Spontaneous symmetry breaking. Goldstone s theorem. The Higgs

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Spontaneous symmetry breaking. Goldstone s theorem. The Higgs

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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Classical Field Theory

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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Classical Field Theory

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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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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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Classical Field Theory

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

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

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

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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