Legendre polynomial
The
and is even or odd depending on the parity of
Mathematica
The Legendre polynomial
be generated in Wolfram Mathematica with π β ( π₯ ) LegendreP[β, x].
Properties
- The Legendre polynomials satisfy the orthonormality condition
Proof of 1
Without loss of generality, assume
π β² < π πΌ = 2 β + β ! β ! β β² ! β« 1 β 1 π β ( π₯ ) π β β² ( π₯ ) π π₯ = β« 1 β 1 ( ( π π π₯ ) β ( π₯ 2 β 1 ) β ) ( ( π π π₯ ) β β² ( π₯ 2 β 1 ) β β² ) π π₯ = [ β β π = 1 ( ( π π π₯ ) β β π ( π₯ 2 β 1 ) β ) ( ( π π π₯ ) β β² + π β 1 ( π₯ 2 β 1 ) β β² ) ] π₯ = 1 π₯ = β 1 + ( β 1 ) π β« 1 β 1 ( π₯ 2 β 1 ) β ( π π π₯ ) β β² + β ( π₯ 2 β 1 ) β β² π π₯ Now the integral term on the final line is zero, since the highest power of
is π₯ and π₯ 2 β β² . Each of the sum terms contains at least one β β² + β > 2 β β² factor and is hence zero. Thus for ( π₯ 2 + 1 ) the integral is zero. For the case of β β β β² β β β β² πΌ = ( 2 β β ! ) 2 β« 1 β 1 [ π β ( π₯ ) ] 2 π π₯ = ( β 1 ) β β« 1 β 1 ( π₯ 2 β 1 ) ( π π π₯ ) 2 β ( π₯ 2 β 1 ) β π π₯ = ( β 1 ) β β« 1 β 1 ( π₯ 2 β 1 ) β ( 2 β ! ) π π₯ = 2 ( 2 β ) ! β« 1 0 ( 1 β π₯ 2 ) β π π₯ Let
, so π₯ = c o s β‘ π , π π₯ = β s i n β‘ π π π , and ( 1 β π₯ 2 ) = s i n 2 β‘ π . Then [ 0 , 1 ] β [ π 2 , 0 ] πΌ = 2 ( 2 β ) ! β« 0 π / 2 s i n 2 β β‘ π ( β s i n β‘ π ) π π = 2 ( 2 β ) ! β« π / 2 0 s i n 2 β + 1 β‘ π π π = 2 ( 2 β ) ( 2 β β ! ) 2 ( 2 β + 1 ) ! which proves ^P1
Footnotes
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2018. Introduction to quantum mechanics, Β§4.1, p. 135 β©