1

An introduction to Supersymmetric Quantum

Mechanics

By Thomas Acton

2014

3rd Year Project

King’s College London

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

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

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.

𝐻1𝜓0 = −ℏ2

2𝑚

𝑑2𝜓0

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

𝑑2𝜓(𝑥)

𝑑𝑥2−

ℏ

√2𝑚[𝑊′(𝑥)𝜓(𝑥) + 𝑊(𝑥)𝜓′(𝑥)] + 𝑊(𝑥)

ℏ

√2𝑚𝜓′(𝑥) + 𝑊2(𝑥)𝜓(𝑥)

= (−ℏ2

2𝑚

𝑑2

𝑑𝑥2−

ℏ

√2𝑚𝑊′(𝑥) + 𝑊2(𝑥)) 𝜓(𝑥)

∴ 𝑉1(𝑥) = 𝑊2(𝑥) −ℏ

√2𝑚𝑊′(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

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

√2𝑚𝑊′(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:

𝐻1𝜓n(1) = 𝐸𝑛

(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.

𝐻2𝐴𝜓n(1) = 𝐸𝑛

(1)(𝐴𝜓n(1)) = 𝐴𝐴†𝐴𝜓n

(1)

𝐻2𝜓n(2) = 𝐸𝑛

(2)𝜓n(2) = 𝐴𝐴†𝜓n

(2)

𝐻1(𝐴†𝜓n(2)) = 𝐸𝑛

(2)(𝐴†𝜓n(2)) = 𝐴†𝐴𝐴†𝜓n

(2)

𝜓𝑛(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 𝜓𝑚(2)

.

𝜓𝑚(2)

(𝑥) = 𝑐𝐴𝜓𝑛(1)

(𝑥)

∫ 𝜓𝑚(2)

𝜓𝑚∗(2)

= |𝑐|2 ∫ 𝜓𝑛∗(1)

𝐴†𝐴𝜓𝑛(1)

= |𝑐|2𝐸𝑛(1) ∫ 𝜓𝑛

(1)𝜓𝑛

∗(1)= 1

∴ 𝑐 =1

√𝐸𝑛(1)

(1.8)

(1.9)

(1.10)

(1.11)

(1.12)

(1.13)

(1.14)

(1.15)

(1.16)

(1.17)

(1.18)

6

∴ 𝜓𝑚(2) = 𝐴

𝜓n(1)

√𝐸𝑛(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:

𝐸𝑛(2)

= 𝐸𝑛+1(1)

𝜓𝑛(2) = 𝐴

𝜓n+1(1)

√𝐸𝑛+1(1)

𝜓𝑛+1(1) = 𝐴†

𝜓n(2)

√𝐸𝑛(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

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†2𝐴2

𝐻3 = 𝐴2𝐴†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) = 𝑊12(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

𝑑𝑥𝑙𝑛𝜓0

(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 = 𝐴1𝐴1† + 𝐸0

(1)= −

𝑑2

𝑑𝑥2+ 𝑉2(𝑥)

𝑉2(𝑥) = 𝑊12(𝑥) + 𝑊1

′(𝑥) + 𝐸0(1)

= 𝑉1(𝑥) + 2𝑊1′(𝑥) = 𝑉1(𝑥) − 2

𝑑2

𝑑𝑥2𝑙𝑛𝜓0

(1)(𝑥)

This method produces the expected results for the energy eigenvalues and wave function:

𝐸𝑛2 = 𝐸𝑛+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

𝜓𝑛(2)

=𝐴1𝜓𝑛+1

(1)

√(𝐸𝑛+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

𝑑𝑥ln (𝜓0

(2))

𝐻2 = 𝐴1𝐴†1 + 𝐸0

(1)= 𝐴†

2𝐴2 + 𝐸0(2)

= 𝐴†2𝐴2 + 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 = 𝐴2𝐴†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) = 𝑊22(𝑥) + 𝑊2

′(𝑥) + 𝐸1(1)

= 𝑉2 + 2𝑊2′ = 𝑉2(𝑥) − 2

𝑑2

𝑑𝑥2𝑙𝑛𝜓0

(2)

𝑉3(𝑥) = 𝑉1(𝑥) − 2𝑑2

𝑑𝑥2𝑙𝑛𝜓0

(1)− 2

𝑑2

𝑑𝑥2𝑙𝑛𝜓0

(2)= 𝑉1(𝑥) − 2

𝑑2

𝑑𝑥2𝑙𝑛 (𝜓0

(1)𝜓0

(2))

This procedure is consistent with the previous results, confirming the relation between the energy

eigenvalues:

𝐸𝑛(3)

= 𝐸𝑛+1(2)

= 𝐸𝑛+2(1)

𝜓𝑛(3)

=𝐴2𝜓𝑛+1

(2)

√(𝐸𝑛+1(2)

− 𝐸0(2)

)

=𝐴2𝐴1𝜓𝑛+2

(1)

√(𝐸𝑛+2(1)

− 𝐸1(1)

)√(𝐸𝑛+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.

𝐻𝑘 = 𝐴†𝑘𝐴𝑘 + 𝐸𝑘−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

𝐴𝑘 =𝑑

𝑑𝑥+ 𝑊𝑘(𝑥)

𝐴𝑘† = −

𝑑

𝑑𝑥+ 𝑊𝑘(𝑥)

𝑊𝑘(𝑥) = −𝑑

𝑑𝑥𝑙𝑛𝜓0

(𝑘)(𝑥)

𝑉𝑘(𝑥) = 𝑉1(𝑥) − 2𝑑2

𝑑𝑥2𝑙𝑛 (𝜓0

(1)… 𝜓0

(𝑘−1))

𝐸𝑛(𝑘)

= 𝐸𝑛+1(𝑘−1)

= 𝐸𝑛+𝑘−1(1)

𝜓𝑛(𝑘)

=𝐴𝑘−1 … 𝐴1𝜓𝑛+𝑘−1

(1)

√(𝐸𝑛+𝑘−1(1)

− 𝐸𝑘−2(1)

) … √(𝐸𝑛+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

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):

𝐻𝑘 = −𝑑2

𝑑𝑥2+ 𝑉1(𝑥; 𝑎𝑘) + ∑ 𝑅(

𝑘−1

𝑖=1

𝑎𝑖)

Next we introduce the wave function 𝜓0(1)(𝑥; 𝑎𝑘) to the general form of the kth Hamiltonian to see

what we can deduce about the energy levels.

𝐻𝑘𝜓0(1)(𝑥; 𝑎𝑘) = −

𝑑2

𝑑𝑥2+ 𝑉1(𝑥; 𝑎𝑘)𝜓0

(1)(𝑥; 𝑎𝑘) + ∑ 𝑅(

𝑘−1

𝑖=1

𝑎𝑖)𝜓0(1)(𝑥; 𝑎𝑘)

−𝑑2

𝑑𝑥2+ 𝑉1(𝑥; 𝑎𝑘)𝜓0

(1)(𝑥; 𝑎𝑘) = 𝐻1(𝑥; 𝑎𝑘)𝜓0(1)(𝑥; 𝑎𝑘) = 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

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), 𝐸𝑛(𝑘)

=

𝐸𝑛+1(𝑘−1)

= 𝐸𝑛+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:

𝐸𝑛(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; 𝑎𝑘) and we

need to manipulate this into an equation for 𝜓𝑘(1)

(𝑥; 𝑎1), (which J. Maluck [10] shows is done through

the application of the operator 𝐴†).

𝜓𝑘(1)

(𝑥; 𝑎1) =𝐴†(𝑥; 𝑎1)𝜓𝑘−1

(1) (𝑥; 𝑎2)

√𝐸𝑘−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

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):

𝐸𝑛 = 𝐴2 − (𝐴 − 𝑛𝛼)2

Using this formula it makes sense to see what the system looks like. To do this we calculate 𝐸𝑛(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 𝐸𝑛(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

𝜓1(1)

(𝑥; 𝑎1) =(−

𝑑𝑑𝑥

+ 𝐴 − 𝐵𝑒𝑥𝑝(−𝑎𝑥)) 𝑁𝑒𝑥𝑝(− ∫ 𝑊(𝑥′)𝑑𝑥′)

√𝐸𝑛−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

𝛼𝑒−𝛼𝑥

𝜓𝑛(𝑦) = 𝑦𝑠−𝑛exp (−1

2𝑦)𝐿𝑛

2𝑠−2𝑛(𝑦)

𝑠 =𝐴

𝛼

(3.18)

(3.19)

(3.20)

(3.21)

14

Bibliography

1. Sean Carroll, Dark Matter, Dark Energy: The Dark Side of the Universe, The Teaching Company

2. H. Miyazawa (1966). "Baryon Number Changing Currents". Prog. Theor. Phys. 36 (6): 1266–1276

3. Gervais, J. -L.; Sakita, B. (1971). "Field theory interpretation of supergauges in dual models".

Nuclear Physics B 34 (2): 632

4. D.V. Volkov, V.P. Akulov, Pisma Zh.Eksp.Teor.Fiz. 16 (1972) 621; Phys.Lett. B46 (1973) 109;

V.P. Akulov, D.V. Volkov, Teor.Mat.Fiz. 18 (1974) 39

5. http://en.wikipedia.org/wiki/Supersymmetry (02/12/14)

6. http://en.wikipedia.org/wiki/Quantum_field_theory (02/12/14)

7. F. Cooper, A. Khare and U. Sukhatme, "Supersymmetry and Quantum Mechanics" Phys. Rept.

251:267-385, (1995)

8. D. J. Griffiths (2006). Introduction to Quantum Mechanics. Reed College: Pearson Education Inc.

ISBN: 978-81-7758-230-7

9. F. Cooper, A. Khare, U. Sukhatme (2001). Supersymmetry in Quantum Mechanics. World

Scientific Publishing Co. ISBN: 981-02-4605-6

10. J. Maluck (2013) “An Introduction to Supersymmetric Quantum Mechanics and Shape Invariant

Potentials”. Amsterdam University College

11. E. Schrödinger, Proc. Roy. Irish Acad. A46 (1940) 9.

12. L. Infeld and T. E. Hull. The factorization method. Rev. Mod. Phys.,23(1):21{68, (1951)

13. L. E. Gendenshtein (1983). Derivation of exact spectra of Schrodinger equation by means

of supersymmetry. JETP Lett, 38(356)

14. E.Nygren (2010) Supersymmetric Quantum Mechanics, University of Bern