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This question (similar other questions) have troubled me for last 5 years. Two months ago I thought I would simply solve the Hydrogen Atom problem and see these Associated Legendre Polynomials come to life by themselves. I haven't solved it yet.

I began by writing the Schrodinger equation for Hydrogen atom. $$ -\frac{\hbar^2}{2m} \nabla^2 \psi + V \psi = E\psi $$ I followed a well know method of separation of variables. $\psi(r,\theta,\phi) = R(r)P(\theta)Q(\phi)$. I plugged it in the above Schrodinger equation. The result was, $$ \underset{\text{function of r}}{\underbrace{\frac{1}{R}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right) + \frac{2mr^2}{\hbar^2}(E-V)}} + \underset{\text{function of x}}{\underbrace{\frac{1}{P}\frac{d}{dx}\left((1-x^2)\frac{dP}{dx}\right) + \frac{1}{(1-x^2)}\underset{\text{function of }\phi}{\underbrace{\frac{1}{Q}\frac{d^2Q}{d\phi^2}}}}} = 0 $$

Here $x = \cos\theta$. Now we can proceed step by step to solve it.

Step 1. $\frac{1}{Q}\frac{d^2Q}{d\phi^2} = -m^2$ where m is an integer (due to the periodic nature of the $Q(\phi)$).

Step 2. This gives us the two following equations, $$ \frac{1}{R}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right) + \frac{2mr^2}{\hbar^2}(E-V) = A $$

$$ \frac{1}{P}\frac{d}{dx}\left((1-x^2)\frac{dP}{dx}\right) - \frac{m^2}{(1-x^2)} = -A $$ where A is some constant. We have not got a clue about this constant at all this point.

I simply am interested in associated Legendre so I went for the later equation.

Step 3. $$ (1-x^2)^2P'' -2x(1-x^2)P' +\left(A(1-x^2) - m^2\right)P = 0 $$ Now this equation can be solved by assuming power series solution of the form, $P = \sum_{j=0}^{\infty} a_j x^j$.

Step 4. When this expansion is plugged into the the Associated Legendre DE I got,

$(j+2)(j+1)a_{j+2}+[A-m^2-2j^2]a_j+[(j-2)(j-3)-A]a_{j-2} = 0$

Remembering that $a_j = 0\ \forall\ j<0$ I found few terms, $$a_2 = -\frac{(A-m^2)}{2\cdot1}a_0$$

$$a_3 = -\frac{(A-m^2-2)}{3\cdot2}a_1$$

$$a_4 = \left[A + \frac{(A-m^2)(A-m^2-8)}{2\cdot1}\right]\frac{a_0}{4\cdot3}$$

$$a_5 = \left[A + \frac{(A-m^2-2)(A-m^2-18)}{3\cdot2}\right]\frac{a_1}{5\cdot4}$$

Now I don't see any pattern emerging through which I could place condition on the constant A. Kindly someone help me this question is driving me crazy.

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The Legendre polynomials $P_n(x)$ are the starting point for most developments. These polynomials were defined in the late 1700's by Legendre when performing potential expansions in $\mathbb{R}^3$: $$ \frac{1}{|x-x_1|}=P_0(\cos\theta)\frac{1}{|x_1|}+P_1(\cos\theta)\frac{|x|}{|x_1|^2}+P_2(\cos\theta)\frac{|x|^2}{|x_1|^3}+\cdots. $$ In Legendre's expansion, $\theta$ is the angle between the vectors at the origin. Legendre was able to show that such an expansion would involve $P_n$, which would be an $n$-th order polynomial. Everything can be reduced to the case where $x_1$ is a unit vector along the $z$-axis. Then, using spherical coordinates, with $\theta$ being the angle from the $z$ axis, $$ \frac{1}{|x-\hat{z}|}=\frac{1}{\sqrt{1-2r\cos\theta+r^2}} = \sum_{n=0}^{\infty}r^{n}P_n(\cos\theta). $$ The Laplacian in spherical coordinates is $$ \nabla^2= \frac{1}{r^2}\left[\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r}+\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\sin\theta\frac{\partial}{\partial\theta}+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\right] $$ Plugging the gravitational potential into $\nabla^2 \frac{1}{|x-\hat{z}|}$ then led to differential equations for the Legendre polynomials $$ \sum_{n=0}^{\infty}n(n+1)r^nP_n(\cos\theta)+\frac{1}{\sin\theta}\frac{d}{d\theta}\sin\theta\frac{d}{d\theta}P_{n}(\cos\theta) = 0. $$ The substitution $x=\cos\theta$ then leads to $$ \frac{d}{d\theta}=-\sin\theta\frac{d}{dx} \\ \sin\theta\frac{d}{d\theta}=-\sin^2\theta\frac{d}{dx}=-(1-x^2)\frac{d}{dx} \\ \frac{1}{\sin\theta}\frac{d}{d\theta}\sin\theta\frac{d}{d\theta}=\frac{d}{dx}(1-x^2)\frac{d}{dx} $$ Therefore, the Legendre polynomials $P_n(x)$ satisfy $$ \frac{d}{dx}(1-x^2)\frac{d}{dx}P_n(x)+n(n+1)P_n(x)=0,\;\;\; n=0,1,2,3,\cdots. $$ In 1816 Olinde Rodrigues discovered an elegant way to write these polynomials: $$ P_n(x) = \frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n. $$ So you have the eigenvalues $n(n+1)$ and eigenfunctions that form an orthogonal system. Because the polynomials are dense in $L^2(-1,1)$, these polynomials form a complete orthogonal basis of $L^2(-1,1)$. The normalization constants required for the $P_n$ are not difficult to compute using the Rodrigues formula and integration by parts.

By playing with the generating function $$ \frac{1}{\sqrt{1-2rx+r^2}} = \sum_{n=0}^{\infty}r^nP_n(x), $$ the higher order Legendre functions were discovered. For example, differentiating $m$ times in $x$ gives $$ \frac{(1)(3)(5)(\cdots)(2m+1)r^{m}}{(1-2rx+r^2)^{m/2}}=\sum_{n=m}^{\infty}r^{n}P_n^{(m)}(x). $$ The polynomials $P_n^{(m)}$ are the associated Legendre polynomials. If you start with the Legendre equation and differentiate it $m$ times, you end up with a new differential equation for the functions $y=P_n^{(m)}$ of the form $$ (1-x^2)y''-2(m+1)xy'+(n(n+1)-m(m+1))y = 0. $$ It is this form of the associated Legendre equation where you can get a series solution. And this equation is equivalent to the equation you want to study. This equation is no longer in Sturm-Liouville form. However, the substitution $$ y = (1-x^2)^{-m/2}g $$ puts the equation into standard form with respect to $g$: $$ \frac{d}{dx}\left((1-x^2)\frac{dg}{dx}\right)-\frac{m^2}{1-x^2}g+n(n+1)g=0. $$ In this way, the standard solutions of the associated Legendre equation were found to be $$ P_n^m(x) = (-1)^m\frac{1}{2^nn!}(1-x^2)^{m/2}\frac{d^{n+m}}{dx^{n+m}}(x^2-1)^{n}. $$ So, quite remarkably, your constant $A$ is found to not depend on $m$; A depends only on $n$ because $$ \left[-\frac{d}{dx}(1-x^2)\frac{d}{dx}+\frac{m^2}{1-x^2}\right]P_{n}^{m}(x)=n(n+1)P_{n}^{m}(x),\;\;\; n=m,m+1,m+2,\cdots. $$ (Note that for $n < m$, the functions $P_n^m$ are $0$.) So the values of your separation constant $A$ are $n(n+1)$ for $n=0,1,2,3,\cdots$, but the only valid values for a given $m$ are $n(n+1)$ for $n=m,m+1,m+2,\cdots$. For a fixed $n$, there are non-zero solutions for $m=0,1,2,3\ldots,n-1$. And is this structure that imposes itself on the orbits of the electron. So, finally, your radial equation is known. $A=n(n+1)$ is required, which comes out of $\frac{d}{dr}\left(r^2\frac{d}{dr}\right)r^n=n(n+1)r^n$.

This analysis was firmly established by the mid to late part of the 19th century, well before Quantum Mechanics. Many attempts have been made to replace this analysis, and to make it more intuitive or abstract, but it seems to me that nothing beats the original story or the closed-form solutions. Quantum ladder operators are really nice, though, I must admit. Of course the radial equation for the Hydrogen isotope was not studied in that time because no standard equations would have led to it. That's why the radial equation has no name attached to it, other than the radial Hydrogen equation.

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  • $\begingroup$ There appears 2 errors in the solution answer. Right above 5th equation from, should be functions y=(1-x**2) to power -m/2 times P,n,(m). This can be seen by substituting in actual assoc Legendre polynomial solutions. 3rd equation from end should be d/dx, not dx/dx. Do correct me if I am wrong. Also it would be a big help to many, I believe, if that 5th equation from end had a more explicit and easily understood, description of how attained. $\endgroup$ Mar 26 '18 at 16:27
  • $\begingroup$ @RobertDiller : Thank you for pointing out the errors. I probably won't add the details of the differentiation because the editor is really annoying when there are too many equations. $\endgroup$ Mar 26 '18 at 19:00
  • $\begingroup$ @RobertDiller : I would appreciate if you would check that the errors you spotted have been properly corrected now. Thanks in advance! $\endgroup$ Mar 26 '18 at 22:45
  • $\begingroup$ Still an error to correct, in the line: "functions $y = P_n^m$ of the form". It should be: "functions $y = (1 - x^2)^{-m/2} P_n^m$ of the form". The 2nd equation after this line, $y = (1-x^2)^{-m/2} g$ is the same equation, with $g = P_n^m$. $\endgroup$ Mar 31 '18 at 17:23
  • $\begingroup$ @RobertDiller : Do you mean where I wrote functions $y=P_n^{(m)}$? There I was indicating the $m$-th derivative of $P_n$. I'm a little confused about the correction that is needed. I may be causing some confusion by use of $P_n^{(m)}$ and $P_n^m$, but I think this is standard. $\endgroup$ Mar 31 '18 at 18:49
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Very sorry about continued confusion. Upon further insight and other reference indicates that the error is in the wording for P_n^(m), not the equation y=P_n^(m), as I had thought. The error is the sentence: The polynomials P_n^(m) are the associated Legendre polynomials. It should be changed to be: The polynomials P_n^(m) are the mth derivatives in x of the Legendre polynomials, P_n. It might be worthwhile to add after the y differential equation that the degree of y is n-m via that y degree is degree of P_n being n and subtracting m as the derivative number.

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