Dy x
ð Þ ¼ λy x
ð Þ:
ð1:64Þ
According to a general principle of SOLDE, it has two linearly independent
solutions. In the case of (1.61), we choose exponential functions for those solutions
described by
e
ikx and e
Àikx k 6 ¼ 0
ð
Þ:
This is because the above functions do not change a functional form with respect to
the differentiation and we ascribe solving a differential equation to solving an
algebraic equation among constants (or parameters). In the present case, λ and
k are such constants.
The parameter k could be a complex variable, because λ is allowed to take a
complex value as well. Linear independence of these functions is ensured from a
nonvanishing Wronskian, W. That is,
W ¼
e
ikx
e
Àikx
e
ikx
À Á 0
e
Àikx
À
Á 0
¼
e
ikx
e
Àikx
ike
ikx
Àike
Àikx
¼ Àik À ik ¼ À2ik:
ð1:65Þ
If k 6 ¼ 0, W 6 ¼ 0. Therefore, as a general solution, we get
y x
ð Þ ¼ ae
ikx
þ be
Àikx k 6 ¼ 0
ð
Þ,
ð1:66Þ
where a and b are (complex) constant. We call two linearly independent solutions
e
ikx and e
Àikx (k 6 ¼ 0) a fundamental set of solutions of a SOLDE. Inserting (1.66)
into (1.61), we have
λ À k
2
À
Á
ae
ikx
þ be
Àikx
À
Á ¼ 0:
ð1:67Þ
For (1.67) to hold with any x, we must have
λ À k
2
¼ 0, i:e:, λ ¼ k
2
:
ð1:68Þ
Using BCs (1.62), we have
ae
ikL
þ be
ÀikL
¼ 0 and ae
ÀikL
þ be
ikL
¼ 0:
ð1:69Þ
Rewriting (1.69) in a matrix form, we have
e
ikL
e
ÀikL
e
ÀikL
e
ikL
a
b
¼
0
0
:
ð1:70Þ
For a and b in (1.70) to have nonvanishing solutions, we must have
1.3 Simple Applications of Schrödinger Equation
15