Labelle P. Supersymmetry DeMYSTiFied

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216

Supersymmetry Demystified

Using the antisymmetry of the structure constants, this can be written as



μν





= F

μν

+ f

bac



μν

This step was made because now we can use Eq. (10.14) to write this as



μν





= F

μν







μν

which is the inﬁnitesimal limit of

μν

= exp







μν

(10.23)

This shows that the ﬁeld strength transforms in the adjoint representation! Note

however that in Eq. (10.23), F

μν

is a column vector with N

− 1 entries, in contrast

with the F

μν

of Eq. (10.22) which is an N by N matrix.

As we see, unlike the abelian case, the ﬁeld strength of a nonabelian gauge theory

is not gauge-invariant, and we cannot simply use −F

μν

/4 as a kinetic term (of

course we can’t, it’s a matrix!). On the other hand, the trace of this quantity is

gauge-invariant, and the kinetic term is taken to be

kin

=−



μν



The factor of 1/2 is chosen so that the trace reduces to a sum of kinetic terms with

conventional normalization, i.e.,

−



μν



=−

μν





=−

μν

(10.24)

where Eq. (10.12) was used.

Note that the ﬁeld strength F

μν

contains the coupling constant g, as we can see

from Eq. (10.21). It is sometimes convenient to have all dependence on the coupling

constant made explicit in the lagrangian. This can be done by rescaling the gauge

ﬁeld in the following way:

→

CHAPTER 10 Supersymmetric Gauge Theories

217

in which case the ﬁeld strength is rescaled to

μν

→

μν

(10.25)

where the new ﬁeld strength does not contain the coupling constant anymore. The

kinetic energy of the gauge bosons then is

kin

=−



μν



=−

μν

(10.26)

EXERCISE 10.2

a. Verify that the generators of Eq. (10.13) obey the normalization condition

[Eq. (10.12)].

b. Construct explicitly the matrices representing the generators, of SU(2) in the

adjoint representation, and verify that they obey Eqs. (10.10) and (10.15).

10.4 The QCD Lagrangian in Terms

of Weyl Spinors

Since most of us have been taught the standard model in terms of Dirac spinors, it

may be useful to pause and see what the lagrangian of a nonabelian gauge theory

looks like in terms of Weyl spinors. We will use the example of QCD for a single

quark species (i.e., a single ﬂavor). This is particularly interesting because quarks

and antiquarks belong to two nonequivalent representations of SU(3)

, so it will

be instructive to see how this works out in terms of Weyl spinors.

Let us then write the QCD lagrangian in terms of left-chiral Weyl spinors. The

derivation is almost identical to the one we followed in Section 4.6 to write the

QED lagrangian in terms of left-chiral spinors.

We start with the gauge-invariant lagrangian for a quark, i.e.,

QCD

= 



iγ

− m





where 

actually represents a column vector of the three-color quark states in the

fundamental representation 3. To be more explicit, this is

QCD

= 



iγ

∂

− g

− m





(10.27)

218

Supersymmetry Demystified

where g

is the strong coupling constant, and

≡

As noted in the preceding section, the λ

are the eight Gell-Mann matrices. Of

course, in the standard model, there are no explicit mass terms, all masses arising

from spontaneous symmetry breaking.

What we now want to do is to write Eq. (10.27) in terms of the left-chiral quark

and antiquark spinors χ

and χ

. We have already shown in Section 4.5 [see Eq.

(4.14)] that





iγ

∂

− m





= iχ

†

¯σ

∂

+i χ

†

¯σ

∂

− m(χ

· χ

+ ¯χ

· ¯χ

)

The two extra terms coupling the Weyl spinors to the gauge ﬁeld can be obtained

from Eq. (4.12) by replacing ∂

with ig

Terms containing A

=−g

†T

− g

†

¯σ

= g

†

¯σ

− g

†

¯σ

where we have used Eqs. (A.5) and (A.36) on the ﬁrst term. Note that the spin

matrices ¯σ

and σ

commute with the gauge ﬁeld matrix A

because the former

acts on the spin degrees of freedom, whereas the latter acts in color space.

We see that there is a new twist compared with the situation we faced with the

QED lagrangian: The transpose affects the gauge ﬁeld A

because it is a matrix.

Using the fact that the Gell-Mann matrices are hermitian, we may write

= A

(λ

)

= A

(λ

)

∗

so the QCD lagrangian may be written

QCD

= i χ

†

¯σ



∂

−

(λ

)

∗



+i χ

†

¯σ



∂

+ig



−m(χ

· χ

+ ¯χ

· ¯χ

) (10.28)

The fact that a complex conjugation must be taken on the term containing the

gauge ﬁeld in the kinetic energy of the antiquark tells us that this ﬁeld transforms

CHAPTER 10 Supersymmetric Gauge Theories

219

in the representation

3. As usual, we may introduce gauge-covariant derivatives D

acting on the quark ﬁelds and write Eq. (10.28) as

QCD

= iχ

†

¯σ

+i χ

†

¯σ

− m(χ

· χ

+ ¯χ

· ¯χ

)

with the understanding that when acting on a 3 representation, such as χ

, the

covariant derivative is

= ∂

whereas, when acting on the representation

3,itis

= ∂

−

(λ

)

∗

We see that writing the lagrangian in terms of the quark and antiquark Weyl

spinors makes it explicit that they transform under the nonequivalent representations

3 and

10.5 Free Supersymmetric

Nonabelian Gauge Theories

Now we are ready to describe free supersymmetric nonabelian gauge theories (free

in the sense that we don’t consider interaction terms beyond those dictated by gauge

invariance). We know that in an SU(N ) gauge theory, there are N

− 1 gauge ﬁelds

because that’s the number of generators. To make the theory supersymmetric, we

will therefore need the same number of left-chiral gauginos λ

and auxiliary ﬁelds

. Our starting point for our lagrangian thus is

L =−

μν

μνa

+i λ

a†

¯σ

(10.29)

Be careful about not confusing the D

of the covariant derivative with the auxiliary

ﬁelds D

! And keep in mind that the index a on λ labels the N

− 1 gauginos

forming the adjoint representation, not their spinor components.

Note that we cannot write a Fayet-Illiopolos term, i.e., a term linear in the

auxiliary ﬁeld, in the case of a nonabelian gauge theory because the D

are not

gauge-invariant.

220

Supersymmetry Demystified

As a ﬁrst guess for the supersymmetric transformations that leave this lagrangian

invariant (up to total derivatives, as always), it makes sense to try the transformations

that we worked out in the abelian case, Eq. (10.9), applied separately to each of the

− 1 ﬁelds F

μν

,λ

and D

Unfortunately, things are not so simple. We have one more constraint now: The

SUSY transformations must not conﬂict with gauge transformations. To see if there

is such a conﬂict, we ﬁrst need to know in what representation the gauginos and

auxiliary ﬁeld transform, if at all. To answer this, consider a generic ﬁeld X (which

could be bosonic or fermionic) whose transformation under SUSY is, symbolically,

→ X

+ δ

(10.30)

If we want this transformation (together with the SUSY transformations of all the

other ﬁelds in the theory) to be a symmetry of the lagrangian, it better still hold if

we ﬁrst perform a gauge transformation of the ﬁeld X before applying it! And for

this to be true, the two sides of Eq. (10.30) obviously must transform the same way

under a gauge transformation.

Let us see what this implies. As our ﬁrst guess for the transformation of each of

the gauginos λ

, let’s take the exact same expression we had in the abelian case,

namely,

δλ

¯σ

ζ F

μν

+ ζ D

(10.31)

For this to be valid in the nonabelian case, we are forced to conclude that the

gauginos and the auxiliary ﬁelds must transform like the ﬁeld strengths under

gauge transformations, i.e., in the adjoint representation. With that choice, the

SUSY transformation (10.31) does not conﬂict with the gauge transformation, in

the sense that supersymmetry and gauge transformations commute. Of course, it

does not mean that the guess (10.31) is the correct SUSY transformation for the

lagrangian (10.29); that’s another issue. For now, we are simply making sure that

what we’ll try for the SUSY transformations does not conﬂict with the gauge

symmetry.

Now let’s see if the other transformations we had obtained in the abelian case

cause any problem. Consider next the transformation of the auxiliary ﬁeld. The

transformation in the abelian case suggests that we take

δ D

=−iζ

†

¯σ

∂

+i (∂

)

†

¯σ

Is that acceptable? Well, the auxiliary ﬁelds transform in the adjoint representation,

as we just saw. Does the right-hand side transform the same way? The answer is

CHAPTER 10 Supersymmetric Gauge Theories

221

clearly no, because of the derivatives acting on the gauginos, which will pick up

a derivative of the gauge parameter on transformation. But we know how to ﬁx

that from elementary particle physics; we simply have to replace the derivative

with a covariant derivative! However, since the gauginos transform in the adjoint

representation, we need to use this representation to deﬁne the covariant derivative

acting on them.

At this point, expressions become quite awkward if we show all the SU(N )

indices explicitly, so let’s drop the indices on the gauginos λ. All we need to do is to

think of each λ as being a column vector with N

− 1 entries. With this simpliﬁed

notation, the covariant derivative acting on a gaugino is simply

λ =



∂

+ig



In case you are interested, with explicit SU(N ) indices, this equation becomes

= ∂

+ig





Therefore, our guess for the transformation of the auxiliary ﬁelds is [hiding once

more the SU(N ) indices]:

δ D =−iζ

†

¯σ

λ +i(D

λ)

†

¯σ

What about the SUSY transformation of the gauge ﬁelds A

? The transformation

in the abelian case suggests that we take

δ A

= ζ

†

¯σ

+ λ

a†

¯σ

ζ (10.32)

However, at ﬁrst sight, this seems completely wrong. The gauginos transform in

the adjoint representation, whereas the gauge ﬁelds don’t. But, there is no need to

despair. Recall that what really matters is that the transformed ﬁeld A

+ δ A

has

the same gauge transformation as A

itself. Let’s have a closer look.

Consider contracting both sides of Eq. (10.32) with T

. The matrix A

= A

transforms as in Eq. (10.19):

→ A



= UA

†



∂



†

(10.33)

whereas λ = λ

transforms as

λ → λ



= U λU

†

222

Supersymmetry Demystified

Since δ A

contains only the gauginos, we also have

δ A

→ U δ A

†

(10.34)

Using Eqs. (10.33) and (10.34), we ﬁnd that the quantity A

+ δ A

transforms as

+ δ A

→ UA

†



∂



†

+U δ A

†

= U



+ δ A

†



∂



†

which is precisely the gauge transformation of A

itself! This tells us that the SUSY

transformation (10.32) does not conﬂict with the gauge symmetry.

To summarize, our ﬁrst guesses for the SUSY transformations of the ﬁelds are

δλ =

¯σ

ζ F

μν

+ ζ D

δ D =−iζ

†

¯σ

λ +i(D

λ)

†

¯σ

δ A

= ζ

†

¯σ

λ + λ

†

¯σ

Remains the task of checking whether these transformations indeed leave the

lagrangian (10.29) invariant. It turns out that they do! However, we won’t prove

this here because the demonstration is almost identical to the one we carried out

in detail for the abelian case. The interested reader can ﬁnd the details in Ref. 31.

Instead, it is more interesting to turn our attention to supersymmetric interactions

between vector and chiral multiplets.

10.6 Combining an Abelian Vector Multiplet

With a Chiral Multiplet

Now we consider building supersymmetric theories that couple chiral and vector

multiplets. We start with a free abelian gauge multiplet and a single free chiral

multiplet. By free chiral multiplet, we mean that we will not include any self-

interaction among the ﬁelds of the chiral multiplet or any mass terms for χ and φ.

We will come back to the question of interactions of the chiral multiplet at the end

of this section.

We want to build a supersymmetric theory in which the ﬁelds of the chiral mul-

tiplet interact with the ﬁelds of the vector multiplet. The simplest way to couple the

CHAPTER 10 Supersymmetric Gauge Theories

223

two multiplets is to take the chiral multiplet to have a U(1) charge. In other words,

under the U (1) gauge transformation, we take the ﬁelds of the chiral multiplet to

transform as

X ⇒ exp[iq(x)]X (10.35)

where X stands for any of the three ﬁelds φ,χ,orF, and q is their common charge.

As before, we take the spinor and auxiliary ﬁelds of the abelian gauge multiplet λ

and D to be gauge-invariant, whereas the gauge ﬁeld A

transforms as usual.

In order for the lagrangian of the chiral multiplet to be gauge-invariant, we must

replace all the derivatives acting on the ﬁelds of that multiplet by gauge-covariant

derivatives; i.e., we must make the replacement

∂

X → D

X =



∂

+iq A



where X stands again for any of the ﬁelds of the chiral multiplet. We therefore have

L = (D

φ)

†

φ) + iχ

†

¯σ

χ + F

†

F −

μν

+i λ

†

¯σ

∂

λ +

+ ξ D

(10.36)

This is our starting point. Note that the gauge ﬁeld A

is already coupled to the

ﬁeld of the chiral multiplet through the gauge-covariant derivative. We see that we

get three new interaction terms arising from gauging the ﬁelds:

int

)

= iqφ A

∂

†

−iqφ

†

∂

φ −qχ

†

¯σ

χ (10.37)

which provide interactions between the gauge ﬁeld A

and the ﬁelds of the chiral

multiplet. We have included a label 1 because there will be a second interaction

lagrangian.

To be as general as possible, we now look for gauge-invariant interactions com-

bining the ﬁelds of the two multiplets that do not arise from gauging the derivatives.

Since the ﬁelds of the chiral multiplets transform as Eq. (10.35), gauge-invariant

combinations can be formed if we multiply any two of them as long as we apply a

hermitian conjugate to one of them. Since our goal is to couple these ﬁelds with the

ﬁelds F

μν

,λ, and D, which have dimensions 2, 3/2, and 2, we need only to retain

combinations of φ,χ, and F with dimension 5/2 at most. The only combinations

of ﬁelds in a chiral multiplet that are gauge-invariant and of dimension less than or

equal to 5/2 are

†

φφ

†

224

Supersymmetry Demystified

and their hermitian conjugates. We now combine these terms with the gauge-

invariant ﬁelds F

μν

,λ, and D to get interactions. Keeping only terms of dimension 4

or less, we get (we are not worrying about matching indices and Lorentz invariance

for now)

μν

†

φ Dφ

†

φφ

†

χλ + h.c. φ

†

φλ + h.c. (10.38)

The ﬁrst two terms are of dimension 4, whereas the last two have dimension 7/2.

Now let’s worry about Lorentz invariance. The ﬁrst term contains two Lorentz

indices that we must contract with something. We cannot use derivatives because the

terms are already of dimension 4. The only dimensionless quantities with Lorentz

indices that we have at our disposal are σ

and ¯σ

, but since F

μν

†

φ does not

contain any spinor between which to sandwich the Pauli matrices, we are out of

luck, and we must drop that term.

The last term in Eq. (10.38) has a similar problem. In order to get a Lorentz

invariant, we must introduce a second spinor to dot with λ, which produces an

expression of dimension 5, so we must drop that term as well.

In the third term of Eq. (10.38), the two spinors can be combined in a Lorentz-

invariant way using the spinor dot products introduced in the Chapter 1, so our ﬁnal

form is

†

χ ·λ + φ ¯χ ·

λ (10.39)

Recall that in spinor dot products, the order does not matter, which is why we did

not list, say, φ

†

λ · χ as a different possibility.

The second term of Eq. (10.38) is already Lorentz-invariant, so it is ﬁne as it

stands. Thus we ﬁnally get the following interaction lagrangian

int

)

= c



†

χ ·λ + φ ¯χ ·



+ c

†

φ D (10.40)

where c

and c

are are arbitrary constants to be determined. We have assumed that

is real. Of course, there is no a priori reason why this should be the case, but it

will simplify our calculation, and it will turn out to be general enough for building

a supersymmetric lagrangian. For the same reason, we will also take c

to be real

in the rest of our calculations.

Our total interaction lagrangian therefore is the sum of Eqs. (10.37) and (10.40).

EXERCISE 10.3

We have taken the ﬁelds of the chiral multiplet to be charged. We also could

have considered them to be neutral. In that case, what would be the most general

CHAPTER 10 Supersymmetric Gauge Theories

225

(renormalizable, gauge-invariant, and real) interaction lagrangian mixing ﬁelds of

both multiplets?

The goal now is to see if it is possible to ﬁnd values of c

and c

for which the

theory will be supersymmetric. Of course, we know that the free lagrangians for the

vector and chiral multiplets are supersymmetric, so we need only to worry about

the interaction terms.

A priori, there is no reason to believe that the vector and chiral multiplets should

both transform with a common SUSY parameter ζ . In fact, we will ﬁnd that we

cannot make the theory supersymmetric if we use a common parameter for both

supermultiplets. So let’s use ζ for the parameter of the transformation of the chiral

multiplet and aζ for the vector multiplet, where a is an arbitrary constant that we

will take to be real (with hindsight, we will see that this was indeed sufﬁciently

general). Then the transformation we take for the ﬁelds of the vector multiplet are

simply

δ A

= a(ζ

†

¯σ

λ + λ

†

¯σ

ζ)

δλ =

μν

¯σ

ζ + aDζ

δλ

†

=−

μν

†

¯σ

+ aζ

†

δ D =−iaζ

†

¯σ

∂

λ +ia(∂

†

) ¯σ

whereas the transformations of the ﬁeld of the chiral multiplet have to be modiﬁed

by replacing the partial derivatives by covariant derivatives, i.e.,

δφ = ζ · χ

δφ

†

= ¯χ ·

δχ =−i(D

φ)σ

iσ

∗

+ Fζ

δχ

†

=−i(D

φ)

†

iσ

+ F

†

δF =−iζ

†

¯σ

δF

†

= i (D

χ)

†

¯σ

These are our “guesses” for the SUSY transformations. It turns out that the

transformation of the auxiliary ﬁeld will have to be modiﬁed to make our theory

supersymmetric.