1. 1
An introduction to Supersymmetric Quantum
Mechanics
By Thomas Acton
2014
3rd
Year Project
King’s College London

2. 2
Abstract
This report will focus on the aim of teaching an undergraduate student, with only a basic
understanding in quantum mechanics (QM), the core concepts of Supersymmetric Quantum
Mechanics (SUSY QM). To do this standard QM will be expanded on and the factorization methods
explained. After this the structure of the Hamiltonian hierarchy will be explained and great emphasis
will be laid on the relation between the energy eigenstates. Later the shape invariance condition will
be explained and how it is a powerful tool, which allows the user to algebraically work out all the
energy eigenstates and wavefunctions. To further demonstrate how shape invariance works, an
example using the Morse potential will be worked though.

3. 3
Introduction
One of the yet unattained goals of Physics is to obtain a unified theory of all the basic forces of
nature: The strong, electroweak, and gravitational interactions. Supersymmetry (SUSY) is considered
the best framework we currently have at achieving that unification.
A very basic form of supersymmetry was first proposed in 1966 by Hironari Miyazawa. His theory
related mesons to baryons but was mostly ignored at the time for not including spacetime [2]
. 5 years
later - in 1971, two groups independently “rediscovered” supersymmetry: J. L. Gervais and B. Sakita
[3]
followed by Yu. A. Golfand and E. P. Likhtman. Then the year after another pair also contributed
to the field: D.V. Volkov and V.P. Akulov [4]
.
Unlike Miyazawa, who was looking at SUSY in terms of hadronic physics [2]
, the other 3
groups were interested in it for its applications in QFT (Quantum Field Theory) [5]
, which is a new
way of approaching spacetime and fundamental fields. QFT establishes a relationship between
elementary particles of different quantum nature (bosons and fermions) by treating them as excitations
of an underlying field [6]
.
At face, SUSY relates the different elementary particles: bosons and fermions [1]
. To achieve this,
each particle is linked with a particle from the other group, referred to as its superpartner, whose spin
differs by a half-integer (the superpartner of a boson is a fermion and visa-versa). In a theory with
perfectly unbroken supersymmetry, each pair of superpartners shares the same mass and internal
quantum numbers besides spin, which is something we should be able to look for experimentally. It
was expected that supersymmetric particles would be found at CERN in the LHC, but so far there has
been no evidence of them.
Supersymmetry differs from currently known symmetries in physics as it manages to establish a
symmetry between classical and quantum physics, as a result it has led to new discoveries and created
new fields of study, such as Supersymmetric Quantum Mechanics (SUSY QM) which will be the
focus of this report. SUSY QM adds the SUSY superalgebra on top of quantum mechanics. Once the
algebraic structure is understood, the results fall out and you never need return to the origin of the
Fermi-Bose symmetry [7]
.

4. 4
SUSY QM Basics
In normal quantum mechanics the Hamiltonian is normally expressed in the form of equation (1.1) [8]
:
𝐻 = −
ℏ2
2𝑚
𝑑2
𝑑𝑥2
+ 𝑉(𝑥)
SUSY QM differs as it involves pairs of partner Hamiltonians which are closely related to each other.
By starting off with the requirement that the ground state wave function satisfy 𝐻𝜓0(𝑥) = 0 [9]
we
arrive at the following conclusion in the form of equation (1.3) which will be useful later.
𝐻1 𝜓0 = −
ℏ2
2𝑚
𝑑2
𝜓0
𝑑𝑥2
+ 𝑉1(𝑥)𝜓0(𝑥) = 0
∴ 𝑉1(𝑥) =
ℏ2
2𝑚
𝜓0
′′
(𝑥)
𝜓0(𝑥)
As I have shown, with little to no knowledge about the system, and using conventional quantum
mechanics we are able to learn something about the potential 𝑉1. The next big step towards obtaining
a SUSY QM theory is to factorize the Hamiltonian in a method introduced by Schrödinger [11]
, then
later generalised by Infeld and Hull [12]
.
𝐻1 = 𝐴†
𝐴
Where 𝐴†
and 𝐴 are now defined as the operators:
𝐴†
= −
ℏ
√2𝑚
𝑑
𝑑𝑥
+ 𝑊(𝑥)
𝐴 =
ℏ
√2𝑚
𝑑
𝑑𝑥
+ 𝑊(𝑥)
Here 𝑊(𝑥) refers to the superpotential. Via combinations of equation (1.4), (1.5) and (1.6) [10]
and
then finally equation (1.1) the first potential, 𝑉1 can be written in terms of the superpotential, 𝑊(𝑥)
(equation (1.7)).
𝐻1 𝜓(𝑥) = (−
ℏ
√2𝑚
𝑑
𝑑𝑥
+ 𝑊(𝑥)) (
ℏ
√2𝑚
𝑑
𝑑𝑥
+ 𝑊(𝑥)) 𝜓(𝑥)
= −
ℏ2
2𝑚
𝑑2
𝜓(𝑥)
𝑑𝑥2
−
ℏ
√2𝑚
[𝑊′(𝑥)𝜓(𝑥) + 𝑊(𝑥)𝜓′(𝑥)] + 𝑊(𝑥)
ℏ
√2𝑚
𝜓′(𝑥) + 𝑊2(𝑥)𝜓(𝑥)
= (−
ℏ2
2𝑚
𝑑2
𝑑𝑥2
−
ℏ
√2𝑚
𝑊′(𝑥) + 𝑊2(𝑥)) 𝜓(𝑥)
∴ 𝑉1(𝑥) = 𝑊2(𝑥) −
ℏ
√2𝑚
𝑊′
(𝑥)
It can be seen that equation (1.7) satisfies the Riccati equation and hence using his methods can be
solved to show:
(1.1)
(1.2)
(1.3)
(1.4)
(1.6)
(1.5)
(1.7)

5. 5
𝑊(𝑥) = −
ℏ
√2𝑚
𝜓0
′
(𝑥)
𝜓0(𝑥)
The next step in constructing the SUSY QM theory is to obtain the partner Hamiltonians. 𝐻2 is
obtained by defining it using the operators 𝐴 and 𝐴†
, but switching the order in which they appear.
𝐻2 = 𝐴𝐴†
Using the same method as we did to obtain the result for 𝑉1, can write the partner Hamiltonian in the
form as equation (1.1), and though the combination of equations (1.5), (1.6), (1.9) and (1.10) the
potential, 𝑉2, referred to as the supersymmetric partner can be worked out in terms of the
superpotential, 𝑊(𝑥).
𝐻2 = −
ℏ2
2𝑚
𝑑2
𝑑𝑥2
+ 𝑉2(𝑥)
𝑉2(𝑥) = 𝑊2(𝑥) +
ℏ
√2𝑚
𝑊′
(𝑥)
From what I have shown so far, it should be obvious, there is a deep relationship between partner
Hamiltonians and we will now investigate the link between the energy eigenvalues and introduce the
idea that the operators 𝐴†
and 𝐴 can be thought of as analogues to the creation and annihilation
operators of the standard quantum mechanical harmonic oscillator. So far, it should be clear that the
following is true:
𝐻1 𝜓n
(1)
= 𝐸 𝑛
(1)
𝜓n
(1)
= 𝐴†
𝐴𝜓n
(1)
Then by multiplying by 𝐴, remembering the order in which we apply the operators is important, we
are able to make the crucial link between the two Hamiltonians.
𝐻2 𝐴𝜓n
(1)
= 𝐸 𝑛
(1)
(𝐴𝜓n
(1)
) = 𝐴𝐴†
𝐴𝜓n
(1)
𝐻2 𝜓n
(2)
= 𝐸 𝑛
(2)
𝜓n
(2)
= 𝐴𝐴†
𝜓n
(2)
𝐻1(𝐴†
𝜓n
(2)
) = 𝐸 𝑛
(2)
(𝐴†
𝜓n
(2)
) = 𝐴†
𝐴𝐴†
𝜓n
(2)
𝜓 𝑛
(𝑚)
refers to the nth
excited state of the mth
waveform. From equations (1.12) and (1.13) it is clear
that 𝐴𝜓n
(1)
(𝑥) is an eigenfunction of 𝐻2 and so can be expressed in the form of equation (1.16). Next,
by assuming 𝜓n
(1)
(𝑥) is normalised and using equation (1.12) we are able to work out the constant of
normalisation, 𝑐 and therefore 𝜓 𝑚
(2)
.
𝜓 𝑚
(2)
(𝑥) = 𝑐𝐴𝜓 𝑛
(1)
(𝑥)
∫ 𝜓 𝑚
(2)
𝜓 𝑚
∗(2)
= |𝑐|2
∫ 𝜓 𝑛
∗(1)
𝐴†
𝐴𝜓 𝑛
(1)
= |𝑐|2
𝐸 𝑛
(1)
∫ 𝜓 𝑛
(1)
𝜓 𝑛
∗(1)
= 1
∴ 𝑐 =
1
√ 𝐸 𝑛
(1)
(1.8)
(1.9)
(1.10)
(1.11)
(1.12)
(1.13)
(1.14)
(1.15)
(1.16)
(1.17)
(1.18)

6. 6
∴ 𝜓 𝑚
(2)
= 𝐴
𝜓n
(1)
√ 𝐸 𝑛
(1)
If we recall that 𝐴𝜓0
(1)
= 0, and that we set 𝐸0
(1)
= 0 as one of the conditions long ago we can clearly
see that 𝐻2 must have no zero energy ground state. In fact, we can now determine that 𝑚 = 𝑛 − 1 and
show the relation between the spectra of the partner Hamiltonians:
𝐸 𝑛
(2)
= 𝐸 𝑛+1
(1)
𝜓 𝑛
(2)
= 𝐴
𝜓n+1
(1)
√ 𝐸 𝑛+1
(1)
𝜓 𝑛+1
(1)
= 𝐴†
𝜓n
(2)
√ 𝐸 𝑛
(2)
After this derivation it should be a lot clearer how 𝐴†
and 𝐴 convert between 𝐻1 and 𝐻2. This is the
underlying concept of SUSY. The partner Hamiltonians change between bosons and fermions, we can
see that every eigenstate of 𝐻1 has a corresponding eigenstate with the same energy in 𝐻2 (apart from
the ground zero energy state). Thus we have described a bosonic Hamiltonian, whose SUSY partner is
fermionic with the same energy (mass).
Using this approach it should now also be obvious that by knowing all the eigenstates for 𝐻1 we
already know all the eigenstates for 𝐻2 or visa-versa (except we would not know the solution for the
ground state if we started with 𝐻2). Further yet, we can show visually; during the conversion between
Hamiltonians the operators act in a way which creates or destroy nodes within the eigenfunction. It
can be seen how the ground states of each potential each only have 1 node, yet different energy
eigenvalues. This method can be useful when imagine the upper limit on how many Hamiltonians can
be created from the original Hamiltonian (with limited allowed energy eigenvalues), since each time a
new partner is made a node is destroyed and all ground states must possess at least one node to exist.
(1.19)
Figure 1.1: The potential 𝑉(𝑥) = 0 (left) and its supersymmetric partner potential 𝑉(𝑥) =
2𝑐𝑜𝑠𝑒𝑐2
(𝑥) (right). The first 3 energy eigenvalues are shown on the left potential, while the
partner potential only shows the first 2.
(1.20)
(1.21)
(1.22)

7. 7
Hamiltonian Hierarchy
The next step is to simplify and generalise the previous methods to produce a processes for creating
new SUSY partner Hamiltonians. For example; we can reconstruct 𝐻2 by redefining a new set of
operators, then using the same principle as equations (1.4) and (1.9) we redefine 𝐻2 and are able to
construct a new partner Hamiltonian 𝐻3 off of that.
𝐻2 = 𝐴†
2 𝐴2
𝐻3 = 𝐴2 𝐴†
2
This process can be applied recursively, each time containing one less bound state than the previous
Hamiltonian. The process can be repeated as many times as 𝐻1 has bound states. Using this logic a
hierarchy of Hamiltonians can be constructed which all possess the same structure. To completely
understand this process we will begin by revisiting and expanding on the previous work. For
simplicity, and for the rest of the report we will now define:
ℏ = 2𝑚 = 1
Recalling before that we set 𝐻𝜓0(𝑥) = 0, we now opt to simply shift the Hamiltonians to obtain the
zero energy ground state for each Hamiltonian within the equation, which leaves us with 𝐻1(𝑥) and
𝑉1(𝑥) in the form of:
𝐻1(𝑥) = 𝐴1
†
𝐴1 + 𝐸0
(1)
= −
𝑑2
𝑑𝑥2
+ 𝑉1(𝑥)
𝑉1(𝑥) = 𝑊1
2(𝑥) − 𝑊1
′(𝑥) + 𝐸0
(1)
Where 𝐸0
(1)
is the ground state energy. Next the operators and superpotential are simplified to:
𝐴1 =
𝑑
𝑑𝑥
+ 𝑊1(𝑥)
𝐴1
†
= −
𝑑
𝑑𝑥
+ 𝑊1(𝑥)
𝑊1(𝑥) =
𝜓0
(1)′
(𝑥)
𝜓0
(1)
(𝑥)
= −
𝑑
𝑑𝑥
𝑙𝑛𝜓0
(1)
(𝑥)
We keep applying these new rules to 𝐻2 to retrieve equation (2.8) and by combining equation (2.7)
with equation (2.9) reach its final form:
𝐻2 = 𝐴1 𝐴1
†
+ 𝐸0
(1)
= −
𝑑2
𝑑𝑥2
+ 𝑉2(𝑥)
𝑉2(𝑥) = 𝑊1
2(𝑥) + 𝑊1
′(𝑥) + 𝐸0
(1)
= 𝑉1(𝑥) + 2𝑊1
′(𝑥) = 𝑉1(𝑥) − 2
𝑑2
𝑑𝑥2
𝑙𝑛𝜓0
(1)
(𝑥)
This method produces the expected results for the energy eigenvalues and wave function:
𝐸 𝑛
2
= 𝐸 𝑛+1
(1)
(2.1)
(2.2)
(2.3)
(2.4)
(2.5)
(2.5)
(2.6)
(2.7)
(2.8)
(2.9)
(2.10)

8. 8
𝜓 𝑛
(2)
=
𝐴1 𝜓 𝑛+1
(1)
√(𝐸 𝑛+1
(1)
− 𝐸0
(1)
)
Through the redefinition of the operators 𝐴2
†
and 𝐴2 (pay close attention to the subscript) and the
order of their application, we are able to provide further insight into the structure of the hierarchy of
Hamiltonians, starting with equation (2.15). It is important to remember that: 𝐸0
(2)
= 𝐸1
(1)
, and that we
are shifting the spectrum of the Hamiltonian to account for the ground state zero energy of 𝐻2.
𝐴2
†
= −
𝑑
𝑑𝑥
+ 𝑊2(𝑥)
𝐴2 =
𝑑
𝑑𝑥
+ 𝑊2(𝑥)
𝑊2(𝑥) = −
𝑑
𝑑𝑥
ln(𝜓0
(2)
)
𝐻2 = 𝐴1 𝐴†
1 + 𝐸0
(1)
= 𝐴†
2 𝐴2 + 𝐸0
(2)
= 𝐴†
2 𝐴2 + 𝐸1
(1)
This allows us to continue our investigation in the hierarchy of Hamiltonians by applying what we
know to be true about 𝐻3 from equation (2.2).
𝐻3 = 𝐴2 𝐴†
2 + 𝐸1
(1)
= −
𝑑2
𝑑𝑥2
+ 𝑉3(𝑥)
We are able to work out 𝑉3(𝑥) by combining the usual form of the potential, with equations (2.9) and
(2.15). This can then be simplified using the composition laws of logarithms.
𝑉3(𝑥) = 𝑊2
2
(𝑥) + 𝑊2
′
(𝑥) + 𝐸1
(1)
= 𝑉2 + 2𝑊2
′
= 𝑉2(𝑥) − 2
𝑑2
𝑑𝑥2
𝑙𝑛𝜓0
(2)
𝑉3(𝑥) = 𝑉1(𝑥) − 2
𝑑2
𝑑𝑥2
𝑙𝑛𝜓0
(1)
− 2
𝑑2
𝑑𝑥2
𝑙𝑛𝜓0
(2)
= 𝑉1(𝑥) − 2
𝑑2
𝑑𝑥2
𝑙𝑛 (𝜓0
(1)
𝜓0
(2)
)
This procedure is consistent with the previous results, confirming the relation between the energy
eigenvalues:
𝐸 𝑛
(3)
= 𝐸 𝑛+1
(2)
= 𝐸 𝑛+2
(1)
𝜓 𝑛
(3)
=
𝐴2 𝜓 𝑛+1
(2)
√(𝐸 𝑛+1
(2)
− 𝐸0
(2)
)
=
𝐴2 𝐴1 𝜓 𝑛+2
(1)
√(𝐸 𝑛+2
(1)
− 𝐸1
(1)
)√(𝐸 𝑛+2
(1)
− 𝐸0
(1)
)
From equations (2.19) and (2.20) we can see how the solution of the third partner Hamiltonian can be
expressed in terms of the solutions to the previous Hamiltonians. This form of proof by induction
shows that it is indeed now possible for all Hamiltonians in the chain to be expressed this way. This
process can be written in a general way, for any (𝑛 = 𝑘) < 𝑚, where m is the number of bound states
in 𝐻1.
𝐻 𝑘 = 𝐴†
𝑘 𝐴 𝑘 + 𝐸 𝑘−1
(1)
= −
𝑑2
𝑑𝑥2
+ 𝑉𝑘(𝑥)
(2.11)
(2.12)
(2.13)
(2.14)
(2.15)
(2.16)
(2.17)
(2.18)
(2.19)
(2.20)
(2.21)

9. 9
𝐴 𝑘 =
𝑑
𝑑𝑥
+ 𝑊𝑘(𝑥)
𝐴 𝑘
†
= −
𝑑
𝑑𝑥
+ 𝑊𝑘(𝑥)
𝑊𝑘(𝑥) = −
𝑑
𝑑𝑥
𝑙𝑛𝜓0
(𝑘)
(𝑥)
𝑉𝑘(𝑥) = 𝑉1(𝑥) − 2
𝑑2
𝑑𝑥2
𝑙𝑛 (𝜓0
(1)
… 𝜓0
(𝑘−1)
)
𝐸 𝑛
(𝑘)
= 𝐸 𝑛+1
(𝑘−1)
= 𝐸 𝑛+𝑘−1
(1)
𝜓 𝑛
(𝑘)
=
𝐴 𝑘−1 … 𝐴1 𝜓 𝑛+𝑘−1
(1)
√(𝐸 𝑛+𝑘−1
(1)
− 𝐸 𝑘−2
(1)
) … √(𝐸 𝑛+𝑘−1
(1)
− 𝐸0
(1)
)
This is a an extremely powerful tool when analysing SUSY QM problems, as it enables us to know
the all solutions for many Hamiltonians algebraically by just knowing that they are related via
supersymmetry to a Hamiltonian whose solutions are already known or can be easily worked out.
(2.22)
(2.23)
(2.24)
(2.25)
(2.26)
(2.27)

10. 10
Shape Invariance
Next we will introduce the Shape invariance condition, a condition introduced by Gendensthein [13]
,
which helped to categorize fully analytically solvable potentials whose energy eigenvalues and wave
functions are explicitly known. The condition states that if the SUSY partner potentials differ only by
parameters; which is to say that if the second potential is a function with parameters 𝑎1, and the first
potential can be written with parameters 𝑎2 plus a remainder as a function of 𝑎1, then they are
categorised to be shape invariant potentials [10]
. This condition is stated as:
𝑉2(𝑥; 𝑎1) = 𝑉1(𝑥; 𝑎2) + 𝑅(𝑎1)
Here 𝑎1, 𝑎2 are sets of parameters, which are related though a function 𝑎2 = 𝑓(𝑎1). Most importantly
for the shape invariance condition to be satisfied the remainder must be independent of variable 𝑥.
We will now show that between our ability to create a hierarchy of Hamiltonians, which all possess
the same structure, and the shape invariance condition we are able to solve every well known exactly
solvable potential. Again we will be using: ℏ = 2𝑚 = 1 for simplicity. We start off by constructing
our chain of Hamiltonians by applying the shape invariance condition. Equation (3.2) is trivial, and
(3.3) is produced directly through the application of equation (3.1).
𝐻1 = −
𝑑2
𝑑𝑥2
+ 𝑉1(𝑥; 𝑎1)
𝐻2 = −
𝑑2
𝑑𝑥2
+ 𝑉2(𝑥; 𝑎1) = −
𝑑2
𝑑𝑥2
+ 𝑉1(𝑥; 𝑎2) + 𝑅(𝑎1)
𝐻3 = −
𝑑2
𝑑𝑥2
+ 𝑉2(𝑥; 𝑎2) + 𝑅(𝑎1) = −
𝑑2
𝑑𝑥2
+ 𝑉1(𝑥; 𝑎3) + 𝑅(𝑎2) + 𝑅(𝑎1)
To produce the 3rd
Hamiltonian we have subtracted the remainder 𝑅(𝑎1) from 𝐻2, refactorised the
Hamiltonian in the form of 𝐻3 and then reapplied the original remainder 𝑅(𝑎1). It is essential that
𝑅(𝑎1) not be left out or the ground energy state would not be zero and normalisation would not be
possible. The parameter 𝑎3 is related though the relation: 𝑎3 = 𝑓(𝑎2) = 𝑓𝑓(𝑎1). At this point it
should be clear this process is infinitely repeatable and we can generalise in the form of equation
(3.5):
𝐻 𝑘 = −
𝑑2
𝑑𝑥2
+ 𝑉1(𝑥; 𝑎 𝑘) + ∑ 𝑅(
𝑘−1
𝑖=1
𝑎𝑖)
Next we introduce the wave function 𝜓0
(1)
(𝑥; 𝑎 𝑘) to the general form of the kth
Hamiltonian to see
what we can deduce about the energy levels.
𝐻 𝑘 𝜓0
(1)
(𝑥; 𝑎 𝑘) = −
𝑑2
𝑑𝑥2
+ 𝑉1(𝑥; 𝑎 𝑘)𝜓0
(1)
(𝑥; 𝑎 𝑘) + ∑ 𝑅(
𝑘−1
𝑖=1
𝑎𝑖)𝜓0
(1)
(𝑥; 𝑎 𝑘)
−
𝑑2
𝑑𝑥2
+ 𝑉1(𝑥; 𝑎 𝑘)𝜓0
(1)
(𝑥; 𝑎 𝑘) = 𝐻1(𝑥; 𝑎 𝑘)𝜓0
(1)
(𝑥; 𝑎 𝑘) = 0
Unsurprisingly (I am almost certain by construction), the 1st
Hamiltonian is contained within equation
(3.6) (highlighted red), which we set to always equal zero earlier in the report. From this we are able
to conclude that the remainders are closely linked with the energy levels.
(3.5)
(3.6)
(3.7)
(3.1)
(3.2)
(3.3)
(3.4)

11. 11
𝐸0
(𝑘)
= ∑ 𝑅(
𝑘−1
𝑖=1
𝑎𝑖)
This is a useful tool as it allows us to know the ground energy levels of all Hamiltonians in the
hierarchy simply by comparing shapes with partner potentials. Recalling equation (2.26), 𝐸 𝑛
(𝑘)
=
𝐸 𝑛+1
(𝑘−1)
= 𝐸 𝑛+𝑘−1
(1)
, which shows the rule for converting between energy levels, we can deduce that the
entire energy spectrum of 𝐻1 is now known – just by being able to calculate the ground states of the
other Hamiltonians in the chain:
𝐸 𝑛
(1)
= ∑ 𝑅(𝑎𝑖)
𝑛
𝑖=1
Without forgetting that 𝐸0
(1)
= 0, and in cases where this is not true we simply need to remember to
shift all the energy levels by the appropriate value. At this point it should be clear that all we need is a
reliable way to calculate the first excited state, or the 2nd
Ground state to calculate the entire energy
spectrum. To achieve this we recall that the kth
Hamiltonian has the ground state 𝜓0
(1)
(𝑥; 𝑎 𝑘) and we
need to manipulate this into an equation for 𝜓 𝑘
(1)
(𝑥; 𝑎1), (which J. Maluck [10] shows is done through
the application of the operator 𝐴†
).
𝜓 𝑘
(1)
(𝑥; 𝑎1) =
𝐴†(𝑥; 𝑎1)𝜓 𝑘−1
(1)
(𝑥; 𝑎2)
√ 𝐸 𝑘−1
(1)
This proves what was stated earlier, that for any potential satisfying shape invariance after the first
ground state and function relating the variables, 𝑓(𝑎) is known, the entire energy spectrum is known.
As an example let’s use the Morse potential. The superpotential is given in the form of equation
(3.11), from Supersymmetry in Quantum Mechanics [9].
𝑊(𝑥) = 𝐴 − 𝐵𝑒𝑥𝑝(−𝛼𝑥)
Next, from equations (1.7) and (1.11) we can write the first two partner potentials:
𝑉1(𝑥; 𝑎1) = 𝐴2
+ 𝐵2
exp(−2𝛼𝑥) − 2𝐵 (𝐴 +
𝛼
2
) exp(−𝛼𝑥)
𝑉2(𝑥; 𝑎1) = 𝐴2
+ 𝐵2
exp(−2𝛼𝑥) − 2𝐵 (𝐴 −
𝛼
2
) exp(−𝛼𝑥)
To satisfy the shape invariance condition of equation (3.1) we rewrite 𝑉1(𝑥) in the new form of
equation (3.14) using the function 𝑓(𝑎1) = 𝐴 − 𝛼.
𝑉1(𝑥; 𝑎2) = (𝐴 − 𝛼)2
+ 𝐵2
exp(−2𝛼𝑥) − 2𝐵 (𝐴 −
𝛼
2
) exp(−𝛼𝑥)
By substituting (3.13) and (3.14) into (3.1) we are left with the remainder in the form of equation
(3.16):
𝑅(𝑎1) = 𝐴2
− (𝐴 − 𝛼)2
(3.8)
(3.9)
(3.10)
(3.11)
(3.12)
(3.13)
(3.14)
(3.15)

12. 12
And so using this, we are able to work out all the energy eigenstates of the system, as explained
earlier. The energy eigenstate of the nth
excited level for the first Hamiltonian is now known via
application of equation (3.17):
𝐸 𝑛 = 𝐴2
− (𝐴 − 𝑛𝛼)2
Using this formula it makes sense to see what the system looks like. To do this we calculate 𝐸 𝑛
(1)
for
𝑛 = 0,1,2,3 with some dummy values on the variables 𝐴, 𝐵, 𝛼. Upon doing so we confirm that 𝐸0
(1)
=
0.
To go a step further I have represent this visually. I plotted the superpotential, partner potentials and
the energy levels in figure 3.1 below. By inspection we can now see how 𝐸1
(1)
, labelled E1, is the
ground state energy level of the potential 𝑉2(𝑥) as the line labelled E0 is not contained within the
potential 𝑉2(𝑥).
Next we are interested in the wavefunctions. We already know from equation (3.10) once we know
one we can construct the rest. Equation (3.17) shows how the ground state wavefunction is calculated
[14]
, and using equations (3.10), (1.5) and (3.17) we arrive at equation (3.18).
𝜓0(𝑥) = 𝑁𝑒𝑥𝑝 (− ∫ 𝑊(𝑥′)𝑑𝑥′)
𝑛 𝐸 𝑛
(1)
0 0
1 5.76
2 10.24
3 13.44
(3.16)
Table 3.1: The first 4 energy eigenvalues.
Figure 3.1: A graph showing the superpotential, partner potentials and the first 4 energy levels.
eigenvalues.
(3.17)

13. 13
𝜓1
(1)
(𝑥; 𝑎1) =
(−
𝑑
𝑑𝑥
+ 𝐴 − 𝐵𝑒𝑥𝑝(−𝑎𝑥)) 𝑁𝑒𝑥𝑝(− ∫ 𝑊(𝑥′)𝑑𝑥′)
√ 𝐸 𝑛−1
(1)
Above the Morse potential has already been substituted in for 𝑊(𝑥). This can be simplified for the
Morse potential by making a suitable substitution:
𝑦 =
2𝐵
𝛼
𝑒−𝛼𝑥
𝜓 𝑛(𝑦) = 𝑦 𝑠−𝑛
exp(−
1
2
𝑦)𝐿 𝑛
2𝑠−2𝑛
(𝑦)
𝑠 =
𝐴
𝛼
(3.18)
(3.19)
(3.20)
(3.21)

14. 14
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