Solutions to Problems in Peskin and Schroeder,
An Introduction To Quantum Field Theory
Homer Reid
November 16, 2002
Chapter 2
Problem 2.1
(a)
The Lagrangian density is
L
=

1
4
F
μν
F
μν
.
Since
F
ij
=

F
ji
there are really only 6 terms in the sum:
=

1
2
bracketleftBig
F
01
F
01
+
F
02
F
02
+
F
03
F
03
+
F
12
F
12
+
F
13
F
13
+
F
23
F
23
bracketrightBig
.
Also,
F
ij
=

F
ij
if one of
i
or
j
is zero, and
F
ij
= +
F
ij
if both
i, j
are nonzero,
so we may write
= +
1
2
bracketleftbig
F
2
01
+
F
2
02
+
F
2
03

F
2
12

F
2
13

F
2
23
bracketrightbig
=
1
2
bracketleftbig
(
d
0
A
1

d
1
A
0
)
2
+ (
d
0
A
2

d
1
A
2
)
2
+ (
d
0
A
3

d
1
A
3
)
2

(
d
1
A
2

d
2
A
1
)
2

(
d
1
A
3

d
3
A
1
)
2

(
d
1
A
2

d
2
A
1
)
.
2
bracketrightbig
.
Evidently we have
∂
L
∂A
ν
= 0
∂
L
∂
(
∂
μ
A
ν
)
=
braceleftBigg
(
∂
μ
A
ν

∂
ν
A
μ
)
,
μν
= 0

(
∂
μ
A
ν

∂
ν
A
μ
)
,
μν
negationslash
= 0
.
=

F
μν
=
F
νμ
.
1
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 2
2
Hence the EulerLagrange equation is
0 =
∂
μ
∂
L
∂
(
∂
μ
A
ν
)
=
∂
μ
F
νμ
=
∂
μ
∂
ν
A
μ

∂
μ
∂
μ
A
ν
.
Identifying
E
i
=

F
0
i
and
B
k
=

epsilon1
ijk
F
ij
,
the EulerLagrange equations be
come that
∂
1
E
1
+
∂
2
E
2
+
∂
3
E
3
= 0
(
∇ ·
E
= 0)
epsilon1
ijk
∂
i
B
j
=
∂
0
E
k
parenleftbigg
∇ ×
B
=
∂
E
∂t
parenrightbigg
.
which are two of Maxwell’s equations.
(b)
First let’s review this whole Noether theorem business. Suppose we make
a change, parameterized by a small parameter
α,
under which the fields in the
Lagrangian undergo the transformations
φ
i
(
x
)
→
φ
prime
i
(
x
) =
φ
i
(
x
) +
α
Δ
φ
i
(
x
)
(1)
where the Δ
φ
i
(
x
) are fields themselves. This transformation does not affect the
equations of motion if the Lagrangian of the system is only modified by the
addition of the divergence of a fourvector, i.e. if
L → L
prime
=
L
+
α∂
μ
J
μ
(
x
)
.
(2)
On the other hand, computing directly from (1) we can also write the trans
formed Lagrange density as
L → L
prime
=
L
+
summationdisplay
i
braceleftbigg
∂
L
∂φ
i
(
α
Δ
φ
i
) +
∂
L
∂
(
∂
μ
φ
i
)
(
α∂
μ
Δ
φ
i
)
bracerightbigg
=
L
+
α
summationdisplay
i
braceleftbigg
∂
L
∂φ
i
(Δ
φ
i
) +
∂
μ
parenleftbigg
∂
L
∂
(
∂
μ
φ
i
)
Δ
φ
i
parenrightbigg

Δ
φ
i
∂
μ
∂
L
∂
(
∂
μ
φ
i
)
bracerightbigg
(where we used the chain rule,
∂
(
fg
) =
f∂g
+
g∂f
)
=
L
+
α
summationdisplay
i
braceleftbigg
∂
μ
parenleftbigg
∂
L
∂
(
∂
μ
φ
i
)
Δ
φ
i
parenrightbigg
+ Δ
φ
i
parenleftbigg
∂
L
∂φ
i

∂
μ
∂
L
∂
(
∂
μ
φ
i
)
parenrightbiggbracerightbigg
=
L
+
α∂
μ
summationdisplay
i
parenleftbigg
∂
L
∂
(
∂
μ
φ
i
)
Δ
φ
i
parenrightbigg
where we used the EulerLagrange equation in going to the last line. Comparing
with (2) we see that
∂
μ
summationdisplay
i
∂
L
∂
(
∂
μ
φ
i
)
Δ
φ
i
=
∂
μ
J
μ
Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 2
3
or
∂
μ
j
μ
= 0
,
j
μ
=
J
μ

summationdisplay
i
Δ
φ
i
∂
L
∂
(
∂
μ
φ
i
)
.
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 Quantum Field Theory, Lagrangian mechanics, Noether's theorem, Lagrangian, Euler–Lagrange equation, Homer Reid, Schroeder Problems