How Do You Apply the Raising Operator to Find Y^{2}_{2}(\theta, \phi)?

[Rahmuss]

Homework Statement

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.

Homework 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)}.

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

 

[nrqed]

Homework Statement

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.

Homework 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)}.

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

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?

 

[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.

 

[nrqed]

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.

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!

 

[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}