d
2 x t
ð Þ
dt 2 þ ω
2 x t
ð Þ ¼ 0:
ð2:2Þ
In (2.2) we set ω positive, namely,
ω ¼
ffiffiffiffiffiffiffiffi
s=m
p
,
ð2:3Þ
where ω is called an angular frequency of the oscillator.
If we replace ω
2 with λ, we have formally the same equation as (1.61). Two
linearly independent solutions of (2.2) are the same as before (see Example 1.1); we
have e
iωt and e
Àiωt (ω 6 ¼ 0) as such. Note, however, that in Example 1.2 we were
dealing with a quantum state related to existence probability of a particle in a
potential well. In (2.2), on the other hand, we are examining a position of harmonic
oscillator undergoing a force of a spring. We are thus considering a different
situation.
As a general solution we have
x t
ð Þ ¼ ae
iωt
þ be
Àiωt ,
ð2:4Þ
where a and b are suitable constants. Let us consider BCs different from those of
Examples 1.1 or 1.2 this time. That is, we set BCs such that
x 0
ð Þ ¼ 0 and x
0 0
ð Þ ¼ v 0 v 0 > 0
ð
Þ:
ð2:5Þ
Notice that (2.5) gives initial conditions (ICs). Mathematically, ICs are included
in BCs (see Chap. 10). From (2.4) we have
x t
ð Þ ¼ a þ b ¼ 0 and x
0 0
ð Þ ¼ iω a À b
ð
Þ¼v 0 :
ð2:6Þ
Then, we get a ¼ À b ¼ v 0 /2iω. Thus, we get a simple harmonic motion as a
solution expressed as
x t
ð Þ ¼
v 0
2iω
e
iωx
À e
Àiωx
À
Á ¼
v 0
ω
sin ωt:
ð2:7Þ
From this, we have
E ¼ K þ V ¼
1
2
mv
2
0 :
ð2:8Þ
In particular, if v 0 ¼ 0, x(t) 0. This is a solution of (2.1) that has the meaning
that the particle is eternally at rest. It is physically acceptable as well. Notice also that
unlike Examples 1.1 and 1.2, the solution has been determined uniquely. This is due
to the different BCs.
32
2 Quantum-Mechanical Harmonic Oscillator