Legendre polynomial

The $ℓ$ th Legendre polynomial $𝑃 ℓ$ for $ℓ \in ℕ 0$ is a polynomial of degree $ℓ$ given by the Rodrigues' formula¹ #m/def/fun

𝑃 ℓ (𝑥) = 1 2 ℓ ℓ! (𝑑 𝑑 𝑥) ℓ (𝑥 2 - 1) ℓ

and is even or odd depending on the parity of $ℓ$ .

Mathematica

The Legendre polynomial $𝑃 ℓ (𝑥)$ be generated in Wolfram Mathematica with LegendreP[ℓ, x].

Properties

The Legendre polynomials satisfy the orthonormality condition

\int 1 - 1 𝑃 ℓ (𝑥) 𝑃 ℓ' (𝑥) 𝑑 𝑥 = (2 2 ℓ + 1) 𝛿 ℓ ℓ'

Proof of 1

Without loss of generality, assume $𝑙' < 𝑙$

𝐼 = 2 ℓ + ℓ! ℓ! ℓ'! \int 1 - 1 𝑃 ℓ (𝑥) 𝑃 ℓ' (𝑥) 𝑑 𝑥 = \int 1 - 1 ((𝑑 𝑑 𝑥) ℓ (𝑥 2 - 1) ℓ) ((𝑑 𝑑 𝑥) ℓ' (𝑥 2 - 1) ℓ') 𝑑 𝑥 = [ℓ \sum 𝑗 = 1 ((𝑑 𝑑 𝑥) ℓ - 𝑗 (𝑥 2 - 1) ℓ) ((𝑑 𝑑 𝑥) ℓ' + 𝑗 - 1 (𝑥 2 - 1) ℓ')] 𝑥 = 1 𝑥 = - 1 + (- 1) 𝑛 \int 1 - 1 (𝑥 2 - 1) ℓ (𝑑 𝑑 𝑥) ℓ' + ℓ (𝑥 2 - 1) ℓ' 𝑑 𝑥

Now the integral term on the final line is zero, since the highest power of $𝑥$ is $𝑥 2 ℓ'$ and $ℓ' + ℓ > 2 ℓ'$ . Each of the sum terms contains at least one $(𝑥 2 + 1)$ factor and is hence zero. Thus for $ℓ \neq ℓ'$ the integral is zero. For the case of $ℓ \neq ℓ'$

𝐼 = (2 ℓ ℓ!) 2 \int 1 - 1 [𝑃 ℓ (𝑥)] 2 𝑑 𝑥 = (- 1) ℓ \int 1 - 1 (𝑥 2 - 1) (𝑑 𝑑 𝑥) 2 ℓ (𝑥 2 - 1) ℓ 𝑑 𝑥 = (- 1) ℓ \int 1 - 1 (𝑥 2 - 1) ℓ (2 ℓ!) 𝑑 𝑥 = 2 (2 ℓ)! \int 10 (1 - 𝑥 2) ℓ 𝑑 𝑥

Let $𝑥 = c o s 𝜃$ , so $𝑑 𝑥 = - s i n 𝜃 𝑑 𝜃$ , $(1 - 𝑥 2) = s i n 2 𝜃$ , and $[0, 1] \to [𝜋 2, 0]$ . Then

𝐼 = 2 (2 ℓ)! \int 0 𝜋 / 2 s i n 2 ℓ 𝜃 (- s i n 𝜃) 𝑑 𝜃 = 2 (2 ℓ)! \int 𝜋 / 2 0 s i n 2 ℓ + 1 𝜃 𝑑 𝜃 = 2 (2 ℓ) (2 ℓ ℓ!) 2 (2 ℓ + 1)!

which proves ^P1

#state/develop | #lang/en | #SemBr

2018. Introduction to quantum mechanics, §4.1, p. 135 ↩

Legendre polynomial

Properties

Footnotes