This can be seen from the propagator, e.g. for U(1) [QED] we have
Z
Z
1
1 2
=
d4 xAµ (x)(η µν ∂ 2 − ∂ µ ∂ ν )Aν (x)
d4 x − Fµν
4
2
with the propagator
2
(ηµρ ∂ −
We constrain the gauge directions, by introducing a gauge fixing term at each
point x
Ga (A) = 0
Now define ∆F P [A] by [FP = Faddeev-Popov]
∂µ ∂ρ )DFρν (x
− y) =
iδµν δ(x
− y)
i∆µν
F
≡
DFµν
However ηµρ ∂ 2 − ∂µ ∂ρ cannot be inverted, as multiplying both sides of the
∆F P [A]
Z
[dU]δ[Ga (AU )] = 1
equation by ∂ µ gives
[δ[ ] is a product of δ functions at each space-time point.]
0 6= i∂ ν δ(x − y)
where Aµ is the gauge transformed A, i.e.
A contradiction!
Alternatively we can say that the operator has zero eigenvalues. (In the
canonical formalism, tackle this with the Gupta-Bleuler method.)
To complete the functional integral, we must ‘factor’ out these gauge transformations
i.e. “
Z
[dA] ∼
Z
[dĀ][dU] ”
i
AUµ = U(Aµ + ∂µ )U †
g
and [dU] is the Haar measure, invariant under group transformations, i.e. if
U ′ = U0 U, then [dU ′ ] = [dU].
∆F P [A] is invariant under gauge transformations as
Z
∆F P [AU0 ]−1 =
[dU]δ[Ga (AU0 U )]
U ′ = U0 U
Z
′
=
[dU ′ ]δ[Ga (AU )]
= ∆F P [A]−1
“gauge orbits” related by
gauge transformations
gauge inequivalent Aµ s
‘gauge orbits’ (related by gauge transformations)
gauge inequivalent Aµ s (assumed to interest each gauge orbit only once)
105
Thus we have
Z
Z
Z
R 4
R 4
[dA]ei d xLY M (A) =
[dA] ∆F P [A] [dU]δ[Ga (AU )] ei d xLY M (A)
{z
}
|
1
Z
Z
R 4
′
=
[dU] [dA′ ]∆F P [A′ ]δ[Ga (A′ )]ei d xLY M (A )
106
upon interchanging integrals, and then changing variables A′ = AU , ∆F P
Now
and LY M being gauge invariant.
We see that we have factored out the (infinite) gauge transformation ‘volume’,
R
[dU]. Now throw it away (it cancels in ratios anyway).
Z
[dA]ei
R
d4 xLY M (A)
∼
Z
[dA]∆F P [A]δ[Ga (A)]ei
R
d4 xLY M (A)
(90)
Ga (AU ) = Ga (Aµ + DµA Λ)
δGa A b
(D Λ) + O(Λ2 )
= Ga (A) +
δAbµ µ
which gives
∆F P [A]−1 G=0 =
=
This is now a well defined integral, but we now need to evaluate ∆F P [A]
µ
or
where
∆F P [A]−1 =
Z
δGa A b [dU ] = [dΛ] around Λ ∼ 0
(Dµ Λ) ]
[dΛ]δ[Ga (A) +
b
δAµ
G=0
Z
1
1
h
i dN xδ(N ) (M x) =
δG
det M
det δA
DµA Z
G=0
[dU]δ[Ga (AU )]
a
a
∆F P [A]|G=0 ∼ det
U
As this factor is multiplied by δ[G (A)], the only solution of G (A ) =
δG A
D
δAµ µ
(92)
Ga (A + O(Λ)) = Ga (A) + O(Λ) = 0 is Λ = 0 or U = I. See the picture,
where we have assumed that we have a unique solution (otherwise Gribov
Thus we have
copies may exist).
Z
So for infinitesimal Λ, then U(Λ) = 1 + igΛata and
i
Aµ = Abµ tb
AUµ = UAµ U † + U∂µ U †
g
= Aaµ + ∂µ Λa − igif abc Abµ Λc + . . .
⇒ AaU
µ
[dA]ei
R
d4 xLY M (A)
∼
Z
[dA] det
R 4
δG A
Dµ δ[Ga (A)]ei d xLY M (A)
δAµ
(93)
f
We now wish to re-write the extra terms so as to have an Lef
YM
≡ Aaµ + (DµA Λ)a + . . .
1. Replace
where
Ga (A) → Ga (A) − ω a (x) [Nothing changes in derivation]
where ω a (x) is any scalar function. The LHS of eq. (93) is independent
(DµA )ab
ab
c ab
= δ ∂µ − ig(T )
Acµ
c ab
(T )
≡ −if
cab
is the covariant derivative in the adjoint representation (cf eq. (83))
107
(91)
1
R
4
a2
of ω a , so can integrate over a Gaussian e−i 2ξ d xω to give
Z
R 4 a2
R 4
i
i
a
2
[dω]δ[Ga (A) − ω a (x)]e− 2ξ d xω = e− 2ξ d x[G (A)]
108
Note: need ξ → 0 to recover original δ-function constraint, Ga (A) = 0
The gauge field propagator as given from the quadratic piece of the action
in A is then
Z
Z
1
1
d4 xLA,quad
=
d4 xAaµ (x)(η µν ∂ 2 − (1 − )∂ µ ∂ ν )δ ab Abν (x)
YM
2
ξ
2. Introduce anticommuting (Grassmann) fields c, c̄
Z
R
δG
δG A
DA c
−i d4 xc̄ δA
µ µ
det
Dµ = [dc̄dc]e
δAµ
(under Lorentz transformations) but are anticommuting. They can be
i.e. diagonal in group space, so these indices can be suppressed (if wished)
1
ηµρ ∂ 2 − (1 − )∂µ ∂ρ DFρν (x − y) = iδµν δ(x − y)
ξ
used in internal lines in computing Feynman diagrams – the Fadeev-
or
c and c̄ (independent) cannot be physical fields as they are scalars
Popov ghosts.
µν
i∆µν
F (x − y) ≡ DF (x − y) =
Z
d4 p −ip·(x−y) −i
e
(2π)4
p2 + iǫ
So finally we have
Z
[dA]ei
R
d4 xLY M (A)
∼
Z
[dA][dc̄dc]ei
R
f
d4 xLef
Y M (A,c,c̄)
(94)
with
f
Lef
Y M (A, c, c̄) = −
1
δG A
1 a 2
F
− Ga (A)2 − c̄
D c
4 µν
2ξ
δAµ µ
(95)
pµ pν
η µν − (1 − ξ) 2
p
(97)
[check
(see example sheet)
1
δρµ ∂ 2 − (1 − )∂ µ ∂ρ DFρν (x − y)
ξ
Z
d4 p −ip·(x−y)
1
−i
pρ pν
=
e
−δρµ p2 + (1 − )pµ pρ 2
η ρν − (1 − ξ) 2
4
(2π)
ξ
p + iǫ
p
Z
d4 p −ip·(x−y) −i
2 µν
µ ν
e
−p η + (1 − ξ)p p
=
(2π)4
p2 + iǫ
1
1
+ (1 − )pµ pν − (1 − ξ)(1 − )pµ pν
ξ
ξ
= iη µν δ(x − y)
1 −1+ξ+ 1 −1 = 0
pµ pν coefficient: 1 − ξ + 1 − ξ
ξ
The obvious gauge condition (Lorentz) is Ga (A) = ∂ · Aa so
f
Lef
Y M (A, c, c̄)
δGa
δAbµ
1 a2
1
= − Fµν
− (∂.Aa )2 + ∂ µ c̄a DµAab cb
4
2ξ
= ∂ µ δ ab or
(96)
]
Note the different sign to the KG equation
ξ = 0 Landau gauge [∂.A = 0 as have δ function again]

Feynman gauge
U(1) 
[Simple looking propagator]
ξ = 1
Feynman − t′ Hooft gauge SU(N) 
(98)
109
110
Note also that for QED as f abc → 0, then DµA → ∂µ is independent of gauge
field A, so that the Faddeev-Popov ghosts decouple, and hence do not appear
in the action.
Finally: Physical amplitudes being gauge invariant do not depend on the
gauge fixing term (and Faddeev-Popov ghosts).
The generating functional is
¯ ζ]
Z[J, η̄, η, ζ,
Z
R 4
a aµ
a a
a a
a a
a a
= N [dA][dη̄dη][dc̄dc]ei d x[L0 (A,ψ,ψ̄,c,c̄)+Lint (A,ψ,ψ̄,c,c̄)+Jµ A +η̄ ψ +ψ̄ η +ζ̄ c +c̄ ζ ]
Z
δ
δ
δ
δ
δ
= N exp i d4 zLint −i , −i , i , −i , i
Z0gluon [J]Z0quark [η̄, η]Z0ghost [ζ̄, ζ]
δJ
δη̄ δη
δζ̄ δζ
Now use analogies to our previous scalar and fermion theories
22
Feynman Rules for Gauge Theories
Including fermions, our final Lagrangian in the Lorentz gauge is
1
1 a2
/ − m) ψ +∂ µ c̄a DµAab cb
− (∂ µ Aaµ )2 + ψ̄(iD
L = − Fµν
|{z}
4
2ξ
|{z}
22.1
Propagators
22.1.1
gluon fields
(99)
quark/
electron
gluon/
photon
SUc (3)/U (1) gauge groups
We must now find the Feynman rules
1
Z0gluon [J] = e− 2
νb
d4 xd4 yJ µa (x)i∆ab
F µν (x−y)J (y)
i∆µνab
(x − y) ≡ DFµνab (x − y) =
F
p
a, µ
where
22.1.2
b, ν
= i∆µνab
(p) =
F
1
1
Lint = − g 2f abe f cde Aaµ Abν Acµ Adν − gf abc (∂µ Aaν − ∂ν Aaµ )Abµ Acν
4
2
+g ψ̄γ µ ta ψAaµ − gf abc (∂µ c̄a )cb Acµ
111
p2
pµ pν
−i
η µν − (1 − ξ) 2 δ ab
+ iǫ
p
quark fields
4
4
i
ij
j
Z0quark [η̄, η] = e− d xd yη̄ (x)iSF (x−y)η (y)
Z
d4 p −ip·(x−y)
i
iSFij (x − y) =
e
δ ij
(2π)4
p/ − m + iǫ
R
and
pµ pν ab
d4 p −ip·(x−y) −i
µν
e
δ
η
−
(1
−
ξ)
(2π)4
p2 + iǫ
p2
Z
so
L = L0 + Lint
1
1
L0 = − (∂µ Aaν − ∂ν Aaµ )(∂ µ Aνa − ∂ ν Aµa ) − (∂µ Aµa )2
4
2ξ
+ψ̄(i∂/ − m)ψ + ∂ µ c̄a ∂µ ca
R
112