Associated Legendre function/Proofs: Difference between revisions

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==Orthonormality Proof==
{{subpages}}
It is demonstrated that the associated Legendre functions are orthogonal in two different ways and their normalization constant for each is derived.


This section demonstrates that the Associated Legendre Functions are orthogonal and derives their normalization constant.
==Theorem (orthogonality relation 1)==  
 
:<math>\int\limits_{-1}^{1}P_{l}^{m} \left( x\right) P_{k}^{m} \left( x\right)
===Theorem===
 
<math>\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
dx =\frac{2}{2l+1} \frac{\left( l+m\right) !}{\left( l-m\right) !} \delta
_{lk}. </math>
_{lk}, </math>
 
where:  
[''Note: The proof uses the more common <math>P_{l}^{m} </math> notation, rather than'' 
:<math>P_{l}^{m} \left( x\right) =\frac{\left( -1\right) ^{m} }{2^{l} l!} \left(
<math>P_{l}^{\left( m\right)} </math>]
 
Where:  
<math>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(
1-x^{2} \right) ^{\frac{m}{2} } \frac{d^{l+m} }{dx^{l+m} } \left[ \left(
x^{2} -1\right) ^{l} \right], </math>
x^{2} -1\right) ^{l} \right], \quad 0\leq m\leq l.</math>
<math>0\leq m\leq l.</math>


===Proof===
===Proof===


The Associated Legendre Functions are regular solutions to the  
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  
general Legendre equation: <math>\left( \left[ 1-x^{2} \right] y^{'} \right) ^{'} +\left( l\left[ l+1\right]
-\frac{m^{2} }{1-x^{2} } \right) y=0</math>
, where <math>z^{'} =\frac{dz}{dx}. </math>
 
This equation is an example of a more general class of equations  
known as the [[Sturm-Liouville theory | Sturm-Liouville equation]]s. Using Sturm-Liouville  
known as the [[Sturm-Liouville theory | Sturm-Liouville equation]]s. Using Sturm-Liouville  
theory, one can show that
theory, one can show the orthogonality of functions with the same superscript ''m'' and different subscripts:
<math>K_{kl}^{m} =\int\limits_{-1}^{1}P_{k}^{m} \left( x\right) P_{l}^{m}
:<math>
\left( x\right)  dx </math>
K_{kl}^{m} =\int\limits_{-1}^{1}P_{k}^{m} \left( x\right) P_{l}^{m}
vanishes when
\left( x\right)  dx = 0 \quad\hbox{if}\quad k \ne l .
<math>k\neq l.</math>
</math>
However, one can find  
However, one can find  
<math>K_{kl}^{m} </math>
<math>K_{kl}^{m} </math>
directly from the above definition, whether or not  
directly from the above definition, whether or not  
<math>k=l:</math>
<math>k=l</math>.
 
This involves evaluating the overlap integral directly from the definition of the associated Legendre functions given in the main article. Indeed, inserting the definition of the function twice:
<math>K_{kl}^{m} =\frac{1}{2^{k+l} \left( k!\right) \left( l!\right) }
:<math>
\int\limits_{-1}^{1}\left\{ \left( 1-x^{2} \right) ^{m} \frac{d^{k+m}
K_{kl}^{m} =\frac{1}{2^{k+l}\; k! \; l! }
}{dx^{k+m} } \left[ \left( x^{2} -1\right) ^{k} \right] \right\} \left\{
\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]
\frac{d^{l+m} }{dx^{l+m} } \left[ \left( x^{2} -1\right) ^{l} \right]
\right\}  dx. </math>
\right\}  dx.  
</math>


Since
Since ''k'' and ''l''  occur symmetrically, one can without loss of generality assume  
<math>k</math> and  
that ''l'' &ge; k.  Use the well-known integration-by-parts equation
<math>l</math> occur symmetrically, one can without loss of generality assume  
:<math>
that  
\int_{-1}^1 u\; v'\; dx = \left. u\,v\right|_{-1}^1 - \int_{-1}^{1} v u' \;dx
<math>l\geq k.</math>
</math>
Integrate by parts
''l'' + ''m'' times, where the curly brackets in the integral indicate the factors, the first being  
<math>l+m</math>
''u'' and the second ''v''&rsquo;. For each of the first ''m'' integrations by parts,  
times, where the curly brackets in the integral indicate the  
''u'' in the <math> uv|_{-1}^1</math> term contains the factor (1&minus;x<sup>2</sup>),
factors, the first being  
so the term vanishes. For each of the remaining ''l'' integrations,  
<math>u</math>
''v''  in that term contains the factor (''x''<sup>2</sup>&minus;1)
and the second  
<math>v'.</math>
For each of the first  
<math>m</math>
integrations by parts,  
<math>u</math>
in the  
<math>\left. uv\right| _{-1}^{1} </math>
term contains the factor  
<math>\left( 1-x^{2} \right) </math>;
so the term vanishes. For each of the remaining  
<math>l</math>
integrations,  
<math>v</math>
in that term contains the factor  
<math>\left( x^{2} -1\right) </math>;  
so the term also vanishes. This means:
so the term also vanishes. This means:
 
:<math>
<math>K_{kl}^{m} =\frac{\left( -1\right) ^{l+m} }{2^{k+l} \left( k!\right)
K_{kl}^{m} =\frac{\left( -1\right) ^{l+m} }{2^{k+l} \; k!\;
\left( l!\right) } \int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l}
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^{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]
\frac{d^{k+m} }{dx^{k+m} } \left[ \left( x^{2} -1\right) ^{k} \right]
\right]  dx. </math>
\right]  dx.  
</math>


Expand the second factor using Leibnitz' rule:
Expand the second factor using Leibnitz' rule:


<math>\frac{d^{l+m} }{dx^{l+m} } \left[ \left( 1-x^{2} \right) ^{m}
:<math>\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]
\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(
\right] =\sum\limits_{r=0}^{l+m}
l+m-r\right) !}  \frac{d^{r} }{dx^{r} } \left[ \left( 1-x^{2} \right) ^{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)
\right] \frac{d^{l+k+2m-r} }{dx^{l+k+2m-r} } \left[ \left( x^{2} -1\right)
^{k} \right]. </math>
^{k} \right].  
</math>


The leftmost derivative in the sum is non-zero only when  
The leftmost derivative in the sum is non-zero only when ''r'' &le; 2''m''
<math>r\leq 2m</math>
(remembering that ''m'' &le; ''l''). The other derivative is non-zero only when ''k'' + ''l'' + 2''m'' &minus; ''r'' &le; 2''k'', that is, when ''r'' &ge; 2''m'' + ''l'' &minus; ''k''. Because ''l'' &ge; ''k'' these two conditions imply that the only non-zero term in the sum occurs when ''r'' = 2''m'' and ''l'' = ''k''.  
(remembering that  
<math>m\leq l</math>
). The other derivative is non-zero only when  
<math>k+l+2m-r\leq 2k</math>,  
that is, when
<math>r\geq 2m+(l-k).</math>
Because  
<math>l\geq k</math>
these two conditions imply that the only non-zero term in the  
sum occurs when  
<math>r=2m</math>
and  
<math>l=k.</math>
So:
So:
 
:<math>
<math>K_{kl}^{m} =\frac{\left( -1\right) ^{l+m} }{2^{2l} \left( l!\right) ^{2}
K_{kl}^{m} =\ (-1)^{l}\ \delta_{kl} \; \frac{(-1)^{l+m} }{2^{2l}\, (l!)^{2}}  
} \frac{\left( l+m\right) !}{\left( 2m\right) !\left( l-m\right) !} \delta
\binom{l+m}{2m}
_{kl} \int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l} \frac{d^{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^{2m} } \left[ \left( 1-x^{2} \right) ^{m} \right] \frac{d^{2l}
}{dx^{2l} } \left[ \left( 1-x^{2} \right) ^{l} \right]  dx. </math>
}{dx^{2l} } \left[ \left( 1-x^{2} \right) ^{l} \right]  dx,
 
</math>
To evaluate the differentiated factors, expand  
where &delta;<sub>''kl''</sub> is the [[Kronecker delta]] that shows the orthogonality of functions with ''l'' &ne; ''k''. The factor <math>(-1)^{l}</math> at the front of <math>K_{kl}^{m}</math> comes from switching the sign of x<sup>2</sup>-1 inside (x<sup>2</sup>-1)<sup>l</sup>.
<math>\left( 1-x^{2} \right) ^{k} </math>
To evaluate the differentiated factors, expand (1&minus;x&sup2;)<sup>''k''</sup>
using the binomial theorem:  
using the binomial theorem:  
<math>\left( 1-x^{2} \right) ^{k} =\sum\limits_{j=0}^{k}\left(
:<math>
\begin{array}{c}
\left( 1-x^{2} \right) ^{k} =\sum\limits_{j=0}^{k} \binom{k}{j}  
k \\ j
( -1)^{k-j} x^{2(k-j)}.   
\end{array}
</math>
\right) \left( -1\right) ^{k-j} x^{2\left( k-j\right) }.  </math>
The only term that survives differentiation 2''k''
The only thing that survives differentiation  
times is the ''x''<sup>2''k''</sup>
<math>2k</math>
term, which after differentiation gives
times is the  
:<math>
<math>x^{2k} </math>
(-1)^k \, \binom{k}{0}\, 2k! = (-1)^{k}\, (2k)! \, .
term, which (after differentiation) equals:
</math>  
<math>\left( -1\right) ^{k} \left(
Therefore:
\begin{array}{c}
:<math>
k \\ 0
K_{kl}^{m} =\ (-1)^{l}\delta _{kl}\; \; \frac{1}{2^{2l}\; (l!) ^{2} } \frac{(2l)!\,(l+m)!}{(l-m)!}  
\end{array}
\int\limits_{-1}^{1}(x^{2} -1)^{l}  dx  
\right) \left( 2k\right) !=\left( -1\right) ^{k} \left( 2k\right) !</math>. Therefore:
\qquad\qquad\qquad\qquad\qquad\qquad (1)
 
</math>  
<math>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 </math> ................................................. (1)
 
Evaluate  
Evaluate  
<math>\int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l}  dx </math>
:<math>
\int\limits_{-1}^{1}(x^{2} -1)^{l}  dx </math>
by a change of variable:  
by a change of variable:  
<math>x=\cos \theta \Rightarrow dx=-\sin \theta d\theta\;and\; 1-x^{2} =\sin \theta. </math>   
:<math>
Thus, <math>\int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l}  dx
x=\cos \theta\; \Longrightarrow\; dx=-\sin \theta d\theta\quad\hbox{and}\quad 1-x^{2} =(\sin \theta)^2.  
=\int\limits_{0}^{\pi }\left( \sin \theta \right) ^{2l+1}  d\theta. </math> [To eliminate the negative sign on the second integral, the limits are switched
</math>   
from <math> \pi \rightarrow  0 \; </math> to <math>\; 0 \rightarrow \pi </math> , recalling that <math> \; -1 = \cos (\pi) \;</math> and <math>\; 1 = \cos (0) \; </math>].
Thus,  
:<math>\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,  
</math>  
where we recall that
:<math>  -1 = \cos\,\pi \quad\hbox{and}\quad  1 = \cos\,0.
</math>
The limits were switched from
:<math> \pi \rightarrow  0 \; \quad\hbox{and}\quad 0 \rightarrow \pi </math> ,
which accounts for one minus sign .
Integration of
:<math>
\frac{d(\sin^{n-1}\theta\cos\theta)}{d\theta} = (n-1)\sin^{n-2}\theta - n\sin^{n}\theta
</math>  
gives
:<math>
\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. 
</math>
Since
:<math>\left. -\sin^{n-1} \theta \cos \theta \right| _{0}^{\pi } =0 \quad\hbox{for}\quad n > 1,</math>  
it follows that
:<math>\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 
</math>
for ''n'' &gt; 1.  


A [http://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions table of standard trigonometric integrals] shows:
<math>\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.  </math>
Since <math>\left. -\sin \theta \cos \theta \right| _{0}^{\pi } =0,</math> <math>\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  </math>
for
<math>n\geq 2.</math>
Applying this result to  
Applying this result to  
<math>\int\limits_{0}^{\pi }\left( \sin \theta \right) ^{2l+1}  d\theta  </math>
:<math>\int\limits_{0}^{\pi }\left( \sin \theta \right) ^{2l+1}  d\theta  </math>
and changing the variable back to  
and changing the variable back to ''x''
<math>x</math>
yields:  
yields:  
<math>\int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l}  dx =\frac{2\left(
:<math>\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
l+1\right) }{2l+1} \int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l-1}  dx
</math>
</math>
for <math>l\geq 1.</math>
for ''l'' &ge; 1.
Using this recursively:
Using this recursively:
:<math>
\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 </math>
Noting:
:<math>\int\limits_{-1}^{1}\  dx\ =\ 2</math>
and


<math>\int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l}  dx =\frac{2\left(
:<math> \frac{2l }{2l+1} \frac{2\left( l-1\right) }{2l-1} \frac{2\left( l-2\right)
l+1\right) }{2l+1} \frac{2\left( l\right) }{2l-1} \frac{2\left( l-1\right)
}{2l-3} ...\frac{2}{3} \ =\ \frac{2^ll! }{(2l+1)(2l-1)(2l-3)...3}\ =\ \frac{2^ll! }{\frac{(2l+1)! }{2^ll!}}\ =\frac{2^{2l} \left(
}{2l-3} ...\frac{2\left( 2\right) }{3} \left( 2\right) =\frac{2^{l+1}
l!\right) ^{2} }{\left( 2l+1\right) !}</math>
l!}{\frac{\left( 2l+1\right) !}{2^{l} l!} } =\frac{2^{2l+1} \left(
l!\right) ^{2} }{\left( 2l+1\right) !}. </math>


Applying this result to (1):
:<math>
\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) !}</math>


<math>K_{kl}^{m} =\frac{1}{2^{2l} \left( l!\right) ^{2} } \frac{\left(
Applying this result to equation (1):
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}
:<math>K_{kl}^{m} =\delta _{kl}\; \frac{1}{2^{2l}\; (l!)^{2} } \frac{(
\frac{\left( l+m\right) !}{\left( l-m\right) !} \delta _{kl}. </math>
2l)!\;(l+m)!}{(l-m)!}\; \frac{2^{2l+1} \;(l!)^{2} }{( 2l+1)!} = \delta _{kl}\,\frac{2}{2l+1}
QED.
\frac{( l+m) !}{( l-m) !}
\qquad\qquad \mathbf{QED}.
</math>
 
Clearly, if we define new associated Legendre functions  by a constant times the old ones,
:<math>
\bar{P}^m_l(x) \equiv \sqrt{ \frac{2l+1}{2}\; \frac{(l-m)!}{(l+m)!} }\; P^m_l(x)
</math>
then the overlap integral becomes,
:<math>
K^m_{kl} = \int\limits_{-1}^{1} \bar{P}^m_k(x) \bar{P}^m_l(x) \;dx = \delta_{kl},
</math>
that is, the new functions are normalized to unity.


===Comments===
===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 theory|Sturm-Liouville equation]]s.


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 possible to demonstrate their orthogonality using principles associated with operator calculus. Let us write
 
:<math>
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
P^m_{l}(x) = w(x)^{1/2} \; \nabla^m P_l(x)
</math>
where
:<math>
:<math>
\nabla_x \equiv \frac{d}{dx}.
\nabla \equiv \frac{d}{dx} \quad \hbox{and}\quad w(x) \equiv (1-x^2)^m.
</math>
</math>
Indeed, let ''w(x)'' be a function with ''w''(1) = ''w''(&minus;1) = 0, then
Clearly, the case ''m'' = 0 is,
:<math>
:<math>
\langle w g | \nabla_x f\rangle = \int_{-1}^1 w(x)g(x)\nabla_x f(x) dx   
P^{0}_l(x) = (1-x^2)^{0/2}\; \nabla^0 P_l(x) = P_l(x).
= \left[ w(x)g(x)f(x) \right]_{-1}^{1}  - \int_{-1}^1 \Big(\nabla_x w(x)g(x)\Big) f(x) dx  
</math>
= - \langle \nabla_x (w g) |  f\rangle
 
The proof given above starts out by implicitly proving the anti-Hermiticity of &nabla;.
Indeed, noting that ''w(x)'' is a function with ''w''(1) = ''w''(&minus;1) = 0 for ''m'' &ne; 0, it follows from partial integration that
:<math>
\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 (w g) \;|\; f\;\rangle
</math>
</math>
Hence
Hence
:<math>
:<math>
\nabla_x^\dagger = - \nabla_x \;\Longrightarrow\; \left(\nabla_x^\dagger\right)^{l+m}  = (-1)^{l+m} \nabla_x^{l+m}
\nabla^\dagger = - \nabla \;\Longrightarrow\; \left(\nabla^\dagger\right)^{m}  = (-1)^{m} \;\nabla^{m}.
</math>
</math>
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 <math> P_l</math>, 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'' &le; ''l'')
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 &Pi;<sub>''p''</sub> of order ''p'' lower than ''l''. In bra-ket notation 
:<math>
:<math>
\langle w \nabla_x^m P_k | \nabla_x^m P_l\rangle\quad\hbox{with}\quad w\equiv (1-x^2)^m,
\langle\; \Pi_p \;|\; P_l \;\rangle = 0\quad \hbox{if}\quad O\left[\Pi_p\right] \equiv p < l.
</math>
</math>
then
Knowing this,
:<math>
:<math>
\langle w \nabla_x^m P_k | \nabla_x^m P_l\rangle =
\int\limits_{-1}^{1}P_{l}^{m} \left( x\right) P_{k}^{m} \left( x\right)
(-1)^m \langle \nabla_x^m  (w \nabla_x^m P_k) |  P_l\rangle
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 .
</math>
</math>
The bra is a polynomial of order ''k'', and since ''k'' &le; ''l'',  the bracket is non-zero only if ''k'' = ''l''.
The bra is a polynomial of order ''k'', because
:<math>
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,
</math>
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'' &le; ''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.
 
==Theorem (orthogonality relation 2)==
 
<math>\int_{-1}^{1}\frac{P_{l}^{m}(x) P_{l}^{n} (x)}{(1-x^{2})} dx = \delta_{mn}\frac{(l+m)!}{m(l-m)!}
, m \neq 0 </math>.
 
where:
 
<math>P_{l}^{m}(x) = \frac{(-1)^{m}}{(2^{l} l!)} (1-x^{2})^{m/2} \frac{d^{l+m}}{dx^{l+m}} (x^{2}-1)^{l}
, 0 \leqq m \leqq l</math>.
 
===Proof===
 
Let:
 
<math>T_{l}^{mn} = \int_{-1}^{1}\frac{P_{l}^{m}(x) P_{l}^{n} (x)}{(1-x^{2})} \ dx
= \frac{(-1)^{m+n}}{(2^{l} l!)^2}  \int_{-1}^{1} \{(1-x^{2})^{\frac{(m+n)}{2}-1}
\frac{d^{l+m}}{dx^{l+m}} (x^{2}-1)^{l}\} \{\frac{d^{l+n}}{dx^{l+n}} (x^{2}-1)^{l}\} dx</math>.
 
The functions (x<sup>2</sup>-1)<sup>k</sup> are even functions; so their
j<sup>th</sup> order derivatives are even or odd functions according as j is even or odd. 
Therefore, if m or n is even but the other is odd, then one of the two factors in curly braces in the
preceding expression is an even function, and the other is an odd function.  This makes the entire
integrand an odd function.  When integrated between the limits that are negatives of each other,
it yields 0, as it should, since <math>\delta_{mn} = 0</math>.  So we need to consider further
only even integrands, for which m and n are either both even or both odd.  In this case all exponents
in the integrand are non-negative integers (except when m = n = 0).
 
The integrand can be integrated by parts <math> \ell+n</math> times, the first
factor in curly braces and its derivatives being identified as
u and the second factor as v' in the formula:
 
<math>\int_{-1}^{1} uv' dx = uv \Big|_{-1}^{1} - \int_{-1}^{1} vu' dx</math>.
 
Since m and n occur symmetrically, we can assume without loss
of generality that <math>n \leqq m</math>. For the first n-1 integrations
by parts, the uv term vanishes at the limits because u includes
at least one factor of (1-x<sup>2</sup>). The same also is true for the
n<sup>th</sup> integration, unless n = m, in which case the uv term is:
 
<math>S_{l}^{m} = \frac{(-1)^{m+1}}{(2^{l} l!)^{2}} \frac{d^{m-1}}{dx^{m-1}}\{(1-x^{2})^{m-1}
\frac{d^{l+m}}{dx^{l+m}}(x^{2}-1)^{l}\} \{\frac{d^{l}}{dx^{l}}(x^{2}-1)^{l} \} \Big|_{-1}^{1}</math>.
 
For the remaining <math>\ell</math> integrations the uv term vanishes at the
limits because v includes at least one factor 1-x<sup>2</sup> (and perhaps u does also). The result is:
 
<math>T_{l}^{mn} = \delta_{mn}S_{l}^{m} + \frac{(-1)^{l+m}}{(2^{l} l!)^{2}} \int_{-1}^{1}
(x^{2}-1)^{l} \frac{d^{l+n}}{dx^{l+n}}\{(1-x^{2})^{\frac{(m+n)}{2}-1} \frac{d^{l+m}}{dx^{l+m}}(x^{2}-1)^{l}\}
dx</math>.
 
The highest power of x in the binomial expansion of (x<sup>2</sup>-1)<sup>l</sup>
is <math>2\ell</math>; after <math>\ell+m</math> derivatives of it are taken, the highest power
is <math>\ell-m</math>. The highest power of x in the expansion of <math>(1-x^{2})^{\frac{(m+n)}{2}-1}</math>
is m+n-2; so the highest power in the expression in curly braces
is <math>\ell+n-2</math>. After <math>\ell+n-2</math> derivatives of it are taken, the highest
power is 0; that is, the expression is a constant. Taking the
remaining 2 derivatives causes the integrand to vanish. Therefore:
 
<math>T_{l}^{mn} = \delta_{mn}S_{l}^{m}
= \delta_{mn}\frac{1}{(2^{l} l!)^{2}} \frac{d^{m-1}}{dx^{m-1}}\{(x^{2}-1)^{m-1}
\frac{d^{l+m}}{dx^{l+m}} (x^{2}-1)^{l}\} \{\frac{d^{l}}{dx^{l}}(x^{2}-1)^{l}\} \Big|_{-1}^{1}</math>.
 
Since this function is odd, we need evaluate it only at its upper
limit x = 1 and double the result.
 
An expression of the form <math>(x^{2}-1)^{k} = (x-1)^{k}(x+1)^{k}</math> can be differentiated using Leibnitz's rule. Only one term in the sum is of interest here, namely the one in which (x-1)<sup>k</sup> is differentiated exactly k times so that no factors of (x-1) remain.  If it is differentiated fewer times, then it vanishes at x = 1.  If it is differentiated more times, then it vanishes everywhere.  So, ignoring terms with factors of (x-1), we have (from right to left in the preceding expression):
 
<math>\frac{d^{l}}{dx^{l}}(x^{2}-1)^{l} = [\frac{d^{l}}{dx^{l}}(x-1)^{l}](x+1)^{l} = l!
(x+1)^{l}</math> , which equals <math> 2^{l} \ell!</math> when x = 1:
 
<math>\frac{d^{l+m}}{dx^{l+m}}(x^{2}-1)^{l} = \frac{(l+m)!}{l!m!} [\frac{d^{l}}{dx^{l}} (x-1)^{l}]
[\frac{d^{m}}{dx^{m}}(x+1)^{l}]</math>
 
::<math>= \frac{(l+m)!}{l!m!} [l!] [\frac{l!}{(l-m)!} (x+1)^{l-m}]</math>
 
::<math>= (l+m)! \frac{l!}{m!(l-m)!} (x+1)^{l-m}</math>
 
<math>\frac{d^{m-1}}{dx^{m-1}}\{(x^{2}-1)^{m-1} \frac{d^{l+m}}{dx^{l+m}}(x^{2}-1)^{l}\}</math>
 
::<math>= (l+m)! \frac{l!}{m!(l-m)!} \frac{d^{m-1}}{dx^{m-1}}\{(x-1)^{m-1}(x+1)^{l-1}\}</math>
 
::<math>= (l+m)! \frac{l!}{m!(l-m)!} (m-1)! (x+1)^{l-1}</math>
 
::<math>= (l+m)! \frac{l!}{m(l-m)!} (x+1)^{l-1}</math> , which equals <math>2^{l-1}(l+m)! \frac{l!}{m(l-m)!} </math> when x = 1.
 
<math>T_{l}^{mn} = 2\delta_{mn}\frac{1}{(2^{l} l!)^{2}} \{2^{l-1}(l+m)! \frac{l!}{m(l-m)!}\}
\{2^{l} l!\} = \delta_{mn}\frac{(l+m)!}{m(l-m)!}</math>
 
which has been doubled to include the lower limit x = -1. QED.
 
When n = m = 0:
 
<math>T_{l}^{00} = \int_{-1}^{1} \frac{P_{l}^{0}(x) P_{l}^{0} (x)}{(1-x^{2})} dx</math>
 
::<math>= \frac{1}{(2^{l} l!)^{2}} \int_{-1}^{1} \{(1-x^{2})^{-1} \frac{d^{l}}{dx^{l}}(x^{2}-1)^{l}\}
\{\frac{d^{l}}{dx^{l}}(x^{2}-1)^{l}\} dx </math>
 
::<math>= \frac{1}{(2^{l} l!)^{2}} \int_{-1}^{1} \frac{1}{2}[(x+1)^{-1} - (x-1)^{-1}]
\{\frac{d^{l}}{dx^{l}}(x^{2}-1)^{l}\}^{2} dx</math>.
 
When expanded in powers of (x+1) and (x-1) by using Leibnitz's
rule, the expression in curly braces includes exactly one term,
(x-1)<sup>l</sup>, having no factor (x+1) and exactly one term, (x+1)<math>^{l}</math>, having no factor (x-1). Therefore in a Laurent series expansion of the integrand about either x = 1 or x = -1 all terms except
one have non-negative exponents, and their integral is finite.
The one remaining term in the integrand is a constant multiple
of 1/(x+1) or 1/(x-1); so its integral logarithmically diverges
at the limits, making T<math>_{l}^{00}</math> infinite.

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More information relevant to Associated Legendre function.

It is demonstrated that the associated Legendre functions are orthogonal in two different ways and their normalization constant for each is derived.

Theorem (orthogonality relation 1)

where:

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:

However, one can find directly from the above definition, whether or not . This involves evaluating the overlap integral directly from the definition of the associated Legendre functions given in the main article. Indeed, inserting the definition of the function twice:

Since k and l occur symmetrically, one can without loss of generality assume that l ≥ k. Use the well-known integration-by-parts equation

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 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:

Expand the second factor using Leibnitz' rule:

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

where δkl is the Kronecker delta that shows the orthogonality of functions with lk. The factor at the front of 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:

The only term that survives differentiation 2k times is the x2k term, which after differentiation gives

Therefore:

Evaluate

by a change of variable:

Thus,

where we recall that

The limits were switched from

,

which accounts for one minus sign . Integration of

gives

Since

it follows that

for n > 1.

Applying this result to

and changing the variable back to x yields:

for l ≥ 1. Using this recursively:

Noting:

and

Applying this result to equation (1):

Clearly, if we define new associated Legendre functions by a constant times the old ones,

then the overlap integral becomes,

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

where

Clearly, the case m = 0 is,

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

Hence

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 Πp of order p lower than l. In bra-ket notation

Knowing this,

The bra is a polynomial of order k, because

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 kl, 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.

Theorem (orthogonality relation 2)

.

where:

.

Proof

Let:

.

The functions (x2-1)k are even functions; so their jth order derivatives are even or odd functions according as j is even or odd. Therefore, if m or n is even but the other is odd, then one of the two factors in curly braces in the preceding expression is an even function, and the other is an odd function. This makes the entire integrand an odd function. When integrated between the limits that are negatives of each other, it yields 0, as it should, since . So we need to consider further only even integrands, for which m and n are either both even or both odd. In this case all exponents in the integrand are non-negative integers (except when m = n = 0).

The integrand can be integrated by parts times, the first factor in curly braces and its derivatives being identified as u and the second factor as v' in the formula:

.

Since m and n occur symmetrically, we can assume without loss of generality that . For the first n-1 integrations by parts, the uv term vanishes at the limits because u includes at least one factor of (1-x2). The same also is true for the nth integration, unless n = m, in which case the uv term is:

.

For the remaining integrations the uv term vanishes at the limits because v includes at least one factor 1-x2 (and perhaps u does also). The result is:

.

The highest power of x in the binomial expansion of (x2-1)l is ; after derivatives of it are taken, the highest power is . The highest power of x in the expansion of is m+n-2; so the highest power in the expression in curly braces is . After derivatives of it are taken, the highest power is 0; that is, the expression is a constant. Taking the remaining 2 derivatives causes the integrand to vanish. Therefore:

.

Since this function is odd, we need evaluate it only at its upper limit x = 1 and double the result.

An expression of the form can be differentiated using Leibnitz's rule. Only one term in the sum is of interest here, namely the one in which (x-1)k is differentiated exactly k times so that no factors of (x-1) remain. If it is differentiated fewer times, then it vanishes at x = 1. If it is differentiated more times, then it vanishes everywhere. So, ignoring terms with factors of (x-1), we have (from right to left in the preceding expression):

, which equals when x = 1:

, which equals when x = 1.

which has been doubled to include the lower limit x = -1. QED.

When n = m = 0:

.

When expanded in powers of (x+1) and (x-1) by using Leibnitz's rule, the expression in curly braces includes exactly one term, (x-1)l, having no factor (x+1) and exactly one term, (x+1), having no factor (x-1). Therefore in a Laurent series expansion of the integrand about either x = 1 or x = -1 all terms except one have non-negative exponents, and their integral is finite. The one remaining term in the integrand is a constant multiple of 1/(x+1) or 1/(x-1); so its integral logarithmically diverges at the limits, making T infinite.