Supersymmetry. Theory, Experiment, and Cosmology

410 A review of the Standard Model and of various notions of quantum field theory

the first corresponding to the requirement of the scalar field being massless, the second introducing a renormalization scale µ. These determine δm2 and δλ. It then follows that

or alternatively, the renormalized coupling is expressed in terms of the bare coupling as

Thus, the beta function, which describes the dependence of the renormalized coupling with respect to the renormalization scale, reads

More generally, in a theory involving fields of various spins, the e ective potential

In this expression, M φ are field-dependent masses and the supertrace is defined as:

J

A.5.4 Conformal anomaly

As we have seen in the previous example, renormalization introduces a scale µ. The question of scale invariance must therefore be revisited at the quantum level.

For example, we see from (A.255) that the logarithmic correction breaks dilatation invariance: indeed, under φ → φ = eαφ, the e ective potential is modified as

414 A review of the Standard Model and of various notions of quantum field theory

where the last term is nonzero because on the surface at infinity Σ∞, d3σµ k3 and

If we change the variable kµ into k µ = kµ + pµ2 , we may prove that the contribution of Fig. A.10(a) to (p1 + p2)λT µνλ vanishes. Similarly for Fig. A.10(b) with k µ = kµ + pµ1 (and exchange of µ and ν). Hence

(p1 + p2)λTµνλ = − 8π12 µνλσ(p1 + p2)λ(p1 − p2)σ = 4π12 µνλσpλ1 pσ2 .

This corresponds to an anomalous conservation law for the axial current18:

which corresponds to triangle diagrams similar to Fig. A.10 with one axial current and two photons (Fµν is the photon field strength).

We may now return to a discussion of axial anomalies in the Standard Model. Since SU (2) and U (1)Y have an axial character, there is a danger to have an anomaly that would destroy the symmetry at the quantum level. For example, the diagrams of Fig. A.11 may lead respectively to a (a) SU (2) (b) mixed U (1)Y − SU (2) and

(c) U (1)Y anomaly. However, the SU (2) anomaly (a) has an overall group factor

Tr ta{tb, tc} = 12 δbcTr(ta) = 0,

17More precisely, d3σµF (k) = lim|K|→+∞ KKµ iS3(|K|)F (|K|) where S3(K) = 2π2K3.

18We note in this computation the important rˆole played by chirality and the matrix γ5. If we were working in dimensional regularization where there is no linear divergence, we would encounter a di culty defining γ5 in 4 + dimensions. Proper treatment [84] leads to the same result.

Thus, the Standard Model is free of axial anomalies and its symmetries remain good symmetries at the quantum level. It remains to understand the origin of this somewhat miraculous cancellation (see Chapter 9).

Further reading

•P. Ramond, Field theory: a modern primer, 2nd edition, Frontiers in physics, Addison-Wesley publishing company 1990.

•S. Coleman, Secret symmetry in Laws of hadronic matter, Proceedings of the 1973 International School of Subnuclear Physics, Erice, ed. A. Zichichi, pp. 138-223.

Exercises

Exercise 1 Experimentally, the neutron and proton are not degenerate in mass: mnc2 − mpc2 1.29 MeV. We have seen that isospin, which is a symmetry of strong interactions, imposes the same mass. Is the observed mass di erence an e ect of electromagnetic or weak interactions?

Hint: If the main e ect was electromagnetic, one would have mp > mn since the proton, being charged, has a larger electromagnetic mass. Thus the e ect comes mainly from weak interactions which are at the origin of quark masses: the d quark is heavier than the u and thus the neutron (ddu) is heavier than the proton (uud). Let us note that most of the nucleon mass comes from the strong interaction that confines the quarks. Quark masses only account for a small fraction of the nucleon mass.

Exercise 2 Prove (A.48).

Hint: Expand ∂µ U U −1 = ∂µ1l = 0.

416 A review of the Standard Model and of various notions of quantum field theory

Exercise 3 Polar decomposition: any complex matrix may be written as the product of a hermitian matrix and a unitary matrix. Let Λ be a general complex matrix. Since ΛΛ† is hermitian, we can introduce a unitary matrix V such that V † ΛΛ† V = D2, a real diagonal matrix. We note H ≡ V DV †, a hermitian matrix, and U ≡ H−1Λ.

(a)Prove that U = HΛ†−1.

(b)Show that U is unitary.

Exercise 4 We define two-point functions Π33µν , Π3µνB , and ΠBBµν for the SU (2) × U (1) gauge fields A3µ and Bµ as in (A.192).

(a)Express the corresponding quantities A and F in terms of the original quantities in (A.193). One will show in particular that

(b) Use these relations to obtain from (A.189) the value of δr, as given in (A.204). Hints:

(a)To prove (A.285), use Aγγ = AγZ = 0 to obtain two relations between A3B , ABB and A33.

Exercise 5 Compute explicitly the leading contribution to the electroweak parameter1 in the Standard Model (i.e. the first term in (A.205)).

Exercise 6

(a)Prove that ∂µΦ has scaling dimension d + 1 if Φ is a field of scaling dimension d.

(b)Prove (A.216).

Hints:

(a)∂ Φ (x ) = ∂ eαdΦ(x) = eα(d+1)∂µΦ(x).

µ∂(e−αxµ)

(b)We have from (a) the infinitesimal transformation: δ∂λφ = α(2+xµ∂µ)∂λφ. Hence, for example,

Exercise 7 We consider the theory described by the action (A.214) setting for simplicity the fermion field to zero and we derive some of the properties of the energy–momentum Θµν .

(a) Check that Θµν is conserved if and only if T µν is. Prove that the generators P µ and M µν are identical, whether constructed from T µν or Θµν .

(b) We write

Θµν = T µν − 16 ∂λ∂τ Xλτ µν , Xλτ µν = gλµgτ ν − gλτ gµν φ2.

(c) ∂µ∂τ Aτ µ = 0 and ∂τ Aτ 0 = ∂iAi0 does not contribute to an integral over space.

Exercise 8 We wish to show that infinitesimal coordinate transformations x µ = xµ − ξµ(x) which belong to the conformal group of a d-dimensional spacetime (d > 2) are such that ξµ is at most quadratic in x. We assume that spacetime is flat with metric

ηµν .

(a) Deduce from (A.232) that

[ηµν + (d − 2)∂µ∂ν ] (∂ρξρ) = 0.

(b) Conclude that ξ is at most quadratic in x.

Exercise 9 We study in this exercise the model of unification of [180]. This model provides a unification of electroweak interactions within the simple gauge group SO(3). It does not require the introduction of neutral currents but of extra leptons, and was not confirmed by experiment. But it provides the simplest example where the electromagnetic U (1) symmetry is part of a larger nonabelian symmetry which allows us to consider the magnetic charge as a topological charge (see Section 4.5.3 of Chapter 4).

We consider a gauge theory with gauge group SO(3) SU (2). In the standard basis for the generators of the adjoint representation (T a)bc = −i abc, the gauge fields are Aaµ, a = 1, 2, 3. We want to interpret these fields as the intermediate vector boson Wµ± and the photon Aµ:

418 A review of the Standard Model and of various notions of quantum field theory

boson and to the photon. Show that the gauge coupling g must be equal to the electromagnetic coupling e.

3.In the Georgi–Glashow model (1972b), one introduces charged E+ and neutral E0 leptons besides the electron and the neutrino in the following representations of SO(3):

•(EL+, νL sin θGG + NL0 cos θGG , e−L ) 3

•(ER+, NR0 , e−R ) 3

•NL0 sin θGG − νL cos θGG 1

•νR 1.

Explain why by writing explicitly the couplings to the gauge bosons. Show that the coupling constant to the intermediate vector boson is gW = e sin θGG .

4.One introduces a real scalar field φa, a = 1, 2, 3, in the vector representation T a with the following action (with summation over repeated indices):

4.Dµφ± = ∂µφ± ieAµφ± ± ieWµ±φ0, Dµφ0 = ∂µφ0 − ieWµ+φ− + ieWµ−φ+.

Since 12 DµφaDµφa e2φ20 Wµ+W −µ, SO(3) is broken down to U (1) and the photon field remains massless. Out of the scalar degrees of freedom, φ± provide WL± and φ0 is the physical Higgs.