Lecture 20

Spontaneous Breaking of Non

Abelian Gauge Symmetries

We will now consider the case when the spontaneously broken non abelian symmetry is
gauged. As we saw for the case of abelian gauge symmetry, the spontaneous breaking
of the symmetry will be realized in the sense of the Anderson-Higgs mechanism, i.e. the
NGBs would not be in the physical spectrum, and the gauge bosons associated with the
broken generators will acquire mass. We will derive these results carefully in what follows.

20.1 The Generalized Anderson-Higgs Mechanism

We consider a lagrangian invariant under the gauge transformations

a (x) ta
φ(x) → eiα φ(x) , (20.1)

where ta are the generators of the group G, and the gauge fields transform as they should.
If we consider infinitesimal gauge transfomations and write out the field φ(x) in its groups
components, we have

φi (x) → δij + iαa (x) (ta )ij φj (x)

(20.2)

In general, we consider representations where the φi (x) fields in (20.2) are complex. But
for the purpose of our next derivation, it would be advantegeous to consider their real
components. So if the original representation had dimension n, we now have 2n compo-
nentes in the real fields φi (x). If this is the case, then the generators in (20.2) must be
imaginary, since the αa (x) are real paramenter functions. This means we can write them
as

1
2LECTURE 20. SPONTANEOUS BREAKING OF NON ABELIAN GAUGE SYMMETRIES

taij = i Tija , (20.3)

where the Tija are real. Also, since the ta are hermitian, we have

†
taij = taij , (20.4)

we see that

Tija = −Tjia , (20.5)

so the T a are antisymmetric. In general, the lagrangian of the gauge invariant theory for
a scalar field in terms of the real scalar degrees of freedom would be1

1 
µ


L= Dµ φi D φi − V φi , (20.6)
2

where the repeated i indices are summed. We can write the covariant derivatives above
as

Dµ φ(x) = ∂µ − igAaµ (x)ta φ(x) = ∂µ + gAaµ (x)T a φ(x) ,

(20.7)

where we omitted the group indices for the fields and the generators. We are interested
in the situation when the potential in (20.6) induces spontaneous symmetry breaking.
To see how this affects the gauge boson spectrum we must examine in detail the scalar
kinetic term:

1 1 1
Dµ φi Dµ φi = ∂µ φi ∂ µ φi + g 2 Aaµ Abµ T a φ i T b φ i





2 2 2
(20.8)
gAaµ a
T φ i ∂ µ φi ,


+

where we used the notation

T a φ i = Tija φj ,


(20.9)
1
Here we concentrate on the scalar sector of L since it is here that SSB of the gauge symmetry arises.
We can imagine adding fermion terms to L coupling them both to the gauge bosons through the covariant
derivative, as well as Yukawa couplings between the fermions and the scalars. Of course, all these terms
must also respect gauge invariance.
20.1. THE GENERALIZED ANDERSON-HIGGS MECHANISM 3

and as usual repeated group indices i, j are summed. If the potential V (φi ) has a non triv-
ial minimum then, the vacuum expectation value (VEV) of the fields φi at the minimum
is

h0|φi |0i = hφi i ≡ φ0 i
, (20.10)

which says that we are signling out directions in field space which may have non trivial
VEVs. Then the terms in L quadratic in the gauge boson fields, i.e. the gauge boson
mass terms, can be readily read off (20.8):

1 2 a bµ
Lm = Mab Aµ A , (20.11)
2

where the gauge boson mass matrix is defined by

2
≡ g 2 T a φ0 T b φ0



Mab i i
. (20.12)

Since the T a ’s are real, the non zero eigenvalues of Mabv

2
are definite positive. We can
clearly see now that if

T a φ0 = 0, (20.13)

then the associated gauge boson Aaµ remains massless. That is, theunbroken generators,
which as we saw in the previous lecture, do not have NGBs associated with them, do not
result in a mass term for the corresponding gauge boson. On the other hand, if

T a φ0 6= 0 , (20.14)

then we see that this results in a gauge boson mass term. The generators satisfying
(20.14) are of course the broken generators which result in massless NGBs. However, just
as we saw for the abelian case, these NGBs can be removed from the spectrum by a gauge
transformation. To see how this works we consider the last term in (20.8), the mixing
term. This is

Lmix. = gAaµ T a φ0 i ∂ µ φi .


(20.15)

Thus, we see that if the associated generator is broken, i.e. (20.14) is satisfied, then
there is mixing of the corresponding gauge boson with the massless φi fields, the NGBs.
It is clear that, just as in the abelian case, we can eliminate this term by a suitable
gauge transformation on Aaµ . This would still leave the mass term unchanged, but would
4LECTURE 20. SPONTANEOUS BREAKING OF NON ABELIAN GAUGE SYMMETRIES

= + +

(a) (b) (c)

Figure 20.1: Contributions to the gauge boson two point function in the presence of
spontaneous gauge symmetry breaking. Diagram (a) includes the tree level as well as
loop diagrams, all of which are transverse contributions. Diagram (b) is the contribution
from the gauge boson mass term. Diagram (c) depicts the contribution from the massless
NGBs.

completely eliminate the NGBs mixing in (20.15) from the spectrum. But even if we
leave the NGBs in the spectrum, and we still have to deal with the mixing term (20.15),
we can still see that the gauge boson two point function remains transverse, a sign that
gauge invariance is still respected despite the appearance of a gauge boson mass. This is
depicted in Figure 20.1.
In order to obtain diagram (c) we need to derive the Feynman rule resulting from the
mixing term Lmix (20.15). In momentum space this becomes

q a
= g T a φ0 i q µ ,


(20.16)

where the NGB momentum is flowing out of the vertex (its sign changes if it is flowing
into the vertex). The contributions to diagram (a) are transverse as they come from
either the leading order propagator or the loop corrections to it, both already shown
to be transverse. Then the two point function for the gauge boson in the presence of
spontaneous symmetry breaking is

iδab
Πµν = Π(a) 2 a b




µν + iMab gµν + g T φ0 i qµ g T φ 0 i
(−q ν
q2
q µ aν
= Π(a) 2


µν + iMab gµν − , (20.17)
q2

where to obtain the second line we used (20.12). Then, just as we saw for the abelian
case, we see that the gauge boson two point function is transverse even in the presence of
gauge boson masses.
20.1. THE GENERALIZED ANDERSON-HIGGS MECHANISM 5

Example 1: SU (2)
In this first example we gauge the SU (2) of the first example in the previous lecture. The
lagrangian

† 1 a aµν
L = Dµ φ Dµ φ − V (φ† φ) − Fµν F , (20.18)
4

with the covariant derivative on the scalar field is2

Dµ φ(x) = ∂µ − igAaµ (x)ta φ(x) ,

(20.19)

where the SU (2) generators are given in terms of the Pauli matrices as

σa
ta = , (20.20)
2
with a = 1, 2, 3. Since they transform according to

a (x)ta
φ(x)j → eiα jk φk (x) , (20.21)

with j, k = 1, 2, then that are doublets of SU (2). Since each of the φj (x) are complex
scalar fields, we have four real scalar degrees of freedom. We will consider the vacuum

 
0
hφi = √v
, (20.22)
2

such that, as required by imposing a non trivial minimum, we have

v2
hφ† φi = , (20.23)
2
where the factor of 2 above is chosen for convenience. We are particularly interested in
the gauge boson mass terms. These can be readily obtained by substituting the vacuum
value of the field in the kinetic term. This is

†
Lm = Dµ hφi Dµ hφi ,
g 2 a bµ
 

ab 0
= A A 0 v t t , (20.24)
2 µ v
2
We have gone back to complex scalar fields for the remaining of the lecture.
6LECTURE 20. SPONTANEOUS BREAKING OF NON ABELIAN GAUGE SYMMETRIES

where we used (20.22) in the second line. But for the case of SU (2) we can use the fact
that

σ a , σ b = 2δ ab ,

(20.25)

which translates into

1
ta , tb = δ ab , .

(20.26)
2

Then, if we write

1 a bµ a b 1 b aµ b a
Aaµ Abµ ta tb = A A t t + Aµ A t t
2 µ 2
1 a bµ  a b 1
= Aµ A t , t = Aaµ Aaµ , (20.27)
2 4

where in the last euality we used (20.26). Then we obtain

1
Lm = g 2 v 2 Aaµ Aaµ , (20.28)
8

which results in a gauge boson mass of

gv
MA = . (20.29)
2

Notice that all three gauge bosons obtain this same mass. It is interesting to compare
this result with what we obtained in the previous lecture for the spontaneous breaking
of a global SU (2) symmetry using the same vacuum as in (20.22). In that case, we saw
that all generators were broken, i.e. there are three massless NGBs in the spectrum and
the SU (2) is completely (spontaneously) broken in the sense that none of its generators
leaves the vacuum invariant. In the case here, where the SU (2) symmetry is gauged, we
see that all three gauge bosons get masses. This is in fact the same phenomenon: none
of the gauge symmetry leaves the SU (2) vacuum (20.22) invariant. However, the end
result is three massive gauge bosons, not three massless NGBs. We argued in our general
considerations above that, just as for the abelian case before, the NGBs can be removed
by a gauge transformation. Let us see how this can be implemented. We consider the
following parametrization of the SU (2) doublet scalar field:
20.1. THE GENERALIZED ANDERSON-HIGGS MECHANISM 7

 
0
a (x)ta /v
φ(x) = eiπ   , (20.30)
v+σ(x)
√
2

where σ(x) and π a (x) with a = 1, 2, 3 are real sclalar fields satisfying

hσ(x)i = 0 = hπ a (x)i , (20.31)

so that this choice of parametrization is consistent with the vacuum (20.22). Clearly, the
potential will not depend on the π a (x) fields

m2 † λ †
2
V (φ† φ) = − φ φ+ φφ , (20.32)
2 2

The minimization results in3

m2
hφ† φi = , (20.33)
2λ

which results in

m2
v2 = . (20.34)
λ

Replacing this in the potential (20.32) we obtain

√
mσ = 2λv . (20.35)

And of course, the implicit result of having

mπ1 = mπ2 = mπ3 = 0 . (20.36)

But how do we get rid of the massless NGBs ? If we define the following gauge transfor-
mation

a (x)ta /v
U (x) ≡ e−iπ (20.37)
3
√
Notice the different factor in the denominator of the second term. This is due to the factor of 2 in
the definition of the vacuum.
8LECTURE 20. SPONTANEOUS BREAKING OF NON ABELIAN GAUGE SYMMETRIES

under which the fields transform as

 
0
φ(x) → φ0 (x) = U (x) φ(x) =   ,
v+σ(x)
√
2
(20.38)
i
Aµ → A0µ = U (x) Aµ U −1 (x) − ∂µ U (x) U −1 (x) ,


g

where we used the notation Aµ = Aaµ ta . It is clear from the first transformation above,
that φ0 (x) does not depend on the π a (x) fields. Thus, the gauge transformation (20.38)
has removed them from the spectrum completely. However, the number of degrees of
reedom is the same in boths gauges. We had three transverse gauge bosons (i.e. 6 degrees
of freedom) and four real scalar fields. In this new gauge we have three massive gauge
bosons (i.e. 9 degrees of freedom) plus one real scalar, σ(x). The total number of degrees
of freedom is always the samee. The gauge were the NGBs diissapear of the spectrum is
called the unitary gauge.

Example 2: SU (3)
We now consider the caso with G = SU (3), and the vacuum is chosen to be

 
0
hφi =  0  . (20.39)
√v
2

The SU (3) generators ta , with a = 1, · · · , 8, where explicitely written in the previous

lecture. They generaly satisy the group algebra and normalization

[ta , tb ] = if abc tc
1 ab
Tr ta tb =
 
δ .
2

In addition to these, we can write the anticommutator as

1
ta , tb = δ ab + dabc tc ,

(20.40)
3
20.1. THE GENERALIZED ANDERSON-HIGGS MECHANISM 9

where the dabc are totally symmetric constants4 . We want to compute the gauge boson
masses. Just as we did for the SU (2) case in (20.24), the gauge boson mass terms come
from the kinetic terms with φ replaced by the vacuum (20.39).

 
2 0

† g a bµ
= Dµ hφi Dµ hφi = Aµ A 0 0 v ta tb  0 


Lm
2
v
 
0
g 2 a bµ
1 a b
= Aµ A 0 0 v t ,t  0  (20.41)
2 2
v

wher in the second line we used the same trick as in SU (2) to write the anticommutator.
Then, using (20.40) we obtain

 
2 0
g
n1 o
Lm = Aa Abµ 0 0 v δ ab + dabc tc  0  . (20.42)
4 µ 3
v

In order to proceed further in the computation if the gauge boson masses we need to know
the values of the dabc constants for SU (3). The non zero values are

1
d118 = d228 = d338 = −d888 = √
3
1
d448 = d558 = d668 = d778 = − √
2 3
1
d146 = d157 = −d247 = d256 = d344 = d355 = −d366 = −d377 = . (20.43)
2

However, when it comes to the masses of the gauge bosons for a, b = 1, 2, 3 there is a
simpler way to compute them. We know from the previous lecture that these geerators
can be written as

 
i 1 σi 0
t = , (20.44)
2 0 0 0

with the σ i the Pauli matrices for i = 1, 2, 3. We also know that they annihilate the
vacuum defined in (20.39), that is
4
In general, this expression is valid for SU (N ) groups, with the 3 replaced by N . For the particular
case of N = 2 the dabc vanish.
10LECTURE 20. SPONTANEOUS BREAKING OF NON ABELIAN GAUGE SYMMETRIES

ti hφi = 0 , (20.45)

for i = 1, 2, 3. Thus, just by looking at the first line in (20.41) we can see that the gauge
bosons A1µ , A2µ and A3µ will be massless, that is5

MA1 = MA2 = MA3 = 0 . (20.46)

Once again, it is interesting to compare with the spontaneous breaking of the global SU (3)
symmetry studied in the previous lecture. There we saw that the first three generators
of SU (3) still left the vacuum invariant, and therefore there were no massless NBGs
associated with them. Here, the fact that the SU (2) subgroup of SU (3) defined by
the generators in (20.44) still preserve the vacuum (20.39) results in those gauge bosons
remaining massless. It is in this sense that the SU (2) remains unbroken.
Conversely, we expect that the broken generators, i.e. those that do not annihilate the
vacuum (20.39), will be associated with massive gauge bosons, as they were associated
with the existence of massless NGBs in the case of the global symmetry. To obtain their
masses we first observe that the off diagonal mass terms in (20.42), i.e. those for a 6= b
are zero, even for a, b = 4, 5, 6, 7, 8. To see this, we write the corresponding mass terms as


2 0
g
La6m=b = Aaµ Abµ 0 0 v dabc tc  0  .


(20.47)
4
v

The terms contributing are those with d’s in the last line of (20.43). For these, the
generators tc appearing in (20.47) correspond to c = 4, 5, 6 and 7. But these generators,
when acting on the vacuum in (20.47) do not leave it invariant. They will result in a
column vector that is orthogonal to the vacuum in (20.39), i.e. will give zero when it
hits the row vector (0 0 v). Then, all these non diagonal mass terms vanish. Finally
we consider the diagonal mass terms, i.e. a = b = 4, 5, 6, 7, 8. Their contributions come
both from the δ ab as well as from the d’s in the second line in (20.43). Let us explicitely
compute the case of a = b = 4. We have

 
2 0
a=b=4 g
n1 o
Lm = A4 A4µ 0 0 v 448 8
+d t  0  . (20.48)
4 µ 3
v

When substituting for the values of d448 and for t8 (see previous lecture) we obtain
5
It is clear that all the elements of the 3 × 3 mass matrix will vanish, both diagonal and non diagonal.
20.2. THE ELECTROWEAK STANDARD MODEL 11

a=b=4 1 2 2 4 4µ
Lm = g v Aµ A , (20.49)
8

which results in

gv
MA4 = . (20.50)
2

However, since all the dab8 with a = b 6= 8 have the same value and only t8 contributes,
then it is clear that we will have

gv
MA4 = MA5 = MA6 = MA7 = . (20.51)
2

Finally, taking into account the different value of d888 , we obtain

gv
MA8 = √ . (20.52)
3

These completes the spectrum of gauge bosons: five massive, and 3 massless ones. Just as
we expect from the fact that the spontaneous symmetry breaking induced by the vacuum
(20.39) respects SU (2). We say that the spontanous symmetry breaking pattern is

SU (3) → SU (2) . (20.53)

20.2 The Electroweak Standard Model

For an example of spontaneous breaking of a gauge symmetry we look no further than to
the standard model (SM) of particle physics. What is broadly called the SM is a quantum
field theory that describes all the interactions of all the known elementary particles, with
the exception of a quantum description of gravity. These are the strong interactions as
described by quantum chromodynamics (QCD), an unbroken SU (3) gauge theory; and
the electroweak interactions as described by the gauge symmetry SU (2)L × U (1)Y . It is
the latter that is spontaneously broken to give rise to three massive weak gauge bosons
and a massless photon responsible for the electromagnetic interactions. Since the QCD
gauge symmetry is not spontaneously broken by the Anderson-Higgs mechanism, we will
ignore it in what follows, it is a “spectator” interaction that does not change when the
electroweak symmetry is broken to electromagnetism.
12LECTURE 20. SPONTANEOUS BREAKING OF NON ABELIAN GAUGE SYMMETRIES

20.2.1 Why SU (2)L × U (1)Y ?

The electroweak gauge group is a product of an SU (2) gauge symmetry and a U (1) one.
How do we know this is the electroweak gauge symmetry group ? And, what is the
meaning of the subscripts L and Y ? Although we will not attempt to go through all the
fascinating history of experimental evidence that goes into building the electroweak SM,
it is worthwhile to understand the main points that result in the building of this strange
looking, yet amazingly successful description of fundamental physics.
Let us review the main evidences leading to the gauge structure of the electroweak theory.

• Weak Interactions (Charged): Weak decays, such as β decays n → p e− ν̄e or µ− →

νµ ν̄e e− among many others, are mediated by charged currents. Let us look at the
case of muon decay. It is very well described by a four fermion interaction, i.e.
with a non renormalizable coupling GF , the Fermi constant. In fact, all other weak
interactions can be described in this way with the same Fermi constant ( to a very
good approximation, more later). The relevant Fermi lagrangian is

GF
LFermi = √ µ̄L γµ νL ēL γ µ νe ,



(20.54)
2

where we already included the fact that the charged weak interactions only involve
left handed fermions. That is, the phenomenolgically built Fermi lagrangian above
tells us that the weak decay of a muon is described by the product of two charged
vector currents coupling only left handed fermions. The fact that only left handed
fermions participate in the charged weak interactions is an experimentally estab-
lished fact, observed in all charged weak interactions. This is done by a variety of
experimental techniques. Fors instance, in the case of muon decay, the angular distri-
bution of the outgoing electron is very different if this is left or right handed. Precise
measurements (performed over decades of increasingly accurate experiments) have
concluded that the outgoing electron is left handed only. The different couplings
involving left and right handed fermions require parity violation. Moreover, the
charged weak interactions require maximal parity violation: only one handedness
participate. If we assume that the non renormalizable four fermion interaction is
the result of integrating out a gauge boson with a renormalizable interaction, this
would point to the need of 2 charged gauge bosons.This is schematically shown in
Figure 20.2. Assuming that mµ  MW , we integrate out the massive vector gauge
boson to obtain

GF g2
√ = 2
, (20.55)
2 8MW

where g is the renormalizable coupling of the gauge bosons to fermions in diagram

20.2. THE ELECTROWEAK STANDARD MODEL 13

− −
e−
W−
− e−
e

−
e

(a) (b)

Figure 20.2: Diagram (a) is the Feyman diagrams associated with the four fermion Fermi
lagrangian (20.54). Diagram (b) shows the corresponding exchange of a massive charged
gauge boson, Wµ± .

(b). The charged vector gauge bosons, W ± were discovered in the 1980s and studied
with gerat detail ever since.

• Weak Neutral Currents: In addition to the charged currents described by (20.54),

we have know since experimental evidence first appeared in the 1970s, that there
are also neutral weak currents. These were first observed by neutrino scattering
off nucleons. Normally, the charged currents would result in νe N → e− N 0 , with
N and N 0 protons and neutrons. This is just a crossed diagram of β decay. But
the reaction ν N → ν N was also observed. Many other reactions involving neutral
currents have been observed since then. They also violate parity. However, they do
not do so maximally. This menas that the neutral current, or the vector gauge boson
that we need to integrate out to obtain them at low energies, couple differently to
lef and right handed fermions but, unlike the charged currents, the do couple to
right handed fermions. The neutral vector gauge boson, Z 0 , was also discovered in
the 1980s and its properties studied with great precision.

• Electromagnetism: Of course, we know that the electromagnetic interactions are

described by a quantum field theory, QED, mediated by a neutral massless vector
gauge boson, the photon. One important feature to remember is that the photon
coupling in QED is parity invariant. No parity violation is present in QED.

The elements described above suggest that we need: 4 gauge bosons for a unified
description of the weak and electromagnetic interactions. Three of them are massive,
one (the photon) must remain massless after spontaneous symmetry breaking. The SM
gauge group is then G = SU (2) × U (1) which matches the number of gauge bosons.
However, we know that two of these only couple to left handed fermions. whereas one
of the massive ones (the neutral) couples differently to left and right handed fermions.
Finally, the photon must remain massless and its couplings parity invariant. The choice
of gauge group is then
14LECTURE 20. SPONTANEOUS BREAKING OF NON ABELIAN GAUGE SYMMETRIES

G = SU (2)L × U (1)Y , (20.56)

where the three gauge bosons couple to left handed fermions only, and the U (1)Y is not
identified with the U (1)EM , the abelian gauge symmetry responsible for electromagnetism.
As we will see below, two of the SU (2)L gauge bosons will result in the Wµ± . On the other
hand to obtain the Z 0 and the photon we will need to carefully choose the pattern of
spontaneous symmetry breaking of the SM gauge group in (20.56) down to U (1)EM .

20.2.2 Spontaneous Breaking of SU (2)L × U (1)Y → U (1)EM

All matter in the SM (i..e. fermions and scalars) will have some transformation property
under the SM gauge group G. This means that we need to assign to each particle a charge
under the abelian group factor, the U (1)Y . This is called hypercharge, since it is not quite
the electric charge.
We first consider a scalar field Φ in the fundamental representation of SU (2) and with
hypercharge U (1)Y ,

YΦ = 1/2 . (20.57)

The scalar doublet can be written as

 
φ+
Φ= , (20.58)
φ0

where φ+ and φ0 are complex scalar fields, resulting in four real scalar degrees of freedom6 .
Under a SU (2)L × U (1)Y gauge transformation, the Higgs doublet transforms as

a (x)ta
Φ(x) → eiα eiβ(x)YΦ , Φ(x) , (20.59)

where ta are the SU (2)L generators (i.e. Pauli matrices divided by 2), αa (x) are the three
SU (2)L gauge parameters, β(x) is the U (1)Y gauge parameter, and it is understood that
the U (1)Y factor of the gauge transformation contains a factor of the identity I2×2 after
the hypercharge YΦ . Thus, the covariant derivative on Φ is given by

 
Dµ Φ(x) = ∂µ − igAaµ (x)ta − ig 0 Bµ (x)YΦ I2×2 Φ(x) . (20.60)

6
At this point, the labels “+” and “0” are just arbitrary, since we have not even defined electric
charges But these label will be consistent in the future, after we have done this.
20.2. THE ELECTROWEAK STANDARD MODEL 15

Here, Aaµ (x) is the SU (2)L gauge boson, Bµ (x) the U (1)Y gauge boson, and g and g 0 are
their corresponding couplings. The lagrangian of the scalar and gauge sectors of the SM
is then

† 1 a aµν 1
L = Dµ Φ Dµ Φ − V (Φ† Φ) − Fµν F − Bµν B µν , (20.61)
4 4
a
where Fµν is the usual SU (2) field strength built out of the gauge fields Aaµ (x) and Bµν is
the U (1)Y field strength given by the abelian expression

Bµν = ∂µ Bν (x) − ∂ν Bµ (x) . (20.62)

As usual, we consider the potential

2
V (Φ† Φ) = −m2 Φ† Φ + λ Φ† Φ ,


(20.63)

which is minimized for

m2 v2
hΦ† Φi = ≡ . (20.64)
2λ 2
In order to fulfil this, we choose the vacuum

 
0
hΦi = √v
. (20.65)
2

Just as in the previous examples of SSB of non abelian gauge symmetries, the next
question is what is the symmetry breaking pattern, i.e. what gauge bosons get what
masses, if any. In particular, we want one of the four gauge bosons in G to remain
massless after imposing the vacuum hΦi in (20.65). This means that there must be a
generator or, in this case, a linear combination of generators of G that annihilates hΦi,
leaving the vacuum invariant under a G transformation. This combination of generators
must be associated with the massless photon in U (1)EM , the remnant gauge group after
the spontaneous breaking. One trick to identify this combination of generators is to
consider the gauge transformation defined by

α1 (x) = α2 (x) = 0
α3 (x) = β(x) . (20.66)

The exponent in the gauge transformation has the form

16LECTURE 20. SPONTANEOUS BREAKING OF NON ABELIAN GAUGE SYMMETRIES

   
3 3 β(x) h 1 0 1 0 i
iα (x)t + iβ(x)YΦ I2×2 = i +
2 0 −1 0 1
 
iβ(x) 1 0
= . (20.67)
2 0 0

Then we see that this combination

t3 + YΦ hΦi = 0 ,


(20.68)

indeed annihilates the vacuum, leaving it invariant. Thus, we suspect that this linear
combination of SU (2)L × U (1)Y generators must be associated with the massless photon.
We will come back to this point later.
We now go to extract the gauge boson mass terms from the scalar kinetic term in (20.61).
This is

†
Lm = Dµ hΦi Dµ hΦi
 
1 a a 0

bµ b 0 µ

0
= (0 v) gAµ t + g YΦ Bµ gA t + g YΦ B . (20.69)
2 v

For the product of the two SU (2) factors we will use the trick in (20.27). Then, the
only terms we need to be careful about are the mixed ones: one SU (2) times one U (1)Y
contribution. There are two of them, and each has the form

σ3 1 v2 0 3 µ
 
1 0
(0 v) g g 0 YΦ =− g g Aµ B , (20.70)
2 2 v 2 4

where in the second equality we used YΦ = 1/2. We then have

1 v 2 n 2 1 1µ 2 2 2µ 2 3 3µ 02 µ 0 3 µ
o
Lm = g Aµ A + g Aµ A + g Aµ A + g Bµ B − 2gg Aµ B . (20.71)
2 4

From this expression we can clearly ee that A1µ and A2µ acquire masses just as we saw in
the pure SU (2) example. It will be later convenient to define the linear combinations

A1µ ∓ iA2µ
Wµ± ≡ √ , (20.72)
2
20.2. THE ELECTROWEAK STANDARD MODEL 17

which allows us to write the first two terms in (20.71) as

g 2 v 2 + −µ
LW
m = Wµ W . (20.73)
4

These two states have masses

gv
MW = . (20.74)
2

On the other hand, the fact that A3µ and Bµ have a mixing term prevents us from reading
off masses. We need to rotate these states to go to a bases without mixing, a diagonal
basis. In order to clarify what needs to be done, we can write the last three terms in
(20.71 in matrix form

1 v2 3 −g g 0
  
g2 A3µ
Lneutral = (A Bµ ) , (20.75)
m
2 4 µ −g g 0 g 02 Bµ

where the task is to find the eigenvalues and eigenstates of the matrix above. It is clear
that one of the eigenvalues is zero, since the determinant vanishes. Then the squared
masses of the physical neutral gauge bosons are

Mγ2 = 0
(20.76)
2
v
MZ2 = (g 2 + g 02 )
4

The eigenstates in terms of A3µ and Bµ , the original SU (2)L and U (1)Y gauge bosons
respectively, are

1
g 0 A3µ + gBµ


Aµ ≡ p (20.77)
g 2 + g 02

1
gA3µ − g 0 Bµ .


Zµ ≡ p (20.78)
g 2 + g 02

Alternatively, we could have obtained the same result by defining an orthogonal rotation
matrix to diagonalize the interactions above. That is, rotating the states by
18LECTURE 20. SPONTANEOUS BREAKING OF NON ABELIAN GAUGE SYMMETRIES

    
Zµ cos θW − sin θW A3µ
= , (20.79)
Aµ sin θW cos θW Bµ

results in diagonal neutral interactions if we have

g g0
cos θW ≡ p , sin θW ≡ p , (20.80)
g 2 + g 02 g 2 + g 02

where θW is called the Weinberg angle. It is useful to invert (20.79) to obtain

A3µ = sin θW Aµ + cos θW Zµ (20.81)

Bµ = cos θW Aµ − sin θW Zµ . (20.82)

Using these expressions for A3µ and Bµ we can replace them in the covariant derivative
acting on the scalar doublet Φ. Their contribution fo Dµ is

−igA3µ t3 − ig 0 YΦ Bµ = −iAµ g sin θW t3 + g 0 cos θW YΦ − i g cos θW t3 − g 0 sin θW YΦ Zµ

(20.83)
  g  
= −ig sin θW t3 + YΦ Aµ − i t3 − (t3 + YΦ ) sin2 θW Zµ ,
cos θW

where it is always understood that the hypercharge YΦ is always multiplied by the identity,
and in the last identity we used the fact that

g 0 cos θW = g sin θW , (20.84)

and trigonometric identities. We can conclude that is Aµ is to be identified with the

photon field, then its coupling must be e times the charged of the particle it is coupling
to (e.g. −1 for an electron. Thus we must impose that

e = g sin θW , (20.85)

and that the charge operator, acting here on the field Φ coupled to Aµ is defined as

Q = t3 + YΦ . (20.86)
20.2. THE ELECTROWEAK STANDARD MODEL 19

Then we can read the photon coupling to the doublet scalar field Φ from

 
φ+
−i e Aµ Q Φ(x) = −i e Aµ Q . (20.87)
φ0

Substituting YΦ = 1/2 we have

      
φ+ 1 0 φ+ φ+
Q = = , (20.88)
φ0 0 0 φ0 0

which tells us that the top complex field in the scalar doublet has charge equal to 1 (in
units of e, the proton charge), whereas the bottom component has zero charge, justifying
our choice of labels. On the other hand, we see that fixing Q to be the electromagnetic
charge operator, completely fixes the couplings of Zµ to the scalar Φ. This is now, from
(20.83),

g  
−i Zµ t3 − Q sin2 θW Φ . (20.89)
cos θW

We will see below that the choice of fixing the Aµ couplings to be those of electromag-
netism, fixes completely the Zµ to all elementary couplings, giving a wealth of predictions.

20.2.3 Gauge Couplings of Fermions

The SM is a chiral gauge theory, i.e. its gauge couplings differ for different chiralities. To
extract the left handed fermion gauge couplings, we look at the covariant derivative

Dµ ψL = ∂µ − igAaµ ta − ig 0 YψL Bµ ψL ,


(20.90)

where YψL is the left handed fermion hypercharge. On the other hand, since right handed
fermions do not feel the SU (2)L interaction, their covariant derivative is given by

Dµ ψR = ∂µ − ig 0 YψR Bµ ψR ,


(20.91)

with YψR its hypercharge. Using the covariant derivatives above, we can extract the
neutral and charged couplings. We start with the neutral couplings, which in terms of
the gauge boson mass eigenstates are the couplings to the photon and the Z.

Neutral Couplings: From (20.90), the neutral gauge couplings of a left handed fermions
are
20LECTURE 20. SPONTANEOUS BREAKING OF NON ABELIAN GAUGE SYMMETRIES

−igt3 A3µ − ig 0 YψL Bµ ψL = ig sin θW t3 + YψL Aµ ψL

g 2 t3 − ig 02 YψL


− i p Zµ ψL , (20.92)
g 2 + g 02

where on the right hand side we made use of (20.81) and (20.82). Now, we know that the
photon coupling should be

−ie QψL , (20.93)

with QψL the fermion electric charge operator. Thus, we must identify

QψL = t3 + YψL , (20.94)

as the fermion charge. We can use our knowledge of the fermion charges to fix their
hypercharges. As an example, let us consider the left handed lepton doublet. For the
lightest family, this is written in the notation

 
νeL
L= . (20.95)
e−
L

The action of t3 on L is

  
3 1/2 0 νeL
t L =
0 −1/2 e−
L
   3 
(1/2) νeL tνe νeL
= ≡ L , (20.96)
(−1/2) e−L t3eL e−
L

where in the last equality we defined t3νeL = 1/2 and t3eL as the eigenvalues of the operator
t3 associated to the electron neutrino and the electron. Then, we have

    
1/2 + YL 0 νeL (1/2 + YL ) νeL
QL L = = . (20.97)
−1/2 + YL 0 e−
L (−1/2 + YL ) eL

But we know that the eigenvalue of the charge operator applied to the neutrino must
be zero, as well as that the eigenvalue of the electron must be −1. Thus, we obtain the
hypercharge of the left handed lepton doublet
20.2. THE ELECTROWEAK STANDARD MODEL 21

1
YL = − , (20.98)
2

which is fixed to give us the correct electric charges for the members of the doublet L.
We can do the same with the right handed fermions. These, however do not have t3 in
the covariant derivative (see (20.91) ). Then, for e−
R , the right handed electron, we have
3
that teR = 0, which means that, since

Qe− = −1 , (20.99)
R

then the right handed electron’s hypercharge is equal to it:

Ye− = −1 . (20.100)
R

Similarly, the right handed electron neutrino has zero electric charge, which results in

YνR = 0 . (20.101)

Now that we fixed all the lepton hypercharges by imposing that they have the QED
couplings to the photon, we can extract their couplings to the Z as predictions of the
electroweak SM. From (20.92) we have

g
−i g cos θW t3 − g 0 sin θW Yψ Zµ ψ = −i cos2 θW t3 − sin2 θW Yψ Zµ ψ



cos θW
g
t3 − sin2 θW Qψ Zµ ψ , (20.102)


= −i
cos θW

where the initial expressios makes use of cos θW and sin θW in terms of g and g 0 , in the
first equality we used that tan θW = g 0 /g and, in the final equality, we used that in general
Qψ = t2 + Yψ , independently of the fermion chirality, as long as we generalize (20.94) for
right handed fermions using t3ψR = 0. For instance, from (20.102) we can read off the
lepton couplings of the Z boson. These are,
22LECTURE 20. SPONTANEOUS BREAKING OF NON ABELIAN GAUGE SYMMETRIES

g 1
νe L : −i
cos θW 2
g  1 
e−
L : −i − + sin2 θW
cos θW 2
(20.103)
g  
e−
R : −i − sin2 θW
cos θW
νeR : 0.

From the couplings above, we see that every lepton has a different predicted coupling to
the Z. These are, of course, three level predictions. Measurements of these Z couplings
have been performed with subpercent precision for a long time, and the SM predictions
for the fermion gauge couplings have passed the tests every time. Another, interesting
point, is that right handed neutrinos have no gauge couplings in the SM: no Z coupling,
certainly no electric charge and no QCD couplings. Thus, from the point of view of the
SM, the right handed neutrino need not exist.

Charged Couplings:
We complete here the derivation of the gauge couplings of leptons by extracting their
charged couplings.These come from the SU (2)L gauge couplings, as we see from

0 Wµ+
 
g
A1µ t1 A2µ t2


−ig + = −i √ , (20.104)
2 Wµ− 0

which then involve only left handed fermions. Then, from the gauge part of the left
handed doublet kinetic term

LL = L̄i6DL , (20.105)

we obtain their charged couplings

0 Wµ+
  
g νeL
Lch. = (ν̄eL ēL ) γ √µ
L
2 Wµ− 0 eL
(20.106)
g n o
= √ ν̄eL γ µ eL Wµ+ + ēL γ µ νeL Wµ− ,
2

where we can see that, as required by hermicity, the second term is the hermitian conjugate
±
of the first. The Fermi lagrangian can be obtained from Lch. L by integrating out the W
gauge bosons.
20.2. THE ELECTROWEAK STANDARD MODEL 23

We now briefly comment on the electroweak gauge coulpings of quarks. Just as for leptons,
we concentrate on the first family. The left handed quark doublet is

 
uL
qL = , (20.107)
dL

We know that, independently of helicity, the charges of the up and down quarks are
Qu = +2/3 and Qd = −1/3. Then we have

 

3
 +2/3 0
QqL = t + YqL = , (20.108)
0 −1/3

which results in

1
YqL = . (20.109)
6

The hypercharge assignments for the right handed quarks are again trivial and given by
the quark electric charges. We have

2 1
YuR = + , YdR = − . (20.110)
3 3

With these hypercharge assignments we can now write the quark couplings to the Z.
Using (20.102) we obtain

g 1 2
uL : −i − sin2 θW
cos θW 2 3
g  1
2 1
dL : −i − + sin θW
cos θW 2 3
g  2 
uR : −i − sin2 θW
cos θW 3
g  1 
dR : −i sin2 θW . (20.111)
cos θW 3

Once again, we see that each type of quark has a different coupling to the Z. All of these
predictions have been tested with great precision, confirming the SM even beyond leading
order.
The charged gauged couplings of left handed quarks are trivial to obtain: they are dictated
by SU (2)L gauge symmetry and therefore there must be the same as those of the left
handed leptons in (20.106). So we have
24LECTURE 20. SPONTANEOUS BREAKING OF NON ABELIAN GAUGE SYMMETRIES

g n o
¯L γ µ uL W − .
Lch.
q = √ ū L γ µ
d L W +
µ + d µ (20.112)
2

20.2.4 Fermion Masses

We have seen that SSB leads to masses for same of the gauge bosons, preserving gauge
invariance. We now direct our attention to fermion masses. In principle these terms

Lfm = mψ ψ̄L ψR + h.c. , (20.113)

are forbidden by SU (2)L × U (1)Y gauge invariance since thy are not invariant under

a a
ψL → eiα (x)t eiβ(x)YψL ψL
ψR → eiβ(x)YψR ψR .

But the operator

ψ̄L Φ ψR , (20.114)

is clearly invariant under the SU (2)L gauge transformations, and it would be U (1)Y
invariant if

−YψL + YΦ + YψR = 0 . (20.115)

Since YΦ = 1/2, this form of the oprator will work for the down type quarks and charged
leptons. For instance, since YL = −1/2 and YeR = −1, the operator

Lme = λe L̄ Φ eR + h.c., (20.116)

is gauge invariant since the hypercharges satisfy (20.115). In (20.116) we defined the
dimensionless coupling λe which will result in a Yukawa coupling of electrons to the Higgs
boson. To see this, we write Φ(x) in the unitary gauge, so that
20.2. THE ELECTROWEAK STANDARD MODEL 25

 
0
Lme = λe (ν̄eL ēL )   eR + h.c.
v+h(x)
√
2
v 1
= λe √ ēL eR + λe √ h(x) ēL eR + h.c. , (20.117)
2 2

where the first term is the electro mass term resulting in

v
me = λe √ , (20.118)
2

and the second term is the Yuawa interaction of the electron and the Higgs boson h(x).
We can rewrite (20.117) as

me
Lme = me ēL eR + h(x) ēL eR + h.c. , (20.119)
v
from which we can see that the electron couples to the Higgs boson with a strength equal
to its mass in units of the Higgs VEV v. Similarly, for quarks we have that the operator

Lmd = λe q̄L Φ dR + h.c., (20.120)

is gauge invariant since YqL = 1/6 and YdR = −1/3 satisfy (20.115). Them we obtain

md
Lmd = md d¯L dR + h(x) d¯L dR + h.c. , (20.121)
v
and where the down quark mass was defined as

v
md = λd √ . (20.122)
2

As we can see, it will be always the case that fermions couple to the Higgs boson with
the strength mψ /v. Thus, the heavier the fermion, the stronger its coupling to the Higgs.
Finally, in order to have gage invariant operators with up type right handed quarks we
need to use the operator

Lmu = λu q̄L Φ̃ uR + h.c. , (20.123)

where we defined
26LECTURE 20. SPONTANEOUS BREAKING OF NON ABELIAN GAUGE SYMMETRIES

 v+h(x) 
√
2
Φ̃(x) = iσ 2 Φ(x)∗ =   , (20.124)
0

where in the last equality we are using the unitary gauge. It is straightforward7 to prove
that Φ̃(x) is an SU (2)L doublet with YΦ̃ = −1/2, which is what we need so as to make
the operator in (20.123) invariant under U (1)Y . Then we have

mu
Lmu = mu ūL uR + h(x) ūL uR + h.c. , (20.125)
v

with

v
mu = λu √ . (20.126)
2

The fermion Yukawa couplings are parameters of the SM. In fact, since there are three
families of quarks their Yuakawa couplings are in general a non diagoinal three by three
matrix. This fact has important experimental consequences. On the other hand, we
could imagina having something similar if we introduce a right handed neutrino. This
however, might be beyond the SM, since this state does not have any SM gauge quantum
numbers. Overall, the SM is determined by the paremeters v, g, g 0 and sin θW in the
electroweak gauge sector, plus all the Yukawa couplings in the fermion sector leading to
all the observed fermion masses and mixings.

Additional suggested readings

• An Introduction to Quantum Field Theory, M. Peskin and D. Schroeder, Chapter
20.

• The Quantum Theory of Fields, Vol. II, by S. Weinberg, Chapter 21.

• Quantum Field Theory , by M. Srednicki, Chapters 84 and 86.

• Quantum Field Theory in a Nutshell, by A. Zee. Chapter IV.6.

7
Only need to use that σ 2 σ 2 = 1, and that σ 2 (σ a )∗ σ 2 = −σ a .