Associated Legendre function/Proofs

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

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}, }

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], \quad 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, \quad\hbox{where}\quad z' \equiv\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 = 0 \quad\hbox{when}\quad k \ne 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 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 k and l occur symmetrically, one can without loss of generality assume that l ≥ k. Indeed, 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 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 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:

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,} 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( n-1\right) }{n} \int\limits_{0}^{\pi }\sin ^{n-2} \theta d\theta } for 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 n\geq 2.} Applying this result 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 \int\limits_{0}^{\pi }\left( \sin \theta \right) ^{2l+1} d\theta } and changing the variable back 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 x} yields: 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 =\frac{2\left( l+1\right) }{2l+1} \int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l-1} dx } for 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 1.} Using this recursively:

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 =\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):

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) !} \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

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 \nabla_x \equiv \frac{d}{dx}. }

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

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 \langle w g | \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 (w g) | f\rangle }

Hence

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 \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 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} , 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)

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 \langle w \nabla_x^m P_k | \nabla_x^m P_l\rangle\quad\hbox{with}\quad w\equiv (1-x^2)^m, }

then

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 \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.