The Chemistry Maths Book, Second Edition

13.4 The Legendre equation 379

We have

Then

P

3

2

(x) 1 = 1 15(1 1 − 1 x

2

)xP

3

2

(cos 1 θ) 1 = 1151 sin

2

1 θ 1 cos 1 θ

P

3

3

(x) 1 = 1 15(1 1 − 1 x

2

)

322

P

3

3

(cos 1 θ) 1 = 1151 sin

3

1 θ

0 Exercise 18

Orthogonality and normalization

We saw in Section 12.6 that the solutions of the Schrödinger equation for the particle

in a box are orthogonal functions (equation (12.58)). Orthogonality is a property of

the solutions of a class of second-order linear differential equations called Sturm–

Liouville equations that includes the Legendre equation. The Legendre polynomials

are orthogonal in the interval− 11 ≤ 1 x 1 ≤ 11 :

(13.25)

EXAMPLE 13.8Show thatP

1

is orthogonal (a) toP

2

and (b) toP

3

.

We have

Then

(a)

In this case, P

1

(x)is an odd function of xwhereas P

2

(x)is an even function.

The integrand is odd and the integral is zero (see Section 5.3).

=−











−−





























=

1

2

3

4

1

2

3

4

1

2

0

ZZ

−

+

−

+

−

+

=−=

1

1

12

1

1

3

1

1

1

2

3

1

2

3

P x P x dx x x dx

x

() () ( )

442

42

−

















x

Px Px x Px x x

12

2

3

3

1

1

2

31

1

2

()=, () (= − ,) () (= 53 − ).

Z

−

+

′

=≠′

1

1

PxP xdx 0 l l

ll

() () when

P

3

12

3

2

Px x x (cos ) sin ( cosθθθ=− 51 )

3

12122

3

2

() ( ) (=− 151 −)

/

yPx x x y x== −,′=−,y x′′=, ′′′

3

32

1

2

53

3

2

() ( ) ( 51 ) 15 yy = 15