WOLFRAM|DEMONSTRATIONS PROJECT

Supersymmetry for the Square-Well Potential

​
square well
n=0
n=1
n=2
n=3
supersymmetric
partner
n=1
n=2
n=3
The most elementary problem in quantum mechanics considers a particle of mass
m
in a one-dimensional infinite square well of width
a
("particle in a box"). The Schrödinger equation can conveniently be written in the modified form
-
2
ℏ
2m

ψ
n
(x)-
2
π
2
ℏ
2m
2
a
ψ
n
(x)=
E
n
ψ
n
(x)
in
0⩽x⩽a
, such that the ground state energy is rescaled to
E
0
=0
. The eigenstates are then given by
E
n
=
2
π
2
ℏ
2m
2
a

2
(n+1)
-1],
ψ
n
(x)=
2
a
sin[(n+1)πx/a]
. The quantum number
n
is now equal to the number of nodes in the wavefunction. For simplicity, let
2
ℏ
2m
=1
and
a=π
. The Schrödinger equation then simplifies to
H
ψ
n
(x)=
E
n
ψ
n
(x)
with
H=-
2
d
d
2
x
-1
,
E
n
=n(n+2)
,
ψ
n
(x)=
2/π
sin[(n+1)x]
,
n=0,1,2,…
.
The first step is to define the superpotential
W(x)=-
ψ
0
'(x)
ψ
0
(x)
=-cotx
and two ladder operators
A=
d
dx
+W(x)
and
+
A
=-
d
dx
+W(x)
. The original Hamiltonian is then given by
H≡
H
1
=
+
A
A
. The operator obtained by reversing
A
and
+
A
,
H
2
=A
+
A
, is called the supersymmetric-partner Hamiltonian. More explicitly,
H
1
=-
2
d
d
2
x
+
V
1
(x)
and
H
2
=-
2
d
d
2
x
+
V
2
(x)
, where
V
1
(x)=
2
W(x)
-W'(x)
and
V
2
(x)=
2
W(x)
+W'(x)
. It can then be shown that if
ψ
n
(x)
is an eigenfunction of
H
1
with eigenvalue
E
n
then
A
ψ
n
(x)
is an eigenfunction of
H
2
with the same eigenvalue:
H
2
A
ψ
n
(x)=
E
n
A
ψ
n
(x)≡const
ε
n
ϕ
n
(x)
. We denote the eigenfunction of
H
2
by
ϕ
n
(x)
call its eigenvalue
ε
n
. For unbroken supersymmetry,
ε
n
=
E
n
. Note that
A
ψ
0
(x)=0
, meaning that the ground state of
H
1
has no superpartner. Correspondingly, we find
H
1
+
A
ϕ
n
(x)=const
E
n
ψ
n
(x)
. (The constants provide normalization factors.) Note that the operator
A
removes one of the nodes of the wavefunction
ψ
n
(x)
as it converts it into
ϕ
n
(x)
. Conversely,
+
A
adds a node.
In this Demonstration, you can plot any of the lowest four square-well eigenfunctions
ψ
n
(x)=
2/π
sin[(n+1)x]
,
n=0,1,2,3
on a scale with each origin at the corresponding eigenvalue
E
n
=n(n+2)
. On the right are the corresponding eigenfunctions of the supersymmetric partner Hamiltonian
H
2
, moving in the potential well
V
2
(x)=2
2
cot
x+1
(compared to
V
1
(x)=-1
). The first three normalized supersymmetric eigenstates are given by
ϕ
1
(x)=
8/3π
2
sin
x
,
ε
1
=
E
1
;
ϕ
2
(x)=
16/π
cosx
2
sin
x
,
ε
2
=
E
2
;
ϕ
3
(x)=
32/15π
[2+3cos(2x)]
2
sin
x
,
ε
3
=
E
3
.
In particle physics, supersymmetry has been proposed as a connection between bosons and fermions. Although this is a beautiful theory, there is, as yet, no experimental evidence that Nature contains supersymmetry. If it does exist, it must be a massively broken symmetry. It is possible that the Large Hadron Collider will find supersymmetric partners of some known particles.