QM question, angular momentum operator and eigen functions

[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 I am doing for revision
i think 1) is = A*R(r)*sin(theta)exp[i*phi] and the eigenvalue is [STRIKE]h[/STRIKE]

 

[gabbagabbahey]

i think 1) is = A*R(r)*sin(theta)exp[i*phi] and the eigenvalue is [STRIKE]h[/STRIKE]

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?

 

[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

 

[indie452]

i still don't really understand what i need to do...
when you say <br /> L_{z}\Phi(\phi)=m\Phi(\phi)<br /> 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.

 

[indie452]

how about this?

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

 

[gabbagabbahey]

how about this?

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

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.

 

[indie452]

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

 

[indie452]

does this eigen value answer the question?

 

[gabbagabbahey]

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

 

[indie452]

i got the eigen function being

\frac{1}{\sqrt{2\pi}}e^im\phi

 

[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?