358 Chapter 12Second-order differential equations. Constant coefficients
EXAMPLE 12.11Show that the functions ψ
1and ψ
2satisfy the orthogonality
condition (12.70).
We have
and this is zero becausee
2π
i1 = 1 1.
0 Exercise 27
The functions (12.69) can be interpreted as representating clockwise rotation of the
particle whenn 1 > 10 and anticlockwise rotation whenn 1 < 10 (see Section 8.6 for the
rigid rotor). They are complex functions when n 1 ≠ 10 , and for some purposes it is more
convenient to have solutions of the Schrödinger equation that are real functions.
The convention is to use the trigonometric functions obtained by means of Euler’s
formula. We have
and the combinations
(12.71)
are the alternative orthonormal set. The symmetries of these functions are illustrated
in Figure 12.8.
We note that these symmetries are also the symmetries around the molecular axis
of the molecular orbitals of a linear molecule, withn 1 = 10 for σ-orbitals,n 1 = 1 ± 1 for
π-orbitals, n 1 = 1 ± 2 forδ-orbitals, and so on. They are also the symmetries of the
normal modes of vibration of a circular drum.
0 Exercise 28
1
2
1
2
1
2
1
2
()cos ()sinψψ θθψψ
nn nnn
i
+= , −= n
−−ππ
ψ θθθ
±=±
n() (cosninsin )
1
2 π
==−
1
2
11
2
1
022ππ
ππi
e
i
e
iiθZZZ
0212022021
2
1
2
πππππ
ψψθθθ θ
θθ*( ) ( )deed
ii==
−eed
iθθ
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.n=0
n=± 1 n=± 2
Figure 12.8