Associated Legendre function/Proofs

It is demonstrated that the associated Legendre functions are orthogonal and their normalization constant is 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 given in the main article. The 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 the 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 functions given in the main article, whether or not k = l. Indeed, inserting the definition of the functions twice:

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

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:

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

where δ_kl is the Kronecker delta that shows the orthogonality of functions with l ≠ k. The factor $(-1)^{l}$ at the front of $K_{kl}^{m}$ comes from switching the sign of x²-1 inside (x²-1)^l. 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:

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

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,

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

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 . Integration of

${\frac {d(\sin ^{n-1}\theta \cos \theta )}{d\theta }}=(n-1)\sin ^{n-2}\theta -n\sin ^{n}\theta$

gives

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

Since

$\left.-\sin ^{n-1}\theta \cos \theta \right|_{0}^{\pi }=0\quad {\hbox{for}}\quad n>1,$

it follows that

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

for n > 1.

Applying this result to

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

and changing the variable back to x yields:

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

for l ≥ 1. Using this recursively:

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

Noting:

$\int \limits _{-1}^{1}\ dx\ =\ 2$

and

${\frac {2l}{2l+1}}{\frac {2\left(l-1\right)}{2l-1}}{\frac {2\left(l-2\right)}{2l-3}}...{\frac {2}{3}}\ =\ {\frac {2^{l}l!}{(2l+1)(2l-1)(2l-3)...3}}\ =\ {\frac {2^{l}l!}{\frac {(2l+1)!}{2^{l}l!}}}\ ={\frac {2^{2l}\left(l!\right)^{2}}{\left(2l+1\right)!}}$

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

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 proof presented above assumes only that the reader is familiar with basic calculus and it is therefore accessible to the widest possible audience. However, as mentioned, their orthogonality also follows from the fact that the associated Legendre equation belongs to a family known as the Sturm-Liouville equations.

It is possible to demonstrate their orthogonality using principles associated with operator calculus. Let us write

$P_{l}^{m}(x)=w(x)^{1/2}\;\nabla ^{m}P_{l}(x)$

where

$\nabla \equiv {\frac {d}{dx}}\quad {\hbox{and}}\quad w(x)\equiv (1-x^{2})^{m}.$

Clearly, the case m = 0 is,

$P_{l}^{0}(x)=(1-x^{2})^{0/2}\;\nabla ^{0}P_{l}(x)=P_{l}(x).$

The proof given above starts out by implicitly proving the anti-Hermiticity of ∇. Indeed, noting that w(x) is a function with w(1) = w(−1) = 0 for m ≠ 0, it follows from partial integration that

$\langle \;w\,g\;|\;\nabla f\;\rangle \equiv \int \limits _{-1}^{1}\;w(x)\,g(x)\;{\big (}\nabla f(x){\big )}\;dx=\left.w(x)\;g(x)f(x)\right|_{-1}^{1}-\int \limits _{-1}^{1}{\Big (}\nabla w(x)\,g(x){\Big )}\,f(x)\;dx=-\langle \;\nabla (wg)\;|\;f\;\rangle$

Hence

$\nabla ^{\dagger }=-\nabla \;\Longrightarrow \;\left(\nabla ^{\dagger }\right)^{m}=(-1)^{m}\;\nabla ^{m}.$

To demonstrate orthogonality of the associated Legendre polynomials, we use a result from the theory of orthogonal polynomials. Namely, a Legendre polynomial of order l is orthogonal to any polynomial Π_p of order p lower than l. In bra-ket notation

$\langle \;\Pi _{p}\;|\;P_{l}\;\rangle =0\quad {\hbox{if}}\quad O\left[\Pi _{p}\right]\equiv p<l.$

Knowing this,

$\int \limits _{-1}^{1}P_{l}^{m}\left(x\right)P_{k}^{m}\left(x\right)dx\equiv \langle \;w\,\nabla ^{m}P_{k}\;|\;\nabla ^{m}P_{l}\;\rangle =(-1)^{m}\langle \;\nabla ^{m}\{w\,\nabla ^{m}P_{k}\}\;|\;P_{l}\;\rangle .$

The bra is a polynomial of order k, because

$O\left[\nabla ^{m}P_{k}\right]=k-m,\quad O\left[w(x)\right]=2m\;\Longrightarrow \;O\left[w(x)\,\nabla ^{m}P_{k}(x)\right]=k+m\;\Longrightarrow \;O\left[\nabla ^{m}\{w(x)\,\nabla ^{m}P_{k}(x)\}\right]=k,$

where it was used that m times differentiation of a polynomial lowers its order by m and that the order of a product of polynomials is the product of the orders. Since we assumed that k ≤ l, the integral is non-zero only if k = l. Hence it follows readily that the associated Legendre polynomials of equal superscripts and non-equal subscripts are orthogonal. However, the hard work (given above) of computing the normalization for the case k = l remains to be done.