Labelle P. Supersymmetry DeMYSTiFied

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266

Supersymmetry Demystified

where we have included the indices that the quantity C must necessarily carry. This

operator satisﬁes Eq. (11.53) if we take C

˙ac

to be independent of the Grassmann

variables. Our goal now is to see if it is possible to choose an expression C

˙ac

such

that Eq. (11.54) anticommutes with Q

∂

˙a

˙ac

, Q

= 0

Using the explicit representation of Q

[Eq. (11.35)], we get

∂

˙a

˙ac

, i

∂

∂θ

(σ

)

= 0

∂

˙a

∂

∂θ

∂

˙a

,(σ

)

˙ac

∂

∂θ

˙ac

,(σ

)

= 0 (11.55)

The ﬁrst anticommutator is identically zero. The last anticommutator also vanishes

on the condition that C has no spacetime dependence (so that it commutes with

This leaves two nontrivial anticommutators. The second and third anticommutators

of Eq. (11.55) cancel out on the condition that (see Exercise 11.6)

˙ab

=−

(σ

)

b ˙a

=−

(σ

)

b ˙a

∂

(11.56)

EXERCISE 11.6

Prove that the second and third anticommutators of Eq. (11.55) cancel out if C

˙ab

given by Eq. (11.56). You may, of course, just plug in the result in Eq. (11.56), but it

is more interesting (and rewarding) to obtain the answer directly from Eq. (11.55).

In doing it this way, you may assume that C

˙ab

does not depend on θ so that it

commutes with ∂/∂θ

to pull it out of the third anticommutator. Of course, if that

assumption had not worked, we would have had to consider a more general ansatz

for C

˙ab

CHAPTER 11 Superspace Formalism

267

Using this in Eq. (11.54), we ﬁnally obtain

˙a

∂

˙a

−

(σ

)

c ˙a

∂

(11.57)

Of course, we could have started with a constraint containing a derivative with

respect to θ

instead of a derivative with respect to

˙a

. Imposing that this second

constraint anticommutes with all the supercharges then leads to the expression

∂

∂θ

−

(σ

)

∂

(11.58)

The two differential operators we just obtained can be used to write down con-

straints that are consistent with SUSY transformations. Our next step obviously is

to look for the most general superﬁelds that satisfy

˙a

(x,θ,

θ) = 0 (11.59)

These superﬁelds are called left-chiral superﬁelds for reasons that will become clear

in Chapter 12, hence the subscript LC. The ﬁelds satisfying

= 0 (11.60)

are called right-chiral superﬁelds. However, most supersymmetric theories, includ-

ing the minimal supersymmetric standard model, do not use right-chiral superﬁelds,

so we will focus on left-chiral superﬁelds in Chapter 12.

Using some of the tricks of Exercise 11.5, it is straightforward to raise the indices

to obtain

=−

∂

∂θ

( ¯σ

)

∂

˙a

=−

∂

˙a

( ¯σ

)

˙ab

∂

(11.61)

We may then deﬁne dot products between the constraints, i.e.,

D · D ≡ D

D ·

D ≡

˙a

268

Supersymmetry Demystified

that we will use later on. An important observation is that on any superﬁeld S,we

have

D · D S = D

D · D S = 0

and

˙a

D ·

D S =

˙a

D ·

D S = 0 (11.62)

This follows from the fact that three constraints of the same type necessarily contain

either a Grassmann variable squared or the second derivative with respect to a

Grassmann variable.

Later we will need the anticommutation relations satisﬁed by the constraints.

Obviously,

, D

}={D

, D

}={

˙a

}={

˙a

}=0

whereas a short calculation gives

}=−i(σ

)

∂

(11.63)

11.8 Quiz

1. What is the result of the integral



dθ

2. What is the derivative

∂

∂θ





equal to?

3. Consider a general function of three Grassmann variables F(θ

,θ

).Ifwe

expand in all three Grassmann variables, how many independent terms do we

generate?

4. How many supercharges did we have to introduce?

5. What is the result of 

∂

CHAPTER 12

Left-Chiral Superfields

As mentioned at the end of Chapter 11, a superﬁeld satisfying the constraint (11.59)

is called a left-chiral superﬁeld. In this chapter we will only discuss left-chiral

superﬁelds that are scalars under Lorentz transformations. We will denote these

superﬁelds by . The reason why such a superﬁeld is called left-chiral even though

it is a scalar under Lorentz transformations is that one of its component ﬁelds is a

left-chiral spinor, as we will see explicitly shortly. In Chapter 13 we will encounter

a left-chiral superﬁeld that is not a scalar.

12.1 General Expansion of

Left-Chiral Superfields

One obvious observation we can make about left-chiral superﬁelds is that since the

constraint does not involve the coordinates θ

, they can be arbitrary functions of

that variable. Of course, a superﬁeld depending only on θ would automatically be

left-chiral, but we want to be as general as possible by introducing a dependence

270

Supersymmetry Demystified

on x and

θ as well. The dependence of  on these two variables clearly must be

related in a nontrivial manner.

To proceed, the trick is to construct a new coordinate, let’s call it y

, that is

in principle a function of both x

and

θ and that satisﬁes by itself the constraint

equation

˙a

θ) = 0 (12.1)

If we succeed in identifying such a coordinate, then any function of y

and θ will

be a left-chiral superﬁeld  because it will trivially satisfy

˙a

(y

,θ) = 0

The solution of Eq. (12.1) is surprisingly simple. Since the differential operator

˙a

contains derivatives with respect to x

and

˙a

[see Eq. (11.57)], the simplest

guess is to write y

as x

plus something proportional to

θ. We therefore try the

ansatz

= x

(12.2)

where K is an unknown expression that we must now ﬁnd. We have included the

indices on K that it must carry for the equation to be consistent. Note that y

is an

ordinary (i.e., commuting) variable, so K

must be a fermionic quantity (which is

also obvious from the fact that it carries a spinor index). Let us now assume that

does not depend on x

and

˙a

so that

˙a

= 0. Of course, if this doesn’t

work, we will have to come back and drop this assumption. Note that we have

chosen to place θ

˙a

to the left of K so that we won’t have to worry about picking

up a minus sign as we move the partial derivative

∂

˙a

through K

Now we impose Eq. (12.1). A quick calculation reveals that

˙a

= K

˙a

−

(σ

)

c ˙a

so we must take

˙a

(σ

)

c ˙a

Putting this back into Eq. (12.2), we ﬁnally get

= x

(σ

)

CHAPTER 12 Left-Chiral Superfields

271

We can move the

to the far right to get

= x

−

θσ

(12.3)

The most general left-chiral superﬁeld is therefore a general function of y

and

θ, i.e.,

(y

,θ)

We will choose to take  to be a scalar under Lorentz transformations, although it

is possible to write left-chiral ﬁelds with different Lorentz properties. We will meet

such left-chiral superﬁelds in Chapter 13.

It is straightforward to see that the most general right-chiral superﬁeld [that obeys

the constraint (11.60)] is a function of the variables

θ and z

with

≡ x

θσ

Note that if we take the hermitian conjugate of a left-chiral superﬁeld, we obtain



†

,θ) = (y

μ†

θ) = (z

θ)

which is a right-chiral superﬁeld! So we obtain the important result that for any

left-chiral superﬁeld ,



†

= 0 (12.4)

Let us now see what the component ﬁelds of a left-chiral superﬁeld are. To ﬁnd

this out, we simply expand in the Grassmann variables appearing in the ﬁeld, as we

did in Section 11.4. We will do this in two steps. We ﬁrst expand in θ only, leaving

untouched. This, of course, gives us exactly the same result as Eq. (11.25),

namely,



(y,θ) = φ(y) + θ · χ(y) +

θ ·θ F (y) (12.5)

We see that the ﬁrst ﬁeld, φ, is a scalar because the superﬁeld itself is taken to be

a scalar. Since θ · θ is a Lorentz invariant, F is also a scalar ﬁeld. Finally, since θ

is a left-chiral spinor, χ also must be a left-chiral spinor. Those are the component

ﬁelds of the left-chiral superﬁeld . The fact that one of the component ﬁelds of 

272

Supersymmetry Demystified

is a left-chiral spinor, whereas there is no right-chiral spinor, is the reason why the

superﬁeld is called a left-chiral superﬁeld. Keep in mind, however, that  itself is

a Lorentz scalar!

We are not quite done. We still haven’t fully expanded in terms of the Grassmann

variables because y

depends on them. We do that now. The Taylor expansion of

the component ﬁelds is straightforward. Consider ﬁrst the scalar ﬁeld:

φ(y) = φ



−

θσ



= φ(x) −

θσ

θ∂

φ(x) −

θσ

θθσ

θ∂

∂

φ(x) (12.6)

The last line is an exact result, not an approximation, because higher-order terms

automatically vanish since they contain Grassmann variables squared. Note that

θσ

θ is a commuting quantity, so we may move it as a block anywhere we want.

The last term on the right of Eq. (12.6) can be simpliﬁed using Eq. (A.56), which

allows us to write

θσ

θθσ

θ∂

∂

φ(x) =

μν

θ ·θ

θ ·

θ∂

∂

φ(x)

θ ·θ

θ ·

θ∂

∂

φ(x)

≡

θ ·θ

θ ·

φ(x)

where

 is the d’Alembertian operator. Using this result, Eq. (12.6) can be written

φ(y) = φ



−

θσ



= φ(x) −

θσ

θ∂

φ(x) −

θ ·θ

θ ·

φ(x) (12.7)

The expansion of χ(y) is identical except that we need only to keep the ﬁrst two

terms in the expansion with respect to θ because χ is already dotted with θ. Thus

we get

(y) = χ

−

θσ

θ) = χ

(x) −

θσ

θ∂

(x)

CHAPTER 12 Left-Chiral Superfields

273

On the other hand, since F is multiplied by θ ·θ , we may replace F(y) directly by

F(x) because all higher-order terms are automatically zero.

Substituting the expansions of φ(y), χ (y), and F(y) into Eq. (12.5), we ﬁnally

get that a left-chiral superﬁeld has the following expansion in terms of its component

ﬁelds:

 =φ(x) −

θσ

θ∂

φ(x) −

θ ·θ

θ ·

φ(x)

+ θ ·χ(x) −

θσ

θθ · ∂

χ(x) +

F(x)θ · θ

(12.8)

Recall that θ ·∂

χ stands for θ

· ∂

. Obviously, a left-chiral superﬁeld does not

contain all the terms that the most general superﬁelds have (see Exercise 11.2).

The fact that the constraint we have used is preserved by supersymmetry (SUSY)

tells us that applying an arbitrary SUSY transformation to Eq. (12.8) will generate a

result that will be of the same form; i.e., there will be no new type of terms produced

(e.g., no terms with

θ ·

θ and no θ will be generated by a SUSY transformation).

We see that a general left-chiral superﬁeld contains ﬁelds with the same Lorentz

transformation properties as the ones we needed to build the Wess-Zumino la-

grangian: a left-chiral spinor ﬁeld and two complex scalar ﬁelds. However, to really

show that we can identify the components ﬁelds of  with the ﬁelds of the Wess-

Zumino lagrangian, we need to show that they transform the same way under SUSY.

We will prove that this is indeed the case, a result that we anticipated by using the

same names for the three ﬁelds that we used in Chapters 5 and 6. The beauty of the

superﬁeld formalism is that we will recover the SUSY transformations of all three

component ﬁelds from the single transformation of the superﬁeld.

12.2 SUSY Transformations of the

Component Fields

By deﬁnition, the SUSY-transformed superﬁeld 



can be written in terms of the

transformed component ﬁelds as





= φ



(x) −

θσ

θ∂



(x) −

θ ·θ

θ ·

φ



(x)

+θ ·χ



(x) −

θσ

θθ · ∂



(x) +



(x)θ · θ (12.9)

All we did was rewrite Eq. (12.8) with a prime on all the ﬁelds.

274

Supersymmetry Demystified

Now we calculate the explicit SUSY-transformed superﬁeld using Eq. (11.48).

The effect of

is simply to translate all the ﬁeld in spacetime by x

→ x

+ a

so we will only consider the effect of the supercharges. We have, from Eq. (11.32),





=  +



ζσ

θ −θσ



∂

 + ζ · ∂ +

ζ ·

∂ (12.10)

So now we have two different expressions for the SUSY-transformed ﬁeld. Equa-

tion (12.9) gives us the transformed superﬁeld in terms of the transformed compo-

nent ﬁelds φ



(x), χ



(x), and F



(x). On the other hand, the right-hand side of Eq.

(12.10) also gives the transformed superﬁeld in terms of the inﬁnitesimal parame-

ters ζ ,

ζ , and the original ﬁelds φ(x), χ(x), and F(x) [after using Eq. (12.8) for ].

Setting the two results equal to one another will give us the SUSY transformations

of all the component ﬁelds.

The ﬁrst step in evaluating the right-hand side of Eq. (12.10) is to calculate the

partial derivatives of  with respect to x

, θ

, and

˙a

. The derivative with respect

to x

is straightforward:

∂

 = ∂

φ −

θσ

θ∂

∂

φ −

θ ·θ

θ ·

θ∂

∂

φ +θ · ∂

−

θσ

θθ · ∂

∂

χ +

θ ·θ∂

F (12.11)

The derivatives with respect to the Grassmann variables require some care. We

get

∂

∂θ

=−

(σ

)

∂

φ −

θ ·

φ +χ

−

(σ

)

θ ·∂

−

(∂

)θσ

θ +θ

F (12.12)

where we have used the result of Exercise 11.3 for the derivative of θ · θ. On the

other hand, we ﬁnd [after using identity (A.50)]

∂

∂

˙a

( ¯σ

)

˙ab

∂

φ −

θ ·θ

˙a

φ +

( ¯σ

)

˙ab

θ ·∂

χ (12.13)

What we must do now is to substitute Eqs. (12.11), (12.12), and (12.13) into

the right-hand side of Eq. (12.10). Note that all terms with three θ or three

automatically vanish because they necessarily contain the square of a Grassmann

CHAPTER 12 Left-Chiral Superfields

275

coordinate. This leaves us with





=  +



ζσ

θ −θσ



∂

φ +



ζσ

θ −θσ



θσ

θ∂

∂



ζσ

θ −θσ



θ ·∂

χ +

ζσ

θθσ

θθ · ∂

∂

χ +

ζσ

θ∂

F θ · θ

−

ζσ

θ∂

φ −

ζ · θ

θ ·

φ +ζ · χ −

ζσ

θθ · ∂

χ −

ζ · ∂

χθσ

+ζ · θ F +

ζ ¯σ

θ∂

φ −

θ ·θ

ζ ·

φ +

ζ ¯σ

θθ · ∂

χ (12.14)

All we have to do now is to replace  and 



by Eqs. (12.8) and (12.9), and we will

be able to read off the transformations of the ﬁelds (after a bit of work, of course).

The trick is to realize that terms with different powers of the Grassmann variables

and

˙a

are independent.

For example, consider the terms with no Grassmann variables appearing on the

two sides of Eq. (12.14). Setting them equal, we ﬁnd the following result:



= φ + ζ · χ (12.15)

which is exactly the transformation law for φ that we obtained in Chapter 6 [see

Eq. (6.99)]!

Now consider the terms that are linear in θ and have no

θ.Weget

θ ·χ



= θ ·χ −

θσ

ζ∂

φ +ζ · θ F +

ζ ¯σ

θ∂

Using Eq. (A.49), we may write

ζ ¯σ

θ =−θσ

ζ , which shows that the last term is

actually equal to the second term, giving us (showing now explicitly all the indices):



= θ

−i θ

(σ

)

∂

φ +θ

which yields



= χ

−i (σ

)

∂

φ +ζ

F (12.16)

This is indeed the same as in Eq. (6.99), although it is not obvious at ﬁrst sight.

Recall that in Chapter 6, all the spinors had lower undotted indices, so we need to

express the ζ spinor on right-hand side of Eq. (12.16) in terms of its lower undotted