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Light-Matter InteractionPhysics and Engineering at the Nanoscale$

John Weiner and Frederico Nunes

Print publication date: 2012

Print ISBN-13: 9780198567653

Published to Oxford Scholarship Online: December 2013

DOI: 10.1093/acprof:oso/9780198567653.001.0001

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(p.251) Appendix F Properties of the Legendre Functions

(p.251) Appendix F Properties of the Legendre Functions

Source:
Light-Matter Interaction
Publisher:
Oxford University Press

The Legendre functions are a family of solutions to the Legendre differential equation. This equation determines the angular behavior of many physical problems including the scalar Helmholtz wave equation in optics, Maxwell’s equations in classical electrodynamics, and the Schrodinger wave equation in quantum mechanics. We discuss here the properties of these functions common to all these physical situations.

The Legendre equation itself is

(F.1)
( 1 x 2 ) P n ( x ) 2 x P n ( x ) + n ( n + 1 ) P n ( x ) = 0
where n = 0, 1, 2, … In most physics problems x = cos θ, and the Legendre equation takes the form
(F.2)
1 sin θ d d θ ( sin θ d P n ( cos θ ) d θ ) + n ( n + 1 ) P n ( cos θ ) = 0
or more explicitly, carrying out the derivative,
(F.3)
d 2 P n ( cos θ ) d θ + cot θ d P n ( cos θ ) d θ + n ( n + 1 ) P n ( cos θ ) = 0
where the derivative is with respect to the polar coordinate θ rather than x. We have already encountered this latter form in Eqs 2.100, 2.101, 2.102 when studying the angular solutions of Laplace’s equation.

F.1 Generating Function

The Legendre polynomials can always be obtained from a generating function

(F.4)
g ( t , x ) = ( 1 2 x t + t 2 ) 1 / 2 = n = 0 P n ( x ) t n , | t | < 1

The left side of this equation can be cast into a power series in t n such that left side and right side, also a power series in t n can be equated term by term.

(F.5)
( 1 2 x t + t 2 ) 1 / 2 = n = 0 k = 0 [ n / 2 ] ( 1 ) k ( 2 n 2 k ) ! 2 2 n 2 k n ! k ! ( n k ) ! ( 2 x ) n 2 k ) t n = n = 0 P n ( x ) t n

(p.252) where the upper limit on the sum, [n/ 2], means n/ 2 when n is even and (n − 1)/2 when n is odd. Therefore we can obtain an expression for each P n(x) in terms of a sum over the index k.

(F.6)
P n ( x ) = k = 0 [ n / 2 ] ( 1 ) k ( 2 n 2 k ) ! 2 n k ! ( n k ) ! ( n 2 k ) ! x n 2 k

Using Eq. F.6 function we find, for example,

P 0 ( x ) = 1 P 1 ( x ) = x P 2 ( x ) = 1 2 ( 3 x 2 1 ) P 3 ( x ) = 5 2 ( x 3 3 x )

F.2 Recurrence Relations

Families of functions that are solutions to a differential equation often exhibit recurrence relations that express how a given solution P n(x) is related in some simple way to its neighbors, P n +1(x) and Pn− 1(x). These recurrence relations are often another, and simpler way to spawn all the needed members of the family once one member is known. In the case of the Legendre functions we can find a recurrence relation by starting from the generating function, Eq. F.4, and differentiating it.

(F.7)
g ( t , x ) t = x t ( 1 2 x t + t 2 ) 3 / 2 = n = 0 n P n ( x ) t n 1

Using Eq. F.4 we can write this last expression as the sum of four terms, each of which is a power series in t.

(F.8)
( 1 2 x t + t 2 ) n = 0 n P n ( x ) t n 1 + ( t x ) n = 0 P n ( x ) t n = 0
or
(F.9)
n = 0 n P n ( x ) t n 1 n = 0 x ( 2 n + 1 ) P n ( x ) t n + n = 0 ( n + 1 ) P n ( x ) t n + 1 = 0

Now we seek to rewrite the various terms so that they are all sums of a power series in t n. We can do this by adjusting the indices n. Thus,

(F.10)
n = 0 n P n ( x ) t n 1 is equivalent to n = 0 ( n + 1 ) P n + 1 t n
and
(F.11)
n = 0 ( n + 1 ) P n ( x ) t n + 1 is equivalent to n = 0 n P n 1 ( x ) t n
and therefore we can write
(F.12)
( 2 n + 1 ) x P n ( x ) = ( n + 1 ) P n + 1 ( x ) + n P n 1 ( x )

(p.253) Equation F.12 is the recurrence relation for Legendre polynomials. It is a prescription for expressing a given P n(x) in terms of the adjacent polynomials in the series, P n−1(x) and P n+1(x). For example, it is easy to remember the first two Legendre polynomials, P 0(x)= 1, P 1 = x. Using Eq. F.12 we find immediately P 2 = 1/2(3x 2 −1). Then, with P 1(x) and P 2(x) in hand we find P 3(x)= 1/2(5x 3 3x). Thus, in principle any member of the series may be found from the recurrence relation.

F.3 Parity

It appears by inspection that even-index Legendre polynomials are even and odd-index polynomials are odd. In general, the parity of the polynomials can be demonstrated by recourse to the generating function, Eq. F.4.

g ( t , x ) = g ( t , x ) n = 0 P n ( x ) t n = n = 0 P n ( x ) ( t ) n = n = 0 P n ( x ) ( t ) n t n

Equating equal powers of t n we have

(F.13)
P n ( x ) = ( 1 ) n P n ( x )
which shows that, in general, even-index Legendre polynomials are even and odd-index polynomials are odd in the x → −x parity operation.

F.4 Orthogonality and Normalization

The Legendre equation, Eq. F.1, can be regarded as an operator-eigenfunctioneigenvalue equation, Ôψ − λψ = 0, where Ô the operator is

(F.14)
O ^ = d d x ( 1 x 2 ) d d x

The eigenfunctions ψ are the Legendre polynomials P n(x) and the eigenvalues λ = n(n + 1). Viewed in this way, it can be shown that the “operator” is Hermitian and therefore the eigenfunctions are orthogfinal and the eigenvalues real. Therefore the Legendre polynomials are orthogfinal

(F.15)
1 1 P n ( x ) P m ( x ) d x = 0 m n
and in fact they form a complete set, spanning the space of x from –1 to +1. The normalization factor comes from the determination of the integral when m = n. We start with the generating function, square it and write
(F.16)
( 1 2 t x + t 2 ) 1 = [ n = 0 P n ( x ) t n ] 2

(p.254) Then we integrate both sides and note that all the PnPm cross terms will vanish by virtue of Eq. F.15.

(F.17)
1 1 d x 1 2 t x + t 2 = n = 0 1 1 [ P n ( x ) ] 2 d x t 2 n

The integral on the left can be readily evaluated by making a change of variable, y =1 + 2t + t 2 and dy = 2tdx. Making the substitution in Eq. F.17, together with the change in limits x = 1 → y = (1 + t)2 and x =1 → y = (1 − t)2 we have

(F.18)
1 1 d x 1 2 t x + t 2 = 1 t l n ( 1 + t 1 t ) = n = 0 1 1 [ P n ( x ) ] 2 d x t 2 n

Now the ln expression can be expanded in a power series

(F.19)
1 t ln ( 1 + t 1 t ) = 2 n = 0 t 2 n 2 n + 1
and therefore
(F.20)
2 n = 0 t 2 n 2 n + 1 = n = 0 1 1 [ P n ( x ) ] 2 d x t 2 n

Finally, equating each term in the power series t 2n on each side of Eq. F.20, we have

(F.21)
1 1 [ P n ( x ) ] 2 d x = 2 2 n + 1
which is the normalization condition.

Using the conventifinal expression that summarizes orthonormality, we write

(F.22)
1 1 P m ( x ) P n ( x ) d x = 2 δ m n 2 n + 1