SUSY AND EXTRA DIMENSIONS LECTURE NOTESpages.physics.cornell.edu/~ajd268/Notes/SUSYandExtraDim.pdf · 2014-05-14 · 1 SUSY AND EXTRA DIMENSIONS LECTURE NOTES Lecture notes based

1

SUSY AND EXTRA DIMENSIONSLECTURE NOTES

Lecture notes based on a course given by Fernando Quevedo.We begin with discussing the mathematical tools involved

in Supersymmetry. We then move on to discussing the theoryin detail with a focus on N = 1. Lastly, we move

on to the basics behind theories of extra dimensions.

Presented by: Fernando Quevedo

LATEX Notes by: JEFF ASAF DROR

2013

CORNELL UNIVERSITY

Contents

1 Introduction 41.1 Physical Motivations of SUSY . . . . . . . . . . . . . . . . . . . . . . . . 4

1.1.1 What do we know so far? . . . . . . . . . . . . . . . . . . . . . . 41.1.2 Problems with the SM . . . . . . . . . . . . . . . . . . . . . . . . 61.1.3 Beyond SM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Supersymmetry Algebra and Representations 72.1 Poincare Symmetry and Spinors (Review) . . . . . . . . . . . . . . . . . 72.2 Properties of the Lorentz Group . . . . . . . . . . . . . . . . . . . . . . . 82.3 Representations of SL(2,C) . . . . . . . . . . . . . . . . . . . . . . . . . 92.4 Invarient Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.5 Generators of SL(2,C) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.6 Product of Spinors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.7 Connection to Dirac Spinors . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 SUSY Algebra 143.1 History of SUSY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.2 Representations of the Poincare Group . . . . . . . . . . . . . . . . . . . 18

4 N = 1 Supersymmetry Representations 214.1 Massless Multiplets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.2 Massive Multiplets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

5 Extended Supersymmetry 305.1 Representations of N > 1 SUSY . . . . . . . . . . . . . . . . . . . . . . . 31

5.1.1 Massless Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.A Important Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

6 Superfields and Superspace 406.1 Properties of θα, θα coordinates . . . . . . . . . . . . . . . . . . . . . . . 436.2 Superfields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446.3 Chiral Superfields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 516.4 Vector Superfields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

6.4.1 Wess-Zumino Gauge . . . . . . . . . . . . . . . . . . . . . . . . . 54

2

CONTENTS 3

6.5 Transformations of Superfields . . . . . . . . . . . . . . . . . . . . . . . . 556.5.1 Chiral Superfields . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

6.A Derivatives and spinor charges . . . . . . . . . . . . . . . . . . . . . . . . 566.B Useful Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

7 Four Dimensional Supersymmetric Lagrangians 587.1 N = 1 Global Supersymmertry . . . . . . . . . . . . . . . . . . . . . . . 58

7.1.1 Chiral Superfields . . . . . . . . . . . . . . . . . . . . . . . . . . . 587.1.2 Miraculous Calculation in Detail . . . . . . . . . . . . . . . . . . 637.1.3 Vector Superfields . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

7.2 Action as a Superspace Integral . . . . . . . . . . . . . . . . . . . . . . . 697.3 Non-abelian generalization . . . . . . . . . . . . . . . . . . . . . . . . . . 707.4 Non-Renormalization Theorems . . . . . . . . . . . . . . . . . . . . . . . 717.5 N = 2, 4 Global SUSY . . . . . . . . . . . . . . . . . . . . . . . . . . . . 757.6 Supergravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

8 Supersymmetry Breaking 798.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 798.2 F and D Breaking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

8.2.1 F -Term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 818.2.2 D Term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

8.3 SUSY breaking in N = 1 supergravity . . . . . . . . . . . . . . . . . . . 86

9 The MSSM 879.1 Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 879.2 Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 889.3 Supersymmetry Breaking . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

9.3.1 SUSY Sector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

9.3.2 Messenger Sector . . . . . . . . . . . . . . . . . . . . . . . . . . . 919.4 Hierarchy Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

Chapter 1

Introduction

The standard reference is Wess and Bagger. However, this book is very dry. You cantrust their equations is right, however there aren’t many physical motivations in the book.There are plenty of review articles that we can use as well.

1.1 Physical Motivations of SUSY

1.1.1 What do we know so far?

• Quantum Mechanics and Special Relativity - together give QFT.

We have a field. Excitations of this field give particles. From this we have learnedthat there are two completely different types of particles, bosons and fermions.They are basically unrelated and behave very differently from each other. Bosonshave integer spin while fermions have half integer spin.

• An example of a very successful quantum field theory is the Standard Model (SM).

– The SM describes the strong and electroweak interactions which are mediatedby s = 1 particles (gluons, photons, W±, Z).

– The matter fields have spin 1/2 and these are the quarks and leptons

– The Higgs boson is a spin 0 particle

– The graviton is spin 2

• This is just one example of a QFT. There are many QFT’s.

• The Basic Principle is symmetry

– Spacetime symmetries (Lorentz, Poincare, General Coordinate Transforma-tion)

– Internal symmetries which transform the fields themselves without transform-ing space and time. In the SM the symmetry isGSM = SU(3)C︸︷︷︸

strong

×SU(2)L × U(1)︸︷︷︸Electroweak

4

1.1. PHYSICAL MOTIVATIONS OF SUSY 5

• Importance of symmetries are that

– Label particles: mass, spin, charge, colour, etc.

– Determine interactions between particles (through the “gauge principle”)

– As an example consider

L = ∂µφ∂µφ∗ − V (|φ|2) (1.1)

where φ is a scalar field. This Lagrangian is invariant under the transformation,φ → eiαφ; α → const. This is called a global symmetry. If we want α to bea function of spacetime, α = α(x) then the potential is still invarient but thekinetic term isn’t,

∂µφ→ eiα (i (∂µα)φ+ ∂µφ) (1.2)

but if we modify and have

Dµφ ≡ ∂µφ+ iAµφ→ eiα(∂µφ+ i (∂µα)φ+ iA′µφ

)(1.3)

This is equal to Dµφ if A′µ = Aµ − ∂µα. We get a coupling AµφAµφ∗.

Similarly for L = ψγµDµψ, we get a Dirac coupling, ψAµψ.

– Another property of symmetries that is very important is that symmetries canhide. If we take again the same Lagrangian,

L = DµφDµφ∗ − V (|φ|) (1.4)

If V is a potential as ∼ |φ|2 + |φ|4 then we have a symmetry about φ →eiαφ with the vacuum sharing the same symmetry as the potential. We have〈φ〉 = 0. The symmetry is manifest. However, if we have the mexican hatpotential then you have the same symmetry, φ → eiαφ but the minimum isat 〈φ〉 6= 0. The symmetric point is not the minimum. The minimum is notinvariant under this phase rotation. You will not see the symmetry at all. Thesymmetry will be hidden. This is very important since the real symmetries ofnature may be much bigger then what we see because we live in a world thatbreaks that symmetry. This phenomena is known as spontaneous symmetrybreaking. In the SM we have

SU(2)L × U(1)〈φ〉∼102GeV−−−−−−−−→ U(1)

The symmetries can hide from us, but if we are clevor enough we can stilluncover them.

6 CHAPTER 1. INTRODUCTION

1.1.2 Problems with the SM

• Quantum Gravity

• Why? There are many why questions in the SM that we don’t know. For examplewhy is the gauge group of the Standard Model is GSM = SU(3) × SU(2) × U(1)?Why there are three families of quarks and leptons? Why is it that we live in 3 + 1dimensions? etc.

• We have MEW ∼ 102GeV,MPlanck ∼ 1019GeV and so

MEW

MPlanck

. 10−15 1 (1.5)

This is the standard hierarchy problem.

• There is a second hierarchy problem that is even more difficult. There is nowevidence that our universe is accelerating due to a vacuum energy which is aboutzero,

Λ ∼ 10−120 (1.6)

which is completely different then the energy of the vacuum calculated by QFT.

MΛ

MEW

∼ 10−15 1 (1.7)

This is known as the cosmological constant problem.

1.1.3 Beyond SM

• Experiments such as the LHC are looking for new particles

• We may have more symmetries

– Internal: G→ GSM → SU(3)× U(1)

– Spacetime symmetries

∗ More dimensions

∗ Supersymmetry

∗ Beyond QFT - string theory (needs supersymmetry)

Chapter 2

Supersymmetry Algebra andRepresentations

2.1 Poincare Symmetry and Spinors (Review)

Poincare: Given xµ,

x′µ = Λµνx

ν︸︷︷︸Lorentz

+

Translation︷︸︸︷aµ (2.1)

The Lorentz transforms obey,

ΛTηΛ = η, η =

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

(2.2)

We work with the isochronous Lorentz group. Which is the Lorentz group connected tothe identity which is denoted

SO(3, 1)↑ (2.3)

though we will often just omit the up arrow in our notation. The algebra consists ofgenerators, Mµν , P σ such that

[P µ, P ν ] = 0 (2.4)

[Mµν , P γ] = i (P µηνγ − P νηµγ) (2.5)

[Mµν ,Mρσ] = i (Mµσηνρ +Mνρηµσ −Mµρηνσ −Mνσηµρ) (2.6)

e.g. A representation for the generators are

(Mρσ)µν = i (ηρµδσν − ησµδρν ) (2.7)

As an excercise you can multiply these matrices and see if they satisfy the algebra.

7

8 CHAPTER 2. SUPERSYMMETRY ALGEBRA AND REPRESENTATIONS

2.2 Properties of the Lorentz Group

SO(3, 1) ∼ SU(2)× SU(2) (Locally) since we can define

Ji ≡1

2εijkMjk (rotations) (2.8)

Ki ≡M0i (Lorentz Boosts) (2.9)

where i, j, k ∈ 1, 2, 3. We can now define

Ai ≡1

2(Ji + iKi) (2.10)

Bi ≡1

2(Ji − iKi) (2.11)

which satisfy

[Ai, Aj] = iεijkAk (2.12)

[Bi, Bj] = iεijkBk (2.13)

[Ai, Bj] = 0 (2.14)

We have two SU(2) algebras. However, A and B are non-Hermitian in general. So Aand B do not truely represnt spin operators. However note the fact that representationsare labeled by “spins” (A,B) but physical spin J = A + B. Under Parity (x0 → x0,xi → −xi) we have Ji → Ji, Ki → −Ki (this is found by looking at equations 2.8 and2.9 and by how many spatial indices each of J and K have). Under Parity we them haveA↔ B. This is one important property of SO(3, 1).

Another important property is that

SO(3, 1) ∼= SL(2,C) (2.15)

SL(2,C) is the group of 2× 2 matrices with unit determinant. The symbol ∼= representshomomorphic. It is not isomorphic because the map is not one-to-one, but in fact itstwo-to-one as we will see.

Since

x ≡ xµeµ =

x0

x1

x2

x3

(2.16)

and

x ≡ xµσµ =

(x0 + x3 x1 − ix2

x1 + ix2 x0 − x3

)(2.17)

where σµ are the Pauli matrices. The matrix and vector carry the same information. Theintereseting thing is that under SO(3, 1),

|x|2 = x20 − x2

1 − x22 − x2

3 (2.18)

2.3. REPRESENTATIONS OF SL(2,C) 9

is invarient but under SL(2,C)x→ x′ = NxN † (2.19)

where N ∈ SL(2,C) and

det x = det x′ = x20 − x2

1 − x22 − x2

3 (2.20)

However this map is not one-to-one but two-to-one since:

N = ±(

1 00 1

)−→ Λ =

1 0 0 00 1 0 00 0 1 00 0 0 1

(2.21)

SL(2,C) is closer the algebra and hence defines the representations of the algebra and isa universal covering group. This is the group that is simply connected.

2.3 Representations of SL(2,C)

The fundamental repersentation is

ψ′α = N βα ψβ (2.22)

where α, β = 1, 2 with N being a 2× 2 matrix. The conjugate representation is

χ′α = N∗ βα χβ (2.23)

with α, β = 1, 2. These two represetnations are independant. There are two contravariantrepresentations

ψ′α = ψβ(N−1

) α

β(2.24)

χα = χβ(N−1,∗) α

β(2.25)

2.4 Invarient Tensors

We already know thatηµν = (ηµν)−1 (2.26)

is invarient under SO(3, 1). Here in SL(2,C) the invarient tensor is

εαβ =

(0 1−1 0

)= −εαβ (2.27)

It is invarient because

ε′αβ = ερσN αρ N β

σ = detNεαβ = εαβ (2.28)


Now we can raise and lower indices as we wish with εαβ. In particular,

ψα = εαβψβ (2.29)

χα = εαβχβ (2.30)

where we have defined εαβ to be the same as εαβ. The contravarient representations arenot indepdent. They can be obtained from the fundamental by raising and lowering theindices with εαβ.

The other object that can be considered here is something that has mixed SL(2,C)and SO(3, 1) indices,

(σµ)αα , σµ = 1,σ (2.31)

We know that(xµσ

µ)αα → N βα (xνσ

ν)βγ N∗ γα = Λ ν

µ xνσµαα (2.32)

From here we see thatσµαα = N β

α σνβγ (Λµν)−1N∗ γ

α (2.33)

which is invarient. Similar relations hold for σ,

(σµ)αα ≡ εαβεαβσµββ

= (1,−σ) (2.34)

2.5 Generators of SL(2,C)

(σµν) βα ≡

i

4(σµσν − σν σµ) β

α (2.35)

(σµν)αβ ≡i

4(σµσν − σνσ)αβ (2.36)

Both of σµν and σµν satisfy the Lorentz algebra which implies that

ψα →(e−

i2ωµνσµν

) β

αψβ (2.37)

χα →(e−i2ωµν σµν

)αβ

(2.38)

ψ is the left handed spinor. You can check that (A,B)→ (1/2, 0). χα is a right handedspinor, and you can check that (A,B) → (0, 1/2). Parity exchanges A and B whichimplies that (0, 1/2)↔ (1/2, 0).

For (1/2, 0) : Ji = 12σi and Ki = − i

2σi. For the (0, 1/2) we have Ji = 1

2σi and Ki = i

2σi

There are some useful relations for our generators and the Pauli matrices,

σµσν + σν σµ = 2ηµν1 (2.39)

Trσµσν = 2ηµν (2.40)

(σµ)α,α(σµ)β,β = 2δβαδβα (2.41)

σµν =1

2iεεµρσσρσ (self-dual) (2.42)

2.6. PRODUCT OF SPINORS 11

2.6 Product of Spinors

We have the product of left-handed spinors,

χψ ≡ χαψα = −χαψα (2.43)

Similarly for right handed spinors,

χψ ≡ χαψα = −χαψα (2.44)

In particular,ψ2 = ψαψα = εαβψβψα = ψ2ψ1 − ψ1ψ2 (2.45)

So the square of the spinor is the difference above. This is one motivation to choosethese numbers to be anticommuting. If they were commuting we would have ψ2 = 1.We choose the components of ψ to be anti commuting numbers, or Grassman numbers:ψ1ψ2 = −ψ2ψ1. Which gives

χψ = ψχ (2.46)

We also have

ψαψβ =1

2εαβ (ψψ) (2.47)

which leads the Fierz identity,

(θψ) (θψ) = −1

2(ψψ) (θθ) (2.48)

andψσµνχ = − (χσµνψ) (2.49)

We define the dagger such that(θψ)† = θψ (2.50)

and(ψσµχ)† = χσµψ (2.51)

(ψα)† ≡ ψα (2.52)

ψα = ψ∗β(σ0)βα

(2.53)

where the lower identity is just a way to control the indices (σ0 = 1)

2.7 Connection to Dirac Spinors

We define

γµ =

(0 σµ

σµ 0

)(2.54)


thenγµ, γν = 2ηµν (2.55)

That means that these Gamma matrices defined in this way are a representation of theDirac algebra (Clifford Algebra for Dirac matrices). We also have

γ5 ≡ iγ0γ1γ2γ3 =

(−1 00 1

)(2.56)

This matrix has eigenvalues of ±1 which is defined as the chirality. We work in the Weylrepresentation, where the chirality of a state is manifest.

Furthermore, you can define,

Σµν ≡ i

4γµν =

(σµν 00 σµν

)(2.57)

These are the generators of the Lorentz group. Dirac spinors,

Ψ ≡(ψαχα

)(2.58)

Notice

γ5Ψ =

(−ψαχα

)(2.59)

So we can define projection operators,

PR/L ≡1

2

(1± γ5

)(2.60)

such that

PLΨ =

(ψα0

)(2.61)

PRΨ =

(0χα

)(2.62)

Finally the conjugate is given by

Ψ ≡(χα ψα

)= Ψ†γ0 (2.63)

We have the column conjugate,

Ψc ≡(χαψα

)= CΨT (2.64)

where

C ≡(εαβ 0

0 εαβ

)(2.65)

2.7. CONNECTION TO DIRAC SPINORS 13

takes particles to anti-particles. If Ψ satisfies the Dirac equation for a given charge thenΨc satisfies the same equation with opposite charge.

We can have majorana spinors, ψα = χα where

Ψ =

(ψαψα

)= Ψc (2.66)

We can always write a Dirac spinor as a sum of Majorana spinors.

Ψ = ΨM1 + iΨM2 (2.67)

We can ask the question whether there is a spinor that is both Weyl and Majorana in4 dimensions. This is not possible since if it’s Weyl then one of its components is zero,however by definition Majorana particles have the same value for both left and rightcomponents. Hence these spinors are zero.

Chapter 3

SUSY Algebra

3.1 History of SUSY

• In the 1960’s there was a lot of confusion. There were many things going on at thistime. People were putting particles in muliplets through their isospin. These com-bined particles of the same spin into multiplets of SU(3)f . People began thinkingwhy not mix particles from different spins into their own larger multiplets.

• In 1967: Coleman-Mandula released a paper that introduced what they called No-go theorem. They tried to look for the most general symmetries for scatteringamplitudes. They concluded that the most general symmetry of the S-matrix is thedirect product of

Poincare︸︷︷︸Pµ,Mµν

⊗Ta︷︸︸︷

internal (3.1)

• 1971: Golfand and Licktman extneded Poincare algebra to include spinor genera-tors, Qα, α = 1, 2.

Ramond, Nevuer + Schwarc, Gervais + Sakita discussed SUSY in 2D (from stringtheory)

• 1973: Volkov and Akulov were trying to understand why the neutrino was a masslessparticle. They tried to think of neutrinos as Goldstone particles, m = 0. Theydiscovered how to break supersymmetry.

• 1974: Wess + Zumino began SUSY field theories in 4D. This paper is consideredthe breakthrough of SUSY and implied how it can be used in field theories.

In 1975 there was was an interesting paper by Haag Lopussaski and Sohnus wherethey took studied the Coleman-Mandela(CM) theorem and generalized the theorem toinclude spinor generators. We will now build the algebra as these people did about 30years ago. The Poincare generators are P µ,Mµν . In principle there can be any other

14

3.1. HISTORY OF SUSY 15

generators of any number of indices but they are ruled out by the CM theorem. ForSUSY we will add generators in the (1

2, 0) representation which we call

QAα

A=1,2,...Nα=1,2 (3.2)

and we will also add operators in the (0, 12) representation given by

QAα

A=1,2,...,Nα=1,2 (3.3)

We will not prove it, however these are the only generators you can add. The proof isgiven in Weinberg. These operators will obey different types of relations we are used to,since typically Lie groups obey commutation relations. However, spinor relations obeywhat’s known as a Graded Algebra,

OaOb − (−1)ηaηbObOa = iCcabOc (3.4)

for which ηa = 0 for Oa bosonic generator and ηa = 1 for Oa fermionic generator. ηa iscalled a grading. For bosons we have the standard commutator. For 2 fermions we havean anticommutator.

The SUSY generators are P µ,Mµν , QAα , Q

Aα , A = 1, ..., N . N = 1 is called a simple

SUSY. N > 1 is called extended SUSY.We start with N = 1. We know

[P µ, P ν ] (3.5)

[P µ,Mρσ] (3.6)

[Mµν ,Mρσ] (3.7)

We need to find

1. [Qα,Mµν ]

2. [Qα, Pµ]

3. Qα, Qβ

4.Qα, Qα

5. [Qa, Ti]

where Ti are internal symmetry generators (e.g. U(1)). Because of CM we know that theTi commute with the P µ,Mµν but we don’t know that holds for the spinor generators.

We begin by trying to find the first one. To find the commutator of Q and M weneed to use the fact that Qα is a spinor. Hence it must transform under the spinorrepresentation of the Lorentz Group. We know that

Q′α =(e−

i2ωµνσµν

) β

αQβ (3.8)

≈(

1− i

2ωµνσ

µν

) β

α

Qβ (3.9)

16 CHAPTER 3. SUSY ALGEBRA

But Q is not only a spinor it is also as an operator. Q as an operator also transformsunder Lorentz Transformation. As an operator it transforms as follows

Q′α = U †QαU ,U = e−i2ωµνMµν

(3.10)

Note that we wrote the Lorentz transformations in two different ways. This is the stan-dard way to derive the relations. You can similarly do this to get the algebra from theMµν ’s and P µ’s. We can write

Q′α =

(1 +

i

2ωµνM

µν

)Qα

(1− i

2ωµνM

µν

)(3.11)

Comparing with our other expression for Q′α we have

− i

2ωµν (σµν) β

α Qβ =i

2ωµν [Mµν , Qα]− 1

4ωµνωρσM

µνQαMρσ (3.12)

Dropping terms proportional to ω2 we have

[Qα,Mµν ] = (σµν) β

α Qβ (3.13)

where we have used the fact that the equation must hold for all ωµν .We now need to derive the second relation, the one with P µ and Qa. The commutator

[Qα, Pµ] must be linear in the P ’s , M ’s and Q’s. We know already that the Qα is

anticommuting and P µ is commuting so the right hand side must also be anticommuting.We cannot have just M ’s or just P ’s. The only thing you can write down that satisfiesthe conditions is

[Qα, Pµ] = cσµααQ

α (3.14)

where c is a constant. The only problem left is to fix the constant. Still we need to knowthe complex conjugate of this expression which gives,[

Qα, P µ]

= c∗ (σµ)αβ Qβ (3.15)

Exercise! (Hint: take adjoint and use (Qα)† = Qα and (σµQ)†α =(Qσµ

)α

).To find the constant we use the Jacobi identity which says that

[A, [B,C]] + [C, [A,B]] + [B, [C,A]] = 0 (3.16)

Taking A = P µ , B = P ν , and C = Qα gives

[P µ, [P ν , Qα]] +

:0

[Qα, [Pµ, P ν ]] + [P ν , [Qα, P

µ]] = 0 (3.17)

−cσναα[P µ, Qα

]+ cσµα,α

[P ν , Qα

]= 0 (3.18)

|c|2 σναα (σµ)αβ Qβ − |c|2 σµαα (σν)αβ Qβ = 0 (3.19)

|c|2 (σν σµ − σµσν) βα︸︷︷︸

6=0

Qβ = 0 (3.20)

3.1. HISTORY OF SUSY 17

and hence c = 0 and we have[Qα, P

µ] = 0 (3.21)

This can be derived insightfully by just realizing that Qα is a spinor which are invariantunder spatial translations, and hence the commutation relation must be zero.

We need to derive two more relations. First we need Qα, Qβ. Because of the indexstructure of the Q’s we must have

Qα, Qβ

= k (σµν) βα Mµν (3.22)

We need to find the constant k. The left hand side commutes with P µ (as we just found).The right side does not commute wtih P µ (the Lorentz generators do not commute withP µ). The only way this can be true is if

Qα, Qβ = 0 (3.23)

To finish we need to findQα, Qβ

. Qα is a (1/2, 0) object while Qβ is a (0, 1/2)

spinor. So on the right hand side we must have a (1/2, 1/2) object which is P µ. So

tσµααPµ (3.24)

There is no way to fix t. All the games that we played above will not work here. So byconvention we take t = 2 and we have the final relation

Qα, Qβ

= 2σµ

αβPµ (3.25)

So the anticommutator of two spinors gives a translation. The away to think about thisis Qα |fermion〉 = |boson〉 and Qα |boson〉 = |fermion〉 and if we act on a boson twiceas

QαQα |boson〉 = [boson〉 ( translated) (3.26)

There is one thing we have yet to discuss which is with the internal symmetries,[Qα, Ti]. Usually as in the case for the CM theorem,

[Qα, Ti] = 0 (3.27)

except that the supersymmetry algebra has what is called an automorphism which changes

Qα → eiλQα (3.28)

Qα → e−iλQα (3.29)

where you multiply Q by a phase and Q by an opposite phase, the whole algebra doesn’tchange since in the commutation and anticommutation relations we found above Q onlyhas non-trivial commutation relations when it comes with an Q. This is called R sym-metry (it is a U(1) symmetry). We have

[Qα, R] = Qα (3.30)[Qα, R

]= Qα (3.31)


3.2 Representations of the Poincare Group

We now try to get a better understanding of state labelling. First consider the non-relativistic labelling of particles with spin. Recall that the rotation group, Ji : i = 1, 2, 3satisfies

[Ji, Jj] = iεijkJk (3.32)

with a Casimir operator (an operator that commutes with all the generators)

J2 =3∑i=1

J2i (3.33)

The Casimir operator labels the irreducible representations by the eigenvalues, j(j + 1)of J2. Within these representations, you can diagonalize one of the operators, J3 to getthe eigenvalues −j,−j + 1, ..., j − 1, j. The states can then be labelled as |j, j3〉.

This is not sufficient for relativistic physics since here we want to relativistic operators,such as P µ and Mµν to label our states. Recall that in the Poincare group there are twoCasimir operators. One of these involves the Pauli-Lubanski vector,

Wµ =1

2εµνρσP

νMρσ (3.34)

where ε0123 = −ε0123 = +1. The Poincare Casimirs are then given by

C1 = P µPµ, C2 = W µWµ (3.35)

It is straightforward to check that these both commute with the generators of the Poincaregroup. We can label our states with the eigenvalues of our two Casimirs,

|m,w〉 (3.36)

However, that is not all. Just as we used j3 to label our states above, we can also lookfor generators that commute with the momenta. [Q 1: Don’t thes generators need also tocommute with Wµ?]. These tell you which are sensible labels for your states. The largestgroup that leave your momenta invarient are known as the Little Group. Consider amassive particle. You can always go to the rest frame and then you have

P µ = (m, 0, 0, 0) (3.37)

In this case the little group is equal to the rotation group in three dimensions. In thisframe it is easy to calculate what the Pauli-Lubanski vector is,

W0 =1

2ε0νρσP

νMρσ (3.38)

Since the momentum is zero in this frame we have W0 = 0 in the rest frame. Now considerWi. In this case we have

Wi =1

2εiνρσP

νMρσ (3.39)

3.2. REPRESENTATIONS OF THE POINCARE GROUP 19

As before the only non-trivial component of Pµ is P0 so

Wi =1

2εi0jkP

0M jk = ... = −mJi (3.40)

where the above requires some playing with the epsilon symbols. So we have W 2 = m2J2.Clearly, Ji and J2 commute with Wµ, Pµ,W

2, and P 2. So the w in the labelling of thestates can be interpreted as the j of the mutliplet and we can use ji as our quantumnumber. So for massive particles we can label the representations of the Poincare groupby,

|m, j; pµ, j3〉 (3.41)

For massless particles, it is a bit more complicated. You can choose the momenta tobe

P µ = (E, 0, 0, E) (3.42)

Naively anything that rotations x and y is the Little Group. However, this is not quiteall. The group is actually bigger then that. The symmetry transformations of the Pauli-Lubanski vector define the Litte Group.[Q 2: Why?] One can go ahead and calculate thecomponents of the Pauli-Lubanski vector in this case,

W0 = EJ3 (3.43)

W1 = −E(J1 −K2) (3.44)

W2 = −E(J2 +K1) (3.45)

W3 = −EJ3 (3.46)

Now Wi forms a “Euclidean group in 2 dimension”,

[W1,W2] = 0 (3.47)

[W3,W1] = −iEW2 (3.48)

[W3,W2] = iEW1 (3.49)

These in principle defines the Little Group. The difficulty is that it has an infinitedimensional representation. That means that the particle has an extra label. However,we don’t know of any particle with this property. The way to fix that is to just considera finite subset. The solution for this is to set W1 = W2 = 0. Then since W0 and W3

commute we have SO(2) rotations as a subgroup. By looking at equations (3.43)-(3.46)it’s easy to see that Wµ = λPµ, where we call λ the helicity of the particle. The statesare then labelled as

|0, 0; pµ, λ〉 (3.50)

We often omit the first two zeros for brevity. Under CPT the states transform to |pµ,−λ〉.This can be understood as follows. Parity inverts the momenta and time reveral invertsthe spin and momenta of the particle. So the net result is a flip in the helicity. Therelation,

e2πiθJz |pµ, λ〉 = e2πiλ |pµ, λ〉 = ± |pµ, λ〉 (3.51)

hence we require λ = 0,±12,±1, .. e.g. λ = 0. This is a consequence of the Lorentz Group

being double covered. In the SM we have


• λ = 0→ Higgs

• λ = ±12→ quarks and leptons

• λ = ±1→ photons, gluons

• λ = ±2→ graviton

Notice that we defined all the particles that we know of in terms of the massless multiplets.The reason for this is that all the particles are fundamentally massless but gain massthrough the Higgs mechanism.

Chapter 4

N = 1 SupersymmetryRepresentations

We want to look for mutliplets to better understand our theories and find which statesare degenerate. For example, in Hydrogen finding the different multiplets (correspondingto different L values) gives a deeper understanding of the system then just knowing thatsome states have given energies. It tells you why the energy states are structured in theway that they are. As a relevant example consider the SU(2) multiplets of the weakinteraction. Here the mutliplets combine the electron and the neutrino into a doublet(since the weak force is broken their symmetry isn’t manifest in the end). As a lastexample consider the SU(3)C multiplets of QCD. Here the different colored particles areput in a multiplet (the red, green, and blue up quark for example).

First we need to look for the Casimirs of the SUSY algebra. The Casimir, C1 = PµPµ,

is still a Casimir for SUSY. This is easy to see since P µ commutes with the SUSYgenerators. However, the second Casimir that we were using, is no longer a Casimir forSUSY. That means that spin is no longer a good label within a multiplet. That meansyou can have particles of different spin within a multiplet. C2 is no longer a Casimir. Butthere is another operator we can define instead. The new Casimir is a bit complicated.It is given by

C2 ≡ CµνCµν (4.1)

where

Cµν = BµPν −BνPµ (4.2)

Bµ = Wµ −1

4Qα (σµ)αβ Qβ (4.3)

The corresponding quantum number for this Casimir is often called the Superspin.Before we start building the mutliplets of SUSY we first prove a general result for all themutliplets.

The claim is as follows: In any supersymmetric multiplet the number of bosons (nB)is equal to the number of fermion (nF ),

nB = nF (4.4)

21

22 CHAPTER 4. N = 1 SUPERSYMMETRY REPRESENTATIONS

From this statement arises the idea that for every particle in the SM there exists acorresponding particle with opposite spin.

The proof is as follows. Consider an operator, (−1)F (abbreviated (−)F ) known asthe fermion number which is defined through

(−)F |B〉 , (−)F |F 〉 = − |F 〉 (4.5)

where |B〉 is defined as a boson and |F 〉 a fermion. Its easy to see that (−)F anticommuteswith Qα. Consider the action of the operator on a fermion:

(−)FQα |F 〉 = (−)F |B〉 = |B〉 (4.6)

but−Qα(−)F |F 〉 = + |B〉 (4.7)

since an analogous result can be shown for the operator acting on a fermion we have(−)F , Qα

= 0 (4.8)

We first define the trace operator in the multiplet space. It is given by

Tr O =∑n

〈n|O|n〉 (4.9)

where the sum is over all the states in the multiplet (bosons and fermions)Now consider the following object (here by trace we mean summing over all states

within a multiplet),

Tr

(−)FQα, Qβ

= Tr

(−)FQαQβ + (−)F QβQα

(4.10)

We now use the anticommuting property we showed above for the first terms and thecyclic property of the trace for the second,

Tr

(−)FQα, Qβ

= Tr

−Qα(−)F Qβ + (−)FQαQβ

(4.11)

= 0 (4.12)

On the other hand the computation can be done more explicitly usingQα, Qβ

=

2(σµ)αβPµ,

Tr

(−)FQα, Qβ

= Tr

(−)F2(σµ)αβPµ

(4.13)

We fix P µ to be the same for a single multiplet (we are using it as one of the labels ofthe multiplets) so we have

Tr

(−)FQα, Qβ

= (σµ)αβPµTr

(−)F2

(4.14)

All the preceding objects are clearly non-zero and thus we know that

Tr(−)F = 0 (4.15)

4.1. MASSLESS MULTIPLETS 23

but

Tr(−)F =

nB∑i=1

〈Bi|(−)F |Bi〉+

nF∑j=1

〈Fj|(−)F |Fj〉 (4.16)

= nB − nF (4.17)

since this is equal to zero we have proven our result,

nB = nF (4.18)

This is true for any mutliplet. We call these multiplets a supermultiplet.

4.1 Massless Multiplets

We now try to find what types of particles are in multiplets in SUSY. Without SUSY allobjects in the same mutliplet have the same spin. As well will see this will no longer bethe case with supersymmetry.

Consider the massless supermultiplet. We work in the frame that,

Pµ = (E, 0, 0, E) (4.19)

For this we have the first Casimir to be zero,

C1 = PµPµ = 0 (4.20)

Plugging in this value of Pµ one can show that

C2 = 2m2YiYi = 0 (4.21)

where

Yi ≡ Ji −1

4mQσiQ (4.22)

= Ji −1

4mQβσ

βαi Qα (4.23)

So the representation is labelled by two zeros, which is not very enlightening. The statesin the representation are only labelled by their momenta and helicity,

|P µ, λ〉 (4.24)

The difference from multiplets in the SM is that now we can have states with different λin the same multiplet as will see.


We now see how to find the states in a multiplet. We start with the algebra. Recallthat

Qα, Qα

= 2σµααPµ (4.25)

= 2(Eσ0 + Eσ3

)αα

(4.26)

= 2E

(2 00 0

)αα

(4.27)

and so we can see from here thatQ1, Q1

= 4E but the rest of the anticommutators

are zero. In particular, Q2, Q2

= 0 (4.28)

That means for any state in the representation

〈P µ, λ|Q2Q2|pµλ〉 = |Q2 |P µ, λ〉|2 = 0 (4.29)

and hence Q2 = 0 in this representation. Now consider Q1:

Q1, Q1 = 4E (4.30)

in this multiplet.That means that we can define

a ≡ Q1

2√E

, a† ≡ Q1

2√E

(4.31)

which gives a, a†

= 1 , (4.32)

a, a =a†, a†

= 0 (4.33)

This an algebra that we are already familiar with, the algebra of creation and annihilationoperators. We can use these operators as we do in QM to build the representation byacting with these operators on the vacuum. We also know how they act on particles withspin.

We now consider the commutator of the annhilation operator with J3:[a, J3

]=

1√2E

1

2

[Q1,M

12 −M21]

(4.34)

=1√2E

1

2

(σ12) β

1 Qβ − (σ21) β1 Qβ

(4.35)

= − 1√2E

i

4

(σ1σ2 − σ2σ1

) β

1Qβ (4.36)

=1√2E

1

2(σ3) 1

1 Q1 (4.37)

=1

2a (4.38)

4.1. MASSLESS MULTIPLETS 25

That means that if we take

J3 (a |p, λ〉) =

(aJ3 − 1

2(σ3) 1

1

)a |p, λ〉 (4.39)

= (aλ− 1

2a) |p, λ〉 (4.40)

= (λ− 1

2) (a |p, λ〉) (4.41)

So the state a |aλ〉 is an eigenstate of J3 with helicity λ− 12.

By the same arguement one can show that a† |p, λ〉 has helicity λ + 12. So to build

a representation we start with a state that we call “the vacuum”. The quotes here arebecause this is not the real vacuum it is just the state of minimum helicity.

|Ω〉 ≡ |p, λ〉 (4.42)

By equation 4.41 we see that,

a† |Ω〉 = |p, λ+1

2〉 (4.43)

and since a† anticommutes with itself,

a†a† |Ω〉 = 0 (4.44)

So the whole multiplet consists of

|p, λ〉 , |p, λ+1

2〉 (4.45)

As usual we must add the CPT conjugates of this, so that means that we have|p,±λ〉 , |p,±(λ+

1

2)〉

(4.46)

As an example conside λ = 0. Then we have a λ = 0 scalar and a λ = 12

fermion. Thisis usually called the “Chiral Multiplet”. The name arises from having one chiral fermionand a scalar. Examples of such multiplets are

λ = 0 scalar λ = 12

fermionsquark quarkslepton leptonHiggs Higgsino

We can also have “Vector or Gauge multiplets” with λ = 12, 1. Examples include

λ = 12

fermion (“gauginos”) λ = 1 bosonphotino photongluino gluonWino WZino Z


For now we skip λ = 1, 12

case but we discuss it more later. The last example is theλ = 3

2, 2 multiplet,

λ = 32

fermion λ = 2 bosongravitino graviton

We cannot have states with λ > 2. This is non-trivial but we will address this point later.

4.2 Massive Multiplets

We have,Pµ = (m, 0, 0, 0) (4.47)

We start by considering the Casimirs,

P 2 = m2 (4.48)

and one can show thatC2 = 2m2Y iYi (4.49)

where

Yi ≡ Ji −1

4m

(QσQi

)=Bi

m(4.50)

These Yi satisfy the SU(2) algebra (Ex),

[Yi, Yj] = iεijkYk (4.51)

The eigenvalues of Y 2 are y(y + 1). Thus we need to add two labels to our multiplets,one for each Casimir.

To find the rest of our labels we again consider the anticommutator,

Qα, Qα = 2(σµαα)Pµ (4.52)

In the eigenspace of massive particles at rest this takes the form,

Qα, Qα = 2mδαα (4.53)

This is the same result we had for massless particles with a different scaling factors anda non-zero anticommutator for both Q1 and Q2. Thus we need two sets of creation andannihilation operators,

a1,2 =1√2m

Q1,2 (4.54)

a†1,2 =1√2m

Q1,2 (4.55)

and ap, a

†q

= δpq (4.56)

4.2. MASSIVE MULTIPLETS 27

We can define the “vacuum state”, |Ω〉 such that

ap |Ω〉 = 0 (4.57)

With this we see that

Yi |Ω〉 = Ji |Ω〉 −1

4mQσiQ |Ω〉 = Ji |Ω〉 (4.58)

So on acting on the vacuum the Yi label is the same as the spin label, j.The state is labelled by the mass, spin, momentum, and projected spin angular mo-

mentum,|Ω〉 ≡ |m, y;P µ, j3〉 (4.59)

For the rest of the mutliplet we can use the same trick as before

a1 |j3〉 = |j3 −1

2〉 (4.60)

a†1 |j3〉 = |j3 +1

2〉 (4.61)

However for a2 we have σ322 = −1 which gives

a2 |j3〉 = |j3 +1

2〉 (4.62)

a†2 |j3〉 = |j3 −1

2〉 (4.63)

That means that if we act on |Ω〉 by a†1,

a†1 |Ω〉 = k1 |m, j = y +1

2;P µ, j3 +

1

2〉+ k2 |m, j = y − 1

2;P µ, j3 +

1

2〉 (4.64)

where k1 and k2 are Clebsh-Gordan coefficients. Similarly we have,

a†2 |Ω〉 = k3 〈m, j = y − 1

2; pµ, j3 −

1

2〉+ k4 〈m, j = y +

1

2;P µ, j3 −

1

2〉 (4.65)

Acting on these states again with the same creation operator gives zero. So the onlyother thing we can do is

a†2a†1 |Ω〉 = −a†1a

†2 |Ω〉 (4.66)

This is a spin j object. In total we have,

2 states |m, j = y;P µ, j3〉 → 2(2y + 1)

1 state |m, j = y +1

2;P µ, j3〉 → 2(y +

1

2) + 1

1 state |m, j = y − 1

2;P µ, j3〉 → 2(y − 1

2) + 1


Above we asssumed y 6= 0. This case is in fact a bit simpler. The state annihilatedby aα is,

|Ω〉 = |m, j = 0; pµ, j3 = 0〉 (4.67)

then we have,

a†1/2 |Ω〉 = |m, j = 1/2; pµ, j = ±1/2〉 (4.68)

a†1a†2 |Ω〉 = |m, j = 0; pµ, j3 = 0〉 (4.69)

The last state appears to be the same as |Ω〉 but it is not. |Ω〉 is a scalar while a†1a†2 |Ω〉

is a pseudoscalar.To see this recall that under parity we have, (A,B) ↔ (B,A). So under parity we

have Q↔ Q up to a phase,

P−1QαP = ηP(σ0)αβQβ = ηPσ

0αβεβγQγ (4.70)

Similarly,P−1QαP = η∗P σ

0αβε

βγQγ (4.71)

This is consistent with the algebra of the Q′s:Qα, Qα

= 2σµααPµ (4.72)

such that under parity,P µ → (P0,−p). This is easy to show for a particular value of α, αas we show for α = 1, α = 2.

P−1Q1Q2P =(σ0)

1βεβγQγ

(σ0)

2βεβγQγ (4.73)

= −Q2Q1 (4.74)

and soP−1

Q1, Q2

P = −

Q2, Q1

= −

Q1, Q2

(4.75)

On the other hand the right hand side gives,

P−1(2σµ

12Pµ)P =

(P 0 + P 3 P 1 − iP 2

P 1 + iP 2 P 0 − P 3

)12

(4.76)

= 2(P 1 − iP 2

)(4.77)

= −2σµ12Pµ (4.78)

This is consistent with the transformation of the anticommutator above as expected.Also,

P−1P−1QαPP = P−1(σ0)αβεβα(σ0)αβεβγQγ (4.79)

=

(1 00 1

)(0 −11 0

)(1 00 1

)(0 −11 0

) γ

α

Qγ (4.80)

= −Qα (4.81)

4.2. MASSIVE MULTIPLETS 29

Now

|Ω′〉 ≡ a†1a†2 |Ω〉 = |m, j = 0;P µ, 0〉 (4.82)

is not annhiliated by the a′s but by the a†’s instead. These states are related by switchinga↔ a†. So under Parity,[Q 3: Show this]

|Ω〉 ↔ |Ω′〉 (4.83)

We can then construct states that are Parity eigenstates,

|±〉 = |Ω〉 ± |Ω′〉 (4.84)

which have Parity ±1 (|+〉 is a scalar and |−〉 is a pseudoscalar.For massive N = 1 multiplet y = 0,

|Ω〉 = |m, j = 0; pµ, 0〉 , a1,2 |Ω〉 = 0 (4.85)

Chapter 5

Extended Supersymmetry

We have seen the basics of supersymmetry with,

QAα A = 1, ...,N (5.1)

QAα A = 1, ...,N (5.2)

The algebra for this extended version is the same as the N = 1 except forQAα , Q

Bα

= 2σµααPµδ

AB (5.3)

QAα , Q

Bβ

= εαβZ

AB (5.4)

ZAB is the new ingredient of extended supersymmetry and known as central charges.These central charges commute with all the operators,[

ZAB, P µ]

=[ZAB,Mµν

]=[ZAB, QA

α

]=[ZAB, ZCD

]=[ZAB, Ta

]= 0 (5.5)

These charges provide what is called the Abelian invariant subalgebra of internal sym-metries. For the rest of the internal symmetries we have the standard,

[Ta, Tb] = iCabcTc (5.6)

Now recall earlier that we had an R-symmetry which was just a U(1) phase underwhich the algebra was invariant. We define R - Symmetry to be elements of internalsymmetry group G, [

QAα , Ta

]= SABQ

Bα (5.7)

If ZAB = 0 then the R-symmetry group is just U(N). You can see that if you multiplythe Q’s by a unitary matrix U then the algebra is invariant. However if ZAB 6= 0 thenthe R symmetry group is just a subgroup of U(N) and you must study it case by case.

30

5.1. REPRESENTATIONS OF N > 1 SUSY 31

5.1 Representations of N > 1 SUSY

5.1.1 Massless Particles

Choose the momentum to be Pµ = (E, 0, 0, E). Then we use,

QAα , QB,α

= 4E

(1 00 0

)αα

δAB (5.8)

As before we can chooseQA

2 = 0 (5.9)

and define

aA ≡ QA1

2√E

, a†,A =

QA1

2√E

(5.10)

which satisfy, aA, aB,†

= δAB (5.11)

The states are

States helicity # of States“vacuum” |Ω〉 λmin 1

a†,A |Ω〉 λmin + 1/2 N

a†,Aa†,B |Ω〉 λmin + 1 N(N+1)2

=

(N2

)a†,Aa†,Ba†,C |Ω〉 λmin + 3/2

(N3

)...

......

a†,Na†,N−1...a†,1 |Ω〉 λmin +N/2 1

Total number of states within one multiplet is

N∑n=0

(Nn

)= 2N (5.12)

So the number of states in a multiplet can be very large.We now consider some examples,

1. N = 2:

(a) λmin = 0:

λ = 0

λ = 1/2 λ = −1/2

λ = 1

32 CHAPTER 5. EXTENDED SUPERSYMMETRY

[Q 4: Don’t we always need to include the −λ states in each multiplet...?] Wesee the N = 2 multiplet is a combination of 2 N = 1 multiplets. This is afamous multiplet and is known as the N = 2 vector multiplet.

(b) λmin = 1/2

λ = −1/2

λ = 0 λ = 0

λ = 1/2

This is called the N = 2 hypermultiplet.

N = 4

(a) λmin = −1:# of States λ

1 −14 −1/26 04 1/21 1

2. N = 8. The particle content is:

# of States λ1 ±28 ±3/228 ±156 ±1/270 0

Below are a few remarks:

• In general λmax − λmin = N/2

• One can show that if we want a theory that is renormalizable you cannot haveobjects with helicity greater than 1. Thus we have a constaint on possible valuesof N . If we at most want |λ| = 1 then λmax − λmin is at most 2 and thus we havethe condition,

N ≤ 4 (5.13)

• “Strong Belief”: There are no massless particles with helicity greater than 2 since|λ| > 1/2 massless particles at low momentum must couple to conserved currents(e.g. ∂µjµ = 0 for E&M where λ = ±1). For λ = ±2 the conserved current is ∂µTµν .The corresponding conserved charge is the 4-momentum. Beyond this there are no


more conserved currents to couple to. Since this are no more conserved currentsthere are no more conserved particles. This arguement has also been extended toSUSY which allowed λ = ±3/2 spin particles. If this is true then that implies that

λmax − λmin ≤ 4 (5.14)

which implies that

N ≤ 8 (5.15)

This theory was a promising candidate for a grand unified theory as it unites allthe possible states in one multiplet. A real United States.

• Another arguement for seeing that N = 8 is a maximum is that if N > 8 then wehave more then 1 “graviton”. This would be very strange since we would have atheory with different gravities.

• However, N > 1 SUSY are in general non-chiral. The SM is a chiral theory.However as we will argue now, for all the extended supersymmetry the multipletsare generically are non-chiral.

The idea is that all these multiplets are λ = ±1 particles. Any gauge bosonstransform in a particular representation of the corresponding group, the adjointrepresentation. The adjoint representation is a real representation and non-chiral.This implies that the corresponding particles within the same multiplet (the λ =±1/2 particles) will transform under the same representation and hence will notbe chiral. We know that the SM particles live in complex representations (thefundamental representations).

For the N = 2 hypermultiplet,

λ = −1/2

λ = 0 λ = 0

λ = 1/2

the helicity 1/2 comes together with the helicity −1/2. Hence they are innatelynon-chiral again. This kills all N > 1. We are only left with N = 1 and N = 0.For instance in N = 1, (

1/20

)(5.16)

• However in N = 1 it has a prediction already: at least one extra particle for eachSM particle with the same mass since its in the same multiplet. This prediction isclearly violated! The only way out is that supersymmetry is a broken symmetry(at low energies, E ∼ 102GeV).


Recall that for the massless case we found that the number of states in a multipletis 2N . To derive this relation we didn’t need to use the central charges at all. This isbecause in the massless case, Q2 = 0 which by

QAα , Q

Bβ

= εαβZ

AB (5.17)

implies that ZAB = 0.We now address the massive case. We have,

P µ = (m, 0, 0, 0) (5.18)

and we have QAα , QB,β

= 2σµ

αβPµδ

Aβ = 2m

(1 00 1

)δAβ (5.19)

Now the central charges do not need to be zero. There are two cases,

1. ZAB = 0

2. ZAB 6= 0

We consider each case separately. First consider ZAB = 0:As for N = 1 we now have,

aAα ≡QAA√

2m(5.20)

We have 2N creation and annihiliation operators and we have

22N states (5.21)

each of dimension (2y + 1)1.For example for N = 2,

y = 0 spin # of states|Ω〉 0 1

a†,Aα |Ω〉 1/2 4a†a† |Ω〉 0(3states), 1(3states) 6a†a†a† 1/2 4

a†a†a†a† |Ω〉 0 1

where we don’t bother putting labels on the a†’s but just take them to be combinationsof different values of A,α. The detailed structure of these multiplets is not particularlyimportant for us but more to give a flavor of how the larger multiplets look like. Here wehave 16 = 22×2 states as expected.

Now consider the second case, ZAB 6= 0:

H ≡(σ0)βα

QAα − ΓAα , Qβ,A − Γβ,A

≥ 0 (5.22)

1Recall that the degenerate states with spin j are −j,−j + 1, ..., j.


where this number is greater then zero since we are multiplying a number by its complexconjugate. We define

ΓAα = εαβUABQγβ

(σ0)γβ

(5.23)

where is an arbitrary unitary matrix. Using the SUSY algebra one can show

H = 8mN − 2Tr(ZU † + UZ†

)≥ 0 (5.24)

[Q 5: Show.] Now any matrix can be written in the form,

Z ≡ HV (5.25)

where H = H† is positive and V V † = 1. Note we can isolate H,√Z†Z =

√HV V †H† = H (5.26)

This is known as the polar decomposition. Recall that we are free to choose the matrixU . We choose U = V . This sets

ZU † + UZ† = H +H = 2H (5.27)

and soH = 8mN − 4TrH = 8mN − 4Tr

√Z†Z ≥ 0 (5.28)

where by square root of a Z†Z we mean a matrix such that when multiplied by itself isequal to Z†Z2. Thus we have,

m ≥ Tr√Z†Z

2N(5.29)

This is a bound on the mass of the particles and is usually known as the BPS bound.The bound is saturated when

m =Tr√Z†Z

2N(5.30)

These are known as BPS states. When this condition holds we have H = 0 which impliesthat

QAα − ΓAα = QA

α − εαβV ABQγ,

(σ0)γβ

(5.31)

We have a combination of the Q’s and Q’s that is equal to zero. Recall that for themassless case we had Q2 = 0 which changed the number of states from 22N → 2N . TheBPS states are have the same constraint and we have 2N states. The correspondingmultiplet is “shorter”. This is the main property that distinguishes these states.

As an example consider the case that N = 2. The central charge is antisymmetric sothe most general matrix can be written,

ZAB =

(0 q1

−q1 0

)(5.32)

2The matrix square root isn’t unique but that need not bother us here.


where the degree of freedom of the central charge is called a charge and denoted by q1

(the reason for subscript 1 will become clear soon). The BPS condition is

m ≥ 1

42q1 =

q1

2(5.33)

This can be generalized for any other N . Consider N > 2 with an even N . If N is evenyou can always bring a matrix to the form,

Z =

0 q1

−q1 0 00 q2

−q2 0. . .

. . . . . .

0 0 qN−qN 0

(5.34)

The same analysis we just did can be done for each block. The BPS condition will holdblock by block and we have,

m ≥ qi2

(5.35)

for all i. Some blocks may satisfy this condition and some may not.If k < N/2 of the qi = 2m then we will have

2N − 2k (5.36)

creation operators and so we will have 22(N−k) states.

• For k = 0 this implies we have 22N states (the standard long multiplet)

• For 0 < k < N/2 we have 22(N−k) states and are usually called short multiplets

• For k = N/2 we have 2N states and is known as an ultra-short multiplet

This is another use of superlatives in SUSY. We used super, hyper, and now ultra. Weare competing with the comics which try to invent names for their heros.

Earlier considered the case with k = 1 and N = 2 so we had an ultra-short multiplet.Ultra-short multiplets play an important role due to the size of the multiplets. We nowhave a few comments

1. BPS states and bounds were started in soliton (e.g. monopoles) solutions of Yang-Mills systems. Here bound is an energy bound.

2. BPS states, (ultra) short states, tend to be stable. This is because the mass ofthese particles is small (its at the bottom of the lower bound). This stops particlesfrom decaying.

5.A. IMPORTANT COMMUTATORS 37

3. For BPS states we have mass ∼ charge. This condition is found in whats knownas extremal black holes.

4. BPS states have been cruicial for undertanding strong-weak coupling duality (aka“s duality”) in both field and string theory. In particular D-branes in string theoryare BPS states.

5.A Important Commutators

[Q 6: Under Construction]

[aα, J

2]

=1

4εijkεi`m

[M jkM `m

](5.37)

=1

4

[aα,M

jkM jk]−[aα,M

jkMkj]

(5.38)

=1

2

[aα,M

jkM jk]

(5.39)

=1

2

(σjk) β

αQβM

jk +M jkaαMjk − aαM jkM jk

(5.40)

=1

2

(σjk) β

αaβM

jk +M jk(σjk) β

αaβ

(5.41)

Now(σjk) β

αand M jk live in difference spaces and so we have,

[aα, J

2]

=1

2

(σjk) β

α

aβM

jk +M jkaβ

(5.42)

=1

2

(σjk) β

α

2M jkaβ +

(σjk) γ

βaγ

(5.43)

Now we have,

σjkσjk =i2

4 · 4(σjσk − σkσj

) (σjσk − σkσj

)(5.44)

= i2 · i2

4εjk`εjkmσ

`σm (5.45)

=1

2σ`σ

` (5.46)

=1

2

(σ2

1 + σ22 + σ2

3

)(5.47)

=3

2(5.48)


Thus we can finally write,[aα, J

2]

=(σjk) β

αM jkaβ +

3

4aα (5.49)

=1

2εjk`

(σ`) β

αM jkaβ +

3

4aα (5.50)

=(σ`) β

αJàβ +

3

4aα (5.51)

Taking the hermitian conjugate of this equation gives,[J2, a†α

]=(σ`) β

αa†βJ` +

3

4a†α (5.52)

=(σ`) β

αJà†β− 3

4a†α (5.53)

These formula are quite general and holds for acting on states of any spin. Supposefor example we act on a spinless object, i.e. one that obyes M jk |j = 0〉 = 0. In this case,

J2a†α |j = 0〉 =

3

4a†α + a†αJ

2

|j = 0〉 (5.54)

=3

4aα |j = 0〉 (5.55)

So the state a†α |j = 0〉 is a j = 12

object! a†α changes the j value of the state.To better understand the results we now expand the J` term:

σ`J` = σ1J1 + σ2J2 + σ3J3 (5.56)

= −1

2

(σ1 − iσ2

)J+ −

1

2

(σ1 + iσ2

)J− − σ3J3 (5.57)

= −(

0 01 0

)J+ −

(0 10 0

)J− −

(1 00 −1

)J3 (5.58)

= −(J3 J−J+ −J3

)(5.59)

and

σà†J` =

(a†

1J3 a†

2J−

a†1J+ −a†

2J3

)(5.60)

So, (σ`) β

1a†βJ` |j, j3〉(

σ`) β

2a†βJ` |j, j3〉

=

(j3a†1|j, j3〉+ A−a

†2|j, j3 − 1/2〉

A+a†1|j, j3 + 1/2〉 − j3a

†2|j, j3〉

)(5.61)

where

A+ ≡√j(j + 1)− j3(j3 + 1) (5.62)

A− ≡√j(j + 1)− j3(j3 − 1) (5.63)

5.A. IMPORTANT COMMUTATORS 39

Now consider,

〈j, j3|a†α|j, j3〉 =1

j(j + 1)〈j, j3|a†αJ2|j, j3〉 (5.64)

=1

j(j + 1)〈j, j3|a†β(σ`) β

α J` +3

4a†α + a†αJ

2|j, j3〉 (5.65)(〈j, j3|a†2|j, j3 − 1/2〉〈j, j3|a†1|j, j3 + 1/2〉

)= 0 (5.66)

As a second example consider acting on a spinor. In this case M jk = σjk and,(σjk)βα

(σjk)δγ

=1

4εjk`εjkm

(σ`) β

α(σm) δ

γ (5.67)

=1

2

(σ`) β

α

(σ`) δ

γ(5.68)

= δ δα δ

βγ −

1

2δ βα δ δ

γ (5.69)

and hence for the spinor representation,(σjk) β

α

(M jk

) δ

γ

(aβ) χ

δ=(aγ) χ

α−(aα) χ

δ(5.70)

and so we have, [aα, J

2] χ

γ=(aγ) χ

α−(aα) χ

γ+

3

4

(aα) χ

γ(5.71)

=(aγ) χ

α− 1

4

(aα) χ

γ(5.72)

Acting on a spin 1/2 state doesn’t return back an eigenstate. Instead it gives some linearcombination of states.

Chapter 6

Superfields and Superspace

So far we have just dealt with supermultiplets in terms of one-particle states. In thecase of field theory that is not enough. We need to talk about the interaction of theparticles and have a field theory that is supersymmetric. We have only the charactersof the game but we need to know how to play. For that we have to recall what we knowfrom standard field theories. The particles are described by fields, φ(xµ) with xµ as thefour dimensional spacetime coordinates. The fields can be many types depending onhow they transform under the Poincare group. What we want to do now is to make asupersymmetric generalization of φ(xµ). The fields will be a function of coordinates anda function of what we call superspace. We want:

Φ (X) (6.1)

with Φ transformaing under Super-Poincare and X are coordinates in superspace. Wefirst need to define what superspace is. We use symmetries to define this space. Weknow that every continuous group defines a manifold. There is a mapping from a groupG to a manifold M such that the elements of the group, g = eiαaT

aare labelled by the

coordinates of the space, αa.

G →MG (6.2)g = eiαaTa

→ αa (6.3)

We have as many parameters as generators of the group so the dimensionallity of thespace is the same as the dimension of the Lie group,

dimG = dimMG (6.4)

We now consider some examples,

1. G = U(1): g = eiαQ where α ∈ [0, 2π]⇒MU(1) = S1. The manifold is a circle.

2. G = SU(2): g =

(p q−q∗ p∗

)with the condition that det g = 1 or |p|2 + |q2|2 = 1.

40

41

If we parametrize p and q such that,

p = x1 + ix2 (6.5)

q = x3 + ix4 (6.6)

then the condition is just∑4

i=1 x2i = 1 and hence MSU(2) = S3. The manifold is a

3-sphere. This is also why SU(2) is the covering group of SO(3).

3. G = SL(2,C): g = HV ; where H = H† and V = SU(2) by the polar decomposition.Since we have an SL(2,C) matrix we also require detH = 1. So you can see thatthe manifold corresponding the SL(2,C) is the manifold corresponding to SU(2)times the manifold corresponding to H, but H we can write,

H = xµσµ =

(x0 + x3 x1 + ix2

x1 − ix2 x0 − x3

)(6.7)

for some xµ. [Q 7: Shouldn’t we have a negative on all the xi’s?]. detH = 1 impliesthat

(x0)2 −3∑i

(xi)2

= 1 (6.8)

This is a hyperbole in 4 dimensions and is like R3. So the corresponding manifoldfor SL(2,C),

MSL(2,C) = R3 × S3 (6.9)

since each of the two manifolds is simply connected we know that SL(2,C) is simplyconnected.

More generally we can have coset. The coset of G/H is found by taking every elementg ∈ G and identifying it with gh, where h ∈ H. We again consider a few examples,

1. G/H = U(1)1×U(1)2U(1)1

. In this case an element of G and H can be written as

g = ei(α1Q1+α2Q2) (6.10)

h = eiβQ1 (6.11)

thengh = ei[(α1+β)Q1+α2Q2] = g (6.12)

Pictorially we have,

0 2πα10

2π

α2

××××××××

42 CHAPTER 6. SUPERFIELDS AND SUPERSPACE

We set all the values of α1 on the line (marked by ×) to be equivalent to some valueand the total group is just,U(1)2.

2. SU(2)/U(1) = S2

3. We can also define Minkowski space-time as a coset,

Poincare

Lorentz=ωµν , aµωµν

→ aµ = xµ = Minkowski (6.13)

Note that since the Lorentz and the Poincare transformations do not commute thisrelation is not trivial but it can be shown to be true.

4. We now come to the most important example for us. N = 1 superspace is given by,

Super Poincare

Lorentz=

ωµν , aµ, θα, θα

ωµν

=aµ = xµ, θαθα

(6.14)

An element of the Poincare group should be of the form,

gPoincare = exp i (ωµνMµν + aµPµ) (6.15)

The elements of the Super Poincare group is then

g = expi(ωµνMµν + aµPµ + θαQα + θαQ

α)

(6.16)

Note that normally have algebra in terms of anticommutators for the Q’s and not commu-tators. Using the Grassman numbers we can rewrite our algebra. Consider for example,

Qα, Qα

= 2σµααPµ (6.17)

By multiplying from the left by θα and from the right by θα this can be rewritten,[θαQα, Qβ θ

β]

= 2θα(σµαβ

)θβPµ (6.18)[θQ, θQ

]= 2θσµθPµ (6.19)

To expand split the group elements we need to the BCH formula,

eAeB = eA+B+ 12

[A,B]+... (6.20)

Returning back to our definition of superspace. We have xµ which is a Minkowskivariables and two anti-commuting numbers, θ and θ which represent extra dimensions.We have 4 anticommuting dimensions.

6.1. PROPERTIES OF θα, θα COORDINATES 43

6.1 Properties of θα, θα coordinates

Consider first one single θ parameter.

f(θ) = f0 + f1θ (6.21)

The most general function of an anticommuting function is a linear function. A Mathe-maticians paradise!

We can take derivatives,df

dθ= f1 (6.22)

We define an integral such that ∫dθdf

dθ= 0 (6.23)

In particular, ∫dθ = 0 (6.24)

Since the most general function will be linear, so the other integral we need to define,∫θdθ ≡ 1 (6.25)

This normalization is done by convention. We impose this not to be zero to keep ourdiscussion non-trivial. Then the Delta function in this space is

δ(θ) = θ (6.26)

since ∫dθθf(θ) = f(0) (= f0) (6.27)

and ∫dθf(θ) =

∫dθf0 + f1θ = f1 (6.28)

So we see that ∫dθf(θ) =

df(θ)

dθ(6.29)

So the integral and derivative are the same in this space; this is one of the curiousproperties of Grassman numbers.

So far we have focused on a single Grassman number. Now we move on to multipleθ’s. Recall that we defined

θθ ≡ θαθα (6.30)

θθ ≡ θαθα (6.31)


According this notation we have,

θαθβ = −1

2εαβθθ (6.32)

θαθβ =1

2εαβ θθ (6.33)

So we can take derivatives,

∂θβ

∂θα= δ β

α (6.34)

∂θβ

∂θα= δ β

α (6.35)

We can also take integrals. We know that∫dθ1dθ2θ2θ1 = 1 (6.36)

Using equation 6.32 we can write θ1θ2 = 12θθ and so∫

1

2dθ1dθ2︸︷︷︸d2θ

θθ = 1 (6.37)

where we defined the diffential for two dimensional Grassman variables above. Further-more, you can write,

d2θ = −1

4dθαdθβεαβ (6.38)

d2θ =1

4dθαdθβεαβ (6.39)

We can also write, ∫d2θd2θ (θθ)

(θθ)

= 1 (6.40)

Extending the fact that the derivatives and integrals are the same we have,∫d2θ =

1

4εαβ

∂

∂θα∂

∂θβ(6.41)∫

d2θ = −1

4εαβ

∂

∂θα∂

∂θβ(6.42)

6.2 Superfields

N = 1 superspace is given by xµ, θα, θα

(6.43)

6.2. SUPERFIELDS 45

To define a superfield recall that a scalar field, φ(xµ) is a function of xµ and it transformsunder the Poincare group. For example under translations:

φ→ e−iaµPµφeia

µPµ (6.44)

where Pµ are the generators of translations. But φ is also a function of xµ. So we canalso write with another representation of the momentum operator, Pµ. Any function ofxµ can be transformed as,

φ(xµ)→ eiaµPµφ(xµ) = φ(xµ + aµ) (6.45)

(1 + iaνPν)φ(xµ) = φ(xµ) + aν∂νφ(xµ) (6.46)

⇒ Pµ = −i∂µ (6.47)

We’re doing something you’ve done since before you knew how to ride a bike.We have two expressions for the same φ so we can find exactly how φ transforms:

(1− iaµPµ)φ(xµ) (1 + iaµPµ) = (1 + iaνPν)φ(xµ) (6.48)

and so we have,

iaµ [φ, Pµ] = iaµPµφ (6.49)

[φ, aµPµ] = −iaµ∂µφ (6.50)

We now follow these same steps but now for superfields.For a general “scalar” superfield we write down every possible term using a exact

Taylor expansion:1 we have,(take a deep breath and prepare for this)

S(x, θ, θ) =

scalar︷︸︸︷φ(x) +θ

LSF︷︸︸︷ψ(x) +θ

RPF︷︸︸︷χ(x) +θθM(x) + θθN(x)

+(θσµθ

)Vµ(x)︸︷︷︸

Vec. field

+ (θθ)(θλ(x)

)+(θθ)

(θρ(x)) + (θθ)(θθ)D(x) (6.51)

where for convenience we write, LSF (RSF) as left spinor field (right spinor field). This isthe most general representation of supersymmetry. As we will see later this representationis reducible, however its not easy to find out into what. This can be contrasted with thePoincare group where we had the irreducible representations, φ, ψα, Aµ, ....

Recall that when building a Poincare invariant Lagrangian we use products of repre-sentations of the Poincare group to build the most general Poincare invariant Lagrangian.Otherwise, it is really hard to make Poincare invariant terms! We want to go on to do

1By scalar we do not demand that each constituents of the the terms making up the field be scalarbut that each term is. In other words,

Sθ = θS , Sθ = θS

for any Grassman number θ.


the same thing here. We want to build the most general SUSY invariant Lagrangian. Todo this we will use representations of supersymmetry.

Under Poincare transformations we have, S → eiaµPµS. We now move on to how to

transform our scalar superfield under superspace transformations (we already know howit transforms under Poincare tranformations)

S(x, θ, θ)→ e−i(εQ+εQ)Sei(εQ+εQ) (6.52)

S(x, θ, θ)→ ei(εQ+εQ)S (6.53)

where ε is a infinitesimal 2 component Grassman object and we denote the diffential formof the spinor charge operator Q and the abstract form as Q (notice that the first andsecond equation above use different forms of the supercharge operator). Recall that fortranslations we knew that eia

µPµφ must transform to give φ(xµ + aµ). However, hereit isn’t quite obvious what the transformation will be. We allow the possibility thatsuperspace transformations will also shift the real space coordinates. The most generalvector quantity that’s linear in the ε’s gives the transformation,

xµ → xµ −(icεσµθ + h.c.

)(6.54)

where c is a constant that we need to fix. We could not have a transformation that’sa function of xµ due to the assumed homogeneity of space-time. We have a generaltransformation,

S(xµ − icεσµθ, θ + ε, θ + ε) (6.55)

Doing a transformation in θ but keeping ε = 0 gives,

(1 + iεQ)S(xµ, θ, θ

)= S

(xµ, θ, θ

)−(icεασµ

αβθβ)∂µS(xµ, θ, θ) + εα∂αS

(xµ, θ, θ

)(6.56)

iεαQαS(xµ, θ, θ) = εα−icσµ

αβθβ∂µ + ∂α

S(xµ, θ, θ) (6.57)

or

iQα = −icσµαβθβ∂µ +

∂

∂θα(6.58)

(6.59)

Repeating the procedure only this time taking ε = 0 and keeping nonzero ε gives thedifferential forms of our generators:

Qα = −i ∂∂θα− cσµ

αβθβ

∂

∂xµ(6.60)

Qα = i∂

∂θα+ c∗θβσµβα∂µ (6.61)

Pµ = −i∂µ (6.62)

6.2. SUPERFIELDS 47

These objects are representations of the generators. To fix c we have to impose thecommutation relation:

Qα, Qα

= 2σµααPµ (6.63)

It is a straightforward exercise to show that there are no restrictions on the imaginarypart of c (which we set to zero) but just on the real part which must be 12.

We now know how functions transform under SuperPoincare. We still have to findthe analogue to the commutator in equation 6.50 which tells us how the fields transform.Comparing two transforms of S we have,

exp(i(εQ+ εQ)

)S exp

(−i(εQ+ εQ)

)= exp

(i(εQ+ εQ)

)S (6.64)

δS = i[S, εQ+ εQ

]= i(εQ+ εQ

)S (6.65)

To find how each of the components from equation 6.51 transform we just need to ap-ply the differential form of the SUSY generators to the superfield. We write down thetransformations of each component. We work through a few components explicitly.

Consider the action of Q and Q on the first part of S: the scalar field φ(x). Wewant to know what the contribution to the scalar field will be after transformation. Thuswe apply the differential operator and then only keep scalar terms (the other terms willcontribute to the transformation of other components). There is no hope of any term butthe first three to contribute to the scalar field as they all have higher powers of θ or θ:3

i

εα(−i ∂∂θα− σµ

αβθβ∂µ

)+ εα

(i∂

∂θα+ θβσµβα∂µ

)φ+ θψ + θχ+ ...

(6.66)

= i(−iεαψα + iεαχα

)(6.67)

= εψ + εχ (6.68)

where we only kept terms without θ or θ dependence and used εαχα = −εχ. Thus wehave,

δφ = εψ + εχ (6.69)

Notice that here we see that a scalar field, φ, goes to fermions under supersymmetry asexpected.

2Note that to derive this relationship you must use the Grassman numbers product rule,

∂

∂θ(fg) =

∂f

∂θg − f ∂g

∂θ

3Here we need the relations:

∂αθβ = δ β

α , ∂αθβ = δ β

α , ∂αθα = −θα∂α


We now transform the fermion field. We have,

i


αβθβ∂µ

)+ εα

(i∂


)φ+ θθM + θσµθVµ + ...

(6.70)

= iεαθβ (σµ)βα ∂µφ+ 2εαθαM + εασµβαVµθβ (6.71)

= θ (iσµε∂µφ+ 2εM + σµεVµ) (6.72)

so

δψ = σµε (i∂µφ+ Vµ) + 2εM (6.73)

As a last example we explore the transformations of the D term as it will turn out to beparticularly important in supersymmetry. We have,

i


αβθβ∂µ

)+ εα

(i∂


)θ2θρ+ θ2θλ+ ...

(6.74)

= −iεασµαβθ2θβ θα∂µλ

α − iεα (σµ)βα θ2θβθα∂µρ (6.75)

= iθ2θ2∂µ

(1

2εασµ

αβλβ − 1

2ρβ (σµ)βα εα

)(6.76)

We leave the rest of the transformations as an exercise. We list them all below,

δφ = εψ + ψχ (6.77)

δψ = 2εM + σµε (i∂φ+ Vµ) (6.78)

δχ = 2εN − εσµ (i∂µφ− Vµ) (6.79)

δM = ελ− i

2∂µψσε (6.80)

δN = ερ+i

2εσµ∂µχ (6.81)

δVµ = εσµλ+ ρσµε+i

2∂νψσµσνε− εσνσµ∂νχ (6.82)

δλ = 2εD +i

2(σνσµε) ∂µVν + i (σµε) ∂µM (6.83)

δρ = 2εD − i

2(σν σµε) ∂µVν + iσµε∂µN (6.84)

δD =i

2∂µ(εσµλ− ρσµε

)(6.85)

Note the transformation of D is by a total derivative. We will use this property soon.To find supersymmetry invariant terms we want to find the type of operations we

can apply to a superfield (the representation of supersymmetry) such that it remains asuperfield:

• If S1 and S2 are superfields then S1S2 is also a superfield. Proof: Suppose S = S1S2

6.2. SUPERFIELDS 49

then

δS = i[S1S2, εQ+ εQ

](6.86)

= i(S1S2

(εQ+ εQ

)− S1

(εQ+ εQ

)S2 + S1

(εQ+ εQ

)S2

−(εQ+ εQ

)S1S2

)(6.87)

= i(S1

[S2, εQ+ εQ

]+[S1, εQ+ εQ

]S2

)(6.88)

= S1i(εQ+ εQ

)S2 + i

(εQ+ εQ

)S1S2 (6.89)

= i(εQ+ εQ

)(S1S2) (6.90)

where in the last step we used the fact that Q and Q are differential operators andobey the product rule.

• Linear combinations of superfields are superfields since,[αS1 + βS2, εQ+ εQ

]= α

(εQ+ εQ

)S1 + β

(εQ+ εQ

)S2 (6.91)

=(εQ+ εQ

)(αS1 + βS2) (6.92)

• ∂µS is a superfield since,[∂µS, εQ+ εQ

]= ∂µ

[S, εQ+ εQ

](6.93)

= ∂µ(εQ+ εQ

)S (6.94)

=(εQ+ εQ

)∂µS (6.95)

• However, ∂αS is not a superfield since,

δ (∂αS) = i[∂αS, εQ+ εQ

](6.96)

= i∂α[S, εQ+ εQ

](6.97)

= i∂α[εQ+ εQ

]S (6.98)

6= i(εQ+ εQ

)S (6.99)

Note the subtle point here. We were able to pull out the ∂α in the second line sinceQ and Q are the abstract objects. However, afterwards the same cannot be donesince then we are in a particular representation of the Q’s namely, the diffentialrepresentation.

We have seen such a situation in the past where the derivative didn’t transform inthe way that we want it too - in building a gauge invariant Lagrangian. We willhave a similar situation here. We define covariant derivatives,

Dα ≡ ∂α + iσµαβθβ∂µ (6.100)

Dα ≡ −∂α − iθβσµβα∂µ (6.101)


These satisfy,

Dα,Qβ =Dα, Qβ

=Dα,Qβ

=Dα, Qβ

= 0 (6.102)

We prove these relations in appendix 6.A

Since these derivatives anticommute with the Q’s they will commute with the εQ’s.That means that will obey [

Dα, εQ+ εQ]

= 0 (6.103)

and so DαS is a superfield.

Also one can show that Dα, Dα

= 2iσµαα∂µ (6.104)

They have very similar properties to the Q’s however they have the very specialproperty that DαS is a superfield. Continuing with the remarks

• S = f(x) is a superfield only if f = const. However, S = φ(x) for example is not asuperfield (since δφ = εψ + εχ, a term with θ’s).

• However this doesn’t mean that by truncating the superfield you can never geta superfield. There are some ways to eliminate some components are still havea superfield. This is equivalent to saying that S is a reducible representation ofSUSY. In general we can impose consistent constraints on S.

We now consider some examples of truncated superfields.

1. Chiral Superfield: Denoted by Φ it is defined such that when acted opon by thesuperderivative we get zero,

DαΦ = 0 (6.105)

Such an object will transform to itself under supersymmetry since D acting on afield is still a superfield.

2. Antichiral Superfield: Denoted by Φ and we impose the constraint,

Dαφ = 0 (6.106)

3. Vector Superfield: V = V †. We note that while we call this a vector superfield it isstill scalar in nature in the sence that all its terms are bosonic. The origin of thisname will become clearer when we discuss this below.

4. Linear Superfield: Denoted by L and obeys the constraints,

DDL = 0 (6.107)

L = L† (6.108)

The Chiral superfield will include the quarks and leptons and the vector superfield willinclude the gauge fields.

6.3. CHIRAL SUPERFIELDS 51

6.3 Chiral Superfields

We have the constraint,DαΦ = 0 (6.109)

To find the components of Φ we will use a trick. We define a new coordinate,

yµ ≡ xµ + iθσµθ (6.110)

If we consider now Φ(y, θ, θ) then the constraint for Φ takes the form,

DαΦ = −∂αΦ(x, θ, θ)− iθβσµβα∂

∂xµΦ(x, θ, θ) (6.111)

= −∂αΦ(y, θ, θ)− ∂Φ(y, θ, θ)

∂yµ∂yµ

∂θα− iθβσµβα

∂

∂yµΦ(y, θ, θ) (6.112)

= −∂αΦ− ∂µΦ (−iθσµ)α − i (θσµ)α ∂µΦ (6.113)

= −∂αΦ (6.114)

where we have used,

∂

∂θα

(θβ (σµ)ββ θβ

)= θβ (σµ)ββ εβα (6.115)

= − (θσµ)α (6.116)

Combining Eq 6.114 with the definition of a superfield we have,

∂αΦ(y, θ, θ) = 0 (6.117)

and so the field doesn’t depend on θ! This gives a chiral field a very simple expansion.So a chiral field is just

Φ(

no θ︷︸︸︷yµ, θ) = φ(yµ) +

convention︷︸︸︷√2 θψ(yµ) + θθF (yµ) (6.118)

The φ(yµ) is a scalar object and will describe the squarks, sleptons, and the Higgs forinstance. The ψ(yµ) will describe the quarks, leptons, and Higgsinos. The F (yµ) is a newobject that will turn out not to be physical and known as an auxiliary field.

φ is complex so it has two components. ψ is complex but it has an index giving atotal of 4 components. F is complex and has two components. F is needed in this regardto ensure that we have an equal number of bosonic and fermionic components.

Now we have a simple expression for Φ in terms of its independent components.However, yµ is not a physically interesting variable. We can go back and rewrite ourexpression in terms of xµ. To do this we expand each component in xµ:

Φ(y, θ) = φ(x) + iθσµθ∂

∂xµφ(x)− 1

2θσµθθσν θ

∂2

∂xµ∂xνφ(x) (6.119)

+√

2θ

(ψ(x) + iθσµθ

∂

∂xµψ(x)

)+ θ2F (x)


To simplify this expression note that

θσµθθσν θ = θασµααθαθβσν

ββθβ (6.120)

=1

4εαβεαβθ2θ2σµαασ

νββ

(6.121)

=1

4θ2θ2σν

ββ(σµ)ββ (6.122)

=1

2θ2θ2ηµν (6.123)

and

θβθασµααθα∂µψβ = −1

2εβασµααθ

α∂µψβ (6.124)

=1

2σµααθ

α∂µψα (6.125)

= −1

2∂µψσ

µθ (6.126)

So we have,

Φ(x, θ, θ) = φ(x) +√

2θψ(x) + θθF (x) + iθσµθ∂µφ−i√2

(θθ) ∂µψσµθ

− 1

4(θθ)

(θθ)φ (6.127)

By switching variables we get back the older components in the general superfield butwith fewer variables. We only have φ, ψ, and F as variables in a chiral superfield. Tofind the transformations of the variables of a chiral superfield we can just compare withthe general superfield transformations we found earlier. This gives,

δφ =√

2εψ (6.128)

δψ = i√

2σµε∂µφ+√

2εF (6.129)

δF =√

2i∂µ (εσµψ) (6.130)

Again we see that fermions go to bosons and bosons go to fermions under SUSY trans-formations. Notice

• δF is just a total derivative. We will need to keep this in mind when we discussLagrangians.

• Products of chiral superfields are also chiral superfields since,

Dα (Φ1Φ2) = (DαΦ1) Φ2 + Φ1 (DαΦ2) = 0 (6.131)

In general any (holomorphic) function of a chiral superfield, f(Φ), will be a chiralsuperfield since,

Dαf(Φ) =∂f(Φ)

∂Φ· DαΦ = 0 (6.132)

6.4. VECTOR SUPERFIELDS 53

• If Φ is chiral, then Φ = Φ† is anti-chiral (which obeys DαΦ = 0)

• ΦΦ†,Φ + Φ† are superfields but don’t have any particle chirality.

6.4 Vector Superfields

For vector superfields we impose the condition,

V (x, θ, θ) = V †(x, θ, θ

)(6.133)

We now write the general vector superfield (this will be a nice exercise for your hand.),

V (x, θ, θ) = C(x) + iθχ(x)− iθχ+i

2θθ (M(x) + iN(x))

− i

2θθ (M(x)− iN(x)) + θσµθVµ(x) + iθθθ

(λ(x) +

i

2σµ∂µχ(x)

)− iθθθ

(λ(x) +

i

2σµ∂µχ(x)

)+

1

2θθθθ

(D − 1

2C

)(6.134)

The important thing about this superfield is it has components C,M,N,D, Vµ real bosoniccomponenets which gives 1 + 1 + 1 + 1 + 4 bosonic components. You have χα, λα complexfermionic components which gives 4+4 components. The field is called a vector superfieldsince it includes the Vµ component. This is the relevant part.

Consider some chiral superfield Λ. The combination i(Λ− Λ†

)is a vector superfield

since i(Λ− Λ†

)is real. Comparing components with the general form of the vector

superfield, (Exercise)

C = i(φ− φ†

)(6.135)

χ =√

2ψ (6.136)

1

2(M + iN) = F (6.137)

Vµ = −∂µ(φ+ φ†

)(6.138)

λ = D = 0 (6.139)

We can use this as a way to define a transformation on vector fields. This will let usgeneralize the standard gauge transformations. Recall that under a gauge transformationa vector transforms as,

Aµ(x)gauge−−−→ Aµ(x)− ∂µα(x) (6.140)

The generalized gauge transformation is

V (x, θ, θ)→ V (x, θ, θ) + i(Λ(x, θ, θ)− Λ†(x, θ, θ)

)(6.141)


From expression 6.138 we have that under the transformation above,

Vµ → Vµ − ∂µ(φ+ φ†

2

)(6.142)

= Vµ − ∂µ Re(φ)︸︷︷︸α

(6.143)

This induces a standard gauge transformation for the vector component of V . Λ isn’t aphysical quantity; it just a convenient superfield. So we can fix the gauge by choosing aparticular value for φ, ψ, and F within Λ to gauge away some of the components of V .This is what is usually called the Wess-Zumino gauge.

6.4.1 Wess-Zumino Gauge

In the Wess-Zumino gauge we have,

VWZ =(θσµθ

)Vµ(x) + iθθθλ(x)− iθθθλ(x) +

1

2θθθθD(x) (6.144)

We started with a standard vector superfield with all the different components. We thenshifted our field which gave the reduced version of V . We now see the prominance of thevector component, which is now the first component of the vector superfield. Vµ(x) willcontain all the gauge particles (photons, W±, Z, gluons). λ(x) will contain the gauginos(photino, Wino, Zino, gluino). D(x) is an extra component or auxillary field.

We will want to find the square of this superfield as well. Due to the anticommutingproperty of θ and θ we only need to keep the first term,

V 2WZ = θα (σµ)αβ θβθβ (σν)βα θαVµVν (6.145)

=1

4(θθ)

(θθ)

(σµ) ββ (σν)β

βVµVν (6.146)

=1

2(θθ)

(θθ)VµV

µ (6.147)

and clearly,V nWZ = 0 ∀n > 2 (6.148)

This is a nice property of using the Wess-Zumino gauge.Even though the Wess-Zumino gauge has some nice properties its important to note

the WZ gauge is not supersymmetric. That means that if we take VWZ and transform itunder supersymmetry it is no longer in the Wess Zumino gauge:

VWZSUSY−−−→ VWZ (6.149)

However, under a combination of SUSY and generalized gauge transformation we canalways do the following:

VWZSUSY−−−→→ VWZ

gen. gauge−−−−−−→ V ′′WZ (6.150)

6.5. TRANSFORMATIONS OF SUPERFIELDS 55

Under a standard gauge transformation we know how matter fields and vector fieldstransform. For example for a U(1) gauge the fields get a phase. We now try to see howsuperfields transform under gauge transformations.

6.5 Transformations of Superfields

6.5.1 Chiral Superfields

Lets first consider non-SUSY transformations. We can have a U(1) gauge:

φ(x)→ eiαqφ(x) (6.151)

Vµ(x)→ Vµ + ∂µα (6.152)

where α(x) is a parameter and q is the charge.

We know how vector superfields transform. We hypotheisize a form for the chiralsuperfield:

V → V − i

2

(Λ− Λ†

)(6.153)

Φ→ eiqΛΦ (6.154)

Since Λ is a chiral superfield the transformed Φ is also a chiral superfield.

We now want to define the field strength superfield; the analogue to (for U(1) gauge)Fµν ≡ ∂µVν − ∂νVµ. We define a new object

Wα ≡ −1

4D2DαV (6.155)

We now consider the properties of this object,

• This is the first superfield we’ve considered so far that carries an index.

• Wα is chiral

• Wα is invariant under generalized gauge transformation

These two properties are easy to show after writing Wα in components,

Wα(y, θ) = −iλα(y) + θαD(y)− i

2(σµσνθ)α Fµν + θθσµ

αβ∂µλ

β (6.156)

where Fµν ≡ ∂µVν − ∂νVµ. This is called the field strength superfield since one of itscomponents includes the field strength tensor from the SM.


6.A Derivatives and spinor charges

Above we made the following claim:

Dα,Qβ =Dα, Qβ

=Dα,Qβ

=Dα, Qβ

= 0 (6.157)

We prove these properties below:

Dα,Qβ =∂α + iσµ

αβθβ∂µ,−i∂β − σνβγ θγ∂ν

(6.158)

The derivative term trivially anticommute. The cross-terms anticommute as well since∂α, θ

β

= 0 and the rest of the factors just pull right out. The same occurs with the

the final term due to the anticommutator,θβ, θγ

= 0.

We now consider a less trivial example,Dα, Qα

=∂α + iσµ

αβθβ∂µ, i∂α + θβσµβα∂µ

(6.159)

The two derivative and θ, θ terms again trivially anticommute giving zero. The possibleconcern here are the cross terms. We have,

σµαβθβ∂µ, ∂α

= σµ

αβθβ∂µ∂α + σµ

αβδβα∂µ − σ

µ

αβθβ∂µ∂α = σµαα∂µ (6.160)

∂α, θβσµβα∂µ

= σµαα∂µ (6.161)

But these two terms have a relative minus sign in the anticommutator hence,Dα, Qα

= 0 (6.162)

By symmetry the other two anticommutators are also zero.

6.B Useful Relations

ψσµχ = ψασµααχα (6.163)

= (ψσµ)α χα (6.164)

= − (ψσµ)α χα (6.165)

= χα (σµ)αβ ψβ (6.166)

θαψβ = −1

2εαβθψ (6.167)

=1

2εβαψθ (6.168)

= −ψβθα (6.169)

6.B. USEFUL RELATIONS 57

So two contravariant spinors anticommute.

θσµθθσν θ =1

2θ2θ2ηµν (6.170)

Consider the transformation yµ = xµ + iθσµθ,

∂

∂θα=∂yµ

∂θα∂

∂yµ+

∂

∂θα= i(σµθ)α

∂

∂yµ+

∂

∂θα(6.171)

∂

∂θα=∂yµ

∂θα∂

∂yµ+

∂

∂θα= −i (θσµ)α

∂

∂yµ+

∂

∂θα(6.172)

∂

∂xµ=

∂

∂yµ(6.173)

Using these relationships we can find the covarient derivatives in these coordinates,

Dyα = ∂α + 2i (σµ)αβ θβ∂µ (6.174)

Dyα = −∂α (6.175)

Chapter 7

Four Dimensional SupersymmetricLagrangians

7.1 N = 1 Global Supersymmertry

We want to find couplings among superfields (Φ, V,Wα) which include particles of theSM. For this to work we need a way to build Lagrangians that are invariant up to a totalderivative under a SUSY transformation.

7.1.1 Chiral Superfields

To find an object L(Φ) such that δL is a total derivative under SUSY transformationswe use the fact that

• For a general superfield, S = ...+ (θθ)(θθ)D(x), the D term transforms as,

δD =i

2∂µ(εσµλ− ρσµε

)(7.1)

• For a chiral superfield, Φ = ...+ (θθ)F (x), the F term transforms as,

δF = ∂µ

(i√

2εσµψ)

(7.2)

Now recall that any holomorphic function of chiral superfield is also a chiral superfield.So we can write the most general Lagrangian with only chiral superfields as,

L = K(Φ,Φ†

) ∣∣∣∣D

+

(W (Φ)

∣∣∣∣F

+ h.c.

)(7.3)

where

∣∣∣∣D

refers to the pulling out only the D term of the corresponding superfield (without

θ, θ) and

∣∣∣∣F

refers to pulling out only the F term of the corresponding general superfield

58

7.1. N = 1 GLOBAL SUPERSYMMERTRY 59

(without θ, θ). The function, K is known as the Kahler potential and is a real functionof Φ and Φ†. The chiral superfield, W , is known as the superpotential.

As an example we consider a renormalizable Lagrangian for the chiral superfield. Thedimension of a superfield, Φ is the same as the dimension of its scalar component, φ.We want to get back to Lagrangians that we recognize from QFT. We take the of thefermionic component and scalar component to be the same as for the Klien Gordan andDirac equations,

[Φ] = [φ] = 1 , [ψ] = 3/2 (7.4)

From the expansion of the chiral superfield,

Φ = φ+√

2θψ + θθF + ... (7.5)

we have,

[θ] = −1

2, [F ] = 2 (7.6)

This already hints that F isn’t a standard bosonic field. Renormalizable Lagrangian haveterms such as,

∂µφ†∂µφ , m2 φ†φ︸︷︷︸

dim=2

, g(φ†φ)2︸︷︷︸

dim=4

(7.7)

Operators with dimension greater then 4 are not renormalizable. In the Kahler potentialthe D and F terms appear in the form,

K = ...+ θ2θ2KD (7.8)

W = ...+ (θθ)WF (7.9)

In this notation K

∣∣∣∣D

= KD). However, in this discussion we are allowed to have dimen-

sionful coefficients infront of our operators so we must have,

[KD] ≤ 4 (7.10)

[WF | ≤ 4 (7.11)

This requires

[K] ≤ 2 (7.12)

[W ] ≤ 3 (7.13)

Thus we cannot for example have terms of the form K = Φ†ΦΦ†Φ. Since K must be realthe only other form it can be is Φ† + Φ, however this leads the a total derivative in theLagranaian and so does not contribute. So in general we have,

K = Φ†Φ , W = α + λΦ +m2

2Φ2 +

g

3Φ3 (7.14)

60 CHAPTER 7. FOUR DIMENSIONAL SUPERSYMMETRIC LAGRANGIANS

The corresponding Lagrangian is known as the Wess Zumino Model,

L = Φ†Φ

∣∣∣∣D

+

((α + λΦ +

m2

2Φ2 +

g

3Φ3

) ∣∣∣∣F

+ h.c.

)(7.15)

We now expand the result in components. Recall the expression for the chiral superfield,

Φ = φ+√

2θψ + θθF + iθσµθ∂µφ−i√2

(θθ) ∂µψσµθ − 1

4(θθ)

(θθ)∂µ∂

µφ (7.16)

Φ† = φ∗ +√

2ψθ + θθF ∗ − iθσµθ∂µφ∗ +i√2

(θθ)θσµ∂µψ −

1

4(θθ)

(θθ)∂µ∂

µφ∗ (7.17)

We need to multiply the two superfields together and we keep only terms with a θ2θ2

coefficient (keeping in mind that θ3 = 0). This procedure gives,

K

∣∣∣∣D

= −1

4(φ∗∂µ∂

µφ+ φ∂µ∂µφ∗ − 2∂µφ∂µφ

∗) + FF ∗ − i

2

(−ψσµ∂µψ + ∂µψσ

µψ)

(7.18)

= ∂µφ∗∂µφ+ FF ∗ + i

(ψσµ∂µψ

)(7.19)

where we used σµαα = σµαα and θαθβ = −12εαβθ2. We now need to calculate the contribution

due to the superpotential.

α

∣∣∣∣F

= 0 (7.20)

λΦ

∣∣∣∣F

= λF (7.21)

m

2Φ2

∣∣∣∣F

= m

(−1

2ψ2 + Fφ

)(7.22)

g

3Φ3

∣∣∣∣F

= g(−ψ2φ+ Fφ2

)(7.23)

We have,

W

∣∣∣∣F

=(λ+mφ+ gφ2

)F − 1

2(m+ 2gφ)ψ2 (7.24)

Above we calculated the contributions to the Lagrangian explicitly. Alternatively wecan do an exact Taylor expansion due to the anticommuting nature of the Grassmanvariables,

W (Φ) = W (φ) + (Φ− φ)︸︷︷︸...+√

2θψ+θθF+...

∂W

∂φ+

1

2(Φ− φ)2︸︷︷︸

...+(θψ)(θψ)+...

∂2W

∂φ2(7.25)

Calculating the derivatives of W with respect the superfield evaluated at θ = θ = 0 (weexpanded around Φ = Φ):

∂W

∂φ= λ+mφ+ gφ2 (7.26)

∂2W

∂φ2= m+ 2gφ (7.27)


where we have defined∂W

∂φ≡ ∂W

∂Φ

∣∣∣∣Φ=φ

(7.28)

Thus we can write the full Lagrangian as,

L = ∂µφ∗∂µφ+ iψσµ∂µψ + FF ∗ +

(∂W

∂φF + h.c.

) ∣∣∣∣F

− 1

2

(∂2W

∂φ2ψψ + h.c.

) ∣∣∣∣F

(7.29)

The first and second terms are the standard scalar field and fermion kinetic terms respec-tively!

The part of the Lagrangian that depends on the auxiliary field F takes the form,

L(F ) = FF ∗ +∂W

∂φF +

∂W ∗

∂φ∗F ∗ (7.30)

Notice that this is quadratic and without any space-time derivatives. This means thatthe field F does not propagate. So we can easily eliminate F using the field equations

∂L(F )

∂F= F ∗ +

∂W

∂φ= 0 (7.31)

∂L(F )

∂F ∗= F +

∂W ∗

∂φ∗F ∗ = 0 (7.32)

which implies that

F = −∂W∗

∂φ∗(7.33)

Substituting the result back in the Lagrnagian gives

L(F ) =

∣∣∣∣∂W∂φ∣∣∣∣2 − 2

∣∣∣∣∂W∂φ∣∣∣∣2 = −

∣∣∣∣∂W∂φ∣∣∣∣2 ≡ −V(F )(φ) (7.34)

This defines the scalar potential since it is the part in the Lagrangian that only dependson the scalar field, φ. From this expression we can easily see that it is a positive definitescalar potential,

V (φ) ≥ 0 (7.35)

In summary we,

L = ∂µφ∗∂µφ+ iψσµ∂µψ −1

2

(∂2W

∂φ2ψψ + h.c.

)− V (φ, φ∗) (7.36)

Remarks:

• This Lagrangian with N = 1 is a particular case of a standard N = 0 Lagrangian.

• This is a theory of a scalar field and a Majorana fermion since the fermion kineticterms are iψσµ∂ψ, and mass terms are m2

(ψψ + ψψ

).


• Suppose you have a W of the form,

W =m

2Φ2 +

g

3Φ3 (7.37)

The scalar field mass term is ∣∣∣∣∂W∂Φ

∣∣∣∣Φ=φ

= m2 |φ|2 (7.38)

The fermionic mass term is

1

2

∂2W

∂φ2ψψ + h.c. =

1

2mψψ + h.c. (7.39)

(we will go through this minimization process more carefully soon) so the massof the fermionic and bosonic fields are determined by the same parameter m.[Q8: Why is the fermionic mass term ψψ and not ψLψR + h.c.?]

• The Yukawa coupling (φψψ) is equal to g. However, g is also the scalar self-coupling(∼ g2 |φ|4).

The Yukawa coupling equally the self-coupling is the source of what is called the “mirac-ulous cancellations in supersymmetry pertrubation theory. Diagarams that look com-pletely different in non-supersymmetry theories have the same coupling in SUSY. Forexample we have,

+φ φ

φ

g2

φ φ

ψ

ψ

g g

If the couplings were arbitrary then you would get completely different contributions fromeach diagram and they each diverge. If the couplings match up then instead they addup to give you a finite contribution. This is the origin to the idea that supersymmetrictheories have much better quantum behavior then standard theories. If this field is theHiggs field then this explains how SUSY can help solve the hierarchy problem.

In general we don’t need to stick to Kahler potentials of the type that we considered.You can have many scalar fields or a non-renormalizable potential. We consider theformer genearlization,

K(Φi,Φj,†) , W (Φi) (7.40)

As we did for W we can expand K around Φi = φi. This gives in components (Ex),

K(Φi,Φj,†) ∣∣∣∣

D

=∂2K

∂φi∂φj∂µφ

i∂µφ∗j (7.41)


The kinetic terms are no longer the standard kinetic terms. This is usually written as

Kij∗∂µφi∂µφ∗j (7.42)

Kij∗ is a metric in the space with coordinates φi which is a complex Kahler-manifold,

gij∗ = Kij∗ (7.43)

This is often called a non-linear sigma model for historical reasons. A Kahler manifoldis that the metric is given in terms of a potential. Not any metric can be written as asecond derivative of an object. This is a property of Kahler manifolds.

7.1.2 Miraculous Calculation in Detail

Shifting the superfields we can reduce the most general superpotential to a form of,

m

2Φ2 +

g

3Φ3 (7.44)

To do a calculation we want to write out the Lagrangian in detail in terms of only thereal fields. We define,

φ ≡ φ1 + iφ2√2

(7.45)

Explicitly we have,

V (φ) =(mφ+ gφ2

) (mφ∗ + gφ∗2

)(7.46)

= m2 |φ|2 +mg(φφ∗2 + φ∗φ2

)+ g2 |φ|4 (7.47)

=m2

2

(φ2

1 + φ22

)+mg√

2

(φ3

1 + φ1φ22

)+g2

4

(φ4

1 + φ42 + 2φ2

1φ22

)(7.48)

The fermion mass/Yukawa term is,

1

2

(∂2W

∂φ2

∣∣∣∣D

+ h.c.

)=

1

2

(m+ 2gφ)ψ2 + (m+ 2gφ∗) ψ2

(7.49)

=1

2

m(ψ2 + ψ2) + 2g (φ1 + iφ2)ψ2 + 2g (φ1 − iφ2) ψ2

(7.50)

Now since we have a Majorana fermion we can make the definition,

Ψ ≡(ψψ

)(7.51)

which gives

1

2

(∂2W

∂φ2

∣∣∣∣D

+ h.c.

)=

1

2

mΨΨ + 2gΨ

(φ1 − iγ5φ2

)Ψ

(7.52)


So we can write the full Lagrangian as 1,

L =1

2∂µφi∂µφi −

m2

2φiφi +

1

2Ψ(i/∂ −m

)Ψ− mg√

2

(φ3

1 + φ1φ22

)− g2

4

(φ4

1 + φ42 + 2φ2

1φ22

)+ gΨ

(φ1 − iγ5φ2

)Ψ (7.53)

Our goal is now to calculate the φ1 propagator to second order. We have (we denoteφ1 with a blue dashed line and φ2 with a red dashed line),

= + + +

+ +

The first two loop diagrams are logarithmically divergent and give natural contributionsto the propagator. The third loop has an amplitude of,

iM =

∫d4`

(2π)4

1

Si(−ig2)

`2 −m2 + iε(7.54)

where S = 1/3 is the symmetry factor.The fourth loop gives the same amplitude and with the same sign and a symmetry

factor of 1.The final fermionic loop has an amplitude of

iM =

fermionic loop↑−1

2

∫d4`

(2π)4

1

S(ig)2Tr

i(/−m)

`2 −m2 + iε

i(/p− /+m)

(`− p)2 −m2 + iε(7.55)

with a symmetry factor of S = 1 and the factor of 1/2 infront is to have a propernormalization (we have an extra 1/2 in the Lagrangian infront the Dirac ).

The trace is

Tr[(/−m)(/− /p+m)

]= −4` · p+ 4`2 + 4m2 (7.56)

= 2((`− p)2 − `2 −m2

)+ `42 + 4m2 (7.57)

= 2(p− `)2 + 2`2 + 2m2 (7.58)

= 2((p− `)2 −m2) + 2(`2 −m2) + 6m2 (7.59)

1If the factor in front the of Dirac kinetic Lagrangian worries you its because we are dealing with aMajorana field.


and so we can write the amplitude as,

iM = −2g2

∫d4`

(2π)4

1

`2 −m2 + iε+

1

(`− p)2 −m2 + iε

+3m2

((`− p)2 −m2 + iε)(`2 −m2 + iε)

(7.60)

Adding up the quadratically divergent one loop contributions we have an amplitude of,

g2

∫d4`

(2π)4

1

`2 −m2 + iε− 1

(`− p)2 −m2 + iε− 6m2

((`− p)2 −m2 + iε)(`2 −m2 + iε)

(7.61)

Notice the the first two contributions cancel in the high energy limit! We have lostthe quadratic divergence. The diagrams scales logarithmically instead giving naturalcorrections to the Higg’s mass.

7.1.3 Vector Superfields

Thus far our discussion was for chiral scalar fields. We now move onto vector (scalar)superfields. We want to relate the couplings of Φi, V,Wα. We first recall what we do innon-SUSY models. We have a kinetic term for instance of the form, ∂µφ∂

µφ† and therewas a symmetry, φ → eiαqφ. If α = α(x) then the kinetic term is no long invariant. Wehave steps,

(a) We introduce the convariant derivative,

Dµφ = ∂µφ− iqAµφ (7.62)

where Aµ is a field that transforms as

Aµ → Aµ + ∂µα (7.63)

Our Lagrangian now takes the form,

L = DµφDµφ∗ + ... (7.64)

and this defines how the gauge fields couple to the scalar for instance.

(b) We then add a kinetic term for Aµ. For a U(1) gauge boson,

− 1

4FµνF

µν (7.65)

where Fµν = ∂µAν − ∂νAµ.

We now follow these steps for supersymmetry only now instead of promoting globaltransformations to local transformations we force supersymmetry invariance.


(a) The kinetic terms arise from the Kahler potential, K = Φ†Φ, and we want to see howit transforms. This term will not be invariant under the transformation (Λ is somechiral superfield),

Φ→ eiqΛΦ (7.66)

because

Φ†Φ→ Φeiq(Λ−Λ†)Φ (7.67)

so in general if Λ is a superfield this will not be invariant. The solution is to introducea gauge field such that the Kahler term is invariant. We introduce V such that

K = Φ†e2qV Φ (7.68)

where V transforms as,

V → V − i

2

(Λ− Λ†

)(7.69)

With this K is invariant since,

K → Φ†e−iqΛ†e2q(V− i

2(Λ−Λ†))eiqΛΦ = Φ†e2qV Φ (7.70)

This is a generalized gauge transformation.

The derivatives in non-SUSY theories tell us how ordinary fields couple to gaugefields. In the same way this prescription will tell us how chiral superfields couple tovector superfields.

(b) We now need to add a kinetic term for V . Recall that Wα is a chiral superfield (notthe completely unrelated superpotential!) given by,

Wα(y, θ) = −iλα(y) + θαD(y)− i

2(σµσνθ)α Fµν + θθσµ

αβ∂µλ

β (7.71)

Lkin =

coupling↑τWαWα

∣∣∣∣F

(7.72)

Since Wα is a chiral superfield so is WαWα. Thus to make this into a Lagrangianwe were forced to take the F term of it. We chose this particular term because aswe saw in an earlier lecture Wα has an Fµν term embedded in it. WαWα is going togeneralize the FµνF

µν that we had earlier.

If we want this term to be renormalizable then τ needs to be a constant. In a moregeneral case we can have it as a function of the chiral fields,

Lkin = [f(Φ)WαWα]

∣∣∣∣F

+ h.c. (7.73)


where we have chosen the coefficient to be a function only of a chiral field Φ since wewant the Lagrangian to be able to take the F term of the object in square bracketswhich will only hold if the term is a chiral field2. f(Φ) is known as the gauge kineticfunction.

(c) We followed the same steps as we did for non-supersymmetric case. Introduced agauge field to enforce gauge invariance and then add a kinetic term. However, in thesupersymmetric case there is something else that we can add that is allowed by allthe symmetries of the theory. This term is known as the Fayet Iliopoulos term:

LFI = ξV

∣∣∣∣D

(7.74)

where ξ is a constant. This term is only present for U(1) gauge theories since forthese theories the gauge field is not charged under U(1) (the photon is chargeless). [Q9: Doesn’t V transform as above?]For non-abelian gauge theories the gauge fields (andtheir corresponding D terms) would transform under the gauge group and thereforehave to be forbidden.

These are all the ingredients of supersymmetric-QED. The renormalizable Lagrangianwill be,

L =(Φ†e2qV Φ

) ∣∣∣∣D

+

(W (Φ)

∣∣∣∣F

+ h.c.

)+

(τWαWα

∣∣∣∣F

+ h.c.

)+ ξV

∣∣∣∣D

(7.75)

The superpotential is quite restricted since it must be invariant under supersymmetricand gauge transformations. If we have a single superfield Φ charged under U(1) then weare forced to have W = 0 since otherwise it wouldn’t be gauge invariant (W (φ) doesn’tinvolve Φ†, only powers of Φ).

As we did for the case of chiral superfields we now write this out in terms of compo-nents. We do this piece by piece [Q 10: Show this!] (we quote the component form andleave the derivations as exercises).

1.

Φ†e2qV Φ

∣∣∣∣D

= FF ∗ + ∂µφ∂µφ∗ + iψσµ∂µψ

+ qV µ

(1

2ψσµψ +

i

2φ∗∂µφ−

i

2φ∂µφ

∗)

+i√2q(φλψ − φ∗λψ

)+q

2

(D +

1

2VµV

µ

)|φ|2 (7.76)

where to derive this expression we have used the Wess Zumino gauge (though thefull expression is gauge invariant so this is a general result). In this gauge V 2 is

2Recall that a function of a chiral superfield is a chiral superfield.


simple and V 3 = 0 so the exponential is cut off at second order and is somewhatmanageable. Notice that this Lagrangian is precisely what we expected for QED.We could have written it in a shorter way but we want to be explicit. It has thestandard kinetic terms for a scalar and fermion field. This is a gauge theory so itmust come out in terms of covariant derivatives. Compare this with,

DµφDµφ∗ =

(∂µ −

i

2qVµ

)φ

(∂µ +

i

2qVµ

)φ∗ (7.77)

= ∂µφ∂µφ∗ − iq

2Vµ (φ∂µφ

∗ − ∂µφφ∗) +1

4q2VµV

µ |φ|2 (7.78)

and

iψσµDµψ = iψσµ(∂µ −

i

2qVµ

)ψ (7.79)

= iψσµ∂µψ +q

2ψσµVµψ (7.80)

so we have,

Φ†e2qV Φ

∣∣∣∣D

= DµΦDµΦ∗ + iψσµDµψ +

New interactions due to SUSY︷︸︸︷i√2q(φλψ − φ∗λψ

)+ FF ∗ +

q

2D |φ|2

(7.81)

It is no coincidence that we recovered our old interactions from non-supersymmetrySUSY. We imposed the Lagrangian to be gauge invariant. By doing this we arerestricting ourselves to a very small number of possible Lagrangian terms.

2. We considered the special case of one chiral superfield,Φ, and hence W (Φ) = 0. Inany case we won’t see any new interactions in this case over before since it doesn’tinvolve any gauge field interactions.

3. Next we consider the term index superfield term,

WαWα

∣∣∣∣F

+ h.c. =1

2D2 − 1

4FµνF

µν + iλσµ∂µλ−i

8FµνF

µν (7.82)

where,Fµν = εµνρσF

ρσ (7.83)

• As we hoped we have a kinetic term for the gauge boson, 14FµνF

µν .

• We have a Dirac kinetic term for the λ field. We associate these fields withthe gauginos.

• The D field only appears on its own and doesn’t propagate in spacetime. It isjust an auxillary field (analogous to F for the chiral field).

7.2. ACTION AS A SUPERSPACE INTEGRAL 69

• The last term ∝ FµνFµν can be shown to be equal to a total derivative(this is

a common exercise in an electromagnetism class as it shows that the Maxwellkinetic term is unique. We should point out though that while this is true inabelian gauge theory, this doesn’t hold in non-abelian gauge theory and thisterm because important an QCD.

4. The last term we need to consider is

ξV

∣∣∣∣D

= ξD (7.84)

As in the case of the chiral superfield we eliminate the auxillary field. The D dependentpart of L is

L(D) =q

2D |φ|2 +

1

2D2 + ξD (7.85)

So the field equation gives,

D = −q2|φ|2 − ξ (7.86)

We can now substitute this back into the D part of the Lagrangian,

L(D) =(q

2|φ|2 + ξ

)D +

1

2D2 (7.87)

= −1

2

(q2|φ|2 + ξ

)2

(7.88)

= −V(D) (7.89)

(we denote scalar potential by V to avoid confusion with the vector superfield)This is apart of the potential of the field φ as it has no spacetime derivatives. In general the totalscalar potential is

V = V(F ) + V(D) (7.90)

=

∣∣∣∣∂W∂φ∣∣∣∣2 +

1

2

(ξ +

1

2q |φ|2

)2

(7.91)

≥ 0 (7.92)

7.2 Action as a Superspace Integral

The Lagrangian we have written can alternatively be written as an integral over super-space. Recall that without supersymmetry we have,

S =

∫d4xL (7.93)


Recall that ∫d2θθ = 1 (7.94)∫d4θθ2θ2 = 1 (7.95)

and any other combination will be zero. We previously wrote the Lagrangian as

L = K

∣∣∣∣D

+

(W

∣∣∣∣F

+ h.c.

)+

(WαWα

∣∣∣∣F

+ h.c.

)(7.96)

Notice that the D term is precisely the term with θ2θ2 and the F term has θ2. So we canwrite this as

L =

∫d4θK +

(∫d2θW + h.c.

)+

(∫d2θWαWα + h.c.

)(7.97)

with the understanding that we set θ = 0 in the superpotential.There is a big difference between the Kahler potetial and the other two terms since

it has an integral over all of superspace while the other two terms have integrals of halfof superspace.

7.3 Non-abelian generalization

Thus far we have only discussed abelian gauge theories. We breifly discuss the non-abeliancase. In this case we have,

eV → e−iΛ†eV eiΛ (7.98)

We still have,Φ→ eiΛΦ (7.99)

however we have,

Wα = −1

8

(DD

) (e−2VDαe2V

)(7.100)

andW ′α = e−2iΛWαe

2iΛ (7.101)

and WαWα turns intoTrWαWα (7.102)

We will not use this formailsm but include it for completeness.The most general action is:

S[K(

Φ†ieqV Φi

),W (Φ), f (Φi) , ξ

]=

∫d4xd4θ (K + ξV ) +

∫d4xd2θ [W + f (Φi)W

αWα + h.c.] (7.103)

where ξ is only non-zero in U(1). Given K,W, f, and ξ our supersymmetric theory isdetermined.

7.4. NON-RENORMALIZATION THEOREMS 71

7.4 Non-Renormalization Theorems

We know that the whole Lagrangian will depend on Kahler potential, super potential,gauge kinetic function and Fayet Iliopoulos term we want to see how they behave underquantum corrections. In other words how K,W, f, ξ behave under quantum corrections.The claim is that

• K(Φ,Φ†

)gets corrections order by order in perturbation theory

• f(Φ) only gets corrections are one loop order

• W (Φ) is nonrenormalizable at any order in perturbation theory. This doesn’t meanthat it doesn’t get corrected. There are corrections but they are non perturbative.

• ξ also doesn’t get corrected (though not as peculiar as ξ has no field dependence)

This was first proved in 1977 by Grisaru, Rocek, Seigel using what’s known as “super-graphs”. These are generalization of Feynman diagrams to superfields. The result thatthey found was that quantum corrections only come as∫

d4xd4θ ... (7.104)

except the one loop corrections of f (remember that except for the Kahler potential eveyterm only has integrals over half of superspace). In 1993 Seiberg(based on string theoryarguements from Witten) was able to prove these results using symmetry arguementsand holomorphicity. We use the second technique to prove the results3.

We can write the action explicitly by,

S =

∫d4xd4θ

[K(Φ,Φ†, eV

)+ ξVU(1)

]+

∫d4xd2θ [W (Φ) + f (Φ)WαWα + h.c.]

(7.105)where we write VU(1) to remember that we only include this extra term in the special caseof a U(1) symmetry.

We introduce two chiral superfields fields called “spurion” fields, X and Y each withcomponents,

X = (x, ψx, Fx) , Y = (y, ψy, Fy) (7.106)

These fields will not propagate however we will use these fields to mimic something thatoccurs in string theory. In string theory there are usually no arbitrary parameters. Allthe parameters such as the Yukawa couplings just come from expectation values of fields.

In this regard we write the action as,

S =

∫d4xd4θ

(K + ξVU(1)

)+

∫d4x

∫d2θ (Y W (Φ) +XWαWα + h.c.) (7.107)

3This arguement follows Wienberg 27.8.


so X is going to mimic the gauge coupling and Y is substituting the Yukawa couplingsand self-couplings of the matter fields by self-couplings of Y . If we set Y = 0 then we arekilling all couplings of the matter fields.

To prove the non-renormalization theorems we use

(i) Symmetries

(ii) Holomorphicity

(iii) Limits of X →∞ and Y → 0.

Symmetries

In our Lagrangian we have a few symmetries

1. The standard symmetries: The gauge symmetries and supersymmetry.

2. We also have an R-symmetry. Recall that this is the name we give for any internalsymmetry that does not commute with supersymmetry. An R symmetry exists forall supersymmetries. In the case of N = 1 this is a U(1) symmetry. This impliesthat different SUSY multiplets don’t have the same charge. You assign the chargesof the superfield corresponding to the charge of the first component of the superfield.However, its important to note that θ is also charged under this symmetry. Thisis because each multiplet must transform differently. [Q 11: Why do the Grassmanvariables need to be charged, I don’t see the connection?]

The charges are as follows,

field Φ V X Y θ θ Wα

U(1)R − charge 0 0 0 2 −1 1 1

The reasoning as follows. Suppose Y has charge 2 (we chose this arbitrarily). Nextsuppose that we give θ a charge of −1. Then the scalar component of Y has charge2, the fermion component has charge 1 and so on. Wα is not an independentsuperfield but comes from V (recall that Wα ∼ D2DV ). Each D behaives like a θ.So the charge for Wα is 2− 1 + 0 = 1.

Now we can see why this assignment is a good assignment for the symmetry. Theterm ∫

d4xd4θ[K + ξVU(1)

](7.108)

is invariant under U(1)R transformations.

W (Φ) is invariant because Φ has charge 0 but d2θ is not invariant and now we seewhy we chose Y to have the particular R charge. We chose it such that d2θY W (Φ)will be invariant.

Wα has charge 1. So the charge of d2θXWαWα is −2 + 1 + 1 = 0. So this is asymmetry of our Lagrangian.

7.4. NON-RENORMALIZATION THEOREMS 73

This symmetry did not exist before. However, by adding the spurion fields X andY we added these symemtries.

3. We also have another symmetry which is called Peccei-Quinn(PQ) symmetry. Thesymmetry is that every field is invariant except that

X → X + ir (7.109)

where r ∈ R. This is a symmetry because XWαWα involves terms like

Re (x)FµνFµν + Im (x)FµνF

µν (7.110)

where x is the scalar component of the spurion field. Shifting the imaginary part ofX by a constant just changes the coefficient of FµνF

µν which is a total derivative.x is called an axion field.

Holomorphicity

Consider the quantum-correction action (Wilsonian action). The action associated withintegrating all momentum greater then some cutoff in the path integral, such that,

exp (Sλ) =

∫|p|>λ

∫Dφ exp (iS) (7.111)

We have,

Sλ =

∫d4x

∫d4θ[J(Φ,Φ†, eV , X, Y,DΦ, ...

)+ ξ

(X,X†, Y, Y †

)VU(1)

]+

∫d4x

∫d2θ [H (Φ, X, Y,Wα)] + h.c. (7.112)

where H is an unknown holomorphic function. The U(1)R invariance will help us con-strain terms in the Lagrangian.

For J(Φ,Φ†, eV , X, Y,D, ...

)the only thing we know is that we have is that Y appears

together with Y † to any possible power. So we don’t have very many restrictions.However for H we do have restrictions. The only object that transform under R-

symmetry are Y (charge 2) and Wα (charge 1). The charge of H must be 2. This Hmust have the form,

H = Y h (X,Φ) + g (X,Φ)WαWα (7.113)

From the PQ symmetry we know that the only thing that transforms is X and ittransforms as a shift. Thus any power of X will not be invariant. We must haveh (X,Φ) = h (Φ). Terms linear in X only contribute a total derivative (per our dis-cussion above) so we have g (X,Φ) = g (Φ) + αX where α is an unknown constant.[Q12: Can α be a function of φ?] We have,

H = Y h(Φ) + (αX + g(φ))WαWα (7.114)

Notice that we have significantly reduced the possibile dependencies. This is starting tolook very similar to the Lagrangian we started with.


Limits

Now we can use the last tool in our repertoire which is the limits, X →∞, and Y → 0.In the Y → 0 limit we have,

Y h(Φ)→ 0 (7.115)

Thus the high energy action has no effect on the real action to any order in perturbationtheory. Which is implies that W is not renormalized! [Q 13: Why?]

The same thing can be done for X. The loop-corrected gauge kinetic function takesthe form,

f(Φ) = αX + g(Φ) (7.116)

Note that the gauge kinetic terms go as xFµFµ ∼ x∂[µAν]∂[µAν]. Each ∂µ gets Fourier

transformed into a 1/x and thus the propagator goes like

∼ 1

x(7.117)

Gauge self couplings go as X3 [Q 14: Why?],

We could the number of x powers in any diagram. It is given by

Nx = VW − IW (7.118)

We also know that the number of loops L goes like

L = IW − VW + 1 (7.119)

Combining these results we haveNx = 1− L (7.120)

For L = 0 (tree level), Nx = 1. The power of x is equal to one and hence αX is the treelevel result (α = 1). For L = 1 we have Nx = 0 which gives the term g(Φ). Thereforethe gauge kinetic term,X + g(Φ) is corrected only at 1 loop.

The Kahler potential, being non-holomorphic, is corrected to all orders. In generalwe also have a FI-term, (

ξ(X,X†, Y, Y †

)VU(1)

) ∣∣∣∣D

(7.121)

Gauge invariance: V → i2

(Λ− Λ†

)implies that ξ must be a constant. Any dependence

on the chiral superfields will destroy this invariance. However, this is not necessarily thesame constant that we started with. ξ can only get contributions from diagrams knownas tadpole diagrams,

7.5. N = 2, 4 GLOBAL SUSY 75

But these diagrams happen to be proportional to all the charges,∑i

qi = Tr(QU(1)

)(7.122)

But if TrQ 6= 0 the theory is inconsistent in the sence that it has what is called “gravi-tational anomalies”. These occur when you have couplings of the type,

gµν

ψαAµ

This diagram is proportional to Tr (Q). If this is different from zero then it tells you thatyou are breaking gauge invariance. If there are no gravitational anomalies then the FIterm is not corrected.

7.5 N = 2, 4 Global SUSY

Recall that for N = 1 the action is dependent on three arbitrary functions and oneparameter,

S = S [K,W, f ; ξ] (7.123)

The most general Lagrangian has N = 0 as in the SM. N = 1 is just a special case ofN = 0. On the same note extended SUSY theories (N > 1) are special cases of N = 1models. Thus have a larger N puts restrictions on the dependencies above.

For N = 2 vector multiplet,

Aµ

λ λ

φ

where the states in blue are a N = 1 vector multiplet(so it comes form an N = 1 vectorsuperfield, V ) and the states in red come from a chiral N = 1 chiral superfield, Φ.

It so happens that this particular case is the simplest. You can write the N = 2action in terms of N = 1 actions. It turns out that W = 0, and K, f can be written interms of a single holomorphic function, F(Φ). This function is called the prepotential.


This is good news since in N = 1 we had no information about K. However, now K willbe derived from a holomorphic function, F . In particular,

f(Φ) =d2FdΦ2

(7.124)

K(Φ,Φ†

)=

1

2i

[Φ†e2V d∂F

∂Φ− h.c.

](7.125)

Full perturbative action is just given in one loop,

F =

Φ2

Φ2 log Φ2

Λ2

(7.126)

where Λ denotes some cutoff. However, we could have some non-perturbative corrections.So in general we have,

F(Φ) = F1 loop + Fnon-pert (7.127)

Perturbative means that F1 loop will be a series in the powers of the couplings∑

n gnan.

Non-perturbative means that something cannot be written in this form. For example∑Ane

−an/g2 since we know this function and all its derivatives vanish at zero. Youcannot get any information about F from Fnon-pert since the Taylor expansion vanshesat zero. Non-perturbative effects can be important and in general need to be understood.

There are also vector and hypermultiplets. These are in general more compilcated.We now consider N = 4. We can write the vector multiplet as, Aµ

xl ψ1

φ1

︸︷︷︸N=2 vector

+

φ2

ψ3 ψ2

φ3

N=2 hyper

The effective action does not depend on any arbitrary functions. There is only one freeparamters,

f = τ =Θ

2π+

4πi

g2(7.128)

This means that Reτ multiplies FµνFµν and Imτ is the coefficient of F µνFµν . This is the

only free parameter in a N = 4 theory.Notice,

• N = r is not only renormalizable but finite

• The β function follows,

β(g) = M2 dg

M2(7.129)

where M2 is the renormalization scale. For N = 1 the β function is

bg3 (7.130)

Depending on the value of b the coupling increases or decreases,

7.6. SUPERGRAVITY 77

E

g

For N = 4 β = 0 and the coupling does not change with energy. This means thatwe have what’s called a conformal field theory. This theory is very well behaved.

• There is something called S-duality which essentially takes

τ → aτ + b

cτ + d(7.131)

where a, b, c, d ∈ Z form what is called a SL(2,Z) group. For particular choices ofthe coefficients,

τ → 1

τ(7.132)

Remember that the imaginary part of τ is the gauge coupling so your changingthe coupilng from weak to strong. This is one of the beautiful properties of thesetheories; you can control the strong coupling from the strong coupling.

• N = 4 has an AdS/CFG duality to a string theory in 5 dimensions.

7.6 Supergravity

We know that in gauge theories that if we set a field φ such that

φ→ eiαφ (7.133)

and α goes from a constant to α(x) then we get a gauge theory. In SUSY we know that

δS = i(εQ+ εQ

)S (7.134)

What happens if we make ε a function of spacetime coordinates? SUSY is a part of awhole spacetime symmetry. If we make ε a function of x then we also change the structureof spacetime. If we make Poincare local we get gravity. If we make SUSY local then weget supergravity.

Gravity is determined by a correcponding metric which can be written in terms oftetrad fields, eaµ which are connected to the metric through

gµν = eaµebνηab (7.135)


In supergravity we have a field, ψµ, called the gravitino (one gravity for each super-symetry). In this sence gravitino is the gauge field of supergravity.

The supergravity action is

SSG =

∫d4x√gR +

(ψµσνDρψσ − ψµσνDρψσ

)εµνρσ + ...

(7.136)

where R is the Einstein-Hilbert term and the second term is called a Rarita-Schwingerterm.

The total action will beS = SSG + S [K,W, f ; ξ] (7.137)

This is similar for global supersymmetry except there are extras:

• Kahler symmetry:

K → K(φ, φ†

)+ h(φ) + h∗ (φ∗) (7.138)

W → eh(φ)W (7.139)

S depends ong = K − log |W |2 (7.140)

• The F-term potential is:

VF−term = eM/M2p

K−1ij∗DiWDj∗W

∗ − 3|W |2

M4p

(7.141)

where

DiW = ∂iW + (∂iK)W

M2p

(7.142)

This is called the Kahler covariant derivative.

Notice that when we send Mp → ∞ we reduce to the standard supersymmetriccase. Further notice that VF−term is not longer positive as it was in supersymmetry.

Chapter 8

Supersymmetry Breaking

8.1 Basics

Recall that when we talk about symmetry breaking in gauge theories we have a field φthat transforms as,

φ→(eiα

aTa) jiφj (8.1)

such that

δφi = iαa (T a) ji φj (8.2)

The symmetry is broken if δφi 6= 0 for the vacuum state. In other words in the vacuumthe field is not invariant under the transformation. Another way to say it is

(αaT a) ji φj (vac) 6= 0 (8.3)

The typical example for the U(1) case is for

φ = ρeiθ (8.4)

then under a transformation eiα,

φ′ = ρeiαeiθ (8.5)

which gives

δρ = 0 (8.6)

δθ = α (8.7)

Now suppose you have a potential of the form,

79

80 CHAPTER 8. SUPERSYMMETRY BREAKING

Any transformation changes the angle θ but leaves the modulus invariant. Taking thestate from vacua to vacua. This field α is the Goldstone boson.

We now move onto supersymmetry. Supersymmetry is broken if the vacuum state(|vac〉) is transformed under supersymmetry, i.e.

Qα |vac〉 6= 0 (8.8)

Let’s consider, Qα, Qβ

= 2 (σµ)αβ Pµ (8.9)

Suppose we multiply both sides by (σν)βα:

(σν)βαQα, Qβ

= 2 (σν)βα (σµ)αβ Pµ (8.10)

= 4P ν (8.11)

Now consider ν = 0. This implies that we have,Qα, Q†α

= 4P 0 → 4E (8.12)

where we have defined Qα,† ≡ (σ0)αβQβ.

Note,

1. The left hand side of our expression is positive definite. Acting with this operatoron any state says that that the energy of that state must be positive. This is aresult we also saw earlier where we found the scalar potential was positive.

2. We can sandwich this expression between the vacuum state:

〈vac|QαQ†α +Q†,αQα|vac〉 ≥ 0 (8.13)

If supersymmetry is broken then we have Qα |vac〉 = 0 and the energy is zero. ThusSUSY being broken is equivalent to having energy of the vacuum greater then zero,

SUSY⇔ E > 0 (8.14)

The energy is the order parameter of supersymmetry breaking in the same way thatthe vev was order parameter of gauge symmetry breaking.

8.2. F AND D BREAKING 81

8.2 F and D Breaking

8.2.1 F -Term

Suppose we have a chiral superfield. We know that the transformation of the supersym-metry of each of the components is of the form,

δφ =√

2εψ (8.15)

δψ =√

2εF +√

2iσµε∂µφ (8.16)

δF = i√

2εσµ∂µψ (8.17)

We want to look on the right hand side if any terms are different from zero for the vacuumthen we know we have supersymmetry breaking. [Q 15: Why do we not care about totalderivatives here?]As we know a scalar can get a vacuum expectation value. However, afermion could not get a VEV as then under Lorentz transforms the VEV itself wouldalso transform, breaking Lorentz invariance. So you cannot have ψ getting a VEV. Wecannot have δφ 6= 0 on the vacuum as this would require,

〈vac|ψ|vac〉 6= 0 (8.18)

Simiarly we cannot have δF 6= 0 as this would too require a fermionic VEV. This onlyleaves δψ. However, ∂µφ is also a vector and cannot attain a VEV by Lorentz invariance.The only term that can get a VEV without breaking Lorentz invariance is the F term inδψ. So we can have SUSY breaking if

〈vac|F |vac〉 6= 0 (8.19)

and we have

〈vac|δφ|vac〉 = 〈vac|δF |vac〉 = 0 (8.20)

〈vac|δψ|vac〉 =√

2ε 〈vac|F |vac〉 6= 0 (8.21)

Recall that in Gauge symmetry breaking ρ was invariant and only θ transformed andδθ corresponded to a Goldstone boson (a scalar). We have an analogous situation here.Carrying this analogy we call ψ as a Goldstone-fermion or a “goldstino”. Note that thegoldstino is not the supersymmetric partner of a Goldstone boson. It is just the fermionthat transforms non-trivially under supersymmetry.

Remember that we were constructing the corresponding Lagrangian we have foundthe the F term of the scalar potential was equal to

VF−Term = K−1ij∗∂W

∂φi

∂W ∗

∂φ∗j∝ FF ∗ (8.22)

It is interesting that in the vacuum F will be different then zero. Furthermore, since thepotential is positive definite we have

VF > 0⇔ 〈vac|F |vac〉 6= 0 (8.23)


So as we mentioned earlier SUSY breaking corresponds to energy of the vacuum beinggreating than zero.

We now consider some forms for the potential.

φ

V (φ)

Here we have 〈φ〉 = 0 and 〈V 〉 = 0. So the potential is both gauge symmetry andsupersymmetric. A potential of the form,

φ

V (φ)

This potential is supersymmetric as we have V = 0 on the vacuum but not gauge invari-ant. A potential of the form,


φ

V (φ)

is gauge symmetric and non-supersymmetric (as is the case in the SM) and the lastpossibility is,

φ

V (φ)

which is non-supersymmetric and not gauge invariant.

Example: O’Raifertaigh Model

We now discuss a toy model with a concrete superpotential with an expectation value ofF to be greater then zero. For that you need three superfields,

Φ1,Φ2,Φ3 (8.24)

The Kahler potential is taken to be the canonical one but a superpotential of the form,

W = gΦ1

(Φ2

3 −m2)

+MΦ2Φ3 (8.25)

and we assume M m. Recall that

F ∗i = −∂W∂φi

= −∂W∂Φi

∣∣∣∣Φi=φi

(8.26)


so we have,

F ∗1 = −g(φ2

3 −m2)

(8.27)

F ∗2 = Mφ3 (8.28)

F ∗3 = 2gφ1φ3 +Mφ2 (8.29)

If we have any of these terms are different from zero on the vacuum then we breaksupersymmetry. You have to cook a good superpotential to be able to do this; you needto find a set of three equations that cannot be solved simultaneously.

To see that these are inconsistent suppose you set the first equation equal to zero.This implies 〈φ3〉 = ±m but then you can’t have 〈F2〉 = 0 unless m = 0. This was aparticular example but can be shown more generally. Since you cannot have

〈F1〉 = 〈F2〉 = 〈F3〉 = 0 (8.30)

it implies that you have supersymmetry breaking.We now want to find the spectrum of the model (i.e. the masses of all the particles).

The potential in this case is just,

V =

∣∣∣∣∂W∂φi∣∣∣∣2 (8.31)

since the second derivative of the Kahler potential is just the identity matrix in this case.This is equal to,

V = g2∣∣φ2

3 −m2∣∣2 +M2 |φ3|2 + |2gφ1φ3 +Mφ2|2 (8.32)

To minimize this potential one needs to treat the real and imaginary parts of φ3

independently but can use the modulus of φ1 and φ2 as a parameter. If m2 < M2

2g2then

this procedure gives a minimum at,

〈φ2〉 = 〈φ3〉 = 0, 〈φ1〉 is arbitrary (8.33)

At this point we have,〈V 〉 = g2m4 > 0 (8.34)

The potential has one flat direction which is called a “modulus”. Since this directionis completely flat it implies that the mass of φ1 is zero. We now compute the mass of theother two. 〈φ1〉 can be any value but we compute the masses of φ2 and φ3 at the pointwhere its zero for simplicity.

We start with the fermions. Recall that the fermion masses are given in terms of theYukawa couplings. Recall that the mass term is,⟨

∂2W

∂φi∂φj

⟩ψiψj (8.35)

So the mass matrix for the fermions is

mfij ≡

⟨∂2W

∂φi∂φj

⟩(8.36)


At 〈φ2〉 = 〈φ3〉 = 0 the mass matrix is very simple. We only have two nonzero terms,

mfij =

0 0 00 0 M0 M 0

(8.37)

The first row equalling zero implies that mψ1 = 0. For ψ2 and ψ3 you need to diagolizethe 2 × 2 matrix. The eigenvalues are ±M . However for the fermion you can alwaysredefine the field using ψ → eiπγ5/2ψ. This leaves the kinetic term invariant but switchesthe sign of the mass term. Thus we have,

mψ2 = mψ3 = M (8.38)

Recall that

δψ1 ∼ 〈F1〉 6= 0 (8.39)

so ψ1 is our Goldstino. It is massless as we predicted earlier.

We now consider the scalars. To find their masses we need to take the second deriva-tives of the potential. The quadratic piece of the potential is given by,

Vquad = −mg2(φ2

3 + φ∗23

)+M2 |φ3|2 +M2 |φ2|2 (8.40)

Its easy to take derivatives with respect to φ2 since only its modulus appears here, how-ever φ3 is more complicated as you have to consider its real and imaginary componentsseparately,φ3 = ma + ib. Taking the second derivatives gives a diagonal matrix with themasses,

mφ2 = M (8.41)

ma =√M2 − 2m2g2 (8.42)

mb =√M2 + 2m2g2 (8.43)

So the spectrum is as follows,

Particle Masses

a = Reφ3

(φ2, ψ2) , ψ3

b = Imφ3

mass = 0

mass = M

(φ1, ψ1)


So now ψ3 doesn’t come with a bosonic particle with the same mass but with two particleswith differenet masses. Recall that F1 is the quantity that breaks supersymmetry by aquantity gm2. This is what end up gives us the splitting. This is an explicit example ofhow SUSY is broken. This splitting which we expect to be in the real world and shouldbe of the order of a TeV energy. We expect to find scalar partners to the fermions on theorder of a TeV scale away from the fermions.

In this model we have one scalar that is heavier the the fermion but there is alsoanother one that is light then the fermion. This is the general behavoir that appears intree level supersymmetry breaking and is a phenomelogical problem since if we had suchlight scalars we should have seen them by now.

Because of the non-renormalization theorems we proved earlier the superpotential attree level is exact to all orders in perturbation theory. Having SUSY unbroken at treelevel implies that it is unbroken to all orders in perturbation theory. Thus we must breakSUSY nonperturbative effects. We discuss this in detail shortly.

8.2.2 D Term

Suppose you have a vector superfield,

V = V (λ,Aµ, D) (8.44)

One can show that [Q 16: check] the only way to have supersymmetry breaking is to usethe gaugino,

δλ ∼ εD (8.45)

with〈D〉 6= 0 (8.46)

So the gaugino is the goldstino. [Q 17: Do exercises!]

8.3 SUSY breaking in N = 1 supergravity

• Supergravity multiplet adds new auxillary fields which we denote Fg. Supersym-metry breaking requires nonzero 〈Fg〉.

• The F term is proportional to,

F ∝ DW =∂W

∂φ+

1

M2Pl

∂K

∂φW (8.47)

[Q 18: skipped 16 min - 21 min about supergravity.]

Chapter 9

The MSSM

We now have all the ingredients to generalize the Standard Model into what is known asthe Minimal Supersymmetry Standard Model.

9.1 Particles

We need Vector superfields that provide the SM SU(3)C × SU(2)L × U(1)Y symmetry.In other words we need a set of vector superfields for SU(3), another set for SU(2) and alast vector superfield for U(1). We also need chiral superfields that represent the matterparticles. We need to have,

• Quarks:

Qi = (3, 2,−1/6)︸︷︷︸left-handed

; uci = (3, 1, 2/3) , dci = (3, 1,−1/3)︸︷︷︸right-handed

The notation is as follows. we want to present each field as a left chiral superfield.To do this instead of writing uR and dR as is usually done in the SM we write theright handed particles as left fields with both a conjugate and a bar. Furthermorewe label our particles with a generation label, i = 1, 2, 3.

• Leptons:Li = (1, 2, 1/2) ; eci = (1, 1,−1) , νci = (1, 1, 0)

We have a newcomer here which is a right handed neutrino. We add this in becausewe know that neutrinos have a mass.

• Higgses:H1 = (1, 2, 1/2) , H2 = (1, 2,−1/2)

H2 is a completely new field. It is required in order to give mass to the Higgs. Recallthat H1 is a chiral superfield. It has the Higgs particle which is a scalar but it alsohas a new fermionic partner called the Higgsino. This Higgsino will contribute toanomalous diagrams (triangle diagrams) such as,

87

88 CHAPTER 9. THE MSSM

The way to cancel this anomaly is to have an extra fermion to with opposite charge.

9.2 Interactions

• We want renormalizable interactinos so we have,

K = Φ†e2qV Φ (9.1)

Further the function infront of the kinetic term (recall that this the the term thatcontains the gauge invariant vector potential functions, ∝ Ga

µνGµνa . Due to the

renormalizability requirement the function is just a constant,

fa(Φ) = τa (9.2)

where

Reτa =4π

g2a

(9.3)

where we have one coupling for each gauge group. a = 1 for the SU(3) charge,a = 2 for the SU(2) charge, and a = 3 for the U(1) charge.

One can show that the running of the couplings with SUSY provides a unificationscale called the GUT scale,

ga

E

SUSY−−−→

ga

EEGUT

• Lastly for the MSSM we take ξ = 0 since otherwise you break charge and coloursymmetries.

9.2. INTERACTIONS 89

• The last ingredient is the superpotential. It needs couplings among the fields wehave written. The most general renormalizable superpotential is given by,

W = y1QH2uc + y2QH1d

c + y3LH1ec + µH1H2 +WBL (9.4)

where,

WBL = λ1LLec + λ2LQd

c + λ3ucdcdc + µ′LH2 . (9.5)

W as the typical terms you would expect. For example QH2uc in components gives

a term Higgs-quark-quark as in the SM. Further, note that each term is singletunder each transformation as required for gauge invariance. For example underSU(3) uc is a 3 and Q is a 3 so you have a singlet, Q and H2 are both doublets sowe again have a singlet, and lastly the hypercharges add up to zero.

The extra terms in WBL break baryon or lepton numbmer. For example we havecouplings of the type,

ud

u e+

uu

d

where we use the same letter for the superfield as for the field and for the corre-sponding superfield we denote them by a tilde. Such diagrams give rise to protondecay, p→ e+ + π0, within seconds assuming natural couplings. These terms are asource of embarrassment for SUSY.

We need some mechanism to forbid such contributions. The easiest thing to do isto impose a global symmetry known as R parity,

R ≡ (−)3(B−L)+2S (9.6)

Notice that R = 1 for all observed particles and R = −1 for all superpartners. Thisforbids all terms in WBL. Having this symmetry has some important implications.

– The lightest superpartner (LSP) is stable since it cannot decay to any particlein the SM.

– Usually, the LSP is neutral (e.g. Higgsino, photino, ... ) and for this reason itis called a “neutrilino”. This is the best candidate for dark matter (WIMP).

– In colliders the supersymmetric particles can only be produced in pairs whichdecay to LSP which gives a signal in the detectors of “missing energy”. Thisis one of the best ways to test supersymmetry.


9.3 Supersymmetry Breaking

Recall that in the SM we have two sectors,

quarksleptons

Observable

Higgs

Symmetry Breaking

Coupled Yukawase.g.Hψψ

In SUSY the situation is a bit different as in general we need 3 sectors,

quarkssquarkssleptons

...

Observable SUSY

Messenger

Its possible to break supersymmetry directly instead as with supersymmetry breakinghowever, this turns out not to work very well. Most of the uncertainty of supersymmetricmodel is not as much from the supersymmetry breaking sector but from the messengersector.

9.3.1 SUSY Sector

We discussed previously that you can break supersymmetry at tree level (and hence at anyorder in perturbation theory), however that can be very bad because you get light particlesthat we should have seen by now. Thus we must break SUSY in a non-perturbative way.

To understand what this means consider the following example. Suppose you have aSUSY breaking sector with the group G = SU(M) with matter such that it is asymptot-ically free,

9.3. SUPERSYMMETRY BREAKING 91

g

EΛ MPl

In these models for instance you can have the gauginos to get a VEV,

〈λλ〉 6= 0 (9.7)

(analogous to Cooper pairs in superconducitivity). The fermions can only get close toeach other at large enough couplings. By dimensional analysis we have,

〈λλ〉 ∼ Λ3 (9.8)

Depending on the model one can show that this can break supersymmetry,

MSUSY ∼ Λ = MPle−a/g2 (9.9)

due to the renormalization group running (a is some constant). So Λ, the natural scaleof the theory, can be hierarchically much smaller then the Planck energy.

9.3.2 Messenger Sector

This is the most model dependent part of supersymmetry. There are several ways oftransmitting the information that SUSY was broken:

1. Gravity mediation: You break SUSY in one sector which doesn’t couple to theobservable sector. However, gravity couples to everybody which in turn transmitsthe breaking to the observable sector. This is perhaps the most popular approach.It has several interesting properties. Since it is gravity mediated we know that themass splitting in the observable sector must be very small if MPl is very large (inthe limit that MPl →∞ there is no gravity).

∆m =something

MPl

(9.10)

By dimensional analysis we must have,

∆m =M2SUSY

MPl

(9.11)

We want ∆m ∼ 1TeV to be able to fix the hierarchy problem. We know thatMPlanck ∼ 1018GeV. Thus we know that,

MSUSY =√MEWMPl ∼ 1011GeV (9.12)


So there is an intermediate scale between the electroweak scale and the Planck scalethat gives SUSY breaking. Furthermore, the gravitino gets a mass given by,

m3/2 =M2SUSY

MPl

∼ TeV (9.13)

(Aside: the gravitino gets a mass from what’s known as the Super-Higgs effect;the gravitino “eats” the goldstino to get a mass. This is different from the super-symmetric version of the Higg’s mechanism where a whole vector superfield eats awhole chiral superfield.)

2. Gauge Mediation: With the total gauge group equal to the gauge group of the SMmultiplied by sector that breaks supersymmetry,

G = (SU(3)× SU(2)× U(1))︸︷︷︸G0

×GSUSY (9.14)

The matter fields are charged under both G0 and GSUSY . In the gravity mediationcase the particles introduced in the SUSY breaking sector were all singlets underthe gauge group of the SM. However, this is not the case here. By dimensionalanalysis we have,

∆m ∼MSUSY (9.15)

which we demand to be of order 1TeV. On the other hand the gravitino mass, m3/2

will still be equal toM2SUSY

MPl

∼ 10−3eV (9.16)

This would imply that the gravitino is the LSP.

3. Anomaly Mediation: The auxillary fields of the supergravity models get a VEV.This is always present but suppressed by loop effects.

In each of these mechanisms we have an effective Lagrangian,

L = LSUSY + LSUSY (9.17)

for the observable sector. The LSUSY terms are called soft breaking terms. They takethe form of,

LSUSY = (Mλλaλa + h.c.)︸︷︷︸

gaugino masses

+

scalar masses︷︸︸︷m2

0φa ∗φa + (Aφφφ+ h.c.)︸︷︷︸

trilinear

(9.18)

Given a model for the messenger sector you need to determine what each of these termsare. When people talk about the MSSM they talk about the supersymmetric part plusthese soft-breaking terms. The are called soft-breaking since they don’t effect the goodultraviolet properties of supersymmetry.

9.4. HIERARCHY PROBLEM 93

9.4 Hierarchy Problem

In the Standard Model we know that

• The gauge particles are massless due to gauge invariance.

• Fermions are also massless since gauge invariance forbids mψψ.

• There is no symmetry that protects bosons from getting a mass.

In SUSY

• Bosons have the same mass as fermions which implies that their mass is also pro-tected by symmetry

• Gives miraculous calculations

The biggest problem that SUSY faces is it doesn’t solve the cosmological constantproblem. You may wonder if there is a way to address both problems at the same time.One possible solution to solve this problem is known as the “Landscape”. The idea is toconsider a potential that has many minima ∼ 10100 with each minima corresponding tosome possible universe. Then since very few universes could support life as we know it,our universe is just one of the very few possible choices making the fine-tuning in somesense natural.

One may wonder if you could use this same idea to solve the hierarchy problem. Thishas been applied for example in a model called Split-SUSY.

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SUSY AND EXTRA DIMENSIONS LECTURE NOTESpages.physics.cornell.edu/~ajd268/Notes/SUSYandExtraDim.pdf · 2014-05-14 · 1 SUSY AND EXTRA DIMENSIONS LECTURE NOTES Lecture notes based