Supersymmetry. Theory, Experiment, and Cosmology

Exercises 19

(b)Consider a Weyl neutrino νL of left-handed chirality and its charge conjugate νRc of opposite chirality. The following mass term

Lm = − 12 mM ν¯L νRc − 12 mM ν¯Rc νL ,

is called a Majorana mass term. Write it in terms of the single field νM ≡ νL + νRc , which is called a Majorana neutrino. Can such a term arise in the Standard Model from gauge symmetry breaking (by a Higgs doublet)?

(c)One introduces a neutral lepton described by a Weyl spinor NR of right-handed chirality. What are its quantum numbers under SU (2) × U (1)? Introduce the Dirac neutrino ν ≡ νL + NR . Rewrite the Dirac mass term

Lm = − 12 mD νν¯

in terms of νL and NR . Can it be obtained by gauge symmetry breaking (with a Higgs doublet) in the context of the Standard Model?

Hints:

(a) CγT C−1

= γ5, γ5

= γT = γ†

, hence

T

= 1−γ5

T

and C

T = 1−γ5 C

T .

Ψ

Ψ

Ψ

R

Ψ

5

5 5

R

2

2

(b)Lm = − 12 mM ν¯M νM , mM φ if φ is an isovector (t3 = 1/2 for νL as well as ν¯Rc ).

(c)NR is a gauge singlet; Lm = − 12 mD ν¯L NR + h.c.; mD φ if φ is an isodoublet (as in the Standard Model).

Exercise 2 Show that there is a cancellation between the tadpole diagrams (A), (B) and (Ψ) in the last section.

Hints:

	mλ	4
			d k		1
(A)	3√

2(2π)4 k2 − m2

(B)	mλ		d	4k	1
	√

2(2π)4 k2 − m2

		mλ					d4k	1
	(Ψ) (−)	4√								.
							(2π)4		k2 − m2
			2

2

The singular rˆole of supersymmetry

This chapter presents basic notions about supersymmetry and supersymmetry breaking. In particular, it underlines the way supersymmetry is a unique symmetry among spacetime symmetries.

2.1Why supersymmetry?

Supersymmetry is a symmetry between boson and fermion fields. Why should we try to search for such a symmetry? One very fundamental reason is the following.

Bosons are the mediators of interaction: their statistics allows for a coherent superposition and thus for a macroscopic force, such as the Coulomb force. On the other hand, fermions are the constituents of matter: their statistics is translated at the macroscopic level into the additive character of matter. It is certainly a very fundamental question to ask whether there exists a symmetry which unifies matter and radiation.

The educated reader already knows that the present models of supersymmetry applied to particle physics do not fulfill this goal. Should this be taken as an indication that they are not yet complete? But let us put this aside and try to identify the general properties of a symmetry between bosons and fermions.

First, the generators of such a symmetry must carry a spinorial index, since they correspond to the transformation of an integer spin field into a spinor field. They are thus not commuting with Lorentz transformations. This is much in contrast with the behavior of generators of any internal symmetry, such as SU (3). In this sense, supersymmetry is necessarily a spacetime symmetry.

The first attempts were faced with a no-go theorem due to Coleman and Mandula [80] which we could rephrase in a non-technical way as:

The only conserved charges which transform as tensors under the Lorentz group are:

•Pµ, the generators of translations;

•Mµν , the generators of Lorentz transformations.

We recall (see Section A.1 of Appendix Appendix A) that the charges obtained in the Noether procedure may be interpreted as the generators of the transformations they are associated with. The Coleman–Mandula theorem implies that the charges associated with symmetries besides those of Poincar´e group are necessarily Lorentz scalars.

Why supersymmetry? 21

Let us illustrate the physical limitations behind this no-go theorem on an example due to E. Witten [374].

We consider two free scalar fields of vanishing mass φ1 and φ2. They are described by the Lagrangian

L = 21 ∂µφ1∂µφ1 + 21 ∂µφ2∂µφ2		(2.1)
and the corresponding equation of motion
φ1 = 0,	φ2 = 0,	(2.2)
implies the conservation ∂µJµ = 0 of the current
Jµ = φ1∂µφ2 − φ2∂µφ1.		(2.3)
But since one also has ∂αφ1,2 = 0 = ∂α∂β φ1,2, the currents
Jµρ = ∂ρφ1∂µφ2 − φ2∂µ∂ρφ1
Jµρσ = ∂ρ∂σ φ1∂µφ2 − φ2∂µ∂ρ∂σ φ1		(2.4)
are also conserved:
∂µJρ = 0,	∂µJρσ = 0.	(2.5)
µ	µ

Thus, following the Noether procedure, the charges

Qρ = d3x J	ρ	,	Qρσ = d3x J	ρσ	(2.6)
	0			0

are conserved.

This does not mean much because φ1 and φ2 are free fields. To the question “can one write an interaction term between φ1 and φ2 such that Qρ and Qρσ remain conserved?”, Coleman and Mandula answer “no”. Indeed, following Witten, let us suppose that it is possible. Then the value of the traceless part of Qρσ in a one (massless) particle state |p of momentum p is

p|Qρσ|p = A pρpσ − 41 p2gρσ

(2.7)

where we have imposed the tracelessness condition gρσ Q

ρσ

= 0.

p

p

If we consider an interaction between two such

massless particles on-shell:

|

1

2 →

ρ

p

σ

+ p

ρ

p

σ

the tensorial charge in the initial state is A(p

) and in the final

|p3 p4 , ρ σ

ρ σ

1

1

2

2

state A(p3p3 + p4p4 ). Thus, if the charge is conserved in interactions, one obtains the

condition:

p1ρp1σ + p2ρp2σ = p3ρp3σ + p4ρp4σ .

(2.8)

This yields, in2

the center of mass of the reaction (p

=

p

, p

=

−

p

, E

= E

=

)

2

= (p1

· p3)

2

or

1

−

2

3

4

1

2

E3 = E4), (p1

E14(1 − cos2 θ) = 0.

(2.9)

22 The singular rˆole of supersymmetry

The scattering angle θ takes specific values: θ = 0 or π. This is only compatible with the analyticity of the scattering amplitudes if the particles did not interact in the first place (θ = 0).

This example, which may be generalized to particles with spin [239], illustrates a very general fact. In a two-particle collision, conservation of Pµ and Mµν leaves a scattering angle undetermined. If there existed another tensorial charge, scattering would only be possible for discrete values of the angle, in contradiction with observation, as well as the principles of quantum field theory (themselves based on observation!).

The obstruction seems rather general. But the hypotheses of Coleman and Mandula were too restrictive. More specifically, they considered the following algebra

[Pµ, Pν ] = 0

[Pµ, Mρσ ] = i (ηµρ Pσ − ηµσ Pρ)
[Mµν , Mρσ ] = i (ηνρ Mµσ − ηνσ Mµρ − ηµρ Mνσ + ηµσ Mνρ)	(2.10)
[T a, T b] = i Cabc T c

where the last line corresponds to a Lie algebra associated with the internal symmetry. They conclude that the generators Ta are Lorentz scalars:

[T a, Pµ] = [T a, Mµν ] = 0.

(2.11)

But we are looking here for a symmetry whose charges/generators are operators acting on the Hilbert space which replace a fermion field by a boson field. They can be

	written as			d3k
	Q =				d3k fij (k, k )aB† j (k )aFi (k)	(2.12)
		ij

where aFi (k) annihilates a fermion of type i and momentum k and a†Bj (k ) creates a boson of type j and momentum k . Due to the (anti)commutation rules of bosons and fermions, Q must have anticommutation relations1.

This is what allows us to evade the conclusions of Coleman and Mandula. In order to see that, let us consider a second example with a complex scalar field φ and a fermion of left chirality ΨL . The free field Lagrangian is

L = ∂µφ ∂µφ + Ψ¯ L iγµ∂µΨL .		(2.13)
One easily checks, using the equations of motion
φ = 0,	γµ∂µΨL = 0,	(2.14)
that the current
Jrµ = ∂ρφ (γργµΨL )r		(2.15)

1In mathematical terms, we must consider graded Lie algebras instead of the usual Lie algebras whose definitions only involve commutators.

The supersymmetry algebra 23

is conserved. Indeed, using the Dirac equation,

		∂µJrµ = (∂µ∂ρφ ) (γργµΨL )r			r = φ ΨLr = 0.
		= (∂µ∂ρφ ) 21 {γρ, γµ}ΨL						(2.16)
Thus Qr =		d3xJr0 is conserved.
The	surprise is that it is now possible to write an interaction term which leaves
intact the conservation of Qr. Indeed, consider
		Lint = −λ2|φ|4 − λ φ		R ΨL + φ			L ΨRc .	(2.17)
			Ψc			Ψ
Then Jrµ receives an extra contribution
		Jrµ = ∂ρφ (γργµΨL )r + iλφ 2 γµΨRc r						(2.18)

but is still conserved. And thus Qr conservation follows.

Although the Coleman–Mandula no-go theorem can be evaded, it imposes severe constraints on the possible ways out.

For example, the anticommutator {Qr, Qs} is an operator which is a symmetric combination of two spin 1/2 objects. It is thus of bosonic type and obeys commutation relations. Moreover, being a symmetric combination of two Lorentz spinors, it is a vector under Lorentz transformations ([(1/2) (1/2)]s = (1)). Coleman–Mandula restrictions then apply to it and impose it to be proportional to the generator of translations Pµ.

One may push this analysis [215] further and show (see for example [342]) that such constraints basically fix the algebra of fermionic generators2. In this sense, supersymmetry is unique.

2.2The supersymmetry algebra

Let us write the supersymmetry charge Qr which, by convention, we choose to be a Majorana spinor3. The basic supersymmetry algebra reads

¯	µ	Pµ		(I)
{Qr, Qs} = 2γrs
[Qr, P µ] = 0				(II)
[Q , M µν ] = iσµν Q			s	(III)
r	rs

2[We will return to this in Chapter 4.]

3As stressed in Appendix B, a Majorana spinor has four real components in the Majorana basis where the gamma matrices are purely imaginary. Useful relations involving Majorana spinors may be found in equations (B.40) and (B.41) of this Appendix.

24 The singular rˆole of supersymmetry
where
Q¯r = QT γ0 r		(2.19)
and
σµν = 1	[γµ, γν ].	(2.20)
4

The relation (III) simply means that Qr transforms as a spinor in spacetime rotations. Before going further let us note that one may define N such supersymmetry

charges: Qi, i = 1, . . . , N . In this case, (I) is replaced by:

i ¯j	ij	µ	(2.21)
{Qr, Qs} = 2δ		γrs Pµ.

We will return to this more general case in Chapter 4, and show that interesting new terms may appear.

The supersymmetry algebra has several important consequences. First from (II) one infers

[Qr, P µPµ] = 0.

(2.22)

Let us then consider two states |b and |f of respective mass mb and mf which are obtained from one another by supersymmetry:

Qr|b = |f .

Then since P µPµ|b = m2b |b and P µPµ|f = m2f |f ,

P µPµ Qr|b = P µPµ|f = m2f |f

= Qr P µPµ|b = m2b Qr|b = m2b |f ,

hence mb = mf .

If supersymmetry is explicit in the particle spectrum, bosons and fermions which are supersymmetric partners must have equal mass. However, we do not observe in nature a boson field with the same mass as the electron. We conclude that, if the Hamiltonian describing the fundamental interactions is supersymmetric, supersymmetry is not explicit in the spectrum: it must be spontaneously broken.

Secondly, we may obtain from (I) an expression for the Hamiltonian of the system. Indeed, (I) reads

{Qr, Qt}γts0 = 2γrsµ Pµ.

Contracting with γsr0 , one obtains

{Qr, Qt} γ02 tr = 2 T r γ0γµ Pµ.

r,t

				Supersymmetry breaking 25
Using γ02 = 1l and Tr(γ0γµ) = 4g0µ, one obtains
r	Qr2 = 4P 0 = 4P0
thus	1		r Qr2.
				(2.23)
H =
		4

In a (globally) supersymmetric theory, the energy of the states is positive. This has some important consequences for supersymmetry breaking as we now see.

2.3Supersymmetry breaking

Let |Ω be the vacuum of the theory. If supersymmetry is a symmetry of the vacuum, that is if there is no spontaneous breaking of supersymmetry,

		Qr|Ω = 0.						(2.24)
In other words, the vacuum is invariant under eiQ. Then
1				r Qr2|Ω = 0
H|Ω =
			4
and the energy of the vacuum vanishes.
Reciprocally, if the vacuum energy vanishes,
1					1
Ω|H|Ω = 0 =		r Ω|QrQr|Ω =					r	Qr|Ω 2
	4					4

and

Qr|Ω = 0, i.e. supersymmetry is a symmetry of the vacuum.

We thus have proven the following statement: supersymmetry is spontaneously broken (i.e. it is not a symmetry of the vacuum) if and only if

Ω|H|Ω = 0.

(2.25)

Thus vacuum energy is the order parameter for global supersymmetry breaking.

The statement obviously di ers from the corresponding one for gauge symmetry breaking. Indeed let us consider a field φ, which is nonsinglet under a gauge group G, and its potential V (φ). We denote by φ the value of φ at the minimum of V (φ). Being space and time independent, it minimizes the kinetic energy (which vanishes) and by definition the potential energy. Thus, the vacuum energy is given by V ( φ ).

(i)If φ = 0 and V ( φ ) = 0, neither gauge symmetry, nor supersymmetry is broken.

(ii)If φ = 0 and V ( φ ) = 0, supersymmetry is spontaneously broken but not gauge symmetry.

26 The singular rˆole of supersymmetry

(iii)If φ = 0 and V ( φ ) = 0, on the contrary gauge symmetry is spontaneously broken but not supersymmetry.

(iv)If φ = 0 and V ( φ ) = 0, both supersymmetry and gauge symmetry are spontaneously broken.

An important consequence of having vacuum energy as the order parameter of supersymmetry breaking is that supersymmetry may be broken in a finite volume. Indeed, we will discuss in the next section supersymmetry breaking in quantum mechanical systems, i.e. systems with only “time” dependence. This puts again supersymmetry on a fundamentally di erent status from an ordinary continuous symmetry. It is well-known that there is in the latter case no possibility of spontaneous symmetry breaking in a finite volume. Through superposition, the groundstate can always be made invariant under the symmetry because mixing between states is always possible in finite volume. It is only in the infinite volume limit that one may define unmixed ground states with spontaneously broken symmetry.

If supersymmetry is a symmetry of the vacuum for any finite value of the volume, then, obviously (the limit of zero being zero), in the infinite volume limit the vacuum energy vanishes and supersymmetry is conserved. The converse is not true however: it may well be that supersymmetry is spontaneously broken for any finite value of the volume and restored in the infinite volume limit.

Let us also stress that the criterion (2.25) is a criterion of spontaneous breaking of global symmetry. When we make supersymmetry local, we need to include gravity. Indeed, the commutation relation (I) imposes to include “local” translations, which are nothing but the local reparametrizations of spacetime which play a central rˆole in general relativity. The local version of supersymmetry is called supergravity, and the inclusion of gravity modifies the criterion for spontaneous supersymmetry breaking. In fact, we will see in Chapter 6 that local supersymmetry may be spontaneously broken even when Ω|H|Ω = 0. This is welcome news since Ω|H|Ω is interpreted as the cosmological constant which is observed to be either zero or very small.

Let us return to the pairing of boson–fermion states discussed in the previous section. In the supersymmetric case, all states are paired except for the vacuum state (since Qr|Ω = 0). Once supersymmetry is spontaneously broken, the supersymmetry algebra remains valid but Qr|Ω = 0. We conclude that all states should be paired. What is then the fermionic state associated with the bosonic state |b since there is no longer a supersymmetric partner |f ? It turns out that, as in any spontaneous breaking of a global symmetry, there appears in the spectrum a massless Goldstone excitation. Because of the fermionic nature of supersymmetry, it is a Goldstone fermion,

˜

usually referred to as the Goldstino. This massless fermion, which we note G, can be created from the vacuum by the supersymmetry current:

˜	(2.26)
Ω|Jrµ|Gs (γµ)rs.

It is described by a Majorana spinor.

Then, any bosonic state |b is paired to (i.e. has the same energy as) the fermionic

˜	˜	˜	˜	↑	˜	↓, the
state |bG	(since G is massless). Since G has two helicity states G				and G

number of bosonic and fermionic zero modes does not seem to match. It is only

| ˜ ↑ ˜ ↓

because we are missing the bosonic state bG G which completes the energy level.

w(x) = − √h¯

	Supersymmetric quantum mechanics	27
	˜	˜
For example, the vacuum state |Ω is now paired with the bosonic state |G ↑		G ↓
˜	˜

and the fermionic states |G ↑ and |G ↓ . The rˆole of this massless Goldstino field is

central in the discussions of supersymmetry breaking.

2.4Supersymmetric quantum mechanics

We turn, in this section, to a topic which may seem to lie somewhat out of our main track. It provides, however, a nice illustration of the notions introduced in the previous sections and it has a wide range of applications. Moreover, some of the methods developed here will be used to discuss later the central issue of supersymmetry breaking.

It proves to be useful to advocate the methods of supersymmetry to study onedimensional potential problems in quantum mechanics. Moreover, since finite volume supersymmetric systems may be assimilated to quantum mechanical systems with a finite number of degrees of freedom, such a study will allow us to discuss the issue of supersymmetry breaking in finite volume. Before being more explicit, let us review a few facts of one-dimensional quantum mechanics.

In one-dimensional potential problems, there is a deep connection between the bound state wave function and the potential. Consider the ground state (of zero energy) wave function ψ0(x) of the Schr¨odinger equation with potential V1(x)

	¯h2
H1ψ0(x) = −		∂x2 ψ0(x) + V1(x) ψ0(x) = 0.					(2.27)
	2m
Then obviously
			h¯2 ψ (x)
V1(x) =			0				(2.28)
			2m		ψ0(x)

and the potential can be reconstructed if the wave function has no node. One can in fact factorize the Hamiltonian in the following form H1 = A+A with

p

A+

p

A =

√

+ iw(x),

=

√

− iw(x)

(2.29)

2m

2m

where p = ih∂¯ x, and the identification

V1(x) = w2(x) −

¯h

w (x).

(2.30)

√

2m

The “superpotential” w(x) is given in terms of the ground state wave function as

ψ (x)

0 . (2.31)

2m ψ0(x)

In other words, the ground state wave function satisfies Aψ0(x) = 0, since this implies H1ψ0(x) = 0.

28 The singular rˆole of supersymmetry

There is obviously a companion Hamiltonian H2 = AA+:

H2 = − ¯h2 ∂2 + V2(x),

2m x

with

V2(x) = w2(x) + √¯h w (x). 2m

Note that the energy eigenstates of the two systems are related. If

H1|ψn(1) = En(1) |ψn(1)

then

H2A|ψn(1) = AH1|ψn(1) = En(1) A|ψn(1) .

(2.32)

(2.33)

Hence, apart from the ground state for which A|ψ0 = 0, the energy eigenstates of the two systems are in one-to-one correspondence (see Fig. 2.1).

One may wonder why we end up with a dissymmetry when H1 and H2 seem to

play symmetric roles. The reason is that the zero energy ground states of H1 and H2 satisfy, respectively, A|ψ0(1) = 0 and A+|ψ0(2) = 0 and thus can be written as

√

x

(1)

2m

ψ0

(x) = N (1) exp −

0

w(y)dy ,

(2.34)

h¯

√

x

(2)

2m

ψ0

(x) = N (2) exp +

0

w(y)dy

(2.35)

¯h

E

H1

H2

E4(1)

A

A+

E3(1)

|ψ3(1)

E2(1)

|ψ2(1)

E1(1)

|ψ1(1)

0

|ψ0

Fig. 2.1 Energy spectrum of the two companion Hamiltonians H1 and H2, respectively (2.27) and (2.32).