2Some examples in one dimension
IB Quantum Mechanics
2.2 Infinite well — particle in a box
The simplest case to consider is the infinite well. Here the potential is infinite
outside the region [
−a, a
], and we have much less to think about. For
|x| > a
,
we must have ψ(x) = 0, or else V (x)ψ(x) would be infinite.
x
V
−a
a
V (x) =
(
0 |x| ≤ a
∞ |x| > a.
We require
ψ
= 0 for
|x| > a
and
ψ
continuous at
x
=
±a
. Within
|x| < a
, the
Schr¨odinger equation is
−
~
2
2m
ψ
00
= Eψ.
We simplify this to become
ψ
00
+ k
2
ψ = 0,
where
E =
~
2
k
2
2m
.
Here, instead of working with the complex exponentials, we use
sin
and
cos
since
we know well when these vanish. The general solution is thus
ψ = A cos kx + B sin kx.
Our boundary conditions require that ψ vanishes at x = ±a. So we need
A cos ka ± B sin ka = 0.
In other words, we require
A cos ka = B sin ka = 0.
Since
sin ka
and
cos ka
cannot be simultaneously 0, either
A
= 0 or
B
= 0. So
the two possibilities are
(i) B = 0 and ka = nπ/2 with n = 1, 3, ···
(ii) A = 0 and ka = nπ/2 with n = 2, 4, ···
Hence the allowed energy levels are
E
n
=
~
2
π
2
8ma
2
n
2
,
where n = 1, 2, ···, and the wavefunctions are
ψ
n
(x) =
1
a
1
2
(
cos
nπx
2a
n odd
sin
nπx
2a
n even
.
These are normalized with
R
a
−a
|ψ
n
(x)|
2
dx.
ψ
1
:
x
V
−a −a
ψ
2
:
x
V
−a −a
ψ
3
:
x
V
−a −a
ψ
4
:
x
V
−a −a
This was a rather simple and nice example. We have an infinite well, and the
particle is well-contained inside the box. The solutions just look like standing
waves on a string with two fixed end points — something we (hopefully) are
familiar with.
Note that
ψ
n
(
−x
) = (
−
1)
n+1
ψ
n
(
x
). We will see that this is a general feature
of energy eigenfunctions of a symmetric potential. This is known as parity.