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:

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