Recurrence relations for Associated Legendre Polynomials

In summary, the student is trying to solve a problem in Bransden and Joachain's QM, but is having trouble with the last two equations. He has managed to solve the first two equations, but is not able to solve the last two. He has tried differentiating with respect to s, but has not been able to get anywhere. He has looked at a book on special functions to try to find a solution, but has not been successful.
  • #1
Clever-Name
380
1

Homework Statement



I'm working on problem 6.11 in Bransden and Joachain's QM. I have to prove 4 different recurrence relations for the associate legendre polynomials. I have managed to do the first two, but can't get anywhere for the last 2

Homework Equations



Generating Function:
[tex] T(\omega, \phi) = \sum_{l} P_{l}^{m}(\omega)s^{l} = \frac{(2m)!(1-\omega^{2})^{m/2}s^{m}}{2^{m}(m!)(1-2\omega s + s^{2})^{m+1/2} }[/tex]
Where m is really [itex]|m|[/itex]

I need to prove the following:

[tex] a)\ (2l+1)\omega P_{l}^{m}(\omega) = (l-m+1)P_{l+1}^{m}(\omega) + (l+m)P_{l-1}^{m}(\omega)[/tex]
[tex] b)\ (2l+1)(1-\omega^{2})^{1/2}P_{l}^{m-1} = P_{l+1}^{m}(\omega) - P_{l-1}^{m}(\omega)[/tex]
[tex] c) (1-\omega^{2})\frac{dP_{l}^{m}(\omega)}{d\omega} = (l+1)\omega P_{l}^{m}(\omega) - (l+1-m)P_{l+1}^{m}(\omega)[/tex]
[tex] d) (1-\omega^{2})\frac{dP_{l}^{m}(\omega)}{d\omega} = -(1-\omega^{2})^{1/2}(l+m)(l-m+1)P_{l}^{m-1}(\omega) + m\omega P_{l}^{m}(\omega)[/tex]

The Attempt at a Solution


I was able to prove A and B by differentiating the generating function with respect to s and doing some rearranging and such, but c and d just aren't working for me. I started by differentiating with respect to omega and here is what I have so far:

First multiply through by [itex](1-2\omega s + s^{2})^{m+1/2}[/itex] and define [itex] c_{m} = \frac{(2m)!s^{l}}{2^{m}(m!)}[/itex]

Ignoring the summation sign, and the functional dependence of P on omega, for readability we have:
[tex]
(1-2\omega s + s^{2})^{m+1/2}P_{l}^{m}s^{l} = c_{m}(1-\omega^{2})^{m/2}[/tex]
Take derivative wrt omega:
[tex]
(m+1/2)(-2s)(1-2\omega s + s^{2})^{m-1/2}P_{l}^{m}s^{l} + (1-2s\omega + s^{2})^{m+1/2}P'_{l}^{m}s^{l} = c_{m}(m/2)(-2\omega)(1-\omega^{2})^{m/2-1}[/tex]
Where P' denotes derivative wrt omega.

Dividing through by [itex](1-2\omega s + s^{2})^{m+1/2}[/itex] and multiplying through by [itex](1-\omega^{2})[/itex] we get:

[tex]
\frac{-(1-\omega^{2})(2m+1)sP_{l}^{m}s^{l}}{1-2\omega s +s^{2}} + (1-\omega^{2})P'_{l}^{m}s^{l} = \frac{-c_{m}m\omega(1-\omega^{2})^{m/2}}{(1-2\omega s + s^{2})^{m/2}}[/tex]

But that last term is just [itex]m\omega P_{l}^{m}s^{l}[/itex], so we arrive at:

[tex]
\frac{-(1-\omega^{2})(2m+1)sP_{l}^{m}s^{l}}{1-2\omega s +s^{2}} + (1-\omega^{2})P'_{l}^{m}s^{l} = -m\omega P_{l}^{m}s^{l}[/tex]

Which is fairly close to c. I've tried subbing in a or b here but it just gets messier and I can't make it simplify.

Any suggestions? Or does anyone know of a textbook or other resource that goes though these derivations in detail?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
edit - vela fixed it for me. Thanks.

[STRIKE]I can't seem to edit my post anymore but there's a mistake in what I wrote, LHS should be s^l not s^m all though my shown work.[/STRIKE]
 
Last edited:
  • #3
I made the change for you.
 
  • #4
Try Lebedev's book for special functions to see how the full solution is built. Of course, there are other books on special functions or orthogonal polynomials in particular which address the recurrance relations.

The thread should have been posted in the mathematics section of HW, since it's a problem in mathematics, not physics.
 
  • #5
Lebedev's book only has the derivation for the Legendre polynomials, little to no mention of recurrence relations, at least as far as I could find. I've looked at at least 20 different books where these functions are relevant (quantum, mathematical physics, books on special functions), and none seem to go into any detail about the recurrence relationships other than to say 'it's easy to prove'.

While I agree this is more of a math problem, anyone working with quantum will have to deal with the legendre polynomials and so might be more familiar with them, hence why I posted here.
 
  • #6
In the English translation of Lebedev published in 1965 start reading from page 192, section 7.12.
 
  • #7
Oh, my bad, completely missed that entire section. Thank you, this seems to have what I've been looking for.
 
  • #8
The fact that ##l## appears in the recurrence relation coefficients suggests you might need to differentiate with respect to ##s## somewhere along the way.
 
  • #9
Clever-Name said:
Generating Function:
[tex] T(\theta, \phi) = \sum_{l} P_{l}^{m}(\omega)s^{l} = \frac{(2m)!(1-\omega^{2})^{m/2}s^{l}}{2^{m}(m!)(1-2\omega s + s^{2})^{m+1/2} }[/tex]
Where m is really [itex]|m|[/itex]

This looks very wrong for a couple of reasons:

(1) You've called your generating function [itex]T(\theta, \phi)[/itex], but I don't see any [itex]\theta[/itex]- or [itex]\phi[/itex]-dependence anywhere.

(2) [itex]l[/itex] is summed over in [itex]\sum_{l} P_{l}^{m}(\omega)s^{l}[/itex], but the RHS of your equation has [itex]l[/itex]-dependence.
 
  • #10
Ah, yes, my mistake. T should be [itex] T(\omega,s)[/itex]. Where [itex] \omega = cos(\theta)[/itex]. It's just a simplification that's made during the derivation. My prof used T(theta,phi) in class by mistake so that's just how I ended up writing it from my notes, it should be omega and s.

As for the summation, another mistake of mine that I failed to catch, the s^l on the right hand side should be s^m, I'm not able to edit it right now.
 
  • #11
You have to be more careful regarding notations, in an exam you may lose points or precious time trying to fix the errors.

EDIT: You may want to write things on paper, photocopy them and attach to your initial post. It may take less time than writing all that LaTex code.
 
Last edited:
  • #12
When I physically write it down I don't make those mistakes, it's typing it out in latex after a long frustrating day of getting nowhere on the problem that caused the mistakes. Simple typos, that's all.
 
  • #13
Clever-Name said:
As for the summation, another mistake of mine that I failed to catch, the s^l on the right hand side should be s^m, I'm not able to edit it right now.

That still doesn't seem quite right. According to Srivastava, you should have

[tex]T(\omega,s) = \sum_{l=m}^{l=0} P_l^m(\omega) s^l = \frac{(2m)! (1-\omega^2)^{ \frac{m}{2} } s^{ \frac{m}{2} }}{2^m m! (1-2\omega s + s^2)^{ \frac{m+1}{2} }}[/tex]

Aside from some the exponents on the RHS being different than yours, the sum goes from [itex]l=m[/itex] to [itex]l=0[/itex], not [itex]l=0[/itex] to [itex]l=\infty[/itex], as usually implied by writing [itex]\sum_l[/itex] without qualifying what values [itex]l[/itex] may have.

Srivastava also says, "The form of the generating function is cumbersome and is seldom used", which may explain why I've never seen it before. So, my question to you is whether you were told use the generating function for these problems, because it may be easier to use the relation to the normal Legendre Polynomials,

[tex]P_l^m(\omega) = (1-\omega^2)^{ \frac{m}{2} } \frac{d^m P_l(\omega)}{d\omega^m}[/tex]

especially if you've already proven certain recurrence relationships for [itex]P_l(\omega)[/itex]
 
  • #14
Hm.. That disagrees with my text, I took a picture and uploaded it, no mistakes by me this time!

All the problem says is "Establish the recurrence relations (6.97) for the associated Legendre functions."

I figured the generating function would be the appropriate method to use.
 

Attachments

  • 2012-09-30 18.17.38.jpg
    2012-09-30 18.17.38.jpg
    24.9 KB · Views: 969
  • #15
As said, check out Lebedev's derivation and compare his inputs/outputs (formulas) with Gradshteyn & Rytzhik or Abramowitz & Stegun.
 
  • #16
I've looked through Lebedev and he derives equation c) that I'm looking for first by using some recursion relations that I am not given in my text or notes. I'm first trying to doing only with what I'm given, but that book has been helpful so far.
 

Related to Recurrence relations for Associated Legendre Polynomials

1. What are recurrence relations for Associated Legendre Polynomials?

Recurrence relations for Associated Legendre Polynomials are mathematical formulas that describe the relationship between different values of the polynomials. These relations allow us to calculate the values of the polynomials for different orders and degrees.

2. Why are recurrence relations important for Associated Legendre Polynomials?

Recurrence relations are important for Associated Legendre Polynomials because they provide a more efficient way of calculating the values of the polynomials compared to other methods such as direct integration. They also allow us to derive new properties and identities of the polynomials.

3. How are recurrence relations derived for Associated Legendre Polynomials?

Recurrence relations for Associated Legendre Polynomials are derived using the properties of the polynomials and the differential equation they satisfy. By manipulating the properties and applying the differential equation, we can obtain the recurrence relations.

4. Can recurrence relations be used for all orders and degrees of Associated Legendre Polynomials?

Yes, recurrence relations can be used for all orders and degrees of Associated Legendre Polynomials. However, the complexity of the relations may increase for higher orders and degrees.

5. Are there different recurrence relations for Associated Legendre Polynomials?

Yes, there are multiple recurrence relations for Associated Legendre Polynomials due to the different ways they can be derived. Some may be more efficient for certain orders and degrees, while others may be more general. It is important to choose the appropriate relation for the desired calculation.

Similar threads

  • Advanced Physics Homework Help
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
7
Views
1K
Replies
25
Views
3K
Replies
10
Views
1K
Replies
1
Views
660
Replies
12
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
Replies
16
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
Back
Top