Labelle P. Supersymmetry DeMYSTiFied

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136

Supersymmetry Demystified

and

|p=



i0 jk

|p

=−



0ijk

|p



ijk

|p

where we have used Eq. (7.2). It turns out that the operator



ijk

is simply the

total angular momentum of the particle, i.e.,



ijk

= L

+ S

(7.8)

EXERCISE 7.2

The differential operator representation

μν

of the Lorentz generators acting on a

spinor is deﬁned as

exp



μν



χ ≡ χ + δ

with δ

χ given in Eq. (6.81). This immediately leads to

μν

= σ

μν

+i (x

∂

− x

∂

)

= σ

μν

+ x

− x

Use this to prove Eq. (7.8). HINT: You may assume that the two sides of Eq. (7.8)

are applied to a one-particle state of deﬁnite momentum so that you can replace the

operators

and

by the actual four-momenta p

and p

We therefore have

|p=m(S

+ L

)|p (7.9)

which, in the rest frame, reduces to

|p=mS

|p

So the Casimir operator W

acting on this state gives

|p=W

|p−W

|p=−m



|p=−m

s(s + 1)|p (7.10)

CHAPTER 7 Applications of the SUSY Algebra

137

We now see that the second property that can be used to specify the massive

representations of the Lorentz group is the spin (or total angular momentum, if we

are not working in the rest frame). Let’s classify all the possible states of a massive

particle at rest. For this, we need a complete set of commuting observables. In

addition to the four-momentum squared (whose eigenvalue is just m

) and the

square of the spin, we also may use the component of the spin along the z axis, S

because this operator obviously commutes with P

and with



(a fact familiar

from quantum mechanics). Note that

= m(L

+ S

) (7.11)

which, in the rest frame of the particle, reduces to W

= mS

Therefore, massive states are labeled in their rest frame by three quantum num-

bers: p, s, and s

, with

|p, s, s

=m|p, s, s



|p, s, s

=−m



|p, s, s



=−m

s(s + 1)|p, s, s



|p, s, s

=S

|p, s, s

=s

|p, s, s



As we all know from quantum mechanics, s

may take 2s + 1 values, ranging from

−s to +s in integer steps.

Now consider massless particles. In this case, we obviously cannot go to the

rest frame of the particle, but we may choose to work in a frame where the four-

momentum is given by

= (E, 0, 0, E) (7.12)

We will denote the corresponding state by |p

. We then have

|p

= (P

, P

)|p

= (E, 0, 0, E)|p

(7.13)

|p

= P

|p

= 0

|p

= P

|p

= E|p

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Supersymmetry Demystified

We can use these results to write

|p

= (P

, 0, 0, P

)|p

(7.14)

The reason why this is a useful thing to do will become clear in a moment.

It turns out that like P

, W

gives zero when acting on a massless one-

particle state. The proof of this result is a bit subtle. Obviously, we cannot simply

let m go to zero in Eq. (7.10) because that equation had been obtained with the

assumption that we could work in the rest frame of the particle. A detailed proof

can be found in Wigner’s original paper

or in Section 2.7.2 of Ref. 28.

The fact that W

gives zero when applied to a one-particle massless state,

together with Eq. (7.3), implies that when acting on such a state, the operator W

is proportional to the four-momentum operator

|p

= hP

|p

(7.15)

(the meaning of the constant h will be elucidated shortly). To prove Eq. (7.15),

consider

|p

= (W

− W

)|p

= (W

− W

)|p

= E(W

− W

)|p

where we have used Eqs. (7.13) and (7.14). But this must be zero because of

Eq. (7.3), so we conclude that

|p

= W

|p

(7.16)

Using this result, we have

|p

= (W

− W

)|p

=−(W

+ W

)|p

(7.17)

Since W

is zero when acting on a massless state, we obtain

|p

= W

|p

= 0 (7.18)

CHAPTER 7 Applications of the SUSY Algebra

139

Therefore, when acting on the state, the action of the operator W

reduces to

|p

= (W

, 0, 0, W

)|p

(7.19)

From Eq. (7.4) we know that W

commutes with P

(because it commutes with

), so we may take |p

to be a common eigenstate of these two operators. This

implies that we may write

|p

= hP

|p

(7.20)

Now we see why we wrote Eq. (7.14) in that form. By comparing with Eq. (7.19),

we now may conclude that

|p

= hP

|p

(7.21)

as we had set out to prove.

From Eq. (7.13) we see that the action of W

on the state is

|p

= (hE, 0, 0, hE)|p

(7.22)

To ﬁnd out what the constant h represents, we need to work out explicitly the

left-hand side of Eq. (7.22). Let’s pick μ = 3. The derivation that led to Eq. (7.9) is

still valid if we simply replace P

|p=m|pwith P

|p

= E|p

. We then obtain

|p

= E(S

+ L

)|p

= Es

|p

where we have assumed that there is no orbital angular momentum. Using this in

Eq. (7.22) implies that h is simply the z component of the spin, i.e.,

h = s

(7.23)

Since our state has its momentum along the z direction, we may, of course, write

h =s ·

p (7.24)

which is nothing other than the helicity of the particle. By the way, the helicity is

often denoted by λ, but we prefer to use h because we will use λ to represent a

left-chiral spinor that appears in supersymmetric gauge theories.

140

Supersymmetry Demystified

We have obtained this result by considering μ = 3 in Eq. (7.22), but we get the

same result if we pick μ = 0, as you are invited to conﬁrm in Exercise 7.3.

In contrast with massive states, massless representations of the Lorentz group

are therefore completely speciﬁed by only two numbers: their energy (which then

speciﬁes their four-momentum) and their helicity h! The reason we don’t have

2s + 1 states, as for massive particles, lies in Eq. (7.18), which reveals that we do

not have any ladder operators to change the z component of the spin. Therefore,

there is only one spin state!

Does this mean that massless particles can be observed with only one helicity?

Well, not quite. Recall that a physical theory also must respect CPT invariance. It

turns out that CPT conjugate states have opposite helicities, so a physical theory

necessarily must contain two states, with helicities h and −h. The states with the

intermediate values of s

that we encounter for massive particles therefore are absent

for massless particles. This is the reason the photon is observed with helicities +1

and −1 or that the graviton is predicted to have helicities +2 and −2 only.

We will now repeat this exercise in the context of SUSY.

EXERCISE 7.3

Prove that we again obtain h =s ·

p if we set μ = 0 in Eq. (7.22).

7.2 Effects of the Supercharges on States

We will now use the SUSY algebra to determine what states are needed to build a

supersymmetric theory. Of course, we have worked out one such theory explicitly,

the Wess-Zumino model, but we would like to see how the results we obtained there

can be generalized.

We will only consider the classiﬁcation of massless supersymmetric states. These

are the only states of relevance if we are interested in the minimal supersymmetric

extension of the standard model because in this theory, as is the case in the standard

model itself, all masses are generated through spontaneous symmetry breaking.

As we saw in the previous section, massless states are identiﬁed by their four-

vector and their helicity h. To emphasize this, we will change our notation from

|p

to |p, h. What we want to now is to work out the effect of the supercharges

on this state.

The supercharges commute with the four-momentum operator but not with the

Lorentz generators, so they can change the helicity of states but not their four-

momentum. Our ﬁrst step, then, is to compute the commutator of the supercharges

with W

. Actually, we really only need the commutator with W

because of

Eqs. (7.16) and (7.18).

CHAPTER 7 Applications of the SUSY Algebra

141

We will ﬁrst compute [Q

, W

] and come back to the hermitian conjugates of

the supercharges later. Using Eqs. (7.1) and (6.83), this is given by

, W

] =



0νρσ

, M

ρσ

]



0νρσ

, M

ρσ

] P



0νρσ

(σ

ρσ

)

where we have used the fact that the supercharges commute with the momentum

operator. We will drop the matrix indices a and b in the next few steps to alleviate

the notation; we just need to keep in mind that on the right-hand side, Q is a column

vector on which the Pauli matrices are acting.

The only components of P

that do not vanish when applied to a massless state

are P

and P

. If we take ν = 0, we get zero because of the Levi-Civita symbol.

This leaves us with

[Q, W

] =



03ρσ

ρσ



03ij

=−



3ij

(7.25)

In Exercise 7.2 we showed that



ijk

= σ

which allows us to write

−



3ij

=−

Reinstating the matrix indices, our ﬁnal result is therefore is

, W

] =−

(σ

)

(7.26)

142

Supersymmetry Demystified

This is also the commutator of Q

with the operator W

because W

= W

with

our choice of metric. The only elements of σ

that are nonzero are

(σ

)

=−(σ

)

= 1 (7.27)

Using this in Eq. (7.26) we obtain

, W

] =−

(7.28)

and

, W

] =

(7.29)

Now let’s ﬁnd out what this is telling us. Recall that massless states satisfy

Eq. (7.21). In particular, the effect of applying W

is simply

|p, h=hE|p, h (7.30)

We’d like to ﬁnd out what is the effect of applying the supercharges Q

and Q

on |p, h. Consider ﬁrst

|p, h (7.31)

It’s clear that this state has four-momentum p

because the supercharges commute

with the momentum operator:

|p, h=Q

|p, h=p



|p, h



Now let’s determine the helicity of Eq. (7.31). Using a trick familiar from quantum

mechanics, we write

|p, h) = [W

, Q

]|p, h+Q

|p, h

E|p, h+Q

hE|p, h

= E



h +



|p, h

CHAPTER 7 Applications of the SUSY Algebra

143

The end result is

|p, h) = E



h +



|p, h

which, from Eq. (7.30), tells us that

|p, h=



p, h +



Therefore the action of the supercharge Q

is to raise the helicity by one-half. Note

that this implies that supercharges transform bosons into fermions and vice versa,

as expected.

From Eq. (7.29) it is clear that Q

has the opposite effect of Q

; i.e., the super-

charge Q

lowers the helicity by one-half:

|p, h=



p, h −



It is natural to expect that the hermitian conjugate Q

†

and Q

†

will have the

opposite effect of Q

and Q

. One indeed can show that this is the case using

the commutator (6.86). The calculation is almost identical to the one we followed

from Eq. (7.25) to Eq. (7.29) except that it involves supercharges with upper dotted

indices. Since Chapter 3 was not prerequisite material for this chapter, we will only

quote the ﬁnal result, leaving the derivation to Exercise 7.4, which you are invited

to do when you have become familiar with the van der Waerden notation. The end

result is



†

, W



†



†

, W



=−

†

(7.32)

By comparing with Eqs. (7.28) and (7.29), we see that, as expected, Q

†

and Q

†

lower and raise the helicity by one-half, respectively.

EXERCISE 7.4

Once you have become acquainted with the van der Waerden notation of Chapter 3,

prove Eq. (7.32) using Eq. (6.86).

144

Supersymmetry Demystified

7.3 The Massless SUSY Multiplets

Now that we know the action of the supercharges on one-particle states, we will

use the information to build irreducible representations of SUSY, also called SUSY

multiplets or supermultiplets for short. To ﬁnd these, we will work out the minimal

set of states connected by the application of the supercharges. These states will

represent the particle content necessary to build supersymmetric theories. Again,

we restrict ourselves to massless states.

Our starting point is the SUSY algebra, Eqs. (6.70), (6.71), and (6.72). Using

once again a frame in which

|p, h=P

|p, h=0

|p, h=P

|p, h=E|h (7.33)

we have

|p, h=



− P



|p, h=



02P



|p, h (7.34)

Recall the anticommutation relations (6.72):

, Q

†

}=(σ

)

(7.35)

Using this and Eq. (7.34), we conclude that

, Q

†

}|p, h={Q

, Q

†

}|p, h={Q

, Q

†

}|p, h=0 (7.36)

whereas

, Q

†

}|p, h=2P

|p, h (7.37)

The ﬁrst relation in Eq. (7.36) is particularly useful. Its expectation value in a state

|p, h is

p, h|{Q

, Q

†

}|p, h=p, h|Q

†

|p, h+p, h|Q

†

|p, h



|p, h



+||Q

†

|p, h||

= 0

CHAPTER 7 Applications of the SUSY Algebra

145

Since we are working with physical states whose norm is positive deﬁnite, this

implies that

|p, h=Q

†

|p, h=0

We therefore only need to concern ourselves with the supercharges Q

and Q

†

What general statements can we make about the states forming a supermultiplet?

They will all have the same four-momentum because the supercharges commute

with the momentum operator, but they will, of course, have different helicities. It

is clear that each supermultiplet has a state of minimum helicity because Q

is the

only operator at our disposal that lowers the helicity, and Q

= 0. Let us then start

with the state in a supermultiplet with the lowest helicity, which we will denote by

|p, h

min



By deﬁnition, acting with Q

on this state gives zero. On the other hand, acting

with Q

†

gives

†

|p, h

min

=



p, h

min



We cannot generate states of higher helicity because (Q

†

)

= 0. So this is it! A

massless supermultiplet contains only two states of the same momentum but with

helicities differing by half a unit (and therefore, one state is a boson, whereas the

second is a fermion).

To be precise, our result applies to the so-called N = 1 SUSY, which involves

only one Weyl spinor of charges Q (and its hermitian conjugate Q

†

). One can

write down a supersymmetric with any number N of spinor charges, and these are

often referred to as extended supersymmetries. We will brieﬂy discuss extended

supersymmetries at the end of this chapter.

As far as N = 1 SUSY is concerned, we therefore need only two states to build

a theory. But it would not be physical because it would not be CPT-invariant.

Recall that CPT changes the sign of the helicity, so the two members of a massless

supermultiplet cannot be CPT conjugate states. A physical supersymmetric theory

then must contain a supermultiplet plus its CPT conjugate, for a total of four states.

To be precise, here we are counting only physical, on-shell states.

In the next section we will discuss in more details the massless supermultiplets.

We will not discuss massive representations. The analysis is only slightly more

involved, but for our purposes, only the massless representations will matter. Of

course, except for the photon and the gluon, all particles are massive, but in the