Labelle P. Supersymmetry DeMYSTiFied

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276

Supersymmetry Demystified

components. Using

= (iσ

)

b˙c

˙c

[see Eq. (A.31)] and

˙c

= ζ

∗

, we can write our

result as



= χ −i σ

(iσ

)ζ

∗

∂

φ +ζ F

with the convention now that the indices on all the spinors are downstairs and

undotted. This is exactly our result in Eq. (6.99), including the term proportional

to the auxiliary ﬁeld F in the transformation of the spinor.

Now let’s consider the terms containing two Grassmann variables θ

and no

in Eq. (12.14). These are



θ ·θ =

F θ · θ −

θσ

ζθ · ∂

χ +

ζ ¯σ

θθ · ∂

The two last terms can be combined using again

ζ ¯σ

θ =−θσ

ζ , giving us



θ ·θ =

F θ · θ −i θσ

ζθ · ∂

The last term can be rewritten with the two Grassmann variables dotted into one

another using the identity (12.47) (which will be derived in Exercise 12.2):



θ ·θ =

F θ · θ +

θ ·θ(∂

χ)σ

from which we can read off the transformation of the auxiliary ﬁeld, i.e.,



= F + i (∂

χ)σ

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

(∂

χ)σ

ζ =−

ζ ¯σ

∂

which leads to

δF =−i

ζ ¯σ

∂

χ (12.17)

in agreement with Eq. (6.99).

Let us pause and summarize what we have accomplished in this section. By

comparing the terms with no Grassmann variables, with a single Grassmann variable

, and with two Grassmann variables θ

on the two sides of Eq. (12.14), we have

established the transformation properties of the ﬁelds φ(x), χ (x), and F(x) under

CHAPTER 12 Left-Chiral Superfields

277

SUSY and have found that they agree with the transformations we established in

Chapters 5 and 6.

However, we are clearly not done. We should consider all the other terms that

appear in Eq. (12.14). Now that we have established the transformation laws of

all the ﬁelds, there is no leeway left: We should show that with the transformation

laws we just established, all the terms on the two sides of Eq. (12.14) agree. This is

obviously quite labor-intensive, and we won’t pursue the demonstration here, but

rest assured that it does work. The courageous reader is invited to complete the

calculation.

12.3 Constructing SUSY Invariants Out of

Left-Chiral Superfields

The key reason for the introduction of superﬁelds is that they make the construction

of SUSY-invariant theories very easy. We will now see how this works in the case

of left-chiral superﬁelds.

The starting point is the seemingly innocuous observation that the F(x) ﬁeld

transforms as a total derivative under SUSY. Since the F(x) ﬁeld is the coefﬁcient

of the θ ·θ/2 term in the expansion of , we can phrase this by saying: The

coefﬁcient of the θ ·θ/2 term in the expansion of a left-chiral superﬁeld (x,θ,

θ)

transforms as a total derivative under SUSY.

We saw in the section on Grassmann calculus that it is possible to “project out,”

i.e., to isolate, the coefﬁcient of θ ·θ/2 of an arbitrary function of the Grassmann

variables simply by integrating over d

θ, so we can say that



θ(x,θ,

θ) (12.18)

transforms as a total derivative under SUSY. Recall that d

θ is a shorthand for

dθ

The fact that Eq. (12.18) transforms as a total derivative tells us that it can be

used as a term in a SUSY-invariant lagrangian density. In other words, the quantity



θ(x,θ,

θ) (12.19)

is invariant under a SUSY transformation.

278

Supersymmetry Demystified

In itself, Eq. (12.19) is not interesting because it just gives the ﬁeld F(x), not

interactions or kinetic terms. What we would like to do is to introduce SUSY-

invariant interactions and kinetic terms using superﬁelds.

The next key observation is that the product of any number of left-chiral su-

perﬁelds is itself a chiral superﬁeld. What we mean by this is that the product of

left-chiral superﬁelds can be written as a superﬁeld that has the same expansion in

the Grassmann variables and that transforms as a single left-chiral superﬁeld under

SUSY.

This will become more clear if we carry out the explicit proof in the special case

of the product of two left-chiral superﬁelds. To be general, we will consider the

product of two distinct superﬁelds that we will call 

and 

. This is preferable to

using 

and 

, which might lead to confusion when we start writing out the

Weyl spinor ﬁelds because χ

and χ

could be mistaken for the components of the

spinors.

If we go back to writing left-chiral superﬁelds in terms of the variables y

introduced in Eq. (12.3), we have



(y,θ) = φ

(y) + θ · χ

(y) +

θ ·θ F

(y) (12.20)

and



(y,θ) = φ

(y) + θ · χ

(y) +

θ ·θ F

(y) (12.21)

Evidently, the product of those two superﬁelds can be cast in the form of a

left-chiral superﬁeld, i.e., with the expansion



≡ 

= φ

(y) + θ · χ

(y) +

θ ·θ F

(y) (12.22)

The three ﬁelds φ

,χ

, and F

are obviously functions of the ﬁelds appearing in



and 

. A short calculation reveals that

(y) = φ

(y)φ

(y)

(y) = φ

(y)χ

(y) + φ

(y)χ

(y)

(y) = φ

(y) F

(y) + φ

(y) F

(y) − χ

(y) · χ

(y) (12.23)

where we have used Eq. (A.54).

On the other hand, it is clear that multiplying an arbitrary superﬁeld with a left-

chiral superﬁeld does not yield a left-chiral superﬁeld. As a special case of this,

CHAPTER 12 Left-Chiral Superfields

279

consider multiplying a left-chiral superﬁeld by its hermitian conjugate 

†



†

 =



†

(y) +

θ · ¯χ(y) +

θ ·

θ F

†

(y)



φ(y) + θ · χ(y) +

θ ·θ F (y)



Obviously, the result is not a left-chiral superﬁeld. For example, it contains a func-

tion of y times

θ ·

θ, which is absent in a left-chiral superﬁeld.

Going back to the product of two left-chiral superﬁelds 



, it is important

to realize that it is not sufﬁcient to show that it has the same Grassmann expan-

sion as a single left-chiral superﬁeld. To really prove that the result, Eq. (12.22),

is a bona ﬁde left-chiral superﬁeld, we must prove that under SUSY, the ﬁelds

, χ

, and F

transform the same way as the components of a left-chiral su-

perﬁeld. In other words, we need to show that the SUSY transformations of these

three ﬁelds are identical to transformations (12.15), (12.16), and (12.17). This turns

out to be an easy proof if we use supercharges to implement SUSY transforma-

tions. In that language, what we need to check is whether the following relation

holds:



exp



−i ζ · Q − i

ζ ·







exp



−i ζ · Q − i

ζ ·







= exp(−i ζ · Q − i

ζ ·

Q)(



)

Let us deﬁne

−i ζ · Q − i

ζ ·

Q ≡ O

Then, to ﬁrst order in ζ , what we need to check is whether



(O

) + (O

)

= O(



) (12.24)

This is actually trivially true because the supercharges Q and

Q are linear operators,

and the left-chiral superﬁelds are bosonic quantities. These two facts allow us to

write

Q(



) = (Q

)

+ 

(Q

)

and similarly for

Q. Therefore, Eq. (12.24) is indeed valid, and this completes the

proof that the product of two left-chiral superﬁeld does transform like a single left-

chiral superﬁeld under SUSY. It is then clear, by induction, that the product of any

number of left-chiral superﬁelds also transforms like a left-chiral superﬁeld.

280

Supersymmetry Demystified

Now that we have convinced ourselves that multiplying left-chiral superﬁelds

produces new superﬁelds of the same nature, the obvious question is: Why should

we care? To answer this, let’s go back to the observation that the auxiliary ﬁeld

F transforms as a total derivative under SUSY. This means that in the product of

any number of left-chiral superﬁelds, the term proportional to θ · θ necessarily

transforms as a total derivative under a SUSY transformation.

Let’s deﬁne the F term of any function of superﬁelds (not necessarily left-chiral)

as the coefﬁcient of the θ ·θ/2 term. The factor of one-half is included because

then the F term can be written very simply in terms of integrations over Grassmann

variables. Let’s write a function of superﬁelds as F(S

, S

,...), where the S

are

general superﬁelds. We use the following notation to represent the F term of that

function:



θF (S

, S

,...) ≡ F(S

, S

,...)



(12.25)

Thus we use the notation |

to mean “extraction of the F term.”

Expression (12.25) is obviously a nontrivial combination of the component ﬁelds

in the superﬁelds S

, S

,... and is not in general invariant under SUSY. However,

if F is a holomorphic function of left-chiral superﬁelds, i.e., a function of the



and not of the 

†

, then it is itself a left-chiral superﬁeld. In that case, its F

term transforms as a total derivative under SUSY, and therefore, Eq. (12.25) is a

supersymmetric lagrangian density!

In the case of a holomorphic function of left-chiral superﬁelds, we will use the

notation W instead of F. The conclusion then is that

W(

,

,...)



(12.26)

is a supersymmetric lagrangian density (i.e., it transforms with a total derivative

under SUSY).

The function W is, of course, the superpotential we introduced in Chapter 8 as

a function of the scalar ﬁelds φ

. Here, it is a function of the superﬁelds 

, which

is the origin of the name superpotential. Now we understand the reason for the

requirement of holomorphicity we had obtained in a roundabout way earlier: If we

multiply by the hermitian conjugate 

†

of any of the superﬁelds, the superpotential

will no longer be a left-chiral superﬁeld, and its F term will not be a supersymmetric

lagrangian density.

It’s hard to overemphasize the power of the result that Eq. (12.26) is a super-

symmetric lagrangian. We can take any number of left-chiral superﬁelds, multiply

them and extract the F term of the result, and we will automatically have created

a SUSY-invariant theory! Compared with the efforts we had to put in to build the

simplest supersymmetric theories in Chapters 5, 6, and 8, this is child’s play! The

CHAPTER 12 Left-Chiral Superfields

281

ease it brings to the construction of supersymmetric theories is one of the reasons

that make the superspace approach to SUSY so enticing.

We will soon show how we can use this observation to recover some of the SUSY-

invariant lagrangians we built in previous chapters. We will not reproduce all of

them this way, but luckily, only a slight generalization of the trick will be needed to

construct all the supersymmetric theories we have found previously at a fraction of

the effort! Moreover, we will be able to construct more complex supersymmetric

lagrangians, including the minimal supersymmetric standard model (MSSM), with

very little work.

Before doing an explicit example, two points deserve to be emphasized. First,

although the trick just described gives us a way to write down an inﬁnite number

of supersymmetric theories, the fact that we want to consider only renormalizable

theories will greatly restrict the choices available to us, as we will discuss shortly.

The second point is that the F term of an arbitrary superﬁeld obviously does

not, in general, transform as a total derivative. For example, the F term of 

†



does not transform as a total derivative because this combination of ﬁelds is not

itself a chiral superﬁeld, as we discussed previously.

This raises the question of whether it is possible to extract a supersymmetric

lagrangian from a product of arbitrary superﬁelds. The answer is yes, and the way

to do this will provide the generalization to which we hinted earlier. We will come

back to this later in this chapter.

Let us now construct explicitly a supersymmetric theory using our newfound

approach. The simplest (nontrivial) case is the F term of the product of two left-

chiral superﬁelds 



. We multiplied two left-chiral superﬁelds in Eq. (12.23),

but the result was expressed in terms of the variable y, and what we need is the F

term of 

(x,θ,

θ)

(x,θ,

θ). However, a second of thought shows that as long

as we are interested only in picking up the F term, we simply may set y = x directly

in the product of left-chiral superﬁelds. In other words,





(y,θ)

(y,θ)...









(y = x,θ)

(y = x,θ)...





(12.27)

This result is very useful because the expansion of left-chiral superﬁelds in terms

of y and θ is much simpler than the expansion in terms of x,θ, and

θ!

EXERCISE 12.1

Prove Eq. (12.27). Hint: Using the relation between y

and x

, the proof is trivial.

Using Eqs. (12.23) and (12.27), we therefore have

(



)



= φ

(x) F

(x) + φ

(x) F

(x) − χ

(x) · χ

(x) (12.28)

282

Supersymmetry Demystified

The key point is that this quantity transforms as a total derivative under SUSY, so

it is a supersymmetric lagrangian density!

If this expression seems vaguely familiar, there is a good reason: These terms

appear as part of the Wess-Zumino lagrangian derived in Chapter 8. We still need

more work before recovering the complete Wess-Zumino lagrangian using this

approach, but for now, note how easy it was to obtain a set of SUSY-invariant

interactions compared with the long calculations we had to go through in Chapter 8.

All we needed to do this time was to multiply two left-chiral superﬁelds and extract

the F term of the result!

We have a supersymmetric action, but we must not forget the other conditions that

a viable lagrangian must have. In particular, it must be a real quantity. As it stands,

the lagrangian density in Eq. (12.28) is not real (recall that φ and F are complex

scalar ﬁelds), so we must add its hermitian conjugate to get a real lagrangian.

In addition, Eq. (12.28) does not have the correct dimension for an action. Before

correcting this, let’s take a moment to open a parenthesis on the dimension of left-

chiral superﬁelds. The explicit expansion of such a superﬁeld is (going back to the

variable y for a moment)

 = φ(y) + χ(y) · θ +

θ ·θ F (y)

The superﬁeld therefore must have the same dimension as a scalar ﬁeld, which is

1. Recall that the auxiliary ﬁeld F has dimension 2, which shows that the Grassmann

variables θ

and θ

have dimension −1/2 . On the other hand, since



θθ· θ = 1

we see that the Grassmann inﬁnitesimals dθ

and dθ

have dimension 1/2! This

is in stark contrast with ordinary variables, for which the inﬁnitesimal element has

the same dimension as the variable itself (e.g., the dimensions of dx and x are the

same).

Now let us look at



θ





As we discussed previously, the superﬁelds have dimension 1, so 



has

dimension 2. But the projection of the F term, which involves integrating over

θ, increases the dimension by 1,sotheF term



θ



has dimension 3. We

therefore have to multiply it by a constant m having dimension of mass in order to

obtain a lagrangian density with the correct dimension.

CHAPTER 12 Left-Chiral Superfields

283

If we allow the presence of n left-chiral superﬁelds, the most general supersym-

metric real lagrangian that is constructed out of the product of two such superﬁelds is



+ h.c. =

(φ

+ φ

− χ

· χ

) + h.c. (12.29)

where the sum over the indices i and j goes from 1 to the number of superﬁelds, and

the constants m

have the dimension of mass. Because the superﬁelds commute

(they are bosonic quantities), the product 



is symmetric under the exchange

of the indices i and j, which allows us to choose the coefﬁcients m

to also be

symmetric, which is why we chose to introduce a factor of 1/2.

So this is it! We have constructed a nontrivial interaction lagrangian density that

is invariant, up to total derivatives, under SUSY. And the way we did it was to simply

multiply two left-chiral superﬁelds and extract the F term. As mentioned earlier,

we have in this way recovered some of the terms that appeared in the Wess-Zumino

lagrangian. This should not be too surprising because left-chiral superﬁelds have

the same ﬁeld content as the Wess-Zumino model, so it was to be expected that

constructing SUSY-invariant actions out of these superﬁelds should lead us back

to that model.

However, there is something even more interesting to mention. The expression

on the left-hand side of Eq. (12.29) is exactly the same form as the ﬁrst term

we included in the superpotential W(φ) in Chapter 8! Except that, obviously, in

Chapter 8 the superpotential was a function of the scalar ﬁelds φ

(x) and φ

(x),

not of superﬁelds. This is, of course, not accidental, and the connection will be

explained in more detail shortly.

We have only recovered some of the interactions present in the Wess-Zumino

model. To obtain the other ones, the next obvious thing to try is the F term of the

product of three left-chiral superﬁelds:







θ



If we write explicitly only the terms containing two factors of θ,wehave



θ ·θ



+ φ



+ φ

θ ·χ

+φ

θ ·χ

+ φ

θ ·χ

+···

θ ·θ



+ φ

− φ

· χ

− φ

· χ

−φ

· χ



+···

284

Supersymmetry Demystified

where we have again made use of Eq. (A.54), and the dots represent terms linear

in or independent of θ. Therefore, the F term is simply





= φ

+ φ

−φ

· χ

− φ

· χ

− φ

· χ

Note that the whole expression is symmetric under the exchange of any two

indices (as we expected because the chiral superﬁelds commute). The most general

combination of three chiral superﬁelds therefore is given by

ijk





(12.30)

where the dimensionless coefﬁcients y

ijk

can be taken to be symmetric under the

exchange of any two indices for the same reason that we could take the coefﬁcients

to be symmetric (and the division by 3! = 6 has the same origin as the division

by 2 accompanying m

). Again, we also must add the hermitian conjugate to get a

real lagrangian.

We see that Eq. (12.30) contains all the interactions terms of the Wess-Zumino

lagrangian that were not present in the F term of the product of two superﬁelds!

And notice that, once more, this is precisely the second term of the superpotential

obtained in Chapter 8 but now expressed in terms of the left-chiral superﬁelds.

All the interactions of the Wess-Zumino lagrangian therefore are contained in

the simple expression

int

(W-Z) =





ijk



+ h.c.





(12.31)

We understand now why this potential is called the superpotential: It is a function

of the superﬁelds that contains all the supersymmetric interactions terms.

I hope you are impressed by how easy it is to construct SUSY-invariant interac-

tions using superﬁelds!

Are there other SUSY-invariant lagrangians we can generate this way? Of course,

since we can multiply an arbitrary number of left-chiral superﬁelds and extract the

F term to get such lagrangians. However, remember that we only want to consider

renormalizable lagrangians, for which the sum of the dimensions of the ﬁelds must

not be larger than 4. This means that we won’t consider the product of more than

three superﬁelds. At ﬁrst, this might seem incorrect because a left-chiral superﬁeld

has dimension 1, so shouldn’t we include products of four superﬁelds as well?

CHAPTER 12 Left-Chiral Superfields

285

But recall that the integration over d

θ necessary to extract the F term increases

the dimension by 1, so even though 

has dimension 4, the F term of 

has

dimension 5.

12.4 Relation Between the Superpotential

in Terms of Superfields and the

Superpotential of Chapter 8

Let us now try to understand the relation between the superpotential we wrote here in

terms of superﬁelds and the superpotential we had in Chapter 8. There is obviously a

connection because the two have exactly the same functional dependence. However,

in Chapter 8 the superpotential was expressed in terms of the scalar component

ﬁeld φ, and even more signiﬁcant, there also were derivatives of the superpotential

appearing in the lagrangian. So what is the precise connection?

Consider calculating the F term of a superpotential W that is a function of n

left-chiral superﬁelds:

W(

,

,... ,

)

To begin, assume that the superpotential is a monomial in the superﬁelds, which

contain at most one power of each superﬁeld. In other words, W is simply

W(

,

,... ,

) = 



···

(12.32)

Of course, some of the superﬁelds can be omitted from the product, but none is

raised to a power. It will be trivial to extend our discussion to an arbitrary polynomial

later on.

To make the discussion more transparent, let us write one of the superﬁelds, let’s

say 

, in terms of its component ﬁelds:

W(

···

) = W





···



+ χ

· θ +

θ ·θ F



···



Now let’s ﬁnd out what will be the F term of this expression or, equivalently,

what is the coefﬁcient of θ ·θ/2.

There are many contributions. One comes from keeping the auxiliary ﬁeld F

of 

, in which case, since it already multiplies θ · θ, we must only keep the φ