Normalization of spherical harmonics

[Pietjuh]

There is this excersise in Griffith's QM text that I can't seem to solve. It's about the calculation of the normalization factor of the spherical harmonic functions using the angular momentum step up operator.

These definitions/results are given:

Y_l^m = B_l^m e^{im\phi} P_l^m (\cos\theta )
L_{+} = \hbar e^{i\phi}\left(\frac{\partial}{\partial\theta} +i\cot\theta\frac{\partial}{\partial \phi}\right)
L_+ Y_l^m = A_l^m Y_l^{m+1}
A_l^m = \hbar \sqrt{l(l+1) - m(m+1)}

The problem is to calculate B_l^m. The approach suggested by Griffiths is to calculate L_+ Y_l^m to get a recurrence relation for B_l^m. This is the point where I get stuck. I guess it has something to do with the deravitive of P_l^m. Anyhow, I'll give you my calculation until the point I got stuck and hope for the best that someone sees a mistake :)

<br /> L_+ Y_l^m = \hbar e^{i\phi}\left(\frac{\partial}{\partial\theta} +i\cot\theta\frac{\partial}{\partial \phi}\right) B_l^m e^{im\phi} P_l^m (\cos\theta ) <br />

= \hbar e^{i\phi} B_l^m \left [ e^{im\phi}\frac{\partial}{\partial\theta}\left(P_l^m (\cos\theta )\right) + i\cot\theta P_l^m (\cos\theta ) \frac{\partial}{\partial\phi} e^{im\phi} \right]<br />

Now I calculated the derivative of P_l^m using the following formula:

(1-x^2)\frac{dP_l^m}{dx} = \sqrt{1-x^2}P_l^{m+1} - mxP_l^m

So using the chain rule \frac{d P_l^m(\cos\theta)}{d\theta} = \frac{d P_l^m}{dx}\frac{dx}{d\theta} with x=\cos\theta I got:

\frac{d}{d\theta} P_l^m = - \left[\frac{1}{\sqrt{1-\cos^2\theta}}P_l^{m+1} - \frac{m\cos\theta}{\sin^2\theta}P_l^m\right]\sin\theta<br /> = m\frac{\cos\theta}{\sin\theta}P_l^m - P_l^{m+1}

Plugging all this in I got the following result:

\hbar B_l^m \left[ (m-1)P_l^m\cot\theta - P_l^{m+1}\right] = A_l^m B_l^{m+1}P_l^{m+1}

but when I try to solve this for B_l^m I get an exploding expression :(

 

[dextercioby]

What do you mean??It should look pretty bad indeed...Do you have the expression in Griffiths to compare your result to...??

Daniel.

 

[Pietjuh]

I' m sorry my english isn't that good!
From the last equation I am trying to iterate B_l^m so I will get B_l^m = A B_l^0

Griffiths says it should be B_l^m = (-1)^m\sqrt{\frac{(2l+1)}{4\pi}\frac{(l-|m|)!}{(l+|m|)!}

 

[dextercioby]

Check pages 678 pp 690 from Cohen-Tannoudji.

If u don't have the book,i'll help you a bit later...

Daniel.

 

[Pietjuh]

I don't have that book. The only QM text I have is the one from Griffiths.

 

[dextercioby]

Well,i'm not going to type 100 formulas for you,i'll give you some hints:
#1
\hat{L}_{-} Y_{l,-l}(\vartheta,\varphi) =0

Okay??

Solve this equation and find
Y_{l,-l} (\vartheta,\varphi)

and tell me what u get.Normelize the solution and give the constant of integration as well.

Then i'll guide through step #2.

Daniel.

 

[Pietjuh]

These are a few results I'm getting but again it seems I'm stuck on the same sort of problem again :(

L_{-} Y_l^{-l} = -\hbar e^{-i\phi}\left(\frac{\partial Y_l^{-l}}{\partial\theta} - \cot\theta \frac{\partial Y_l^{-l}}{\partial \phi}\right)=0

So \frac{\partial Y_l^{-l}}{\partial\theta} = \cot\theta \frac{\partial Y_l^{-l}}{\partial \phi}

\frac{\partial Y_l^{-l}}{\partial \phi} = -il B_l^{-l} e^{-il\phi} P_l^{-l}

\frac{\partial Y_l^{-l}}{\partial \theta} = B_l^{-l} e^{-il\phi}\left(P_l^{-l+1} + l\frac{\cos\theta}{\sin^2\theta}P_l^{-l}\right)

so from this I get

P_l^{-l+1} + l\frac{\cos\theta}{\sin^2\theta}P_l^{-l} = -il\cot\theta P_l^{-l}

But I can't see how I should continue...
Forgive me for my seemingly stupidity

 

[dextercioby]

So
\hat{L}_{-} =\hbar e^{-i\varphi}(-\frac{\partial}{\partial \vartheta}+i\cot\vartheta \frac{\partial}{\partial \varphi}) (1)

Okay??
Y_{l,-l}(\vartheta,\varphi)=P_{l,-l}(\vartheta)e^{-il\varphi}(2)

Apply the operator (1) on the function (2) and give me the differential equation for
P_{l,-l}(\vartheta)

Daniel.

 

[Pietjuh]

L_{-} Y_l^{-l} = \hbar e^{-i(l+1)\phi}\left(\cot\theta P_l^{-l} - \frac{\partial P_l^{-l} } {\partial \theta} \right) = 0

so \frac{\partial P_l^{-l} } {\partial \theta} = \cot\theta P_l^{-l}

 

[dextercioby]

Almost,u missed the "l" when differentiating wrt \varphi.It's not PD anymore (one variable only,\vartheta):

\frac{dP_{l,-l}(\vartheta)}{d\vartheta}=l\cot\vartheta P_{l,-l}(\vartheta)

Solve this ODE.Then write the spherical harmonic...

Daniel.

 

[Pietjuh]

Integrating factor is \frac{1}{\sin^l\theta}, so
P_l^{-l} = C \sin^l\theta

So Y_l^{-l} = C \sin^l\theta e^{-il\phi}

#edit# I have to go now. Much thanks for your help! :)

 

[dextercioby]

Perfect.Now normalize it and tell me what is the normalization constant...Chose the phase equal to 1.

Daniel.

 

[Pietjuh]

\int d\theta d\phi C^2 \sin^{2l}\theta = 2\pi C^2 \int_0^\pi d\theta sin^{2l}\theta

since \int_0^\pi d\theta sin^n \theta = \frac{n-1}{n}\int_0^\pi d\theta sin^{n-2} \theta

the integral is equal to

\frac{2l-1}{2l}\frac{2l-3}{2l-2}\cdots \frac{1}{2} \frac{\pi}{2}
if n>=2 and n is even and this is equal to

\frac{2l-1}{2l}\frac{2l-3}{2l-2}\cdots \frac{2}{3}

if n>=3 and n is odd.
But because 2l is always even and >= 2 we only need the first one,
so

C = \frac{1}{\sqrt{2\pi}} \sqrt{\frac{2l-1}{2l}\frac{2l-3}{2l-2}\cdots \frac{1}{2} \frac{\pi}{2}}

 

[dextercioby]

\int d\theta d\phi C^2 \sin^{2l}\theta = 2\pi C^2 \int_0^\pi d\theta sin^{2l}\theta

since \int_0^\pi d\theta sin^n \theta = \frac{n-1}{n}\int_0^\pi d\theta sin^{n-2} \theta

the integral is equal to

\frac{2l-1}{2l}\frac{2l-3}{2l-2}\cdots \frac{1}{2} \frac{\pi}{2}
if n>=2 and n is even and this is equal to

\frac{2l-1}{2l}\frac{2l-3}{2l-2}\cdots \frac{2}{3}

if n>=3 and n is odd.
But because 2l is always even and >= 2 we only need the first one,
so

C = \frac{1}{\sqrt{2\pi}} \sqrt{\frac{2l-1}{2l}\frac{2l-3}{2l-2}\cdots \frac{1}{2} \frac{\pi}{2}}

How did you get that result??Mine is totally different.

BTW,the site which had Abramowitz & Stegun is not working...

Daniel.

 

[Pietjuh]

The formula \int_0^\pi d\theta sin^n \theta = \frac{n-1}{n}\int_0^\pi d\theta sin^{n-2} \theta was on the backcover of my calculus book and the result I got in my previous post was just a sort of copy from an example in the chapter on partial integration.

 

[dextercioby]

Then why do i get the feeling that my result is correct
C_{l}=\frac{1}{\sqrt{4\pi}}\frac{\sqrt{(2l+1)!}}{2^{l} \ l!}

and yours,which i cannot see how it can be put in a form similar to mine,is wrong...

Daniel.

P.S.BTW,mine coincides with the one given by Cohen-Tannoudji,page 682,#29.

 

[Pietjuh]

I think you are absolutely correct, which makes me feel even more stupid because I can't seem to evaluate integrals properly anymore

I'm going to type it out step by step now, and hopefully you can discover (or myself) where I'm going the wrong way.

\int_0^{2\pi}\int_0^\pi |Y_l^m|^2 \sin\theta d\theta d\phi = 1

\int_0^{2\pi}\int_0^\pi |Y_l^m|^2 \sin\theta d\theta d\phi = <br /> \int_0^{2\pi}\int_0^\pi C^2 \sin^{2l+1}\theta d\theta d\phi = <br /> 2\pi C^2 \int_0^\pi \sin^{2l+1} d\theta = 1<br />

Using the iteration formula \int_0^\pi d\theta sin^n \theta = \frac{n-1}{n}\int_0^\pi d\theta sin^{n-2} \theta I get:

<br /> \int_0^\pi \sin^{2l+1} d\theta = \frac{2l}{2l+1} \frac{2l-2}{2l-1} \cdots \frac{2}{3} \int_0^\pi \sin\theta d\theta = \frac{2l}{2l+1} \frac{2l-2}{2l-1} \cdots \frac{2}{3} 2<br />

So

<br /> C^2 = \frac{1}{4\pi} \frac{2l+1}{2l} \frac{2l-1}{2l-2}\cdots \frac{3}{2}<br />

But I can't seem to rewrite this in your form

 

[dextercioby]

Maybe i can

|C_{l}|^{2}=\frac{1}{4\pi} \frac{(2l+1)!}{(2l)!}=\frac{1}{4\pi}\frac{(2l+1)!}{[(2l)!]^{2}}

Then
|C_{l}|=\frac{1}{\sqrt{4\pi}}\frac{\sqrt{(2l+1)!}}{(2l)!}

Writing
(2l)!=2^{l} \ l!
and chosing the phase
\phi=(-1)^{l}

,we get
C_{l}=(-1)^{l}\frac{1}{\sqrt{4\pi}}\frac{\sqrt{(2l+1)!}}{2^{l} \ l!}

which coincides with the formula #14,page 679,Cohen-Tannoudji.


Daniel.

 

[Pietjuh]

ah ok :)
I never heard of this double factorial, so I was wondering how to write (2l+1)(2l-1)... as a factorial :)