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PHY646 - Theory and the

Even Term 2020 Dr. Anosh Joseph, IISER Mohali

LECTURE 46

Tuesday, April 14, 2020 (Note: This is an online lecture due to COVID-19 interruption.)

Topic: The II: Non-Abelian Models.

The Higgs mechanism extends straightforwardly to systems with non-Abelian gauge symmetry. It is not difficult to derive the general relation by which a set of scalar field vacuum expectation values leads to the appearance of gauge masses. Let us work out the relation and then apply it in a number of examples.

Consider a system of scalar fields φi that appear in a Lagrangian invariant under a symmetry group G, represented by the transformation

a a φi → (1iα t )ijφj. (1)

It is convenient to write φi as real-valued fields, for example, writing n complex fields as 2n real fields. Then the group representation matrices ta must be pure imaginary and, since they are Hermitian, antisymmetric. Let us write a a tij = iTij, (2) so that the T a are real and antisymmetric.

If we promote the symmetry group G to a local symmetry, the covariant derivative on the φi is

a a Dµφ = (∂µ − igAµt )φ a a = (∂µ + gAµT )φ. (3)

Then the kinetic energy term for the φi is

1 1 1 (D φ )2 = (∂ φ )2 + gAa (∂ φ T a φ ) + g2Aa Abµ(T aφ) (T bφ) . (4) 2 µ i 2 µ i µ µ i ij j 2 µ i i

Now let the φi acquire vacuum expectation values

hφii = (φ0)i, (5) PHY646 - and the Standard Model Even Term 2020

and expand the φi about these values. The last term in Eq. (4) contains a term with the structure of a gauge boson mass, 1 ∆L = m2 Aa Abµ, (6) 2 ab µ with the mass matrix 2 2 a b mab = g (T φ0)i(T φ0)i. (7)

This matrix is positive semi-definite, since any diagonal element, in any basis, has the form

2 2 a 2 maa = g (T φ0) ≥ 0 (no sum). (8)

Thus, generically, all of the gauge will receive positive masses. However, it may be that some particular generator T a of G leaves the vacuum invariant

a T φ0 = 0. (9)

In that case, the generator T a will give no contribution to Eq. (7), and the corresponding gauge boson will remain massless. As in the Abelian case, the gauge boson receives a contribution from toe Goldstone bosons, which is necessary to make the amplitude transverse. To compute this contribution, we need the vertex that mixes gauge bosons and Goldstone bosons. This comes from the second term in Eq. (4). When we insert the of the scalar field Eq. (5), this term becomes a a ∆L = gAµ∂µφi(T φ0)i. (10)

This term does not involve all of the components of φ - only those that are parallel a a to a vector T φ0 for some choice of T . These vectors represent the infinitesimal rotations of the vacuum; thus the components φi that appear in Eq. (10) are precisely the Goldstone bosons. We encountered the following diagram for the Abelian case

i  kµkν   kµkν  im2 gµν + (m kµ) (−m kν) = im2 gµν − = im2 gµν − . (11) A A k2 A A k2 A k2

Now, using the fact that these bosons are massless, we can compute the counterpart, for the non-Abelian case X   i   gkµ(T aφ ) − gkν(T bφ ) (12) 0 j k2 0 j j The sum runs over those components j with a nonzero projection onto the space spanned by a the T φ0, or equivalently well, over all j. See Fig. 1. This diagram is therefore proportional to the mass matrix Eq. (7). Combining this expression with the contribution to the vacuum polarization

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Figure 1: The Goldstone boson diagram for the non-Abelian case. It corresponds to P  µ a  i  ν b  j gk (T φ0)j k2 − gk (T φ0)j . from Eq. (6), we find a properly transverse result

 kµkν  im2 gµν − , (13) ab k2

2 where mab is given by Eq. (7). See Fig. 2.

2 µν kµkν  Figure 2: The lading-order contributions to the vacuum polarization amplitude, imab g − k2 .

Worked Example: SO(3) → SO(2) Breaking

Let us now apply this formalism to a specific example of non-Abelian gauge theories. There will be one massive gauge boson for each broken-symmetry generator.

Consider an SO(3) . We introduce 3 real scalars φi and the Lagrangian

1 1 1 λ L = − (F a )2 + (∂ φ − igAa τ a φ )2 + m2φ2 − (φ2)2. (14) 4 µν 2 µ i µ ij j 2 i 4! i

These scalars transform in the fundamental representation of SO(3). The potential is minimized for p |hφ~i| = v = 6m2/λ. (15)

By an SO(3) transformation, we can pick the direction and phase so that hφ3i = v and hφ1i =

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hφ2i = 0. That is, without loss of generality, we take

    * φ1 + 0     φ2 = 0 (16) φ3 v

This vacuum is invariant under H = SO(2) ⊂ G = SO(3), which rotates φ1 and φ2. Since SO(2) has one generator and SO(3) has three, there will be two Goldstone bosons that are eaten to form two massive gauge bosons. To see this explicitly, we can expand the Lagrangian in unitary gauge. We find

1 ∆L = m2 Aa Abµ 2 ab µ 1 = g2(T aφ ) (T bφ ) Aa Abµ 2 0 i 0 i µ 0 1 = g2 (0 0 v) T aT b 0 Aa Abµ. (17) 2   µ v

a b a b Upon using [Aµ,Aµ] = 0, we can symmetrize the matrix product T T . This gives

0 1 ∆L = g2 (0 0 v) {T a,T b} 0 Aa Abµ. (18) 4   µ v

Plugging in the SO(3) generators,

0 0 0   0 0 1 0 −1 0 1   2   3   T = 0 0 −1 ,T = i  0 0 0 ,T = 1 0 0 , (19) 0 1 0 −1 0 0 0 0 0 we see by explicit calculation that

0 a b   (0 0 v) {T ,T } 0 v is only non-zero for a = b = 1 or a = b = 2. Thus,

1 1 L = − (F a )2 + m2 (A1 A1 + A2 A2 ), (20) 4 µν 2 A µ µ µ µ

2 2 2 with mA = g v , which describes two massive gauge bosons (corresponding to generators 1 and 2) and one massless one (corresponding to the generator 3).

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It is interesting to note that this model contains both massive and massless gauge bosons, with the distinction between these bosons created by spontaneous symmetry breaking. If we interpret the massive bosons as W bosons and the massless gauge boson a the , it is tempting to interpret this theory as a unified model of weak and electromagnetic . Georgi and Glashow proposed this model as a serious candidate for the theory of weak interactions. However, Nature chooses a different model, in which the spontaneously symmetry breaking pattern is SU(2)×U(1) → U(1).

References

[1] M. E. Peskin and D. Schroeder, Introduction to Quantum Field Theory, Westview Press (1995).

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