Supersymmetry - School of Physics and Astronomys0948358/mysite/SUSY chapter 3.pdfChapter 3 Supersymmetry In this chapter we will ﬁrst discuss the Coleman-Mandula theorem and the

Chapter 3

Supersymmetry

In this chapter we will first discuss the Coleman-Mandula theorem and the reasons why theextension to graded algebras are required. Then we will present the supersymmetric algebrasand finally their representations for di↵erent types of particles.

3.1 Extension to graded algebras

There are two di↵erent types of symmetries for elementary particles:

1. Space-time symmetries: These symmetries correspond to the transformation of space-time coordinates. For example rotations and translations i.e. Poincare transformations.

2. Internal symmetries: In a field theory, these symmetries are related to the transforma-tions of the fields e.g. a(x) ! 0

a(x) = L ba b(x).

In 1971, a no-go theorem by Coleman and Mandula stated that irreducible multiplets of thePoincare symmetry group cannot contain particles of di↵erent mass or spin. They showedthat the general symmetry of the S-matrix is the direct product of the Poincare symmetrywith an internal symmetry group. In other words, the general Lie algebra of the symmetriesof the S-matrix contains

• The energy-momentum operator Pµ

• The Lorentz rotation generator Mµ⌫

• A finite number of Lorentz scalar operators B` which must belong to the Lie algebraof a compact Lie group.

In 1975, however, Haag, Sohnius and Lopuszanski extended the results of Coleman and Man-dula to include symmetry operations obeying fermi statistics and anticommutator relations

54

3.1. EXTENSION TO GRADED ALGEBRAS 55

defined by {A,B} = AB + BA. They found that space-time symmetries and internal sym-metries can only be related to each other by a fermonic symmetry operator Q of spin-1

2

(not3

2

or higher). They concluded that the multiplets can contain particles of di↵erent spins onlyin the presence of supersymmetry. Supersymmetry (SUSY), by definition, is a symmetry be-tween fermions and bosons. Having said that, since spin is related to behaviour under spacialrotations, it is really a space-time symmetry. SUSY has therefore the property of combininginternal and space-time symmetries in order to attempt to unify all fundamental interactions.

Let Q, Q0 and Q00 represent the fermionic (anticommuting) part of the algebra while X,X 0 and X 00 represent the bosonic (commuting) part. Then, schematically we have

{fermionic, fermionic} = bosonic =) {Q,Q0} = X (3.1)

[bosonic, bosonic] = bosonic =) [X,X 0] = X 00 (3.2)

[fermionic, bosonic] = fermionic =) [Q,X] = Q00 (3.3)

The bosonic operators X are determined by the Coleman-Mandula theorem i.e. they areeither elements of the Poincare algebra P = {Pµ,Mµ⌫} or the elements of the compact Liealgebra A which are Lorentz invariant. The algebra A is a direct sum of a semi-simplealgebra A

1

and an Abelian algebra A2

i.e. A = A1

� A2

.

Let us now explore the compact Lie group and the relation between the internal symme-tries in more detail. Given a compact Lie group one can define a connected Lie gorup whichhas special importance as we shall see. These are groups of transformations T (✓), describedby a finite set of real continuous parameters, ✓a, such that every element of the group isconnected to the identity by a path within the group. The elements of the group must thensatisfy the composition rule

T (✓)T (✓) = T�f(✓, ✓)

�(3.4)

where fa(✓, ✓) is a function of ✓s and ✓s. By taking ✓a = 0 to be the coordinates of theidentity, we get

fa(✓, 0) = fa(0, ✓) = ✓a (3.5)

One can then taylor expand the unitary operator corresponding to this transformation in theneighbourhood of the identity as follows:

U(T (✓)) = 1 + i✓ata +1

2✓b✓ctbc + ... (3.6)

where ta, tbc = tcb are operators which are independent of ✓s and the factor of i exist in orderto make sure that ta is a hermitian. Explicitly,

ta = �i@

@✓aU(T (✓))

��✓=0

(3.7)

56 CHAPTER 3. SUPERSYMMETRY

tbc = (�i)(�i)@2

@✓b@✓cU(T (✓))

��✓=0

(3.8)

which is why tbc is symmetric with respect of its two indices. Suppose U(T (✓)) form anordinary representation of this group so that

U�T (✓)

�U (T (✓)) = U

�T�f(✓, ✓)

��(3.9)

Now expanding fa(✓, ✓) to second order and using equation 3.5 we get

fa(✓, ✓) = f(0, 0)| {z }0

+@f

@✓a|{z}1

✓a +@f

@✓a|{z}1

✓a +1

2

@2f

@✓2|{z}0

✓2 +1

2

@2f

@✓2|{z}0

✓2 +1

2

@2f

@✓b✓c+ ...

=✓a + ✓a + fabc ✓

b✓c + ...

(3.10)

since the presence of ✓2 and ✓2 would violate equation 3.5. Note that the coe�cients fabc ✓

b✓c

are real and there is a summation between indices. The left hand side of equation 3.9 cannow be written as

1 + i✓ata +

1

2✓b✓ctbc + ...

� 1 + i✓a

0ta0 +

1

2✓b

0✓c

0tb0c0 + ...

�

=1 + i✓a0ta0 +

1

2✓b

0✓c

0tb0c0 + i✓ata � ✓a✓a

0tata0 +

1

2✓b✓ctbc + ...

(3.11)

and the RHS

1 + i⇣✓a + ✓a + fa

bc ✓b✓c + ...

⌘ta +

1

2

⇣✓b + ✓b + ...

⌘ �✓c + ✓c + ...

�tbc

=1 + i✓ata + i✓ata + ifabc ✓

b✓cta +1

2

⇣✓b✓c + ✓b✓c + ✓b✓c + ✓b✓c

⌘tbc + ...

(3.12)

Equating, relabelling and using the fact that tbc = tcb, we get

� ✓b✓ctbtc = ifabc ✓

b✓cta +1

2✓b✓ctbc +

1

2✓b✓ctbc (3.13)

Comparing coe�cients results in

tbc = �tbtc � ifabc ta (3.14)

Thus, given the function f(✓, ✓), which determines the structure of the group, and thequadratic coe�cients fa

bc one can calculate the second order terms in U(T (✓)) from thegenerators ta which appear in the first order terms. We also must have tbc = tcb. Then

[tb, tc] = tbtc � tctb =� tbc � ifabc ta + tcb + ifa

cb ta = �ifabc ta + ifa

cb ta

⌘iCabc ta

(3.15)

3.2. INTRODUCING SUPERSYMMETRIC ALGEBRAS 57

where Cabc are a set of constants known as the structure constatns.

Therefore, provided that we know the first order term generators ta, we can derived thecomplete power series U(T (✓)) using an infinite sequence of 3.14.

Now, consider the case where the function f(✓, ✓) obeys

fa(✓, ✓) = ✓a + ✓a (3.16)

i.e. is additive. In this case the structure constants vanish and the generators commute:

[tb, tc] = 0 (3.17)

we then say that the group is commutative or Abelian.

Going back to our discussion of the internal symmetry group, recall that we said B` area finite number of Lorentz scalar operators that belong to the Lie algebra of a compact Liegroup. Hence the following relations can be concluded:

[B`, Bm] = iC k`m Bk (3.18)

[B`, Pµ] = [B`,Mµ⌫ ] = 0 (3.19)

since we are taking a direct product.

3.2 Introducing supersymmetric algebras

Supersymmetric transformations are generated by quantum operators Q. When acting on afermionic state by Q, it will be change to a bosonic state and vice versa, i.e.

Q|fermionici = |bosonici , Q|bosonici = |fermionici (3.20)

i.e. Qs change the statistic and thus the spin of the particles by 1

2

.We expect Q not to be invariant under rotations since the e↵ect of rotations on fermions isdi↵erent with that on the bosons. For example, let U be a unitary operator in Hilbert spacerepresenting some rotation by 360� about an arbitrary axis. We know that

U |fermioni = �|fermioni , U |bosoni = |bosoni (3.21)

Then

UQ|bosoni =UQU�1U |bosoni = UQU�1|bosoni⌘U |fermioni = �|fermioni = �Q|bosoni

(3.22)


and soUQU�1 = �Q (3.23)

Hence, the rotated SUSY generator picks ups minus sign just like a fermionic state underrotation. Our aim is now to generalise this notion and show that Q transforms precisely likea spinor operator under Lorentz i.e. as a spin-1

2

object. We will then prove that Q’s do notcommute with Lorentz transformations. In other words, applying a Lorentz transformationand then a supersymmetry transformations is di↵erent with that in the reverse order.

First, we will decompose the generators Q ito a sum of irreducible representations of thehomogeneous Lorentz group:

Q =X

Q↵1...↵a,↵1...↵b (3.24)

where the Q↵1...↵a,↵1...↵b are antisymmetric with respect to the underlined indices belonging

to irreducible spin-12

(a + b) representations of the Lorentz group. Also, note that the greekindices run from one to two. Since Q’s anticommute, one can conclude that a + b must beodd because of the relations between spin and statistics. The goal is to show that a+ b = 1.For this purpose, we will make the following assumptions:

• The Hilbert space in which the operators Q act, has a positive definite metric:

h..|{Q,Q†}|..i = h..|QQ†|..i+ h..|Q†Q|..i =��Q†|..i

��2

+��Q|..i

��2

� 0 (3.25)

with equality only when Q = 0.

• Both Q and its hermitian conjugate Q belong to the algebra.

Now, consider the anticommutator

{Q↵1...↵a,↵1...↵b , Q ˙�1... ˙�a,�1...�b} (3.26)

when all indices are equal to one, we get the product

Q1...1a

, ˙1... ˙1b

Q˙

1... ˙1a

,1...1b

(3.27)

to be a member of spin-(a+b)representation of the Lorentz group. Therefore,

{Q1...1a

, ˙1... ˙1b

, Q˙

1... ˙1a

,1...1b

} (3.28)

will close into a bosonic element of the algebra with spin (a + b). From Coleman-Mandulatheorem we know that this element is either zero or a component of Pµ. But for Poincarea+ b 1, so for a+ b > 1 the result must be zero. On the other hand, equation 3.25 implies


that Q1...1a

, ˙1... ˙1b

= 0. We also said Q↵1...↵a,↵1...↵b are irreducible under Lorentz transformations

so they must all vanish for a + b > 1. Therefore, a + b = 1 which in turn implies that theodd part of the SUSY algebra is completely made of spin-1

2

fermionic operators Q L↵ and its

conjugate Q M↵ . More precisely, Q L

↵ belongs to the (12

, 0) representation and Q M↵ belongs

to the (0, 12

) representation of the algebra.

The anticommutator relation

{Q L↵ , Q↵M} 2 (

1

2, 0)⌦ (0,

1

2) = (

1

2,1

2) (3.29)

where (12

, 12

) is the vector representation which means that the anticommutator must corre-spond to the 4-vector operator Pµ. So we need to use a mapping to write this in the spinorrepresentation corresponding to (1

2

, 12

) on the right hand side of the equation. Recall that thistype of mapping has already been discussed in the previous chapter, explicitly we can refer toequation 2.68. i.e. the anticommmutator closes into P↵↵ = � µ

↵↵ Pµ as a linear combination.Thus

{Q L↵ , Q↵M} = P↵↵C

LM (3.30)

Where CLM is a finite dimensional hermitian matrix.

Proof: We write (Q↵M )† = Q M↵ and

�Q L

↵

�†= Q↵L. Then if we take the hermitian conjugate

of equation 3.30 we get

{Q L↵ , Q↵M}† =

�Q L

↵ Q↵M�†

+�Q↵MQ L

↵

�†=

�Q↵M

�† �Q L

↵

�†+

�Q L

↵

�† �Q↵M

�†

=Q M↵ Q↵L + Q↵LQ

M↵ = {Q M

↵ , Q↵L} = P↵↵CM

L

(3.31)

but the right hand side is

P↵↵�CL

M

�†= P↵↵

�C ML

�⇤(3.32)

Comparing, we get�C ML

�⇤= CM

L =)�(CM

L )t�⇤

= CML =) C† = C (3.33)

is hermitian. Q.E.D

Therefore, CLM can be diagonalised by a unitary transformation (Spectral theorem). As

well as that, because {Q L1

, Q˙

1L} is positive definite, CLM has positive definite eigenvalues.

Hence we can choose a basis such that

{Q L↵ , Q↵M} = 2P↵↵�

LM (3.34)

One can check that this makes sense by considering

0 <2X

↵=1

{Q L↵ ,

�Q L

↵

�†} = 2�µ↵↵Pµ�

LL = 2Tr (�µPµ)↵↵ = 4P 0. (3.35)


which confirms that E = P 0 is positive definite, as required.

Before we proceed to build the rest of the algebra, let us generalise the ordinary Jacobi

identity introduced in quantum mechanics, to include anticommutators as well as commuta-tors. This relation will be very useful, as we shall see.

(�1)⌘c⌘a [[A,B}, C}+ (�1)⌘a⌘b [[B,C}, A}+ (�1)⌘b⌘c [[C,A}, B} = 0 (3.36)

where ⌘ = 1 for fermionic (odd) operators and ⌘ = 0 for bosonic (even) ones. Then, forexample, for [A,B} = C we would have ⌘c = ⌘a + ⌘b (mod 2). To see this, consider thefollowing:

{fermionic, fermionic} = bosonic ! 1 + 1 ⌘ 0

[bosonic, bosonic] = bosonic ! 0 + 0 ⌘ 0

[fermionic, bosonic] = fermionic ! 1 + 0 ⌘ 1

as expected. So, for a more general operator O = ABC... we have ⌘(0) = ⌘a + ⌘b + ⌘c + ...mod(2). Note that

[O,O0} = OO0 � (�1)⌘(O)⌘(O0)O0O = �(�1)⌘(O)⌘(O0

)[O0, O} (3.37)

We are now ready to prove the generalised Jacobi identity:The combination ABC appears twice in this order, i.e. from the first and second terms. ACBalso appears twice but from the second and third terms in equation 3.36. It is su�cient toprove that the coe�cients of these vanish, since the rest of the terms are simply the cyclicpermutations of these two which implies that the result will be zero.For coe�cient of ABC we have (�1)⌘c⌘a from the first term. As for the second term,

(�1)⌘a⌘b [[B,C}, A} = (�1)⌘a⌘b(�)(�1)⌘a(⌘b+⌘c)[A, [B,C}} (3.38)

so for the coe�cient of ABC we have

(�1)⌘c⌘a � (�1)⌘a⌘b(�1)⌘a(⌘b+⌘c) = 0 (3.39)

one can easily check that this is indeed 0 for all possibilities ⌘ = 0, 1. As for the coe�cientsof ACB, from the second and third terms in 3.36 we have

(�1)⌘a⌘b(�)(�1)⌘a(⌘b+⌘c)[A, [B,C}}+ (�1)(�1)⌘b⌘c(�1)⌘c⌘a [[A,C}, B}=� (�1)⌘a⌘b(�1)⌘a(⌘b+⌘c)(�)(�1)⌘b⌘c [A, [C,B}}� (�1)⌘b⌘c(�1)⌘c⌘a [[A,C}, B} =)(�1)⌘a⌘b(�1)⌘a(⌘b+⌘c)(�1)⌘b⌘c � (�1)⌘b⌘c(�1)⌘c⌘a = (�1)⌘c⌘a(�1)⌘b⌘c

�(�1)2⌘a⌘b � 1

�

=0 8 possibilities.

(3.40)


Q.E.D

Our task is now to prove the SUSY generators commute with momenta. We know that thecommutator could contain (1, 1

2

) and (0, 12

) representations since

[Q↵L, Pµ] 2 (

1

2, 0)

| {z }dim=2

⌦ (1

2,1

2)

| {z }dim=4

= (1,1

2)

| {z }dim=6

� (0,1

2)

| {z }dim=2 4

(3.41)

However, there are no (1, 12

) generators present therefore we can only get a combination ofQ’s on belonging to the (0, 1

2

) representation on the right hand side. So we can deduce that

[P↵↵, Q�L] = ZL

M ✏↵�QM

↵ (3.42)

where ZLM are some set of numbers and P↵↵ = � µ

↵↵ Pµ. We make use of the Jacobi identity:

[P� ˙� , [P↵↵, QL

� ]] + [P↵↵, [QL

� , P� ˙� ] + [Q L� , [P� ˙� , P↵↵]] = 0 (3.43)

which implies, using the fact that momenta commute,

ZLM ✏↵� [P� ˙� , Q

M↵ ]� ZL

M ✏�� [P↵↵, QM˙�

] = 0 (3.44)

We need to find [P↵↵, Q M˙�

]. This can be done by taking the conjugate of equation 3.42 i.e.

[Q�L, P↵↵] = ZL

M ✏↵�Q↵M so that the above equation will now read

ZLM ✏↵�

⇣�ZM

K ✏˙�↵Q

K�

⌘� ZL

M ✏��⇣�ZM

K ✏↵ ˙�QK

↵

⌘

)ZLM ZM

K ✏↵ ˙�

�✏↵�Q�

K + ✏��Q↵K�= 0

)ZZ = 0 ) |Z| = 0

(3.45)

and so ZLM must vanish resulting in [P↵↵, Q�

L] = 0. Hence

[Pµ, QL

↵ ] = 0 (3.46)

and[Pµ, Q↵L] = 0 (3.47)

We will now consider the anticommutator of two SUSY generators with undotted indices.Since

{Q L↵ , Q M

� } 2 (1

2, 0)⌦ (0,

1

2) = (0, 0)A � (1, 0)S (3.48)


By Coleman-Mandula theorem, the antisymmetric part (0, 0) has spin zero and so it corre-spond to a scalar while the symmetric part (1, 0) corresponds to the self-dual part of theLorentz generator Mµ⌫ denoted by M↵� . Let us explore this in more detail. Because theleft hand side of the equation involves spinors, we need to find a mapping for the right handside that takes us from Mµ⌫ to M↵� which is symmetric with respect to exchange of its twospinor indices. As presented in the previous chapter, this mapping is provided by equations2.84 and 2.88. So we expect the anticommutator to close into

{Q L↵ , Q M

� } = ✏↵�XLM_ +M↵�Y

LM (3.49)

then using the Jacobi identity we get

[P�� , {Q↵L, Q�

M}] + {Q↵L, [Q�

M , P�� ]}+ {Q�M , [P�� , Q↵

L]} = 0 (3.50)

but the last two terms vanish since Q and P commute so that

[P�� , {Q↵L, Q�

m}] = 0 (3.51)

implying

[P�� , ✏��XLM_ +M↵�Y

LM ] = 0 (3.52)

but P�� commutates with the scalar XLM_ meaning

[P�� ,M↵�YLM ] = 0 ) Y LM [P�� ,M↵� ]

| {z }non-zero

= 0 ) Y LM = 0 (3.53)

Also, one can prove explicitly why XLM_ must be antisymmetric with respect to L and M :

Q↵LQ�

M +Q�MQ↵

L = ✏↵�XLM (3.54)

Relabelling ↵ $ � and L $ M implies

Q�MQ↵

L +Q↵LQ�

M = ✏�↵XML

)Q↵LQ�

M +Q�MQ↵

L = �✏↵�XML

(3.55)

So XLM_ = �X

ML_ . Q.E.D

Therefore,

{Q↵L, Q�

M} = ✏↵�XLM_ = ✏↵�a

`,LM_ B` (3.56)

where B` is a hermitian element a member of A1

�A2

and and a`,LM

_ is antisymmetric in Land M .


Taking the hermitian conjugate of 3.56, the LHS is

{Q↵LQ�

M}† =�Q↵

LQ�M�†

+�Q�

MQ↵L�†

=�Q�

M�†

+�Q↵

L�†

+�Q↵

L�† �

Q�M�†

=Q˙�MQ↵L +Q↵LQ ˙�M = {Q↵L, Q ˙�M}

(3.57)

and the RHS ⇣✏↵�a

`,LM_ B`

⌘†= ✏↵ ˙�a

⇤`,LM

_B` = ✏↵ ˙�X

†LM_

(3.58)

The next relation we’re going to consider, knowing that B` is a Lorentz scalar, is

[Q L↵ , B`] 2 (

1

2, 0)⌦ (0, 0) = (

1

2, 0) (3.59)

so that one expects the right hand side to be a sum of Q’s:

[Q L↵ , B`] = S`

LMQ M

↵ (3.60)

with S`LM , some complex constants. Again, taking the hermitian conjugate the LHS yields

[Q L↵ , B`]

† =�Q L

↵ B`

�† ��B`Q

L↵

�†

=(B`)† �Q L

↵

�† ��Q L

↵

�†(B`)

†

=B`Q↵L � Q↵LB`

=[B`, Q↵L]

(3.61)

Hence

[B`, Q↵L] = S⇤` ML Q↵M (3.62)

Let us write the Jacobi identity for B`, Q L↵ and Q

˙�M :

[B`, {Q L↵ , Q

˙�M}+ {Q L↵ , [Q

˙�M , B`]}� {Q˙�M , [B`, Q

L↵ ]} = 0

)[B`, 2�µ

↵ ˙�Pµ�

LM ] + {Q L

↵ ,�S⇤` KM Q

˙�K}� {Q˙�M ,�S`

LKQ K

↵ }(3.63)

The first terms is equal to zero since B` and P commute. Then

� 2P↵ ˙��LKS⇤` K

M + 2P↵ ˙��KMS`

LK = 0

) 2P↵ ˙�

hS⇤` L

M � S`LM

i= 0

(3.64)


So that the term in the brackets must vanish

S⇤` LM = S`

LM (3.65)

meaning S`LM is hermitian.

Note that generators XLM_ = a

`,LM_ B` form an invariant subalgebra of A

1

� A2

.Proof:

[B`, {Q L↵ , Q M

� }] + {Q L↵ , [Q M

� , B`]}� {Q M� [B`, Q

L↵ ]} = 0

)[B`, ✏↵�XLM_ ] + {Q L

↵ , S M` KQ K

↵ }+ {Q M� , S L

` KQ K↵ } = 0

)[B`, ✏↵�XLM_ ] + {Q L

↵ , Q K↵ }S M

` K + {Q M� , Q K

↵ }S L` K = 0

)✏↵� [B`, XLM_ ] + ✏↵�X

LK_ S M

` K + ✏�↵XMK_ S L

` K = 0

)✏↵�{[B`, XLM_ ] + S M

` KXLK_ � S L

` KXMK_ } = 0

)[B`, XLM_ ] = S L

` KXMK_ � S M

` KXLK_

(3.66)

i.e. [B`, XLM_ ] closes into a set of generators X

LM_ . As well as that, theX

LM_ are linear com-

binations of the B`. Hence XLM_ form an invariant subalgebra of A = A

1

� A2

.

It is possible to show that generators XLM_ commute with all the generators of A using

the identity

[Q L↵ , {Q M

� , Q�K}] + [Q M� , {Q�K , Q L

↵ }] + [Q�K , Q L↵ , Q M

� ] = 0

) [Q L↵ , 2� µ

�� Pµ�M

K ] + [Q M� , 2� µ

↵� Pµ�LK ] + [Q�K , ✏↵�X

LM_ ] = 0

) ✏↵� [Q ˙�k, XLM_ ] = 0

(3.67)

Now we need to show that XLM_ commute with Pµ using equation 3.56 and the fact that P

and Q commute with each other. We start with

[Pµ, {Q L↵ , Q M

� }] =[Pµ, QL

↵ Q M� +Q M

� Q L↵ ]

=PµQL

↵ Q M� + PµQ

M� Q L

↵ �Q L↵ Q M

� Pµ �Q M� Q L

↵ Pµ

=[Pµ, QL

↵ Q M� ] + [Pµ, Q

M� Q L

↵ ] = 0

(3.68)

So the commutator of the right hand side of equation 3.56 yields

✏↵� [Pµ, XLM_ ] = 0 ) [Pµ, X

LM_ ] = 0 (3.69)


i.e. Pµ commutes with XLM_ . Again, we use the identity

[XAB_ , {Q L

↵ , Q˙�M}] + {Q L

↵ , [Q˙�M , X

AB_ ]}� {Q

˙�M , [XAB_ , Q L

↵ ]} = 0 (3.70)

where the first two terms vanish since XAB_ commutes with P . So we must have

{Q˙�M , [X

AB_ , Q L

↵ ]} = 0 (3.71)

On the other hand, [XAB_ , Q L

↵ ] is in the (12

, 0) representation. In other words the commutatorwill result to a linear combination of the Q’s i.e.

[XAB_ , Q L

↵ ] = TABLRQR

↵ (3.72)

and so equation 3.71 will now read

TABLR{Q ˙�M , Q R↵ } = TABLR

⇣2� µ

↵ ˙�Pµ�

RM

⌘

| {z }6=0

= 0 ) TABLR = 0 (3.73)

Hence XAB_ and Q commute:

[XAB_ , Q L

↵ ] = 0 (3.74)

As well as that,

[XKN_ , X

LM_ ] =

1

2✏�↵[{Q K

↵ , Q N� }, XLM

_ ] = 0 (3.75)

Since XLM_ and Q commute. The above equation can easily be verified:

1

2✏�↵[✏↵�X

KN_ , X

LM_ ] =

1

2✏�↵✏↵� [X

KN_ , X

LM_ ]

=1

2(✏12✏21 + ✏21✏12)[X

KN_ , X

LM_ ] = [X

KN_ , X

LM_ ] 4

(3.76)

What [XKN_ , X

LM_ ] = 0 really means is that X

LM_ is Abelian and so forms an Abelian

(invariant) subalgebra of A . On the other hand, we know that A1

is semi-simple, hence

non-abelian. Therefore, XLM_ is a member of A

2

and commute with all the generator of A

[XLM_ , B`] = 0 (3.77)

This is why they are called central charges i.e. they are in the centre of the supersymmetricalgebra.Using this relation, equation 3.66 yields:

S M` KX

LK_ � S L

` KXMK_ = 0 (3.78)


substituting XLM_ = a

`,LM_ B` we get

S M` Ka

k,LK_ � S L

` Kak,MK

_ = 0 (3.79)

then, from the fact that S M` K is hermitian and a

MK_

k is antisymmetric, we get

S M` Ka

k,KL_ = �a

k,MK_ S⇤` L

K (3.80)

On the other hand, S M` K form a representation of compact Lie algebra A = A

1

� A2

.Proof:

[B`, [Bm, Q `↵ ]] + [Bm, [Q L

↵ , B`]] + [Q L↵ , [B`, Bm]] = 0

)� [B`, SL

m M ] + [Bm, S L` MQ M

↵ ] + [Q L↵ , iC K

`m Bk] = 0

)+ S L` MS M

` KQ K↵ � S L

` MS Mm KQ K

↵ + iC k`m S L

K MQ M↵ = 0

(3.81)

Comparing the coe�cients of Q K↵ implies

(SmS`)LK � (S`Sm)LK = �iC K

`m SLK (3.82)

Hence[S`, Sm] = iC k

m` Sk (3.83)

and the conjugate

[S⇤m , S⇤

` ] = �iC Km` S⇤

k (3.84)

as required by the Lie algebra of the compact Lie group. Q.E.D

Equation 3.80 states that matrices ak intertwine the representations S` with its complex

conjugate �S⇤` . So if central charges exist, they must be of the form X

LM_ = a

`,LM_ B`.

Finally, we need to understand how the Supersymmetry generators transform under Lorentztransformations. We will need to use the fact the Q↵ can be transformed as a spinor and aswell as an operator.

1. Spinor: The Lorentz transformation of a spinor in the (12

, 0) representation is Q0↵ =

e(�i2!µ⌫S

µ⌫L ) �

↵ Q� or for infinitesimal trasformations,

Q0↵ =

✓1� i

2!µ⌫S

µ⌫L

◆ �

↵

Q� (3.85)

where�Sµ⌫L

� �↵

= ��S⌫µL

� �↵

are, as before, a set of 2⇥2 antisymmetric matrices obeyingthe same commutation relations as the Lorentz generators Mµ⌫ .

3.3. IRREDUCIBLE REPRESENTATIONS OF SUPERSYMMETRY ALGEBRA 67

2. Operator: We have, under Lorentz transformations Q0↵ = U(⇤)Q↵U�1(⇤), so that for

infinitesimal transformations we get

Q0↵ =

✓1� i

2!µ⌫M

µ⌫

◆Q↵

✓1 +

i

2!µ⌫M

µ⌫

◆(3.86)

Now we can set equal the results of (1) and (2):

� �↵ Q� �

✓i

2!µ⌫S

µ⌫L

◆ �

↵

Q� = Q↵ � i

2!µ⌫M

µ⌫Q↵ +i

2�!µ⌫Q↵M

µ⌫ +O(!2)

) � i

2!µ⌫

�Sµ⌫L

� �↵

Q� = [Q↵,Mµ⌫ ]

✓i

2!µ⌫

◆ (3.87)

Therefore, comparing the coe�cient of !µ⌫ on both sides of the equation,

[Mµ⌫ , Q↵] =�Sµ⌫L

� �↵

Q� (3.88)

Similarly for a spinor in Q↵ in (0, 12

) we have

[Mµ⌫ , Q↵] =�Sµ⌫R

�˙�

↵Q

˙� (3.89)

where, as we have already proved in the previous chapter, (Sµ⌫R )

˙ba = �[(Sµ⌫

L ) ba ]⇤. Note that

(S12

L ) ba = 1

2

�3

and since �3

is real, we also have (Sµ⌫R )

˙ba = �1

2

�3

.

3.3 Irreducible representations of Supersymmetry algebra

Our aim of this section is to derive the Supersymmetric representation of the graded algebrafor both massless and massive cases. Before we proceed to discuss this in detail, let us buildan overall picture of such representations.

Informally, a representation consists of linear operators on some vector space, with the oper-ators having the same ‘properties’ as the objects they are going to represent. We know thatSUSY transformations related bosons and fermions so one can divide the representationsspace into two parts: a bosonic and a fermionic one i.e.

Representation space = Bosonic subspace + Fermionic subspace (3.90)

The bosonic generators Pµ,Mµ⌫ and B` , which form the even part of the algebra, map thesubspaces into themselves, or rather, onto a proper subspace of the original space.


bosonic(fermionic)Pµ,Mµ⌫ ,B`

��! bosonic(fermionic) (3.91)

However, the Supersymmetry generators Q, which form the odd part of the algebra, map thebosonic subspace into the fermionic one and vice-versa.

bosonic(fermionic)Q

��! fermionic(bosonic) (3.92)

For two subsequent odd mappings, which is the case with the anticommutator relations, wehave

bosonic(fermionic)First Q

��! fermionic(bosonic)Second Q��! bosonic(fermionic) (3.93)

Therefore, it can be said that what {Q, Q} = 2�µPµ really means is that there is a mappingfrom one subspace to the other and back to the original one such that the total result is asif we we had acted with Pµ on the original subspace.There is a theorem that states that the fermionic and bosonic subspaces of the represen-tations space of SUSY algebra have the same dimensions. This is because the result of nomapping has dimensions higher than the original and also in the mappings like QQ + QQno dimensions get lost. We will later on prove rigorously that SUSY representations containequal numbers of bosonic and fermionic states.

We know that the SUSY generators Q commute with energy-momentum 4-vector Pµ. ThusP 2, the mass operator, is a Casimir operator of the algebra. Since [Q,P 2] = 0 we concludethat the irreducible representations of the supersymmetry algebra are of equal mass. How-ever, SUSY multiplets contain di↵erent spins and so W 2 is no longer a Casimir operator i.e.[Q,W 2] 6= 0.

Let us now introduce a fermion number operator NF such that (�)NF has eigenvalue +1and -1 on bosonic and fermionic states respectively.

(�)NF |bosonici = +1 |bosonici , (�)NF |fermionici = �1 |fermionici (3.94)

Therefore,

(�)NFQ↵ = �Q↵(�)NF (3.95)

which can easily be checked:

⇥(�)NFQ+Q(�)NF

⇤|bosonici =(�)NF |fermionici+Q|bosonici

=� |fermionici+ |fermionici = 0(3.96)


Similarly,⇥(�)NFQ+Q(�)NF

⇤|fermionici =(�)NF |bosonici �Q|fermionici

=+ |bosonici � |bosonici = 0(3.97)

as required.

For any finite dimensional representation of the algebra, i.e. the trace is well-defined, wehave

Trh(�)NF {Q A

↵ , Q˙�B}

i= Tr

h(�)NF

⇣Q A

↵ Q˙�B + Q

˙�BQA

↵

⌘i(3.98)

Note that trace is cyclic e.g. Tr(ABCD) = Tr(BCDA) = Tr(CDAB = Tr(DABC) and soTr(AB) = Tr(BA). We get

= Tr⇣�Q A

↵ (�)NF Q˙�B +Q A

↵ (�)NF Q˙�B

⌘= Tr(0) = 0 (3.99)

On the other hand {Q A↵ , Q

˙�B} = 2� µ

↵ ˙�Pµ�AB which implies

Trh(�)NF {Q A

↵ , Q˙�B}

i= 2� µ

↵ ˙��ABTr

⇥(�)NFPµ

⇤= 0 (3.100)

For a non-zero momentum Pµ we must then have

Tr(�)NF = 0 (3.101)

implying that there are equal number of eigenvalues +1 and -1. In other words, there areequal number of bosonic and fermionic states. Q.E.D

3.3.1 Massless case

Now we are ready to build the representation of the supersymmetric algebra for massless par-ticles. We start with P 2 = 0. One can boost to a like-like reference frame Pµ = (E, 0, 0, E).Then, the space-time properties of the state are determined by its energy E and helicity �which is the projection of spin into the direction of motion, here being the 3-direction. Here,we have W 0|E,�i = �P 0|E,�i (see equation 1.143), or because of Lorentz covariance

Wµ|E,�i = �Pµ|E,�i (3.102)

Now, since [Q,Pµ]=0, when Q acts on a state, the energy and momentum remain unchanged.As for helicity, we need to consider the following:

W 0Q↵|E,�i = Q↵W0|E,�i| {z }

E�|E,�i

+[W 0, Q↵]|E,�i (3.103)


but we need to determine [W 0, Q↵]. Consider

[Wµ, Q↵] =1

2✏µ⌫⇢�[M⌫⇢P�, Q↵]

=1

2✏µ⌫⇢�

0

@[M⌫⇢, Q↵]P� +M⌫⇢ [P�, Q↵]| {z }0

1

A

=1

2✏µ⌫⇢�(SL⌫⇢)

�↵ Q�P�

(3.104)

For µ = 0:E

2

⇣✏0123(S

12

) �↵ Q� + ✏0213(S

21

) �↵ Q�

⌘= E(S

12

) �↵ Q� (3.105)

on the other hand

(Sij)�

↵ =1

2✏ijk�

k ) (S12

) �↵ =

1

2✏123

�3 =1

2�3 (3.106)

Therefore, we get from 3.103

W 0Q↵|�, Ei = E

✓�1+

1

2�3

◆ �

↵

Q� |E,�i (3.107)

Noting that when ↵ 6= � the result is zero since �3 =

✓1 00 �1

◆, there can only be two cases:

Either ↵ = � = 1, then

Q1

raises the helicity by 1

2

or ↵ = � = 2 which corresponds to -1 so that

Q2

lowers the helicity by 1

2

As already mentioned, [Mµ⌫ , Q↵] =�Sµ⌫R

�˙�

↵Q

˙� with (Sµ⌫R )

˙ba = �1

2

�3

which only di↵ers tothe previous case by a minus sign. Hence

W 0Q↵|�, Ei = E

✓�1� 1

2�3

◆˙�

↵

Q˙� |E,�i (3.108)

Q˙

1

lowers the helicity by 1

2

and


Q˙

2

raises the helicity by 1

2

Recall that we boosted to a frame where Pµ = (E, 0, 0, E) i.e. Pµ = (E, 0, 0,�E), then

{Q A↵ , Q

˙�B} = 2� µ

↵ ˙�Pµ�

AB (3.109)

where

�µPµ =

✓p0

+ p3

p1

� ip2

p1

+ ip2

p0

� p3

◆=

✓0 00 2E

◆(3.110)

Therefore,

{Q A↵ , Q

˙�B} = 2

✓0 00 2E

◆�AB (3.111)

{Q A↵ , Q B

� } = {Q↵A, Q ˙�B} = 0 (3.112)

in the absence of central charges.

Note that because of the positivity condition 3.25, and the fact that the anticommutatorof Q and Q is only non zero when ↵ = � = 2 (due to 3.111), we must have

Q A1

= Q˙

1A = 0 (3.113)

Rescaling the Q’s

aA =1

2pEQ A

2

a†A =1

2pEQ A

˙

2

=�aA

�† (3.114)

where A runs from 1 to N. So we have N creation and annihilation operators satisfying

{aA, a†B} = �AB

{aA, aB} = {a†A, a†B} = 0

(3.115)

There are N fermionic degrees of freedom. Any irreducible representation of this algebra ischaracterised by a Cli↵ord ground state (vacuum) denoted by |E,�

0

i ⌘ ⌦� where � is thestate of lowest helicity. Thus, applying the annihilation operator on this state

aA⌦� = 0 (3.116)

Other states are generated by successive application of N operators�aA

�†.

a†i |E,�0

i = |E,�0

+1

2, ii (3.117)


anda†ia

†j |E,�

0

i = |E,�0

+ 1, iji (3.118)

Note that this is antisymmetric since they are fermionic operators that anticommute i.e.

a†ia†j = a†ja

†i implying |E,�

0

+ 1, iji = �|E,�0

+ 1, jii.

Therefore in general,

⌦(n)

�+ 12n, A1...An

=1pn!a†An

...a†A1⌦� (3.119)

The states ⌦(n)

�+ 12n, A1...An

have helicity � + 1

2

n and are antisymmetric in A1

...An. We now

have states with the same energy but di↵erent helicity which are�Nn

�-times degenerate. The

state of highest helicity is

a†1

a†2

...a†N |E,�0

i = |E,�0

+N

2, 12...Ni (3.120)

implying that the highest helicity is � = �+ N2

which occurs exactly once. The dimension ofrepresentations is

NX

n=0

✓N

n

◆= (1 + 1)N = 2N (3.121)

using binomial expansion. So we have

helicity: �0

�0

+ 1

2

�0

+ 1 ... �0

+ N2

No. of possible states:�N0

�= 1

�N1

�= N

�N2

�...

�NN

�= 1

One can check that the number of fermions and boson are indeed the same:

N/2X

n=0

✓N

2n

◆�

N/2X

n=0

✓N

2n+ 1

◆= 2N�1 � 2N�1 = 0 4 (3.122)

The results are summarised in the tables below.


hel. vs � -2 -3/2 -1 -1/2 0 1/2 1 3/2

2 13/2 1 11 1 1

1/2 1 10 1 1

-1/2 1 1-1 1 1

-3/2 1 1-2 1

Table 3.1: N=1 case

hel vs. � -2 -3/2 -1 -1/2 0 1/2 1

2 13/2 1 21 1 2 11/2 1 2 10 1 2 1

-1/2 1 2 1-1 1 2 1-3/2 2 1-2 1

Table 3.2: N=2 case



hel vs. � -2 -3/2 -1 -1/2 0 1/2

2 13/2 1 31 1 3 31/2 1 3 3 10 1 3 3 1

-1/2 1 3 3 1-1 3 3 1-3/2 3 1-2 1

Table 3.3: N=3 case

hel vs. � -2 -3/2 -1 -1/2 0

2 13/2 1 41 1 4 61/2 1 4 6 40 1 4 6 4 1

-1/2 4 6 4 1-1 6 4 1-3/2 4 1-2 1

Table 3.4: N=4 case

3.3.2 Massive case

In order to determine the representation of the supersymmetric algebra for a massive particle,we first boost to the rest frame of the particle: Pµ = (M, 0, 0, 0) = Pµ. Now we have,

�µPµ =

✓p0

+ p3

p1

� ip2

p1

+ ip2

p0

� p3

◆=

✓M 00 M

◆(3.123)

and so

{Q A↵ , Q

˙��} = 2M�↵ ˙��AB

{Q A↵ , Q B

� } = {Q˙↵A, Q ˙��} = 0

(3.124)


where A and B run to 1 to N. We can rescale the Q’s such that

a↵A =

1p2M

Q↵A (3.125)

�a↵

A�†

=1p2M

Q↵A (3.126)

Then,1p2M

⇥ 1p2M

{Q↵A, Q

˙�B} = �↵ ˙��AB (3.127)

which implies

{a↵A,�a�

B�†} = �↵

��AB

{a↵A,�a�

B�= {

�a↵

A�†

,�a�

B�†} = 0

(3.128)

Our aim from now on is to prove that a↵A and

�a↵

A�†

are in fact the 2N fermionic creationand annihilation operators.

We will first consider the case where N = 1. Using relations 3.88 and 3.89 we get

[Ji, Q↵] =1

2(�i)

�↵ Q� (3.129)

implying

JiQ↵ = Q↵Ji +1

2

��i� �

↵Q� (3.130)

taking the hermitian conjugate and using (Sµ⌫R )

˙ba = �[(Sµ⌫

L ) ba ]⇤,

[Ji, Q↵] = �1

2

h��i� �

↵

i⇤Q

˙� (3.131)

When i = 3 i.e. J3

, we have from 3.130, and taking the normalisation into accout,

a1

|j3

i = |j3

+1

2i (3.132)

a2

|j3

i = |j3

� 1

2i (3.133)

Similarly for Q˙�

a†1

|j3

i = |j3

� 1

2i (3.134)


a†2

|j3

i = |j3

+1

2i (3.135)

Consider now j=0:

[Q↵, J2] = [Q↵, Ji]Ji + Ji[Q↵, Ji] = +

1

2(�i)

�↵ Q

˙�Ji + Ji

✓1

2�i

◆˙�

↵

Q˙� (3.136)

replacing

JiQ ˙� = Q˙�Ji �

1

2(�i)

�˙�Q� (3.137)

we get

1

2(�i)

˙�↵ Q

˙�Ji +1

2(�i)

˙�↵

✓Q

˙�Ji �1

2(�i)

�˙�Q�

◆

=(�i)˙�

↵ Q˙�Ji �

1

4(�i)

˙�↵ (�i)

�˙�Q�

=(�i)˙�

↵ Q˙�Ji �

3

4Q↵

(3.138)

Therefore,

[Q↵, J2] = �3

4Q↵ + (�i)

˙�↵ Q

˙�Ji (3.139)

Let ↵ = 1 and � = 2 then

[Q↵Q ˙� , J3] =[Q˙

1

Q˙

2

, J3

] = [Q˙

1

, J3

]Q˙

2

+ Q˙

1

[Q˙

2

, J3

]

=1

2(�

3

) �˙

1

Q�Q ˙

2

+1

2Q

˙

1

(�3

)˙�

˙

2

Q˙�

(3.140)

However, the first term only contributes to the sum if � = 1 and the second term only doesso when � = 2

1

2Q

˙

1

Q˙

2

+ Q˙

1

✓1

2

◆⇥ (�1) Q

˙

2

=1

2Q

˙

1

Q˙

2

� 1

2Q

˙

1

Q˙

2

= 0 (3.141)

[a†1

a†2

, J3

] = 0 (3.142)

[Q˙

1

Q˙

2

, J2] =[Q˙

1

, J2]Q˙

2

+ Q˙

1

⇥Q

˙

2

, J2

⇤

=� 3

4Q

˙

1

Q˙

2

+ (�i)↵˙

1

Q↵JiQ ˙

2

� 3

4Q

˙

1

Q˙

2

+ (�i)˙�

˙

2

Q˙

1

Q˙�Ji

=� 3

2Q

˙

1

Q˙

2

+ (�i)↵˙

1

Q↵JiQ ˙

2

+ (�i)˙�

˙

2

Q˙

1

Q˙�Ji

(3.143)


Using

[Q˙

2

, Ji] = Q˙

2

Ji � JiQ ˙

2

=1

2(�i)

˙�˙

2

Q˙� (3.144)

to substitute for�JiQ ˙

2

�, we get

= �3

2Q

˙

1

Q˙

2

+ (�i)↵˙

1

Q↵Q ˙

2

Ji �1

2(�i)

↵˙

1

(�i)˙�

˙

2

Q↵Q ˙� + (�i)˙�

˙

2

Q˙

1

Q˙�Ji (3.145)

Summing over i = 1, 2, 3 results in

�3

2Q

˙

1

Q˙

2

+Q˙

2

Q˙

2

J1

� 1

2⇥ 1⇥ 1⇥ Q

˙

2

Q˙

1

+ Q˙

1

Q˙

1

J1

�iQ˙

2

Q˙

2

J2

� 1

2⇥ (�i)⇥ (+i) Q

˙

2

Q˙

1

+ iQ˙

1

Q˙

1

J2

+Q˙

1

Q˙

2

J3

� 1

2⇥ 1⇥ (�1) Q

˙

1

Q˙

2

+ (�1) Q˙

1

Q˙

2

J3

=Q˙

2

Q˙

2

J� + Q˙

1

Q˙

1

J+

(3.146)

where J� decreases J3

by 1 and J+

increases it by 1. On the other hand, Q˙

2

Q˙

2

acting on anystate gives zero. This is because {Q

˙

2

, Q˙

2

} = 0 ) Q˙

2

Q˙

2

= �Q˙

2

Q˙

2

, can only be if Q˙

2

Q˙

2

= 0.Similarly for Q

˙

1

Q˙

1

. Hence,

[Q˙

1

Q˙

2

, J2] = 0 (3.147)

Recall

[Q↵, J2] = �3

4Q↵ + (�i)

˙�↵ Q

˙�Ji

)J2Q↵ = Q↵J2 +

3

4Q↵ � (�i)

˙�˙

1

Q˙�Ji

)J2Q˙

1

|⌦i = Q˙

1

J2 +3

4Q

˙

1

� (�i)˙�

˙

1

Q˙�Ji

(3.148)

Also

(�i)˙�

˙

1

Q˙�Ji =(�

1

)˙�

˙

1

Q˙�J1 + (�

2

)˙�

˙

1

Q˙�J2 + (�

3

)˙�

˙

1

Q˙�J3

=Q˙

2

J1

� iQ˙

2

J2

+ Q˙

1

J3

=Q˙

2

(J1

� iJ2

)| {z }J�

+Q˙

1

J3

(3.149)

Let the ground state be |⌦i = |m, 0, 0i, then

J2Q˙

1

|⌦i =✓Q

˙

1

J2 +3

4Q

˙

1

Q˙

2

� Q˙

2

J� � Q˙

1

J3

◆|⌦i (3.150)


The first term and the last two terms vanish since both j and j3

are zero for this state. Butthe left hand side of the above equation is

J2 Q˙

1

|⌦i| {z }

spin j state

:= j(j + 1) Q˙

1

|⌦i| {z }

(3.151)

So here we must have

j(j + 1)Q˙

1

|⌦i = 3

4Q

˙

1

|⌦i (3.152)

which immediately implies that j = 1

2

. On the other hand, we had we also had, for generalj, Q

˙

1

|j3

i = |j3

� 1

2

i (ignoring the normalisation). Here we have a state with j = 0 ) j3

= 0

therefore,⇣a†1

|⌦i⌘has j = 1

2

and j3

= �1

2

i.e.

a†1

|⌦i = |m,1

2,�1

2i (3.153)

Similarly for Q˙

2

J2Q˙

2

|⌦i = Q˙

2

J2 +3

4Q

˙

2

� (�i)˙�

˙

2

Q˙�Ji (3.154)

and

(�i)˙�

˙

2

Q˙�Ji =(�

1

)˙�

˙

2

Q˙�J1 + (�

2

)˙�

˙

2

Q˙�J2 + (�

3

)˙�

˙

2

Q˙�J3

=Q˙

2

J1

+ iQ˙

2

J2

� Q˙

2

J3

=Q˙

2

(J1

+ iJ2

)� Q˙

2

J3

(3.155)

resulting in 3.154 to be equal to✓Q

˙

2

J2 +3

4Q

˙

2

� Q˙

2

J+

+ Q˙

2

J3

◆|⌦i (3.156)

and because the total j of the state is zero while a†2

|j3

i = |j3

+ 1

2

i we conclude

a†2

|⌦i = |m,1

2,+

1

2i (3.157)

It is important to mention that a†1

and a†2

are grassmann variables, i.e. they satisfy thefollowing:

a†↵a†� = �a†�a

†↵ (3.158)

and{a†

2

, a†2

} = {a†1

, a†1

} = 0 (3.159)


so

a†2

a†2

= �a†2

a†2

) a†2

a†2

|⌦i = 0 (3.160)

Similarly,

a†1

a†1

|⌦i = 0 (3.161)

Therefore, it only remains to consider a†2

a†1

|⌦i which is an object with the same spin valueas ⌦ i.e. if ⌦ has spin j then this state is also a spin j object.

For the case of j = 0 we then have:

|⌦i = |m, j = 0, Pµ, j3

= 0i (3.162)

a†1,2|⌦i = |m, j =

1

2, Pµ, j

3

= ±1

2i (3.163)

a†1

a†2

|⌦i = |m, j = 0, Pµ, j3

= 0i (3.164)

So we have two states of spin 0 and one of spin 1

2

.However, 3.162 and 3.164 look the same. The questions now is that how do we know thesestates are actually di↵erent di↵erent. They are di↵erent since if we apply parity the formertransforms as a scalar while the latter transforms as a pseudovector.

Also note that a†1

a†2

|⌦i belongs to the 1

2

⌦ 1

2

= 0�1 representation, but since a†1

a†2

= �a†2

a†1

isantisymmetric, we only take the antisymmetric part of the representation i.e. (0, 0) meaning

that acting on |⌦i by a†1

a†2

does not change the spin of the state so it remains zero.

Let us now consider the case where j 6= 0 :

• We know that a†2

(a†1

) acting on the state will increase (decrease) the third componentof spin, J

3

by 1

2

. (Operator)

• Also by the addition of angular momenta j1

⌦ j2

= |j1

� j2

|, ..., |j1

� j2

|. (Spinor)

Here we have the product of a spinor (a†) and a state which can be a spinor or otherwise

1

2⌦ j = |j � 1

2|, j + 1

2(3.165)

a†1

|⌦i = k1

|m, j +1

2, Pµ, j

3

� 1

2i+ k

2

|m, j � 1

2, Pµ, j

3

� 1

2i (3.166)


a†2

|⌦i = k3

|m, j +1

2, Pµ, j

3

+1

2i+ k

4

|m, j � 1

2, Pµ, j

3

+1

2i (3.167)

Where the k’s are the Clebsch-Gordon coe�cients.

Recall that a†1

and a†2

are Grassmann variables so a†2

a†1

= �a†2

a†1

i.e. antisymmetric whichimplies that we only take the antisymmetric part of the representation 1

2

⌦ 1

2

= 0� 1. Hence

a†2

a†1

|⌦i = �a†2

a†1

|⌦i which is a spin j object like ⌦ (3.168)

The multiplet has spins (j, j + 1

2

, j � 1

2

, j) or explicitly

2⇥ |m, j, Pµ, j3

iTotal number of states

��! 2⇥ (2j + 1) = 4j + 2

|m, j � 1

2, Pµ, j

3

i ��! 2 (j +1

2) = 2j + 2

|m, j +1

2, Pµ, j

3

i ��! 2 (j � 1

2) = 2j

(3.169)

Note that if the two spin j state are fermions, the other two states are boson (since they di↵erby 1

2

) and vice-versa. This is in agreement with what we proved before that the number ofbosons and fermions in a multiplet is equal.For the case of N = 1 we have

Spin ⌦0

⌦ 12⌦1

⌦ 32

0 2 11/2 1 2 11 1 2 1

3/2 1 22 1

Table 3.5: Massive particle N=1 case

In general, it can be said that, all the states are generated by successive application of a†’son the ground state characterised by |m, s

0

, s3

i. Thus a typical state may be written as

| i =�a A1↵1

�†...�a An↵n

�† |m, s0

, s3

i (3.170)

which is totally antisymmetric.

Maximal spin: Knowing thatA = 1, ..., N and a†2

increases the spin by 1

2

we get a†1

1

...a†1

N|m, s

0

i


which mean that smax

= s0

+ N2

. This is because there are N of a†2

’s.

Minimal spin: We must have smin

� 0 therefore

(0 for N

2

� s0

s0

� N2

Otherwise(3.171)

Top state: One can reach this state y applying all 2N fermionic operators once. Its spin isthe same as that of the original state i.e. s

0

. This is because the result of applying all N ofa†1

’s, each of them decreasing the spin by 1

2

is then cancelled by applying all N of a†2

’s, eachincreasing it by 1

2

. Hence the net result will be the original spin.

Bibliography

[1] Steven Weinberg, The quantum theory of fields: Foundations. volume 1 CambridgeUniversity Press, Cambridge, 2005.

[2] Mark Allen Srednicki, Quantum field theory, Cambridge University Press, Cambridge,2007

[3] Julius Wess and Jonathan Bagger, Supersymmetry and supergravity, Princeton UniversityPress, Princeton, N.J., 2nd ed., rev. and expanded edition, 1992.

[4] Martin F. Sohnius, Introducing supersymmetry, Cambridge, 1985.

[5] Wu-Ki Tung, Group theory in physics, World Scientific, Philadelphia, 1985.

[6] Steven Weinberg, The quantum theory of fields: Supersymmetry. volume 3 CambridgeUniversity Press, Cambridge, 2005.

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