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Supersymmetry. Theory, Experiment, and Cosmology

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BPS states 79

Associated with the electric–magnetic duality, there is an interesting SL(2, Z) symmetry. To make it explicit, we must pay more attention to the θ-term (4.73). In the Prasad–Sommerfield limit (V = 0), the Georgi–Glashow Lagrangian (4.74) may be written

L =

1

Imτ

F aµν + iF˜aµν Fµνa

+ iF˜µνa +

1

DµΦaDµΦa,

(4.114)

32π

 

2

where we have rescaled the gauge fields by the coupling constant and

 

 

 

 

τ =

θ

+

 

4πi

.

 

 

 

(4.115)

 

 

 

 

 

 

 

 

 

 

 

 

2 π

 

g2

 

 

 

 

Since there is periodicity under θ → θ + 2π, the transformation τ → τ + 1 should be a symmetry. Moreover, when θ = 0, the duality invariance ((4.113) with gm = g) reads τ → −1. These two transformations generate the group SL(2, Z) of projective transformations:

+ b

, a, b, c, d Z, ad − bc = 1.

(4.116)

τ → + d

It has been noted by [370] that, once one restores a nonzero vacuum angle θ, the monopole charge is shifted by an amount proportional to θ. To see this, we introduce the charge N associated with the dyon collective coordinate χ introduced earlier.

Following (4.102), we have

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

N = d3x

1

 

δL

Diφa

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ga δ∂0Aia

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

d3xDiφaF0ai +

 

θg

d3xDiφa

1

 

 

 

 

=

 

 

 

 

 

 

ijkFjka

 

ga

8π2a

2

 

1

 

 

 

 

 

 

 

 

 

 

θg

φa

1

 

 

=

 

d3x∂i (φaF0ai) +

 

 

d3x∂i

 

ijkFjka .

(4.117)

ga

8π2a

2

This gives, using (4.88),

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

θg

 

 

 

 

 

 

 

 

 

 

 

 

N =

 

Qe +

 

Qm.

 

 

 

 

 

(4.118)

 

 

 

 

 

 

g

8π2

 

 

 

 

 

Because n is associated with a gauge transformation about φa, and a SO(3) SU (2)/Z2 rotation of 2π is identity, we have the condition e2πN = 1. Hence N is an integer ne and the charge is (Qm = 4π/g for a monopole)16

Qe = neg −

θg

(4.119)

2π nm.

16In the moduli space approximation, this gives Q = −ig∂χ (gθ/2π)nm.

80 The supersymmetry algebra and its representations

We see that the BPS mass formula for a dyon (4.106) is now given by

M = a |ne + nmτ | ,

 

(4.120)

and the SL(2, Z) transformation acts on this integers as

 

a b

1

 

(nm, ne) (nm, ne) c d

.

(4.121)

Further reading

M. Sohnius, Introducing supersymmetry, Phys. Rep. C 128 (1985) 39–204.

J. A Harvey, Magnetic monopoles, Duality, and Supersymmetry, 1995 Trieste Summer School in High Energy Physics and Cosmology, p. 66–125.

Exercises

Exercise 1 Consider the quantity T µν = ξασµν αβ ξβ , where σµν αβ is defined in (B.24) of Appendix B. Show that (σµν )αγ = (σµν )γα. Infer that Tµν belongs to the representation (1,0) of the Lorentz group.

This is what allowed us to write the decomposition (4.3). Hint: Tµν = ξα (σµν )αγ ξγ and [(1/2, 0) × (1/2, 0)]s = (1, 0).

Exercise 2 Use a Jacobi identity to prove that the matrices br defined in (4.9) provide a representation of the internal symmetry algebra:

[br, bs] = iCrstbt.

Hints: Use [[Br, Bs], Q] + [[Bs, Q], Br] + [[Q, Br], Bs] = 0.

Exercise 3 We show in this exercise that the central charges Zij introduced in Section 4.1 commute with all generators of the supersymmetry algebra; hence their name.

1. Prove that the subalgebra spanned by the Zij is invariant:

[Zij , Br] = (br)i k Zkj + (br)j k Zik.

(4.122)

2. One decomposes the Zij on the internal symmetry generators:

Zij = arij Br. (4.123)

r

Use the relevant Jacobi identity to prove the following relation:

arij (br )k l = 0. (4.124)

r

Exercises 81

3.

Using in (4.124) the fact that Br is hermitian, prove that

 

[Zij , Q] = 0, Z

 

¯k

= 0,

 

 

, Q

 

 

[Zij , Zlm] = 0.

ij

 

α˙

 

4.

Deduce that

 

 

 

 

 

[Zij , Br] = 0.

 

 

 

Hence Zij commutes with all generators.

 

 

 

 

5.

Prove the following relation:

 

 

 

 

(br)i kaskj + (br)j kasik = 0.

k

(4.125)

(4.126)

(4.127)

This relation shows that the N × N antisymmetric matrix as is invariant under the internal symmetry group G. This imposes a symplectic structure on the semisimple part of the corresponding algebra.

Using the fact that br is hermitian, equation (4.127) can be written as:

 

 

=

 

 

(b )k

 

 

(4.128)

(b )

kas

as

j

.

r i

kj

 

ik

r

 

 

k

 

 

 

k

 

 

 

 

This is known as the intertwining relation since it shows that the matrix as intertwines the representation br with its conjugate −br .

Hints:

1.

Use the Jacobi identity [{Q, Q}, Br]−{[Q, Br], Q}+{[Br, Q], Q} = 0.

 

 

¯k

r

 

 

l

 

 

¯k

 

 

2.

 

{Q, Qjβ }, Qγ˙

+

{Q, Qγ˙ }, Q+

{Qγ˙ , Q}, Q= 0.

 

3.

From (4.124),

 

ar

 

(br) k = 0; and from (4.122),

 

 

 

 

#

ij

 

 

 

 

 

 

 

 

 

 

 

k

r

k

r

 

 

 

[Zij , Zlm] = r

(br)i

almZkj + (br)j

almZik

= 0.

4.The subalgebra spanned by the Zij is an invariant abelian subalgebra. It must therefore belong to the abelian factor since the internal symmetry group G is the product of a semisimple group (no abelian invariant subgroup) and an abelian factor.

5.Write explicitly (4.122) using (4.126).

Exercise 4 In the construction of Section 4.3.1, show that there are 2N−1 bosonic states (integer helicity) and 2N −1 fermionic states (half integer helicity).

Hint:

 

 

 

 

[N/2]

N

 

[N/2]1

N

 

 

 

 

 

 

 

 

 

2k + 1 ,

 

(1 ± 1)N = k=0 2k

±

k=0

 

[N/2]

N

 

[N/2]1

 

N

 

 

hence

 

2k

=

 

 

2k + 1 = 2N−1.

k=0

k=0

82 The supersymmetry algebra and its representations

Exercise 5 Write equation (III) of Section 2.2 of Chapter 2 in two-component notation:

[Qα, M µν ] = i (σµν )α β Qβ ,

 

Q¯α˙ , M µν = i σµν )α˙

β˙ Q¯β˙ .

(4.129)

¯

¯

 

≡ M12 and J, as

Deduce the spin components of Q

and Q, i.e. the values of J3

indicated in Section 4.3.2.

Hint: Use (B.23) and (B.24) of Appendix B, as well as the explicit forms of σµ and σ¯µ.

Exercise 6 Prove that massless representations are only compatible with vanishing central charges. For this purpose, choose the frame of Section 4.3.1 where Pµ = (E, 0, 0, E) and show that Q2ar annihilates any state |Φ .

Hint: See the footnote 6 of Section 4.3.1; if Q2ar annihilates any state |Φ , then zr = 0 from (4.25).

Exercise 7 Show that the N = 2 supersymmetry transformations (4.37) include the N = 1 supersymmetry transformations of the vector and chiral supermultiplets of parameters ηα1 , η¯1α˙ (given in equation (C.99) of Appendix C).

Hint: Using the fact that all fields belong to the adjoint representation and writing λ ≡

λ1

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

, ψ

, φ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2, we obtain from the transformation

 

λ

(A+iB)/ 2, F ≡ (F1 +iF2))/

 

1

):

 

 

 

 

 

 

 

 

 

law (C.99) of Appendix C (with parameter η ≡ η

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

λ

2

 

 

 

 

1

¯2

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

δS A = η

 

+ η¯

λ

 

¯2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

δS B = −i η

1

λ

2

 

 

 

1

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

− η¯

 

λ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

δ λ2 =

 

η1 F

i(σµη¯1) D A + (σµη¯1) D B,

 

 

 

 

 

 

 

 

 

 

S

α

 

 

2 α

 

 

¯2

 

 

α

 

µ

 

 

 

 

α

µ

 

 

 

 

 

¯1

 

 

 

 

 

¯1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

σ

µ

 

 

 

 

 

 

1

 

µ

Dµλ

2

 

1

λ

1

1

+ ig B, η

1

λ

1

1

,

 

δS F1 = −iη

 

Dµλ

 

− iη¯ σ¯

 

 

− g A, η

 

 

η¯

λ

 

 

+ η¯

λ

 

δS F2 = η1σµDµλ¯2

η¯1σ¯µDµλ2

ig A, η1λ1

+ η¯1λ¯1

 

g B, η1λ1

 

 

η¯1λ¯1 ,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

δS Aµ = η1σµλ¯1 − η¯1σ¯µλ1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

δS λα1 = −ηα1 D − i σµν η1

 

α Fµν ,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

δ

 

 

 

1

 

µ

 

 

¯1

 

 

 

1

σ¯

µ

D

 

1

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

S D = iη σ Dµλ + ¯

 

λ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

µ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

which corresponds to the transformations (4.37) of parameter η1, once one uses the four-component symplectic Majorana notation for spinors and the form (4.35) for the auxiliary fields.

Exercise 8 Consider in electromagnetism the energy density 12 (E2 + B2) and Poynting vector 41π E B. By expressing them in terms of E + iB, show that they are invariant under the duality transformation (4.66).

Hint: |E + iB|2 = (E2 + B2), (E + iB) (E + iB) = 2i E B.

ˆa

a

a

is

Exercise 9 Compute the winding number n in (4.81) when φ

→ rˆ

as r → ∞ r

the unit vector in space).

 

 

 

Hint: n = 1.

 

 

 

Exercises 83

Problem 1 We study in this problem the two-dimensional supersymmetric model of Section 4.5.1. Two-dimensional supersymmetry plays also an important rˆole in string models because the string, being a one-dimensional object, follows a two-dimensional world-sheet in its motion: string theory can be understood as a two-dimensional quantum theory on the world-sheet (see Section 10.1 of Chapter 10).

1.We first discuss spinors in d = 2 dimensions. Since the spacetime dimension falls in the class d = 2 mod 8, we can impose at the same time the Majorana and Weyl conditions. Moreover, the Lorentz group is simply SO(2) U (1) (or rather SO(1, 1) in the Lorentz case that we consider here). Thus its representations are characterized by a single quantum number. The spinor representation has dimension 2; it can be decomposed into spinors of chirality ±1: a Majorana–Weyl spinor is thus a one-dimensional object and chirality is the additive quantum number just mentioned.

More precisely, we consider the two-dimensional gamma matrices:

ρ0 =

0 −i

, ρ1 =

0 i

,

(4.130)

 

i 0

 

i 0

 

 

which satisfy

µ, ρν } = 2gµν

 

 

 

 

 

 

(4.131)

(the metric being (+, −)). We define also the two-dimensional analog of the γ5 matrix:

 

ρ3 =

ρ0

ρ1 =

1

0 .

(4.132)

 

 

 

0

1

 

Then a two-dimensional spinor is written ψ =

ψ

where ψ± is the ± chirality

ψ+

component:

 

 

 

 

 

 

 

 

ψ+

ψ+

 

(4.133)

 

ρ3

ψ

=

ψ.

(a) Show that C = −ρ0 is the matrix that intertwines ρµ with −ρTµ :

µT C1 = −ρµ.

(4.134)

¯ 0 c ¯T

One defines as usual ψ = ψ ρ and ψ = . Show that the Majorana condition ψc = ψ is simply a reality condition with the gamma matrix basis chosen here.

(b) Show that, for Majorana spinors ψ and χ,

 

 

 

¯

 

 

 

 

 

 

 

 

ψχ = χψ,¯

 

 

 

 

 

 

¯

 

µ

 

 

µ

ψ,

(4.135)

¯

ψρ

 

χ = −χρ¯

 

µ

ρ

ν

χ = χρ¯

ν

ρ

µ

ψ,

 

ψρ

 

 

 

 

 

which are the analogs of the four-dimensional formulas (B.40) of Appendix B.

84The supersymmetry algebra and its representations

2.We consider the action

S =

d2x

1

(µφ)2 +

1

ψi/∂ψ¯

+

1

F 2 + F W (φ)

1

W (φ)ψψ¯

, (4.136)

2

 

2

 

2

 

2

where φ, F are real scalar fields, ψ is a Majorana spinor and W = W (φ), W = dW/dφ. Solving for the auxiliary field F (no kinetic term), one recovers the action (4.56).

(a)Show that S is invariant under the global supersymmetry transformations (compare with the four-dimensional analog (3.10) of Chapter 3)

δφ = εψ,¯

δψ = −iρµε∂µφ + F ε, (4.137) δF = −iερ¯ µµψ,

where ε is an anticommuting Majorana spinor.

(b)How many o -shell and on-shell bosonic degrees of freedom in the theory? How many fermionic?

(c)Determine the Noether current jµ associated with the supersymmetry transformation:

jµ = ν φρν ρµψ + iW ρµψ.

(4.138)

3. (a) Using the chiral decomposition introduced in 1 for ψ and jµ, show that the

chiral components of the supersymmetry charge are respectively:

 

 

 

 

 

 

 

Q+ = dx [+φ ψ+ − W (φ)ψ]

 

 

 

 

where ± ≡ ∂0 ± ∂1.

Q=

dx [φ ψ+ W (φ)ψ+]

 

 

(4.139)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(b) Show that the energy–momentum tensor can be written as

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

1

 

 

 

 

 

i

 

 

 

 

Tµν = µφ∂ν φ −

 

gµν σ φ∂σ φ +

 

 

W 2gµν +

 

 

ψρ¯ µν ψ − gµν

 

2

2

2

 

 

 

 

1

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

×

 

ψiγ¯ σ σ ψ −

 

W ψψ¯

.

 

 

 

 

 

 

 

 

(4.140)

 

2

2

 

 

 

 

 

 

 

 

Use the equations of motion to check that it is conserved: µTµν = 0.

 

(c) Use the canonical commutation relations

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[φ(t, x), ∂0φ(t, y)] = (x − y)

 

 

 

 

 

 

 

 

 

+(t, x), ψ+(t, y)} = δ(x − y)

 

 

(4.141)

 

 

 

 

 

(t, x), ψ(t, y)} = δ(x − y)

 

 

 

 

in order to check how the di erent fields behave under the translation gener-

ator Pµ =

dx T0µ:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[φ(x), P ] = i∂

µ

φ(x), [ψ

+

(x), P

µ

] = i∂ ψ

+

(x), [ψ

(x), P

µ

] = i∂ ψ

(x).

µ

 

 

 

 

 

 

 

 

 

µ

 

 

 

 

 

µ

 

(4.142)

 

 

 

 

Exercises 85

4. (a)

Prove that

 

 

 

 

Q±2 = P± ≡ P0 ± P1.

(4.143)

(b)

Show that

 

 

 

 

{Q+, Q} =

dx2W [φ(x)] xφ(x) ≡ Z.

(4.144)

(c)Compute Z for the kink state when V (φ) = W 2(φ)/2 = λ(φ2 − φ20)2/2.

(d)Expressing Q+ ± Qfor a static solution (0φ = 0), show that one of these combinations vanishes for the kink (antikink), i.e. for a solution of (4.52).

Hints: 1(b) Remember that spinors anticommute: hence transposition of a quadratic combination of spinor fields introduces a minus sign.

¯ µ

2(a,c) δψ = iερ¯ µφ + F ε¯.

In order to compute the Noether current, it is advisable to introduce a spacetime dependent ε(x) spinor parameter and to make use of the relation

δS = d2x∂µεj¯ µ.

This allows us to take full account of the anticommuting nature of ε. One finds the following variation for the Lagrangian L:

L = µ µφεψ¯

1

ερ¯

ν ρµψ∂ν φ −

i

(F + 2W ερµψ

2

2

+µε¯[ν φρν ρµψ + iW ρµψ] .

(b)O -shell, two scalar (φ, F ) and two spinor (ψ±) degrees of freedom. On-shell, one must take into account the special nature of 1+1 dimensional spacetimes. Indeed, the free field equations of motion µµφ = 0 and µµψ = 0 are

solved by φ = φ+(x + t) + φ(x − t), ψ+ = ψ+(x + t), ψ= ψ(x − t). Hence one scalar and one spinor left-moving degrees of freedom. And the same for right-movers.

3(c)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

P0 =

dx

1

 

1

 

1

 

 

i

 

 

i

ψ1ψ+ iW ψ+ψ

 

(0φ)2 +

 

 

(1φ)2 +

 

 

W 2 +

 

ψ+1

ψ+

 

 

2

2

2

2

2

P1 =

dx ∂0φ ∂1φ +

 

i

ψ+1ψ+ +

i

ψ1

ψ.

 

 

 

 

 

 

 

 

 

 

2

2

 

 

 

 

To compute [ψ±(x), P0], use the equations of motion:

 

 

 

 

 

 

 

 

∂ ψ± = ±W ψ .

 

 

 

 

 

4 (a) P± =

 

 

 

21 (±φ)2 + 21 W 2

+

 

 

 

 

 

 

 

 

 

 

 

 

dx

 

 

± iψ±xψ± + iW ψ+ψ.

 

 

 

(b) Note that

the commutator [

 

 

 

φ(x), ∂

 

φ(y)] at equal time vanishes.

&

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

'

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) Z = 2

 

 

+φ0

2V (φ)

= 8λφ03/3.

 

 

 

 

 

 

 

 

(d) For a

 

 

−φ0

"

 

 

±

Q

=

 

 

 

±

W )(ψ+

 

ψ

).

 

 

 

 

 

 

 

 

+

 

 

 

 

dx(xφ

 

 

 

 

static

solution, Q

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5

The minimal supersymmetric model

As a part of the general e ort to implement supersymmetry at the level of fundamental interactions, the minimal supersymmetric extension of the Standard Model is motivated and described here at tree level. The presentation of its phenomenology requires a discussion of radiative corrections which play here a central rˆole ; it is therefore postponed to Chapter 7. Of great interest for astrophysics is the lightest supersymmetric particle (LSP) if it is stable: as a weakly interacting massive particle (WIMP), it provides a candidate for dark matter. The last section of this chapter is devoted to these considerations.

5.1Why double the number of fundamental fields?

5.1.1First attempts

If supersymmetry rules the microscopic world of elementary particles, then one should be able to classify the known particles in supersymmetric multiplets or supermultiplets. As is customary (cf. baryons in the SU (3) classification), some elements of the puzzle, i.e. some components of the supermultiplets might be missing and experimental e ort should be looking for them.

In the years 1976–1977, P. Fayet undertook such an e ort, to come to the conclusion that half of the jigsaw puzzle is presently missing. This is such a sweeping statement (although not such an implausible one: after all, the prediction of antiparticles which also led to an almost doubling of the spectrum of known particles, was later confirmed by experiment) that it is worth following P. Fayet’s early e orts to put known particles in supermultiplets.

Obviously, the mass spectrum is not a reliable tool since supersymmetry is (at best) spontaneously broken and thus not visible in the particle spectrum. However, photon and neutrino are a boson–fermion pair of massless particles and it is tempting to associate their charged electroweak partners W± and e±. Indeed a model could be built along these lines [146].

This does not, however, solve the problem at hand: what to do with the other two neutrinos? how to relate the colored particles: the bosons (gluons) are in a di erent representation of SU (3) than the fermions (quarks)? etc.

yH3 2

Why double the number of fundamental fields? 87

Table 5.1

BOSONS

FERMIONS

gauge field

gaugino

 

Higgs

Higgsino

squark

quark

slepton

lepton

 

 

 

5.1.2Squarks, sleptons, gauginos, higgsinos, and the Higgs sector

These considerations lead P. Fayet [148,149] to double the number of fields and introduce a supersymmetric partner to each field of the Standard Model, as shown in Table 5.11. Obviously, this is a very large step backward in our goal of unifying matter and radiation: we end up introducing new forms of matter and of radiation.

There is an extra complication: we need to enrich the gauge symmetry breaking sector of the theory and to introduce two Higgs doublets. We will give two reasons for this.

First, we must pay attention to the gauge anomalies of the new theory. In the Standard Model, the compensation of chiral anomalies is ensured by the field content: as recalled at the end of Section A.6 of Appendix Appendix A, the dangerous triangle diagrams of Figure A.11, which jeopardize the local gauge symmetry at the quantum level, add up to zero when one sums over all possible fermions (three colors of quarks, leptons) in the loop. In the theory that we consider now, we have introduced new fermions: gauginos and Higgsino. As far as chiral anomalies are concerned, gauginos are not a problem since they couple vectorially. On the other hand, a unique Higsino introduces anomalies: for example, a triangle diagram with three U (1)Y vector fields attached to a Higgsino loop is proportional to yH3 = (+1)3 = +1 and is nonvanishing. Introducing a second Higgs doublet of opposite hypercharge solves the problem: yH3 1 +

= (1)3 + (+1)3 = 0.

This second doublet is also needed for the purpose of breaking correctly the SU (2)× U (1) gauge symmetry. We have seen in Chapter 4 that a massive vector supermultiplet consists of one real scalar field, one massive vector field and two Majorana spinors. This can be understood as well by reference to the Higgs mechanism: we start in the Hamiltonian with a massless vector supermultiplet (one vector and one Majorana spinor) and a chiral supermultiplet (two real scalars and one Majorana spinor); one of the scalars provides the longitudinal massive vector degree of freedom and we find in the spectrum precisely the massive vector supermultiplet just described.

The Z0 and W± supermultiplets thus require three longitudinal degrees of freedom Z0L, WL± and three real scalars, respectively H0 and H±. A single complex doublet provides four degrees of freedom, which is clearly insu cient. Two complex doublets H1 and H2 have a total of eight degrees of freedom: the six above plus a scalar h0 and a pseudoscalar A0 which form a chiral supermultiplet (with a Majorana fermion). Thus a supersymmetric version of the Higgs mechanism for breaking gauge invariance requires the presence of two complex doublets.

1One uses the denomination photino, wino, zino, and gluino for the supersymmetric partner of the photon, W, Z, or gluon.

88 The minimal supersymmetric model

It will prove to be useful in what follows to remember this supersymmetric limit in which mH0 → MZ , mH± → MW and mh0 → mA0 .

5.2Model building

5.2.1Chiral supermultiplets and their supersymmetric couplings

The Standard Model is chiral, in the sense that left-handed and right-handed components transform di erently under the gauge interactions. When discussing supersymmetry, it is however more convenient to describe all fermions by spinors of the same chirality, say left-handed. If necessary, we will use charge conjugation in order to do so; for example we will trade uR for ucL . One thus introduces, for each generation of quarks and leptons:

(SU (3), SU (2), Y )

Q =

U

 

3, 2

;

1

 

 

 

D

 

3

 

 

 

L =

N

(1, 2 ; 1)

E

 

 

 

 

 

 

 

 

 

 

 

4

 

U c

 

 

3, 1

;

 

 

 

 

3

 

 

 

 

 

 

 

 

2

 

 

 

Dc

 

 

3, 1

;

 

 

 

 

 

 

 

 

 

3

 

 

 

Ec

 

 

(1, 1

;

2)

 

 

where the capital letter indicates the superfield. For example U contains the lefthanded quark uL and the squark field uL ; whereas U c contains the left-handed antiquark ucL and the antisquark uR (respectively, charge conjugates of the right-handed quark uR and of the squark uR ). Note that uL and uR are completely independent scalar fields and that the index refers to the chirality of their supersymmetric partners.

Moreover we have just seen that we must introduce two Higgs doublets:

H1 =

H10

 

(1, 2 ; 1)

H1

H2 =

H2+

 

(1, 2 ; 1) .

H20

Since by definition, the Minimal SuperSymmetric Model (MSSM) has a minimal field content, the former fields form the minimal set that we are looking for.

Their supersymmetric couplings are described by the superpotential, which is an analytic function of the (super)fields. We may form quadratic terms using the Higgs fields only

W (2) = −µ H1 · H2 = µ H2 · H1

(5.1)