2Some examples in one dimension
IB Quantum Mechanics
2.1 Introduction
In general, we are going to consider the energy eigenvalue problem for a particle
in 1 dimension in a potential V (x), i.e.
Hψ = −
~
2
2m
ψ
00
+ V (x)ψ = Eψ.
In other words, we want to find the allowed energy eigenvalues.
This is a hard problem in general. However, we can consider the really easy
case where
V
(
X
) =
U
, where
U
is a constant. Then we can easily write down
solutions.
If U > E, then the Schr¨odinger equation is equivalent to
ψ
00
− κ
2
ψ = 0,
where κ is such that U − E =
~
2
κ
2
2m
. We take wlog κ > 0. The solution is then
ψ = Ae
κx
+ Be
−κx
.
On the other hand, if U < E, then the Schr¨odinger equation says
ψ + k
2
ψ = 0,
where k is picked such that E − U =
~
2
k
2
2m
. The solutions are
ψ = Ae
ikx
+ Be
ikx
.
Note that these new constants are merely there to simplify our equations. They
generally need not have physical meanings.
Now why are we interested in cases where the potential is constant? Wouldn’t
that be just equivalent to a free particle? This is indeed true. However, knowing
these solutions allow us to to study piecewise flat potentials such as steps, wells
and barriers.
x
V
x
V
x
V
Here a finite discontinuity in
V
is allowed. In this case, we can have
ψ, ψ
0
continuous and
ψ
00
discontinuous. Then the discontinuity of
ψ
00
cancels that of
V , and the Schr¨odinger equation holds everywhere.
In this chapter, we will seek normalizable solutions with
Z
∞
−∞
|ψ(x)|
2
dx.
This requires that
ψ
(
x
)
→
0 as
x → ±∞
. We see that for the segments and
the end, we want to have decaying exponentials
e
−κx
instead of oscillating
exponentials e
−ikx
.