1
X
d
2 X
dx 2 ¼
1
v 2
1
T
d
2 T
dt
2
¼ Àk
2 ,
ð9:14Þ
where k is an undetermined (possibly complex) constant. For the x component, we
get
d
2 X
dx 2 þ k
2 X ¼ 0:
ð9:15Þ
Remember that k is supposed to be a complex number for the moment (see Example 1.1).
Modifying Example 1.1 a little bit such that (9.15) is posed in a domain [0, L] and
imposing the Dirichlet conditions such that
X 0
ð Þ ¼ X L
ð Þ ¼ 0,
ð9:16Þ
we find a solution of
X x
ð Þ ¼ a sin kx,
ð9:17Þ
where a is a constant. The constant k can be determined to satisfy the BCs; i.e.,
kL ¼ mπ or k ¼ mπ=L m ¼ 1, 2, Á Á Á
ð
Þ :
ð9:18Þ
Thus, we get real numbers for k. Then, we have a solution
T t
ð Þ ¼ b sin kvt ¼ b sin ωt:
ð9:19Þ
The overall solution is then
ψ x, t
ð Þ ¼ c sin kx sin ωt:
ð9:20Þ
This solution has already appeared in Chap. 8 as a stationary solution. The readers
are encouraged to derive these results.
In a three-dimensional case, we have a wave equation
∂
2 ψ
∂x
2
þ
∂
2 ψ
∂y
2
þ
∂
2 ψ
∂z
2
¼
1
v 2
∂
2 ψ
∂t
2
:
ð9:21Þ
In this case, we also assume
ψ x, t
ð Þ ¼ X x
ð ÞY y
ð ÞZ z
ð ÞT t
ð Þ:
ð9:22Þ
Similarly we get
342
9 Light Quanta: Radiation and Absorption