Quantum field theory: spin one

276 Pages·2005·1.303 MB·English

by  Srednicki M.

Quantum Field Theory
Part III: Spin One
Mark Srednicki
Department of Physics
University of California
Santa Barbara, CA 93106
[email protected]
This is a draft version of Part III of a three-part introductory textbook on
quantum field theory.
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Part III: Spin One
54) Maxwell’s Equations (3)
55) Electrodynamics in Coulomb Gauge (54)
56) LSZ Reduction for Photons (5, 55)
57) The Path Integral for Photons (8, 56)
58) The Feynman Rules for Spinor Electrodynamics (45, 57)
59) Tree-Level Scattering in Spinor Electrodynamics (48, 58)
60) Spinor Helicity for Spinor Electrodynamics (50, 59)
61) Scalar Electrodynamics (58)
62) Loop Corrections in Spinor Electrodynamics (51, 59)
63) The Vertex Function in Spinor Electrodynamics (62)
64) The Magnetic Moment of the Electron (63)
65) Loop Corrections in Scalar Electrodynamics (61, 62)
66) Beta Functions in Quantum Electrodynamics (52, 62)
67) Ward Identities in Quantum Electrodynamics I (22, 59)
68) Ward Identities in Quantum Electrodynamics II (63, 67)
69) Nonabelian Gauge Theory (32, 58)
70) Group Representations (69)
71) The Path Integral for Nonabelian Gauge Theory (69)
72) The Feynman Rules for Nonabelian Gauge Theory (71)
73) The Beta Function for Nonabelian Gauge Theory (70, 72)
74) BRST Symmetry (71)
75) Chiral Gauge Theories and Anomalies (72)
76) Anomalies in Global Symmetries (75)
77) Anomalies and the Integration Measure for Fermion Fields (76)
78) Background Field Gauge (73)
79) Gervais–Neveu Gauge (78)
80) The Feynman Rules for N N Matrix Fields (10)
×
81) Tree-Level Scattering in Quantum Chromodynamics (60, 79, 80)
82) Wilson Loops, Lattice Gauge Theory, and Confinement (28, 73)
83) Chiral Symmetry Breaking (76, 82)
84) Spontaneous Breaking of Gauge Symmetries (61, 70)
85) R Gauge for Spontaneously Broken Abelian Gauge Theory (84)
ξ
2
86) R Gauge for Spontaneously Broken Nonabelian Gauge Theory (85)
ξ
87) The Standard Model: Electroweak Gauge and Higgs Sector (86)
88) The Standard Model: Lepton Sector (87)
89) The Standard Model: Quark Sector (88)
90) Electroweak Interactions of Hadrons (83, 89)
91) Neutrino Masses (88)
Not yet completed:
92) Solitons and Magnetic Monopoles (84)
93) Instantons and the Vacuum Angle (77)
94) Supersymmetry (89)
95) Unification (91, 94)
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Quantum Field Theory Mark Srednicki
54: Maxwell’s Equations
Prerequisite: 3
The photon is the quintessential spin-one particle. The phenomenon of
emission andabsorbtionofphotonsbymatter isacriticaltopicinmanyareas
of physics, and so that is the context in which most physicists first encounter
a serious treatment of photons. We will use a brief review of this subject
(in this section and the next) as our entry point into the theory of quantum
electrodynamics.
Let us begin with classical electrodynamics. Maxwell’s equations are
E = ρ , (1)
∇·
B E˙ = J , (2)
∇× −
E+B˙ = 0 , (3)
∇×
B = 0 , (4)
∇·
where E is the electric field, B is the magnetic field, ρ is the charge den-
sity, and J is the current density. We have written Maxwell’s equations in
Heaviside-Lorentz units, and also set c = 1. In these units, the magnitude of
the force between two charges of magnitude Q is Q2/4πr2.
Maxwell’s equations must be supplemented by formulae that give us the
dynamics of the charges and currents (such as the Lorentz force law for
point particles). For now, however, we will treat the charges and currents as
specified sources, and focus on the dynamics of the electromagnetic fields.
The last two of Maxwell’s equations, the ones with no sources on the
right-hand side, can be solved by writing the E and B fields in terms of a
4
scalar potential ϕ and a vector potential A,
E = ϕ A˙ , (5)
−∇ −
B = A . (6)
∇×
The potentials uniquely determine the fields, but the fields do not uniquely
determine the potentials. Given a particular ϕ and A that result in a par-
ticular E and B, we will get the same E and B from any other potentials ϕ
′
and A that are related by
′
ϕ = ϕ+Γ˙ , (7)
′
A = A Γ , (8)
′
−∇
where Γ is an arbitrary function of spacetime. A change of potentials that
does not change the fields is called a gauge transformation. The fields are
gauge invariant.
All this becomes more compact and elegant in a relativistic notation.
Define the four-vector potential
Aµ = (ϕ,A) ; (9)
Aµ is also called the gauge field. We also define the field strength
Fµν = ∂µAν ∂νAµ . (10)
−
Obviously, Fµν is antisymmetric: Fµν = Fνµ. Comparing eqs.(5), (9), and
−
(10), we see that
F0i = Ei . (11)
Comparing eqs.(6), (9), and (10), we see that
Fij = εijkB . (12)
k
The first two of Maxwell’s equations can now be written as
∂ Fµν = Jµ , (13)
ν
5
where
Jµ = (ρ,J) (14)
is the charge-current density four-vector. If we take the four-divergence of
eq.(13),weget∂ ∂ Fµν = ∂ Jµ. Theleft-handsideofthisequationvanishes,
µ ν µ
because ∂ ∂ is symmetric on exchange of µ and ν, while Fµν is antisymmet-
µ ν
ric. We conclude that we must have
∂ Jµ = 0 ; (15)
µ
that is, the electromagnetic current must be conserved.
The last two of Maxwell’s equations can be written as
ε ∂ρFµν = 0 . (16)
µνρσ
Plugging in eq.(10), we see that eq.(16) is automatically satisfied, since the
antisymmetric combination of two derivatives vanishes.
Eqs.(7) and (8) can be combined into
Aµ = Aµ ∂µΓ . (17)
′
−
Setting F µν = ∂µAν ∂νAµ and using eq.(17), we get
′ ′ ′
−
F µν = Fµν (∂µ∂ν ∂ν∂µ)Γ . (18)
′
− −
The last term vanishes because derivatives commute; thus the field strength
is gauge invariant,
F µν = Fµν . (19)
′
Next we will find an action that results in Maxwell’s equations as the
equations of motion. We will treat the current as an external source. The
action we seek should be Lorentz invariant, gauge invariant, parity and time-
reversal invariant, and no more than second order in derivatives. The only
candidate is S = d4x , where
L
R
= 1FµνF +JµA . (20)
L −4 µν µ
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The first term is obviously gauge invariant, because Fµν is. After a gauge
transformation, eq.(17), the second term becomes JµA , and the difference
′µ
is
Jµ(A A ) = Jµ∂ Γ
′µ − µ − µ
= (∂ Jµ)Γ ∂ (JµΓ) . (21)
µ µ
− −
The first term in eq.(21) vanishes because the current is conserved. The
second term is a total divergence, and its integral over d4x vanishes (assum-
ing suitable boundary conditions at infinity). Thus the action specified by
eq.(20) is gauge invariant.
Setting Fµν = ∂µAν ∂νAµ and multiplying out the terms, eq.(20) be-
−
comes
= 1∂µAν∂ A + 1∂µAν∂ A +JµA (22)
L −2 µ ν 2 ν µ µ
= +1A (gµν∂2 ∂µ∂ν)A +JµA ∂µK , (23)
2 µ − ν µ − µ
where K = 1Aν(∂ A ∂ A ). The last term is a total divergence, and can
µ 2 µ ν− ν µ
be dropped. From eq.(23), we can see that varying Aµ while requiring S to
be unchanged yields the equation of motion
(gµν∂2 ∂µ∂ν)A +Jµ = 0 . (24)
ν
−
Noting that ∂ Fµν = ∂ (∂µAν ∂νAµ) = (∂µ∂ν gµν∂2)A , we see that
ν ν ν
− −
eq.(24) is equivalent to eq.(13), and hence to Maxwell’s equations.
7
Quantum Field Theory Mark Srednicki
55: Electrodynamics in Coulomb Gauge
Prerequisite: 54
Next we would like to construct the hamiltonian, and quantize the elec-
tromagnetic field.
There is an immediate difficulty, caused by the gauge invariance: we have
too many degrees of freedom. This problem manifests itself in several ways.
For example, the lagrangian
= 1FµνF +JµA (25)
L −4 µν µ
= 1∂µAν∂ A + 1∂µAν∂ A +JµA (26)
−2 µ ν 2 ν µ µ
doesnot contain the timederivative ofA0. Thus, this field hasno canonically
conjugate momentum and no dynamics.
To deal with this problem, we must eliminate the gauge freedom. We do
this by choosing a gauge. We choose a gauge by imposing a gauge condition.
This is a condition that we require Aµ(x) to satisfy. The idea is that there
should be only one Aµ(x) that results in a given Fµν(x) and also satisfies the
gauge condition.
One possible class of gauge conditions is nµA (x) = 0, where nµ is a
µ
constant four-vector. If n is spacelike (n2 > 0), then we have chosen axial
gauge; if n is lightlike, (n2 = 0), it is lightcone gauge; and if n is timelike,
(n2 < 0), it is temporal gauge.
Another gauge is Lorentz gauge, where the condition is ∂µA = 0. We
µ
will meet a family of closely related gauges in section 62.
Inthis section, we willpick Coulomb gauge, also known asradiation gauge
or transverse gauge. The condition for Coulomb gauge is
A(x) = 0 . (27)
∇·
8
We can impose eq.(27) by acting on A (x) with a projection operator,
i
i j
A (x) δ ∇ ∇ A (x) . (28)
i → ij − 2 j
(cid:18) ∇ (cid:19)
We construct the right-hand side of eq.(28) by Fourier-transforming A (x)
i
to A (k), multiplying A (k) by the matrix δ k k /k2, and then Fourier-
i i ij i j
−
transforming back to position space.
e e
Now let us write out the lagrangian in terms of the scalar and vector
potentials, ϕ = A0 and A , with A obeying the Coulomb gauge condition.
i i
Starting from eq.(26), we get
= 1A˙ A˙ 1 A A +J A
L 2 i i − 2∇j i∇j i i i
+ 1 A A +A˙ ϕ
2∇i j∇j i i∇i
+ 1 ϕ ϕ ρϕ . (29)
2∇i ∇i −
In the second line of eq.(29), the in each term can be integrated by parts;
i
∇
in the first term, we will then get a factor of ( A ), and in the second
j i i
∇ ∇
term, we will get afactor of A˙ . Bothofthese vanish by virtue ofthe gauge
i i
∇
condition A = 0, and so both of these terms can simply be dropped.
i i
∇
If we now vary ϕ (and require S = d4x to be stationary), we find that
L
ϕ obeys Poisson’s equation,
R
2ϕ = ρ . (30)
−∇
The solution is
ρ(y,t)
ϕ(x,t) = d3y . (31)
4π x y
Z | − |
This solution is unique if we impose the boundary conditions that ϕ and ρ
both vanish at spatial infinity.
Eq.(31)tellsusthatϕ(x,t)isgivenentirely intermsofthechargedensity
atthesame time, andsohasnodynamics ofitsown. Itisthereforelegitimate
to plug eq.(31) back into the lagrangian. After an integration by parts to
turn ϕ ϕ into ϕ 2ϕ = ϕρ, the result is
i i
∇ ∇ − ∇
= 1A˙ A˙ 1 A A +J A + , (32)
L 2 i i − 2∇j i∇j i i i Lcoul
9
where
1 ρ(x,t)ρ(y,t)
= d3y . (33)
Lcoul −2 4π x y
Z | − |
We can now vary A , and find that each compenent obeys the massless Klein-
i
Gordon equation, with the projected current as a source,
∂2A (x) = δ ∇i∇j J (x) . (34)
− i ij − 2 j
(cid:18) ∇ (cid:19)
For a free field (J = 0), the general solution is
i
A(x) = dk ε∗λ(k)aλ(k)eikx +ελ(k)a†λ(k)e−ikx , (35)
λX=±Z h i
f
where k0 = ω = k , dk = d3k/(2π)32ω, andε (k)andε (k) arepolarization
+
| | −
vectors. The polarization vectors must be orthogonal to the wave vector
f
k. We will choose them to correspond to right- and left-handed circular
polarizations; for k = (0,0,k), we then have
ε (k) = 1 (1, i,0) ,
+ √2 −
ε (k) = 1 (1,+i,0) . (36)
− √2
More generally, the two polarization vectors along with the unit vector in the
k direction form an orthonormal and complete set,
k ε (k) = 0 , (37)
λ
·
ε (k) ε (k) = δ , (38)
λ′ · ∗λ λ′λ
k k
ε (k)ε (k) = δ i j . (39)
∗iλ jλ ij − k2
λ=
X±
The coefficients aλ(k) and a†λ(k) will become operators after quantization,
which is why we have used the dagger symbol for conjugation.
In complete analogy with the procedure used for a scalar field in section
3, we can invert eq.(35) and its time derivative to get
a (k) = +iε (k) d3x e ikx∂↔A(x) , (40)
λ λ − 0
·
Z
a†λ(k) = −iε∗λ(k)· d3x e+ikx∂↔0A(x) , (41)
Z
10