Labelle P. Supersymmetry DeMYSTiFied

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146

Supersymmetry Demystified

minimal supersymmetric standard model, we build the theory using massless multi-

plets and then incorporate their masses through the spontaneous symmetry breaking

of the SU(2)

×U (1)

group, as is done in the standard model.

7.4 Massless SUSY Multiplets and the MSSM

In this section we discuss only the multiplets of N = 1 SUSY. Let us ﬁrst consider

a state with h

min

= 0, which corresponds to a scalar ﬁeld. Then SUSY requires the

presence of an h = 1/2 state as well, which is a right-handed Weyl spinor. Since for

a massless spinor handedness and chirality are equivalent, this is also a right-chiral

spinor. CPT invariance further requires the introduction of a second h = 0 state as

well as an h =−1/2 state. The two h = 0 states can be assembled into a complex

scalar ﬁeld, whereas the two fermion states correspond to a left-chiral Weyl spinor

and its antiparticle. Of course, it is arbitrary which state is called the particle and

which is the antiparticle, so we could just as well talk of a right-chiral particle

together with its left-chiral antiparticle. In any case, the end result is that we obtain

a scalar ﬁeld plus a Weyl spinor and their antiparticles.

This type of multiplet is referred to as a chiral multiplet because the spin 1/2

fermion appears in only one chiral state (with the antifermion having the opposite

chirality, of course). This is obviously the multiplet appearing in our model in

Chapter 5. Note that the states we discuss here are the on-shell, physical states. Off

shell, there are, of course, more degrees of freedom, but our discussion pertains

only to physical states.

In the minimal supersymmetric standard model (MSSM) all the known fermions

are taken to be members of chiral multiplets. Therefore, each fermion is paired up

with a spin 0 supersymmetric partner. The accepted notation is to add a preﬁx s

to the names of the known fermions to designate their spin 0 partners. So each

fermion is paired up with a sfermion. The quarks are paired up with squarks, and

leptons are paired up with sleptons. This terminology is even used with individual

particle names, the scalar partner of the electron being called the selectron, whereas

the supersymmetric partner of the top quark is referred to as the stop! This whim-

sical notation sometimes causes problems (The scharm, anyone? What about the

sstrange?).

The Higgs ﬁeld is also part of a chiral multiplet, being, of course, the scalar

component. Actually, the Higgs content of the MSSM is more complex than in

the standard model, as we will discuss in detail in Chapter 15, and requires the

introduction of two SU (2) doublets of complex Higgs ﬁelds, as opposed to the

single doublet of the standard model. In any case, all these scalar ﬁelds belong to

chiral multiplets and therefore have spinor superpartners. The convention for the

names of spin 1/2 supersymmeric partners of standard model bosons is to add the

CHAPTER 7 Applications of the SUSY Algebra

147

sufﬁx ino at the end of the name. The spin 1/2 partners of the Higgs ﬁelds therefore

are referred to as Higgsinos.

Next, let’s consider supermultiplets with h

min

= 1/2. SUSY requires a second

state with h = 1. CPT then requires states with h =−1 and h =−1/2. We can

think again of the two fermion states as a left-chiral Weyl spinor together with

its antiparticle. The states h =±1 obviously correspond to the two helicity states

of a massless vector boson. This type of multiplet is therefore referred as a vector

multiplet or sometimes as a gauge multiplet when the vector particle is a gauge ﬁeld.

All the gauge bosons of the standard model belong to vector multiplets. Following

the convention described earlier, their spin 1/2 superpartners are generically called

gauginos. More speciﬁcally, the superpartner of the photon, gluons, and weak-

interaction bosons are, respectively, called the photino, the gluinos, the winos, and

the zino. I am not making this up!

For h

min

= 1, the four states have h =±1, ±3/2. Again, we have the two helicity

states of a massless vector particle. As for the massless spin 3/2 state, it is sometimes

referred to as a Rarita-Schwinger ﬁeld.

For h

min

= 3/2, we get four states with helicities h =±3/2, ±2. The massless

spin 2 particle is, of course, the famous graviton, and its spin 3/2 partner is called the

gravitino. These are, of course, crucial elements of supergravity (SUGRA) models

and string theory.

7.5 Two More Important Results

We have seen how the algebra can be used to construct the massless irreducible

representations of N = 1 SUSY without ever writing a lagrangian! We will now

prove two additional important results using only the algebra.

First, we will prove that the number of boson states must be equal to the number

of fermion states in a supermultiplet. The proof will be valid for massive as well as

massless representations and for any value of N .

The ﬁrst step is to introduce the fermion number operator N

, which is deﬁned,

as its name suggests, to give the number of fermions present in a state when it

is applied to that state. Now consider the operator (−1)

. Applied to a bosonic

state, this gives +1, and applied to a fermionic state, it gives −1. If we sum the

expectation value of this operator over all the states belonging to a supermultiplet,

we will get for an answer the number of bosons minus the number of fermions

appearing in this multiplet. Let’s carry out this calculation.

First, since the supercharges change a fermion into a boson, or vice versa, we

have the obvious relation

(−1)

=−Q

(−1)

(7.38)

148

Supersymmetry Demystified

The same relation holds if we replace Q

by Q

†

. This implies that the product of

two supercharges commutes with (−1)

The trick now is to sum the expectation value of

(−1)

, Q

†

}

over all the states belonging to a supermultiplet. In other words, we will take

the trace of this operator over the members of a supermultiplet. A few simple

manipulations then give



(−1)

, Q

†

}



= Tr



(−1)

†

+ (−1)

†



= Tr



(−1)

†

+ Q

†

(−1)



= Tr



−Q

(−1)

†

+ Q

†

(−1)



where we have used Eq. (7.38) in the last line. Now we use the invariance of the

trace under cyclic permutation of the arguments, Tr(ABC) = Tr(CAB), to obtain



−Q

(−1)

†

+ Q

†

(−1)



= Tr



−Q

†

(−1)

+ Q

†

(−1)



= 0

On the other hand, using the anticommutator (6.72), we also have that



(−1)

, Q

†

}



= Tr



(−1)



Since we have proved that the left-hand side vanishes identically, independent of

the four-momentum, we must have

Tr(−1)

= 0

in a SUSY supermultiplet, which implies that the number of fermions must be

equal to the number of bosons. This is a general result and a well-known property

of supersymmetric theories.

As a second proof, we will work out a condition on the expectation value of the

hamiltonian in supersymmetric theories. This leads to a result that is a hallmark of

SUSY, a result that will play a signiﬁcant role in our discussion of SUSY breaking

in Chapter 14.

The hamiltonian is the operator P

. To isolate it in Eq. (6.72), we may use a

simple trick. The trick becomes more obvious if we expand out the sum over Lorentz

CHAPTER 7 Applications of the SUSY Algebra

149

indices in Eq. (6.72):

, Q

†

}=(σ

)

H − (σ

)

where σ

is simply the 2 ×2 identity matrix. What sets apart σ

from the Pauli

matrices is that its trace is not zero. We therefore can isolate the hamiltonian by

taking the trace of both sides, which means summing over a and b with the condition

a = b. This leads to

, Q

†

}+ {Q

, Q

†

}=2H

Consider now calculating the expectation value of the two sides in an arbitrary state

|ψ:

ψ|Q

†

|ψ+ψ |Q

†

|ψ+ψ |Q

†

|ψ+ψ |Q

†

|ψ=2ψ|H |ψ 

(7.39)

Each term on the left-hand side is semipositive deﬁnite because it represents the

norm squared of a state. For example,

ψ|Q

†

|ψ=



|ψ



Since the left-hand side of Eq. (7.39) is semipositive deﬁnite, we have

ψ|H |ψ≥0

If we consider the special case of the state being the vacuum, then Eq. (7.39) gives

the vacuum expectation value (vev) of the hamiltonian:

0|Q

†

|ψ+0|Q

†

|0+0|Q

†

|0+0|Q

†

|0=20|H|0 (7.40)

and we ﬁnd again that

0|H|0≥0 (7.41)

with the inequality saturated if the supercharges annihilate the vacuum. If they do

not annihilate the vacuum, then we have a spontaneous breaking of SUSY, and we

get a strict inequality, i.e.,

0|H|0 > 0 (7.42)

150

Supersymmetry Demystified

Therefore, a necessary and sufﬁcient condition for SUSY to be spontaneously

broken is that the vacuum expectation value of the hamiltonian is nonzero (and

necessarily positive). We will come back to this issue in Chapter 14.

7.6 The Algebra in Majorana Form

Many references give the SUSY algebra in Majorana form instead of the Weyl

form we just worked out. We will now show how to translate all the commutation

relations involving the supercharges into the Majorana language.

The ﬁrst step is to group the supercharges into a Majorana spinor. Recall from

Eq. (4.20) that a Majorana spinor can be built out of a left-chiral Weyl spinor χ and

the hermitian conjugate χ

†

via





iσ

†T



We therefore can assemble the supercharges into a Majorana supercharge

≡



iσ

†T



which, written explicitly, is

⎛

⎜

⎝

†

−Q

†

⎞

⎟

⎠

(7.43)

Let us deﬁne the components Q

, Q

,... through

≡

⎛

⎜

⎝

⎞

⎟

⎠

Then all the anticommutators among the supercharges can be shown to be repro-

duced by the single expression

}=(γ

)

(7.44)

CHAPTER 7 Applications of the SUSY Algebra

151

Here, the bar over Q

represents the usual Dirac adjoint, so

stands for

†

)

If we deﬁne the matrices γ

μν

≡

(γ

− γ

)

then the commutation relation of the supercharges with the Lorentz generators are

, M

μν

] = (γ

μν

)

It is not difﬁcult to show that this expression reproduces both Eqs. (6.83) and (6.86)

using the fact that γ

μν

may be written as

μν



¯σ

μν

0 σ

μν



(7.45)

We will now obtain the SUSY charges from the supersymmetric currents, fol-

lowing Eq. (6.24). But ﬁrst let’s review the general approach.

EXERCISE 7.5

Verify that Eq. (7.44) reproduces the anticommutators (6.70), (6.71), and (6.72).

7.7 Obtaining the Charges from

Symmetry Currents

To see explicitly how to construct the currents, consider the theory of a single

complex scalar ﬁeld. If the ﬁeld is varied by an amount δφ, the lagrangian varies

δL(φ, φ

†

) =

∂L

∂φ

δφ +

∂L

∂(∂

φ)

δ(∂

φ) +

∂L

∂φ

†

δφ

†

∂L

∂(∂

†

)

δ(∂

†

)

This is simply an application of the chain rule from ordinary multivariable calculus

(a similar trick is used to determine the equations of motion of the ﬁelds).

If this transformation is a symmetry of the theory, it leaves the action invariant,

which, as we have discussed previously, implies either that the lagrangian density

152

Supersymmetry Demystified

is invariant or that it changes by a total derivative. To be general, let’s assume that it

changes by a total derivative, which we will write as ∂

, with K

some function

of the ﬁelds. We then have

∂L

∂φ

δφ +

∂L

∂(∂

φ)

∂

(δφ) +

∂L

∂φ

†

δφ

†

∂L

∂(∂

†

)

∂

(δφ

†

) = ∂

(7.46)

where we have used δ(∂

φ) = ∂

(δφ) and δ(∂

†

) = ∂

(δφ

†

). Now, assuming that

φ and φ

†

obey the equations of motion, we have

∂L

∂φ

= ∂



∂L

∂(∂

φ)



(7.47)

and

∂L

∂φ

†

= ∂



∂L

∂(∂

†

)



(7.48)

Substituting Eqs. (7.47) and (7.48) into Eq. (7.46), we get

∂



∂L

∂(∂

φ)

δφ +

∂L

∂(∂

†

)

δφ

†



= ∂

(7.49)

The quantity in brackets is called the Noether current j

, i.e.,

≡

∂L

∂(∂

φ)

δφ +

∂L

∂(∂

†

)

δφ

†

so Eq. (7.49) may be written as

∂

= ∂

Therefore, the quantity

≡ ( j

− K

) (7.50)

is conserved, in the sense that it satisﬁes

∂

= 0

CHAPTER 7 Applications of the SUSY Algebra

153

Of course, in a theory where there are several ﬁelds, such as the supersymmetric

theory we built in Chapters 5 and 6, we have to add terms such as Eqs. (7.47) and

(7.48) for each ﬁeld. We will illustrate this later in this chapter.

Now we get to the key point. The claim is that the charges generating the trans-

formations are given by the spatial integral of the zeroth component of the currents:

Q =



x J

(7.51)

We won’t prove that the charges are indeed given in general by Eq. (7.51) (see any

textbook on quantum ﬁeld theory for a derivation).

As an aside, the Q are often described as conserved charges because they are

time-independent.

As already emphasized, it is important to keep in mind that the currents, and hence

the charges constructed via the currents, are functions of the quantum ﬁelds appear-

ing in the theory, so they are themselves quantum ﬁeld operators. Once the charges

are found using Eq. (7.51), their commutation (or anticommutation) relations can

be worked out explicitly using the equal time commutation/anticommutation of the

ﬁelds.

Let’s now apply this to SUSY!

7.8 Explicit Supercharges as

Quantum Field Operators

Now let us build the SUSY current and ﬁnd explicit expressions for the supercharges

in terms of the ﬁelds of the theory.

In the case of SUSY, Eq. (7.51) is replaced by

Q ·ζ +

Q ·

ζ =



SUSY

We work with the free supersymmetric lagrangian, which we will repeat here:

L = ∂

φ∂

†

+ χ

†

i ¯σ

∂

χ +FF

†

154

Supersymmetry Demystified

First, let’s ﬁnd the SUSY Noether currents. In order to do this, we need

∂L

∂(∂

φ)

δφ = (∂

†

)δφ (7.52)

= (∂

†

)χ ·ζ (7.53)

∂L

∂(∂

χ)

δχ = χ

†

i ¯σ

δχ (7.54)

= χ

†

¯σ

(∂

φ)σ

iσ

∗

(7.55)

∂L

∂(∂

†

)

δφ

†

= (∂

φ) ¯χ ·

On the other hand,

∂L

∂(∂

∂L

∂(∂

†

)

∂L

∂(∂

†

)

= 0

The Noether current is therefore

= (∂

†

)χ ·ζ + (∂

φ)χ

†

¯σ

iσ

∗

+ (∂

φ) ¯χ ·

To ﬁnd ∂

, we simply have to add up the total derivatives we dropped when

we calculated the supersymmetric transformation of our lagrangian. These total

derivatives can be found by comparing Eqs. (5.14) and (5.15) to the equations

preceding them, giving us (recall that we showed that the constant C appearing in

those equations was equal to −i)

∂

= ∂



¯χ ·

ζ∂

φ −i(∂

†

)ζ

iσ

i ¯σ

χ +(∂

†

)ζ · χ



(7.56)

where we have used χ

†

iσ

∗

= ¯χ ·

ζ and ζ

iσ

χ =−ζ · χ .SoK

is simply the

expression inside the parenthesis. Using the fact that the order in spinor dot products

does not matter, the supersymmetric current is simply

SUSY

= j

− K

= (∂

φ)χ

†

¯σ

iσ

∗

− (∂

†

)ζ

iσ

¯σ

χ (7.57)

CHAPTER 7 Applications of the SUSY Algebra

155

We can then use this to ﬁnd the charges:

ζ · Q +

Q ·

ζ = ζ

(−iσ

)Q + Q

†

iσ

∗





(∂

φ)χ

†

¯σ

iσ

∗

− (∂

†

)ζ

iσ

¯σ



On the left-hand side we chose to write

ζ ·

Q as

Q ·

ζ . This was simply in order to

get ζ

∗

to appear on the right of the σ

matrix, as in the current.

The matrix ¯σ

is simply the identity matrix. Since the inﬁnitesimal parameters

ζ and ζ

∗

may be treated as independent, we can read off the charges Q and Q

†

.We

ﬁnd

Q =



x(∂

†

(x))σ

χ(x) (7.58)

and

†



xχ

†

(x)∂

φ(x)σ

(7.59)

which is simply the hermitian conjugate of Q.

This is what we were looking for: explicit expressions for the supercharges

as quantum ﬁeld operators. It is then easy to verify explicitly that these charges

indeed generate the SUSY transformations of the ﬁelds given in Eq. (6.52). This

will be left as an exercise. All we will need are the equal time commutators and

anticommutators familiar from quantum ﬁeld theory:



φ(x, t),

†

(y, t)



= i δ

(x −y)

{χ

(x, t), χ

†

(y, t)}=δ

(x −y) (7.60)

where a dot indicates a derivative with respect to time. All other commutators and

anticommutators are equal to zero. For example, [φ,

φ] = 0.

The commutators of the supercharges with the Poincar´e generators also can be

veriﬁed, but this requires that we ﬁrst work out P

and M

μν

as quantum ﬁeld opera-

tors. We will not pursue this here because, although straightforward, the calculation

is a bit lengthy and is in any case similar to usual quantum ﬁeld theory calculations.

The interested reader will ﬁnd the detailed calculation in Ref. 31.

EXERCISE 7.6

Obtain the commutators (6.52) using the quantum ﬁeld operator representations of

the supercharges (7.58) and (7.59) and the equal time commutators (7.60).