# Angular Momentum Raising Operator

1. Nov 27, 2007

### Rahmuss

1. The problem statement, all variables and given/known data
In Problem 4.3 you showed that

$$Y^{1}_{2}(\theta , \phi) = -\sqrt{15/8\pi} sin\theta cos\theta e^{i\phi}$$

Apply the raising operator to find $$Y^{2}_{2}(\theta , \phi)$$. Use Equation 4.121 to get the normalization.

2. Relevant equations

[Eq. 4.121] $$A^{m}_{l} = \hbar \sqrt{l(l + 1) - m(m \pm 1)} = \hbar \sqrt{(l \mp m)(l \pm m +1)}$$.

3. The attempt at a solution

So, I think my problem, in part, is that I don't know what they mean when they say "Use Equation 4.121 to get the normalization." So, with that in mind I did try something; but it seems too simple:

$$L_{+} Y^{1}_{2}(\theta , \phi) = \epsilon \sqrt{\frac{(2l + 1)(l - |(m+1)|)\fact}{4\pi(l + |(m+1)|)\fact}} e^{i(m+1)\phi} P^{(m+1)}_{l}(x)$$

So, everywhere I have just $$m$$, I add one to it, which really just gives me the formula for $$Y^{2}_{2}$$. Then solving it is simple. So I must not be understanding the question properly. And I don't see what equation 4.121 has to do with anything. I'm used to noramlization being something like:

$$1 = \int^{\infty}_{-\infty}|\psi|^{2} dx$$

Last edited: Nov 27, 2007
2. Nov 27, 2007

### nrqed

First step:

$L_+ Y^1_2$ is equal to what? It's equal to some constant times $Y_2^2$, right? What is that constant?

3. Nov 27, 2007

### Rahmuss

You mean the whole:

$$\hat{Q}f(x) = qf(x)$$

If that's what you mean, then I guess it would just be:

$$L_{+}Y^{1}_{2} = lY^{2}_{2}$$

??? Is that right?
Earlier on the chapter discusses

$$L^{2}f_{t} = \lambda f_{t}$$

But it doesn't apply the $$L_{+}$$ operator.

Last edited: Nov 27, 2007
4. Nov 27, 2007

### nrqed

Of course, Y^1_2 is NOT an eigenstate of L_+ so you can't use the eigenvalue equation.

What I mean is : use equation 4.121 to figure out the constant! That's what 4.121 is for! To figure out the constant generated when applying the raising or lowering operators!

5. Nov 27, 2007

### Rahmuss

Ah, ok. I understand (I hope).

So, with Eq. 4.121 goes to:

$$A^{2}_{2} = \hbar \sqrt{2(2+1) - 2(2\mp 1}$$ ---->

$$A^{2}_{2} = \hbar \sqrt{6 - 2} = 2\hbar$$

The other leads to zero, so I guess I can throw that out.

So, if I understood correctly the first step is done?

$$L_{+} Y^{1}_{2} = 2\hbar Y^{2}_{2}$$ ?

Or do I just say:

$$\frac{L_{+} Y^{1}_{2}}{2\hbar} = Y^{2}_{2}$$