# Spontaneous symmetry breaking. Goldstone s theorem. The Higgs

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## Transcript Of Spontaneous symmetry breaking. Goldstone s theorem. The Higgs

Classical Field Theory

Spontaneous symmetry breaking. Goldstone’s theorem. The Higgs mechanism.

Spontaneous symmetry breaking. Goldstone’s theorem. The Higgs mechanism.

Symmetry of laws versus symmetry of states.

Let us have a quick look at some of the classical ﬁeld theoretic underpinnings of “spontaneous symmetry breaking” (SSB) in quantum ﬁeld theory. Quite generally, SSB can be a very useful way of thinking about phase transitions in physics. In particle physics, SSB is used, in collaboration with the “Higgs mechanism”, to give masses to gauge bosons (and other elementary particles) without destroying gauge invariance. We will explore some of this in due time. To begin to understand spontaneous symmetry breaking in ﬁeld theory we need to reﬁne our understanding of “symmetry”, which is the goal of this section. The idea will be that there are two related kinds of symmetry one can consider: symmetry of the “laws” of motion governing the ﬁeld, and symmetries of the “states” of the ﬁeld.

So far we have been studying “symmetry” in terms of transformations of a (scalar) ﬁeld which preserve the Lagrangian. For our present aims, it is good to think of this as a “symmetry of the laws of physics” in the following sense. The Lagrangian determines the “laws of motion” of the ﬁeld via the Euler-Lagrange equations. As was pointed out in one of the problems, symmetries of a Lagrangian are also symmetries of the equations of motion. This means that if ϕ is a solution to the equations of motion and if ϕ˜ is obtained from ϕ via a symmetry transformation, then ϕ˜ also satisﬁes the same equations of motion. Just to be clear, let me cite a very elementary example. Consider the massless KG ﬁeld described by the Lagrangian density:

L = − 1 ∂αϕ∂αϕ.

(1)

2

It is easy to see that this Lagrangian admits the symmetry

ϕ˜ = ϕ + const.

(2)

You can also easily see that the ﬁeld equations

∂α∂αϕ = 0

(3)

admit this symmetry in the sense that if ϕ is solution then so is ϕ+const. Thus a symmetry of a Lagrangian is also a symmetry of the ﬁeld equations and we will sometimes call it a symmetry of the law governing the ﬁeld.

PROBLEM: It is not true that every symmetry of the ﬁeld equations is a symmetry of the Lagrangian. Consider the massless KG ﬁeld. Show that the scaling transformation ϕ˜ = (const.)ϕ is a symmetry of the ﬁeld equations but is not a symmetry of the Lagrangian.

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Spontaneous symmetry breaking. Goldstone’s theorem. The Higgs mechanism.

If the Lagrangian and its ﬁeld equations are the “laws”, then the solutions of the ﬁeld

equations are the “states” of the ﬁeld that are allowed by the laws. The function ϕ(x)

is an allowed state of the ﬁeld when it solves the ﬁeld equations. A symmetry of a given

“state”, ϕ0(x) say, is then deﬁned to be a transformation of the ﬁelds, ϕ → ϕ˜[ϕ], which

preserves the given solution

ϕ˜[ϕ0(x)] = ϕ0(x).

(4)

Since symmetry transformations form a group, such solutions to the ﬁeld equations are sometimes called “group-invariant solutions”.

Let us consider an elementary example of group-invariant solutions. Consider the KG ﬁeld with mass m. Use inertial Cartesian coordinates. We have seen that the spatial translations xi → xi + const. form a group of symmetries of the theory. It is easy to see that the corresponding group invariant solutions are of the form:

ϕ = A cos(mt) + B sin(mt),

(5)

where A and B are constants. Another very familiar type of example of group-invariant solutions you will have seen by now occurs whenever you are ﬁnding rotationally invariant solutions of PDEs.

PROBLEM: Derive the result (5).

An important result from the geometric theory of diﬀerential equations which relates symmetries of laws to symmetries of states goes as follows. Suppose G is a group of symmetries of a system of diﬀerential equations ∆ = 0 for ﬁelds ϕ on a manifold M , (e.g., G is the Poincar´e group). Let K ⊂ G be a subgroup (e.g., spatial rotations). Suppose we are looking for solutions to ∆ = 0 which are invariant under K. Then the ﬁeld equations ∆ = 0 reduce to a system of diﬀerential equations ∆ˆ = 0 for K-invariant ﬁelds ϕˆ on the reduced space M/K.
As a simple and familiar example, consider the Laplace equation for functions on R3,

∂x2ϕ + ∂y2ϕ + ∂z2ϕ = 0.

(6)

The Laplace equation is invariant under the whole Euclidean group G consisting of translations and rotations. Consider the subgroup K consisting of rotations. The rotationally invariant functions are of the form

ϕ(x, y, z) = f (r), r = x2 + y2 + z2.

(7)

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Rotationally invariant solutions to the Laplace equation are characterized by a reduced ﬁeld f satisfying a reduced diﬀerential equation on the reduced space R+ given by

1 d (r2 df ) = 0.

(8)

r2 dr dr

This is the principal reason one usually makes a “symmetry ansatz” for solutions to ﬁeld equations which involves ﬁelds invariant under a subgroup K of the symmetry group G of the equations. It is not illegal to make other kinds of ansatzes, of course, but most will lead to inconsistent equations or equations with trivial solutions.
Having said all this, I should point out that just because you ask for group invariant solutions according to the above scheme it doesn’t mean you will ﬁnd any! There are two reasons for this. First of all the reduced diﬀerential equation may have no (or only trivial) solutions. Second it may be that there are no ﬁelds invariant under the symmetry group you are trying to impose on the state. As a simple example of this latter point, consider the symmetry group ϕ → ϕ + const. we mentioned earlier for the massless KG equation. You can easily see that there are no functions which are invariant under that transformation group. And I should mention that not all states have symmetry - indeed the generic states are completely asymmetric. States with symmetry are special, physically simpler states than what you expect generically.
To summarize, ﬁeld theories may have two types of symmetry. There may be a group G of symmetries of its laws – the symmetry group of the Lagrangian (and ﬁeld equations). There can be symmetries of states, that is, there may be subgroups of G which preserve certain states.

The “Mexican hat” potential

Let us now turn to a class of examples which serve to illustrate the preceding remarks and which we shall use to understand spontaneous symmetry breaking. We have actually seen these examples before.

We start by considering the real KG ﬁeld with the double-well potential:

L0 = − 1 ∂αϕ∂αϕ − (− 1 a2ϕ2 + 1 b2ϕ4).

(9)

2

2

4

As usual, we are working in Minkowski space with inertial Cartesian coordinates. This

Lagrangian admits the Poincar´e group as a symmetry group. It also admits the symmetry

ϕ → −ϕ, which forms a 2 element discrete subgroup Z2 of the symmetry group of the

Lagrangian. In an earlier homework problem we identiﬁed 3 simple solutions to the ﬁeld

equations for this Lagrangian:

a

ϕ = 0, ± ,

(10)

b

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where a and b are constants. These solutions are highly symmetric: they admit the whole
Poincar´e group of symmetries, as you can easily verify. Because Z2 is a symmetry of the Lagrangian it must be a symmetry of the ﬁeld equations, mapping solutions to solutions.
You can verify that this is the case for the solutions (10). Thus the group consisting of the
Poincar´e group and Z2 form a symmetry of the law governing the ﬁeld. The 3 solutions in (10) represent 3 (of the inﬁnite number of) possible solutions to the ﬁeld equations – they are possible states of the ﬁeld. The states represented by ϕ = ± ab have Poincar´e symmetry, but not Z2 symmetry. In fact the Z2 transformation maps between the solutions ϕ = ± ab . The state represented by ϕ = 0 has both the Poincar´e and the Z2 symmetry. The solution ϕ = 0 thus has more symmetry than the states ϕ = ± ab .
Let us consider the energetics of these highly symmetric solutions ϕ = const. In an inertial reference frame with coordinates xα = (t, xi), the conserved energy in a spatial
volume V for this non-linear ﬁeld is easily seen (from Noether’s theorem) to be

E=

dV

1 ϕ2
,t

+

1 ϕ,iϕi,

1 a2ϕ2

+

1 b2ϕ4

.

(11)

V

2

2

2

4

You can easily check that the solutions given by |ϕ| = 0, ab are critical points of this energy functional. While it might be intuitively clear that these solutions ought to represent

global minima at ϕ = ± ab and a local maximum at ϕ = 0, it is perhaps not so easy to see this explicitly without some further analysis. We can investigate this as follows. Let

us consider the change in the energy to quadratic-order in a displacement u = u(t, x, y, z)

from equilibrium in each case. We assume that u has compact support for simplicity. We

write ϕ = ϕ0 + u where ϕ0 is a constant and expand E to quadratic order in u. We get

V a4

12

i

22

3

a

E = − 4 b2 +

dV
V

2 (u,t + u,iu, ) + a u

+ O(u ), when ϕ0 = ± b

(12)

and

E=

dV

1 (u2
,t

+

u,iui, )

1 a2u2

+ O(u3),

when ϕ0 = 0.

(13)

V

2

2

Evidently, as we move away from ϕ = ± ab the energy increases so that the critical points ϕ = ± ab represent local minima. The situation near ϕ = 0 is less obvious. One thing is

for sure: by choosing functions u which are suitably “slowly varying”, one can ensure that

the energy becomes negative in the vicinity of the solution ϕ = 0 so that ϕ = 0 is a saddle

point if not a local maximum. We conclude that the state ϕ = 0 – the state of highest

symmetry – is unstable and will not be seen “in the real world”. On the other hand we expect the critical points ϕ = ± ab to be stable. They are in fact the states of lowest energy and represent the possible ground states of the classical ﬁeld. Evidently, the lowest energy

is doubly degenerate.

Because the physical ground states have less symmetry than possible, one says that the symmetry group Z2 × Poincar´e has been “spontaneously broken” to just the Poincar´e

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group for the ground state. This terminology is useful, but can be misleading. The theory retains the full symmetry group as a symmetry of its laws, of course. Compare this with, say, the ordinary Klein-Gordon theory with the Lagrangian

L = − 1 ∂αϕ∂αϕ − m2ϕ2.

(14)

2

You can easily check that the solution ϕ = 0 is the global minimum energy state of the theory and that it admits the full symmetry group Z2 × Poincar´e. There is evidently no spontaneous symmetry breaking here.

Let us now generalize this example by allowing the KG ﬁeld to become complex, ϕ: M → C, with Lagrangian

L

=

1 − ∂αϕ

∂αϕ∗

(− 1 a2|ϕ|2

+

1 b2|ϕ|4).

(15)

2

2

4

We assume a ≥ 0, b ≥ 0. This Lagrangian still admits the Poincar´e symmetry, but the discrete Z2 symmetry has been enhanced into a continuous U (1) symmetry. Indeed, it is pretty obvious that the transformation

ϕ → eiαϕ, α ∈ R

(16)

is a symmetry of L. If you graph this potential in (x, y, z) space with x = (ϕ), y = (ϕ) and z = V , then you will see that the graph of the double well potential has been extended into a surface of revolution about z with the resulting shape being of the famous “Mexican hat” form. From this graphical point of view, the U (1) phase symmetry of the Lagrangian specializes to symmetry of the graph of the potential with respect to rotations in the x-y plane.

Let us again consider the simplest possible states of the ﬁeld, namely, the ones which admit the whole Poincar´e group as a symmetry group. These ﬁeld conﬁgurations are necessarily constants, and you can easily check that constant solutions must be critical points of the potential viewed as a function in the complex plane. So, ϕ = const. is a solution to the ﬁeld equations if and only if

b2 ϕ|ϕ|2 − 1 a2ϕ = 0.

4

2

There is an isolated solution ϕ = 0, and (now assuming b > 0) a continuous family of

solutions characterized by

a

|ϕ| = .

(17)

b

The solution ϕ = 0 “sits” at the local maximum at the top of the “hat”. The solutions

(17) sit at the circular set of global minima. As you might expect, the transformation (16)

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maps the solutions (17) among themselves. To see this explicitly, write the general form

of ϕ satisfying (17) as

ϕ = a eiθ, θ ∈ R.

(18)

b

The U (1) symmetry transformation (16) then corresponds to θ → θ + α. The U (1) trans-

formation is a symmetry of the state ϕ = 0. Thus the solution ϕ = 0 has more symmetry

than the family of solutions characterized by (17).

The stability analysis of these highly symmetric states of the complex scalar ﬁeld generalizes from the double well example as follows. (I will spare you most of the details of the computations, but you might try to ﬁll them in as a nice exercise.) In an inertial reference frame with coordinates xα = (t, xi), the conserved energy for this non-linear ﬁeld is easily seen (from Noether’s theorem) to be

E=

d3x

1 |ϕ,t|2

+

1 ϕ,i

ϕ∗, i

1 a2|ϕ|2

+

1 b2|ϕ|4

.

(19)

2

2

2

4

You can easily check that the solutions given by |ϕ| = 0, ab are critical points of this energy functional and represent the lowest possible energy states. As before, the maximally

symmetric state ϕ = 0 is unstable. The circle’s worth of states (17) are quasi-stable in the

following sense. Any displacement in ﬁeld space yields a non-negative change in energy.

To see this, write

ϕ = ρeiΘ,

(20)

where ρ and Θ are spacetime functions. The energy takes the form

E=

d3x

1 ρ2
,t

+

1 ρ,i

ρ,i

+ 1 ρ2(Θ2
,t

+

Θ,i

Θ,i

)

1 a2ρ2

+

1 b2ρ4

.

(21)

2

2

2

2

4

The critical points of interest lie at ρ = ab , Θ = const. Expanding the energy in displacements (δρ, δΘ) from equilibrium yields

1 a4

3 12 1

i 1 a2 2

i

22

E

=

− 4

b2

+

dx

2 δρ,t +

2 δρ,i δρ,

+ 2

b

(δΘ,t + δΘ,i δΘ, ) + a δρ

(22)

Evidently, all displacements except δρ = 0, δΘ = const. increase the energy. The displace-

ment δρ = 0, δΘ = const. does not change the energy, as you might have guessed. The

states (17) are the lowest energy states – the ground states. Thus the lowest energy is in-

ﬁnitely degenerate – the set of ground states (17) is topologically a circle. That these stable

states have less symmetry than the unstable state will have some physical ramiﬁcations

which we will understand after we take a little detour.

Dynamics near equilibrium

A signiﬁcant victory for classical mechanics is the complete characterization of motion near stable equilibrium in terms of normal modes and characteristic frequencies of vibration. It is possible to establish analogous results in classical ﬁeld theory via the process of

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Spontaneous symmetry breaking. Goldstone’s theorem. The Higgs mechanism.

linearization. This is even useful when one considers the associated quantum ﬁeld theory: one can interpret the linearization of the classical ﬁeld equations as characterizing particle states in the Fock space built on the vacuum state whose classical limit corresponds to the ground state about which one linearizes. If this seems a little too much to digest, that’s ok – the point of this section is to make it easier to swallow.
Let us begin again with the simplest example: the real KG ﬁeld with the double-well potential. Suppose that ϕ0 is a given solution to the ﬁeld equations. Any “nearby” solution we will denote by ϕ and we deﬁne the diﬀerence to be δϕ.

δϕ = ϕ − ϕ0.

(23)

The ﬁeld equation is the non-linear PDE:

ϕ + a2ϕ − b2ϕ3 = 0.

(24)

Using (23) we substitute ϕ = ϕ0 + δϕ. We then do 2 things: (1) we use the fact that ϕ0 is a solution to the ﬁeld equations; (2) we assume that δϕ is in some sense “small” so we

can approximate the ﬁeld equations in the vicinity of the given solution ϕ0 by ignoring quadratic and cubic terms in δϕ. We thus get the ﬁeld equation linearized about the

solution ϕ0:

δϕ + (a2 + 3b2ϕ20)δϕ = 0.

(25)

This result can be obtained directly from the variational principle.

PROBLEM: Using (23) expand the action functional to quadratic order in δϕ. Show that this approximate action, viewed as an action functional for the displacement ﬁeld δϕ, has (25) as its Euler-Lagrange ﬁeld equation.
Evidently, the linearized equation (25) is a linear PDE for the displacement ﬁeld δϕ. If the given solution ϕ0 is a constant solution the linearized PDE is mathematically identical to a Klein-Gordon equation for δϕ with mass given by (a2 + 3b2ϕ20). This is in fact the physical interpretation in the vicinity of one of the ground states we have been studying: the dynamics of the ﬁeld is approximately that of a free KG ﬁeld with mass 2a.
From the way the linearized equation is derived, you can easily see that any displacement ﬁeld δϕ constructed as an inﬁnitesimal symmetry of the ﬁeld equations will automatically satisfy the linearized equations. Indeed, this fact is the deﬁning property of an inﬁnitesimal symmetry. Here is a simple example.

PROBLEM: Consider time translations ϕ(t, x, y, z) → ϕ˜ = ϕ(t + λ, x, y, z). Compute the inﬁnitesimal form δϕ of this transformation as a function of ϕ and its derivatives. Show

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that if ϕ solves the ﬁeld equation coming from (9) then δϕ solves the linearized equation (25).

These results easily generalize to the U (1)-invariant complex scalar ﬁeld case, but a new and important feature emerges which leads to an instance of (the classical limit of) a famous result known as “Goldstone’s theorem”. Let us go through it carefully.

Things will be most transparent in the polar coordinates (20). The Lagrangian density

takes the form

L

=

1 − ∂αρ

∂αρ

1 ρ2∂αΘ

∂αΘ

+

1 a2ρ2

1 b2ρ4.

(26)

2

2

2

4

The EL equations take the form

ρ − ρ∂αΘ ∂αΘ + a2ρ − b2ρ3 = 0,

(27)

∂α(ρ2∂αΘ) = 0.

(28)

Two things to notice here. First, the symmetry under Θ → Θ + α is apparent – only derivatives of Θ appear. Second, the associated conservation law is the content of (28).

Let us consider the linearization of these ﬁeld equations about the circle’s worth of equilibria ρ = ab . We can proceed precisely as before, of course. But it will be instructive to perform the linearization at the level of the Lagrangian. To this end we write

a ρ = + δρ;
b

we leave Θ as is since there is no equilibrium choice used for it. We expand the Lagrangian to quadratic order in δρ and Θ since that corresponds to linear ﬁeld equations. We get

1

α

1 a2

α 1 a4 2 2

3

L = − 2 ∂αδρ ∂ δρ − 2 b ∂αΘ ∂ Θ − 4 b2 − a δρ + O(δϕ ).

Evidently, in the neighborhood of equilibrium the complex scalar ﬁeld can be viewed as 2 real ﬁelds – no surprises so far. The main observation to be made is that one of the ﬁelds (δρ) has a mass m = a and one of the ﬁelds (Θ) is massless.

To get a feel for what just happened, let us consider a very similar U (1) symmetric theory, just diﬀering in the sign of the quadratic potential term in the Lagrangian. The Lagrangian density is

L

=

1 − ∂αϕ

∂αϕ∗

µ2|ϕ|2

1 b2|ϕ|4.

(29)

2

4

There is only a single Poincar´e invariant critical point, ϕ = 0, which is a global minimum of the energy and which is also U (1) invariant, so the U (1) symmetry is not spontaneously

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broken in the ground state. In the vicinity of the ground state the linearized Lagrangian takes the simple form*
L = − 1 ∂αδϕ ∂αδϕ∗ − µ2|δϕ|2 + O(δϕ3). 2
Here of course we have the Lagrangian of a complex-valued KG ﬁeld δϕ with mass µ; equivalently, we have two real scalar ﬁelds with mass µ.
To summarize thus far: With a complex KG ﬁeld described by a potential such that the ground state shares all the symmetries of the Lagrangian, the physics of the theory near the ground state is that of a pair of real, massive KG ﬁelds. Using instead the Mexican hat potential, the ground state of the complex scalar ﬁeld does not share all the symmetries of the Lagrangian – there is spontaneous symmetry breaking – and the physics of the ﬁeld theory near equilibrium is that of a pair of scalar ﬁelds, one with mass and one which is massless.
To some extent, it is not too hard to understand a priori how these results occur. In particular, we can see why a massless ﬁeld emerged from the spontaneous symmetry breaking. For Poincar´e invariant solutions – which are constant functions in spacetime – the linearization of the ﬁeld equations involves:
(1) the derivative terms, which being quadratic in the ﬁelds, and given the ground state is constant, are the same in the linearization as for the full ﬁeld equations;
(2) the Taylor expansion of the potential V (ϕ) to second order about the constant equilibrium solution ϕ0.
Because of (1), the nature of the mass terms comes from the expansion (2). Because of the symmetry of the Lagrangian, and because of the spontaneous symmetry breaking, we are guaranteed that through each point in the set of ﬁeld values there will be a curve (with tangent vector given by the inﬁnitesimal symmetry) along which the potential will not change. Taylor expansion about the ground state in this symmetry direction will yield only vanishing contributions because the potential has vanishing derivatives in that direction. Thus the broken symmetry direction(s) deﬁnes the direction(s) in ﬁeld space which correspond to massless ﬁelds. This is the essence of the (classical limit of the) Goldstone theorem: to each broken continuous symmetry generator there is a massless ﬁeld.

The Abelian Higgs model

The Goldstone result in conjunction with minimal coupling to an electromagnetic ﬁeld yields a very important new behavior known as the “Higgs phenomenon”. This results

* We do not use polar coordinates which are ill-deﬁned at the origin.

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from the interplay of the spontaneously U (1) symmetry and the local gauge symmetry. We start with a charged self-interacting scalar ﬁeld coupled to the electromagnetic ﬁeld; the Lagrangian density is

L

=

1 −F

αβ Fαβ

Dαϕ∗Dαϕ

V

(ϕ).

(30)

4

We will again choose the potential so that spontaneous symmetry breaking occurs:

V (ϕ) = − 1 a2|ϕ|2 + 1 b2|ϕ|4.

(31)

2

4

To see what happens in detail, we return to the polar coordinates (20). The Lagrangian

takes the form:

L

=

1 −F

αβ Fαβ

1 ∂αρ∂αρ

1 ρ2(∂αΘ

+

eAα)(∂αΘ

+

eAα)

+

1 a2ρ2

1 b2ρ4.

(32)

4

2

2

2

4

The Poincar´e invariant ground state(s) can be determined as follows. As we have observed, a Poncar´e invariant function ϕ is necessarily a constant. Likewise, it is too not hard to see that the only Poincar´e invariant (co)vector is the the zero (co)vector Aα = 0. Consequently, the Poincar´e invariant ground state, as before, is speciﬁed by
a ρ = b , Θ = const. Aα = 0. (33)

As before the U (1) symmetry of the theory is not a symmetry of this state. As before, we

want to expanding to quadratic order about the ground state. To this end we write

a

Aα = 0 + δAα,

ρ = + δρ, b

Θ = Θ0 + δΘ,

where Θ0 is a constant. We also deﬁne

1 Bα = δAα + e ∂αδΘ.

Ignoring terms of cubic and higher order in the displacements (δA, δρ, δΘ) we then get

L

1 (∂αBβ

∂β Bα )(∂ α B β

∂βBα)

1

ae 2 BαBα

4

2b

(34)

1 ∂αδρ

∂αδρ

1 a2δρ2

2

2

PROBLEM: Starting from (30) derive the results (32) and (34).

As you can see, excitations of ρ around the ground state are those of a scalar ﬁeld with mass a, as before. To understand the rest of the Lagrangian we need to understand the Proca Lagrangian:

Lp

=

1 − (∂αBβ

∂β Bα )(∂ α B β

∂βBα)

1 κ2BαBα.

(35)

4

2

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