Quantum Field Theory
1
R. Clarkson Dr. D. G. C. McKeon
2
January 13, 2003
1
Notes taken by R. Clarkson for Dr. McKeon’s Field Theory (Parts I and II) Class.
2
email:
2
Contents
1 Constraint Formalism 7
1.1 Principle of Least action: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2 Hamilton’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.3 Poisson Brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.4 Dirac’s Theory of Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.5 Quantizing a system with constraints . . . . . . . . . . . . . . . . . . . . . . 24
2 Grassmann Variables 27
2.1 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
2.2 Poisson Bracket . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.3 Quantization of the spinning particle. . . . . . . . . . . . . . . . . . . . . . . 32
2.4 General Solution to the free Dirac Equation . . . . . . . . . . . . . . . . . . 50
2.5 Charge Conjugation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
2.6 Majorana Spinors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
2.7 Time Reversal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
3 Bargmann-Wigner Equations 61
4 Gauge Symmetry and massless spin one particles 71
4.1 Canonical Hamiltonian Density . . . . . . . . . . . . . . . . . . . . . . . . . 74
5 (2
nd
) Quantization, Spin and Statistics 83
10.4 Radiative Corrections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
10.5 Divergences at Higher orders . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
10.5.1 Weinberg’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
10.6 Renormalization Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
10.6.1 Euler’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
10.6.2 Explicit Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . 198
11 Spontaneous Symmetry Breaking 207
11.1 O(2) Goldstone model: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
11.2 Coleman Weinberg Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . 215
11.3 One loop Effective Potential in λφ
4
model . . . . . . . . . . . . . . . . . . . 216
11.4 Dimensional Regularization . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
11.5 Spontaneous Symmetry Breaking in Gauge Theories . . . . . . . . . . . . . . 225
12 Ward-Takhashi-Slavnov-Taylor Identities 229
12.1 Dimensional Regularization with Spinors . . . . . . . . . . . . . . . . . . . . 234
12.1.1 Spinor Self-Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
12.2 Yang-Mills Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
12.3 BRST Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240
12.4 Background Field Quantization . . . . . . . . . . . . . . . . . . . . . . . . . 242
CONTENTS 5
13 Anomalies 249
13.1 Anomalies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255
14 Instantons 257
14.1 Quantum Mechanical Example . . . . . . . . . . . . . . . . . . . . . . . . . . 257
14.2 Classical Solutions to SU(2) YM field equations in Euclidean Space . . . . . 261
6 CONTENTS
Chapter 1
Constraint Formalism
1.1 Principle of Least action:
(t))dt (L = Lagrangian) (1.1.1)
i.e. if q
i
(t) = q
classical
i
(t) + ε δq(t).
As S = S(ε) has a minimum at ε = 0,
dS(0)
dε
= 0 (1.1.2)
{δq(t
i
) = 0 }
{δq(t
f
) = 0}
7
8CHAPTER1.CONSTRAINTFORMALISM
dS(0)
dε
=0
=
t
f
t
i
dt
δq
i
and,asδq
i
isarbitrary,
∂L
∂q
i
=
d
dt
∂L
∂˙q
i
(1.1.3)
1.2Hamilton’sEquations
(Legendretransforms)
p
i
=
∂L
∂˙q
i
(1.2.1)
H=p
i
˙q
A
dq
i
+
∂H
∂p
i
B
dp
i
(1.2.3)
=p
i
d˙q
i
cancels
+˙q
i
dp
i
−
∂L
∂q
i
dq
i
B
dq
i
(1.2.4)
From (1.2.3) and (1.2.4), we have:
˙q
i
=
∂H
∂p
i
(1.2.5)
˙p
i
= −
∂H
∂q
i
(1.2.6)
1.3. POISSON BRACKETS 9
1.3 Poisson Brackets
A = A(q
i
, p
i
) (1.3.1)
B = B(q
i
˙p
i
= {p
i
, H} (1.3.5)
We can generalize this:
d
dt
A(q
i
(t), p
i
(t)) = {A, H} (1.3.6)
−→ Suppose in defining
p
i
=
∂L
∂ ˙q
i
(1.3.7)
we cannot solve for ˙q
i
in terms of p
i
. i.e. in
H = p
i
˙q
i
p
m
2
−
kq
2
2
=
p
2
2m
−
kq
2
2
i.e. can solve for ˙q in terms of p here.
Suppose we used the cartesian coord’s to define system
L =
m
2
˙a
2
+
m
2
˙
b
b = a
Dynamical Variables: a, b, λ
1
, λ
2
∂L
∂λ
i
=
d
dt
∂L
∂
˙
λ
i
= 0 =
˙a −b i = 1
˙
b + a i = 2
(λ
1
, λ
2
) → Lagrangian Multipliers
The trouble comes when we try to pass to Hamiltonian;
p
i
∂p
i
∂ ˙q
j
=
∂
2
L
∂ ˙q
i
∂ ˙q
j
cannot be inverted.
In terms of Lagrange’s equations:
d
dt
∂L
∂ ˙q
i
(q
i
, ˙q
i
) =
∂L
∂q
i
=
∂
2
= −
∂
2
L
∂ ˙q
i
∂q
j
˙q
j
+
∂L
∂q
i
(1.3.9)
∗ → q
i
, ˙q
i
specified at t = t
o
∴ q
i
(t
0
+ δt) = q
i
(t
0
)
(t
0
)
(δt)
3
+ . . .
Can solve for ¨q
i
(t
0
) if we can invert
∂
2
L
∂ ˙q
i
∂ ˙q
j
.
Thus, if a constraint occurs, ¨q
i
(t
0
) cannot be determined from the initial conditions using
1.4. DIRAC’S THEORY OF CONSTRAINTS 11
Lagrange’s equations. i.e. from our example,
λ
i
(t
1
= p
λ
2
= 0), then we
can define
H
0
= p
i
˙q
i
− L(q
i
, ˙q
i
) Constraints hold here (1.4.1)
(So, for our example, (scaling m = 1),
L =
1
2
(˙a
2
+
˙
b
2
) + λ
1
(˙a −b) + λ
p
λ
i
˙
λ
i
−L
= p
a
(p
a
− λ
1
) + p
b
(p
b
− λ
2
) −
1
2
(p
a
− λ
1
)
2
P B
≈ 0 zero if χ
i
= 0 (“weakly” equal to zero)
where H = H
0
+ c
i
χ
i
.
Sept. 15/99
So, we have:
H
0
= p
i
˙q
i
− L(q, ˙q) (1.4.3)
H = H
0
+ c
i
χ
i
(q
i
, p
i
. They can be divided into First class and Second class
constraints.
first class constraints → label φ
i
(1.4.6)
second class constraints → label θ
i
(1.4.7)
For a first class constraint, φ
i
, [φ
i
, χ
j
] ≈ 0 = α
ij
k
x
k
(for all j).
θ
i
is second class if it is not first class.
We know that
H = H
0
+ c
i
χ
i
+ b
j
θ
j
]
= [φ
i
, H
0
] + a
j
[φ
i
, φ
j
] + φ
j
[φ
i
, a
j
] + b
j
[φ
i
, θ
j
] + θ
j
[φ
i
, φ
j
] +
≈0
φ
j
[θ
i
, a
j
] +b
j
[θ
i
, θ
j
] + θ
j
≈0
[θ
i
, b
j
]
≈ 0
≈ [θ
, γ
i
} = {Θ
i
}
H = H
0
+ a
i
φ
i
+ b
i
θ
i
+ c
i
γ
i
(1.4.8)
Provided {φ
i
, γ
j
} ≈ 0, then the condition
d
dt
Θ
i
≈ 0 (1.4.9)
Thus i, j must be even. Hence there are always an even number of 2
nd
class constraints.
Now we define the Dirac Bracket.
[A, B]
∗
= [A, B] −
i,j
[A, θ
i
]d
−1
ij
[θ
j
, B] (1.4.10)
Properties of the Dirac Bracket
1.
[θ
i
, B]
∗
= [θ
i
, B] −
k,l
[θ
i
+ [[B, C]
∗
, A]
∗
+ [[C, A]
∗
, B]
∗
(1.4.13)
3. If A is some 1
st
class quantity, i.e. if [A, χ
i
] ≈ 0 for any constraint χ
i
, then
[A, B]
∗
= [A, B] −
ij
≈0
[A, θ
i
] d
−1
ij
[θ
= 0 by (1). Thus, in H, we can set θ
i
= 0 before
computing [C, H]
∗
.
i.e.
If we want to find
dC
dt
we can use [C, H]
∗
and take H to be just H = H
0
+ a
i
φ
i
+
=0
b
i
θ
i
(Provided we exchange P.B. for Dirac B.).
Thus if we use the Dirac Bracket, we need not determine b
i
. If we include the gauge condition,
we can treat
[Θ
j
, B] (1.4.20)
Sept. 17/99
So, so far:
L =
1
2
(˙a
2
+
˙
b
2
− a
2
− b
2
) + λ
1
(˙a − b) + λ
2
(
˙
b + a)
p
a
=
∂L
∂ ˙a
b
˙
b +
0
p
λ
1
˙
λ
1
+
0
p
λ
2
˙
λ
2
−
1
2
(˙a
2
+
˙
b
b
− λ
2
) −
1
2
((p
a
− λ
1
)
2
+ (p
b
− λ
2
)
2
− a
2
− b
2
) +
λ
1
(p
a
− λ
1
2
) + λ
1
b − λ
2
a
1.4. DIRAC’S THEORY OF CONSTRAINTS 15
dp
λ
1
dt
= [p
λ
1
, H] ≈ 0
p
a
− λ
1
− b = 0
p
b
− λ
2
+ a = 0
Secondary constraints
(tertiary constraints don’t arise)
θ
1
]
(No first class constraints → no gauge condition).
H = H
0
+ c
i
θ
i
→ c
i
fixed by the condition
˙
θ
i
= [θ
i
, H] ≈ 0
- or could move to Dirac Brackets, and let θ
i
= 0.
Need
:
d
ij
=
0 0 1 0
1
= p
a
− λ
1
− b
θ
(1)
2
= p
b
− λ
2
+ a
d
(1)
12
= [θ
(1)
1
, θ
(1)
2
]
= 2
∴ d
(1)
ij
=
2
, Y ] −[X, θ
(1)
2
]
1/2
d
−1
21
[θ
(1)
1
, Y ]
Now we need to eliminate
θ
(2)
1
= p
λ
1
θ
(2)
2
= p
λ
2
16 CHAPTER 1. CONSTRAINT FORMALISM
d
(2)
+ a, p
λ
2
]
= [p
λ
1
, λ
1
]
−
1
2
[p
λ
2
, λ
2
]
= −
1
2
= −d
(2)
21
d
(2)
ij
(2)
ij
)
−1
[θ
(2)
j
, Y ]
∗
We can finally see that
[a, p
a
]
∗∗
= 1
[b, p
b
]
∗∗
= 1
and all other fundamental Dirac Brackets are zero. i.e.
[a, p
b
] = 0
as
p
a
− λ
1
− b = 0
b
a
So,
H = p
a
b − p
b
a
da
dt
= [a, H]
∗∗
= [a, p
a
b − p
b
a]
∗∗
= b
db
dt
= [b, H]
∗∗
= −a
If we have gauge conditions & first class constraints φ
i
:
H = H
0
+ a
= [A, H]
∗
≈ [A, H
0
]
∗
+ a
i
[A, φ
i
]
∗
a
i
→ Arbitrariness
γ = 0 must intersect q
i
(t) at one & only one point.
→ “Gribov Ambiguity” (to be avoided).
18 CHAPTER 1. CONSTRAINT FORMALISM
Sept. 21/99
Relativistic Free Particle
S ∝ arc length from x
µ
(τ
i
) to x
µ
(τ
f
2
(1.4.21)
p
µ
=
∂L
∂ ˙x
µ
= −
m ˙x
µ
√
˙x
2
(1.4.22)
Constraints
p
µ
p
µ
=
m
2
˙x
µ
˙x
µ
˙x
2
0 = p
i
= γ(p
2
− m
2
) (Pure Constraint!)
˙x
µ
=
∂H
∂p
µ
dx
µ
dτ
= κ(2p
µ
) (1.4.24)
1.4. DIRAC’S THEORY OF CONSTRAINTS 19
This arbitrariness in ˙x
µ
is a reflection of the fat that in S, τ is a freely chosen parameter,
i.e.
S = −m
dτ
dx
µ
dτ
dx
µ
dτ
dx
µ
dτ
(1.4.27)
−→ Now let κ =
1
2
dτ
dτ
. Thus, (insert κ into (1.4.24))
dx
µ
dτ
=
dτ
dτ
p
µ
and equate this with (1.4.26) (1.4.28)
dx
µ
dτ
2
0 =
d
dτ
∂L
∂ ˙x
µ
−
∂L
∂x
µ
Thus,
d
dτ
−m ˙x
µ
√
˙x
2
= 0
m¨x
µ
√
˙x
2
= 0
Now identify τ with the arc length along the particle’s trajectory:
20 CHAPTER 1. CONSTRAINT FORMALISM
.
In this case, the equation of motion becomes
m¨x
µ
= 0
The corresponding action is
S =
m
2
τ
f
τ
i
dτ ˙x
2
← absence of
√
means this is not invariant under τ → τ (τ
).
d
dt
∂L
∂ ˙x
µ
−
∂L
∂x
µ
2
H =
p
µ
p
µ
2m
1.4. DIRAC’S THEORY OF CONSTRAINTS 21
Other gauge choice:
τ = x
4
= t(breaks Lorentz invariance)
Work directly from the action:
S = −m
dτ
dt
dτ
dt
dτ
−
dr
dτ
dr
dτ
If we’ve chosen τ = t, then
S = −m
∂
˙
r
=
mv
√
1 −v
2
(momentum)
H = p ·
˙
r −L
= p ·v − L
but p
2
=
m
2
v
2
1−v
2
∴ p
2
(1 −v
2
) = m
2
v
2
= 1 −
p
2
m
2
+ p
2
=
m
2
+ p
2
− p
2
m
2
+ p
2
∴
√
1 −v
2
=
m
2
p
2
+ m
2
√
1 − v
2
=
mc
2
1 −
v
2
c
2
22 CHAPTER 1. CONSTRAINT FORMALISM
Note:
E
2
= p
2
+ m
2
(E
2
− p
2
) − m
2
= 0
∴ p
µ
= (p, E)
˙x
2
e
+ m
2
e
dτ
e = e(τ) , x
µ
= x
µ
(τ)
m
2
→ 0 is well defined in S. As
d
dτ
∂L
∂ ˙e
=0
−
∂L
∂e
= 0
So,
−
2
= −m
dτ
√
˙x
2
Can see that e field can be eliminated → really just a Lagrange multiplier that insures
S = 0 when m = 0.
Note
:
The action
S = −
1
2
dτ
˙x
2
e
+ m
2
e
(1.4.32)
1.4. DIRAC’S THEORY OF CONSTRAINTS 23
is invariant under
τ → τ + f(τ) (1.4.33)
from
d
dτ
∂L
∂ ˙e
−
∂L
∂e
= 0
0 =
d
dτ
˙x
µ
e
from
d
dτ
∂L
∂ ˙x
µ
−
∂L
∂x
µ
= 0
p
e
˙e −L
= p
µ
(−p
µ
e) −
−
1
2
˙x
2
e
+ m
2
e
; ˙x
2
= x
µ
x
µ
= −p
2
e
2
1
2
(p
2
− m
2
) (Secondary Constraint)
Both p
e
= 0 (gauge condition e = 1) and p
2
− m
2
= 0 (already discussed) are first calss.
Note: Remember that
S =
d
4
x
√
g g
µν
∂
µ
φ
A
(x)
In 0 + 1 dimensions
S =
dτ
1
e
d
dτ
φ
A
d
dτ
φ
A
=
dτ
(
˙
φ
A
)
2
e
−→
dτ
i
(ˆq, ˆp) |ψ >
phys.
= 0 (1.5.4)
ex. For L = −
1
2
˙x
2
e
+ m
2
e
χ
1
= p
e
χ
2
= p
2
− m
2
(2 constraints)
Quantization conditions will be
[x
µ
, p
µ
(1.5.7)
1.5. QUANTIZING A SYSTEM WITH CONSTRAINTS 25
If Φ(x) =< x|ψ >
phys
then
−i
∂
∂x
µ
2
− m
2
Φ(x) = 0
Classical Motivation for Spin
Brint, deVecchia & How, Nuclear P. 118, pg. 76 (1977)