Associated Legendre function/Proofs

Orthonormality Proof

This section demonstrates that the Associated Legendre Functions are orthogonal and derives their normalization constant.

Theorem

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

[Note: The proof uses the more common $P_{l}^{m}$ notation, rather than $P_{l}^{\left(m\right)}$]

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

Proof

The Associated Legendre Functions are regular solutions to the general Legendre 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 $z^{'}={\frac {dz}{dx}}.$

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 that $K_{kl}^{m}=\int \limits _{-1}^{1}P_{k}^{m}\left(x\right)P_{l}^{m}\left(x\right)dx$ vanishes when $k\neq l.$ However, one can find $K_{kl}^{m}$ directly from the above definition, whether or not $k=l:$

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

Since $k$ and $l$ occur symmetrically, one can without loss of generality assume that $l\geq k.$ Integrate by parts $l+m$ 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 $\left.uv\right|_{-1}^{1}$ term contains the factor $\left(1-x^{2}\right)$; so the term vanishes. For each of the remaining $l$ integrations, $v$ in that term contains the factor $\left(x^{2}-1\right)$; so the term also vanishes. This means:

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

Expand the second factor using Leibnitz' rule:

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

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

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

To evaluate the differentiated factors, expand $\left(1-x^{2}\right)^{k}$ using the binomial theorem: $\left(1-x^{2}\right)^{k}=\sum \limits _{j=0}^{k}\left({\begin{array}{c}k\\j\end{array}}\right)\left(-1\right)^{k-j}x^{2\left(k-j\right)}.$ The only thing that survives differentiation $2k$ times is the $x^{2k}$ term, which (after differentiation) equals: $\left(-1\right)^{k}\left({\begin{array}{c}k\\0\end{array}}\right)\left(2k\right)!=\left(-1\right)^{k}\left(2k\right)!$. Therefore:

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

Evaluate $\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx$ by a change of variable: $x=\cos \theta \Rightarrow dx=-\sin \theta d\theta \;and\;1-x^{2}=\sin \theta .$ Thus, $\int \limits _{-1}^{1}\left(x^{2}-1\right)^{l}dx=\int \limits _{0}^{\pi }\left(\sin \theta \right)^{2l+1}d\theta .$ [To eliminate the negative sign on the second integral, the limits are switched from $\pi \rightarrow 0\;$ to $\;0\rightarrow \pi$ , recalling that $\;-1=\cos(\pi )\;$ and $\;1=\cos(0)\;$].

A table of standard trigonometric integrals shows: $\int \limits _{0}^{\pi }\sin ^{n}\theta d\theta ={\frac {\left.-\sin \theta \cos \theta \right|_{0}^{\pi }}{n}}+{\frac {\left(n-1\right)}{n}}\int \limits _{0}^{\pi }\sin ^{n-2}\theta d\theta .$ 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 $n\geq 2.$ 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 $l\geq 1.$ 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 (1):

$K_{kl}^{m}={\frac {1}{2^{2l}\left(l!\right)^{2}}}{\frac {\left(2l\right)!\left(l+m\right)!}{\left(l-m\right)!}}{\frac {2^{2l+1}\left(l!\right)^{2}}{\left(2l+1\right)!}}\delta _{kl}={\frac {2}{2l+1}}{\frac {\left(l+m\right)!}{\left(l-m\right)!}}\delta _{kl}.$ QED.

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.