# Angular Momentum(Q&M)

1. May 15, 2009

### Winzer

1. The problem statement, all variables and given/known data
What are the possible measurements for Lz

2. Relevant equations
$$\psi(\theta,\phi) = \sqrt{\frac{3}{4 \pi}} sin(\phi) sin(\theta)$$
probability Lz quantum

3. The attempt at a solution
Well I'm sure I can expand $$sin(\phi)= \frac{e^{i \phi}-e^{-i \phi}}{2 i}$$
Getting m=1,-1.
$$\psi(\theta,\phi) = \sqrt{\frac{3}{4 \pi}} sin(\theta) \frac{e^{i \phi}-e^{-i \phi}}{2 i}$$
Should the the probability be the coefficents mod squared?

2. May 15, 2009

### phsopher

Yes, but you have to keep track of normalization; the coefficients should be those of normalized eigenfunctions of Lz. I suggest writing psi in terms of spherical harmonics.

3. May 15, 2009

### waht

you are working with two quantum numbers, If I remember correctly you have add up the normalized coefficients squared by increasing their orbital quantum number l up to m.

4. May 15, 2009

### Winzer

ok. So:
$$\psi=\frac{\sqrt{2}i}{2}\left(Y_{1,1}+Y_{1,-1}\right)$$ with everything sorted
I have equal probability of measuring $$\pm \hbar$$ for $$L_z$$, right?

5. May 15, 2009

### Winzer

Wait, what I typed can't be right. m has to be -1,0,1 since l=1. so in terms of spherical harmonics I should have one more term $$Y_{1,0}$$ and then normalize that. That makes more sense.

6. May 16, 2009

### malawi_glenn

You don't need a Y(1,0) A priori.

If your psi in post 4 equals the psi in Relevant equations in your first post, then it is ok.

You also check the normalizability of your result in post #4 by simply integrating it over $r^2 d\Omega$ and see if you indeed get 1.

Y(1,0) is proportional to cos(theta), and you only have sin(theta), so you should not expect a Y(1,0) term.

7. May 16, 2009

### Winzer

ok.
In post four the coefficents work out since$$\Sigma |c_k|^2=1$$

8. May 16, 2009

### diazona

You could consider the coefficient of $$Y^1_0$$ to be 0 ;-)