1
$\begingroup$

EDIT: See below for answer, I found an alternative approach which is better in my opinion than trying to find a power series solution. I still don't fully understand why $m>l$ i.e. $P_l^m=0$ is not an acceptable solution. Intuitively I guess this results in the wave function being equal to zero and maybe we have to be able to normalise the wave function.

How can I study the convergence of the following series:

$$ P_l^m=(1-x^2)^{m/2}[c_0+c_1x+\sum_{k=2}^{\infty}c_kx^k]$$

where $$c_{k+2}=\frac{((m+k+1)(m+k)-A)}{(k+2)(k+1)} c_k$$

In particular, how is it possible to show that the series only converges for $A=l(l+1)$ where $l$ is a nonnegative integer, $-l\leqslant m\leqslant l$?

This series which comes up when solving the Schrodinger equation for the hydrogen atom represents an associated Legendre polynomial (according to this page).

Note that it is possible to show the series is equivalent to the following;

$$P_l^m= (1-x^2)^{m/2}[c_0 + \sum_{k=1}^{\infty} c_0x^{2k}\prod_{i=1}^{k}\frac{(m+2i-1)(m+2i-2)-A}{2i(2i-1)}] + (1-x^2)^{m/2}[c_1x + \sum_{k=1}^{\infty}c_1x^{2k+1}\prod_{j=1}^{\lfloor{(2k+1)/2}\rfloor}\frac{(m+2j)(m+2j-1)-A}{2j(2j+1)}]$$

which is equivalent to;

$$(1-x^2)^{m/2}(c_0\sum_{k=0}^{\infty} \frac{c_{2k}}{c_0}x^{2k}+c_1\sum_{k=0}^{\infty}\frac{c_{2k+1}}{c_1}x^{2k+1})$$

Alternatively rather than examining the convergence of the series it might be easier to show that the series is equivalent to the following definition of Legendre polynomials:

$$P_l^m = (-1)^m(1-x^2)^{m/2}\frac{d^m}{dx^m}P_l(x)$$

where

$$P_l(x)=\frac{1}{2^l l!}\frac{d^l}{dx^l}[(x^2-1)^l]$$

Instead of proving the values of $l$ and $m$ for which the series converges it might be easier to show that it is equivalent to this definition from which it is quite intuitive that $l$ is a nonnegative integer since it is the order of a derivative and for the same reason $m$ must be an integer. Taking a derivative of order $l+m$ in the definition of $P_{l,m}$ implies that $-l$ is a lower bound on $m$. If $m>l$ then $P_{l,m}=0$ although I'm not sure why this is not possible.

In order to solve the polar differential equation I made the substitution $y(x)=(1-x^2)^{m/2}f(x)$ which gives another differential equation in terms of f(x) which can be solved this using a power series giving the result above.

Notice that;

$$c_k = \frac{((m+k-1)(m+k-2)-A)}{k(k-1)} c_{k-2}, k=2,3,4,5,...$$

It's possible to get a product formula rather than a recursive formula for the coefficients.

$$c_k= c_0\prod_{i=1}^{k/2}\frac{(m+2i-1)(m+2i-2)-A}{2i(2i-1)}, \quad \text{k even}$$

$$c_k= c_1\prod_{i=1}^{\lfloor{k/2}\rfloor}\frac{(m+2i)(m+2i-1)-A}{2i(2i+1)}, \quad \text{k odd}$$

then a series of all the even terms is $c_0 + \sum_{k=1}^{\infty} c_{2k}x^{2k}$ and a series with all the odd terms is $c_1x + \sum_{k=1}^{\infty} c_{2k+1} x^{2k+1}$ where we use the appropriate even/odd formula for $c_{2k}$ and $c_{2k+1}$.

There is some discussion here although I didn't find it very helpful, if someone can provide a more in depth explanation that would be very much appreciated.

$\endgroup$
2

1 Answer 1

1
$\begingroup$

I found a better approach as described here.

When solving the polar equation you get an equation $$(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+(l(l+1)-\frac{m^2}{1-x^2})y=0$$

If you make the substitution $y=(1-x^2)^{m/2}f(x)$ you get a new differential equation;

$$(1-x^2)\frac{d^2f}{dx^2}-2x(m+1)\frac{df}{dx}+(l(l+1)-m(m+1))f(x)=0$$

If you set $m=0$ you get the regular Legendre equation for which the ordinary Legendre polynomial $P_l$ is a solution. If you differentiate the equation above you get again the regular Legendre equation except $f$ is replaced by $f'$ and $m$ is replaced by $m+1$ that means that since $P_l$ solves the equation for m=0, $P_l'$ solves the equation for m=1. You can continue differentiating and the same holds for a given m the solution of the above equation is the mth derivative of the Legendre polynomial. You could prove that it holds for all $m$ by induction. Then since we made a substitution $y=(1-x^2)^{m/2}f(x)$ the solution to the original equation is $$P_l^m=(1-x^2)^{m/2}\frac{d^m}{dx^m}P_l$$ which is called an associated Legendre polynomial. You can deduce from the azimuth equation that $m$ must be an integer as described here, I found this easiest if you express the solution to the azimuth equation in terms of sines and cosines rather than exponentials.

If you set $m=0$ you get that the solution is the Legendre polynomial $P_l$ which involves a derivative of order $l$ therefore $l$ must be a nonnegative integer since it is the order of a derivative. Then if you expand the definition of the associated Legendre polynomial you have a derivative of order $l+m$ which means that $m$ must be bounded below by $-l$. We have deduced that $l$ is a nonnegative integer and that $m$ is an integer bounded below by $-l$.

How can we deduce that $m\le l$? If $m>l$ then $P_l^m=0$. Maybe this is not allowed because we have to be able to normalise the wave function. Is this an allowable solution?

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .