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6.1. Introduction & the supersymmetry algebra 301

If the infinitesimal generators of the symmetries ofa quantum field theory are bosonic, i.e. they obeycommutation relations, then under some reason-

able physical assumption, the corresponding sym-metry algebra is a product of the Poincare algebraand an internal symmetry algebra.

Haag, Lopuszansk and Sohnius[10] showed that if in ad-dition, anticommuting spinor generators are allowed, (ex-tended) supersymmetry is the only possible generalisationof the Poincare algebra.

In extended supersymmetry there are N sets of spinor gener-ators QA , QA, A = 1, 2, , N; which satisfy the followingmodified relations

{QA , QB} = 2AB P{QA , QB} = ZAB (6.15)

{QA, QB} = ZAB .

The generators ZAB commute with all the other generatorsof the extended supersymmetry algebra.

In addition, the spinors QA transform in the N dimensionali.e. defining representation of the group of N dimensional

unitary matrices U(N) (SU(4) for N = 4). For our caseN = 1, there is a U(1) symmetry. The generators Q andQ may be assigned charges +1 and 1 respectively underthis symmetry. If R is the generator of this U(1), we have

[R, Q] = Q

[R, Q] = Q, (6.16)

and [R, P] = 0, [R, M] = 0. This is a chiral symmetry,and is in general anomalous. We shall make use of thissymmetry in our discussion of effective field theories.


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302 6. N = 1 Supersymmetric Gauge Theories

The supersymmetry algebra is a generalisation of a Lie alge-bra, and hence is constrained by (generalised) Jacobi iden-tities. The structure of these identities are as follows.

(1)AC [[A, B} , C} + (1)BA [[B, C} , A}+ (1)CB [[C, A} , B} = 0,

where, is 0 (respectively 1) for bosonic (fermionic) oper-ators, and the mixed bracket notation [, } stands for ananticommutator when both the operators are fermionic anda commutator otherwise.

6.2 Representations of supersymmetry on states

We shall now discuss the irreducible representations of the su-persymmetry algebra, i.e. a collection of bosonic and fermionicstates/fields that transform covariantly under supersymmetry.First we shall discuss particle states as supersymmetry represen-tations, and come back to the representation on (quantum) fieldsin the next lecture.

To begin with let us recall that particle representations of thePoincare algebra are labelled by the eigenvalues of the followingCasimir operators

P2 = PP

with eigenvalue m2 (mass square),

W2 = WW where W = 12PM is the Pauli-Lubansk vector.

Exercise: Show that

1. for massive particles, i.e. m2 = 0, W2 = m2j(j + 1), where j = j+ +j;

2. for massless particles, i.e. m2 = 0, W = P,where is the helicity.


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6.2. Representations of supersymmetry on states 303

Therefore mass and spin/helicity are the quantum numbersthat label particle states. In order to describe which quantumnumbers label representations of supersymmetry, notice that

[P, Q] = [P, Q] = 0.

Hence P2 commute with the new generators, and continue to bea Casimir of the enlarged symmetry. However,

[M, Q] = 0 [M, Q] = 0,and W2 is no longer a Casimir of the supersymmetry algebra. Weneed to make the following modification. Define

B = W 14

Q Q,

C

= B

P

B

P

. (6.17)

Exercise: Show that [C, Q] = 0.

Therefore the second Casimir of the supersymmetry algebrais C2 = CC

. Supersymmetry multiplets are labelled by massand eigenvalue of the operator C2.

We are now ready to construct the representations of the su-persymmetry algebra on particle states, i.e. on asymptotic on-shell physical states.

First, let us consider massive states, i.e. m2 = 0. In this case,one can go to the rest frame and make the choice P = (m, 0).

With this choice, we find that

{Q, Q} = 2P

= 2m

1 00 1

,

or, explicitly in terms of the components

{Q1, Q1} = 2m{Q2, Q2} = 2m{Q1, Q2} = 0 = {Q2, Q1}.


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We have here two pairs of fermionic creation/annihilation opera-tors.

We shall (arbitrarily) choose the dotted spinorial generators

Q, ( = 1, 2), to be the creation operators and the undotted onesannihilation operators. Further we may rescale Qs by 1/

2m to

define conventionally normalised creation/annihilation operators

a =12m

Q, = 1, 2;

a =12m

Q, = 1, 2. (6.18)

Let us define a state | such that

a| = 0, for = 1, 2.

| is a (Clifford) vacuum state with respect to the fermionic cre-ation/annihilation operators.

What are the quantum numbers that label this state? Thanksto our discussion on Casimirs, we know the answer to this question.One quantum number is of course the mass m. To find the otherone:

Exercise: Show that on massive states, in the restframe,

Bi = mLi 1

4m

QiQ m Li

and hence,

C0i = mBi = m2LiCij = 0.

Therefore, C2 = 2C0iC0i = 2m4LiL

i.

It is easy to check that the generators Li obey the angularmomentum algebra. So the eigenvalues of L2 may be labelled by


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( + 1), where can take any half-integral value. Actually, withour choice of |,

Li| = Li|.Hence, acting on |, = j label the spin j = j+ +j. Being aneigenstate of spin, | is labelled by

| = |m;j, j3, j3 = j, j + 1, , j 1, j.

The (Clifford) vacuum | is (2j + 1)-fold degenerate.The excitations over the vacuum | are defined by the

fermionic creation operators a1 and a2. We have

|a1| a2|

a1a2|(6.19)

i.e. a total of 4(2j + 1) states in the supersymmetry mutiplet.

Recall that a Q transforms as a

0, 12

spinor. In par-

ticular, (by a choice of convention), a1 (respectively a2) has L3

eigenvalue +12 (12). The spins of the different states in a massivesupermultiplet are

state | a1| a2| a1a2|spin j3 j3 +

12 j3 12 j3

(6.20)

If j is an integer, the first and the last states are bosonic, andthe second and third ones are fermionic. The statistics is oppositewhen j is a half odd integer. As an example, consider j = 0.There are then four states in the supersymmetry multiplet, twoof these have spin j3 = 0 and the other two have j3 = 12. Thespin zero states may be combined into a scalar and a pseudo-scalarwhile the spin half states describe the degrees of freedom of a Weylfermion.

This matching of bosonic and fermionic degrees of freedom ina multiplet is a remarkable property of supersymmetry. We caneasily prove the following


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Theoerem: Every representation of supersymme-try algebra contains an equal number of bosonic andfermionic states.

Proof: Let us define the operator (1)NF whose eigen-values are +1 on bosonic and 1 on fermionic states.By definition

(1)NFQ = Q(1)NF .

Now taking a trace of the representation, (which we as-sume to be finite dimensional for it to be well defined),we have

tr (1)NF

{Q, Q

}= tr[(

1)NFQQ

+(1)NFQQ}]i.e. tr

(1)NFP

= 0,

where we have used the supersymmetry algebra in theLHS and the cyclic property of trace and the identityinvolving (1)NF and Q in the RHS. For a fixed valueofP in a given multiplet, we then have, tr[(1)NF ] =0, which proves the assertion.

Now let us discuss the supersymmetry representation on mass-less states. We can choose a reference frame such that

P = (E, 0, 0, E)

Exercise: Show that on massless states W0 = E,W3 = E, and hence

B0 = W0 14

QQ

B3 = W3 +1

4Q3Q.


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Also that the only non-vanishing component of C is

C03 = E(B0 B3) = 12

EQ2Q2,

whence, C2 = 0.

In our chosen basis the supersymmetry algebra takes the fol-lowing form

{Q, Q} = 4E

1 00 0

,

or, explicitly in components

{Q1, Q1} = 4E{Q2, Q2} = 0

{Q1, Q2} = 0 = {Q2, Q1}.As in the massive case, let us define | annhilated by the an-nihilation operators a1 Q1 and a2 Q2. In addition, since{Q2, Q2} = 0, we have

|Q2Q2| = 0,i.e, the excitation Q2| is a null state, or Q2 is zero in the oper-ator sense.

This leaves us with only one pair of creation/annihilation op-erators a = 1

2

EQ1 and a

= 12

EQ1, which satisfy {a, a} = 1.

The massless supersymmetry consists of the states| : a non-degenerate state of helicity

a| : a non-degenerate state of helicity + 12 .(6.21)

Notice that the massless supersymmetry multiplet is not a CPTeigenstate. One needs two pairs of irreducible massless multiplets,i.e. four states of helicity , + 12 and , 12 to complete aCPT eigenstate.

We just finished discussing how certain bosonic and fermionicparticle states form representations of supersymmetry. In other


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words, we have a set of states which transforms covariantly undersupersymmetry variation. These are asymptotic on-shell states.In order to construct a quantum field theory, however, we need

to know how general off-shell states form representations of su-persymmetry. This can be done by considering bosonic andfermionic fields and studying their behaviour under supersymme-try variation. A far more economic and elegant approach is interms of what will be called superfields. Different bosonic andfermionic fields that mix under supersymmetry transformationscan be thought of as components of this single superfield. Thisapproach is also advantageous from a practical point of view, asmany properties of supersymmetry are manifest when expressedin terms of superfields. These concepts were introduced by Salamand Strathdee[11].

To do this, however, we need to make a digression to discussthe algebra and calculus of Grassmann variables. To this end, letus introduce spinor parameters , (, = 1, 2) (notice theindex assignment) which satisfy the relations

{, } = 0,{, } = 0, (6.22){, } = 0,

as also [x, ] = 0, [x, ] = 0. These are anticommuting ana-logues of a complex variable z, and are called Grassmann numbers

or variables. The pair (, ) can be taken to parametrise infinites-imal supersymmetry variation

susy = (Q + Q

) (Q + Q). (6.23)

Due to the anticommuting nature of the parameters (, ), thecombinations Q Q and Q Q satisfy the commutationrelations

[Q, Q] = 2()P,

[Q,Q] = 0 = [Q, Q]. (6.24)


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All the algebraic relations of supersymmetry are now expressedin terms of commutators. (We shall also consider the replace-ment P yP, where y is an infinitesimal parameter fortranslation.)

The infinitesimal variations may now be exponentiated to de-fine a finite transformation

G(y,, ) = exp

iyP + Q + Q

.

Notice the dimensions of the parameters [y] = M1 and [] =[] = M1/2.

Parenthetical comments: Actually, we should have written

G(y,, ) = exp

iy P + Q + Q exp i

2M

,

but we left out the Lorentz transformation part. It is consistentto set that part to identity. In other words, we are parametrisinga coset space defined by the quotient of the super-Poincare groupby its Lorentz subgroup.

Notice also that since the anti-commutators of Q, Q are non-zero, the following forms

exp {i (y P + Q)} exp iQexp

iy P + Q exp {iQ}

and the one we gave earlier for G are not all equivalent. It isalso consistent to work with either of the above forms and get thesame results. However the explicit differential operator form forthe generators Q and Q will be different in each case.

Notice that in the above the Grassmann parameters (, )appear in the same footing as the coordinates y. So the fullparameter space is labelled by

(y, , ); = 0, , 3; , = 1, 2.We should think of this space as 4 normal (i.e. bosonic) plus 4anti-commuting (i.e. fermionic) extension of our familiar space-time. (The dimension of the extended space is sometimes writtenas (4|4).) This is called the N = 1 rigid superspace.


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Just as it is advantageous to construct relativistic quantumfield theory in a manifestly Lorentz covariant formalism, it is ofgreat advantage to formulate supersymmetric theories in super-

space.

6.3 Superspace & superfields

We can define functions in superspace these are going to be thesuperfields and differentiate and integrate them with respectto the coordinates (x, , ). To explain the rules on differentia-tion and integration, let us consider the simpler example of a (1|1)dimensional superspace. This has only two coordinates (x, ) and2 = 0. Due to the nilpotence of the coordinate , Taylor expan-sion of a function f(x, ) in superspace in terms of terminate

after the linear term:

f(x, ) = f0(x) + f1(x), (6.25)

where f0(x) and f1(x) are functions of of the commuting coordi-nate x. Using

d

d() = 1

d

d(1) = 0, (6.26)

it follows thatd

d(f(x, )) = f

1(x). (6.27)

(Notice that the dimension [d/d] is M1/2.)

Now since we want the integral of a total derivative to vanish,we define the following rules of integration

d = 0,

d = 1, (6.28)

which gives d

d

df(x, ) =

d f1(x) = 0.


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6.3. Superspace & superfields 311

The integral so defined is invariant under translation of by anarbitrary constant :

d( + ) f(x, + ) =

d [f0(x) + ( + )f1(x)]

=

df1(x)

=

d f(x, ).

It is a curious fact that integration and differentiation in areequivalent!

dd

f(x, ) = f1(x)

d f(x, ) = f1(x)And consistent with these rules [d] = M1/2, unlike in ordinaryspace. Finally, we can define a delta function by

() =

leading to the expected result

d () = 1.

Coming back to our (4|4) dimensional superspace, we have thefollowing rules (

/ and

/):

= ,

= () = ,

=

,

= . (6.29)

One can also raise index of by chain rule of differentiation

=

= () = .


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The following is a compilation of useful results that will come inhandy in our subsequent calculations:

=

() = 2

= 2 (6.30)

2 () = 4

2

= 4

It is straightforward to derive the above.As for integration, we shall define the following convention

d2 = 14

dd =1

2d1d2,

d2 = 14

dd = 1

2d1d2 (6.31)

d4 d2 d2;

so that, d2 = +1,d2 = +1. (6.32)

After this long detour we get back to the transformation gen-erated by

G(x, , ) = exp

i(xP + Q + Q)

.

This generator is unitary since (Q) = Q = Q. If we

consider two such successive transformations, the result is

G(x,, ) G(y,, ) = exp i(x + y i

i

)P + i ( + ) Q + i

+

Q

.


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6.3. Superspace & superfields 313

Exercise: Show the above. Hint: You will need theBaker-Campbell-Hausdorf formula

eA

eB

= exp

A + B +

1

2! [A, B]

+1

3!

1

2[[A, B], B] +

1

2[A, [A, B]]

+

.

The successive applications of superspace transformations gener-ate the following motion is terms of the superspace coordinates

(x,, )G(y,) (x + y + i , + , + ),

which is given by the following differential operators

yP = i y

x

,

Q =

i

,

Q =

+ i

.

Alternatively, we may write5,

P = i

x,

Q = i, (6.33)

Q = i.

The above differential operator representation leads to the anti-commutator

{Q, Q} = 2P,in apparent disagreement with (6.14), due to the extra minus sign.There is, however, no contradiction, as what we witness here isthe difference in the active and passive points of view of symme-try transformations. The differential operator representation is interms of superspace coordinates.

5Since Q and Q are not hermitian operators, we have rescaled them byfactors of i, without anything going wrong.


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We shall now define a general scalar superfield (x,, ) inour (4|4) dimensional N = 1 rigid superspace. A scalar functionsatisfies

(x, , ) = (x,, ),and hence

=

Q + Q

. (6.34)

Let us now consider the Taylor expansion of in powers of and.

(x,, ) = (x) + (x) + (x) + ()m(x)

+()n(x) + ()v(x) + ()(x)

+()(x) + ()()d(x). (6.35)

Each term in the above expansion is a field in physical (Minkowski)spacetime. In particular

Fields Type Bose dof Fermi dof (x), m(x) scalars 4 2 0

n(x), d(x) (x), (x) L-spinors 0 2 4 (x), (x) R-spinors 0 2 4 v(x) vector 4 2 0

(6.36)

The fields {(x), (x), (x), } are called the components ofthe superfield .

Let us remark parenthetically that the above is the most gen-eral possible expansion for a scalar superfield, since e.g. = and ()() = 2(), etc. (See Appendix 6.11 forthis type of manipulations.)

Now consider the supersymmetry variation of the scalar su-perfield (6.34). In terms of the component fields this leads to thefollowing relations

= + ,

= 2m + ()(v + i),

= 2n + ()( v + i),


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6.4. Chiral & vector superfields 315

m = i2

() ,

n = +i

2(), (6.37)

v = + +i

2() i

2(),

= 2d + i()m +

i

2(v),

= 2d + i()n i

2(v),

d =i

2() i

2()

.

Exercise: (i) Derive the above relations. (You willneed to use the Fierz identities given in Appendix6.11.)(ii) Work out

(12 21) = 2i (12 21) .

6.4 Chiral & vector superfields

The result obtained at the end of the last lecture shows that thegeneral scalar superfield forms a basis for an off-shell linear repre-sentation of supersymmetry:

Supersymmetry variations of component fields are propor-

tional to each other (also involving derivatives).

Therefore the supersymmetry algebra closes, i.e. supersym-metry variations involve only those fields present in the mu-tiplet and no others.

However, there are a large number of component fields. It turnsout that the set is not the minimal one. In other words, the scalarsuperfield representation is reducible6.

6It is not fully reducible though. That is, the scalar superfield cannot bewritten as a direct sum of irreducible superfields.


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In an effort to reduce the number of component fields in (6.35),let us try to set one of the spinor fields, say to zero. To makethis consistent with supersymmetry, we should also require that

its supersymmetry variation vanishes, and so on. In the end, wehave to impose the following set of constraints

(x) = 0v(x) = i(x)n(x) = 0

(x) = 0(x) = i2()

d(x) = 14 (x)

(6.38)

leading to a reduced scalar superfield

R = + + ()m + i()

+ i2

()

() 1

4()() .

Exercise: Check the mutual consistency of the con-straints imposed in (6.38). This demonstrates that thereduced scalar superfield R, (with fewer number ofcomponents than the general scalar superfield ), de-fines an off-shell linear representation of the supersym-metry algebra.

Let us now define

y = x + i

. (6.39)

Using Taylor expansion, and some spinor identities, one canrewrite the restricted superfield R as

R(y, ) = (y) + (y) + ()m(y), (6.40)

which shows that the restricted superfield is a function of y and, but has no explicit dependence on . Had it not been for the in the definition of y in (6.39), we could have concluded that Ris independent of , that is

R(y, )!

= 0.


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This is of course not the case. Another problem is that

[, Q] = i,

that is, does not commute with supersymmetry variation.Hence imposing a constraint like R = 0 is not consistent withsupersymmetry. In other words, R is not a superfield. Thank-fully from our experience with tensor analysis, we know what todo in such a situation: define an appropriate covariant derivativeto impose the constraint consistently. The covariant derivativedefined as

D = i (6.41)leads to a consistent way to impose the constraint

D = 0 (6.42)

on the general scalar superfield (6.35) to restrict it to R.

Exercise: Show that [D, Q] = 0, or equivalently{D, Q} = 0. Also show that {D, Q} = 0, and{D, D} = 0.Exercise: Show that Dy

= 0 and D = 0.

The last exercise shows that a superfield constrained by (6.42) isa function of y and only, and has no explicit dependence on .

Definition: A scalar superfield constrained by the (spinorial)

chirality condition D = 0 is called a chiral superfield.We have already found an example of a chiral superfield in

(6.40):(y, ) = (y) +

2(y) + F(y). (6.43)

(In the above, a scaling 2 and a change in notation m Fhas been done to conform with standard notation in the litera-ture.) A chiral superfield has the following

complex scalar dof = 2 bose [] = M, complex L-spinor dof = 4 fermi [] = M3/2, complex scalar F dof = 2 bose [F] = M2.


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Under infinitesimal supersymmetry transformation

=

2

= 2F + 2i

(6.44)F =

2i

.

An important property of a chiral superfield is that the productof two chiral superfields is again a chiral superfield. This followsfrom the chain rule of covariant differentiation.

The notion of an anti-chiral superfield is immediate:Definition: A scalar superfield that satisfies the condition D =0, where

D = + i

, (6.45)

is an anti-chiral superfield.

Exercise: Show that {D, Q} = 0. Also show that{D, Q} = 0, and {D, D} = 0, and finally

{D, D} = 2i.

Exercise: Let y = x i. Show that Dy = 0and D

= 0.

An anti-chiral superfield is therefore a function of y and only.In particular, if (y, ) is a chiral superfield (6.43), (y, ) is ananti-chiral superfield

(y, ) = (y) +

2(y) + F(y).

As before, the product of two anti-chiral superfields is again ananti-chiral superfield. However the product is neither a chiralnor an anti-chiral superfield. Same applies to the sum ( + ).

Exercise: Show that the imposing the conditionsD = 0 and D = 0 simultaneously on a scalarsuperfield reduces it to a constant.


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Exercise: Define

=

==0 = D

==0

F = DD

==0.

Express , and F in terms of the component fields, and F. Compute the transformation laws for , andFusing the differential operator representation forQ, Q.

Exercise: Show that in terms of the variables(y,, ),the covariant derivatives may be written as

D = + 2i

y ,

D = 2i

y,

D = ,D = . (6.46)

Exercise: Compute the expressions for 2, 3 and in the component field expansion.

There is another way to reduce the number of fields in a generalscalar superfield. For a scalar superfield V(x,, ), let us impose

the covariant reality condition:

V(x,, ) = V(x,, ). (6.47)

In components, this leads to the following set of constraints

=

() =

m = n

v = (v)

= ()

d = d

(6.48)


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Therefore, a scalar superfield restricted by the reality conditionhas four real scalars, (which may be combined into two complexscalars), one real vector and two complex Weyl spinors, (or equiv-

alently, one real Majorana spinor). There are altogether eightbosonic and eight fermionic degrees of freedom.

We can already construct an example of a real superfield triv-ially from a chiral superfield. If is a chiral superfield, the sum(+ ) is a real superfield. In fact, this means that a real super-field is not uniquely determined. Given a real superfield V(x,, ),it is possible to construct another one

V(x,, ) V(x,, ) + (y, ) + (y, ). (6.49)This is a gauge freedom in defining a real superfield. If we expanda real superfield in components

V(x,, ) = (x) + (x) + (x) + M(x) + M(x)

()A(x) + i (x) i (x) + 12

D,

and use the freedom in +

(x,, ) = ( + ) +

2( + ) + F + F

+i( ) + i2

()

i2

() 14

( + ),

we find that

= + = 2 R e =

2

M = F

A = i( ) = (2Im ) (6.50) =

12

D = 12

( + ) = 2 R e .


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6.5. More on vector superfields 321

Let us first point out the most interesting aspect of the gaugetransformation in (6.50) above:

A(x) = (2Im (x)) = (x),

where, (x) is a real scalar field. This is therefore the usual abeliangauge transformation for the vector field A(x) in the componentsof the real superfield V. For this reason, a real superfield is alsoknown as a vector superfield.

The more general superfield transformation (6.49) means thatany superfield action invariant under the above abelian gaugetransformation is also independent of several component fields ofV. In particular, we may choose Re , F and to set , m and to zero respectively. This partially fixes the gauge freedom inV

V + + , and is called the Wess-Zumino gauge. After this

gauge fixing, the vector superfield takes the form

VW Z = ()A + i i + 12

D. (6.51)

It should be stressed that the Wess-Zumino gauge imposes norestriction on Im , therefore it does not fix the abelian gaugefreedom of the component vector field A.

There are four real bosonic degrees of freedom in the compo-nent fields of VW Z: (4 1) = 3 from the vector A, and one fromthe real scalar D. Also there are four real fermionic ones from .

6.5 More on vector superfields

Another way to reformulate Wess-Zumino gauge fixing is to ob-serve that, without any loss of generality, a vector superfield maybe decomposed as

V = VW Z + + , (6.52)

where is a chiral superfield.

Exercise: Work out the component expansion forV2W Z and V

3W Z.


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While the Wess-Zumino condition allows one to get rid of thesuperfluous fields in a vector superfield and retain only the relevantones, it is not invariant under supersymmetry variation. In other

words, the WZ gauge condition is not covariant. It is, however,possible to give a covariant description in which only the necessaryfields in a vector superfield are retained. To this end let us definethe following covariant derivatives on a vector superfield.

W = 14

DD

DV(x,, )

= 18

DD

e2V De+2V

,

W = 14

(DD) DV(x,, )

=1

8

(DD)e+2VDe2V . (6.53)Notice that

1. The construction/definition of the W (W) ensures that7

DW = 0

DW = 0

, (6.54)

that is, W (resp. W) is a (anti-)chiral superfield. Since thisfield also carries a spinor index it is a chiral spinor superfield.

2. However, W is not a general chiral superfield, since it sat-isfies

DW = DW . (6.55)

3. The fields W and W are both invariant under the gaugetransformation (6.49).

Exercise: Prove the two statements mentioned above.

It is important to note that the two superfields W and W areinvariant under the full superfield gauge transformation. There-fore even the abelian gauge invariance of the component vector

7Since the Ds anticommute and has only two independent components,D3 = 0.


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6.5. More on vector superfields 323

field A (that the Wess-Zumino gauge did not fix), is no longeravailable. Consequently, the components of these (anti-)chiral su-perfields can be calculated in the Wess-Zumino gauge without any

loss of generality. That is the exercise we shall do now. In order tosimplify the computation, observe that being a chiral superfield,W is a function of y and only, (similarly W is a function of y

and ), with no explicit dependence on (resp. ).

Exercise: Carry out this computation in detail toshow that

W = i(y) + D(y) + iF(y)+()

(y),

W = + i(y) + D(y) i F(y)

+() (y). (6.56)Hint: Start by writing the real superfield in Wess-Zumino gauge VW Z as a function of (y,, ) and useresults from (6.46).

We see that the component fields in the spinor chiral superfieldW are a left-handed spinor , a scalar D and the field strength F.It is therefore also called a field strength supermultiplet. More-over, since is self-dual (6.9), only the self-dual part of Fcontributes to the degrees of freedom in W.

One can generalise this result in which the vector field in the

vector supermultiplet has abelian gauge invariance to one withnon-abelian gauge invariance. In order to do that, however, wehave to rewrite the abelian gauge condition in a way that is gen-eralisable to the non-abelian case. Recall the Wess-Zumino gaugefreedom V V++. With a scaling i, this is equivalentto

eV ei eV e+i. (6.57)Let us now elevate the superfields to Lie algebra valued superfields

V Vij = taij Va, ij = taij a,


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where ta are hermitian generators of some Lie algebra satisfyingta, tb

= ifabctc, (6.58)

and normalised so that

tr

tatb

= C(R) ab,

C(R) =dim R

dim gC2(R). (6.59)

In the above, dim R and C2(R) are the dimension and thequadratic Casimir in the representation R and dim g is the di-mension of the Lie algebra g.

To first order in gauge parameter superfield

V = i( ) + i2 V, ( +

)+ i

12

V, [V, ( )]

+ . (6.60)

The linear term V = i( ) still allows for a choice of WZgauge which, as in the abelian case, does not fix the non-abeliangauge freedom of the component vector fields Aa. Once we fix the

WZ gauge, we have VaW Z = ()Aa + , and also ( + ) =

2iRe (), ( ) = 2i()Im (). Therefore the third termand above in (6.60) vanish in the WZ gauge as they have too manys or s. The relation

VW Z = i(

) +

i

2V, ( + ) (6.61)

implies the usual non-abelian gauge transformation for the compo-nent fields A (non-abelian gauge field), (a spinor in the adjointrepresentation) and D (auxiliary scalar field in the adjoint repre-sentation).

As in the abelian case, one can define the constrained (anti-)chiral spinor superfield W (and W) by

W = 18

D2e2VDe2V ,

W = +1

8D2e+2VDe

2V. (6.62)


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6.6. Wess-Zumino model, supersymmetry breaking 325

These superfields transform homogeneously under gauge transfor-mations:

W

e2iWe+2i, W

e2i

We+2i . (6.63)

Once again one can compute the component fields of W and Win the WZ gauge. Explicitly

W = 14

D2DVW Z+1

2D2(VW ZDVW Z) 1

4D2DV

2W Z, (6.64)

and similarly for W. After some algebra, the result is the ex-pected non-abelian generalisation

W = i(y) + D(y) + iF(y)+()

(y),

W = i(y) + D(y) i F(y)

() (y), (6.65)where, F = A A + i[A, A] is the non-abelian fieldstrength and is the gauge covariant derivative, e.g. = + i[A, ].

Exercise: Show the transformation property (6.63)and work out the component field expressions (6.65).

6.6 Wess-Zumino model, supersymmetry breaking

There are some properties that we saw before, but did not takeproper notice of, are the following. First, the supersymmetry vari-ation of the , i.e. the highest, component of a scalar superfield

d(x) =i

2

(x) (x) (6.66)is a total (spacetime) derivative. Therefore

d4xd(x) = 0. (6.67)


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In case of a vector superfield, the variation is

D(x) = i

(x) (x)

. (6.68)

Therefore, for any vector superfield V(x,, )

2

d4x

d4 V(x,, ), (6.69)

where is a parameter, is invariant under N = 1 supersymmetrytransformation. With this observation, we immediately have away of constructing actions which are invariant under (N = 1global) supersymmetry. As an example, let us consider the chiralsuperfield (y, ), for which is a vector superfield. Therefore,

d4x d

4 (6.70)

is a supersymmetric action.Of course, (6.69) with V a vector superfield is also an example

of a supersymmetric action. However, on dimensional ground wesee that the parameter has mass dimension two, while the action(6.70) involving does not require a dimensionful parameter.Moreover (6.69) is linear in the component fields, while (6.70) isquadratic. Nevertheless, we shall have occassion to come back to(6.69) later.

Secondly, the supersymmetric variation of the , i.e. againthe highest, component of a chiral superfield

F(x) = 2i(x) = 2i(x) (6.71)is a also total derivative. Therefore

d4x

d2(y, ) +

d2(y, )

(6.72)

is invariant under (N = 1 global) supersymmetry for any chiralsuperfield .

Consider, for example the chiral superfield 2 leading to

m

d4x

d2 2(y, ) + h.c., (6.73)


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where, the complex parameter m has mass dimension one. Asbefore we can, and shall, consider the effect of the term

d4

x

d2

(y, ) + h.c., (6.74)

with [] = M2, later in this lecture.

The advantage of the superfield method lies in the fact thatan action written in terms of superfields in manifestly invariantunder supersymmetry transformations. Let us consider the Wess-Zumino model defined by the lagrangian density8

LW Z =

d4

d2

1

2m2 +

1

3g3

+ h.c. (6.75)

When expanded in terms of the component fields, the first termabove contains the canonical kinetic terms for a complex scalarfield and a Weyl fermion , while the others describe interactionsincluding Yukawa interactions (see exercise in lecture 6.4):

LW Z = ()() + FF

m + g2

F

m + g()2

F i (6.76)

+1

2(m + m) + g + g.

Notice that there is no term involving the derivatives of F. It is

therefore an auxiliary field with algebraic equation of motion

F m g2 = 0. (6.77)

This is readily solved and we use this to eliminate F from theaction. After eliminating the auxiliary field, the bosonic part ofthe action SW Z takes the form

LBW Z = ()() V(), (6.78)8The Wess-Zumino action defines the most general unitary renormalisable

supersymmetric theory for a single chiral superfield.


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where the scalar potential

V() = FF = |m|2+(mg+mg)+|g|2()2. (6.79)

The potential is positive definite. Consequently, the hamiltonianand the total energy are also positive definite.

The last property is actually a general property of supersym-metric theories. We can prove this from the supersymmetry alge-bra (6.14). Explicitly, using

{Q1, Q1} = 2P0 + 2P3,{Q2, Q2} = 2P0 2P3,

the hamiltonian

H P0 can be expressed as

H = 14

Q1Q1 + Q1Q1 + Q2Q2 + Q2Q2

. (6.80)

The expectation value for the energy density in any state | isthus

|H| = 14

||Q1|||2 + ||Q1|||2 + ||Q2|||2 + ||Q2|||2

0, (6.81)

always positive definite.This implies that states with vanishing energy density are su-

persymmetric ground states of the theory. They are ground statesbecause with zero energy we reach the minimum possible value ofenergy, and they are supersymmetric because

|H| = 0 Q| = 0Q| = 0 , for all , = 1, 2. (6.82)

In other words, supersymmetry is preserved in ground states withzero energy. Conversely, in ground states with non-zero (positive,as always with supersymmetry) energy, supersymmetry is spon-tanously broken.


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Coming back to the Wess-Zumino model, let us assume thatm and g are real for simplicity. In this case, the scalar p otentialis

V() = |F|2 = m2||2 + mg( + )||2 + g2||4, (6.83)

while the auxiliary field is

F = m+ g2 =

mA + g(A2 B2)

+i (mB + 2gAB) , (6.84)

where, the complex scalar = A + iB is written in terms of tworeal fields. We see that F can be made to vanish for

B = 0 and A = 0, (m/g).

Since F = 0, (which guarantees that the potential Vvanishes), su-persymmetry is unbroken in the Wess-Zumino model. The super-symmetric vacuum states are parametrised by the above choicesof vacuum expectation values for the scalar fields.

The superpotential of the Wess-Zumino model

W() =1

2m2 +

1

3g3 (6.85)

goes to W() (modulo an unimportant constant shift) under thetransformation

mg

which interchanges the two ground states. However, the sign ofW() can be rotated away by a UR(1) symmetry

ei d2 e+2id2. (6.86)

(Note that the above is unlike a commuting variable x for which xand dx transform in the same way. Recall that for anti-commutingvariables, differentiation and intrgration are equivalent opera-tions.) Suppose under this rotation

(y, ) (y, ) = e2in(y, ei), (6.87)


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i.e. has charge n. The condition that the action remains invari-ant, i.e.,

d2

W((y, )) =

d2

W

(y,

=

d2 W

e2in(y, ei

,

implies that the superpotential must have charge +2 under UR(1)rotation

W() e2iW(). (6.88)For = /2, this is a discrete Z4 symmetry, (under which thefermions are multiplied by a factor of i), relating the two groundstates. However since a 2 rotation changes the sign of fermions,a Z2 Z4 remains unbroken.

It is not possible to break supersymmetry spontaneously inthe Wess-Zumino model involving a single chiral superfield. Itturns out that for this one needs at least three chiral superfields.Before exhibiting this model, let us write the general form of aWess-Zumino type action involving many chiral superfields

S =

d4x

d4 ij

i j

d2 W({i}) + h.c.

. (6.89)

The superpotential

W (

{i

}) = w0 + ii + mij ij + gijk ijk +

(6.90)

is a general functional of the chiral superfields only, i.e. it isa functional only of the is and not the

i s. In other words,

W() is an analytic function of .The bosonic potential in this case is

V({i, i }) =

i

|Fi|2 =

i

W

i

i=i

2

. (6.91)

In fact the name superpotential for W() derives from its closerelation to the (bosonic) potential. Let us stress once again that


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the superpotential is an analytic function of the chiral superfieldsi. Analytic functions are also called holomorphic. Therefore thesuperpotential is a holomorphic function. This is not only true of

the bare lagrangian, but also of the effective one, (if supersym-metry is to remain unbroken). There is, however, some subtletyin the interpretation of the last statement, and we shall return tothis point later. At that time, we shall see the power of require-ment of analyticity (or holomorphy), and how, for many theories,it is sufficient to determine the effective superpotential exactlyandstudy its consequence on the dynamics of the theory.

Let us now study a model of spontaneous supersymmetrybreaking. Consider the lagrangian (first proposed by ORaifear-taigh[12] in 1975):

L = d43

i=1i i d21 + m23 + 12 g122+ h.c.

(6.92)Notice that there is a term linear in . This is of the form thatwas mentioned in the beginning of this lecture.

The algebraic equations of motion for the auxiliary fields are

F1 = +1

2g22,

F2 = m3 + g12, (6.93)F3 = m2.

Clearly, F1 and F3 cannot be made to vanish simultaneously. Theresulting scalar potential is

V = |F1|2 + |F2|2 + |F3|2= 2 + (m2 + g)(Re 2)

2 + (m2 g)(Im 2)2+m2|3|2 + 2mg(123 + 123)+g2|1|2|2|2 + 1

4g2|2|22

. (6.94)

This is positive as long as m2 g. The potential (6.94) is min-imised by choosing

2 = 0, 3 = 0, 1 = unconstrained.


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The value of the potential at the minimum is

V(1) = 2,

a positive definite quantity. Therefore supersymmetry is sponta-neously broken.

Exercise: Compute the masses of the bosons andfermions in the ORaifeartaigh model. There is at leastone massless fermion when supersymmetry is sponta-neously broken. This is analogous to the massless bo-son (Goldstone mode) that appears when a global sym-metry is spontaneously broken, and is called a gold-stino).

Also show that the sum of the mass-square of the

bosons equal to that of the fermions. This too is ageneric feature of supersymmetric models, the propertybeing manifest in vacua with unbroken supersymmetry.

For = 0, there is no supersymmetry breaking, but this caseillustrates another important feature of supersymmetric models.Namely, we have

V(1, 2 = 0, 3 = 0) = 0 (6.95)for arbitrary values of 1. There are infinitely many supersym-metric vacua parametrised by the vacuum expectation value of

the field 1. Thus the parameter space that labels supersymmet-ric vacua is the complex 1-plane (see Fig.6.1(a)).

Let us further restrict to m = 0, i.e. we consider the simplemodel of two chiral superfields 1 and 2 with superpotential

W(1, 2) =1

21

22. (6.96)

This corresponds to the bosonic potential

V(1, 2) = g2|1|2|2|2 + 14

g2|2|22

.


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6.7. Lagrangians of supersymmetric gauge theories 333

01(b)(a) O

1

1

Figure 6.1: Moduli space of supersymmetric vacua (a) for poten-tial (6.95) and (b) for superpotential (6.96) with singularity at theorigin.

We see that V(1, 2) vanishes for 2 = 0 with no condition on1. In the vacuum labelled by 1 the field 2 gets an effectivemass

m2 = 2g|1|.The origin of the parameter space is therefore a special point the mass of the field 2 vanishes there. This is a singular point(or a singularity) in the sense that a heavy field becomes masslesshere.

The name for the parameter space that labels supersymmetricvacua is the moduli space of vacua. Fig.6.1(b) is the moduli spaceof vacua for the model described by (6.96).

Exercise: Analyse the behaviour of the moduli spaceof vacua for a theory of three chiral superfields 1, 2

and 3 with superpotential

W(1, 2, 3) = g 123.

The moduli space is shown in Fig.6.2. It consists ofthree branches meeting at a singular point.

6.7 Lagrangians of supersymmetric gauge theories

In the last lecture, we saw how to write actions for chiral su-perfields leading to supersymmetric actions involving scalars and


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1

3 2

Figure 6.2: Moduli space of supersymmetric vacua described bythe superpotential W = g 123. The origin is a singular point.

spin-half fermions. Constructing an action for a gauge theory isalso along expected lines. We have the field strength supermulti-plet which is a chiral spinor superfield W such that DW = 0.Moreover, (in the non-abelian case9), it transforms homogeneously

W e2iWe+2i.

Therefore,

tr (WW) tr

e2iWWe+2i

= tr (WW)

is both gauge and Lorentz invariant. Further, since this is a chiralsuperfield, we can get a gauge invariant candidate term for thelagrangian by integrating over s:

d2 tr (WW). In terms of

the component fieldsd2 tr (WW) = tr

D2 1

2FF i

4FF

2i

.

The problem with the above is that there is no dependence onthe gauge coupling. Also the FF term is imaginary. Both these

9In the abelian case, W is gauge invariant and no trace need to be taken.


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problems are solved by defining the lagrangian

LSY M = 18

Im d2 tr (WW) , (6.97)

where,

=Y M

2+ i

4

g2. (6.98)

Explicitly, in terms of component fields, the above lagrangianreads as follows

LSY M = 1g2

tr

D2 1

2FF i

Y M322

tr

FF

, (6.99)

where, F = 12F is the dual field strength.Notice that in addition to the gauge fields A, the super-

symmetric Yang-Mills theory contains matter fields: fermions and auxiliary fields D, both in the adjoint representation. Thefermions a are supersymmetric partners of the gauge fields, andare called gluinos or gauginos.

As we had mentioned in the last lecture, it is possible to adda term

2

d4x

d4V(x,, ) =

d4xD(x).

However, this is gauge invariant only for an abelian vector su-

perfield V. Therefore such a term may be added only to the la-grangian of the supersymmetric Maxwell theory. It turns out thatthis term leads to spontaneous breaking of supersymmetry[13].The parameter is known as the Fayet-Iliopoulos parameter.

After having constructed actions for chiral superfields (matter)and vector superfields (gauge fields), we shall now discuss the cou-pling of matter to gauge fields. We shall begin with the couplingof matter to an abelian gauge field.

The first point to notice is that we cannot put the matter fieldsin the gauge supermultiplet. This is because all fields in the this


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multiplet must belong to the same representation as the gaugefields, i.e. in the abelian case they must all be charge neutral andbelong to the adjoint in the non-abelian case.

Consider the chiral superfields i, i = 1, 2, , n. Under aglobal U(1) rotation

i i = e2iqii,where, qi and are real constants. (In particular, Dqi = 0,D = 0. Therefore, i

is also a chiral superfield.) The Wess-Zumino type lagrangian

L =

d4 i i

d2 (mijij + gijk ijk + h.c.)

is invariant under this rotation if

mij = 0 whenever qi + qj = 0,gijk = 0 whenever qi + qj + qk = 0.

If we want to gauge this symmetry, i.e. make a function of x,since D(x) = 0, we must promote the function (x) to a chiralsuperfield (x, ) (D = 0) such that i

= e2iqii is again achiral superfield. This makes the superpotential gauge invariant.However, the kinetic term

i i i e2iqi()i

is no longer invariant. This is familiar from gauging of non-supersymmetric theories. In order to restore invariance underlocal phase rotations, we need to introduce a vector superfieldV(x,, ) such that

eV eV = eieVeii.e, V V = V + i( ). (6.100)

The gauge invariant lagrangian is then

L = 12

d2 WW +

d4 i e

2qiVi

d2 (mijij + gijk ijk) + h.c. (6.101)


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Due to the presence of the exponential in the lagrangian, it is notclear if the theory is renormalisable. We can, however, evaluate itin the Wess-Zumino gauge where

e2qVWZ =

1 + 2qVW Z + 2q2V2W Z

.

In componentsd4 e2qVWZ = () () i

+i

2q

(6.102)+ |F|2 + qD,

where,

= + iqA

= + iqA.

The supersymmetric generalisation of QED requires at leasttwo chiral superfields + and so that we may write a gaugeinvariant mass term. Therefore, we have

chiral superfields e2ie,vector superfield e2V e2ie2V e2i.

The lagrangian in superfields is

LSQED = 12

d2 WW +

d4

+e2eV + +

e

2eV

md2 + + d2 + , (6.103)and in components

LSQED = 12

D2 14

FF i

+(+)(+) + ()()i++ i + m

+ + +

+i

2e

++ ++

m +F + F+ + +F + F++|F+|2 + |F|2 + eD

++

. (6.104)


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Exercise: Re-express the above lagrangian in terms

of the Dirac spinor D =

+

. What are the

additional fields in the lagrangian compared to QED?

Notice that the lagrangian of supersymmetric QED (6.104)contains two complex auxiliary fields F and one real auxiliaryfield D. Their (algebraic) equations of motions are

D + e

++

= 0,

F = m.

(The RHS of the first equation above is in case the Fayet-Iliopoulos term (6.69) is present.) These equations can be solved

to determine the classical scalar potential

V() =|F+|2 + |F|2

+

1

2D2

= m2|+|2 + ||2

(6.105)

+1

2e2|+|2 ||2

2.

The two terms above are called the F- and D-term respectively.In the massless case, we only have the D-term. Therefore,

there are infinitely many degenerate vacua labelled by, upto gaugeequivalence,

+ = a = ,for any complex number a.

In any vacuum with a = 0, gauge symmetry is broken by the(super) Higgs mechanism. The gauge superfield becomes massiveby absorbing one chiral superfield degree of freedom from the mat-ter. One of the chiral superfield degrees of freedom, however, stillremains massless. A gauge invariant description of this masslessdegree of freedom can be given in terms of

X = +. (6.106)


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In the vacuum, X = a2 is an arbitrary complex number. Thismeans that there is no superpotential for the chiral superfield X,at least at the classical level:

Wcl(X) = 0.

The above is known as the D-flatness condition.The vacuum expectation value X gives a gauge invariant

parametrisation of the moduli space of classical vacua. This spacehas a singularity at the origin, i.e. at X = 0, corresponding tothe fact that at this point the U(1) gauge symmetry is unbroken,and all the original microscopic degrees of freedom are masslessthere (see Fig.6.1(b)).

The degeneracy of the classical vacua is accidental. Thereis no symmetry that relates different vacua parametrised by dif-

ferent choices for a, indeed they are inequivalent theories. Thisdegeneracy may therefore be lifted in the quantum theory by adynamically generated superpotential Weff(X).

Recall that X is gauge invariant because carry equal andopposite charges

eie X X.This freedom in defining X can formally be extended to a com-plexification of the gauge group UC(1) C, under which

+

+,

1; (

C).

(In the above, we have extended the gauge parameter to takenon-zero complex values.) The moduli space of vacua was de-scribed by setting the (D-term of the) classical superpotential tozero, and quotienting by the gauge symmetry. This turns outto be exactly equivalent to quotienting by the compexified gaugegroup. Thus, the space of chiral superfields modulo the complex-ified gauge group may be parametrised by the gauge invariantpolynomials of chiral superfields[14]. (In a more general situa-tion, there may be relations between the possible gauge invariantpolynomials.)


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6.8 Supersymmetric QCD classical theory

Much of what we learnt in Lecture 6.7 generalises to the case of

non-abelian gauge theories. In the present lecture, we shall discussthat generalisation.Consider a chiral superfield = {i} which belong to some

representation R of a Lie algebra g:

i i =

e2i

i

jj =

e2i

ataji

j , (6.107)

where, ||taij || with a = 1, 2, , dim g, i, j = 1, 2, , dim R, arematrices in the representation R. The gauge invariant kinetic termfor the field is

e2V =

i

e2Vata

ij

j = trR e2Vata .

This is gauge invariant because the tensor product of the repre-sentations R, its conjugate R and the adjoint contains the singlet.

More generally the lagrangian of a Wess-Zumino type modelinteracting with a non-abelian gauge field is

L = 18

Im

d2 tr (WW)

+

d4

Ie2VI

d2 (mIJIJ + gIJ KIJK) + h.c. (6.108)

The mass term is allowed only when RI = RJ. For a singlechiral superfield, this in only possible for SU(2) (of all SU(n)s).Similarly the gIJ K term is allowed if RI RJ RK contains thesinglet. For example, a single chiral superfield in the doublet ofSU(2) can have a mass term but not a cubic self coupling.

Let us now look at the D-terms in the lagrangian

LD = 12g2

tr D2 + D

=1

2g2DaDb tr (tatb) + Daitaji j

=1

2g2C(R)DaDa + Da trR (t

a()) , (6.109)


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6.8. Supersymmetric QCD classical theory 341

where we have used the normalisation (6.59). The equation ofmotion for Da is therefore

Da = g2

C(R) trR (ta()) , (6.110)

which leads to a scalar potential

VD =

g2

C(R)trR (t

a())

2. (6.111)

This potential must vanish in a vacuum which preserves super-symmetry. Thus, for unbroken supersymmetry, we require

trR (ta()) = 0, (6.112)

the D-flatness condition.

The supersymmetric version of QCD that we shall considerhas

Nf chiral superfields in the fundamental representation Ncof the gauge group SU(Nc):

QAi , A = 1, 2, , Nf; i = 1, 2, , Nc.

(The colour index will not always be displayed explicitly.)There is a global U(Nf) flavour symmetry between the Qswhich transform in the Nf representation. The conjugateanti-chiral superfield Q belongs to Nf of U(Nf) and Nc ofSU(Nc). We shall think often think of ||QAi || and ||QA|| asNc Nf matrices.

Nf chiral superfields QA in the anti-fundamental Nc of thegauge group SU(Nc). These transform as Nf of the flavourgroup. The corresponding conjugate fields are denoted byQ.


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6.8. Supersymmetric QCD classical theory 343

Nf < Nc:By a gauge and flavour rotation, we can diagonalise Q.This implies that the constant on the RHS of (6.115) must

vanish. Upto gauge and flavour rotations, the solution tothe D-flatness conditions are therefore given by

Q = Q =

a1 0 00 a2...

. . .

0 aNf0 0...

...0 0

. (6.116)

The gauge invariant composite superfields whose lowestcomponents parametrise the moduli space of vacua are

MAB = QAQB , A, B = 1, 2, , Nf, (6.117)

called the mesonsuperfields. The vacuum expectation valueofMA

B = |aA|2AB. When this is non-zero, i.e. MAB = 0,

the gauge group is broken. The most generic b ehaviour isSU(Nc) SU(Nc Nf). (Notice that given the vev of M,one can determine, upto gauge and flavour rotations, thevevs of Q and Q and vice versa.)

Nf Nc:An arbitrary A-independent constant is now allowed, so uptogauge and flavour symmetry, a solution to the D-flatnessconditions is

Q =

a1 0 0 0 00 a2...

. . .

0 aNc 0 0

,


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Q =

a1 0 0 0 00 a2...

. . .

0 aNc 0 0

,(6.118)

together with the restriction |aA|2 |aA|2 = c (a constantindependent of A).

Once again we have the meson chiral superfields

MAB = QAQB ,

||MA

B ||=

a1a1

a2a2

. . .

aNc

aNc 0

. . .

0

.

This time, however, it is not possible to determine a anda from the knowledge of M. Moreover, there are newgauge invariant composite operators. For example, in thecase Nf = Nc, there are two such operators

B = A1A2...ANcQA11 QANcNc

=1

Nc! A1A2...ANc B1B2...BNcQA1B1 QANcBNc ,

B = A1A2...ANcQA11 QANcNc (6.119)

=1

Nc!A1A2...ANc

B1B2...BNc QA1B1

QANcBNc

,

called the baryons. The moduli space of vacua is parametrisedby M, B and B. However, they are not all independent,but related through the relation

det ||M|| = BB. (6.120)


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6.9. Quantum corrections & effective action I 347

6.9 Quantum corrections & effective action I

In the previous lecture we sketched an argument, (due to Seiberg),

to determine the effective superpotential of a quantum theory. Letus recall that strategy.

1. Ascertain all the symmetries of a theory in the absence ofthe superpotential. Assign suitable (formal) transformationproperties to the coupling constants such that the symmetryremains valid even in the presence of the superpotential.

2. Regard the quantum corrected effective superpotential tobe a holomorphic function not only of the light chiral super-fields, but also of the coupling constants {gI} (and in thecase of supersymmetric QCD).

3. The effective superpotential should match the results of per-turbation theory in the appropriate limit.

These requirements are often stringent enough to determine theeffective superpotential. The superpotential so determined willalso include non-perturbative corrections.

Parenthetically, let us add that the kinetic term of the gaugefields

d2 Im

eff({}, {gI}, ) WW

is also holomorphic. More precisely, eff is holomorphic in its

arguments. However, in many cases it so happens that the gaugesymmetry is broken and the theory is in a Higgs or a confiningphase. In case gauge symmetry remains unbroken and one is in theCoulomb phase, the same arguments may be applied to determineeff.

As an illustration, let us apply these ideas to determinethe quantum effective superpotential of the Wess-Zumino model(6.75). The classical superpotential (6.85) is

W() =1

2m2 +

1

3g3.


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There is a U(1) R-symmetry that we encountered earlier ineqns.(6.86) and (6.88). It rotates the fermionic component of thesuperspace

= ei

,

and under this transformation the action is invariant if

W() e2i W().

Since the superpotential has R-charge +2, , m and g transformunder UR(1) as follows:

(x, ) (x, ei)m e2im (6.124)g

e2ig.

The kinetic term in (6.75) is also invariant another U(1) trans-formation let us call it S-symmetry:

(x, ) ei(x, ). (6.125)

The superpotential (6.85) breaks this invariance. However, we canformally recover it by assigning the following transformation rulesto the couplings

m e2im,g e3ig. (6.126)

We now demand that Weff is invariant under both UR(1) as wellas US(1), and also that it is holomorphic in m and g (and of course). (In case m and g are real, we shall formally regard them ascomplex and set them to their real values at the end.) The mostgeneral form of Weff consistent with these requirements is

Weff(, m , g) = gambc

R ei(2a+2b)Weff(, m , g)S ei(3a2b+c)Weff(, m , g).


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Invariance under the two U(1)s determines (say) a and b in termsof c. The most general form Weff is then

Weff(, m , g) =

n=0

Angn2m3nn

= m2

n=0

An

g

m

n(6.127)

= m2 f

g

m

, (6.128)

where, An are arbitrary constant coefficients and f() is an arbi-trary function defined by the Taylor series above.

It remains to determine the function f, or equivalently thecoefficients An. To that end, notice that for small values of g,perturbation theory is applicable. We compare this with the Tay-lor series

Weffg0= m2

A0 + A1

g

m+

n=2

An

g

m

n

to find that A0 = 1/2 and A1 = 1/3. It follows from the propertiesof superspace Feynman rules10 that the higher terms in the sumcome from the tree level process shown in Fig.6.3. (Loops are notpossible as it would increase powers of g but not of .)

For n 2, the Feynman diagram in Fig.6.3 is not one par-ticle irreducible, and hence it cannot contribute to the effectivesuperpotential. Therefore the effective superpotential is exactlythe same as the classical superpotential (6.85)

Weff(, m , g) = W(, m , g) =1

2m2 +

1

3g3.

In other words, the superpotential of the Wess-Zumino modelis not renormalised by quantum (and non-perturbative) correc-tions. This important result, first derived in perturbation theory,is called the non-renormalisation theorem.

10A complete discussion of superspace perturbation theory is beyond thescope these lectures. However, all we need to know for this is the fact thatthe propagator for chiral superfield at zero momentum (k = 0) m1.


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g

1/m

g g

. . .

(1) (2) (n)

(n+2) (n+1)

Figure 6.3: Superspace Feynman diagram contributing to the sumin Eq.(6.127).

There is, however, a caveat to the above argument. It wasalready known, (through the analysis of perturbation theory ofsuperfields), that in case there are massless chiral superfields,the non-renormalisation theorem is invalidated. Indeed it wasfound[17] that the effective potential in the Wess-Zumino modelwith m = 0 gets a contribution g3(g)23 which comes fromthe two-loop diagram shown in Fig.6.4. This is clearly non-holomorphic.

*

g

g

g

g

g

*

Figure 6.4: A two loop diagram which gives a non-holomorphiccontribution to the 1PI effective superpotential. (A solid line de-notes a propagator while a dot-dashed line stands for a or a propagator.)


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where

(k) =

0 for |k| MU V,(k) for 0 |k| < ,

where MU V is some UV cut-off. The Wilsonian effectiveaction is defined via

[D] eS[{}] =

k

[D(k)] eS[{}]

=

0|k|


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arising from UB(1) is called the baryon number, and this is a goodsymmetry of the quantum theory. The axial UA(1) on the otherhand, acts oppositely on the left- and right-moving quarks, and

is anomalous. We shall normalise so as to assign baryon number+1 (respectively 1) to the superfields Q (Q). The various U(1)quantum numbers of the quark superfields are as follows:

Q Q Q Q

L +1 0 1 0R 0 1 0 +1B +1 1 1 +1A +1 +1 1 1

The anomaly of the axial UA(1) leads to a shift of the -parameter: Y M

Y M

n, where n is the coefficient of the

anomaly term in the conservation law, or in other words, the num-ber of fermion zero modes contributing to the anomaly. In thiscase, n = 2Nf, which leads to

exp(2i)A e2iNf exp(2i).

In addition, we have the R-symmetry of supersymmetry dis-cussed in lectures VI and the present one. There is no superpo-tential, (we are consideing the massless case12), the superfields Qand Q have zero R-charge. If we write

Q = Q + qL + ,

Q = Q + qL + ,the UR(1) transformation properties are as follows (see Eq.(6.87)):

Q Q, Q Q,qL eiqL, qL eiqL,

ei.12The m = 0 case can be thought of as arising from a Higgs mechanism.

This happens naturally if the N = 1 supersymmetric theory is embedded inthe N= 2 theory, which has a superpotential term QQ, where is the Higgssuperfield valued in the adjoint.


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This symmetry is also anomalous, leading to

exp(2i)R e2i(NcNf) exp(2i).

The two anomalous U(1) symmetries may be combined to ananomaly free UR(1), the generator R of which is

R = R + Nf NcNf

A. (6.129)

The following table shows the relevant symmetry properties of thevarious fields.

Field Flavour UB(1) UA(1) UR(1) UR(1)L-quark qL (Nf, 1) +1 +1 1 NcNfsquark Q (Nf, 1) +1 +1 0

NfNcNf

R-quark qR (1, Nf)

1 +1

1

NcNf

squark Q (1, Nf) 1 +1 0 NfNcNfgluino (1, 1) 0 0 +1 +1

meson MAB

(Nf, Nf) 0 +2 0 2NfNc

Nf

scale (1, 1) 02Nf

3NcNf2Nc2Nf3NcNf 0

(6.130)

In the above we have used the relation (6.123) between the QCDscale and the combination of the coupling constant g and the-angle.

At present we are discussing the case Nf < Nc the only

gauge invariant composite fields are the mesons.The meson superfield matrix MA

Btransforms as (Nf, Nf) un-

der global flavour symmetry. However, the effective superpotentialought to be a flavour singlet and hence can only be a (holomor-phic) function of det M, the determinant of the meson matrix13.It can also depend (holomorphically) on or equivalently, thecomplexified QCD scale . We therefore start with

Weff(M, ) = (det M)a b

13The terms trM, trM2, etc. are not invariants of independent flavour ro-tations SU(Nf)SUr(Nf).


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and require that Weff has the right U(1) charges. This fixes a andb:

Weff

(M, ) = CNfNc 3NcNf

det M1/(NcNf)

, (6.131)

where CNfNc is a constant which cannot be determined by thearguments used so far. Assuming that this constant does notvanish, we find that a non-trivial superpotential is generated byquantum corrections. In the next section, we shall determine thevalue of this constant to be

CNfNc = Nc Nf. (6.132)The effective superpotential (6.131) is called the Affleck-Dine-Seiberg superpotential. Notice that it has the correct (mass)dimension. We should also add that the superpotential is non-

perturbative and hence does not violate any perturbative non-renormalisation theorem.

Let us analyse the behaviour of this superpotential. In orderto find the supersymmetric vacua, we set the F-term to zero. TheF-term

FAB =

MABWeff (6.133)

= (Nc Nf)

3NcNf

det M

1/(NcNf) M1

AB

implies that the bosonic potential tends to zero as

MAB

a

tends to infinity (see Fig.6.5).Thus, we come to the strange conclusion that there is no stable

supersymmetric vacuum (at finite vaccum expectation values ofthe meson superfields). This is inspite of the fact that the classicaltheory we started with had an infinitely degenerate vacua.

Although this seems strange, it is the only possibility consis-tent with supersymmetry. The alternative based on expectationfrom ordinary QCD is chiral symmetry breaking and confinement.In this case the quarks would condense in pairs

qLAqLB = 0.


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M

V(M)

Figure 6.5: Runaway potential of the meson fields.

However, this is the F-term of MAB , a non-zero value for whichwould lead to supersymmetry breaking. Therefore a QCD likevacuum with quark pair condensate is unstable compared to anyvacuum that allows for unbroken supersymmetry14.

The gaugino bilinear can, however, acquire a vacuum ex-pectation value consistent with supersymmetry. For example, inpure super-Yang-Mills theory with Nf = 0, the effective superpo-tential is

Weff() = cNc3 = cNc

3 exp

2ieffNc

. (6.134)

Now, is the scalar component of the composite chiral superfieldWW which couples to . The vacuum expectation value can therefore be found by differentiating the effective lagrangian

14Recall that we are considering quark superfields which are massless. Thesituation changes drastically if these are massive. With a mass term tr(mM)for the mesons in the effective superpotential, one can show that a supersym-metric vacuum is possible at a finite value of M. This moves off to infinity asm is taken to zero.


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Leff with respect to F, the F-term of .

= 16i

Fd2 Weff()

= 16i

Weff()

= 322

NccNc

3 exp

2i

Nc

, (6.135)

or, e82/Ncg2 (for Y M = 0), is the non-perturbativegaugino condensate.

Recall that the R-symmetry shifts ei, hence Y M Y M + 2Nc. When is a multiple of /Nc, theta angle shifts by2, and a Z2Nc subgroup of UR(1) remains unbroken. Howeverthe vacuum expectation value

is invariant under

( = ), breaking Z2Nc spontaneously to a Z2 subgroup. Thisleads to Nc inequivalent vacuum states in pure super-Yang-Millstheory parametrised by the vev of gaugino bilinear

= exp

2im

Nc

, m = 0, 1, , Nc 1. (6.136)

Exercise: (Holomorphic decoupling) Consider a the-ory of two chiral superfields given by the superpotential

W(1, 2) =

1

2 m22 + g

212.

Analyse the classical moduli space of this theory andshow that at a generic point the field 2 isheavy. Findthe effective superpotential for the light field1 by in-tegrating out 2. (In the classical theory, this meansthat one ignores the kinetic term for the heavy field,and solves its resulting algebraic equation of motion.Now apply Seibergs analysis to determine the low en-ergy effective superpotential assuming that the only rel-evant degrees of freedom are those of the light field 1.


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Compare the two results.Identify the diagram that renormalises the superpo-tential, (it is actually a tree diagram), and see that it

does not violate the non-renormalisation theorem.

6.10 Quantum corrections & effective action II

We begin this section by determining the constant in the Affleck-Dine-Seiberg superpotential for supersymmetric gauge theorieswith Nf < Nc. Then we shall move on to the case Nf Nc.

The strategy used to find the constants CNfNc in (6.131) is inrelating the theory with Nf flavours to that with Nf 1 flavoursthrough holomorphic decoupling. The idea of decoupling is quite

simple and is illustrated in an exercise in the previous section. Letus start with the theory with Nf flavours and add (by hand) amass term to the last, i.e., Nfth quark superfield MNfNf:

W(Nf)

eff (M, ) = CNcNf

3NcNf

det M

1/(NcNf)+ mMNfNf.

(6.137)The equations of motion for the mesons with only one index Nfare

W(Nf)

MNfA

A=Nf=

CNcNfNc Nf

3NcNfdet M

1/(NcNf) M1

NfA = 0,

(6.138)and similar equations for the components MANf. The potentialis minimised when all these components (along the last row andcolumn) vanish. Therefore, the meson matrix of the theory re-duces to a block-diagonal form with an (Nf 1) (Nf 1) blockand MNfNf as the last diagonal element. Thanks to this, thedeterminant of the of the meson matrix factorises:

det ||M(Nf)|| = det ||(M(Nf1)|| MNfNf. (6.139)


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6.10. Quantum corrections & effective action II 359

It remains to solve for MNfNf from its equation of motion, ob-

tained by minimising W(Nf):

W(N

f)

MNfNf= CNcNf

Nc Nf

3N

cNf

det M

1/(NcNf) M1NfNf

+m = 0.

(6.140)Using (6.139) and (6.140), the solution is

MNfNf =

CNcNf

m(Nc Nf)

3NcNf

det M(Nf1)

1NcNf

NcNfNcNf+1

.

(6.141)Substituting above in (6.137) in turn, one finds the effective su-perpotential

Weff = (NcNf+1)

CNcNfNc Nf

NcNfNcNf+1

m3NcNf(Nf)

det M(Nf1)

1NcNf+1

.

(6.142)This resembles the effective superpotential of the theory withNf 1 flavours. Indeed, the match is exact provided we makethe following identifications

CNc(Nf1) = (Nc Nf + 1)

CNcNfNc Nf

NcNfNcNf+1

,(6.143)

3NcNf+1(Nf1) = m 3NcNf(Nf) . (6.144)

The first of these equations can be rewritten as

CNcNf

Nc Nf

NcNf=

CNc(Nf1)

Nc Nf + 1

NcNf+1, (6.145)

which is a recursion relation for the coefficients in the Affleck-Dine-Seiberg superpotential. It is easy to see that CNcNf = Nc Nfin (6.132) is a solution to (6.145). A more systematic solutionwould need a boundary condition. This comes from the theory


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with Nf = Nc 1, in which it is possible to calculate the effectivesuperpotential exactly from instanton calculation. This yields thevalue CNC(Nc1) = 1.

The requirement (6.144) is also consistent with the running ofthe coupling constant from the renormalisation group equation:

82

g2()= (3Nc Nf) ln

(Nf)

. (6.146)

In the theory described by the effective superpotential (6.137), thenumber of light flavour degrees of freedom is Nf above the scalem. But below this scale it is Nf1. Hence in this regime it obeysthe Eq.(6.146) with Nf Nf 1. By continuity, the couplingconstants should match at the transition scale = m. Demandingthat this happens is precisely the condition (6.144).

In summary, we have determined the effective superpotentialgoverning the dynamics of supersymmetric QCD with small num-ber of flavours Nf < Nc. Curiously, this superpotential does nothave allow for a supersymmetric vacuum at any finite vev of themesons, which are the relevant gauge invariant degrees of freedom.The only exception is pure SYM theory, in which there are nomesons, and which have Nc supersymmetric vacua parametrisedby the vev of the gluino condensate.

Let us now turn to the cases where the number of flavour islarger than the number of colour: Nf Nc. The story is muchmore complex here and we shall have to content ourselves with a

summary of the main results and only a brief mention of some ofthe arguments.

Consider the case of equal number of flavours and colours Nf =Nc first. The effective superpotential (6.131) for Nf < Nc cannotevidently be extrapolated to this case. Indeed, if we look back atthe table (6.130), we find that all the R-charges of the fields Q,Q, M, B and of the scale vanish in this case. It is, therefore,impossible to write a superpotential, which has R-charge +2 interms of these.

Recall that classically this theory has a moduli space of vacuaparametrised by the meson matrix ||M|| and the baryons B, B


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subject to the constraint (6.120): detM = BB. Schematically,the geometry of this space is that of a cone. At large values ofM, we are in a Higgs phase. However, at the origin of this space,all fields are massless and gauge symmetry is restored. Thus theorigin is a singular point. (See Fig.6.6.)

Seiberg argued that quantum mechanically, the relation (6.120)gets modified to

det M BB = 2Nc . (6.147)There is no symmetry that prevents this modification both sideshave the same R-charge (zero) and dimension (2Nc). Moreover,it is consistent with holomorphic decoupling when we add a massterm mMNcNc to the last meson. To see how this works, we intro-duce an auxiliary chiral superfield of R-charge and dimension2, to write a superpotential

Weff =1

2Nc1

BB det M + 2Nc

+ m MNcNc. (6.148)

The equation of motion of implements the relation (6.147). Atthis point we can simply repeat the steps we followed in the caseNf < Nc. The F-flatness conditions for the superpotential can besolved easily. We find that

B = B = 0,

=m2Nc1

det M(Nc1),

MNcNc = 2Nc

det M(Nc1).

Substituting these values in (6.148), one gets the effective super-potential of the theory with Nf = Nc 1. Finally, there are other,more sophisticated, checks in favour of Seibergs argument thatwe are not in a position to get into.

The quantum moduli space defined by Eq.(6.147) is no longersingular (see Fig.6.6). In the asymptotic regions (where the vevof M is large), quantum corrections are small and the modulispace resembles the classical description. The theory is in the


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(a) (b)

singularirty

Figure 6.6: Moduli space of supersymmetric vacua of SQCD with

Nf = Nc (a) classical (6.120) and (b) quantum (6.147).

Higgs phase. However, the situation is quite different for smallvalues of M. There is no singularity, vevs of M never vanishand consequently do not become massless. The gauge theory isconfining. We see that the Higgs and the confining phases arecontinuously connected in the quantum moduli space.

As a matter of fact, there is no gauge invariant way of dins-tiguishing these two phases in a theory with fundamental matter(quarks). The phases are characterised by the p otential (between

a quark-antiquark pair), which grows linearly in the confiningphase and is a constant in the Higgs phase:

VC(r) r, (6.149)VH(r) = constant. (6.150)

However, the former is unstable to pair production. As the sepa-ration between the quark and the antiquark increases the energyof the flux string grows. Beyond a point it is more favourableto break the string by producing a quark-antiquark pair. Thusthe potential, although increasing linearly for small separation,


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saturates beyond a point and becomes a constant.

For the next case Nf = Nc+1, Seiberg argued that the effective

superpotential

Weff =1

2Nc1

det M BAMABBB

(6.151)

is generated. It is easy to see that Weff has the correct dimen-sion and R-charge. One can also check that it is consistent withholomorphic decoupling.

To proceeed beyond this, let us first consider the case of verylarge number of flavour. The -function vanishes at Nf = 3Nc andis positive beyond this value. Therefore, supersymmetric QCD

with number of flavour exceeding 3Nc is not asymptotically free.Clearly this is very different from what we have seen so far. Thistheory is free in the infra-red with the degrees of freedom beingthe gluons and quarks. It is in a phase where the interactionbetween electric type test charges is a non-abelian analogue of theCoulomb interaction the potential V(r) falls off as

VE(r) 1r ln(r)

. (6.152)

This is called the free electric phase.

Let us return to the theories in which Nf is bigger than Nc,

but not by much. The baryons carry Nf Nc Nc indices.These can therefore be viewed as bound states of Nc componentfields. Seiberg suggested that these be thought of as physicalasymptotic states of a dual theory with gauge group SU(Nc). Italso has chiral superfields corresponding to dual quarks QA andantiquarks QA that come in Nf flavours (A = 1, 2, , Nf) and inaddition, gauge singlet chiral superfield TAB coupled to the dual(anti-)quarks through the superpotential

W = QA TABQB.


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The relation between this description and the original one is calledthe nonabelian electric-magnetic duality. In terms of the dualtheory, the -function vanishes at Nf = 3Nc, i.e., at Nf = 3Nc/2.

For lower values of Nf the dual theory is free in the infra-red withthe dual gluons and quarks being the relevant degrees of freedom.The potential between electric charges of the dual theory

VM(r) =ln(r)

r(6.153)

is again Coulomb like. By duality this is the b ehaviour betweenthe magnetic charges of the original SQCD. In the region Nc + 2 Nf 3Nc thetheory is IR-free and is of magnetic, respectively electric, type(in terms of the original microscopic degrees of freedom). In theintermediate region 32Nc Nf Nc, it is a superconformal fieldtheory corresponding to a nontrivial zero of the -function.

6.11 Appendix: Fierz identities

There are some identities which involve spinors. Much of it followsfrom simple angular momentum addition rule.

If we take the product of a spin s1 =12 representation

15 withanother one s2 =

12 , the result is a direct sum of spin-0 and spin-1

representations:

(s1 = 1/2) (s2 = 1/2) = (s(1) = 0) (s(2) = 1),i.e, 2 2 = 1 3.

A simpler fact is2 1 = 2.

Since the Lorentz algebra is (almost) a direct sum of two angu-lar momentum algebras (see (6.5)), the above rules for tensoringrepresentations apply to the Lorentz algebra as well. The defining

representations here are the12 , 0

or (2, 1) representation called

the left-handed spinor and the

0, 12

or (1, 2) representation

called the right-handed spinor .

15The spin- 12

representation is called the defining or fundamental represen-tation of SU(2), as all other representations can be constructed from it.


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There are three basic possibilities in combining these represen-tations.

We can tensor two left-handed spinors(2, 1) (2, 1) = (1, 1) (3, 1).

More explicitly in terms of the spinors

=1

2() 1

2()

, (6.154)

we get a scalar and a rank-2 self-dual antisymmetric tensor.

Notice that the first term in the RHS above is antisymmet-ric (in indices and ) while the second one is symmetric,

as is expected from the representations of the angular mo-mentum algebra. The coefficients 12 and 12 arethe Clebsch-Gordon coefficients relating different represen-tations.

In order to prove (6.154), based on covariance properties onecan argue that

=1

2( ) + 1

2( + ),

=1

2A +

1

2B

.

The unknown coefficients A and B may be determinedby contracting both sides of the above equation by and respectively. For the latter, we need to use the factthat is self-dual (6.9), and that

tr () =1

2( ) + i

2.

Similarly, for two right-handed spinors

(1, 2) (1, 2) = (1, 1) (1, 3).


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6.11. Appendix: Fierz identities 367

More explicitly in terms of the spinors

=1

2() 1

2()

, (6.155)

where on the RHS we find a scalar and a rank-2 anti-self-dualantisymmetric tensor.

Finally, combining a left- and a right-handed spinor

(2, 1) (1, 2) = (2, 2), =

1

2()

, (6.156)

we get a Lorentz vector.

Using these relations, following from basic group theoretic

facts, it is easy to derive the following relations.

=1

2()

= 1

2()

()() = 12

()()

()() = 12

()()

= (6.157) =

()() =

1

2() i()()

()() =1

2()()

()() =1

2()()

These are known as Fierz identities, and are very useful in sim-plifying expressions involving many spinor fields.


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Bibliography

[1] J. Lykken, Introduction to supersymmetry, TASI lectures

1996, (hep-th/9612114).

[2] J. Wess and J. Bagger, Supersymmetry and supergravity, 2ndEd (1992), Princeton University Press, Princeton.

[3] M. Peskin, Duality in supersymmetric Yang-Mills theory,TASI lectures 1996, (hep-th/9702094).

[4] N. Seiberg, The dynamics of N = 1 supersym-metric field theories in four dimensions in Quan-tum fields & strings: A course for mathemati-cians, Eds. P. Deligne et al , also available at

http://medan.math.ias.edu/QFT/spring/index.html .

[5] P. Sohnius, Phys. Rep. 128 (1985) 39.

[6] P. West, Introduction to supersymmetry and supergravity,World Scientific, Singapore.

[7] Y. Golfand and E. Likhtman, JETP Lett. 13 (1971) 323.

[8] J. Wess and B. Zumino, Phys. Lett. B49 (1974) 52.

[9] S. Coleman and Mandula, Phys. Rev. 159 (1967) 1251.

[10] R. Haag, J. Lopuszansk and M. Sohnius, Nucl. Phys. B88(1975) 257.

[11] A. Salam and J. Strathdee, Nucl. Phys. 76 (1974) 477.

[12] L. ORaifeartaigh, Nucl. Phys. B96 (1975) 331.

[13] P. Fayet and Iliopoulos, Phys. Lett. B51 (1974) 461.

[14] M. Luty and W. Taylor, Phys. Rev. D53 (1996) 3399,(hep-th/9506098).


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[15] K. Intriligator and N. Seiberg, Lectures on supersymmet-ric gauge theories and electric-magnetic duality, Nucl. Phys.Proc. Suppl. B45 (1996) 1, (hep-th/9509066).

[16] N. Seiberg, Phys. Lett. B318 (1993) 469.

[17] I. Jack, D. Jones and P. West, Phys. Lett. B258 (1991) 382.

[18] M. Shifman and A. Vainshtein, Nucl. Phys. B277 (1986) 456.


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Chapter 7

Introduction to AdS/CFT Duality

Dileep P. Jatkar

7.1 Introduction

In these notes on the AdS/CFT correspondence[1] I have triedto present the material as it was covered in the lectures givenat the SERC school. The AdS/CFT correspondence (or duality)has been studied in great detail over last few years and there areseveral reviews are available in the literature. Some of the usefulreferences are [2, 3, 4]1. It is not possible to cover the entiresubject in these lecture notes, nor is it my intention to do so. Iwould, in fact, restrict myself to a simpler and modest goal ofstating the correspondence and giving some elementary methods

of doing checks on the predictions of this correspondence.I have assumed that the reader has basic knowledge of field

theory and general relativity. I have tried to keep these notesself-contained by giving simple examples, whenever needed, byborrowing them from either field theory or relativity.

I thank Ashoke Sen for teaching me the AdS/CFT correspon-dence, Charan Aulakh for explaining Kaluza-Klein formalism and

1The list of references is horribly incomplete. The reader is referred to [4]for a larger list of references.


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372 7. Introduction to AdS/CFT Duality

Debashis Ghoshal and Sudhakar Panda for numerous discussionson this topic. I also thank organisers of the SERC School, Kolkatafor kind hospitality.

Standard model of particle physics is a gauge theory withSU(3) SU(2) U(1) gauge group. It gives partial unificationof strong, weak and electromagnetic forces. SU(3) gauge theoryis a theory of strong interactions, also known as quantum chro-modynamics (QCD). Whereas SU(2) U(1) is a gauge theoryfor electroweak interactions. Fermionic matter in this model con-t

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