2Some examples in one dimension
IB Quantum Mechanics
2.4 Potential well
We will consider a potential that looks like this:
x
V
−a
a
−U
The potential is given by
V (x) =
(
−U |x| < a
0 |x| ≥ a
for some constant
U >
0. Classically, this is not very interesting. If the energy
E <
0, then the particle is contained in the well. Otherwise it is free to move
around. However, in quantum mechanics, this is much more interesting.
We want to seek energy levels for a particle of mass
m
, defined by the
Schr¨odinger equation
Hψ = −
~
2
2m
ψ
00
+ V (x)ψ = Eψ.
For energies in the range
−U < E < 0,
we set
U + E =
~
2
k
2
2m
> 0, E = −
~
2
κ
2
2m
,
where
k, κ >
0 are new real constants. Note that these coefficients are not
independent, since U is given and fixed. So they must satisfy
k
2
+ κ
2
=
2mU
~
2
.
Using these constants, the Schr¨odinger equation becomes
(
ψ
00
+ k
2
ψ = 0 |x| < a
ψ
00
− κ
2
ψ = 0 |x| > a.
As we previously said, we want the Schr¨odinger equation to hold even at the
discontinuities. So we need ψ and ψ
0
to be continuous at x = ±a.
We first consider the even parity solutions
ψ
(
−x
) =
ψ
(
x
). We can write our
solution as
ψ =
(
A cos kx |x| < a
Be
−κ|x|
|x| > a
We match ψ and ψ
0
at x = a. So we need
A cos ka = Be
−κa
−Ak sin ka = −κBe
−κa
.
By parity, there is no additional information from x = −a.
We can divide the equations to obtain
k tan ka = κ.
this is still not something we can solve easily. To find when solutions exist, it is
convenient to introduce
ξ = ak, η = aκ,
where these two constants are dimensionless and positive. Note that this
η
has
nothing to do with parity. It’s just that we have run out of letters to use. Hence
the solution we need are solutions to
η = ξ tan ξ.
Also, our initial conditions on k and κ require
ξ
2
+ η
2
=
2ma
2
U
~
2
.
We can look for solutions by plotting these two equations. We first plot the
curve η = ξ tan ξ:
ξ
η
1π
2
3π
2
5π
2
7π
2
9π
2
11π
2
The other equation is the equation of a circle. Depending on the size of the
constant 2ma
2
U/~
2
, there will be a different number of points of intersections.
ξ
η
So there will be a different number of solutions depending on the value of
2ma
2
U/~
2
. In particular, if
(n − 1)π <
2mUa
2
~
2
1/2
< nπ,
then we have exactly n even parity solutions (for n ≥ 1).
We can do exactly the same thing for odd parity eigenstates. . . on example
sheet 1.
For
E >
0 or
E < −U
, we will end up finding non-normalizable solutions.
What is more interesting, though is to look at the solutions we have now. We
can compare what we’ve got with what we would expect classically.
Classically, any value of
E
in the range
−U < E <
0 is allowed, and the
motion is deeply uninteresting. The particle just goes back and forth inside the
well, and is strictly confined in −a ≤ x ≤ a.
Quantum mechanically, there is just a discrete, finite set of allowed energies.
What is more surprising is that while
ψ
decays exponentially outside the well, it
is non-zero! This means there is in theory a non-zero probability of finding the
particle outside the well! We call these particles bound in the potential, but in
fact there is a non-zero probability of finding the particle outside the well.