you should not write out one hydrogen wavefunction which may count as WRONG answer!

Solve for $$L_{z}\Phi(\phi)=m\Phi(\phi)$$ and you will get the answer. Think about it

4. Apr 10, 2010

### indie452

i still dont really understand what i need to do...
when you say $$L_{z}\Phi(\phi)=m\Phi(\phi)$$ is that you saying m is the eigenvalue? i thought that it was hbar.

im just confused cause we didnt try to find the eigenfunctions in lectures.

5. Apr 10, 2010

### indie452

L(z)$$\Phi$$ = m$$\hbar$$*exp[im$$\phi$$]
-i$$\hbar$$*d/d$$\phi$$*$$\Phi$$ = m$$\hbar$$*exp[im$$\phi$$]

so if $$\phi$$ = Aexp[im$$\phi$$]
normalised A = 1/$$\sqrt{}2\pi$$

Last edited: Apr 10, 2010
6. Apr 10, 2010

### gabbagabbahey

That doesn't prove that $Ae^{im\phi}$ is the only eigenfunction. Assume that the state $\psi(r,\theta,\phi)$ is an eigenfunction of the operator $L_z$. Furthermore, assume that $\psi(r,\theta,\phi)$ is separable (i.e. $\psi(r,\theta,\phi)=f(r)g(\theta)h(\phi)$). Now apply the operator $L_z$ to that eigenfunction and solve the differential equation you get.

7. Apr 10, 2010

### indie452

okay i solved the partial diff. eqn. and got th same value for PHI,

8. Apr 11, 2010

### indie452

does this eigen value answer the question?

9. Apr 11, 2010

### gabbagabbahey

I'm not sure exactly what your final answer is, you haven't posted it.

10. Apr 11, 2010

### indie452

i got the eigen function being

$$\frac{1}{\sqrt{2\pi}}$$eim$$\phi$$

11. Apr 11, 2010

### gabbagabbahey

Then no, that is not the general form of the eigenfunction. Your eigenfunction can also have some radial or polar angle dependence, can it not?