# What is the general solution of the Associated Legendre differential equation when $A$ does not equal $\ell(\ell+1)$?

I am sorry if this question isn't clear, I couldn't think of a better way to phrase it. I am a Physics student trying to solve the angular component of the wave function for a particle in a central potential. I am sure that most if not all of you are familiar with the problem. Here is what I get for the polar wave function.

$$\frac{\sin(\theta)}{\Theta(\theta)}\frac{\partial}{\partial \theta}\left(\sin(\theta)\frac{\partial \Theta(\theta)}{\partial \theta}\right) + A\sin^2(\theta) - m^2 = 0$$

This is simply the differential equation for the Associated Legendre polynomial $$P^m_{\ell}$$ specifically where $$A=\ell(\ell+1)$$ for some integer, $$\ell$$ after a change of coordinates $$x \rightarrow \cos(\theta)$$. The original Associated Legendre differential equation is shown below.

$$\left(1-x^2\right) \frac{d^2 P_\ell^m(x)}{dx^2} - 2x \frac{d P_\ell^m(x)}{dx} + \left(\ell(\ell+1))-\frac{m^2}{(1-x^2)}\right)P_\ell^m(x) = 0$$

My problem is that in all references I have read on the subject it isn't clear why $$A$$ must equal $$\ell(\ell+1)$$. I understand that the series solution for the polar equation when transformed to the $$x$$ variable, $$\Theta(\theta) \rightarrow y(x)$$ yields the important recurrence relation:

$$\left(1-x^2\right) \frac{d^2 y(x)}{dx^2} - 2x \frac{d y(x)}{dx} + \left(A-\frac{m^2}{(1-x^2)}\right)y(x) = 0 \\ y(x) = \left(1-x^2\right)^{m/2} f(x) \\ \left(1-x^2\right) \frac{d^2f(x)}{dx^2} -2(m+1)x \frac{df(x)}{dx} + \left(A - m(m+1)\right)f(x) = 0 \\ f(x) = \sum_{k=0}^\infty f_k x^k \\ f_{n+2} = \frac{(n+m)(n+m+1)-A}{(n+1)(n+2)}f_n$$ And specifying that $$A=\ell(\ell+1)$$ for some positive integer, $$l$$ cuts off higher order components of $$x$$ which generates the associated Legendre polynomial; however, $$x=\cos(\theta)$$, $$0\leq\theta\leq\pi$$ so $$-1\leq x\leq 1$$. I cannot find any proof that demonstrates why this series diverges for all $$x$$ in this range if higher order $$x$$ terms are not cut off by $$A$$. $$f(x) = \sum_{k=0}^\infty f_k x^k \rightarrow \mathrm{diverge/converge?}$$ Moreover the the actual solution to $$\Theta(\theta)$$ is $$\left(1-x^2\right)^{\frac{m}{2}} f(x)$$ due to the earlier substitution so I need to determine whether this diverges or converges for all $$-1\leq x\leq 1$$ $$\Theta(\theta) = (1-x^2)^{\frac{m}{2}}\sum_{k=0}^\infty f_k x^k \rightarrow \mathrm{diverge/converge?}$$ Thank you for reading this far and I hope I have explained my problem adequately, I have tried to be concise as I can. I got the impression from a similar question that the answer might have something to with Hilbert space so I have been reading about that topic but I would appreciate whatever pointers I could get in the right direction.

• Welcome to Physics! This seems like more of a question of mathematics rather than a question of physics (even though Legendre functions are common in physics.) I have flagged the question for migration to Mathematics; if it doesn't get moved, and doesn't get a suitable answer here after a few days, you might consider re-posting it over there instead. Sep 29, 2020 at 16:59
• Hi thanks for the comment I have put it in both as I wasn't sure which it would be most appropriate. Thanks though. Sep 29, 2020 at 17:21

What is the general solution of the Associated Legendre differential equation when $$A$$ does not equal $$\ell(\ell+1)$$?

When you allow $$\ell$$ and $$m$$ to be arbitrary complex numbers $$\lambda$$ and $$\mu$$, the general solution is a linear combination of the Legendre functions $$P_\lambda^\mu(x)$$ and $$Q_\lambda^\mu(x)$$, not associated Legendre polynomials.

My problem is that in all references I have read on the subject it isn't clear why $$A$$ must equal $$\ell(\ell+1)$$.

Any complex $$A$$ can be written in the form $$\lambda(\lambda+1)$$ for some complex $$\lambda$$, so there is no loss of generality. This form is chosen because when $$\lambda$$ is an integer (i.e. $$A=0,2,6,12,\dots$$) and $$\mu$$ is an integer, the $$P$$ functions become polynomials.

• Okay, I was under the impression that the eigenstates had angular momentum quantisation because of this point that only the angular momentum eigenstates converge correctly so that quantises the angular momentum operator on this state, but have I got it the wrong way around? the the constant A is set equal to l(l+1) to make it an eigenstate of the total angular momentum operator and the form yields the associated legendre differential equation? Again sorry for lack of clarity and thank you for your help. Sep 29, 2020 at 18:00
• If you don’t take $A=0,2,6,12,\dots$, then the solutions aren’t normalizable. This is where the quantization comes from. So you understood it correctly. Sep 29, 2020 at 19:19
• That helps a lot, thank you! Sep 29, 2020 at 19:58
• Also, do you have a proof that they cannot be normalised or can you recommend any sites/books that have a detailed proof available? Thanks again. Sep 29, 2020 at 20:56
• I don’t have a proof and after searching for a few minutes I couldn’t find one. You could try asking on Math SE. Sep 29, 2020 at 21:19