6Non-abelian gauge theory
III Advanced Quantum Field Theory
6.5 BRST symmetry and cohomology
In the Faddeev–Popov construction, we have introduced some gauge-fixing terms,
and so naturally, the Lagrangian is no longer gauge invariant. However, we would
like to be able to use gauge symmetries to understand our theory. For example,
we would expect gauge symmetry to restrict the possible terms generated by the
renormalization group flow, but we now can’t do that.
It turns out that our action still has a less obvious symmetry arising form
gauge invariance, known as BRST symmetry. This was discovered by Becchi,
Rouet and Stora, and also independently by Tyruin.
To describe this symmetry, we are going to construct an BRST operator.
Since we want to prove things about this, we have to be more precise about
what space this is operating on.
We let
B
be the (complex) space of all polynomial functions in the fields and
their derivatives. More precisely, it is defined recursively as follows:
–
Let Ψ be any of
{A
µ
, c
a
, ¯c
a
, h
a
}
, and
∂
α
be any differential operator (e.g.
∂
1
∂
2
3
). Then ∂
α
Ψ ∈ B.
– Any complex C
∞
function on M is in B.
– If a, b ∈ B, then a + b, ab ∈ B.
We impose the obvious commutativity relations based on fermionic and bosonic
statistics. Note that by construction, this space only contains polynomial
functions in the fields. This is what we are going to assume when we try to prove
things, and makes things much more convenient because we can now induct on
the degree of the polynomial.
However, for a general gauge-fixing function, we cannot guarantee that the
Yang–Mills Lagrangian lives in
B
(even though for the Lorenz gauge, it does).
What we do can be extended to allow for more general functions of the fields,
but this is difficult to make precise and we will not do that.
This
B
is in fact a
Z/
2
Z
-graded algebra, or a superalgebra, i.e. we can
decompose it as
B = B
0
⊕ B
1
,
where
B
0
, B
1
are vector subspaces of
B
. Here
B
0
contains the “purely bosonic”
terms, while B
1
contains the purely fermionic terms. These satisfy
B
s
B
t
⊆ B
s+t
,
where the subscripts are taken modulo 2. Moreover, if
y ∈ B
s
and
x ∈ B
t
, then
we have the (graded-)commutativity relation
yx = (−1)
st
xy.
If
x
belongs to one of
B
0
or
B
1
, we say it has definite statistics, and we write
|x| = s if x ∈ B
s
.
Definition (BRST operator). The BRST operator Q is defined by
QA
µ
= ∇
µ
c Q¯c = ih
Qc = −
1
2
[c, c] Qh = 0.
This extends to an operator on
B
by sending all constants to 0, and for
f, g ∈ B
of definite statistics, we set
Q(fg) = (−1)
|f|
fQg + (Qf)g, Q(∂
µ
f) = ∂
µ
Qf.
In other words, Q is a graded derivation.
There are a couple of things to note:
–
Even though we like to think of
¯c
as the “complex conjugate” of
c
, formally
speaking, they are unrelated variables, and so we have no obligation to
make Q¯c related to Qc.
– The expression [c, c] is defined by
[c, c]
a
= f
a
bc
c
b
c
c
,
where
f
a
bc
are the structure constants. This is non-zero, even though the
Lie bracket is anti-commutative, because the ghosts are fermions.
–
The operator
Q
exchanges fermionic variables with bosonic variables. Thus,
we can think of this as a “fermionic operator”.
– It is an exercise to see that Q is well-defined.
We will soon see that this gives rise to a symmetry of the Yang–Mills action.
To do so, we first need the following fact:
Theorem. We have Q
2
= 0.
Proof. We first check that for any field Ψ, we have Q
2
Ψ = 0.
– This is trivial for h.
– We have
Q
2
¯c = Qih = 0.
– Note that for fermionic a, b, we have [a, b] = [b, a]. So
Q
2
c = −
1
2
Q[c, c] = −
1
2
([Qc, c] + [c, Qc]) = −[Qc, c] =
1
2
[[c, c], c].
It is an exercise to carefully go through the anti-commutativity and see
that the Jacobi identity implies this vanishes.
– Noting that G acts on the ghosts fields by the adjoint action, we have
Q
2
A
µ
= Q∇
µ
c
= ∇
µ
(Qc) + [QA
µ
, c]
= −
1
2
∇
µ
[c, c] + [∇
µ
c, c]
= −
1
2
([∇
µ
c, c] + [c, ∇
µ
c]) + [∇
µ
c, c]
= 0.
To conclude the proof, it suffices to show that if
a, b ∈ B
are elements of definite
statistics such that
Q
2
a
=
Q
2
b
= 0, then
Q
2
ab
= 0. Then we are done by
induction and linearity. Using the fact that |Qa| = |a| + 1 (mod 2), we have
Q
2
(ab) = (Q
2
a)b + aQ
2
b + (−1)
|a|
(Qa)(Qb) + (−1)
|a|+1
(Qa)(Qb) = 0.
We can now introduce some terminology.
Definition
(BRST exact)
.
We say
a ∈ B
is BRST exact if
a
=
Qb
for some
b ∈ B.
Definition (BRST closed). We say a ∈ B is BRST closed if Qa = 0.
By the previous theorem, we know that all BRST exact elements are BRST
closed, but the converse need not be true.
In the canonical picture, as we saw in Michaelmas QFT, the “Hilbert space”
of this quantum theory is not actually a Hilbert space — some states have
negative norm, and some non-zero states have zero norm. In order for things
to work, we need to first restrict to a subspace of non-negative norm, and then
quotient out by states of zero norm, and this gives us the space of “physical”
states.
The BRST operator gives rise to an analogous operator in the canonical
picture, which we shall denote
ˆ
Q
. It turns out that the space of non-negative
norm states is exactly the states
|ψi
such that
ˆ
Q|ψi
= 0. We can think of this
as saying the physical states must be BRST-invariant. Moreover, the zero norm
states are exactly those that are of the form Q|φi = 0. So we can write
H
phys
=
BRST closed states
BRST exact states
.
This is known as the cohomology of the operator
ˆ
Q
. We will not justify this or
further discuss this, as the canonical picture is not really the main focus of this
course or section.
Let’s return to our main focus, which was to find a symmetry of the Yang–
Mills action.
Theorem.
The Yang–Mills Lagrangian is BRST closed. In other words,
QL
= 0.
This theorem is often phrased in terms of the operator
δ
=
εQ
, where
ε
is a
new Grassmannian variable. Since
ε
2
= 0, we know (
δa
)(
δb
) = 0 for any
a, b
. So
the function that sends any
a
to
a 7→ a
+
δa
is well-defined (or rather, respects
multiplication), because
ab + δ(ab) = (a + δa)(b + δb).
So this
δ
behaves like an infinitesimal transformation with
ε
“small”. This is
not true for Q itself.
Then the theorem says this infinitesimal transformation is in fact a symmetry
of the action.
Proof.
We first look at the (
F
µν
, F
µν
) piece of
L
. We notice that for the purposes
of this term, since
δA
is bosonic, the BRST transformation for
A
looks just like
a gauge transformation
A
µ
7→ A
µ
+
∇
µ
λ
. So the usual (explicit) proof that this
is invariant under gauge transformations shows that this is also invariant under
BRST transformations.
We now look at the remaining terms. We claim that it is not just BRST
closed, but in fact BRST exact. Indeed, it is just
Q(¯c
a
f
a
[A]) = ih
a
f
a
[A] − ¯c
a
δf
δλ
c.
So we have found a symmetry of the action, and thus if we regularize the path
integral measure appropriately so that it is invariant under BRST symmetries
(e.g. in dimensional regularization), then BRST symmetry will in fact be a
symmetry of the full quantum theory.
Now what does this actually tell us? We first note the following general fact:
Lemma.
Suppose
δ
is some operator such that Φ
7→
Φ +
δ
Φ is a symmetry.
Then for all O, we have
hδOi = 0.
Proof.
hO(Φ)i = hO(Φ
0
)i = hO(Φ) + δOi.
Corollary.
Adding BRST exact terms to the Lagrangian does not affect the
expectation of BRST invariant functions.
Proof.
Suppose we add
Qg
to the Lagrangian, and
O
is BRST invariant. Then
the change in hOi is
Z
Qg(x) d
d
x O
=
Z
hQ(gO)i d
d
x = 0.
If we want to argue about this more carefully, we should use
εQ
instead of
Q
.
This has some pretty important consequences. For example, any gauge
invariant term that involves only
A
is BRST invariant. Also, changing the
gauge-fixing function just corresponds to changing the Lagrangian by a BRST
exact term,
Q
(
¯c
a
f
a
[
A
]). So this implies that as long as we only care about gauge-
invariant quantities, then all correlation functions are completely independent of
the choice of f .