1
$\begingroup$

The usual form of Legendre's differential equation which I am familiar with, is: $$ \left(1-x^2\right)\frac{\mathrm d^2P}{\mathrm dx^2} - 2x\frac{\mathrm dP}{\mathrm dx} + \ell\left(\ell+1\right)P = 0 \tag{01} $$

No problem, use series solution to solve.

But when I was looking at the solution of Schrodinger's equation for Hydrogen atom, the equation they got was this: $$ \left(1-x^2\right)\frac{\mathrm d^2P}{\mathrm dx^2} - 2x\frac{\mathrm dP}{\mathrm dx} + \left[\ell\left(\ell+1\right)- \frac{m^2}{1-x^2}\right]P = 0 \tag{02} $$

Awesome! Now how on earth do I solve this? with the extra $\frac{m^2}{1-x^2}$ term?

I guess I could multiply by that on both sides and get rid of it from the denominator but then is it still Legendre equation? Have I read something wrong?

Thank you for your help and as always apologies if I missed something obvious and the question is silly!

ANSWER: Here's the link to a step by step solution for the general Legendre equation: http://www.physicspages.com/2011/03/22/associated-legendre-functions/

$\endgroup$
  • 2
    $\begingroup$ Your first equation is valid in cases of spherical symmetry. You'll need to use associated Legendre polynomials for the more general equation: en.wikipedia.org/wiki/Legendre_function $\endgroup$ – zeldredge Sep 6 '15 at 14:54
2
$\begingroup$

After searching for a while I finally found a detailed solution to this general Legendre equation.
So, I guess this question is answered!

The source for the solution is this:
http://www.physicspages.com/2011/03/22/associated-legendre-functions/

EDIT: As of February 12, 2018 the above link doesn't work. Use this instead: https://web.archive.org/web/20170509204654/http://www.physicspages.com:80/2011/03/22/associated-legendre-functions/

$\endgroup$
  • $\begingroup$ You should probably post that specific solution as an answer; otherwise this post will probably be deleted as Not An Answer. Adding the solution would also help future readers of your question. $\endgroup$ – HDE 226868 Sep 6 '15 at 17:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.