Physical Chemistry , 1st ed.

The orthogonality and normality properties of wavefunctions are usually
combined into a single expression termed orthonormality:

* 0 ifmn

mnd1 ifmn (10.28)

Example 10.16

Demonstrate explicitly that for the 1-D particle-in-a-box, 1 is orthogonal

to  2.

Solution

Evaluate the following integral:

2

a



a

0

sin

1

a

x

sin

2

a

x

dx

(The 2/aconstant has been pulled outside the integral, and the limits of in-

tegration are properly set as 0 to a.) Using the integral table in Appendix 1:

2

a



a

0

sin

1

a

x

sin

2

a

x

dx

2

a

(^)   a 0
sin (^1 a^ ^2 a^ )x
2(^1 a^ ^2 a^ )
sin (^1 a^ ^2 a^ )x
2(^1 a^ ^2 a^ )
10.13 Orthogonality 307

 2

a







a 0

Substituting the limits 0 and ainto this expression and evaluating:

 2

a







 2

a







sin (^3 a^ ) 

0

(^6 a^ )

sin (^1 a^ )^0

(^2 a^ )

sin (^3 a^ )^ a

(^6 a^ )

sin (^1 a^ )^ a

(^2 a^ )

sin (^3 a^ )x

(^6 a^ )

sin (^1 a^ )x

(^2 a^ )



2

a

(^)   

2

a

(^)   
sin 0
(^6 a^ )
sin 0
(^2 a^ )
sin (3 )
(^6 a^ )
sin (
)
(^2 a^ )
 2
a
(^)  


 2

a

(^)  


 0

Therefore,

2

a



a

0

sin

1

a

x

sin

2

a

x

dx 0

which is exactly as it should be for orthogonal functions. You should satisfy

yourself that you get the same answer if you evaluate the integral when you

take the complex conjugate of 2 instead of 1.

Orthonormality is a very useful concept. Integrals whose values are ex-
actly 0 or exactly 1 make mathematical derivations much easier, and it is im-
portant that you develop the skill to recognize when integrals are exactly 1
(because the wavefunctions in the integrand are normalized) or exactly 0
(because the wavefunctions in the integrand are orthogonal). Finally, note
that the orthonormality condition requires that no operator be present inside
the integral. If an operator is present, the operation must be evaluated be-
fore you can consider whether the integral can be exactly 0 or 1.

0

(^6 a^ )

0

(^2 a^ )

0

(^6 a^ )

0

(^2 a^ )