not surprising because we haven’t yet
fixed a gauge
the culprit is the mixing term between
A
μ
and
∂
μ
π
—
efA
μ
∂
μ
π
— simplest
way to eliminate this mixing is to go to Landau gauge
efA
μ
∂
μ
π
=

efπ∂
μ
A
μ
+
total
derivative
→
total derivative
in Landau gauge
nice because it is renormalizable by simple power counting
and the ghosts are the same as before and don’t depend on the vector boson
masses
31
the quadratic terms for
A
and
π
are

1
4
(
∂
μ
A
ν

∂
ν
A
μ
)(
∂
μ
A
ν

∂
ν
A
μ
)+
1
2
(
∂
μ
π
+
efA
μ
) (
∂
μ
π
+
efA
μ
)+
A
μ
s
μ
+
πs
∂
μ
(
∂
μ
A
ν

∂
ν
A
μ
) =
∂
ν
π
+
efA
ν
+
s
ν
∂
μ
(
∂
μ
π
+
efA
μ
) =
s
∂
ν
(
∂
ν
π
+
efA
ν
+
s
ν
) = 0
⇒
∂
μ
s
μ
+
s
= 0
the equations can’t be satisfied for general sources which means that the
quadratic terms are not invertible —
not surprising because we haven’t yet
fixed a gauge
the culprit is the mixing term between
A
μ
and
∂
μ
π
—
efA
μ
∂
μ
π
— simplest
way to eliminate this mixing is to go to Landau gauge
efA
μ
∂
μ
π
=

efπ∂
μ
A
μ
+
total
derivative
→
total derivative
in Landau gauge
nice because it is renormalizable by simple power counting
and the ghosts are the same as before and don’t depend on the vector boson
masses
32
the quadratic terms for
A
and
π
are

1
4
(
∂
μ
A
ν

∂
ν
A
μ
)(
∂
μ
A
ν

∂
ν
A
μ
)+
1
2
(
∂
μ
π
+
efA
μ
) (
∂
μ
π
+
efA
μ
)+
A
μ
s
μ
+
πs
∂
μ
(
∂
μ
A
ν

∂
ν
A
μ
) =
∂
ν
π
+
efA
ν
+
s
ν
∂
μ
(
∂
μ
π
+
efA
μ
) =
s
∂
ν
(
∂
ν
π
+
efA
ν
+
s
ν
) = 0
⇒
∂
μ
s
μ
+
s
= 0
the equations can’t be satisfied for general sources which means that the
quadratic terms are not invertible —
not surprising because we haven’t yet
fixed a gauge
the culprit is the mixing term between
A
μ
and
∂
μ
π
—
efA
μ
∂
μ
π
— simplest
way to eliminate this mixing is to go to Landau gauge
efA
μ
∂
μ
π
=

efπ∂
μ
A
μ
+
total
derivative
→
total derivative
in Landau gauge
nice because it is renormalizable by simple power counting
and the ghosts are the same as before and don’t depend on the vector boson
masses
33
•
+
•
•
+
•
•
•
+
· · ·
where
•
is an insertion of the vector boson mass term

m
2
A
μ
A
μ
/
2
—
numerators of landau gauge propagators just projections operator

g
μν
+
k
μ
k
ν
/k
2
k
2
+
i²
trace of Landau
gauge propagator gives 3
3
2
Z
(
m
2
)
(
k
2
+
i²
)
+
1
2
(
m
2
)
2
(
k
2
+
i²
)
2
+
1
3
(
m
2
)
3
(
k
2
+
i²
)
3
+
· · ·
¶
d
4
k
(2
π
)
4
3
2
Z
log
(
k
2
+
i²
)
(
k
2

m
2
+
i²
)
¶
d
4
k
(2
π
)
4
easier to calculate the second derivative with respect to
m
2
∂
2
∂
(
m
2
)
2
3
2
Z
log
(
k
2
+
i²
)
(
k
2

m
2
+
i²
)
¶
d
4
k
(2
π
)
4
=
3
2
Z
1
(
k
2

m
2
+
i²
)
2
d
4
k
(2
π
)
4
34
•
+
•
•
+
•
•
•
+
· · ·
∂
2
∂
(
m
2
)
2
3
2
Z
log
(
k
2
+
i²
)
(
k
2

m
2
+
i²
)
¶
d
4
k
(2
π
)
4
=
3
2
Z
1
(
k
2

m
2
+
i²
)
2
d
4
k
(2
π
)
4
DR
MS
gives

i
3
32
π
2
log
m
2
μ
2
thus
3
2
Z
log
(
k
2
+
i²
)
(
k
2

m
2
+
i²
)
¶
d
4
k
(2
π
)
4
=

i
3
64
π
2
m
4
log
m
2
μ
2
+
3
4
¶
+
A m
2
+
B
corresponding to a contribution to the potential
3
64
π
2
m
4
log
m
2
μ
2
summed over all massive gauge bosons
this is the ColemanWeinberg potential
back to
Mathematica
file
su3.nb
35
SU
(3)
coupled to an octet
φ
3
×
3
matrix
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 Spring '10
 GEORGI
 Quantum Field Theory, ... ..., wA, Quantum chromodynamics, Gauge theory, Dµ Dµ