Φ x 0
ð Þ ¼ iħ
∂ξ t
ð Þ
∂t
=ξ t
ð Þ:
ð1:54Þ
If RHS of (1.54) varied depending on t, Φ(x 0 ) would be allowed to have various
values, but this must not be the case with our present investigation. Thus, RHS of
(1.54) should take a constant value E. For the same reason, LHS of (1.51) should
take a constant.
Thus, (1.48) or (1.51) should be separated into the following equations:
Hϕ x
ð Þ ¼ Eϕ x
ð Þ,
ð1:55Þ
iħ
∂ξ t
ð Þ
∂t
¼ Eξ t
ð Þ:
ð1:56Þ
Equation (1.56) can readily be solved. Since (1.56) depends on a sole variable t, we
have
dξ t
ð Þ
ξ t
ð Þ
¼
E
iħ
dt or d ln ξ t
ð Þ ¼
E
iħ
dt:
ð1:57Þ
Integrating (1.57) from zero to t, we get
ln
ξ t
ð Þ
ξ 0
ð Þ
¼
Et
iħ
:
ð1:58Þ
That is,
ξ t
ð Þ ¼ ξ 0
ð Þ exp ÀiEt=ħ
ð
Þ:
ð1:59Þ
Comparing (1.59) with (1.38), we find that the constant E in (1.55) and (1.56)
represents an energy of a particle (electron).
Thus, the next task we want to do is to solve an eigenvalue equation of (1.55).
After solving the problem, we get a solution
ψ x, t
ð Þ ¼ ϕ x
ð Þ exp ÀiEt=ħ
ð
Þ,
ð1:60Þ
where the constant ξ(0) has been absorbed in ϕ(x). Normally, ϕ(x) is to be normalized after determining the functional form (vide infra).
1.2 Schrödinger Equation
13