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My question concerns the associated or generalized Legendre polynomials. They are labeled by two numbers $m$ and $l$; i.e., $P_l^m(x)$, for $x \in [-1,1]$. Usually one assumes that $m$ and $l$ are both integers, but I suspect that it is possible for them to be half integers. If this is correct, what is the meaning of formulas like

$$P_l^{-m}(x)=(-1)^m \frac{(l-m)!}{(l+m)!}P_l^m(x)$$

or Rodrigues's formula?


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Can you link a reference in which these polynomial are used? – Davide Giraudo Jan 25 '12 at 13:14
The most similar thing I am able to find is this paper Hunter et al. "Fermion quasi-spherical harmonics" J. Phys. A: Math. Gen. 32 795 (1999). – user23621 Jan 25 '12 at 14:16

It is perfectly alright to have (associated) Legendre functions of half-integer order (or in fact, any arbitrary complex order). The key is to define them in terms of Gaussian hypergeometric functions, e.g.

$$P_\ell^m(z)=\frac{(1+z)^{m/2}}{(1-z)^{m/2}}\frac1{\Gamma(1-m)}{}_2 F_1\left({{-\ell\quad\ell+1}\atop{1-m}} \mid \frac{1-z}{2}\right)$$

to use one particular normalization.

For $\ell$ a half integer, one could use the formulae

$$\begin{align*} P_{-\frac12}^m(z)&=\frac2{\pi}K\left(\frac{1-z}{2}\right)\\ P_\frac12^m(z)&=\frac2{\pi}\left(2E\left(\frac{1-z}{2}\right)-K\left(\frac{1-z}{2}\right)\right) \end{align*}$$

where $K(m)$ and $E(m)$ are complete elliptic integrals with parameter $m$, along with the usual recursion relations over $\ell$ and $m$.

For half-integer $m$ we can start the recursion relations with

$$\begin{align*} P_0^{-\frac12}(z)&=\frac2{\sqrt\pi}\sqrt[4]{\frac{1-z}{1+z}}\\ P_0^\frac12(z)&=\frac1{\sqrt\pi}\sqrt[4]{\frac{1+z}{1-z}} \end{align*}$$

and for both $\ell$ and $m$ half-integer, the recursion relations take the initial values

$$\begin{align*} P_{-\frac12}^{-\frac12}(z)&=2\sqrt{\frac2{\pi}}\frac1{\sqrt[4]{1-z^2}}\arcsin\left(\sqrt{\frac{1-z}{2}}\right)\\ P_{-\frac12}^\frac12(z)&=\sqrt{\frac2{\pi}}\frac1{\sqrt[4]{1-z^2}}\\ P_\frac12^{-\frac12}(z)&=\sqrt{\frac2{\pi}}\sqrt[4]{1-z^2}\\ P_\frac12^\frac12(z)&=\sqrt{\frac2{\pi}}\frac{z}{\sqrt[4]{1-z^2}} \end{align*}$$

Note that for arbitrary order, the Rodrigues formula can be suitably interpreted as a Riemann-Liouville differintegral (i.e., differentiation/integration to arbitrary complex order). I might edit this answer for a demonstration later.

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