BRST
This post has been migrated from my old blog, the math-physics learning seminar.
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!