When you have higher order correlation functions, you get contact terms for each additional field. For
m
= 0
,
square
(
0

T
{
φ
(
x
)
φ
(
y
1
)
φ
(
y
n
)
}
0
)
=
(
0

T
{
square
φ
(
x
)
φ
(
y
1
)
φ
(
y
n
)
}
0
)
(13.138)
−
i
summationdisplay
j
δ
(4)
(
x
−
y
j
)
(
0

T
{
φ
(
y
1
)
φ
(
y
j
−
1
)
φ
(
y
j
+1
)
φ
(
y
n
)
}
0
)
(13.139)
Since the correlation functions define the theory, these SchwingerDyson equations allow us to show that
two theories are equal by matching their correlation functions. For example, go back to the canonically
quantized theory and define
Z
ˆ
[
J
] =
(
0

T
{
e
i
integraltext
φJ
}
0
)
(13.140)
This is not a functional integral, just a generating functional for correlation. Expanding out the exponen
tial, it is just a set of correlation functions of the second quantized
φ
and an external current
J
, which we
already know hot to calculate.
But it has the same properties as the functional integral
(
0

T
{
φ
(
x
1
)
φ
(
x
n
)
}
0
)
= (
−
i
)
n
d
n
Z
ˆ
dJ
(
x
1
)
dJ
(
x
n
)

J
=0
(13.141)
13.12
SchwingerDyson equations
159
We would like to show that
Z
=
Z
ˆ
, which will then provide another proof of the path integral.
To do that, take the SD equations and multiply by
J
’s:
square
(
0

T
{
φ
(
x
)
φ
(
y
1
)
φ
(
y
n
)
J
(
y
1
)
J
(
y
n
)
})
=
(
0

T
{
square
φ
(
x
)
φ
(
y
1
)
φ
(
y
n
)
J
(
y
1
)
J
(
y
n
)
})
(13.142)
−
i
summationdisplay
j
δ
(4)
(
x
−
y
j
)
(
0

T
{
φ
(
y
1
)
φ
(
y
j
−
1
)
φ
(
y
j
+1
)
φ
(
y
n
)
J
(
y
1
)
J
(
y
n
)
})
(13.143)
The first term comes from the expansion of
square
∂Z
ˆ
[
J
]
∂J
(
x
)
. For the second term, note that these are the full inter
acting
φ
′
s
which evolve with
H
, or equivalently satisfy
square
φ
=
V
′
(
φ
)
by the Heisenberg equations of motion.
So this term is part of the expansion of
V
′
parenleftBig
∂
∂J
(
x
)
parenrightBig
Z
ˆ
[
J
]
. Integrating the last term over the
δ
function gives
(
0

T
{
φ
(
y
1
)
φ
(
y
j
−
1
)
φ
(
y
j
+1
)
φ
(
y
n
)
J
(
y
1
)
J
(
y
j
−
1
)
J
(
x
)
J
(
y
j
+1
)
J
(
y
n
)
})
(13.144)
This comes from the expansion of
J
(
x
)
Z
[
J
]
. So if we sum up the whole Taylor series, we get
square
∂Z
ˆ
[
J
]
∂J
(
x
)
=
bracketleftbigg
V
′
parenleftbigg
∂
∂J
(
x
)
parenrightbigg
+
J
(
x
)
bracketrightbigg
Z
ˆ
[
J
]
(13.145)
This is a concise form of the SchwingerDyson equations. More importantly, it is a differential equation
for
Z
ˆ
[
J
]
which no longer makes reference to the Hilbert space. We showed this equation holds for
Z
ˆ
[
J
]
.
You can show in the same way that it holds for the path integral
Z
[
J
]
. Therefore, with the same physical
boundary conditions, the two are equivalent. This is another way to derive the path integral.
13.13
BRST invariance
There is a beautiful symmetry called BRST invariance, which is a residual exact symmetry of the
Lagrangian even after gauge fixing. It is particularly useful for studying more complicated gauge theories,
but it is a little easier to understand in the QED case.
Notice that the gauge fixing term
(
∂
μ
A
μ
)
2
is invariant under gauge transformations
A
μ
→
A
μ
+
∂
μ
α
for
square
α
= 0
. So this is a residual symmetry of the entire Lagrangian. As we have already seen, it is awkward
to deal with constrained theories, and constraints on symmetries are no different. So consider adding to
the Lagrangian two new free fields
c
and
d
L
=
L
[
A,φ
i
] +
1
ξ
(
∂
μ
A
μ
)
2
−
d
square
c
(13.146)
So the path integral becomes
Z
O
=
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 Photon, Quantum Field Theory, Lorentz, lorentz invariance