Labelle P. Supersymmetry DeMYSTiFied

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Supersymmetry Demystified

deﬁned by

∂

S ≡

∂S

∂θ

∂

˙a

S ≡

∂S

∂

˙a

Note that, as usual, an upper position of the index on the partial derivative denotes

the derivative with respect to a variable with a lower index, and vice versa.

Note that in Eq. (11.31) we have chosen the undotted indices to go diagonally

downward and the dotted indices to go diagonally upward, following the convention

introduced in Chapter 3. We can write Eq. (11.31) in a more compact form if we

use the notation for spinor dot products and deﬁne

∂

S ≡ ζ · ∂S

˙a

∂

˙a

S ≡

ζ ·

∂S

in which case Eq. (11.31) is given by

S(x +δx,θ + δθ,

θ +δ

θ) ≈ S +



ζσ

θ −

θσ



∂

+ζ · ∂S +

ζ ·

∂S (11.32)

Expanding to ﬁrst order the exponential in Eq. (11.30) and setting this equal to

Eq. (11.32), we get



−ia

−i ζ · Q −i

ζ ·



S =



ζσ

θ −

θσ



∂

+ζ · ∂S +

ζ ·

∂S (11.33)

We can read off the differential operators

P, Q, and Q

†

by separately setting

equal the terms in a

, ζ

, and

˙a

. If we look at the terms proportional to a

,weget

= i ∂

(11.34)

as we had before, in the absence of SUSY.

Recall that ζσ

θ stands for ζ

(σ

)

. Keeping only the terms containing ζ in

Eq. (11.33) and writing out explicitly all the indices, we get

−iζ

S =

(σ

)

∂

S + ζ

∂

CHAPTER 11 Superspace Formalism

257

which implies

= i ∂

−

(σ

)

∂

(11.35)

Using Eq. (11.34), this also could be written as

= i ∂

(σ

)

The terms in

ζ in Eq. (11.33) give the following relation:

−i

˙a

=−

θσ

ζ∂

˙a

∂

˙a

(11.36)

Here, we have a slight problem. Recall that θσ

ζ stands for θ

(σ

)

,sotheζ

does not appear with the same indices in all the terms, preventing us from isolating

˙a

. We are saved by Eq. (A.50), which allows us to write

θσ

ζ =−

ζ ¯σ

θ =−

˙a

( ¯σ

)

˙ab

Using this in Eq. (11.36), we get

−i

˙a

( ¯σ

)

˙ab

∂

˙a

∂

˙a

which yields

˙a

= i

∂

˙a

−

( ¯σ

)

˙ab

∂

(11.37)

However, we also would like to have the corresponding expression with a lower

dotted index, namely,

˙a

, because we then could check that we recover the al-

gebra of the charges derived in Chapter 6, which was written entirely in terms of

components with lower indices. We can proceed in two ways. We can use the 

symbol introduced in Section 3.4 to lower the index of

˙a

, or we may go back to

Eq. (11.36) and write all the terms with the parameter

ζ having an upper index.

Either way, there are some subtleties in the calculation.

We will follow the second approach and leave the demonstration using the 

for Exercise 11.5. We’ll begin by writing again Eq. (11.36) with all the indices

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Supersymmetry Demystified

explicitly shown:

−i

˙a

=−

(σ

)

∂

˙a

∂

˙a

(11.38)

To express the left-hand side in terms of the charges with lower dotted components,

we use Eq. (A.43)

˙a

Using this in Eq. (11.38), we get

−i

˙a

=−

(σ

)

∂

˙a

∂

˙a

(11.39)

Before we can isolate the charge, we need to rewrite the last term with the index

ζ raised. This is a bit more tricky. The identity we need is

˙a

∂

˙a

∂

˙a

(11.40)

which you are invited to prove in Exercise 11.4. We then can move the

ζ to the

right of the derivative if we change the sign (because they are both Grassmann

quantities):

˙a

∂

˙a

=−∂

˙a

(11.41)

We can move ζ in front of the derivative because it is a constant spinor. Things

would not be so simple if it had a

θ dependence. Note that the end result is

˙a

∂

˙a

=−

∂

˙a

which has the opposite sign to what Eq. (A.43) would predict! This is so because,

as demonstrated in Exercise 11.4,

∂

=−∂

∂

= ∂

∂

=−

∂

and these relations have the opposite sign to the usual relations between Grass-

mann quantities with upper and lower indices shown in Eq. (11.26). Incidentally,

CHAPTER 11 Superspace Formalism

259

a consequence of this is that while for components of spinors we have [see Eq.

(A.31)]



= χ



˙a

¯χ

= ¯χ

˙a

for derivatives we have instead



∂

=−∂



˙a

∂

=−

∂

˙a

(11.42)

Thus the  introduces a minus sign when it is used to raise or lower indices of partial

derivatives.

EXERCISE 11.4

Prove that

∂

= ζ

∂

Then argue that this also implies the validity of Eq. (11.40).

Substituting Eq. (11.41) into Eq. (11.39), we get

−i

˙a

=−

(σ

)

∂

−

∂

˙a

from which we ﬁnally obtain

˙a

=−i

∂

˙a

(σ

)

b ˙a

∂

(11.43)

or, using

= i∂

˙a

=−i

∂

˙a

−

(σ

)

b ˙a

EXERCISE 11.5

Rederive Eq. (11.43) from Eq. (11.37) using  to lower the indices. Hint: You will

need Eqs. (A.27) and (11.42).

We have ﬁnally accomplished the goal we had set for ourselves: We have ex-

pressed the charges as differential operators acting on superﬁelds (i.e., functions

260

Supersymmetry Demystified

of superspace). As a nontrivial double-check, let us now compute the algebra of

the charges, which should agree with what we found in Chapter 6 in Eqs. (6.70),

(6.71), and (6.72), which we will repeat here using the van der Waerden notation:

, Q

}=0

{

˙a

}=0

}=(σ

)

(11.44)

The ﬁrst two almost trivially follow from our operators (11.35) and (11.43). The

only nontrivial point is that we must remember to pick up a sign when moving

Grassmann quantities through one another. For example, consider the following

anticommutator:

{∂

} (11.45)

which appears in the anticommutator {Q

, Q

}. In computing anticommutators

of charges represented by differential operators, we must imagine that the whole

expression is applied to some test function of superspace, as we imagine that a

function of x is present when we compute [

p] in quantum mechanics.

Expanding Eq. (11.45), we get

{∂

}=∂

∂

= (∂

) −

∂

= (∂

)

= 0

where in the parenthesis the derivative is applied only on

, and this gives zero

because the dotted upper components are independent of the undotted upper com-

ponents (recall that ∂

= ∂/∂θ

). From this, the ﬁrst anticommutator of Eq. (11.44)

follows trivially. The second one is as straightforward.

For the third one, we need a bit more work, but not much. We need to compute

{∂

,θ

}=∂

+ θ

∂

= (∂

) − θ

∂

+ θ

∂

= δ

(11.46)

CHAPTER 11 Superspace Formalism

261

where δ

is simply the ordinary Kronecker delta, equal to 1 when a = b and 0 when

a = b. Similarly, we ﬁnd

{

∂

˙a

}=δ

˙a

(11.47)

Again, this delta is simply the usual Kronecker delta.

Using these two results together with Eqs. (11.35) and (11.43) (after relabeling

the indices to avoid summing the same letter twice), we ﬁnd (using ∂

=−i

)

i∂

(σ

)

a ˙c

˙c

, −i∂

−

(σ

)

{∂

,θ

}(σ

)

{

˙c

,∂

}(σ

)

a ˙c

where we have used the fact that the components of σ

are simply numbers, so we

are free to move them around as we please. Using {A, B}={B, A}, which applies

to any operator, including Grassmann ones, and applying Eqs. (11.46) and (11.47),

we ﬁnd

(σ

)

˙c

(σ

)

a ˙c

(σ

)

(σ

)

= (σ

)

which is indeed the third relation of Eq. (11.44)!

11.7 Constraints and Superfields

We have almost reached the point where we can start harnessing the full power of

the superspace approach. The basic idea is brilliant and yet rather simple. We have

seen in Eqs. (11.30) and (11.31) how a superﬁeld transforms under SUSY. On the

other hand, we have seen in Section 11.4 how a superﬁeld can be expanded in terms

of a ﬁnite number of component ﬁelds that only depend on spacetime. In addition,

those component ﬁelds have well-deﬁned behavior under Lorentz transformations.

They will play the role of the actual quantum ﬁelds appearing in supersymmetric

lagrangian.

Putting now these two ideas together; i.e., substituting the expansion of a super-

ﬁeld in terms of component ﬁelds into the SUSY transformation [Eq. (11.30)], we

262

Supersymmetry Demystified

will get in a single equation a picture of how all the ﬁelds making up the theory

transform under supersymmetry!

This is a neat trick, but if this were the only motivation for superspace, it would

be a bit of a let-down. After all, the main difﬁculty we had in constructing su-

persymmetric theories was not in writing down the SUSY transformations of the

ﬁelds but in ﬁnding how to build lagrangians that are invariant under SUSY. This

is actually where the superspace approach truly shines: We will see how, using a

few devilish tricks, superﬁelds make the construction of supersymmetric theories

child’s play! The full power of this approach will be demonstrated when we will

be able to write down with almost no effort the SUSY-invariant interactions of the

minimal supersymmetric standard model.

But we must ﬁrst start with less lofty goals. Our ﬁrst step will be to recover the

Wess-Zumino lagrangian from the superﬁeld approach. Unfortunately, we face a

difﬁculty right from the start. The Wess-Zumino lagrangian involves two complex

scalar ﬁelds and one left-chiral Weyl spinor ﬁeld. However, a general superﬁeld

S(x,θ,

θ) has many more ﬁelds, as you found out if you solved Exercise 11.2.

We could build a theory out of a general superﬁeld, but we obviously would get

something more complex than the Wess-Zumino lagrangian.

On the other hand, a superﬁeld F that is a function of only x and θ contains

just the correct number of ﬁelds [see Eq. (11.25)], which have in addition the right

behavior under Lorentz transformations if we take the superﬁeld to be a scalar

(we also could have considered a superﬁeld depending only on x and

θ instead).

Unfortunately, it is not consistent with SUSY to consider a superﬁeld depending

only on x and θ. The reason is simple: If we apply a supersymmetric transformation

to such a superﬁeld, we end up with a transformed ﬁeld F



that will depend on all

the superspace coordinates.

To see this, let’s look again at Eq. (11.32):



(x,θ,

θ) = F(x,θ,

θ) +



ζσ

θ −

θσ



∂

F(x,θ,

θ)

+ζ · ∂F(x,θ,

θ) +

ζ ·

∂F(x,θ,

θ) (11.48)

If we start with a superﬁeld F depending only on x and θ, this simpliﬁes to



(x,θ) = F(x,θ)+



ζσ

θ −

θσ



∂

F(x,θ)

+ζ · ∂F(x,θ)

We see the problem now. Even though our initial superﬁeld F depends only on x

and θ, the transformed superﬁeld F



depends on

θ as well because this superspace

CHAPTER 11 Superspace Formalism

263

coordinate appears explicitly in the transformation. What this tells us is that it is

inconsistent with SUSY to consider a superﬁeld that is a function of x and θ alone.

Let us now explain the problem again from a different angle. Considering a

superﬁeld F that does not depend on

θ is equivalent to starting from a general

superﬁeld S on which the following constraint has been applied:

∂

˙a

S(x,θ,

θ) = 0

so that S is actually a function of x and θ only, i.e.,

S(x,θ,

θ) = F(x,θ) (11.49)

By the way, the choice of differentiating with respect to the Grassmann variable

with an upper index is only one of convenience. It makes life easier to have the

same position of the index as in the derivatives that appear in the supercharges.

Now let’s assume that we don’t already know the answer and ask again if this

constraint is consistent with SUSY transformations or, in other words, whether

SUSY transformations preserve this constraint. To answer this, we must check

whether, if we apply a SUSY transformation to a superﬁeld F independent of

θ,

the SUSY-transformed ﬁeld F



also will satisfy the same constraint:

∂

˙a



= 0

The left-hand side may be written more explicitly as

∂

˙a



∂

˙a



1 −ia

−i ζ · Q −i

ζ ·



∂

˙a

F −i

∂

˙a



+ ζ · Q +

ζ ·



=−i

∂

˙a



+ ζ · Q +i

ζ ·



where in the last step we used the fact that by deﬁnition F obeys the constraint.

Note that in the last line, the derivative with respect to

˙a

acts on everything to its

right, including the superﬁeld.

Recall that we are checking whether or not this is equal to zero, which will tell

us if the constraint is consistent with supersymmetric transformations. Now comes

the crucial observation: If we can “pass” the ∂/∂

˙a

through the parentheses to

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Supersymmetry Demystified

apply it to F, we will automatically get zero because F satisﬁes the constraint, by

assumption. This is therefore the key point: The constraint is consistent with SUSY

if it commutes with

+ ζ · Q +

ζ ·

The operator ∂/∂

˙a

obviously commutes with a

, so we are left with checking

whether the commutator



∂

˙a

,ζ· Q +

ζ ·



vanishes. Recall that the spinor ζ is a constant spinor (independent of both the

spacetime coordinates x and the Grassmann coordinates θ) so that we can pass the

derivative through ζ or

ζ on the condition that we add a minus sign because both

ζ and ∂/∂θ are Grassmann quantities. This allows us to write the commutator as



∂

˙a

,ζ



∂

˙a





−





∂

˙a

=−ζ



∂

˙a

+ Q

∂

˙a



−



∂

˙a

∂

˙a



=−ζ

∂

˙a

, Q

−

∂

˙a

Since the parameters ζ

and

are independent, the constraint is consistent with

SUSY on the condition that it anticommutes with the SUSY supercharges:

∂

˙a

, Q

= 0

∂

˙a

= 0

Using the explicit representations of the SUSY charges (11.35) and (11.37), we

see that the operator ∂/∂

˙a

does indeed anticommute with

but not with Q

because the latter contains

θ. The conclusion is that the constraint

∂

˙a

S = 0isnot

preserved by a supersymmetric transformation and is therefore useless to us.

We already knew that the constraint ∂/∂

˙a

= 0 is not preserved by SUSY. The

reason we went through this whole discussion, pretending all along that we did not

already know the answer, is that now we have a simple criterion to ﬁnd out if a

CHAPTER 11 Superspace Formalism

265

constraint on superﬁelds is consistent with SUSY. Let’s write such a constraint as

˙a

S(x,θ,

θ) = 0

where

˙a

is some differential operator acting in superspace.

What we have dis-

covered is that for this to be a valid constraint, the operator

˙a

must anticommute

with the SUSY supercharges:

{

˙a

, Q

}=0 (11.50)

{

˙a

}=0 (11.51)

What we now want to do is to construct an explicit operator

˙a

(as a differential

operator acting in superspace) that will satisfy those two conditions.

We know that the partial derivative with respect to

˙a

does not work because

only the second condition is fulﬁlled:

∂

˙a

, Q

= 0 (11.52)

∂

˙a

= 0 (11.53)

Still, let’s use this operator as our starting point in constructing

˙a

. After all, it

satisﬁes Eq. (11.53), so that’s not a bad start. The simplest thing we can try is to

add a term to the derivative that will ensure that Eq. (11.52) also will be satisﬁed.

Of course, we don’t want to spoil Eq. (11.53) by doing that, so the added term must

by itself anticommute with

It is clear by looking at the deﬁnition of

[Eq. (11.37)], that it anticommutes

with θ but not with x

θ. So let’s add a term linear in θ to ∂/∂

˙a

. It will make

our life easier if we add a term proportional to θ

(instead of θ

) because the charge

contains a derivative with respect to Grassmannn variables with upper indices.

So let’s try

˙a

∂

˙a

˙ac

(11.54)

The reason for choosing the operator to have a lower dotted index is simply based on convenience because it

will simplify the next few steps. Of course, once we have the components with lower dotted indices, we can

work out the components with any other type of indices.