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Lecture 22: Legendre Polynomials II (the General Eigenvalue Problem, More of Chapter 12 in Boas)
In this lecture we want to finish our discussion of Legendre’s polynomials and associated functions. Of particular interest is the question of treating the polynomials as solutions of an eigenvalue problem, or more generally a Sturm-Liouville Problem. The underlying idea is to prove that, when we solve an eigenvalue problem (now in function space), the corresponding eigenfunctions constitute a complete, orthogonal function set on the specified coordinate space. This is analogous to what we saw for the finite dimensional eigenvalue problem in terms of vectors and matrices. In that case the problem was finite dimensional, involving n -component vectors and n x n matrices. Here we have a continuous free variable and the results are infinite dimensional. We start by defining a general eigenvalue equation in terms of a second order differential operator, , via (first the eigenvalue equation and then the
definition of the operator)
n n n
n n n
x x x d d x p x x q x x dx dx
^
^ (22.1)
where n is the eigenvalue, n is the corresponding eigenfunction, x is a general
weight function, and p x and q x are general functions of x. This equation is
meant to be valid on an interval a x b (which may be infinite). For the Legendre
case we have p x ^ ^1 ^ x^^2 ,^ q x ^ ^ 0,^ x ^1 with a^ 1,^ b ^1. The general problem is
further specified in terms of boundary conditions at x^ a b ,. These may be expressed
in terms of the behavior at the boundaries of n (Dirichlet), n (Neumann) or a
combination. For the Legendre case we required that the n be finite at both a and b.
We proceed by considering the following integral for 2 of the eigenfunctions, completely analogous to what we did in the finite dimensional, matrix case. We have
b m n n m a b m n n m a
dx
d d d d
dx p x x p x x
dx dx dx dx
b^ b n m m n m n n m a a b m n n m a
d d dxp x p x dx dx
d d p x dx dx
^ ^ ^
^
^
(22.2)
where we have performed one integration by parts. The final integrand (in the next-
to-last-line) involves the expression m^ ^^ n ^ n ^ m , which vanishes because the order
of the eigenfunctions does not matter ( i.e ., they are ordinary functions, not operators or matrices, and commute). For all cases of interest the boundary conditions are such as to guarantee that the final expression in this equation vanishes,
0.
b m n n m a
d d p x dx dx
^
^ ^
^
For example, in the Legendre case the eigenfunctions are finite at a 1, b 1 but
p x vanishes at both points. Thus when we consider the corresponding
combination of the right-hand-sides of the original equations, we have
(^0) .
b n m m n a
^ dx x x x (22.4)
From this equation, just as in the matrix eigenvalue problem, we are led to 2
conclusions. First, for non-identical eigenvalues, n (^) m , the corresponding
eigenfunctions must be orthogonal (with respect to the weight function x ),
0.
b m n n m a
^ dx^^ ^ x^ ^ ^ x^ x ^ ^ ^ (22.5)
By considering a single eigenfunction, m n , where
2
b b n n n a a
^ dx^^ ^ x^ ^ ^ x^ x^ ^ dx^ ^ x^ x ^ (22.6)
2 ^
1 2 2 2 1 2 1 1 2 1 1 4 1 2 2 0
2! 1 2!
2 2!
2!
m m
m x u m m x u u (^) m
m dx x m
m du u u m
^
This last expression defines a special function, the Beta function of Chapter 11.7 in Boas (this function was the basis of the original dual models, circa 1970, that eventually led to string theory),
1 1 1 0
p q p^ q^ p^ q
dx x x B p q
p q p q
^ ^ ^
From the properties of the factorial (Gamma) functions we have
2 2
1
1
!^2 1!^2
mm
mn m n mn
m m I m m^ m
I dxP x P x m
The Legendre polynomials are also a complete set of functions on the interval 1 x 1 in the sense that
0
l l l
l
P x P x x x x x
^ ^ ^ ^ ^ ^ (22.14)
and we can express any function on this interval as a sum of Legendre polynomials
0 1
1
l l l
l l
f x c P x
l c dx P x f x
We have not explicitly proved completeness here except by implication: we have not
found any nonzero function on the interval 1 x 1 for which all of the cl vanish.
We can obtain the analogue of Parseval’s Theorem, which we interpreted as one way of stating completeness in the case of Fourier series expansions. Consider the average square value of our generic function f (^) x ,
1 2 1
1 1 ,^0 2
, 0 0
l l l l l l
ll l l l l l l
dx f x dx c c P x P x
c c c l l
(^)
^
The full “length squared” of the function f (^) x is reproduced by the sum of the
squares of the “amplitudes” along the “basis vectors” provided by the Legendre polynomials (including the weight (^1) 2 l (^1) , which you will learn corresponds to one
over the number of states with a given total angular momentum l ).
Note that, as a result of the fact that the Legendre polynomials are orthogonal and of degree l , the expansion of an arbitrary polynomial of degree m will include Legendre polynomials only up to degree m ,
0
Poly ,.
m l l l
x m c P x
(^) (22.17)
As a specific example consider the case discussed in Section 12.9 of Boas, the step function at x 0 ,
x
x
x
^ ^
^ ^
From Eq. (22.15) the coefficients in the Legendre polynomial expansion of this function are given by
1 1
1 0
l 2 l 2 l
l l
c dx P x f x dx P x
(^) (22.19)
0 1 3 5
2 1 0
n n n
x P x P x P x P x
n
n
P x
n n
Next let us return again to Laplace’s equation and include dependence on the
azimuthal angle . Recall that in spherical coordinates we have
2 2 2 2 2 2 2 2
sin
sin sin
r r
r r r r r
Again we separate variables, ^ r ,^^ , ^ R r ^ ^ , and, multiplying through
by r^2 and dividing by , we have
2 2 2 2
sin 0. sin sin
d d d d d r R R dr dr d d d
^ ^ ^
^
In our discussion of the Legendre polynomials above we assumed that was a
constant. Here we want to remove that simplification and consider general, but periodic, behavior on the interval 0 2 . Since the choice of the plane where
0 is arbitrary (just like the choice of the orientation of the z -axis), we expect that
the functions describing physical systems will be smoothly periodic, i.e ., continuous, through this plane. We already know from our study of complex numbers and Fourier transforms that such periodic functions can be faithfully represented by the
complete set of basis functions eim^ with integer m , m . So we should take one of these to represent the characteristic behavior for and then (again!) use
linear superposition in the end to represent the most general behavior. Thus, with this Ansatz, the last term in Laplace’s equation becomes
2 2 2 2 2
sin sin
d m
d
Hence with the same Ansatz for the radial terms as above,
1 d r 2 d R r l l 1 ,
R r dr dr
^
we obtain the Associated Legendre’s equation
^ ^ ^ ^ ^
2 2 2 2 2
2 2 2 2 2 2 2
sin 1 0 sin sin
d d m l l d d d d m x x l l x dx dx x
d x d l l m x x x dx x dx x (^) x
^ ^ ^ ^
^
^
The fact that there are now double poles at x 1 in the coefficient of the zero
derivatives term does not change the fact that x 0 is a regular point (no
singularities in the coefficient functions), while x 1 are regular singular points (only simple poles of order 1 in the first derivative term and order two in the zero derivative term). Thus the analysis goes through much as in the ordinary Legendre case. The boundary conditions are the same and the Sturm-Liouville analysis is
identical except that now q x ^ ^ m^2^^ 1 x^2 , instead of zero. It is straightforward
to verify that the Associated Legendre’s equation is solved by the Associated Legendre function defined by
^
2 2
2 2 2
m^ m m l (^) m l
m m^ l l l m l
d
P x P x
dx
d
x x m l
l dx
m l
^ ^
, , 1 2 , , , , 1 0
, , ,
, , , , ,
cos cos , cos ,
,
cos cos ,
m l m l m
l m l m l m l m
ll mm
l m l m l m
l m l m l m l m l m
Y Y
d d Y Y d Y Y
Y Y
F B Y B d Y F
^
(^)
In terms of the addition theorem of vectors (the angle between two vectors),
r r ˆ^ ˆ^ cos cos cos ^ sin sin ^ cos , we can also write
, ^ ^ , ^ ^ ^ ^ ^
, 0
cos ,
l m l m^ l m^ l^4 l
l
Y Y P
^ ^ ^ ^ ^ (22.34)
which just corresponds to choosing to measure the polar angle with respect to one of the defined vectors instead of the original z -axis so that only m 0 contributes.
The above analysis means that we can expand any function on the sphere as a sum of spherical harmonics. Examples of the spherical harmonics for low eigenvalues are
0,0 1,0 1, 1
2 2,0 2, 1
2 2 2, 2
, cos , sin , 4 4 8 5 15 3cos 1 , cos sin , 16 8 15 sin. 32
i
i
i
Y Y Y e
Y Y e
Y e
To close this discussion let us return again to Laplace’s equation. To find a solution
to the radial part of the equation we try the Ansatz R r ~ r^ . Substituting we find
2 2
1
l l l (^) l l
d d d d
r R r r r r l l
R r dr dr dr dr
l c r
R r
l d r
(^)
^ ^ ^ ^
As suggested earlier we find two solutions, one that is well behaved for r 0 , and one well behaved for r . Depending on the specific boundary conditions, the solution to Laplace’s equation will have the general form (see Section 13.7 in Boas)
1 , , , 0
, , cos , ,
l l l l m l m l m l m l
r c r d r Y
(^) ^ ^ (22.37)
with the constants cl m , and dl m , chosen to match the boundary conditions. For
example, if the function is desired between 2 concentric spheres of radii r 1 and r 2 ,
on which is specified, we will need both the cl,m and the dl,m to match the boundary
conditions on the spheres. If the function is desired inside a sphere of radius r 1 , on
which is specified, and is finite everywhere inside of the sphere (another form of
boundary condition), the dl,m all vanish and we will need only the cl,m to match the
boundary conditions on the sphere. If the function is desired outside of a sphere of
radius r 1 , on which is specified, and is finite everywhere outside of the sphere
(another form of boundary condition), the cl,m all vanish, except perhaps c 0,0, and we will need only the dl,m to match the boundary conditions on the sphere.
