Strictly speaking, we should be careful to assure that (1.81) holds on the basis of
(1.80). It is because we have yet the possibility that k is a complex number. To see it,
we examine zeros of a cosine function that is defined in a complex domain. Here the
zeros are (complex) numbers to which the function takes zero. That is, if f(z 0 ) ¼ 0, z 0
is called a zero (i.e., one of zeros) of f(z). Now we have
cos z
1
2
e
iz
þ e
Àiz
À
Á
; z ¼ x þ iy x, y : real
ð
Þ :
ð1:93Þ
Inserting z ¼ x + iy in cosz and rearranging terms, we get
cos z ¼
1
2
cos x e
y
þ e
Ày
ð
Þþi sin x e
Ày
À e
y
ð
Þ
½
:
ð1:94Þ
For cosz to vanish, both its real and imaginary parts must be zero. Since e
y + e
Ày
> 0
for all real numbers y, we must have cosx ¼ 0 for the real part to vanish; i.e.,
x ¼
π
2
þ mπ m ¼ 0, Æ1, Æ2, Á Á Á
ð
Þ :
ð1:95Þ
Note in this case that sinx ¼ Æ 1 (6 ¼0). Therefore, for the imaginary part to vanish,
e
Ày
À e
y
¼ 0. That is, we must have y ¼ 0. Consequently, the zeros of cosz are real
numbers. In other words, with respect to z 0 that satisfies cosz 0 ¼ 0 we have
z 0 ¼
π
2
þ mπ m ¼ 0, Æ1, Æ2, Á Á Á
ð
Þ :
ð1:96Þ
The above discussion equally applies to a sine function as well.
Thus, we ensure that k is a nonzero real number. Eigenvalues λ are positive
definite from (1.68) accordingly. This conclusion is not fortuitous but a direct
consequence of the form of a differential equation we have dealt with in combination
with the BCs we imposed, i.e., the Dirichlet conditions. Detailed discussion will
follow in Sects. 1.4, 8.3, and 8.4 in relation to the Hermiticity of a differential
operator.
Example 1.2: A Particle Confined Within a Potential Well The results obtained
in Example 1.1 can immediately be applied to dealing with a particle (electron) in a
one-dimensional infinite potential well. In this case, (1.55) reads as
ħ
2
2m
d
2
ψ x
ð Þ
dx 2 þ Eψ x
ð Þ ¼ 0,
ð1:97Þ
where m is a mass of a particle and E is an energy of the particle. A potential V is
expressed as
1.3 Simple Applications of Schrödinger Equation
19
(1.80). It is because we have yet the possibility that k is a complex number. To see it,
we examine zeros of a cosine function that is defined in a complex domain. Here the
zeros are (complex) numbers to which the function takes zero. That is, if f(z 0 ) ¼ 0, z 0
is called a zero (i.e., one of zeros) of f(z). Now we have
cos z
1
2
e
iz
þ e
Àiz
À
Á
; z ¼ x þ iy x, y : real
ð
Þ :
ð1:93Þ
Inserting z ¼ x + iy in cosz and rearranging terms, we get
cos z ¼
1
2
cos x e
y
þ e
Ày
ð
Þþi sin x e
Ày
À e
y
ð
Þ
½
:
ð1:94Þ
For cosz to vanish, both its real and imaginary parts must be zero. Since e
y + e
Ày
> 0
for all real numbers y, we must have cosx ¼ 0 for the real part to vanish; i.e.,
x ¼
π
2
þ mπ m ¼ 0, Æ1, Æ2, Á Á Á
ð
Þ :
ð1:95Þ
Note in this case that sinx ¼ Æ 1 (6 ¼0). Therefore, for the imaginary part to vanish,
e
Ày
À e
y
¼ 0. That is, we must have y ¼ 0. Consequently, the zeros of cosz are real
numbers. In other words, with respect to z 0 that satisfies cosz 0 ¼ 0 we have
z 0 ¼
π
2
þ mπ m ¼ 0, Æ1, Æ2, Á Á Á
ð
Þ :
ð1:96Þ
The above discussion equally applies to a sine function as well.
Thus, we ensure that k is a nonzero real number. Eigenvalues λ are positive
definite from (1.68) accordingly. This conclusion is not fortuitous but a direct
consequence of the form of a differential equation we have dealt with in combination
with the BCs we imposed, i.e., the Dirichlet conditions. Detailed discussion will
follow in Sects. 1.4, 8.3, and 8.4 in relation to the Hermiticity of a differential
operator.
Example 1.2: A Particle Confined Within a Potential Well The results obtained
in Example 1.1 can immediately be applied to dealing with a particle (electron) in a
one-dimensional infinite potential well. In this case, (1.55) reads as
ħ
2
2m
d
2
ψ x
ð Þ
dx 2 þ Eψ x
ð Þ ¼ 0,
ð1:97Þ
where m is a mass of a particle and E is an energy of the particle. A potential V is
expressed as
1.3 Simple Applications of Schrödinger Equation
19