I am a student of physics and, especially in quantum mechanics, we are presented with the Legendre equation: \begin{eqnarray} (1-x^2)y''-2xy'+l(l+1)y=0. \end{eqnarray} Doing some calculations, we conclude that $l\in\mathbb{Z}^+$. My question is to know why this happens in a geometric and / or qualitative way. In other words, I want to know whether, in a qualitative or geometric way, it is possible to conclude that $l\in \mathbb{Z}^+$.
$\begingroup$
$\endgroup$
4
-
4$\begingroup$ You conclude this fact about $l$ … from what? Do you mean just from the bare existence of solutions to the differential equation? $\endgroup$– LSpiceCommented Dec 28, 2018 at 2:14
-
1$\begingroup$ This is a "classic" result. When we look for solutions in $[- 1,1]$, by power series $ \sum_n a_n x^n $ this condition arises naturally to ensure that $ y(x) <\infty $. $\endgroup$– Leonardo S. VieiraCommented Dec 28, 2018 at 2:28
-
$\begingroup$ Solutions exist for any complex $l$, they are the socalled Legendre functions. $\endgroup$– Carlo BeenakkerCommented Dec 28, 2018 at 10:43
-
1$\begingroup$ This question is more appropriate for MathSE. $\endgroup$– Alexandre EremenkoCommented Dec 28, 2018 at 16:27
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
I am not sure if this is the qualitative/geometric interpretation -of the integrality of the $l$ parameter- you are looking for, but if the parameter $l$ is a non-negative integer then the Legendre polynomials $P_l$ and the associated Legendre polynomials $P_l^m$ are proportional to the matrix elements of special cases of Wigner's $D$-matrix, which provides irreducible representations of the Lie groups $SU(2)$ and $SO(3)$.
In other words, the physical significance (or the geometric interpretation if you prefer) of the integrality of the parameter $l$ is directly related to the fact that the corresponding solutions $P_l$ (of legendre's DE) are (bounded) eigenfunctions of the Laplacian on $S^2$ and they provide irreducible, spin-$l$, representations of the Lie group $SO(3)$.
answered Dec 28, 2018 at 17:11
-
2$\begingroup$ . . . and the spin is quantized, and negative $l$ are equivalent to nonnegative $l$ because the involution $l \leftrightarrow 1-l$ switches integers $l<0$ with $l \geq 0$ and preserves the one coefficient $l(l+1)$ that depends on $l$. $\endgroup$ Commented Dec 29, 2018 at 1:08
$\begingroup$
$\endgroup$
Legendre's equation with arbitrary $\ell$ has infinitely many solutions on $(-1,1)$. Solutions make a vector space of dimension $2$. There is a subspace $V_1$ of dimension $1$ which consists of solutions bounded (even continuous) at $x=1$ and and another subspace $V_{-1}$ of dimension $1$ which consists of solutions bounded at $x=-1$. The question is: when (for which $\ell$) we have $V_1=V_2$. In other words, when there exists a solution which is bounded (continuous) on $[-1,1]$. The answer is: when $\ell$ is a non-negative integer.
This is the meaning of "some calculations" you refer to.
Now, this problem arises from problems of mathematical physics: to solve the Laplace equation in $R^3$, or to solve the eigenvalue problem for Laplace's equation. Using spherical coordinates, we separate the variables, and one of the resulting equations in both problems (the equation for the function of the latitude) is Legendre's equation. The condition that solutions must be bounded follows from this source problem: solutions (or eigenfunctions) of the Laplace equation in $R^3$ must be of course bounded on the unit sphere.
answered Dec 28, 2018 at 16:23