V x
ð Þ ¼
0
ÀL x L
ð
Þ ,
1 ÀL > x; x > L
ð
Þ :
&
Rewriting (1.97), we have
d
2
ψ x
ð Þ
dx 2 þ
2mE
ħ
2
ψ x
ð Þ ¼ 0
ð1:98Þ
with BCs
ψ L
ð Þ ¼ ψ ÀL
ð Þ ¼ 0:
ð1:99Þ
If we replace λ of (1.61) with
2mE
ħ
2 , we can follow the procedures of Example 1.1. That
is, we put
E ¼
ħ
2
λ
2m
ð1:100Þ
with λ ¼ k
2 in (1.68). For k we use the values of (1.81) and (1.84). Therefore, with
energy eigenvalues we get either
E ¼
ħ
2
2m
Á
2l þ 1
ð
Þ
2 π
2
4L
2
l ¼ 0, 1, 2, Á Á Á
ð
Þ ,
to which
ψ x
ð Þ ¼
ffiffiffi
1
L
r
cos kx kL ¼
π
2
þ lπ l ¼ 0, 1, 2, Á Á Á
ð
Þ
h
i
ð1:101Þ
corresponds or
E ¼
ħ
2
2m
Á
2n
ð Þ
2 π
2
4L
2
n ¼ 1, 2, 3, Á Á Á
ð
Þ ,
to which
ψ x
ð Þ ¼
ffiffiffi
1
L
r
sin kx kL ¼ nπ n ¼ 1, 2, 3, Á Á Á
ð
Þ
½
Š
ð 1:102Þ
corresponds.
Since the particle behaves as a free particle within the potential well (ÀL x L )
and p ¼ ħk, we obtain
20
1 Schrödinger Equation and Its Application