# QM: More Spherical Harmonics

1. Feb 14, 2009

### rabbit44

1. The problem statement, all variables and given/known data
A system's wavefunction is proportional to sin^2p. What are the possible results of measurements of Lz and L^2? Give the probabilities of each possible outcome.

I'm using p for theta and q for phi.
2. Relevant equations

3. The attempt at a solution
So I believe that the value of L^2 is 6 (as l is 2), and the possible values of Lz are -2 and +2. But I can't find the probabilities. I tried:

|psi> = a|2,2> + b|2,-2>

where the kets are |l,m> and the spherical harmonice are Y(l,m)

<2,2|psi> = int dpdq <2,2|p,q><p,q|psi>
<2,2|psi> = int dpdq Y(2,2)* [aY(2,2) + bY(2,-2)]

And then inserted the spherical harmonic expressions in, but I just get a=a, so I guess this isn't the way.

I also thought about using the ladder operators, applying L+|psi> = 2a|2, -1> but I don't think this is helpful.

Any help wouldbe much appreciated!!

EDIT: Another thing I tried was saying Lz|l,m>=m|l,m>. So I thought if I had a matrix representation of Lz, I could maybe have

Lz (a*, b*) = 2(a*, -b*)

Where ( , ) is a column matrix. But I don't know the matrix representation of Lz or what basis it should be in. Please help!

Last edited: Feb 15, 2009
2. Feb 16, 2009

### malawi_glenn

the wavefunction has NO phi dependence?

3. Feb 16, 2009

### rabbit44

Yeah that was my first reaction. But I don't think it is possible to have no phi dependence and be proportional to sin^2, so I assumed that it meant it was proportional to sin^2 as well as some function of phi. I may be wrong though?

4. Feb 17, 2009

### malawi_glenn

The spherical harmonics are basis functions for angular functions, so one should be able to find a linear combination of them to build up sin^2 theta.

Now if I give you this hint, you should be able to do it.

$$-3\sin ^2 \theta +2 = 3\cos ^2 \theta - 1$$

Also recall that one of the S. harm's does not depend on any angle at all.

5. Feb 17, 2009

### rabbit44

Ahhhh I forgot about that one.

I see now, I just need Y(2,0) and then need to subtract an amount of Y(0,0) to get rid of the constant, and normalise.

Thank you!!!

6. Feb 17, 2009

### malawi_glenn

correct observation :-)

Good luck