Associated Legendre function/Proofs

Orthonormality Proof

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

Theorem

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_{l}^{m} } notation, rather than Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_{l}^{\left( m\right)} } ]

Where: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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], } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\leq m\leq l.}

Proof

The Associated Legendre Functions are regular solutions to the general Legendre equation: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( \left[ 1-x^{2} \right] y^{'} \right) ^{'} +\left( l\left[ l+1\right] -\frac{m^{2} }{1-x^{2} } \right) y=0} , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_{kl}^{m} =\int\limits_{-1}^{1}P_{k}^{m} \left( x\right) P_{l}^{m} \left( x\right) dx } vanishes when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k\neq l.} However, one can find Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_{kl}^{m} } directly from the above definition, whether or not Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k=l:}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l} occur symmetrically, one can without loss of generality assume that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l\geq k.} Integrate by parts Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l+m} times, where the curly brackets in the integral indicate the factors, the first being Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} and the second Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v'.} For each of the first Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m} integrations by parts, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} in the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left. uv\right| _{-1}^{1} } term contains the factor Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( 1-x^{2} \right) } ; so the term vanishes. For each of the remaining Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l} integrations, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v} in that term contains the factor Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( x^{2} -1\right) } ; so the term also vanishes. This means:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r\leq 2m} (remembering that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\leq l} ). The other derivative is non-zero only when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k+l+2m-r\leq 2k} , that is, when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r\geq 2m+(l-k).} Because Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l\geq k} these two conditions imply that the only non-zero term in the sum occurs when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=2m} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l=k.} So:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( 1-x^{2} \right) ^{k} } using the binomial theorem: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2k} times is the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^{2k} } term, which (after differentiation) equals: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( -1\right) ^{k} \left( \begin{array}{c} k \\ 0 \end{array} \right) \left( 2k\right) !=\left( -1\right) ^{k} \left( 2k\right) !} . Therefore:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l} dx } by a change of variable: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\cos \theta \Rightarrow dx=-\sin \theta d\theta\;and\; 1-x^{2} =\sin \theta. } Thus, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi \rightarrow 0 \; } to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \; 0 \rightarrow \pi } , recalling that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \; -1 = \cos (\pi) \;} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \; 1 = \cos (0) \; } ].

A table of standard trigonometric integrals shows: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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.