BRST formalism

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(A draft of an alternate exposition has been added at BRST quantization.)

In theoretical physics, the BRST formalism is a method of implementing first class constraints. The letters BRST stand for Becchi, Rouet, Stora, and (independently) Tyutin who discovered this formalism. It is a sophisticated method to deal with quantum physical theories with gauge invariance. For example, the BRST methods are often applied to gauge theory and quantized general relativity.

1 Classical version
2 Quantum version
- 2.1 Example
3 References

[edit] Classical version

This is related to a supersymplectic manifold where pure operators are graded by integral ghost numbers and we have a BRST cohomology.

[edit] Quantum version

The space of states is not a Hilbert space (see below). This vector space is both Z₂-graded and R-graded. If you wish, you may think of it as a Z₂×R-graded vector space. The former grading is the parity, which can either be even or odd. The latter grading is the ghost number. Note that it is R and not Z because unlike the classical case, we can have nonintegral ghost numbers. Operators acting upon this space are also Z₂×R-graded in the obvious manner. In particular, Q is odd and has a ghost number of 1.

Let H_n be the subspace of all states with ghost number n. Then, Q restricted to H_n maps H_n to H_n+1. Since Q²=0, we have a cochain complex describing a cohomology.

The physical states are identified as elements of cohomology of the operator $Q$ , i.e. as vectors in Ker Q_n+1/Im Q_n. The BRST theory is in fact linked to the standard resolution in Lie algebra cohomology.

Recall that the space of states is Z₂-graded. If A is a pure graded operator, then the BRST transformation maps A to [Q,A) where [,) is the supercommutator. BRST-invariant operators are operators for which [Q,A)=0. Since the operators are also graded by ghost numbers, this BRST transformation also forms a cohomology for the operators since [Q,[Q,A))=0. See the superJacobi identity.

Although the BRST formalism is more general than the Faddeev-Popov gauge fixing, in the special case where it is derived from it, the BRST operator is also useful to obtain the right Jacobian associated with constraints that gauge-fix the symmetry.

The BRST is a supersymmetry. It generates the Lie superalgebra with a zero-dimensional even part and a one dimensional odd part spanned by Q. [Q,Q)={Q,Q}=0 where [,) is the Lie superbracket (i.e. Q²=0). This means Q acts as an antiderivation. See algebra representation of a Lie superalgebra.

Because Q is Hermitian and its square is zero but Q itself is nonzero, this means the vector space of all states prior to the cohomological reduction has an indefinite norm! This means it is not a Hilbert space!

For more general flows which can't be described by first class constraints, see Batalin-Vilkovisky

[edit] Example

For the special case of gauge theories (of the usual kind described by sections of a principal G-bundle) with a quantum connection form A, a BRST charge (sometimes also a BRS charge) is an operator usually denoted $Q$ .

Let the $\mathfrak{g}$ -valued gauge fixing conditions be $G=\xi\partial^\mu A_\mu$ where ξ is a positive number determining the gauge. There are many other possible gauge fixings, but they will not be covered here. The fields are the $\mathfrak{g}$ -valued connection form A, $\mathfrak{g}$ -valued scalar field with fermionic statistics, b and c and a $\mathfrak{g}$ -valued scalar field with bosonic statistics B. c deals with the gauge transformations wheareas b and B deals with the gauge fixings. There actually are some subtleties associated with the gauge fixing due to Gribov ambiguities but they will not be covered here.

Q A = D c

where D is the covariant derivative.

$Qc={i\over 2}[c,c]_L$

where [,]_L is the Lie bracket, NOT the commutator.

Q B = 0

Q b = B

Q is an antiderivation. {Which antiderivation is it? Is it the Lie Algebra acting on B AND b?}

The BRST Lagrangian density

$\mathcal{L}=-{1\over 4g^2}Tr[F^{\mu\nu}F_{\mu\nu}]+{1\over 2g^2}Tr[BB]-{1\over g^2}Tr[BG]-{\xi\over g^2}Tr[\partial^\mu b D_\mu c]$

While the Lagrangian density isn't BRST invariant, its integral over all of spacetime, the action is.

The operator $Q$ is defined as

$Q = c^i \left(L_i-\frac 12 {f_{ij}}^k b_j c^k\right)$

where $c i, b i$ are the Faddeev-Popov ghosts and antighosts, respectively, $L i$ are the infinitesimal generators of the Lie group, and $f i j k$ are its structure constants.