%Date: Sat, 4 Apr 1998 16:14:21 -0500 (EST)
%From: Pavel Etingof
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\topmatter
\title Lecture II-2, Part II: Spontaneous breaking of gauge symmetry
\endtitle
\author {\rm {\bf Edward Witten} }\endauthor
\endtopmatter
\centerline{Notes by Pavel Etingof and David Kazhdan}
In this lecture we will consider gauge symmetry breaking.
{\bf 2.1. Gauge symmetry.}
Recall what gauge symmetry is. We have a spacetime $X=\R_{time}\times
X_0$. We have a compact gauge group $G$.
We have a field theory where a field configuration
is a connection is some principal bundle over $\R^d$ and possibly
some matter fields.
Recall the Hamiltonian approach to gauge theory.
Let $\tilde M_0$ be the space of solutions to the classical
equations of motion. On $\tilde M_0$ we have an action
of the group $\hat G$ of gauge transformations.
Let $M_0\subset \tilde M_0$ be the space of all solutions
where the G-bundle is trivialized in the time direction, and
the connection is trivial in that direction. Such solutions
as usual are completely determined by the pair $A(t_0),\frac{dA}{dt}(t_0)$,
where $t=t_0$ is a space cycle, and initial data for the matter fields.
It is clear that any element of $\tilde M_0$
can be brought to $M_0$ by a gauge transformation, so
$M_0$ still contains all solutions up to gauge transformations.
Suppose that $X_0=\R^{d-1}$. In this case we may consider only trivial
bundles, and connections which vanish at spatial infinity.
In other words, if $\Cal A$
is the space of connections $A$ on the trivial $G$-bundle over $\R^{d-1}$
which vanish at $\infty$ then $M_0$ for pure gauge theory is
$T^*\Cal A$. If matter fields are present, then $M_0$ is a product of
$T^*\Cal A$ with some other space.
Define $\tilde G$ to be
the group of elements of $Maps(\R^{d-1},G)$
which have a limit at infinity, and $\tilde G_0$ to be the subgroup of
$\tilde G$ consisting of functions which tend to 1 at $\infty$.
We have $\tilde G/\tilde G_0=G$. This quotient group is called the group
of constant gauge transformations at $\infty$ and called $G_\infty$.
The group $\tilde G$ acts symplectically on $M_0$.
The physical phase space in gauge theory is the symplectic quotient
$M=M_0//\tilde G_0$. Note that we only divide by $\tilde G_0$ and not by
the whole group $\tilde G$, so that we inherit an action
of the quotient $G_\infty$ on $M$. It is this symmetry group whose
breaking we will discuss.
{\bf 2.2. Breaking of gauge symmetry
and charges at infinity.}
{\bf Definition.} Suppose we have a (classical) gauge theory, and
let $s\in M$ be its vacuum state. Let $H\subset G=G_\infty$ be
the stabilizer of $s$. In this case
we will say that at the vacuum state $s$ the gauge
symmetry is broken from $G$ to $H$.
Thus, by symmetry breaking we mean essentially the same thing as for
global symmetry: there is a
symmetry of the Poisson algebra of functions on $M$ which
does not fix a particular vacuum state.
{\bf Important remark.} The above expression ``the same thing ''
should be taken with great care. There are some fundamental differences
between the two situations, which will become clear below.
They come from the fact that in the situation we are considering here,
(unlike Lecture II1) the physical observables, being gauge invariant
by definition, automatically commute with $G$ and therefore do not, in
general, separate points on $M$; i.e. not every function on $M$ is
``observable''. In other words, the action of
$G$ on the ``theory'' (in the sense of Lecture II1) is trivial
from the beginning.
Let us now compute the action of $G$ (classically).
First of all, we have a moment map $\mu:M_0\to \tilde \g_0^*$, where
$\tilde\g_0$ is the Lie algebra of $\tilde G_0$ -- the algebra of
functions from $\R^{n-1}$ to the Lie algebra $\g$ of $G$ which vanish at
infinity. Thus, for any $\e\in \tilde\g_0^*$ we have a Hamiltonian
$Q(\e)\in C^\infty(M_0)$ defined by $Q(\e)(X)=\mu(X)(\e)$.
In fact, it is easy to compute $Q(\e)$ using Noether formalism. Namely,
$$
Q(\e)=\int_{\R^{d-1}}Tr(\frac{\d A}{dt}\nabla_A \e)d^{d-1}x+
\text{ matter terms },\tag 2.1
$$
On $M$, $Q(\e)=0$ if $\e$ vanishes at infinity. Thus, on $M$ we have
$\nabla_A^*\frac{dA}{dt}=
\text{ matter terms }$. In particular, in pure gauge theory
$\nabla_A^*\frac{dA}{dt}=0$.
Taking this into account, we see that on $M$
$$
Q(\e)=\int_{\R^{d-1}}Tr(\nabla_A (\e\frac{dA}{dt}))d^{d-1}x.\tag 2.2
$$
Using Stokes' formula, we can rewrite (2.2) as
$$
Q(\e)=\lim_{r\to\infty}\int_{S^{n-2}(r)}
*_{d-1}Tr(\e\frac{dA}{dt})=\lim_{r\to\infty}
\int_{S^{n-2}(r)}*_dTr(\e F), \tag 2.3
$$
where $F$ is the curvature of the spacetime connection
corresponding to the given point of $M_0$, and $S^k(r)$ is the k-sphere of
radius $r$. This formula defines the hamiltonians
for the action of $G=G_\infty$ on $M$.
This formula shows that $Q(\e)$ vanishes for all gauge transformations
(not necessarily vanishing at infinity) on a particular state
if $F=o(r^{2-n})$, $r\to\infty$ on that state. However, if
this is not the case, then $Q(\e)$ may be nonzero for a constant $\e$.
{\bf Example.} Consider a $U(1)$ gauge theory with a charged
complex scalar. The fields are a connection $A$ on a hermitian line bundle
and a section $\phi$ of this bundle. The Lagrangian is
$$
\Cal L=\frac{1}{4e^2}\int F^2+\int |D_A\phi|^2d^4x+\int\frac{\l}{8}
(|\phi|^2+v^2)^2d^4x.\tag 2.3
$$
This is the most general renormalizable Lagrangian in these fields in
4 dimensions. Here $e,\l,v$ are parameters and $e^2,\l$ are positive while
$v^2$ can be positive or negative. For simplicity we assume first
that $v^2\ne 0$.
This theory is not believed to exist in the UV, but we will regard it
as an effective theory for some more fundamental theory.
Classically (and quantum mechanically for $e^2,\l<<1$) we have two cases.
1. $v^2>0$; the potential has a single minimum.
2. $v^2<0$; the potential has a circle of minima.
Let us consider how in these two cases the theory behaves in the infrared.
\centerline{\epsfxsize=2in\epsfbox{plot2.eps}}
\centerline{Figure 1. The potential for $v^2>0$.}
\bigskip
{\bf Case 1.} $v^2>0$. In this case the minimum of energy is attained when
$\phi=0$. First consider the case when the gauge coupling vanishes:
$e^2=0$. In this case our theory is a direct product of a pure (free)
abelian gauge theory and the $\phi^4$ theory. Therefore, it has a unique
vacuum, and
the particles which occur at the lower part of the spectrum are
a massless vector, or gauge boson
(coming from gauge theory) and two massive real scalars
(coming for $\phi^4$ theory).
If we turn on small $e^2$ the situation should remain the same.
Indeed, certainly nothing can happen to the massive scalars
(the part of the Hilbert space
with the nonzero charge, where these scalars are, has a mass gap, and
massiveness is an open condition); moreover, their masses must be equal since
there is a $U(1)$ symmetry at infinity (the $Q(\e)$ for constant $\e$)
which prohibits the masses to differ. The fact that $Q(\e)\ne 0$
is clear since this is so at $e^2=0$, when $Q(\e)$ represents the $U(1)$
global symmetry.
Furthermore, the massless vectors cannot become massive.
Indeed, recall that a massless vector means an irreducible
representation of $SO(3,1)$ with $p^2=0$ and spin 1, i.e. the space of
sections of a 2-dimensional equivariant vector bundle over the light cone.
This vector bundle cannot be deformed to an equivariant vector bundle over
the hyperboloid, since the stabilizer group $SO(3)$ of a point on the
hyperboloid does not have an irreducible 2-dimensional representation.
Thus, the quantum theory for small coupling
will have the same particles -- two massive scalars
(the real and imaginary part of $\phi$) and a massless vector
(the gauge boson).
{\bf Remark.} The above argument on non-deformability of a massless vector
fails in 3 and 2 dimensions. For example, in 3 dimensions, the
massless vector is just the space of functions over the cone, which
can be successfully deformed into the space of functions over a hyperboloid.
This actually happens when in pure $U(1)$ gauge theory one introduces
a Chern-Simons term $c\int A\wedge dA$. The theory remains free but
becomes massive, yielding one massive scalar. In the theory we are
considering (for 3 dimensions), this cannot happen dynamically since
the Chern-Simons term is odd under change of orientation, but in other
theories this could happen.
In fact, quantum mechanically
the operator $Q(\e)$ (for a suitable normalization of $\e$)
has integer eigenvalues, and thus defines (in quantum theory)
a $\Z$-grading of the corresponding Hilbert space. In particular,
since $Q(\e)\ne 0$, there are sectors of the Hilbert space which
cannot be reached from the vacuum by applying local operators.
This shows that we have a fundamental violation of Wightman axioms:
the representation of the operator algebra in the physical Hilbert space
is not irreducible. However, the theory still has one vacuum only:
the minimal energy in the sectors with nonzero charge $Q(\e)$ is positive.
\centerline{\epsfxsize=2in\epsfbox{plot1.eps}}
\centerline{Figure 2. The potential for $v^2<0$.}
\bigskip
{\bf Case 2.} $v^2<0$. Let $v^2=-b^2$. Then classically we have a minimum of
energy on the circle $|\phi|=b$. This implies
that any finite energy configuration has the property
$\phi=be^{i\theta_0}$ at infinity, where $\theta_0$ is a constant.
Therefore, by a gauge transformation which has a finite limit
at infinity, we can arrange
that $\phi$ is real and positive: $\phi=b+w$ where $w$ is a new real
variable.
Writing the Lagrangian in terms of the new variables, we will get
something with the following quadratic part:
$$
L_{quadratic}=\frac{1}{4e^2}\int F^2+\int d^4x((dw)^2+M^2w^2)+\int d^4xb^2A^2.
\tag 2.4
$$
It is seen from (2.4) than now all fields are massive. Of course,
Lagrangian (2.4) is not gauge invariant for $A$, since we have already
``spent'' the gauge symmetry on making $\phi$ real.
Thus, infrared limit of the corresponding quantum theory is trivial
for small values of the couplings. In particular, there are no massless gauge
bosons: they have been ``eaten'' by the $\phi$-field. This situation
is called Higgs phenomenon, or spontaneous breaking of gauge symmetry.
Note that in spite of the presence of a circle of zero energy states,
our theory has only one vacuum. In other words, all points of the circle are
regarded as the same state, on the grounds that they are gauge equivalent
to each other and therefore define equivalent realizations
(i.e. give the same expectation values of gauge invariant local operators)
This is a fundamental difference between gauge
and global symmetry breaking. In global symmetry breaking, the points of
the circle represent different vacua (realizations) of the theory,
since there exist non-symmetric operators which have different
expectation values at different point of the circle.
Note also
that the operator $Q(\e)$ doesn't act in the Hilbert space of states,
since classically $Q(\e)$ generates a group which rotates the circle
and permutes the zero energy states. In particular, in this case
local operators act irreducibly in the Hilbert space, and there are
no sectors
which cannot be reached from the vacuum. This is the difference between
case 2 and case 1: in case 1, as you remember, $Q(\e)$ acts in the Hilbert
space nontrivially and defines a splitting into sectors.
{\bf Remark.} If one tries to compute $Q(\e)$ in Case 2 (when the symmetry
is broken) using formula (2.3), the answer will be zero since the integrand
dies rapidly at infinity.
The particles which are found in the infrared in the situation
of Case 2 are, according to (2.4), a massive vector ($A$) and a massive
scalar ($\phi$). There is only one scalar since $\phi$ is now real.
Thus, at the level of representation theory the Higgs phenomenon arising
in Case 2 boils down to a deformation of representations of the Poincare
group: a massless vector plus a massless scalar is deformed to a massive
vector. Recall for comparison that a massless vector separately
cannot be deformed into a massive representation.
Finally, consider the special case $v^2=0$. In this case classically
we have no symmetry breaking as for $v^2>0$, and the particles
are a massless vector and two massless scalar. However, it is not
expected to be the quantum answer, since this configuration is not stable
under perturbations.
{\bf Remark.} If the Lagrangian we start with is not IR free
(say, it is the Lagrangian of an asymptotically free gauge theory)
then the classical analysis we discussed above does not apply
in quantum theory. In this case the infrared behavior of the theory
is difficult to determine. In particular, it could happen that in the
infrared the gauge group of the ultraviolet theory will be replaced with
some completely different group, which is not even a subgroup in the original
group.
{\bf 2.3. Symmetry breaking and gauging.}
In conclusion, let us discuss the connection between global symmetry and gauge
symmetry breaking. Suppose we have a Lagrangian $L$ of a field theory
(say in 4 dimensions) which has a global $U(1)$ symmetry.
A typical example is when the theory contains some scalar fields $\phi_j$
which are sections of hermitian vector bundles,
and $U(1)$ acts by multiplication of these sections
by $e^{in_j\theta}$ This $U(1)$ symmetry can be gauged, by introducing
a $U(1)$ gauge field $A$ and new Lagrangian
$$
L_{gauged}=\frac{1}{4e^2}\int F_A^2+L_A, \tag 2.5
$$
where $L_A$ is $L$ in which all derivatives of $\phi_j$ are replaced
by covariant derivatives.
The statement is that
if $L$ is infrared free
then for small gauge couplings the symmetry breaking behavior
of the theories defined by $L$ and $L_{gauged}$ is usually the same.
Namely, if there is breaking of global symmetry for $L$
then there is breaking of gauge symmetry for $L_{gauged}$ and vice versa.
Indeed, let us consider both cases.
{\bf Case 1}. No global symmetry breaking. In this case
classically the minimum of energy is at $\phi_j=0$, and thus there is a
$U(1)$-invariant vacuum. In quantum theory,
$U(1)$ acts in the Hilbert
space, and there are no massless particles (Goldstone bosons) corresponding
to $U(1)$.
In this case, for small gauge coupling $e$, the matter part
$L_A$ of the Lagrangian almost decouples from the gauge part;
so classically we get a massless gauge boson.
To consider the quantum mechanical situation,
we assume that there are no massless particles
in the ungauged theory. In this case, the above classical answer
is also quantum mechanical for small couplings, by the non-defomability
of representations from massless to massive. However, if massless particles
(say Goldstone bosons corresponding to other symmetries which are broken)
are present, this answer may not be true.
{\bf Case 2.} Global symmetry breaking. In this case at the minimum of energy
some of the $\phi_j$ is not zero. There is no invariant vacuum, and
there is a Goldstone boson corresponding to this symmetry breaking.
In this case, pick a vacuum state and
a component $\phi_j^{(n)}$ which is not zero at this vacuum.
In the gauged theory, we can perform a gauge transformation
which will make this component real and positive.
This shows that if there are no other
massless particles (in particular, no other broken global symmetries),
all fields in the theory will become massive. This happens classically, due
to Higgs mechanism as in Case 2 above, and also quantum mechanically for small
couplings. Thus, we have breaking of gauge symmetry.
\end