Labelle P. Supersymmetry DeMYSTiFied

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336

Supersymmetry Demystified

We now write all three ﬁelds in terms of their real and imaginary parts:

≡

√

+iI

)

which leads to

V =

− 2g

+ 2g



+ I



+ g



√

+ gR

− gI

√

2 g φ

 R





√

+ gR

+ gI

√

2 g φ

 I



From this, one can ﬁnd the masses of the scalar ﬁelds using the approach de-

scribed in the preceding section. If we choose the expectation value φ

 equal to

zero, the result is particularly simple. The masses of all the ﬁelds are (see Exer-

cise 14.1)



− g



+ g

= m

= 0 m

= m

= m (14.13)

The last two results can be summarized by saying that the complex scalar ﬁelds φ

and φ

have masses

= 0 and m

= m

whereas φ

sees its real and imaginary parts acquiring different masses through

SSB.

EXERCISE 14.1

Find the masses of the scalar ﬁelds for φ

 = 0, and then show that by setting

φ

=0, you recover the result in Eq. (14.13).

To ﬁnd the mass of the fermions, we go back to the general expression for the

lagrangian of left-chiral superﬁelds, Eq. (8.62), which shows that the fermion mass

CHAPTER 14 SUSY Breaking

337

terms arise necessarily from the terms in

fermion

=−



i =j

∂

∂φ

· χ

+ h.c. (14.14)

that do not contain any scalar ﬁelds. In other words, the fermion mass terms all

come from the terms in the superpotential, which are either quadratic or bilinear in

the scalar ﬁelds. To use this equation, we must express the superpotential in terms

of the shifted ﬁelds, which, in our case, just means that we replace φ

by φ

+φ



in Eq. (14.6):

W = m φ

+ g φ



− M



+ g φ





− M



The lagrangian (14.14) then is given by

fermion

=−

m χ

· χ

− g φ

 χ

· χ

+ h.c. +···

where we wrote only the terms that will contribute to the mass matrix of the

fermions. We immediately see that the Weyl spinor χ

is massless. The mass matrix

in the basis χ

, χ

M =



0 m

m 2 gφ





(14.15)

so that the squared mass matrix is

= M

†

M =



2 gm φ



2 gm φ

 m

+ 4 g

φ





whose eigenvalues are

= m

+ 2 g

φ



± 2 g φ





φ



+ m

= (



φ



+ m

± gφ

)

(14.16)

so the masses of the two fermions are



φ



+ m

± gφ

 (14.17)

338

Supersymmetry Demystified

With φ

=0, we see that both fermions have a mass equal to m. It is easy to ﬁnd

the spinors that are mass eigenstates when φ

=0 because then the mass matrix

in Eq. (14.15) is simply

M =



0 m

m 0



whose two eigenstates are clearly

χ =

± χ

√

(14.18)

Now, if we compare what we found for the boson masses, Eq. (14.13), with the

fermion masses we just calculated, we see that the degeneracy between one of the

complex scalar ﬁelds and one of the fermions is lifted.

To summarize the situation when φ

=0, there is one complex scalar ﬁeld

and one fermion that are both massless (φ

and χ

), and there is a second pair

of boson and fermion with a mass m [φ

and either combination appearing in

Eq. (14.18)], but the third boson/fermion pair is not degenerate in mass because

the remaining fermion has a mass m, whereas we have two real scalar ﬁelds with

masses



± g

. Therefore, SUSY is spontaneously broken: the lagrangian

is supersymmetric, but the vacuum breaks the symmetry.

Spontaneous symmetry breaking of SUSY or, to be more precise, of global SUSY

(which is what we consider in this book) always introduces fermion massless modes

as opposed to the usual Goldstone bosons. These massless fermion modes therefore

are referred to as goldstinos. In the O’Raifeartaigh model, there is one goldstino,

the spinor χ

. Note that it is the fermion that belongs to the same multiplet as the

auxiliary ﬁeld whose vev is nonzero, i.e., F

. This is a general feature of F-type

SSB.

EXERCISE 14.2

Find the mass spectrum of the bosons and fermions both in the limit g → 0, M = 0

and in the limit M → 0, g = 0. You will observe that the mass degeneracy between

the fermionic and bosonic degrees of freedom reappears. Note that this holds no

matter what φ

 is equal to.

There is unfortunately an aspect of F-type SUSY breaking that makes it unattrac-

tive from a phenomenological point of view. Before addressing this issue, we need

to introduce yet another “superquantity,” the supertrace.

CHAPTER 14 SUSY Breaking

339

14.6 The Supertrace

The supertrace is deﬁned as

STr(M

) ≡



particles

(−1)

(2s + 1)m

particle

where it is understood that the sum is performed over all the physical particles

appearing in the theory (not over the auxiliary ﬁelds), and s denotes the spin of the

particle. It is also understood that there is one term for each real scalar ﬁeld and

one term for each Weyl spinor.

In the case of theories with left-chiral superﬁelds only, there are only scalar ﬁelds

and spinor ﬁelds, so the sum reduces to two terms:

STr(M

) =



scalars

− 2



spinors

(14.19)

This sum obviously vanishes in supersymmetric theories because for each Weyl

spinor, there are two real scalar ﬁelds (one complex scalar ﬁeld) with the same mass.

What is astonishing is that the supertrace still vanishes when SUSY is spontaneously

broken with F-type breaking. We can check that this is indeed the case with the

masses obtained in last section. When we set φ

=0, we obtained the following

squared masses for the bosons:

Squared masses of the bosons = 0, 0, m

, m

− g

, m

+ g

so that



scalars

= 4 m

By the way, the reason it is called a supertrace is that the sum of the squared masses

is simply given by the trace of the squared-mass matrix M

Now, let’s look at the fermions. We have three fermions with masses 0, m, and

m,so



spinors

= 4 m

Again, we could have simply taken the trace of the squared fermion mass matrix

and multiplied it by 2.

340

Supersymmetry Demystified

The end result is that the supertrace (14.19) vanishes indeed. We obtain the same

result if we keep φ

 arbitrary (see Exercise 14.3).

EXERCISE 14.3

Verify that the supertrace still vanishes if we don’t set φ

=0.

The key point to note here is that the reason the supertrace still vanishes even

if we have SSB is that one of the scalar squared masses increased by an amount

relative to its “unbroken” value of m

, whereas another scalar squared mass

decreased by the same amount.

This is an important result and turns out to be a general feature of F-type SUSY

breaking: The heaviest boson state is heavier than all the fermions, and as a con-

sequence of the vanishing of the supertrace, the lightest boson state also must be

lighter than all fermion states. This makes F-type SUSY breaking phenomenologi-

cally unappealing because it implies that each known fermion would have a lighter

scalar partner. For example, this would imply the existence of a charged scalar

particle lighter than the electron, which is ruled out experimentally.

So let us turn our attention to D-type SUSY breaking.

14.7 D-Type SUSY Breaking

For this type of symmetry breaking to occur, a Fayet-Illiopoulos term is necessary.

This means that at least one abelian gauge ﬁeld must be present. We will ﬁrst illus-

trate a simple situation where D=0 cannot hold irrespective of the expectation

values of the auxiliary ﬁelds F

, so we are free to ignore their contribution to the

potential.

The simplest model in which this type of spontaneous symmetry breaking may

occur is a U (1) gauge theory with a single charged left-chiral superﬁeld. If there is

only one charged chiral superﬁeld, there is no superpotential because it is impossible

to write down a function of  alone that is gauge-invariant. As we saw in Chapter 13,

the lagrangian then is (with ξ taken to be real)



†

−2qV





+ FF



+ ξ D



(14.20)

From Eqs. (14.2) and (14.3), we have

D = q



†

φ −



(14.21)

CHAPTER 14 SUSY Breaking

341

and the potential may be written as

V =



|φ|

−



(14.22)

The reason for factoring out the charge is that it makes clear that the effect of the

Fayet-Iliopoulos term depends crucially on whether ξ/q is positive or negative.

Case ξ/q < 0

In this case, it is clear from Eq. (14.21) that there is no ﬁeld φ for which D=0

(because φ

†

φ is semipositive deﬁnite), and thus SUSY is spontaneously broken.

From Eq. (14.22), we ﬁnd that the potential is minimum when φ = 0, in which case

min

Let us now have a look at the mass spectrum to conﬁrm explicitly that SUSY

is spontaneously broken. The ﬁelds of the gauge potential superﬁeld A

and the

photino λ are massless. Since there is no superpotential, there is no mass term for

the left-chiral spinor component of the superﬁeld  either. If we write the complex

scalar ﬁeld in terms of two real scalar ﬁelds as usual, i.e.,

φ =

√

(R +iI)

and retain only the terms that will contribute to the squared-mass matrix, the po-

tential (14.22) is equal to

V (φ) =

|qξ |

+ I

) +···

We see that both real scalar ﬁelds acquire a mass equal to

√

|qξ |. Thus SUSY is

indeed spontaneously broken: The spinor ﬁeld χ is massless, but its scalar partners

are massive.

Notice how different the situation is from F-type breaking: here, all the bosons

have seen their mass increase relative to the “unbroken” value of m = 0. This

is more encouraging from a phenomenological point of view because this would

provide an explanation for why the SUSY partners of the known fermions would

have escaped detection so far. However, the present theory is just a toy model, and

342

Supersymmetry Demystified

the situation turns out to be not so encouraging when we try to apply this approach

to the MSSM, as we will see in Chapter 16.

Let’s ﬁnd out what happens to the supertrace. Obviously, it does not vanish

anymore:

STr(M

) =



scalars

− 2



spinors

= 2 |qξ | (14.23)

There is, however, a generalization of the rule STr(M

) = 0 that is valid for D-type

SSB. The proof is not difﬁcult, but we will simply quote the result for abelian gauge

theories with only one auxiliary ﬁeld acquiring a vev (see Section 7.7 of Ref. 5 for

a derivation):

STr(M

) = 2 D



(14.24)

where the sum is over all the charges of the left-chiral superﬁelds present in the

theory. In the case of F-type breaking, there is no D ﬁeld with a vev, so we recover

the result that the supertrace then is zero. In our example, D=|ξ |, and there is

only one left-chiral superﬁeld, of charge q, so the formula does indeed give

STr(M

) = 2 |qξ |

as we found in Eq. (14.23).

Note that even though SUSY is spontaneously broken, the gauge symmetry

remains unbroken because at the minimum of the potential, φ = 0. We therefore

have a potential that corresponds to Figure 14.2.

Case ξ/q > 0

In this case, the potential is minimized for

|φ|=

(

and the potential at the minimum vanishes, i.e.,

min

= 0

CHAPTER 14 SUSY Breaking

343

SUSY is therefore unbroken, but since the minimum of the potential occurs for a

nonzero expectation value of the ﬁeld, the gauge symmetry is broken spontaneously

(we are in the situation depicted in Figure 14.3). This is clear because the kinetic

energy of the scalar ﬁeld is D

φ D

†

, which contains the term q

†

φ.

Therefore, at the minimum of the potential, the gauge ﬁeld acquires a mass equal

to 2 q ξ . A more detailed analysis reveals that the imaginary part of the scalar ﬁeld

has been eaten by the gauge ﬁeld, whereas the real part acquires a mass equal to

the mass of the gauge boson.

14.8 Second Example of D-Type SUSY Breaking

Now we will consider an example where D=0 and F=0 cannot hold simul-

taneously. The simplest model in which this occurs is supersymmetric QED [see

Eq. (13.55)] with a Fayet-Illiopoulos term. The superpotential is W = m E

Using Eq. (14.3), we ﬁnd

D = e |

− e |

− ξ (14.25)

It is clear that we can ﬁnd (nonzero) values of the scalar ﬁelds to make this vanish,

no matter what the sign of ξ is.

Now consider the expressions for the F

†

and F

†

=−

∂W

∂

=−m

(14.26)

These two can be set equal to zero if we simply take

= 0, but then Dis not

equal to zero! Conversely, if we choose values of the scalar ﬁelds such that D=0,

the expectation values of the F

†

are not zero. Therefore, SUSY is spontaneously

broken.

The potential is given by

V = m







e |

− e |

− ξ



= (m

− eξ)|

+ (m

+ eξ)|



e |

− e |



(14.27)

We will take ξ to be positive. The minimization of the potential depends on the

sign of the quantity m

− eξ . We will only consider m

− eξ>0, in which case,

344

Supersymmetry Demystified

all the terms in Eq. (14.27) are positive, and the potential is obviously minimized

for

= 0, so the gauge symmetry is unbroken. The value of the potential at

its minimum is V

min

The masses of the scalar can be read directly from Eq. (14.27) because there are no

bilinear terms in the two scalar ﬁelds. The squared masses then are simply m

± eξ .

It is also easy to check that the electron mass is unaffected by the spontaneous

breakdown of SUSY and therefore remains equal to m.

We can now check that Eq. (14.24) is indeed satisﬁed. The left-hand side is

equal to

STr(M

) = m

+ eξ + m

− eξ − 2m

= 0

This indeed agrees with the right-hand side because the sum of the charges is equal

to zero!

Now comes the kicker. The MSSM contains the hypercharge abelian gauge

symmetry, so one could have hoped to use a Fayet-Illiopoulos term to generate

spontaneous SUSY breaking, but, as is the case in the standard model, the sum of

the hypercharges of all the particles is zero (which is required to ensure cancellation

of anomalies).

Therefore, in the context of the MSSM, D-type SUSY breaking suffers from the

same problem that we had with F-type breaking: since the supertrace is zero, light

sleptons and squarks are predicted, in conﬂict with experiments.

We therefore must look somewhere else for an explanation of SUSY breaking.

14.9 Explicit SUSY Breaking

Since there is no (yet) known phenomenologically viable mechanism to sponta-

neously break SUSY in the MSSM, the only option left is to break it explicitly,

i.e., by adding by hand terms that violate the symmetry. Of course, if we do this

haphazardly, we will lose the feature that got us excited in SUSY in the ﬁrst place:

the cancellation of quadratic divergences. We have to be more subtle. We must look

for SUSY-breaking terms that will not spoil this cancellation. Such terms are said

to break SUSY softly. To see how this can be done (if it is possible at all), it is

instructive to get back to the Wess-Zumino model and to have a closer look at the

cancellation of divergences we worked out in Chapter 9.

The Wess-Zumino model was given, in Majorana form, in Eqs. (8.69) through

(8.71). There is a lot of symmetry in this lagrangian: All the particles have the same

mass, and that mass also appears as a coupling constant in the cubic interactions.

In addition, the same coupling constant g appears in all interactions. What we want

to understand is which of these features are absolutely necessary for the quadratic

CHAPTER 14 SUSY Breaking

345

divergences to cancel. To do that, let us rewrite the lagrangian with distinct masses

and coupling constants in all the terms. We then write the free lagrangian as

∂

A∂

A −

∂

B∂

B −



(i γ

∂

− m

)

while we represent the interactions by

I 1

=−

− g

−

I 2

=−m

− m

I 3

=−g

A 



I 4

=−ig

B



Note that we have chosen to set the parameter m that appears in L

I 2

equal to the

mass of the fermion, which we can obviously do in all generality because we have

introduced the arbitrary coupling constants g

and g

Unbroken SUSY corresponds to

= m

,...,g

≡ g

Of course, what really matters is the relationship between the coupling constants,

not their absolute values. So we shall ﬁx one of them to be equal to g and then

parametrize the breaking of SUSY by how much we can allow the other coupling

constants to deviate from their supersymmetric value, which is also equal to g.

It is a simple matter to go back to the calculation of the B propagator of Chapter 9

and keep track of the masses and coupling constants appearing in each diagram. In

that chapter, all quadratically divergent integrals were proportional to the integral

. Since we are now keeping track of the masses, let us have a closer look at this

integral. It is given by



(2π)

−ip·(x−y)

− m

+i 

− m

+i 



(2π)

− m

−i 

where m may either be m

, m

,orm

. However, the quadratic divergence arising

from I

is independent of the mass m. The logarithmically divergent contribution and

the ﬁnite piece do depend on m, but since we are concerned only with maintaining

the cancellation of the quadratic divergences, we may safely set to zero the mass

m in I