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 associated Legendre differential 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 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:

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{if}\quad k \ne l . }

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

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} =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:

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}\; 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

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_{-1}^1 u\; v'\; dx = \left. u\,v\right|_{-1}^1 - \int_{-1}^{1} v u' \;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 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} \; 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:

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} \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:

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} =\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:

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} \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

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

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} =\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

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

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 =(-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

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

The limits were 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 \; \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:

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{(n-1) }{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 n ≥ 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 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 l ≥ 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 equation (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} =\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,

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

then the overlap integral becomes,

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^m_{kl} = \int\limits_{-1}^{1} \bar{P}^m_k(x) \bar{P}^m_l(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

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^m_{l}(x) = w(x)^{1/2} \; \nabla^m_x P_l(x) }

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

Clearly, the case m = 0 is,

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^{0}_l(x) = (1-x^2)^{0/2}\; \nabla^0_x 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

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 \equiv \int\limits_{-1}^1\; w(x)\,g(x)\;\big(\nabla_x f(x)\big) \; dx = \left. w(x)\;g(x)f(x) \right|_{-1}^{1} - \int\limits_{-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)^{m} = (-1)^{m} \nabla_x^{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

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\; \Pi_p \;|\; P_l \;\rangle = 0\quad \hbox{if}\quad O\left[\Pi_p\right] \equiv p < l. }

Knowing this,

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 \equiv \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, 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 O\left[\nabla_x^m P_k\right] = k-m, \quad O\left[w(x)\right] = 2m \; \Longrightarrow\; O\left[w(x)\, \nabla_x^m P_k(x) \right] = k+m \; \Longrightarrow\; O\left[ \nabla_x^m \{w(x)\, \nabla_x^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 bracket 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.