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!