Labelle P. Supersymmetry DeMYSTiFied

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236

Supersymmetry Demystified

As when we combined a chiral and abelian gauge multiplet, the transformation

of the nonabelian gauge multiplet must be considered with ζ replaced by −ζ/

√

and we also must add a term to the transformation of the auxiliary ﬁeld F:

δF = previous terms +

√

2gφT

ζ ·

Let’s see what happens if we eliminate the auxiliary ﬁelds D

. The equation of

motion for D

gives

= g



†



Plugging this back into the terms containing D,weget

− gφ

†

Dφ =

− gφ

†

=−



†



†



which gives a contribution to the scalar potential equal to

(φ) =



†



†



(10.60)

If there are several left-chiral multiplets labeled by an index i, we get instead

(φ

) =



†



†



where the index i labels the different scalar ﬁelds, each of which is a multiplet in

the fundamental representation. In other words, each φ

must be thought of as a

column vector with N entries. Thus the complete scalar potential in this case is

V (φ

) =



∂W

∂φ





†



†



(10.61)

CHAPTER 10 Supersymmetric Gauge Theories

237

10.9 Quiz

1. In what representation does the photino transform?

2. When we combine a vector multiplet with a chiral multiplet, what are the

changes that must be made to the SUSY transformations of the component

ﬁelds?

3. What is the Fayet-Illiopoulos term?

4. What is the gauge-covariant derivative for an antiquark left-chiral Weyl

spinor?

5. Why is the auxiliary ﬁeld of a vector multiplet a real scalar ﬁeld, whereas the

auxiliary ﬁeld for a chiral multiplet is a complex scalar ﬁeld?

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CHAPTER 11

Superspace Formalism

So far we have constructed supersymmetric theories the “hard way”: by writing

the most general renormalizable lagrangians that are real and Lorentz-invariant and

then tweaking them to see if we could make them invariant under supersymmetry

(SUSY). There is an easier way to obtain supersymmetric theories: the so-called

superspace approach, pioneered by Salam and Strathdee.

We now discuss this

approach.

The basic idea is the following: We have seen that in the case of spacetime sym-

metries, such as invariance under translation or rotation, the charges can be written

as differential operators instead of functions of the ﬁelds. At ﬁrst glance, it might

seem that this idea can’t be applied to SUSY because the supersymmetric transfor-

mations of the ﬁelds yield other ﬁelds at the same spacetime point. However, we

have discovered that the anticommutation of two supersymmetric transformations

of a ﬁeld produces the same ﬁeld evaluated at a shifted spacetime point! Indeed,

the anticommuator of two SUSY charges gives the four-momentum operator P

whose effect is to translate ﬁelds in spacetime. This is a key aspect of SUSY: It is

a symmetry that mixes in a nontrivial way an internal symmetry with spacetime

symmetries.

240

Supersymmetry Demystified

This relation between the supercharges and spacetime symmetries suggests the

possibility of setting up the SUSY charges as differential operators acting on an

extended spacetime which is referred to as superspace. We will make this idea more

explicit in this chapter.

Until now, we have not used the van der Waerden notation described in Chapter

3; all Weyl spinors were expressed in terms of their lower, undotted components.

However, calculations in superspace would become amazingly cumbersome if we

would insist on sticking to spinors with lower undotted indices only. Besides, most

SUSY references make extensive use of the van der Waerden notation, so it’s a

good idea to get used to manipulating expressions written in that notation. If you

haven’t read Chapter 3 yet, grab a cup of coffee, put on some relaxing music, and

go over it before coming back to the next section.

11.1 The Superspace Coordinates

For this idea to work, a nontrivial step ﬁrst must be taken. To understand this,

let’s go back to the simple example of the charges P

. The corresponding unitary

transformation U is [see Eq. (6.18)]

U (a) ≡ exp(ia

)

This operator is what we sometimes called U

in Chapter 6. The notation U(a)

is preferable here because we will be considering transformations depending on

several parameters, and it becomes awkward to write the unitary transformations

with several subscripts.

The quantity a

is obviously the inﬁnitesimal four-vector parameterizing the

transformation. But there is something special about this four-vector: It is a dis-

placement four-vector which appears in the argument of the transformed ﬁeld, as

is clear if we write the transformed ﬁeld explicitly [see Eqs. (6.8) and (6.15)]:

φ(x



) = U (a)φ(x) U

†

(a)

= φ(x

+ a

) (11.1)

In particular, this implies that if we start with the ﬁeld evaluated at the spacetime

origin (in some given frame) φ(0), we can generate the ﬁeld at any position φ(x)

by applying a transformation U with parameter a

= x

φ(x) = U(x)φ(0) U

†

(x)

= exp





φ(0) exp



−ix



(11.2)

CHAPTER 11 Superspace Formalism

241

We would like to be able to view the SUSY charges as producing a translation

of the ﬁelds too. The parameters that are dotted with the supercharges (ζ and

ζ )

are Grassmann quantities, which we now want to treat as displacement vectors.

Therefore, we must extend the spacetime dependence of the ﬁelds to include a

dependence on four Grassmann coordinates in addition to the usual four spacetime

coordinates. This extended spacetime is what is referred to as superspace. In this

context, superspace must not be seen as a physical space but simply as a mathe-

matical trick that simpliﬁes the construction of supersymmetric theories, as we will

see.

The extra four Grassmann coordinates are commonly denoted by θ

,θ

, and their

hermitian conjugates. Since the SUSY parameter ζ is a left-chiral Weyl spinor, it

is natural to take θ

and θ

to form a left-chiral Weyl spinor that we will denote

simply by θ:

θ ≡





We also deﬁne

θ =





Recall that, by deﬁnition,

= (θ

)

†

and

= (θ

)

†

But the Grassmann coordinates are not quantum ﬁelds, so we may replace the

dagger acting on the components by a simple complex conjugation, i.e.,

(θ

)

†

= θ

1∗

and (θ

)

†

= θ

2∗

We now introduce superﬁelds, which are functions of all eight coordinates x

θ, and

θ. Consider a general superﬁeld

(x,θ,

θ)

that we take to be a scalar under Lorentz transformations (later we will encounter

superﬁelds that are not scalars).

Now we will apply a SUSY transformation to this superﬁeld. Our SUSY trans-

formation will be a function of the inﬁnitesimal parameters a

, ζ , and

ζ ,aswehave

242

Supersymmetry Demystified

used so far. Thus let us write this as

U (a,ζ,

ζ)(x,θ,

θ)U

†

(a,ζ,

ζ) = (x



,θ



) (11.3)

This is our starting point. Our goal is to obtain representations for the SUSY charges

as differential operators acting in superspace.

This is a nontrivial exercise. In principle, what we would like to do is to follow

the steps we took in Chapter 6 to work out the differential operator representations

of the Poincar´e generators from the transformations of the coordinates:

Coordinate transformations ⇒ charges as differential operators

Unfortunately, we don’t know how the superspace coordinates transform under

SUSY, only the SUSY algebra. What can we do? The trick is that we can actually

invert the step described in Eq. (6.33); i.e., we can go from the algebra directly to

the transformation of the coordinates:

Algebra ⇒ coordinate transformations

Once we have the coordinate transformations, it is a simple matter to ﬁnd the

differential operator representation of the charges.

Let us ﬁrst see how this works in the simple context of ordinary translations.

11.2 Example of Spacetime Translations

Our goal, then, is to ﬁgure out how coordinates transform under translations using

only the algebra of the components of the momentum operator as input. Of course,

the approach here will be overkill when applied to simple translations, but it is

nevertheless instructive to see the technique applied in a familiar and simple context

before forging ahead with SUSY.

Let us go back to the action of the four-momentum charges P

on a scalar

ﬁeld shown in Eqs. (11.1) and (11.2). Note that P

here is a quantum ﬁeld oper-

ator, not a differential operator. We cannot freely move the P

through the quan-

tum ﬁeld φ(x), but we are free to move a

or x

through the charges P

with

impunity.

Now, let us write again Eq. (11.1) but use Eq. (11.2) for φ(x), giving

U (a) U (x)φ(0) U

†

(x) U

†

(a) = φ(x



)

CHAPTER 11 Superspace Formalism

243

The trick is to realize that we can write the transformed ﬁeld as arising from a single

transformation produced by the operator U (x



U (a) U (x)φ(0) U

†

(x) U

†

(a) ≡ U(x



)φ(0) U

†



)

which yields

U (a) U (x) = U (x



)

Writing the unitary operators in terms of the charges P

,wehave

exp(ia

) exp(ix

) = exp(ix

μ

) (11.4)

In order to relate x



to x and a, we must simply combine the exponentials on the

left-hand side using the Baker-Campbell-Hausdorff formula:

exp(A) exp(B) = exp



A + B +

[A, B] +···



Using this for the left-hand side of Eq. (11.4) and setting equal the arguments of

the exponents, we ﬁnally get

i(a

+ x

) P

−

, P

] +···=ix

μ

This is what we were seeking: a way to determine the transformed coordinates in

terms of the old coordinates using only the algebra of the charges. Of course, in

the case of the momentum operator, the algebra is trivial, i.e., [P

, P

] = 0, which

yields

μ

= x

+ a

Once we have established how the coordinates are transformed by the symmetry

operation, it is simple to ﬁnd the differential operator representation of the four-

momentum as we did in Chapter 6. The only subtle point to keep in mind is that the

action of the differential operator is deﬁned through [see Eqs. (6.25) and (6.26)]

exp



−ia



φ(x) = φ(x + δx) (11.5)

From here it is trivial to obtain

= i ∂

244

Supersymmetry Demystified

We will now repeat these steps in the case of supersymmetric transformations

applied to superﬁelds.

11.3 Supersymmetric Transformations of the

Superspace Coordinates

We now apply the approach of the preceding section to SUSY. Our starting point

is Eq. (11.3). The SUSY equivalent of Eq. (11.2) is obviously

(x,θ,

θ) = U(x,θ,

θ)(0) U

†

(x,θ,

θ) (11.6)

On the other hand, we have, by deﬁnition,

(x



,θ



) = U (x



,θ



)(0) U

†



,θ



) (11.7)

Inserting Eqs. (11.6) and (11.7) into Eq. (11.3), we are led to

U (a,ζ,

ζ)U (x,θ,

θ) = U(x



,θ



) (11.8)

This is what we are going to use to ﬁgure out how the shifted coordinates are

related to the original coordinates after a SUSY transformation with parameters a,ζ,

and

ζ .

It might seem that we are being a bit masochistic here. One might be tempted

to conclude that in analogy with the result for translations in real space x



= x +

a, the corresponding result here simply should be x



= x + a, θ



= θ +ζ , and



θ +

ζ . However, the situation is more subtle than this because we can, using

, build quantities out of θ and ζ that will transform like four-vectors, so the

transformation of x could turn out to contain more than just a. In fact, we know

that the transformation of x has to be modiﬁed by SUSY because we have already

established that the commutator of two SUSY transformations yields a shift in

spacetime. Therefore, x



will depend on θ and ζ in a nontrivial way.

The next step is to write down explicit expressions for U in terms of the charges.

What should we take for the supersymmetric equivalent of a

or x

? The role

of the momentum operator is played by the supercharges Q and

Q, whereas the

coordinates x

should be replaced by θ and

θ. Since we need Lorentz invariants to

use in the exponential of U , the equivalent of x

Q ·θ +

Q ·

CHAPTER 11 Superspace Formalism

245

On the other hand, the role of the a

is played by ζ and

ζ , so the equivalent of

Q ·ζ +

Q ·

Since the commutator of two supersymmetric transformations contains the gener-

ators P

, we need to consider both translations and supersymmetric transformations

together to work out the transformations of the coordinates in superspace.

However, when we get to the point of actually writing out U , we face a difﬁculty:

There are several possible choices for U that are not equivalent. For example, we

could choose

U (x,θ,

θ) = exp(ix · P) exp(iθ · Q) exp(i

θ ·

Q) (11.9)

but we could as well choose

U (x,θ,

θ) = exp(ix· P) exp(i

θ ·

Q) exp(i θ · Q) (11.10)

U (x,θ,

θ) = exp(ix· P + i

Q ·

θ +iQ· θ) (11.11)

These three choices are not equivalent because the charges Q and

Q do not commute

(however,

commutes with everything, so other choices with a · P in different

positions are all equivalent to one of the three expressions given above). There is no

right or wrong choice here, but the details of the following steps depend on which

expression we pick. It turns out that the rest of the discussion is simpler if we use

Eq. (11.9) or Eq. (11.10), but most references on SUSY use the more symmetric

Eq. (11.11) as a starting point. We will follow the majority of the literature and

consider Eq. (11.11).

Note that the operator (11.11) is unitary. To show this, we ﬁrst need to evaluate

†

= exp



−ix· P − i (

Q ·

θ)

†

−i (Q ·θ)

†



= exp



−ix· P − iQ· θ −i

Q ·



where we have used the fact that P

is hermitian as well as the identities (A.44)

to apply the hermitian conjugate on the spinor dot products. It is now clear that

†

U = U

†

U = 1 because the arguments of the exponentials in U and U

†

differ