Supersymmetry. Theory, Experiment, and Cosmology

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φ = −2λ2φ 2φ − λΨcR ΨL ,

φ = −2λ2φ 2φ − λΨcR ΨL ,

iγµ∂µΨL = 2λφ ΨRc ,

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Supersymmetric quantum mechanics 29

where N (1), N (2) are normalization constants. Obviously, only one of them (if any) is a normalizable state. We made the assumption that it is ψ0(1)(x) and we labelled it simply ψ0(x). We will return later to the case where none of these states is normalizable.

Let us now make explicit the presence of the supersymmetry algebra. Deﬁne

0 A+

Q1 = A 0

√

σ1 + w(x)σ2,

iA+

Q2 = iA −

√

σ2 − w(x)σ1.

(2.36)

One may easily check the algebra

{Q1, Q1} = {Q2, Q2} = 2H,

{Q1, Q2} = 0,

[Q1, H] = 0 = [Q2, H],

(2.37)

with

A+A

H =

0 AA+

h¯

+ w2

(x) − σ3

√

w (x)

(2.38)

One recognizes the supersymmetry algebra, this time with a single dimension (time). The Hamiltonian (2.38) is found for example in the following one-dimensional physical system: an electron moving along the z-axis in a x-dependent magnetic ﬁeld

aligned along the z-axis. Indeed, the corresponding Hamiltonian reads:

H =

(p − eA)2

div A +

|e|h¯

σ.B,

(2.39)

with

px = py = 0, pz = p,

Ax = Az = 0, Ay = −

w(x).

(2.40)

The state of the system is described by a two-component spinor Ψ =

, 0

corresponding to spin up and 1

to spin down along the z-axis. Since [σ3, H] = 0,

the spin is conserved in interactions. Note that in the one-dimensional models that we consider, the notion of spin is purely artiﬁcial (there is no rotation group). In fact, fermions may be bosonized.

Since the Q’s obey anticommutation relations, one may interpret them, as well as A and A+, as fermion type operators. We may call the eigenstates of H1, which include the ground state |ψ0 , the bosonic states of the system. The eigenstates of H2 are obtained from the previous ones by the action of A and are thus interpreted as the fermion degrees of freedom. The ground state of the total system is described by

Ψ0(x) =	ψ (x)
Ψ0(x) =	00	.

30 The singular rˆole of supersymmetry

We can go further and write a Lagrangian formulation of the system. There is a single spacetime variable – the time t – and the Lagrangian reads [78, 328]:

dx(t) 2

− [W (x)]2 + ihψ¯ †

dψ

¯h

ψ†, ψ .

L =

−

√

W (x)

(2.41)

We have introduced a fermion variable ψ in L because, in this form, L is invariant under a supersymmetry transformation which relates x and ψ (see Exercise 2). We need an explicit matrix representation of the fermion variable ψ which realizes the standard anticommutation relations:

{ψ(t), ψ(t)} = {ψ†(t), ψ†(t)} = 0, {ψ(t), ψ†(t)} = 1.

(2.42)

One may simply choose

ψ(t) =	0 1	, ψ†(t) =	0 0	.	(2.43)
	0 0		1 0

The corresponding Hamiltonian reads

	p2		2		¯h
H =		+ [W (x)]		+	√		W (x) [ψ†, ψ].	(2.44)
	2m
						2m

which coincides with (2.38) if one uses [ψ†, ψ] = −σ3 and one makes the identiﬁcation4:

w(x) = W (x).				(2.45)
It is interesting to note that the fermion number operator
0 0	1
NF = ψ†ψ = 0 1	=		1 − σ3	(2.46)
		2

is conserved through interactions, as can be checked using the ψ equation of motion.


Once again, one ﬁnds that states of the form			α	are of the boson type whereas
			0
states of the form	0	are of the fermion type. For future reference, we note that
	β

σ3 = (−)NF .

Let us now use such models to discuss the issue of spontaneous supersymmetry breaking, following the seminal paper by [371]. We consider systems described by the Hamiltonian (2.38). We assume that |w(x)| → ∞ when |x| → ∞ in order to have a discrete spectrum.

4Note that, by analogy of (2.41) with equation (1.33) of Chapter 1, what we should call the superpotential in this section is the function W (x), not w(x).

Supersymmetric quantum mechanics 31

In the weak coupling limit h¯ → 0, V = w2. Hence the number of supersymmetric vacua is the number of zeros of w. Let us then consider corrections of order h¯. Assuming that w has a simple zero at x = x0, we write

w(x) =

(x − x0)2 .

ω(x − x0) + O

(2.47)

Then the Hamiltonian (2.38) reads

1 2

H =

mω

(x − x0)

− h¯

(2.48)

One recognizes in the ﬁrst two terms (bosonic in the sense of the Hamiltonian (2.44)) the Hamiltonian of a harmonic oscillator with frequency ω: the zero point energy is ¯hω/2. The third term has eigenvalues ¯hω/2. The ground state thus corresponds to the eigenvalue σ3 = +1 (a boson state) and has vanishing energy5: ¯hω/2 − ¯hω/2 = 0. One may check that the cancellation persists to higher orders in (x − x0). One can also note that the fermion state (σ3 = −1) has energy ¯hω/2 + hω/¯ 2 = ¯hω.

Let us now consider the exact spectrum of the theory and not rely on a perturbative analysis. A supersymmetric ground state satisﬁes:

Q1|ψ = Q2|ψ = 0.

(2.49)

Since H = Q21 = Q22, it su ces that Q1|ψ = 0. Multiplying the ﬁrst of equations (2.36) by σ1, one obtains

h¯ ∂ψ

= −w(x)σ3ψ.

(2.50)

√

∂x

The solution of this ﬁrst order di erential equation is

ψ(x) = exp −

ψ(0).

√2m

w(y)σ3

(2.51)

¯h

For this to be the ground state of the theory with quantized energy levels, it has to be normalizable, which imposes that w(x) has opposite signs when x → +∞ and

x	→ −∞	. An equivalent statement is to require that w has an odd number of zeroes.
			3	. Then	x	dy w(y) = λx	4	/4, and the choice
	Let us take for example w(x) = λx				0
σ3ψ(0) = + ψ(0) corresponds to a normalizable					wave function. Note that this corre-

sponds to the spectrum described earlier, with bosonic degrees of freedom (including the vacuum) associated with σ3 = +1.

On the other hand, if w has an even number of zeros, then it is not possible to write a supersymmetric vacuum which is a normalizable state: supersymmetry is spontaneously broken. If w has no zero, the perturbative analysis already told us so. But if w has zeros (in even number) this contradicts the perturbative analysis above: supersymmetry is broken at a nonperturbative level. One speaks of dynamical breaking of supersymmetry. We will study in more detail this type of breaking in Chapter 8.

5This thus results from a cancellation between the bosonic terms and the fermionic terms in (2.44).

32 The singular rˆole of supersymmetry

2.5Witten index

It is a remarkable fact that the issue of spontaneous supersymmetry breaking was settled in the preceding section by discussing the behavior of a function at inﬁnity or the number of its zeroes. Such a property is, to some extent, stable under continuous deformations of the parameters of the theory: it is, in some sense, topological. This has been formalized by E. Witten [373] and connected with higher mathematics – the [14, 15] index formula. Let us have a closer look at this from the physics point of view.

The problem is whether one can ﬁnd a supersymmetric ground state, i.e. a normalizable state of vanishing energy. This question can be settled in a ﬁnite volume: if the energy of the state vanishes for all ﬁnite values of the volume, it vanishes as well in the inﬁnite volume limit. Since ﬁnite volume ﬁeld theory amounts to quantum mechanics with a ﬁnite number of degrees of freedom, the framework described in the preceding section is su cient to address the problem.

For example, the spectrum of the theory has necessarily – if there is a supersymmetric ground state – the form shown in Fig. 2.1. Apart from the ground state, bosonic and fermionic levels are paired. If we deform continuously the parameters of the theory, levels move up or down, they might coalesce, reach zero energy or on the contrary emerge from zero energy, but they remain paired. Thus the di erence between the number of bosons nB and the number of fermions nF at a given level remains unchanged. This is of particular importance at the zero energy level where we expect to ﬁnd the ground state. Witten thus deﬁnes the index (a topological notion) as:

I = nB(E=0) − nF(E=0) .	(2.52)
Introducing the operator
ΓF = (−)NF	(2.53)

which commutes with the Hamiltonian and is equal to +1 for a boson and −1 for a fermion, we have

I = TrΓF = Tr(−)NF

(2.54)

where Tr means that we sum over all states. Clearly, if I = 0, there exists one or more states with vanishing energy: supersymmetry is a symmetry of the vacuum. If one is looking for spontaneous symmetry breaking, one should thus restrict one’s attention to theories with vanishing Witten index. However, I = 0 is obviously only a necessary condition.

Let us take a few examples for the sake of illustration. First, consider as “superpotential” w(x) a monotonic function such that w(x0) = 0. Then, we can deform the parameters of the theory in order to have

w(x) = λ(x − x0).

(2.55)

If λ > 0, then an analysis similar to the one of the preceding section that led to (2.51) shows that the corresponding state is a boson (σ3|ψ = +|ψ ). If λ < 0, it is a fermion (σ3|ψ = −|ψ ). In any case, I = 0 and there is no chance to ﬁnd the spontaneous breaking of supersymmetry. On the other hand, if w(x) = λ(x−x1)(x−x2) with x1 = x2, then we ﬁnd one boson state (by analogy with the preceding case, this

Witten index 33

corresponds to the zero where w is increasing) and one fermion state (corresponding to the zero where w is decreasing). Thus I = 0: one may have spontaneous supersymmetry breaking. We see that I is directly related to the parity of the number of zeroes of w(x) (or just as well to the comparative sign between w(+∞) and w(−∞)).

As another example, we may consider the Wess–Zumino model discussed in Chapter 1. To compute the Witten index, one may use perturbation theory since one is allowed to change continuously the parameters. The potential of the model is given by equation (1.35) in Chapter 1:

V = |mφ + λφ2|2.

(2.56)

Assuming m large and λ small in order to stay within perturbation theory, we have two ground states φ0 = 0 and φ0 = −m/λ. On the other hand, we have seen that all ﬁelds (bosons and fermions) are massive, of mass m. We thus have two bosonic states of vanishing energy, the two vacua, and I = 2. This remains true when one reaches the nonperturbative regime as well as when m goes to zero6.

The previous example should make clear which deformations of the theory are permissible. If we add a term of order φ6 to the potential (2.56), we obviously change the asymptotic behavior of the energy and there is no reason for the Witten index to remain unchanged: states may be brought in or sent out to inﬁnity. Thus, the Witten index is only invariant under changes which do not modify the asymptotic (large ﬁeld) regime: adding a new coupling or setting an existing one to zero may require some caution.

We end this section by showing how the Witten index can be formally computed using functional methods. One needs to introduce a regulator β and write the regulated

Witten index as
I(β) = Tr(−)NF e−βH/h¯ .	(2.57)

We will let the regulator β go to zero at the end of the computation.

If we consider the general system (2.38) with the two companion Hamiltonians H1 and H2, then we have:

I(β) = Tr e−βH1/h¯ − e−βH2/h¯ .

(2.58)

This may be written explicitly as

I(β) = Tr σ3

[dpdx]

hw¯ /√2m]/h¯

e−β[p

/(2m)+w

−σ

2πh¯

[dpdx]

]/h¯ sinh

e−β[p

/(2m)+w

√

πh¯

[dpdx]

e−β[p

/(2m)+w

]/h¯

√

w ,

(2.59)

πh¯

where, in the last line, we have used an expansion in β since we will send it to zero. If we take w(x) = λx2n+1, we obtain I = 1 whereas for w(x) = λx2n (n integer),

the integrand is odd under x → −x and I = 0. We recover that it is only for an odd number of zeros (here coinciding) that the Witten index vanishes, which allows for the possibility of spontaneous supersymmetry breaking.

6In this limit, the fermion becomes massless. However, because I = 0, this is not a Goldstino ﬁeld.

34 The singular rˆole of supersymmetry

Further reading

•E. Witten, Introduction to Supersymmetry, in Proceedings of the International School of Subnuclear Physics, Erice 1981, ed. by A. Zichichi, Plenum Press, 1983.

•F. Cooper, A. Khare and U. Sukhatme, Supersymmetry and quantum mechanics, Phys. Rep. C 251 (1995) 267–385.

Exercises

Exercise 1

(a)Show that, for every spinor ﬁeld Ψ,

ΨcR ΨL ΨL = 0.

(b)Write the equations of motion for the theory described by the Lagrangians (2.13) and (2.17).

(c)Deduce from these equations of motion that the current Jrµ deﬁned in (2.18) is conserved.

Hints:

(a)ΨL is expressed in terms of two independent spinor components, ΨcR is expressed in terms of the same components; hence the left-hand side is a cubic monomial of only two independent spinor components: two are necessarily identical

and their product vanishes (because spinor components anticommute: ΨrΨr = −ΨrΨr = 0).

Exercise 2 Using the Euler equations for x(t) and ψ(t) derived from the Lagrangian L in (2.41), show that L is invariant (up to a total derivative) under the supersymmetry transformations:


δx = ε†ψ + ψ†ε,
δψ = ε α	dx
			+ βW (x) ,		(2.60)
	dt
δψ† = ε† α		dx
				+ β W (x) ,	(2.61)
		dt

where ε is a spinor, parameter of the supersymmetry transformation, and α, β are complex numbers.

Exercises 35

Hints: The equations of motion read:

d2x

¯h

+ 2W W +

√

[ψ†, ψ] = 0,

dt2

−ih¯

dψ

¯h

+ 2

√

W ψ = 0,

dψ†

− 2

¯h

−ih¯

√

W ψ† = 0.

Using these equations,

δL =

ε†ψ + ψ†ε + ih¯ a

+ βW ψ†ε .

(2.62)

Compare the transformation (2.60) in one dimension with the supersymmetry transformations in four dimensions, e.g. (3.26) of Chapter 3 with F = − (dW/dφ) .

Exercise 3 Consider a quantum mechanical model with “superpotential” w(x) = λ(x2+ a2). For a2 > 0, supersymmetry is broken at tree level since, at the classical level (¯h → 0), V = w2(x) > 0. For a2 < 0, w has an even number of zeros and the arguments of Section 2.4 show that supersymmetry is spontaneously broken. The purpose of this exercise is to show that there exists a change of parameters which interpolates between the two situations.

(a) Deﬁne Q± = (Q1 ± iQ2) /2. Express Q± in terms of A and A+. Show that

Q2	= Q2	= 0
+	−
Q+Q− + Q−Q+ = H.			(2.63)

(c)Determine P± such that

˜ = eP± xQ e−P± x

Q± ±

correspond to the “superpotential” w˜(x) = λ(x2 − a2).

(d) Using (c) and the criterion deﬁned in (b), conclude that, if					H corresponding
2	+ a	2	) has no zero-energy eigenstates, then	˜
to w(x) = λ(x				H corresponding to

w˜(x) = λ(x2 − a2) has no zero-energy eigenstates.

Hints: See [373], Section 3.

Exercise 4 Show that the cancellation of the ground state energy still holds when one considers terms of order (x − x0)2 in (2.47).

Hints: See [339], equation (14).

Exercise 5 We have studied in Section 2.4 only cases where the spectrum is discrete. Supersymmetric quantum mechanical methods prove to be useful also with continuum spectra. We illustrate this by studying the relations that exist between the reﬂection and transmission coe cients of the two companion Hamiltonians H1 and H2.

36 The singular rˆole of supersymmetry

In order to have a continuum spectrum, we must assume that the potential remains ﬁnite as x → −∞ or x → +∞. We thus deﬁne w(x → ±∞) ≡ w±.

(a)One considers an incident plane wave eikx of energy E coming from −∞. The scattering on the potentials V1,2 induces reﬂection and transmission with respective coe cients R1,2 and T1,2. One thus has:

ψ1,2(k, x			→ −∞)	→ eikx + R1,2e−ikx,
ψ	1,2	(k , x	+ )	→	T	1,2	eik x.	(2.64)
	1,2		→ ∞	→		1,2

Writing the “supersymmetry” relations between the wave functions of the two Hamiltonian, show that

w− + ik

, T

w+ − ik

(2.65)

w− − ik

and express k and k in terms of E, w+ and w−.

(b)Deduce that the companion potentials have identical reﬂection and transmission probabilities. In which case is T1 = T2?

(c)Infer that, if one of the potentials is constant, the other one is necessarily reﬂectionless. Using w(x) = µ tanh αx, show that the potential V (x) = λ sech2αx is reﬂectionless.

Hint: See Cooper et al., (1995), p. 278 .

Exercise 6 The methods of supersymmetric quantum mechanics are useful to discuss the problem of localization of four-dimensional gravity in models with extra dimensions. In the ﬁve-dimensional model of Randall and Sundrum [319], the Kaluza– Klein modes of the ﬁve-dimensional graviton (the mediator of ﬁve-dimensional gravity) obey a standard Schr¨odinger equation (2.27):

Hψ = −∂x2ψ(x) + V (x)ψ(x),

where the coordinate x parametrizes the ﬁfth dimension (a factor h¯2/(2m) has been absorbed) and

V (x) =		15		3		δ(x),	(2.66)
				−
	4(	x	+ R)2	−	R
		\| \|

where R is a constant (R is called the AdS5 curvature radius in this model).

(a)The potential V (x) is called the volcano potential. Draw it to understand this name.

(b)Identify the function w(x) corresponding to this potential.

(c)Deduce from it the form of the ground state wave function (Kaluza–Klein zero mode of the ﬁve-dimensional graviton). Show that it is localized around x = 0. This zero mode is identiﬁed as the four-dimensional graviton ﬁeld: four-dimensional gravity is thus localized around x = 0 in the ﬁfth dimension.

Hints: (b) w(x) = (3/2) (2θ(x) − 1)1/(|x| + R); (c) ψ0(x) 1/(|x| + R)3/2.

Basic supermultiplets

We present here, in a hopefully nontechnical fashion, the two basic supermultiplets: (spin 0/spin 12 ) and (spin 1/spin 12 ). The former involves a complex scalar ﬁeld and a spinor ﬁeld chosen to be real (Majorana) or to have a given helicity (Weyl). It is called a chiral supermultiplet. The latter involves a real vector ﬁeld and a Majorana spinor ﬁeld. It is called a vector supermultiplet. If it is associated with a gauge symmetry, the vector is a gauge ﬁeld and its supersymmetric partner is called a gaugino. We will review the form of the supersymmetry transformations for both multiplets and insist on the need for auxiliary ﬁelds. Moreover, in both cases, we will discuss the issue of supersymmetry breaking.

This chapter is self-contained. The more theoretically oriented reader may, however, ﬁnd it useful to read in parallel Appendix C where the use of superspace allows for a more systematic approach. Appendix C uses two-component spinors and it might be a useful exercise to translate its results back into four-component spinors as used in this chapter.

3.1Chiral supermultiplet

3.1.1A ﬁrst look

We consider here, as in the Wess–Zumino model of Chapter 1, a chiral supermultiplet
which consists of:		√
•
	a complex scalar ﬁeld φ(x) = (A(x) + iB(x))/ 2;
•	a Majorana spinor ﬁeld Ψ(x).

Obviously, the scalar ﬁeld consists of two real degrees of freedom: A(x) and B(x). On the other hand, the Majorana spinor has four real components. Its equation of motion – the Dirac equation – ﬁxes two of these degrees of freedom. To see this, consider for simplicity a planar wave of energy k0 > 0 described by the spinor u(k)

which satisﬁes		(k/ − m)	u(k) = 0.		(3.1)

In the center of mass, kµ = (m, 0, 0, 0) and (3.1) becomes
		γ0 − 1 u(k) = 0.
	−	0	0
In the Dirac representation γ0		1 = 0	−2	. We are thus left with two degrees

of freedom.

38 Basic supermultiplets

Thus, the number of bosonic and fermionic degrees of freedom of the chiral supermultiplet seem to be only equal on-shell (that is, using the equations of motion).

Let us now consider the free ﬁeld Lagrangian

	1		Ψ¯ (iγµ∂µ − m) Ψ.
L = ∂µφ	∂µφ − m2φ φ +			(3.2)
		2
It is invariant under the inﬁnitesimal supersymmetry transformation
δA = ε¯Ψ,	δB = iεγ¯ 5Ψ			(3.3)
δΨr = − [iγµ∂µ (A + iBγ5) + m (A + iBγ5)]rs εs				(3.4)

where εs is the (Majorana) spinor parameter of the transformation1.

There is, however, an undesirable problem with the transformations (3.3–3.4).

According to the supersymmetry algebra (I), one has
	¯	= ε¯1r ε2s		¯
[¯ε1Q, ε¯2Q] = ε¯1Q, Qε2		= ε¯1r ε2s		Qr, Qs
= 2 (¯ε			µ		(3.5)
= 2 (¯ε
	1γµε2) P

where we have used the relation ε¯2Q = Qε2 (see Appendix B equation (B.40)) valid for Majorana spinors. The commutators of two supersymmetry transformations of parameters ε1 and ε2 is a translation of vector aµ = 2(¯ε1γµε2) and one should have

[δ1, δ2] Ψr = 2 (¯ε1γµε2) (i∂µΨr)			(3.6)
whereas the transformations (3.3)–(3.4) yield
[δ1, δ2] Ψr = 2i (¯ε1γµε2) ∂µΨ +	i	γµ (iγν ∂ν − m) Ψ r .	(3.7)

	2

This coincides with (3.6) only when we make use of the Dirac equation. In other words, the algebra of supersymmetry only closes on-shell in this formulation.

3.1.2Auxiliary ﬁelds

Luckily, by introducing auxiliary ﬁelds, J. Wess and B. Zumino (1974b) have provided

us with a formulation of supersymmetry where the algebra closes o -shell. Following

√

them, let us add a complex scalar ﬁeld F (x) = (F1(x) + iF2(x))/ 2 and consider the Lagrangian

L	= ∂µφ ∂ φ +	1	Ψ¯	(iγµ∂	µ −	m) Ψ +	Laux
		2
	µ
Laux = F F + m (F φ + F φ ) .							(3.8)

Since F has no kinetic term, there is no dynamical degree of freedom associated with it. One may solve for it using its equation of motion:

F =	−	mφ .	(3.9)

Hence Laux = −m2φ φ and one recovers the Lagrangian (3.2): the theory is identical to the previous one on-shell (that is, making use of the equations of motion).

1Let us note here the canonical dimension of ε. Since scalar ﬁelds (A, B) have canonical dimension 1 and spin 1/2 ﬁelds (Ψ) have canonical dimension 3/2, ε has dimension −1/2.

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