Talk:Associated Legendre function/Addendum

I added a proof of orthogonality and a derivation of the normalization constant for the first equation in the Orthogonality relations section in on the main page. Dan Nessett 16:42, 11 July 2009 (UTC)

Comments on proof

1. 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 given in the Addendum.

2. When as an intermediate the ordinary Legendre polynomials P_l are introduced, 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. We meet (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.

Then, knowing this, the hard work (given in the Addendum) of computing the normalization constant remains.

--Paul Wormer 15:13, 12 July 2009 (UTC)