Labelle P. Supersymmetry DeMYSTiFied

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286

Supersymmetry Demystified

component of all the other superﬁelds multiplying 

. So this contribution is equal



···

θ ·θ F

···φ



which may be written as



···1 ···φ



θ ·θ F

(12.33)

In other words, we set all the superﬁelds equal to their scalar component except



, which is set equal to 1, and we multiply the whole thing by the auxiliary ﬁeld

and θ ·θ/2. But Eq. (12.33) is simply

∂W(φ

···φ

)

∂φ

θ ·θ F

(12.34)

where W is now seen as a function of the scalar components of the superﬁelds (and

is obviously not set equal to 1 in W. If we extract the F term of this, we ﬁnally

get



···

θ ·θ F

···φ





∂W(φ

···φ

)

∂φ

(12.35)

To keep things simple, we have assumed that the superﬁeld 

appears linearly

in the superpotential. However, it’s clear that Eq. (12.35) remains valid no matter

what the exponent of 

is. If 

appears raised to the power k, we will get the

following k contributions:

k W



···1 ···φ



θ ·θ F

k−1

which is indeed still given by

∂W(φ

···φ

)

∂φ

θ ·θ F

because φ

appears as φ

in W.

In this expression, the index i is not summed over because we have only con-

sidered replacing one of all the superﬁelds by its F term. But, clearly, we also

CHAPTER 12 Left-Chiral Superfields

287

must include the contributions coming from keeping the F term of each of the

other superﬁelds, and this done simply by summing over all possible values of the

index i.

We have only considered one type of contribution to the F term of the super-

potential. There is a second source of F terms that comes from keeping the term

linear in θ (and containing a Weyl spinor) from two different superﬁelds. Let’s then

choose to keep the spinor terms from the superﬁelds 

, and 

(with i = j), and

of course, we set all the other superﬁelds equal to their scalar components φ. In this

case, we get a contribution equal to



···θ ·χ

···φ



= W



···1 ···1 ···φ



θ ·χ

=−W



···1 ···1 ···φ



θ ·θχ

· χ

where we have used Eq. (A.54) for the last step. In terms of partial derivatives, this

is obviously

−

θ ·θ

∂

W(φ

···φ

)

∂φ

· χ

The F term of this expression is simply



···θ ·χ

···φ





=−

∂

W(φ

···φ

)

∂φ

· χ

(12.36)

The indices i and j are not summed over because we have only considered the

contribution from a speciﬁc (distinct) pair of superﬁelds. Now we want to include

the contributions from all pairs. First, let’s still consider i = j . If we sum over all

values of i and j while excluding i = j, we will double count each contribution

because

∂

W(φ

···φ

)

∂φ

∂

W(φ

···φ

)

∂φ

We therefore must divide our total result by two, so that the generalization of

Eq. (12.36) is

−

∂

W(φ

···φ

)

∂φ

· χ

(12.37)

288

Supersymmetry Demystified

with now i and j summed over, excluding the case i = j. Now consider i = j and

assume again that 

appears raised to the power k. Then it is clear that the number

of combinations of the form χ

· θχ

· θ we can make is k(k − 1)/2, so Eq. (12.37)

is still valid for i = j!

We have exhausted the possible sources of F terms from a general superpotential.

To summarize, the supersymmetric lagrangian obtained from the superpotential is

given by the sum of Eqs. (12.35) and (12.37), to which we must add the hermitian

conjugate to obtain a real lagrangian:

W(

···

)



∂W(φ

···φ

)

∂φ

−

∂

W(φ

···φ

)

∂φ

· χ

+ h.c.

(12.38)

which is exactly the expressions we obtained in Chapter 8 [see Eq. (8.60)].

12.5 The Free Part of the Wess-Zumino Model

We have reconstructed the interactions appearing in the Wess-Zumino model, but

the free part,

free

= ∂

φ∂

†

+i χ

†

¯σ

∂

χ + F

†

F (12.39)

is missing. We have dropped the index i on the ﬁelds because the free part does not

mix ﬁelds belonging to different superﬁelds, so there is no possibility of confusion.

In other words, we might as well consider a single superﬁeld; if there are several

of them, we simply add a lagrangian of the form of Eq. (12.39) for each one.

Let’s repeat here the expression for a left-chiral superﬁeld given in Eq. (12.8):

 = φ(x) −

θσ

θ∂

φ(x) −

θ ·θ

θ ·

φ(x)

+θ ·χ(x) −

θσ

θθ · ∂

χ(x) +

F(x)θ · θ (12.40)

For the rest of this chapter, all the component ﬁelds will be functions of x, not of y.

Keeping this in mind, we will stop writing explicitly the x dependence to alleviate

the notation.

It is clear that we cannot get the free Wess-Zumino lagrangian by multiplying

 with itself because it does not contain the hermitian conjugates of the ﬁelds. For

example, we see that two terms in Eq. (12.40) contain a derivative of φ, but we

CHAPTER 12 Left-Chiral Superfields

289

don’t have any term with a derivative of φ

†

. To obtain such a term, we obviously

need to consider the hermitian conjugate of Eq. (12.40):



†

= φ

†

θσ

θ∂

†

−

θ ·θ

θ ·

φ

†

θ · ¯χ +

θσ

θ ·∂

¯χ +

†

θ ·

θ (12.41)

where we have used Eqs. (A.52) and (A.44) and have relabeled the dummy Lorentz

index μ → ν so that it does not get confused with the Lorentz index appearing

in .

To get a kinetic term for φ, the obvious thing to try is the product 

†

. Let us

for now only write the terms containing factors of φ and φ

†

and two derivatives:



†

 =−

θ ·θ

θ ·



†

φ +φφ

†



θσ

θθσ

θ∂

φ∂

†

+··· (12.42)

where we are not showing the x dependence of the ﬁelds to simplify the expression.

Using Eq. (A.56), the last term can be simpliﬁed to

θσ

θθσ

θ∂

φ∂

†

θ ·θ

θ ·

θη

μν

∂

φ∂

†

θ ·θ

θ ·

θ∂

φ∂

†

Substituting this into Eq. (12.42), we get



†

 =−

θ ·θ

θ ·



†

φ +φφ

†

− 2 ∂

φ∂

†



+···

Recall that this is a lagrangian density, so we are free to do integrations by parts.

Doing integrations by parts on the ﬁrst two terms, we see that all three terms have

exactly the same form and therefore can be summed to



†

 =

∂

φ∂

†

θ ·θ

θ ·

θ +··· (12.43)

where the dots represent all the terms that do not contribute to the the kinetic term

of the ﬁeld φ. We have thus established that the kinetic term of φ appears as part

of the coefﬁcient of θ ·θ

θ ·

θ/4 in the product 

†

!

The coefﬁcient of this combination of Grassmann variables is known as the D

term because in the superﬁelds describing gauge multiplets that we will consider in

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Supersymmetry Demystified

Chapter 13, this coefﬁcient happens to be the auxiliary ﬁeld D that we encountered

in Chapter 10.

We therefore have that the D term of 

†

 is given by

(

†

)



= ∂

φ∂

†

+···

where the dots represent other terms to which we will soon turn our attention. Of

course, this D term can be projected out using integrations over the Grassmann

coordinates, using [see Eq. (11.29)]



θd

θθ · θ

θ ·

θ = 1

which shows that the D term of 

†

 is given by

(

†

)





θ d

θ

†



= ∂

φ∂

†

+··· (12.44)

Note that projecting out the D term increases the dimension by 2. Indeed, 

†

 is

of dimension 2, but its D term is of dimension 4. This is, of course, obvious from

the fact that dθ and d

θ are of dimension 1/2.

Our result in Eq. (12.44) should make you wonder what other expressions are

contained in the D term of 

†

. Let’s ﬁnd out! Multiplying Eqs. (12.40) and

(12.41), and writing only the terms containing two θ and two

θ,weget



†

 =

θ ·θ

θ ·



∂

†

∂

φ + FF

†



θσ



θ ·χ

θ ·∂

¯χ −

θ · ¯χθ · ∂



(12.45)

where we have combined right away the φ term using our hard work from Eqs.

(12.42) and (12.43), and we have relabeled a Lorentz index ν → μ.

The last two terms can be combined after an integration by parts. Indeed, we

may rewrite the ﬁrst term in the second parentheses as

θ ·χ

θ ·∂

¯χ =−θ ·(∂

χ)

θ · ¯χ

CHAPTER 12 Left-Chiral Superfields

291

so that Eq. (12.45) becomes



†

 =

θ ·θ

θ ·



∂

†

∂

φ + FF

†



−i θσ

θ · ¯χθ · ∂

χ (12.46)

What we would like to do now is to rewrite the last term with the Grassmann

variables appearing in an overall factor θ ·θ

θ ·

θ so that we can easily read off the

D term.

First, we will need the identity (see Exercise 12.2)

θσ

λθ ·∂

χ =−

θ ·θ(∂

χ)σ

λ (12.47)

to group together the two θ and where λ is a left-chiral spinor. In the same exercise,

you are invited also to prove

θ ·λ(∂

χ)σ

θ =−

θ ·

θ(∂

χ)σ

λ (12.48)

Using those two results, the last term of Eq. (12.46) is

−i θσ

θ · ¯χθ · ∂

χ =−

θ ·θ

θ ·

θ(∂

χ)σ

¯χ

EXERCISE 12.2

Prove Eqs. (12.47) and (12.48). Hint: Use Eqs. (A.54) and (A.55).

Finally,



†

 =

θ ·θ

θ ·



∂

†

∂

φ + FF

†

−i (∂

χ)σ

¯χ



+···

where the dots now represent terms that do not contribute to the D term. Finally,

we extract the D term to obtain



†





= ∂

†

∂

φ + FF

†

−i (∂

χ)σ

¯χ (12.49)

This is not exactly the free part of the Wess-Zumino lagrangian of Chapter 8, where

the spinor term was written as iχ

†

¯σ

∂

χ, which using the bar notation, is equivalent

to i ¯χ ¯σ

∂

χ. But using the identity (A.50) once again, we may write

−i (∂

χ)σ

¯χ = i ¯χ ¯σ

∂

292

Supersymmetry Demystified

so that Eq. (12.49) becomes



†





= ∂

†

∂

φ + FF

†

+i ¯χ ¯σ

∂

χ (12.50)

which is exactly the free part of the Wess-Zumino lagrangian!

12.6 Why Does It All Work?

We have just seen how amazingly simple it is to obtain the complete Wess-Zumino

lagrangian in the superﬁeld approach: The F term of the most general function of

left-chiral superﬁelds gives us the interaction terms, whereas the D term of 

†



gives us the free lagrangian.

This is all nice and easy, but it’s only because of our hard work in Chapters 5,

6, and 8 that we know that these lagrangians are supersymmetric. And it was only

after we worked out the transformation of the auxiliary ﬁeld F and showed that it

transforms with a total derivative that we could prove that the F term of the product

of any number of left-chiral superﬁelds also transforms with a total derivative.

What we need is some criterion to tell us how to extract supersymmetric la-

grangians from combinations of superﬁelds without having to check the invariance

under SUSY by hand or to work out the SUSY transformations of the component

ﬁelds explicitly. For example, we would like to understand why we need to extract

the F term of a product of left-chiral superﬁelds instead of keeping, say, the term

independent of Grassmann variables. Or why is it that for 

†

 we need the D term

instead of the F term.

We will now see how we could have predicted these results from the very be-

ginning, even before working out any explicit transformation of the component

ﬁelds. Let’s start with a single left-chiral superﬁeld. Recall that the dimensions of

the ﬁelds φ, χ, and F are

[φ] = 1[χ] =

[F] = 2

Recall also that the inﬁnitesimal parameter ζ of a SUSY transformation has dimen-

sion −1/2. The transformation of F contains one factor of ζ and therefore must be

of the form

δF = ζ × something of dimension 5/2

CHAPTER 12 Left-Chiral Superfields

293

for the dimensions to match. The “something” must be a function that is linear

in one of the other ﬁelds appearing in the theory. Clearly, we must use χ to get

something with a half-integer dimension. And now the punch line: In order to get

something of dimension 5/2 out of χ,wehave to apply a partial derivative to it.

Therefore, the variation of F has to be of the following form

δF  ζ∂

= ∂

(ζ χ )

based only on considerations of dimensions. Of course, this is not complete; we

need something else that has a Lorentz index and which is dimensionless in order

to make the ﬁnal result a scalar quantity, like F, and this something else turns out

to be ¯σ

, as we saw previously. For the present discussion, however, Pauli matrices

are irrelevant, so we will ignore them.

The main point is that using only dimensional considerations, we could have

predicted that F must transform as a total derivative. It is instructive to go through

the same type of argument for the two other ﬁelds of the multiplet, φ and χ , and

see that only F has to transform as a total derivative. For example, consider the

transformation of χ. From dimensional analysis, we have

δχ = ζ × something of dimension 2

This time, we have two choices for the “something”: We could use either ∂

φ or

F. So we expect the transformation to be a linear combination of these two terms,

which is obviously not a total derivative. The reason that we have two possibilities

now is that there is a ﬁeld of lower dimension than χ, the scalar ﬁeld φ, but also a

ﬁeld of higher dimension, the auxiliary ﬁeld F.

We see that the key point that sets apart the transformation of the ﬁeld F is

that it is the ﬁeld of highest dimension, so its transformation necessarily must

contain another ﬁeld of lesser dimension, which forces us to include a derivative

in the transformation. The conclusion is that the ﬁeld of highest dimension in a

supermultiplet must transform as a total derivative!

Finally, since the product of any number of left-chiral superﬁelds is itself a left-

chiral superﬁeld, we know that the F term of such a product necessarily transforms

as a total derivative under SUSY. Consequently, the F term of the product of any

number of left-chiral superﬁelds provides a supersymmetric lagrangian density

(which is invariant up to total derivatives).

Now, what about superﬁelds, or products of superﬁelds, that are not left-chiral?

It remains true that the ﬁeld with the highest dimension must transform as the total

derivative of a ﬁeld of lower dimension. For a general superﬁeld, the ﬁeld of highest

dimension is the ﬁeld that multiplies θ ·θ

θ ·

θ/4. In gauge superﬁelds, this will

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Supersymmetry Demystified

be the auxiliary ﬁeld D that we encountered in Chapter 10. By extension, we call

the quantity that multiplies this particular combination of Grassmann variables in

a general superﬁeld the D term of that superﬁeld.

Consequently, the D term of any type of superﬁeld or of any combination of

superﬁelds also transforms as a total derivative. Therefore, the D term of any com-

bination of superﬁelds is a supersymmetric lagrangian density. This is how we

could have known ahead of time that the D term of 

†

 necessarily would give a

valid candidate for a supersymmetric lagrangian, even without any of the work of

the previous chapters. Isn’t that neat?

An obvious question that you may be asking yourself is why don’t we use

the D term of the superpotential as well as its F term? After all, we have just

argued that the D term of any product of superﬁelds is a valid candidate for a

supersymmetric lagrangian density! The answer is that the D term of a product of

left-chiral superﬁelds is uninteresting because it is necessarily a total derivative. We

are not saying that it



transforms with a total derivative but that it is a total derivative

to start with, which vanishes if we carry out the spacetime integation. This can be

seen explicitly in Eq. (12.8), where the D term is

−

φ(x)

Since the product of any number of left-chiral superﬁelds is necessarily of the form

of Eq. (12.8) with some equivalent ﬁelds φ

, χ

, and F

, we can tell that the D

term of any product of left-chiral superﬁelds is a total derivative. This is the reason

we do not bother with the D term of the superpotential.

One last aside before moving on to the next chapter. We have seen that the D term

of 

†

 is already of dimension 4, so if we are only interested in renormalizable

theories, we cannot consider D terms of any other combinations, such as 

†



†

.

However, it is natural to think of the supersymmetric theories we have constructed

so far as effective ﬁeld theories valid only at a energies below some physical cutoff

 (see Chapter 1). Indeed, we know that a theory without gravity cannot be truly

fundamental and that reason alone indicates that any supersymmetric theory without

gravity must be treated as an effective ﬁeld theory (here, we are only discussing

theories with global SUSY, i.e., with the SUSY parameter ζ spacetime independent,

not supergravity theories, in which SUSY is gauged). In that spirit, one should

include nonrenormalizable interactions.

Consider, then, an effective ﬁeld theory whose ﬁeld content is a set of left-chiral

superﬁelds 

,... ,

. The most general supersymmetric lagrangian contains two

types of terms: the F term of a superpotential, which is holomorphic in the ﬁelds



, and the D term of a function containing at least one hermitian conjugate of a

CHAPTER 12 Left-Chiral Superfields

295

superﬁeld. This second function is referred to as the K

ahler potential K(

,

†

In a renormalizable theory, the K¨ahler potential simply contains the sum of the

kinetic terms, i.e.,

ren

(

,

†

) = 

†



(12.51)

The topics of supergraviy and nonrenormalizable supersymmetric interactions are

beyond the scope of this book, so we will not discuss the K¨ahler potential any

further.

12.7 Quiz

1. Describe the component ﬁelds of a left-chiral superﬁeld.

2. Why do we not consider the D term of the superpotential when building

supersymmetric lagrangians?

3. What is the dimension of the D term of 

†



†

 ?

4. Why do we impose the superpotential to be a holomorphic function of left-

chiral superﬁelds?

5. What is the K¨ahler potential?