Associated Legendre function/Proofs

It will be demonstrated that the associated Legendre functions are orthogonal and their normalization constant will be derived.

Theorem

$\int \limits _{-1}^{1}P_{l}^{m}\left(x\right)P_{k}^{m}\left(x\right)dx={\frac {2}{2l+1}}{\frac {\left(l+m\right)!}{\left(l-m\right)!}}\delta _{lk},$

where:

$P_{l}^{m}\left(x\right)={\frac {\left(-1\right)^{m}}{2^{l}l!}}\left(1-x^{2}\right)^{\frac {m}{2}}{\frac {d^{l+m}}{dx^{l+m}}}\left[\left(x^{2}-1\right)^{l}\right],\quad 0\leq m\leq l.$

Proof

The associated Legendre functions are regular solutions to the associated Legendre differential equation:

$\left(\left[1-x^{2}\right]y'\right)'+\left(l\left[l+1\right]-{\frac {m^{2}}{1-x^{2}}}\right)y=0,$

where the primes indicate differentiation with respect to x.

This equation is an example of a more general class of equations known as the Sturm-Liouville equations. Using Sturm-Liouville theory, one can show the orthogonality of functions with same superscript m and different subscripts:

$K_{kl}^{m}=\int \limits _{-1}^{1}P_{k}^{m}\left(x\right)P_{l}^{m}\left(x\right)dx=0\quad {\hbox{if}}\quad k\neq l.$

In the case k = l it remains to find the normalization factor of the associated Legendre functions such that the "overlap" integral

$K_{kl}^{m}=1.$

One can evaluate the overlap integral directly from the definition of the associated Legendre polynomials given in the main article, whether or not k = l. Indeed, insert twice the definition:

$K_{kl}^{m}={\frac {1}{2^{k+l}\;k!\;l!}}\int \limits _{-1}^{1}\left\{(1-x^{2})^{m}{\frac {d^{k+m}}{dx^{k+m}}}\left[(x^{2}-1)^{k}\right]\right\}\left\{{\frac {d^{l+m}}{dx^{l+m}}}\left[\left(x^{2}-1\right)^{l}\right]\right\}dx.$

Since k and l occur symmetrically, one can without loss of generality assume that l ≥ k. Use the well-known integration-by-parts equation

$\int _{-1}^{1}u\;v'\;dx=\left.u\,v\right|_{-1}^{1}-\int _{-1}^{1}vu'\;dx$

l + m times, where the curly brackets in the integral indicate the factors, the first being u and the second v’. For each of the first m integrations by parts, u in the $uv|_{-1}^{1}$ term contains the factor (1−x²), so the term vanishes. For each of the remaining l integrations, v in that term contains the factor (x²−1) so the term also vanishes. This means:

$K_{kl}^{m}={\frac {\left(-1\right)^{l+m}}{2^{k+l}\;k!\;l!}}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}{\frac {d^{l+m}}{dx^{l+m}}}\left[\left(1-x^{2}\right)^{m}{\frac {d^{k+m}}{dx^{k+m}}}\left[\left(x^{2}-1\right)^{k}\right]\right]dx.$

Expand the second factor using Leibnitz' rule:

${\frac {d^{l+m}}{dx^{l+m}}}\left[\left(1-x^{2}\right)^{m}{\frac {d^{k+m}}{dx^{k+m}}}\left[\left(x^{2}-1\right)^{k}\right]\right]=\sum \limits _{r=0}^{l+m}{\binom {l+m}{r}}{\frac {d^{r}}{dx^{r}}}\left[\left(1-x^{2}\right)^{m}\right]{\frac {d^{l+k+2m-r}}{dx^{l+k+2m-r}}}\left[\left(x^{2}-1\right)^{k}\right].$

The leftmost derivative in the sum is non-zero only when r ≤ 2m (remembering that m ≤ l). The other derivative is non-zero only when k + l + 2m − r ≤ 2k, that is, when r ≥ 2m + l − k. Because l ≥ k these two conditions imply that the only non-zero term in the sum occurs when r = 2m and l = k. So:

$K_{kl}^{m}=\delta _{kl}\;{\frac {(-1)^{l+m}}{2^{2l}\,(l!)^{2}}}{\binom {l+m}{2m}}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}{\frac {d^{2m}}{dx^{2m}}}\left[\left(1-x^{2}\right)^{m}\right]{\frac {d^{2l}}{dx^{2l}}}\left[\left(1-x^{2}\right)^{l}\right]dx,$

where δ_kl is the Kronecker delta that shows the orthogonality of functions with l ≠ k. To evaluate the differentiated factors, expand (1−x²)^k using the binomial theorem:

$\left(1-x^{2}\right)^{k}=\sum \limits _{j=0}^{k}{\binom {k}{j}}(-1)^{k-j}x^{2(k-j)}.$

The only term that survives differentiation 2k times is the x^2k term, which after differentiation gives

$(-1)^{k}\,{\binom {k}{0}}\,2k!=(-1)^{k}\,(2k)!\,.$

Therefore:

$K_{kl}^{m}=\delta _{kl}\;{\frac {1}{2^{2l}\;(l!)^{2}}}{\frac {(2l)!\,(l+m)!}{(l-m)!}}\int \limits _{-1}^{1}(x^{2}-1)^{l}dx\qquad \qquad \qquad \qquad \qquad \qquad (1)$

Evaluate

$\int \limits _{-1}^{1}(x^{2}-1)^{l}dx$

by a change of variable:

$x=\cos \theta \;\Longrightarrow \;dx=-\sin \theta d\theta \quad {\hbox{and}}\quad 1-x^{2}=(\sin \theta )^{2}.$

Thus,

$\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx=(-1)^{2l+1}\int \limits _{\pi }^{0}\left(\sin \theta \right)^{2l+1}d\theta =\int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l+1}d\theta ,$

where we recall that

$-1=\cos \,\pi \quad {\hbox{and}}\quad 1=\cos \,0.$

The limits were switched from

$\pi \rightarrow 0\;\quad {\hbox{and}}\quad 0\rightarrow \pi$ ,

which accounts for one minus sign and further for integer l: (−1)^2l =1 . A table of standard trigonometric integrals shows:

$\int \limits _{0}^{\pi }\sin ^{n}\theta d\theta ={\frac {\left.-\sin \theta \cos \theta \right|_{0}^{\pi }}{n}}+{\frac {(n-1)}{n}}\int \limits _{0}^{\pi }\sin ^{n-2}\theta d\theta .$

Since

$\left.-\sin \theta \cos \theta \right|_{0}^{\pi }=0,$ $\int \limits _{0}^{\pi }\sin ^{n}\theta d\theta ={\frac {\left(n-1\right)}{n}}\int \limits _{0}^{\pi }\sin ^{n-2}\theta d\theta$

for n ≥ 2.

Applying this result to

$\int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l+1}d\theta$

and changing the variable back to x yields:

$\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx={\frac {2\left(l+1\right)}{2l+1}}\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l-1}dx$

for l ≥ 1. Using this recursively:

$\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx={\frac {2\left(l+1\right)}{2l+1}}{\frac {2\left(l\right)}{2l-1}}{\frac {2\left(l-1\right)}{2l-3}}...{\frac {2\left(2\right)}{3}}\left(2\right)={\frac {2^{l+1}l!}{\frac {\left(2l+1\right)!}{2^{l}l!}}}={\frac {2^{2l+1}\left(l!\right)^{2}}{\left(2l+1\right)!}}.$

Applying this result to equation (1):

$K_{kl}^{m}=\delta _{kl}\;{\frac {1}{2^{2l}\;(l!)^{2}}}{\frac {(2l)!\;(l+m)!}{(l-m)!}}\;{\frac {2^{2l+1}\;(l!)^{2}}{(2l+1)!}}=\delta _{kl}\,{\frac {2}{2l+1}}{\frac {(l+m)!}{(l-m)!}}\qquad \qquad \mathbf {QED} .$

Clearly, if we define new associated Legendre functions by a constant times the old ones,

${\bar {P}}_{l}^{m}(x)\equiv {\sqrt {{\frac {2l+1}{2}}\;{\frac {(l-m)!}{(l+m)!}}}}\;P_{l}^{m}(x)$

then the overlap integral becomes,

$K_{kl}^{m}=\int \limits _{-1}^{1}{\bar {P}}_{k}^{m}(x){\bar {P}}_{l}^{m}(x)\;dx=\delta _{kl},$

that is, the new functions are normalized to unity.

Comments

The orthogonality of the Associated Legendre Functions can be demonstrated in different ways. The presented proof assumes only that the reader is familiar with basic calculus and is therefore accessible to the widest possible audience. However, as mentioned, their orthogonality also follows from the fact that the equation they solve belongs to a family known as the Sturm-Liouville equations.

It is also possible to demonstrate their orthogonality using principles associated with operator calculus. For example, the proof starts out by implicitly proving the anti-Hermiticity of

$\nabla _{x}\equiv {\frac {d}{dx}}.$

Indeed, let w(x) be a function with w(1) = w(−1) = 0, then

$\langle wg|\nabla _{x}f\rangle =\int _{-1}^{1}w(x)g(x)\nabla _{x}f(x)dx=\left[w(x)g(x)f(x)\right]_{-1}^{1}-\int _{-1}^{1}{\Big (}\nabla _{x}w(x)g(x){\Big )}f(x)dx=-\langle \nabla _{x}(wg)|f\rangle$

Hence

$\nabla _{x}^{\dagger }=-\nabla _{x}\;\Longrightarrow \;\left(\nabla _{x}^{\dagger }\right)^{l+m}=(-1)^{l+m}\nabla _{x}^{l+m}$

The latter result is used in the proof. Knowing this, the hard work (given above) of computing the normalization constant remains.

When m=0, an Associated Legendre Function is identifed as $P_{l}$, which is known as the Legendre Polynomial of order l. To demonstrate orthogonality for this limited case, we may use a result from the theory of orthogonal polynomials. Namely, a Legendre polynomial of order l is orthogonal to any polynomial of lower order. In Bra-Ket notation (k ≤ l)

$\langle w\nabla _{x}^{m}P_{k}|\nabla _{x}^{m}P_{l}\rangle \quad {\hbox{with}}\quad w\equiv (1-x^{2})^{m},$

then

$\langle w\nabla _{x}^{m}P_{k}|\nabla _{x}^{m}P_{l}\rangle =(-1)^{m}\langle \nabla _{x}^{m}(w\nabla _{x}^{m}P_{k})|P_{l}\rangle$

The bra is a polynomial of order k, and since k ≤ l, the bracket is non-zero only if k = l.