We can check the solutions for all values of x by substituting the different solutions into Schrödinger’s equation. For example, let us put the
ground-state solution into the equation and show that it holds. First,
we calculate the second derivative of the wavefunction. In taking these
derivatives note that the two terms are obtained and can be rewritten in
terms of the wavefunction.
ψ 0 (x) = N 0 e
−x 2 /2α 2
ψ 0 (x) = N 0 e
−x 2 /2α 2
(11.16)
(11.17)
(11.18)
Substitution of the second derivative into Schrödinger’s equation yields:
(11.19)
(11.20)
ψ
α
α
0
2
2
4
2
2
0
2
2
2
( )
x x
m
k
m
E
−
+
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟ +
−
⎛
⎝
Z
Z
⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
= 0
−
−
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
+
Z
2
0
2
4
2
2
2
1
2
m
x
x
k x
ψ
α
α
( )
x x
x
E
x
2
0
0 0
ψ
ψ
( )
( )
=
x
x
=
−
0
2
4
1
( )
ψ
α
α α
2
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
d
d
d
d
2
2
0
0
2
2
0
2
2
x
x
N
x
x e
N
x
ψ
α
α
( )
/
=
−
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟ =
−
e e
x
x
−
−
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
2
2
2
2
4
2
1
/ α
α
α
d
d
d
d
x
x
N
x
e
Ne
x
x
x
ψ
α
α
α
0
0
2
0
2
2
2
2
2
2
( )
/
/
=
=
−
⎛
⎝
−
−
⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟
CHAPTER 11
VIBRATIONAL MOTION
227
0
(a)
U
ϩx
Ϫx
v ϭ 1
v ϭ 2
ψν
v ϭ 3
v ϭ 0
0
(b)
ϩx
Ϫx
v ϭ 1
v ϭ 2
ψ
2
ν
v ϭ 3
v ϭ 0
0
(c)
ϩx
Ϫx
Figure 11.3 Wavefunctions and their energies of the simple harmonic oscillator.
9781405124362_4_011.qxd 4/29/08 9:11 Page 227