BRST and Lie Algebra Cohomology

We saw in previous posts that gauge-fixing is intimately related to BRST cohomology. Today I want to explain the underlying mathematical formalism, as it is actually something very well-known: Lie algebra cohomology. Let \(\mathfrak{g}\) be a Lie algebra and \(M\) a \(\mathfrak{g}\)-module. We will construct a cochain complex that computes the Lie algebra cohomology with values in \(M\), \(H^i(\mathfrak{g}, M)\). Out of thin air, we define
\[ C^\ast(\mathfrak{g}, M) = M \otimes \wedge^\ast \mathfrak{g}^\ast. \]
The grading is just the grading induced by the grading on \(\wedge^\ast \mathfrak{g}^\ast\), which we identify with the BRST ghost number. Let \(e_i\) be a basis for \(M\) and \(T_a\) be a basis for \(\mathfrak{g}\), with canonical dual basis \(S^a\). The differential is defined on generators to be
\[ d e_i = \rho(T_a) e_i \otimes S^a \]
\[ d S^a = \frac{1}{2} f^a_{bc} S^b \wedge S^c \]
where \(\rho: \mathfrak{g} \to \mathrm{End}(M)\) is the representation and \(f^a_{bc}\) are the structure constants of the group. This differential is then extended to satisfy the graded Leibniz rule, and is easily verified to satisfy \(d^2 = 0\) (this is just the Jacobi identity). The Lie algebra cohomology is just the cohomology of this cochain complex. Essentially by definition, we see that
\[ H^0(\mathfrak{g}, M) = \{m \in M \ | \ \xi \cdot m = 0 \ \forall \ \xi \in \mathfrak{g} \}, \]
i.e. \(H^0(\cdot) = (\cdot)^\mathfrak{g}\) is the invariants functor. In fact, this can be taken to be the defining property of Lie algebra cohomology:

Theorem \(H^k(\mathfrak{g}, M) = R^k (M)^\mathfrak{g}\).

Returning to field theory, we see (modulo some hard technicalities!) that, roughly, \(\mathfrak{g}\) is the Lie algebra of infinitesimal gauge transformations, and \(M\) is the algebra of functions on the space of all connections. The ghost and anti-ghost fields can then be seen to be the multiplication and contraction operators. To wit, we can take \(c^a\) to be the operator
\[ c^a: f \mapsto S^a \wedge f \]
and take \(\bar{c}^a\) to be the operator
\[ \bar{c}^a: f \mapsto \frac{\partial}{\partial S^a} f  = T_a \lrcorner f.\]
Then we have
\[ [c^a, \bar{c}^b] = \delta^{ab} \]
so that \(\bar{c}\) is indeed the antifield of \(c\).

BRST

Finally, I want to discuss gauge-invariant of the gauge-fixed theory. (!?) We saw in the previous posts that if we have a gauge theory with connection \(A\) and matter fields \(\psi\), in order to derive sensible Feynman rules we have to introduce a gauge-fixing function \(G\) as well as Fermionic fields \(c, \bar{c}\), the ghosts. (Note: last time I used \(\eta, \bar{\eta}\) for the ghosts but I want to match the more standard notation, so I've switched to \(c, \bar{c}\)).

Usually it is convenient to use the gauge-fixing function \(G(A) = \partial^\mu A_\mu\). Under an infinitesimal gauge-transformation \(\lambda\), \(A\) transforms as
\[ A \mapsto -\nabla \lambda, \]
so \(G(A)\) transforms as
\[ G(A) \mapsto G(A) - \partial^\mu \nabla_\mu \lambda. \]
Hence the term in the Lagrangian involving the ghosts is
\[ -\bar{c}^a \partial^\mu \nabla_\mu^{ab} c^b, \]
and our gauge-fixed Lagrangian is
\[ \mathcal{L} = -\frac{1}{4} |F|^2 + \bar{\psi}(iD\!\!\!/-m)\psi + -\frac{|\partial^\mu A_\mu|^2}{2\xi}
 - \bar{c}^a \nabla_\mu^{ab} c^b \]
Introducing an auxiliary filed \(B^a\), this is of course equivalent to
\[ \mathcal{L} = -\frac{1}{4} |F|^2 + \bar{\psi}(iD\!\!\!/-m)\psi + \frac{\xi}{2} B^a B_a
 + B^a \partial^\mu A_{\mu a} - \bar{c}^a \nabla_\mu^{ab} c^b. \]
Now, there are two questions one might ask: (1) how can we tell that this is a gauge-theory? i.e., what remains of the original gauge symmetry? and (2) does the resulting theory depend in any way on the choice of gauge-fixing function?

The answer to both of these questions is BRST symmetry. The field \(c\) is Lie-algebra valued, so we could think of it as being an infinitesimal gauge transformation. Rather, for \(\epsilon\) a constant odd variable, \(\epsilon c\) is even and an honest infinitesimal gauge transformation. Under this transformation, we have
\[ \delta_\epsilon A = -\nabla (\epsilon c) = -\epsilon \nabla c. \]
Then we define a graded derivation \(\delta\) by
\[ \delta A = - \nabla c. \]
We have a grading by ghost number, where \(\mathrm{gh}(A) = 0, \mathrm{gh}(\psi) = 0, \mathrm{gh}(c) = 1, \mathrm{gh}(\bar{c}) = -1\). We would like to extend \(\delta\) to a derivation of degree \(+1\) that squares to 0. First, we should figure out what \(\delta c\) is. We compute:
\begin{align}
0 &= \delta^2 A \\
&= \delta(-\nabla c) \\
&= -\partial \delta c - (\delta A) c - A (\delta c) + (\delta c) A - c (\delta A) \\
&= -\partial \delta c + (\nabla c) c + c (\nabla c) - [A, \delta c] \\
&= -\nabla(\delta c) + \nabla(c^2).
\end{align}
From this, we see that \(\nabla(\delta c) = \nabla(c^2)\), so we can set
\[ \delta c = c^2 = \frac{1}{2}[c, c]. \]
Then \(\delta^2 c = 0\) is just the Jacobi identity for the group's Lie algebra! Finally, we would like to extend \(\delta\) to act on \(\psi\), \(B\), and \(\bar{c}\) so that \(\delta \mathcal{L} = 0\), and \(\delta^2 = 0\). Since the action on \(A\) is by infinitesimal gauge transformation, this leaves the curvature term of \(\mathcal{L}\) invariant. Similarly, the \(\psi\) term is invariant if we simply take
\[ \delta \psi = c \cdot \psi \]
where dot denotes the infinitesimal gauge transformation. Using the known rules for \(\delta\), we find that
\[ \delta \mathcal{L} = \frac{\xi}{2} \left(\delta B B + B \delta B \right) + \delta B \cdot \partial^\mu A_\mu
 - B \cdot \partial^\mu \nabla_\mu c - \delta\bar{c} \cdot \partial^\mu \nabla_\mu c \]
By comparing coefficients, we find (together with what we've already computed)
\begin{align}
\delta A &= -\nabla c \\
\delta \psi &= c \cdot \psi \\
\delta c &= \frac{1}{2}[c,c] \\
\delta \bar{c} &= B \\
\delta B &= 0.
\end{align}
This is the BRST differential. Now, suppose that \(\mathcal{O}(A, \psi)\) is a local operator involving the physical fields \(A\) and \(psi\). Then by construction,\(delta O\) is the change of \(O\) under an infinitesimal gauge transformation. Hence, we find

An operator \(\mathcal{O}\) is gauge invariant \(\iff \delta\mathcal{O} = 0\).

Now, suppose the functional measure \(\mathcal{D}A \mathcal{D}\psi \mathcal{D}B \mathcal{D}c \mathcal{D}\bar{c}\) is gauge-invariant, i.e. is BRST closed. (This assumption is equivalent to the absence of anomalies, but we'll completely ignore this in today's post.) Then we have
\[ \langle \delta \mathcal{O} \rangle = 0 \]
for any local observable \(\mathcal{O}\). This just follows from integration by parts (this is where we have to assume the measure is \(\delta\)-closed). Now, why is this significant? First, this tells us that the space of physical observables is
\[ H^0(C^\ast_{\mathrm{BRST}}, \delta) \]
where \(C^\ast_{\mathrm{BRST}}\) is the cochain complex of local observables, graded by ghost number.

Now, the real power of the BRST formalism is the following. We find that the gauge-fixed Lagrangian can be written as
\[ \mathcal{L}_{gf} = \mathcal{L}_0 +\delta \left(\bar{c} \frac{B}{2} + \bar{c}\Lambda\right) \]
where \( \Lambda = \partial^\mu \nabla_\mu A \) is our gauge-fixing function, and \(\mathcal{L}_0\) is the original Lagrangian without gauge-fixing. Now the point is, any two choices of gauge fixing differ by terms which are BRST exact, and hence give the same expectation values on the physical observables \(H^0\). So we have restored gauge invariance, while obtaining a gauge-fixed perturbation theory!

Fadeev-Popov Ghosts, continued

Last time I sketched how we can represent an integral over a submanifold \(M \subset \mathbb{R}^n\) by an integral of the form
\[ \int_{\mathbb{R}^n} f(x) \delta(G(x)) \exp\left(\bar{\eta}G(x+\eta) \right) d\eta d\bar{\eta} dx. \]
Here, \(\eta, \bar{\eta}\) are Fermionic variables called Fadeev-Popov ghosts, which are introduced to cancel an unwanted determinant factor. The function \(G(x)\) singles out the submanifold \(M\) as \(M = G^{-1}(0)\).

Now suppose that we start with a vector (or affine) space \(V\), which is acted on by a group \(H\). We would like to undstand integrals over the quotient \(V / H\) in terms of integrals over \(V\). Suppose there is some function \(G(x)\) on \(V\) satisfying the following property:

For each level \(w\) of \(G\), the subspace \(M_w := G^{-1}(w)\) intersects the orbits transversely, and furthermore every \(H\)-orbit intersects \(M_w\) exactly once. (*)

We call such a function a gauge-fixing function, and a level \(w\) a gauge-fixing. By assumption, we have \(V/H \cong M_w\) for any \(w\). Hence using the integral we derived last time, we can integrate over \(M_w\) for any particular choice of \(w\), and this ought to be the same as integrating over \(V/H\). The problem, however, is that in the QFT setting it's not clear what the Feynman rules should be for such a path integral. The final trick is that since the answer should be independent of \(w\), and by integrating over all possible \(w\) we obtain a Lagrangian from which we can derive sensible Feynman rules.

We have some integral

\[Z = \int \delta(G(x) - w) \exp\left\{\frac{i}{\hbar}S(x) + \bar{\eta}dG \eta\right\} dx d\eta d\bar{\eta} \]
which is independent of \(w\). So we add a Gaussian weight an integrate over \(w\):
\begin{align}
Z' &= \int \delta(G(x) -w) \exp\left\{\frac{i}{\hbar}S(x) + \bar{\eta}dG \eta - \frac{1}{2\xi} |w|^2\right\}
 dx dw d\eta d\bar{\eta}\\
&= \int \exp\left\{\frac{i}{\hbar}S(x) + \bar{\eta}dG \eta - \frac{1}{2\xi} |G(x)|^2 \right\} dx d\eta d\bar{\eta}.
\end{align}
Here, \(\xi\) is an arbitrary real positive constant, and we denote the new integral by \(Z'\) to indicate that it differs from the old path integral \(Z\) by (at most) an overall constant. Now the important thing is that the new action appear in the integrand of \(Z'\) is gauge-fixed and hence there is no problem whatsoever in deriving sensible, meaningful Feynman rules. The gauge-fixing term \(|G(x)|^2\) serves to make the action non-degenerate, so that propagators are well-defined, while the term involving the Fermions \(\eta, \bar{\eta}\) generates new Feynman rules that "cancel" the superfluous degrees of freedom due to gauge redundancy.

The question remains, what if we choose some other gauge-fixing function? i.e., what happens if we perturb \(G(x)\) to some new function satisfying property (*)? We'll answer this using the BRST formalism.

Fadeev-Popov Ghosts

Today I want to review the Fadeev-Popov procedure, with a view toward BRST and eventually BV.

Gauge-Invariance and Gauge-Fixing

First we'll review the Fedeev-Popov method, to motivate the introduction of ghosts. Suppose we have a gauge theory involving a \(G\)-connection \(A\) and some field \(\phi\) charged under \(G\). Under a gauge transformation \(g(x)\), \(\phi\) transforms as
\[ \phi \mapsto g \cdot \phi. \]
We would like that the covariant derivative transforms in the same way, i.e.
\[ \nabla \phi \mapsto g \nabla \phi. \]
In terms of the connection 1-form \(A\), the covariant derivative is
\[ \nabla = d + A. \]
Let \(\nabla'\) denote the gauge-transformed covariant derivative, and \(\phi'\) the gauge-transformed field. Then we want
\[ \nabla' \phi' = g \nabla \phi. \]
We compute
\begin{align}
\nabla' \phi' &= (d + A')(g \phi) \\
&= dg \phi + g d\phi + A'(g\phi) \\
&= gg^{-1}dg \phi + gd\phi + g g^{-1} A' g\phi \\
&= g(d\phi + g^{-1} A' g \phi + g^{-1} dg \phi \\
&= g(d\phi + A\phi)
\end{align}
Comparing terms, we see that
\[ A = g^{-1} A' g + g^{-1} dg, \]
so upon re-arranging we have
\[ A' = g A g^{-1} - dg g^{-1} \]

This causes a problem: at critical points of the action, the Hessian of the action is degenerate in directions tangent to the gauge orbits. This means that the propagator is not well-defined, and there is no obvious way to derive the Feynman rules for perturbation theory. The solution is to take the quotient by gauge-transformations. To do this, we pick some gauge-fixing function \(G(A)\) which ought to be transverse to the orbits. Then we can restrict to the space \(G(A) = 0\), on which the Hessian of the action is non-degenerate, leading to a well-defined propagator. Formally, the path integral is
\[ Z = \int_{\{G(A) = 0\}} \exp{\frac{i}{\hbar} S[A, \phi]} \mathcal{D}A \mathcal{D}\phi  \]
Formally, this suggests that the path integral should be something like
\[ Z = \int \delta(G(A)) \exp{\frac{i}{\hbar} S[A, \phi]} \mathcal{D}A \mathcal{D}\phi, \]
but this is not quite right! To understand the source of the problem, we'll first study the finite-dimensional case and then use this to solve the problem in infinite-dimensions.

The Fadeev-Popov Determinant

Suppose we are on \(\mathbb{R}^n\), and we would like to integrate a function \(f(x)\) over a submanifold \(M\) defined by \(M = G^{-1}(0)\) for some smooth function \(G: \mathbb{R}^n \to \mathbb{R}^k\). Naively, we might expect that the answer is
\[ \int_M f(x) \stackrel{?}{=} \int f(x) \delta(G(x)) dx. \]
To see why this is not correct, write the delta function as
\[ \delta(G(x)) = \frac{1}{(2\pi)^k}\int e^{ip\cdot G(x)} d^k p. \]
We can regularize this by taking the limit as \(\epsilon \to 0\) of
\[ \frac{1}{(2\pi)^k}\int \exp\left\{ip\cdot G(x) -\frac{\epsilon}{2} |p|^2\right\} d^k p \]
This integral is Gaussian, so we obtain explicitly
\[ \left(\frac{2\pi}{\epsilon} \right)^{\frac{k}{2}} \exp\left\{-\frac{1}{2\epsilon} |G(x)|^2 \right\}. \]
So our original guess becomes
\[ \left(\frac{1}{2\pi \epsilon}\right)^\frac{k}{2} \int f(x) \exp\left\{ -\frac{1}{2\epsilon} |G(x)|^2 \right\} dx .\]
As \(\epsilon \to 0\), this integral localizes on the locus \(\{G(x) = 0\}\), as desired, but does not give the right answer! To see this, let \(u\) be a coordinate on \(M = G^{-1}(0)\) and \(v\) coordinates normal to \(M\). Then we have
\[ G(x) = G(u,v) = v^T H(u) v + o(|v|^3) \]
where \(H(x)\) is the Hessian of \(|G|^2\) at the point \(x = (u, 0)\). So the integral becomes (as \(\epsilon \to 0\))
\begin{align}
I_\epsilon &= \left(\frac{1}{2\pi \epsilon}\right)^\frac{k}{2} \int\int
 f(u, v) \exp\left\{ -\frac{1}{2\epsilon} v^T H(u) v \right\} du dv \\
&= \int_M \frac{f(u)}{\sqrt{\det H(u)}} du.
\end{align}
This is not correct. We have to account for the determinant of the Hessian. Now, the Hessian is given by
\begin{align} H_{ij} &= \frac{1}{2} \frac{\partial^2 |G|^2}{\partial v^i \partial v^j} \\
&= \frac{\partial}{\partial v^i} \left( G^a \partial_j G^a \right) \\
&=  \left(\partial_i G^a \partial_j G^a + G^a \partial_{ij} G^a \right) \\
&=  \partial_i G^a \partial_j G^a
\end{align}
where we have used the fact that \(G = 0\) on \(x = (u, 0)\). Hence we see that
\[ \det H = (\det A)^2 \]
where \(A\) is the \(k \times k\) matrix with entries \(\partial_i G^a\). Hence
\[ \sqrt{\det H} = \det A. \]
Now there is a straightforward way to eliminate the determinant. We introduce Fermionic coordinates \(\eta^i, \theta^i\), \(i = 1, \ldots, k\). Then by Berezin integration, we have
\[ \int e^{\eta^i G^i(0, \theta^j)} d\theta d\eta = \det A. \]
So in the end, we find
\[ \int_M f(x) d\mu = \int_{\mathbb{R}^n}
f(x) \delta(G(x)) \exp\left(\eta \cdot G(x+ \theta) \right)
 dx d\theta d\eta. \]

Equations of Motion and Noether's Theorem in the Functional Formalism

First, let us recall the derivation of the equations of motion and Noether's theorem in classical field theory. We have some action functional \(S[\phi]\) defined by some local Lagrangian:
\[ S[\phi] = \int L(\phi, \partial \phi) dx. \]
The classical equations of motion are just the Euler-Lagrange equations
\[ \frac{\delta S}{\delta \phi(x)} = 0
 \iff \partial_\mu \left( \frac{\partial L}{\partial(\partial_\mu\phi)} \right)
= \frac{\partial L}{\partial \phi} \]

Now suppose that \(S\) is invariant under some transformation \(\phi(x) \mapsto \phi(x) + \epsilon(x) \eta(x)\), so that \(S[\phi] = S[\phi+\epsilon \eta]\). Here we treat \(\eta\) as a fixed function but \(\epsilon\) may be an arbitrary infinitesimal function. The Lagrangian is not necessarily invariant, but rather can transform with a total derivative:
\[ L(\phi+\epsilon \eta) = L(\phi)
+ \frac{\partial L}{\partial (\partial_\mu \phi)} \eta \partial_\mu \epsilon
+ \epsilon \partial_\mu f^\mu \]
For some unknown vector field \(f^\mu\) (which we could compute given any particular Lagrangian). So let's compute
\begin{align}\delta_\epsilon S &= \int \delta_\epsilon L \\\
&= \int \frac{\partial L}{\partial (\partial_\mu \phi)}
\eta \partial_\mu \epsilon + \epsilon \partial_\mu f^\mu \\\
&= \int \partial_\mu \left(f^\mu - \frac{\partial L}{\partial (\partial_\mu \phi)} \eta \right) \epsilon
\end{align}
Let us define the Noether current \(J^\mu\) by
\[ J^\mu = \frac{\partial L}{\partial (\partial_\mu \phi)} \eta - f^\mu. \]
Then the previous computation showed that
\[ \frac{\delta S}{\delta \epsilon} = -\partial_\mu J^\mu. \]
If \(\phi\) is a solution to the Euler-Lagrange equations, then the variation \(dS\) vanishes, hence we obtain:

Theorem (Noether's theorem) The Noether current is divergence free, i.e.
\[ \partial_\mu J^\mu = 0.\]

Functional Version

First, we derive the functional analogue of the classical equations of motion. Consider an expectation value
\[ \langle \mathcal{O(\phi)} \rangle
= \int \mathcal{O}(\phi) e^{\frac{i}{\hbar} S} \mathcal{D}\phi \]
We'll assume that \(\phi\) takes values in a vector space (or bundle). Then we can perform a change of variables \(\psi = \phi + \epsilon\), and since \(\mathcal{D}\phi = \mathcal{D}\psi\) we find that
\[ \int \mathcal{O}(\phi+\epsilon) \exp\left(\frac{i}{\hbar} S[\phi] \right) \mathcal{D}\phi \]
is independent of \(\epsilon\). Expanding to first order in \(\epsilon\), we have
\[ 0 = \int \left(\frac{\delta\mathcal{O}}{\delta \phi}
 + \frac{i \mathcal{O}}{\hbar} \frac{\delta S}{\delta \phi} \right)
 \exp \left( \frac{i}{\hbar} S \right) \mathcal{D}\phi  \]
So we find the quantum analogue of the equations of motion:
\[ \left\langle \frac{\delta \mathcal{O}}{\delta \phi} \right\rangle
+ \frac{i}{\hbar} \left\langle \mathcal{O} \frac{\delta S}{\delta \phi} \right\rangle = 0\]

Next, we move on to the quantum version of Noether's theorem. Suppose there is a transformation \(Q\) of the fields leaving the action invariant. Assuming the path integral measure is invariant, we obtain
\[ \left\langle QF \right \rangle + \frac{i}{\hbar} \left\langle F QS \right\rangle = 0\]
To compare with the classical result, consider \(Q\) to be the (singular) operator
\[ Q = \frac{\delta}{\delta \epsilon(x)} \]
Then by the previous calculations,
\[ Q S = -\delta_\mu J^\mu, \]
so we obtain
\[ \left\langle \frac{\delta \mathcal{O}}{\delta \epsilon(x)} \right\rangle
= \frac{i}{\hbar} \left\langle \mathcal{O} \partial_\mu J^\mu \right\rangle.  \]
This is the Ward-Takahashi identity, the quantum analogue of Noether's theorem.

The 1PI Effective Action

In this post I'd like to try to understand the 1PI effective action that is often of interest. Suppose we have a QFT in some bosonic field \(\phi(x)\) taking values in a vector space (this is important). Then its vev \(\phi_{cl}(x) := \langle \phi(x) \rangle\) is just an ordinary (but possibly distributional) field on spacetime. The question is, what is the field equation satisfied by \(\phi_{cl}\)? I.e., if we average over quantum effects by replacing all fields by their vevs, what is the action that governs this (now completely classical) theory? The 1PI effective action answers exactly this question.

Consider the generating functional
\[ Z[J] = \int e^{-S[\phi]+\langle \phi, J \rangle} \mathcal{D}\phi \]
Then for a given source \(J\), define the \(J\)-vev of \(\phi(x)\) to be
\[ \phi_J(x) = \frac{\partial \log Z[J]}{\partial J}. \]
Now let's take \(\Gamma\) to be the Legendre transform of \(\log Z\) with respect to \(J\):
\[ \Gamma[\phi_J] = \langle J, \phi_J \rangle - \log Z[J] \]
Then we compute:
\[ \frac{\partial \Gamma}{\partial \phi_J} = J + \frac{\partial J}{\partial \phi_J} \phi_J
-\frac{\partial \log Z[J]}{\partial J} \frac{\partial J}{\partial \phi_J} = J. \]

Now consider the situation without a background source, i.e. \(J = 0\). Then \(\phi_0 = \phi_{cl}\) and we find
\[ \frac{\partial \Gamma}{\partial \phi_{cl}} = 0 \]
Hence, \(\phi_{cl}\) satisfies the Euler-Lagrange equations associated to the functional \(\Gamma\). Note that from the Legendre transform, \(\Gamma\) takes quantum effects (i.e. Feynman diagrams with loops) into account, even though the field and the equations are purely classical!

By studying these equations, we might find instanton solutions (or solitons in Lorentz signature).

Now for the name. Some combinatorics and algebra (which I will skip!) show that \(\Gamma[\phi_{cl}]\) is itself a generating functional for certain correlation functions, then 1PI correlation functions:
\[ \frac{\partial^n \Gamma}{\partial \phi(x_1) \cdots \partial \phi(x_n)} = \langle \phi(x_1) \cdots \phi(x_n) \rangle_{1PI}. \]
The 1PI subscript means that the RHS is computed in perturbation theory by summing over only the connection 1PI (1 particle irreducible) Feynman diagrams.

Warning: As usual, there are regularization issues, both in the UV and IR. UV divergences can be solved by a cutoff (if we only care about effective field theory), but IR divergences are much more technical. For this reason (and others), it is sometimes preferable to try to understand the low energy dynamics by studying the Wilsonian effective action. As the Wilsonian effective action does not take IR modes into account, it can avoid many of the difficulties of the 1PI effective action.

Spontaneous Symmetry Breaking in QFT

In this post I want to try to understand symmetry breaking and the origin of the moduli space of vacua. Most of this can be found in the lectures by Witten in vol. 2 of Quantum Fields and Strings.

Non-Example: Quantum Mechanics

The main point of confusion for me is that my quantum intuition comes from ordinary quantum mechanics. However, this turns out to be incredibly misleading because for most reasonable quantum mechanical systems, spontaneous symmetry breaking cannot occur. In fact, we'll see that even in field theory, the question of whether spontaneous symmetry breaking can occur is intimately related to the geometry of spacetime. Since quantum mechanics is QFT in \(0+1\) dimensions (i.e., the spatial part of spacetime is just a point), spontaneous symmetry breaking is forbidden.

Consider a particle in one spatial dimension, with Hamiltonian
\[ H = -\frac{\hbar^2}{2} + (a^2-x^2)^2. \]
The classical ground states are given by the stationary solutions \(x(t) = \pm a\). Hence we might expect that the quantum Hamiltonian has a degenerate ground state, i.e. the eigenspace of the lowest eigenvalue has dimension greater than one. However, this is not the case!

Sketch of proof: Define a function \(E(\phi)\) on the unit sphere in \(L^2(\mathbb{R})\). If \(\phi\) is a global minimum, then it necessarily satisfies \(H\psi = E_0 \phi\) where \(E_0\) is the lowest eigenvalue of \(H\). On the other hand, \(E(\phi) = E(|\phi|)\) so \(|\phi|\) is also a global minimum, hence satisfies \(H|\phi| = E_0|\phi|\). This equation is elliptic, so by elliptic regularity \(|\phi|\) must be at least \(C^1\), and hence \(\phi(x)\) has constant phase, so we might as well take \(\phi(x)\) to be real and positive. Any other ground state \(\psi\) would have these properties, and hence \((\phi,\psi) \neq 0\). Hence the eigenspace is 1-dimensional and the ground state is not degenerate.

Now by the Stone-von Neumann theorem, there is a unique irreducible representation of the canonical commutation relations on a separable Hilbert space, and by the argument above there is a unique (up to scale) vector \(|\Omega\rangle\) which is the ground state of the Hamiltonian, called the vacuum vector. So for QM, we find a unique representation on \(\mathcal{H}\) together with a unique vacuum vector \(|\Omega\rangle \in \mathcal{H}\). The point I want to stress here is that the Poisson algebra of observables together with the Hamiltonian determine the data \(\mathcal{H}, |\Omega\rangle)\) in an essentially unique way, so there is no ambiguity in quantization and no further choices need to be made.

QFT in Finite Volume

Now we'll argue that a similar symmetry breaking phenomenon should be expected whenever the spatial part of spacetime has finite volume. We'll have to use formal path integral arguments, so of course this won't be totally rigorous. Suppose we have two representations \(\mathcal{H}_\pm\) with vacua \(|\Omega\rangle_\pm\). Then we consider the direct sum \(\mathcal{H} = \mathcal{H}_+ \oplus \mathcal{H}_-\). Then by construction, we should have
\[ (\Omega_+, e^{-tH} \Omega_-) = 0 \]
since \(|\Omega\rangle_\pm\) are orthogonal eigenstates of \(H\). On the other hand, we can compute this inner product using the Feynman-Kac formula. The semi-classical approximation to the path integral yields
\[ (\Omega_+, e^{-tH} \Omega_-) = C \exp\left(- \frac{S(t) V}{\hbar} \right) \]
Where \(S(t)\) is the classical least action and \(V\) is the volume of space. If \(V < \infty\), the right hand side is non-zero, contradicting our assumptions! Hence the vacuum is non-degenerate. So we find that QFT with finite spatial volume is much like QM, at least as far as symmetry breaking is concerned.

Note that this argument is essentially just a formal manipulation of the path integral, so you should expect a result of this form independent of the particular regularization scheme used to define the path integral.

QFT in Infinite Volume

Now we consider the case \(V = \infty\), i.e. a non-compact space. Then the preceding argument fails spectacularly, as does the Stone-von Neumann theorem. So there is no guarantee of a unique irreducible representation of the algebra, and no guarantee of a unique vacuum vector.
So we see that the situation is significantly more complicated. We can expect a moduli space \(\mathcal{M}\) of vacua of the theory, and the low energy effective theory is described by a \(\sigma\)-model with target \(\mathcal{M}\). I'll try to discuss this in more detail in follow-up posts.

Gaussian Integrals: Wick's Theorem

We saw in the last update that the generating function \(Z[J]\) can be expressed as
\[ Z[J] = e^{\frac{1}{2} J \cdot A^{-1} J} \]
(at least as long as we've normalize things so that \(Z[0] = 1\). Now the wonderful thing is that this is something we can compute explicitly:
\[ Z[J] = \sum_{n = 0}^{\infty} \frac{(\frac{1}{2} A^{-1}_{ij} J^i J^j)^n}{n!}
= \sum_{n=0}^\infty \frac{(A^{-1}_{ij} J^i J^j)^n}{2^n n!} \]

For example, in the one-dimensional case (taking \(A = 1\)) we get
\[ Z[J] = \sum_{n=0}^\infty \frac{J^{2n}}{2^n n!} \]
On the other hand, by the definition of the generating function we have
\[ Z[J] = \sum_{n=0}^\infty \frac{\langle x^n \rangle}{n!} J^n \]
Comparing coefficients, we find
\[ \frac{\langle x^{2n} \rangle}{(2n)!} = \frac{1}{2^n n!} \]
so that
\[ \langle x^{2n} \rangle = \frac{(2n)!}{2^n n!}. \]
Let's give a combinatorial description. Given \(2n\) objects, in how many ways can we divide them into pairs? If we care about the order in which we pick the pairs, then we have
\[ {2n \choose 2}{2n - 2 \choose 2} \cdots {2n-(2n-2) \choose 2} = \frac{(2n)!}{2^n} \]
Of course, there are \(n!\) ways of ordering the \(n\) pairs, so after dividing by this (to account for the overcounting) we get exactly the expression for \(\langle x^{2n} \rangle\). This is the first case of Wick's theorem.

Now consider the general multidimensional case. Given \(I = (i_1, \cdots, i_{2n})\), we define a contraction to be
\[ \langle x^{j_1} x^{k_1} \rangle \cdots \langle x^{j_n} x^{k_n} \rangle \]
where \(j_1, k_1, \cdots, j_n, k_n\) is a choice of parition of \(I\) into pairs.

Theorem (Wick's theorem, Isserlis' theorem) The expectation value
\[ \langle x^{i_1} \cdots x^{i_{2n}} \rangle \]
is the sum over all full contractions. There are \((2n)!/ 2^n n!\) terms in the sum.

Proof This follows from our formula for the power series of the generating function. The reason is that the coefficient of  \(J^I\) in \((\frac{1}{2} A^{-1}_{ij} J^i J^k)^n\) is exactly given by summing products of \(A^{-1}_{ij}\) over partitions of \(I\) into pairs, and the \(n!\) in the denominator takes care of the overcounting.

Next up: perturbation theory and Feynman diagrams.

Introduction to Gaussian Integrals

As a warm-up for more serious stuff, I'd like to discuss Gaussian integrals over \(\mathbb{R}^d\). Gaussian integrals are the main tool for perturbative quantum field theory, and I find that understanding Gaussian integrals in finite dimensions is an immense aid to understanding how perturbative QFT works. So let's get started.


The Basics

Let \(A\) be some \(d \times d\) symmetric positive definite matrix. We are interested in the integral
\[ \int_{-\infty}^\infty \exp(-\frac{x \cdot Ax}{2}) dx. \]
Out of laziness, I will suppress the limits of integration and just write this as
\[ \int e^{-S(x)} dx. \]
where \(S(x) = x \cdot Ax / 2\). Now for a function \(f(x)\), we define the expectation value \(\langle f(x) \rangle\) to be
\[ \langle f(x) \rangle_0 = \int f(x) e^{-S(x)} dx \]
Occasionally, we might care about the normalized expectation value
\[ langle f(x) \rangle = \frac{\langle f(x) \rangle_0}{\langle 1 \rangle_0} = \frac{1}{\langle 1 \rangle_0} \int f(x) e^{-S(x)} dx. \]
We mostly care about asymptotics, so we will typically think of a function \(f(x)\) as being a polynomial (or Taylor series). So what we're really interested in is
\[ \langle x^I \rangle = c\int x^I e^{-S(x)} dx, \]
where \(I\) is a multi-index.

The Partition Function

Let us define \(Z[J]\) by
\[ Z[J] = \int e^{-S(x) + J \cdot x} dx. \]

Now the great thing is that
\[ \langle x^I \rangle = \left. \frac{d^I}{dJ^I} \right|_{J = 0} Z[J], \]
so that once we know \(Z[J]\), we can calculate anything. So let's try to compute it. We have

\begin{align}
(Ax - J) \cdot A^{-1} (Ax - J) &= (Ax - J) \cdot (x - A^{-1} J) \\\
&= x \cdot Ax - x \cdot J - J \cdot x + J \cdot A^{-1} J \\\
&= x \cdot Ax - 2 x \cdot J + J \cdot A^{-1} J.
\end{align}
So we see that
\[ -\frac{1}{2} x \cdot A x + J \cdot x = \frac{1}{2} J \cdot A^{-1} J -\frac{1}{2} (x-A^{-1}J) \cdot A(x - A^{-1} J). \]
So, after a change of variales \(x \mapsto x - A^{-1} J\) we find
\[ Z[J] = e^{\frac{1}{2} J \cdot A^{-1} J} Z[0]. \]
Now the argument in the exponential is
\[ \frac{1}{2} A^{-1}_{ij} J^i J^j \]
So we find that
\[ \langle x^i x^j \rangle = \frac{d^2}{dx^i dx^j} Z[J]|_{J = 0} = A^{-1}_{ij}. \]

Now we are ready to prove Wick's theorem and discuss Feynman diagrams, which we'll do in the next post.

Path Integrals 3: Recovering the Spectrum from Asymoptotics

In my previous posts on path integrals, I described (rather tersely) how the path integral, suitably defined and interpreted, can be used to compute the Schwartz kernel of the operators \(e^{iHt}\) (Lorentzian signature) and \(e^{-Ht}\) (Euclidean signature).

Suppose that we understand the spectrum of \(H\) completely (nb: for a given system described by \(H\), this is the goal). For example, suppose we know that the spectrum of \(H\) consists of discrete eigenvalues \(E_n, n = 0, \cdots\) with corresponding eigenvectors \(|n\rangle\),
\[ H|n\rangle = E_n|n\rangle. \]
(For simplicity, I assume there is no continuous spectrum and that the eigenvalues are nondegenerate.) Then we have
\[ e^{-iHt} = \sum_n e^{-i E_n t} \langle n|n\rangle \]
and
\[ e^{-Ht} = \sum_n e^{-E_n t} \langle n|n\rangle \]
Now the second expression turns out to be very useful. Assume the eigenvalues are ordered so that
\[ E_0 < E_1 < \cdots \]
Then we can write
\[ e^{-Ht} = e^{-E_0 t} |0\rangle + \sum_{n \geq 1} e^{-(E_n-E_0)t}|n\rangle \]
Now suppose that \(v\) is some vector which is close to the ground state, in the sense that
\[ \langle v|0\rangle \neq 0 \]
(This is obviously a generic condition, so if we just pick \(v\) randomly we can expect this to be true.) Then we can consider
\[ e^{-Ht} v = e^{-E_0 t} v_0 |0\rangle + \sum_{n \geq 1} e^{-(E_n-E_0)t} v_n|n\rangle \]
Now for \(n \geq 1\), \(E_n-E_0\) is strictly positive, and so for large \(t\) all of the higher terms are exponentially damped. So, we have the asymptotic
\[ e^{-Ht} v \sim e^{-E_0 t }v_0|0\rangle \]
Next comes the really interesting part. Multiply on the right by a position-representation eigenbra \(\langle x|\):
\[ \langle x|e^{-Ht} v \sim e^{-E_0 t} v_0 \langle x|0\rangle \]
Now \(v_0\) is an irrelevant constant, so we might as well take it to be 1 (rescale \(v\) as necessary). The expression \(\langle x|0\rangle\) is exactly the ground state wavefunction in the position representation! Call it \(\psi_0(x)\). So to conclude: the large-t asymptotic of the expression \(\langle x|e^{-Ht}v\) is (up to an overall constant) given by \(e^{-E_0 t} \psi_0(x)\), hence we can recover both the ground state energy and the ground state wavefunction. But the value of this expression is exactly given by the Euclidean path integral. So we have a correspondence:

Asymptotics of Euclidean path integral \(\leftarrow\rightarrow\) The spectrum of \(H\).

Coming next: instantons.

Path Integrals 2: Euclidean Path Integrals and Heat Kernels

Consider the heat equation
\[ \frac{\partial \psi}{\partial t} = -\hat{H} \psi \]
We find similarly
\[ \langle x_N|U_t|x_0\rangle = \int \exp \sum_{j=0}^{N-1} ik_j(x_j - x_{j+1}) -\Delta t H(x_j, k_j) dx dk. \]

Now take \(H(x,k) = k^2/2m + V(x)\) and complete the square:

\[ ik_j(x_j - x_{j+1}) - \Delta t k_j^2/2m - \Delta tV(x_j) + ik_{j+1}(x_{j+1} - x_{j+2}) - k_{j+1}^2/2m - V(x_{j+1}) \]

\begin{align}
 ik_j a_j - \Delta t k_j^2/2m &= -(-2mi k_j a_j / \Delta t + k_j^2) \Delta t/2m \\
&= -(k_j^2 -2mi k_j a_j/\Delta t -m^2 a_j^2/(\Delta t)^2 + m^2 a_j^2/(\Delta t)^2) \Delta t/2m \\
&= -(k_j -mia_j/\Delta t)^2 \Delta t/2m -ma_j^2/2\Delta t
\end{align}

Combining things together, we have that the heat kernel is given by
\[ \langle y|e^{-t \hat{H}}|x\rangle = \int e^{-S_{\textrm{euc}}} \mathcal{D}x \]

That is, Schwartz kernel of time evolution operator is given by the oscialltory Lorentzian signature path integral, whereas the heat kernel is given by the exponentially decaying path integral (better chance of being well-defined). Most importantly, the heat kernel contains most of the essential information about the spectrum of \(\hat{H}\), which is really all we need in order to understand the dynamics.

See ABC of Instantons. (I never understood the title of Nekrasov's "ABCD of Instantons" until I found this classic).

Path Integrals 1: Feynman's Derivation

Consider the Hilbert space \(\mathcal{H} = L^2(\mathbb{R})\) with Lebesgue measure and a Hamiltonian \(H = T(k) + V(x)\) (a sum of kinetic and potential energy). Then the quantum hamiltonian \(\hat{H}\) acts as

\begin{align}
(\hat{H}\psi)(y) &= \frac{1}{2\pi} \int e^{ik(y-x)} T(k) \psi(x) dx dk + V(y)\psi(y) \\
&= \frac{1}{2\pi}\int e^{ik(y-x)} H(k,x) \psi(x) dx dk \\
\end{align}

Now consider the Schrodinger equation
\[ \frac{\partial \psi}{\partial t} = i \hat{H} \psi. \]

We would like to obtain a formula for the solution operator \(U_t = e^{-i \hat{H} t}\). Let us consider its Schwartz kernel \(\langle{y}|U_t|{x}\rangle\). Let \(N\) be a large integer so that \(\Delta t = t/N\) is "small". Then we can write

\[ U_t = U_{\Delta t}^N, \]

Now consider a single term:

\begin{align}
 \langle{y}|U_{\Delta t}|x\rangle &\simeq \langle{y}|1-i\Delta{t}\hat{H}|x\rangle \\
&= \delta(y-x) -i \Delta t \langle x|\hat{H}|y\rangle \\
&= \frac{1}{2\pi}\int e^{ik(y-x)}( 1 - i \Delta t H(k,x)) dk \\
&\simeq \frac{1}{2\pi}\int e^{ik(y-x) - i \Delta t H(k,x)} dk\\
\end{align}
Now we have (taking \(x_0 = x\) and \(x_N = y\))
\begin{align}
\langle y|U_t|x\rangle &= \int dx_1 \cdots dx_{N-1} \\
& \ \times \langle x_N|U_{\Delta t}|x_{N-1}\rangle \cdots \langle x_1|U_{\Delta t}|x_0\rangle \\

&= \frac{1}{(2\pi)^N} \int dx_1 \cdots dx_{N-1} dk_0 \cdots dk_{N-1} \\
& \ \times \ e^{ik_N(x_N-x_{N-1}) - i \Delta t H(k_{N-1},x_{N-1})} \cdots
e^{-ik_N(x_1-x_0) - i \Delta t H(k_0,x_0)} \\
&= \frac{1}{(2\pi)^N} \int dx_1 \cdots dx_{N-1} dk_0 \cdots dk_{N-1} \\
& \ \times \ \exp \sum_{j=0}^{N-1} ik_j(x_{j+1} - x_j) - i \Delta t H(k_j,x_j)

\end{align}
Note this kind of integral seems like it should have a natural interpretation in symplectic or contact geometry, since the expression \(dxdk/(2\pi)^N\) is very nealy the Liouville measure. This is the most general form of the path integral.

Now assume that \(H(k,x) = k^2/2m + V(x)\). Then the \(k\)-dependent terms have the form
\[ \int e^{ik(y-x) - i\Delta t k^2/2m}. \]
Complete the square
\begin{align}
 ik(y-x) -i\Delta t k^2/2m &= -\frac{i \Delta t}{2m} (k^2 - \frac{2m}{\Delta t}k(y-x)) \\
&= -\frac{i \Delta t}{2m} (k^2 - \frac{2m}{\Delta t}k(y-x) + \frac{m^2}{\Delta t^2}(y-x)^2 -\frac{m^2}{\Delta t^2}(y-x)^2) \\
&= -\frac{i \Delta t}{2m}(k - \frac{m}{\Delta t}(y-x))^2 + \frac{i m}{2 \Delta t}(y-x)^2
\end{align}
Now, using that
\[ \int e^{-ak^2} = \sqrt{\frac{\pi}{a}} \]
We have
\[ \int e^{-\frac{i \Delta t}{2m}(k-\frac{m}{\Delta t}(y-x))^2} = \sqrt{\frac{2\pi m}{i \Delta t}} \]
Putting it altogether, we get the more familiar version of the path integral,
\begin{align} \langle y|U_t|x\rangle &\simeq C^N \int dx_1 \cdots dx_{N-1} \\
& \ \times \ \exp i\sum_{j=0}^{N-1} \frac{m}{2\Delta t}(x_{j+1}-x_j)^2 - V(x_j) \Delta t
\end{align}
where
\[ C = \frac{1}{2\pi} \sqrt{\frac{2 \pi m}{i \Delta t}} = \sqrt{\frac{m}{2\pi i \Delta t}} \]