Elektromagnetisme, noter og formelsamling

Transcription

Notes for Elementary Particle Physics
Dennis Hansen
January 15, 2013
Scalar
General
1
1
L = − ∂ µ φ̂∂µ φ − m2 φ̂2 + Ω0
2
2
0 = −∂ 2 + m2 φ̂(x)
ˆ
h
i
˜ â(k)eikµ xµ + â† (k)e−ikµ xµ
φ̂(x) = dk
∂L
∂ φ̇
h
i
0 = φ̂(x, t), φ̂(x0 , t)
±
h
i
0
0 = Π̂(x, t), Π̂(x , t)
±
h
i
3
0
0
iδ (x − x ) = φ̂(x, t), Π̂(x , t)
Π̂(x, t) ≡
[â(p), â(q)] = 0
±
kkout k
√ dΩCM
dLIPS2 (pin ) =
16π 2 s
~ 2
k
out
dσ
|T |
=
2
dΩCM
4 k~
pin kCM s 16π
1
2
|T | dLIPS2 (pin )
dΓ =
2ωk
†
â (p), â† (q) = 0
3
â(p), â† (q) = (2π) 2ωp δ 3 (p − q)
2
Spinor
s = − (pin1 + pin2 )
2
L = Ψ i∂ − m Ψ
0 = i∂ − m Ψ
ˆ
X
˜
Ψ(x, t) = dp
bs (p)us (p)eipx + d†s (p)vs (p)e−ipx
2
u = − (pin1 − pout2 )
(~a · ~σ ) ~b · ~σ = ~a · ~b I + i~σ · ~a × ~b
s=±
ˆ
Ψ(x, t) =
t = − (pin1 − pout1 )
˜
dp
X
b†s (p)us (p)e−ipx + ds (p)v s (p)eipx
{γ µ , γ ν } = − 2g µν I4
s=±
n
o
3
bs (p), b†s0 (p0 ) = (2π) 2ωp δ (3) (p − p0 )δss0
n
o
3
ds (p), d†s0 (p0 ) = (2π) 2ωp δ (3) (p − p0 )δss0
γµγν =
µν
a
b = − (ab) I4 + i2aµ S bν
X
s
X
Photons and QED
s
1
L = − F µν Fµν + J µ Aµ
4
ˆ
i
Xh
0
˜
A (x) = 0 , A(x) = dp
∗λ aλ (p)eipx + λ a†λ (p)e−ipx
λ=±
LQED
1
= − F µν Fµν + Ψ (i
D − m) Ψ
4
1 µ ν
1
{γ , γ } + [γ µ , γ ν ]
2
2
X
us (p)us (p) = − p
+m
vs (p)v s (p) = − p
−m
ν
µν
µ∗
λ (p)λ0 (p) = g
λ
X
∗λi (p)λj (p) = δij −
λ
pi pj
p2
p · λ (p) = 0
p2 = − m2 = −ω 2 + p2
Contents
1 Preface
3
2 Elements of analytical mechanics
4
2.1
Lagrangian formalism
. . . . . . . . . . . . . . . . . . . . . . . . . .
4
2.2
Hamiltonian formulation . . . . . . . . . . . . . . . . . . . . . . . . .
4
3 Relativistic classical eld theory
3.1
5
Lagrangian density and the eld action . . . . . . . . . . . . . . . . .
4 Lorentz transformations
5
9
4.1
Lorentz transformation of coordinates and derivatives . . . . . . . . .
9
4.2
Unitary transformations in QM . . . . . . . . . . . . . . . . . . . . .
10
4.3
Lie algebra of the Lorentz group
14
4.4
Lorentz transformation of scalar elds
4.5
Generalization of the Lorentz transformation
4.6
Representation of the Lorentz group
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . .
. . . . . . . . . . . . .
16
. . . . . . . . . . . . . . . . . .
16
5 Continuous symmetries and Noether's theorem
5.1
Energy-momentum tensor
15
. . . . . . . . . . . . . . . . . . . . . . . .
6 Free quantum elds
18
21
22
6.1
Canonical quantization . . . . . . . . . . . . . . . . . . . . . . . . . .
22
6.2
The Klein-Gordon equation and free elds . . . . . . . . . . . . . . .
22
7 Free spin-0 elds
23
7.1
Lorentz transformation of elds and operators . . . . . . . . . . . . .
26
7.2
Simple inner products and expectation values of states . . . . . . . .
27
8 Interaction and scatting in spin-0 theories
28
8.1
The interaction picture, Dyson expansion
. . . . . . . . . . . . . . .
29
8.2
Interaction picture in spin-0 quantum eld theories . . . . . . . . . .
32
9 Scattering theory
38
9.1
Cross sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
9.2
Mandelstam variables
44
. . . . . . . . . . . . . . . . . . . . . . . . . .
10 Free spin-½ elds
45
10.1 Spinor representations of the Lorentz group
. . . . . . . . . . . . . .
45
. . . . . . . . . . . . . . . .
50
10.3 Canonical quantization of the Dirac eld . . . . . . . . . . . . . . . .
55
10.2 Bispinors, the Majorana and Dirac eld
10.4 Quantization in terms of
b, d
operators . . . . . . . . . . . . . . . . .
10.5 The free Dirac eld and its Hamiltonian
10.6 Conserved charges
60
. . . . . . . . . . . . . . . .
61
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
CONTENTS
11 Interaction and scattering in spin-½ theories
63
11.1 Interaction picture in spin-0 quantum eld theories . . . . . . . . . .
63
11.2 Scattering in spin-½ theory . . . . . . . . . . . . . . . . . . . . . . . .
67
12 Photon elds and Quantum Electrodynamics
12.1 Maxwell's equations and Lagrangian formulation
69
. . . . . . . . . . .
69
. . . . . . . . . . . . . . .
70
12.3 Interactions, spinor QED . . . . . . . . . . . . . . . . . . . . . . . . .
74
12.4 Massive gauge bosons
. . . . . . . . . . . . . . . . . . . . . . . . . .
75
12.5 The Higgs mechanism for massive gauge bosons . . . . . . . . . . . .
76
12.2 The free photon eld and its quantization
13 Non-abelian gauge theories
77
13.1 Extending the symmetry groups to non-abelian
. . . . . . . . . . . .
77
. . . . . . . . . . . . . . . . . . . . . . .
78
. . . . . . . . . . . . . . . . . . . . . . . . .
79
13.4 The Standard Model . . . . . . . . . . . . . . . . . . . . . . . . . . .
81
13.2 Quantum Chromodynamics
13.3 Electroweak interaction
A Nice-2-know relations
82
A.1
Delta functions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A.2
Dirac adjoint/barred identities
A.3
Traces of
γ -matrices
82
. . . . . . . . . . . . . . . . . . . . .
82
. . . . . . . . . . . . . . . . . . . . . . . . . . .
83
B γ -matrices
83
C Quantum Mechanics
84
C.1
Basic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84
C.2
Commutator relations
85
. . . . . . . . . . . . . . . . . . . . . . . . . .
D Collection of Feynman rules
86
D.1
Scalar elds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
86
D.2
Spinor elds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
86
D.3
Photon elds
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
87
D.4
QED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
87
E Mandelstam variables
88
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1
PREFACE
1 Preface
Theese notes are intended for the master course 'Elementary particle physics' at
the University of Copenhagen. First we will have a recap of the classical mechanics
and classical eld theory that Quantum Field Theory uses to some extent.
Then
comes a bit about Lorentz transformations, and its impact on the physics.
Then
something about symmetries in eld theories, especially Noether's theorem and how
to use it.
Next we begin the actual QFT by looking at scalar QFT in terms of
the spin-0 bosons, free eld theory rst, but then how interactions works in QFT.
We then describe spin-½ particles/antiparticles, which are fermions, in the free eld
QFT, and then the interactions. We next is free photon elds and spinor interaction
in QED, and shortly say something about the Higgs mechanism for massive bosons
and their Feynman rules. Finaly, we shortly say something about non-abelian eld
theories (Yang-Mills theories) in general, and give a very short example of QCD,
before discussing the electroweak theory and the Higgs mechanism in the theory,
and briey the Standard Model. In the appendix, some useful stu can be found;
relations, derivations, standard quantum mechanics relations etc.
Suggestions for improvements and corrections are welcome by email [email protected]
- this is pre-release version 0.3 and is subject to typos and one should read the lines
critical - I may have fucked up big time!
Change-log:

Added lots of useful formulas and equations on the front page.

Added QM formulas and Feynman rules, and something more about Mandelstam variables.

Finished main text!

Added some real calculation examples.
Needs to be done:

Add more examples, especially in QED and QCD.

Write more about non-abelian gauge theories and in greater details.

Find typos.
Dennis Hansen
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2
ELEMENTS OF ANALYTICAL MECHANICS
2 Elements of analytical mechanics
2.1 Lagrangian formalism
The cumber stone of the Lagrangian formalism is the Euler-Lagrange equations
d
dt
where there are
so there are
k
n
∂L
∂ q̇i
−
∂L
= 0,
∂qi
generalized coordinates
n = 3N − k
Euler-Lagrange
qi
and
equations in total, where
N
n
generalized velocities
q̇i ,
and
is the number of particles and
is the number of (holonomic) constrains. The Lagrangian is given by
Lagrangian
L=T −V ,
where
T
is the sum of kinetic energy of the particles of the system and
of the potentials of the particles (the force may be conservative so
may be a velocity dependent potential so the generalized force is
V
is a sum
Fi = −∇i V ,
Qi =
or
The Euler-Lagrange equations gives second order dierential equations, and the
generalized coordinates
system is said to have
qi
n
constitutes the
V
∂V
d ∂V
− ∂q
+ dt
( ∂ q̇i )).
i
n
n dimensional conguration space, and the
Conguration space
degrees of freedom. The Euler-Lagrange equations follows
d'Alembert's principle or can be derived from Hamilton's variational principle of the
action integral
ˆ
ˆ
t2
S=
Action
t2
Ldt =
L (q(t), q̇(t), t) dt ,
t1
t1
i.e. that the action has to be stationary along the correct trajectory in conguration
space to rst order (occasionally known as the 'The principle of least action')
ˆ
tion
t2
δS = 0 = δ
Principle of least ac-
L (q(t), q̇(t), t) dt.
t1
We have that for a generalized coordinate
qi , the generalized canonical momentum
Canonical momentum
is given by
pi =
∂L
,
∂ q̇i
with the usual physical interpretation.
2.2 Hamiltonian formulation
The Hamiltonian of a system with
freedom is
n = 3N − k
N
particles and
k
constraints so the degrees of
Hamiltonian
is given by
H(q, p, t) ≡ q̇i pi − L(q, q̇, t) ,
is a Legendre transformation of the Lagrangian, with
momentum and
qi
pi = ∂L/∂ q̇i as the canonical
as the canonical coordinate. From the Legendre transformation
it is seen that the Euler-Lagrange equations has the equivalent formulation given in
Hamilton's equations,
Hamilton's equations
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3
RELATIVISTIC CLASSICAL FIELD THEORY
q̇i =
∂H
,
∂pi
∂H
,
∂qi
ṗi = −
∂L
∂H
=−
.
∂t
∂t
The Hamiltonian formulation now consists of
that spans the
2n
2n rst order dierential equations,
dimensional phase space of the canonical variables.
Hamilton's
Phase space
equations can also be obtained from variation of the action,
ˆ
ˆ
t2
t2
q̇i pi − Hdt = 0 .
Ldt = δ
δS = δ
t1
t1
If the generalized coordinates doesn't depend on time, then
H
is the total energy
of the system and is given by
H = T + V.
There are several other shortcuts to go from the Lagrangian to the Hamiltonian
under certain circumstances, which can be seen in chapter 8.2 of [Goldstein].
A cyclic coordinate
we have
ṗi =
− ∂H
∂qi
qi
is a coordinate that
= 0 ⇒ pi = constant,
H
Cyclic coordinate
doesn't depend explicitly on. Hence
so the corresponding canonical momentum
is conserved.
3 Relativistic classical eld theory
In classical eld theory one treats the spatial variables as labels, just as time is just
a label in ordinary Lagrangian or Hamiltonian formulation, but where the spatial
coordinates remains function to be solve for. Hence it is very easy to obtain a fully
covariant formulation of the equations of motion, and it follows very naturally from
the notation used.
φ(t, x, y, z) =
coordinates
Now we will end up having to solve our equation for a eld
φ(xµ ) (and perhaps some extra variables) in stead of the generalized
qi .
One can say that
q
and
φ
is the same thing, but we have that
continuously labeling of the coordinates, whereas
q
classical eld theory we have that the coordinates
φ
is a
is discretely labeled. Hence in a
xµ
is completely independent.
We could also imagine that the physics demand that the eld is a vector eld,
φ,
φν .
This is for example the case
i
with the the electromagnetical elds, where we have E
= (Ex , Ey , Ez ), but in a fully
so that the eld may have several components
covariant formulation of classical electrodynamics, we have that it is better formulate
stu in the 4-vector potential
Aµ .
3.1 Lagrangian density and the eld action
We can go from the denition of the the Lagrangian to the Lagrangian density, which
is the function that meets the requirement that
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Lagrangian density
3
ˆ
L=
Ld3 xi ,
where we integrate over some region of
R3 .
We have that we can extend the denition of the action integral quite easily to a
structure with much better properties than the former denition. We take
ˆ
Field action
ˆ
S=
Ld4 xµ ,
Ldt =
where we integrate over some region of the space-time (Minkowski space
M=
R1,3 ).
Now, to the good properties: Since the innitesimal volume of the Minkowski
space is invariant
d4 xµ
is Lorentz invariant, we have that the action is the same
in all inertial frames and is Lorentz invariant. Hence, the variation gives the same
result in all inertial frames, and we can be sure that the eld equations are covariant
(given that
L
and other quantities used are Lorentz invariant).
Example 1
.
(Show that the action is Lorentz invariant)
We start by show-
4 µ is invariant. We have by changing coordinates that
ing that the measure d x
µ 0
0
0
∂(x ) 4 µ
d4 xµ = ∂(x
= 1d4 xµ = d4 xµ , since the determinant of the Jacobi maµ0 ) d x
µ ∂(x ) trix = 1 because the corresponding transformation matrix belongs to the
∂(xµ0 )
Lorentz group.
Hence we have for a region
R
of the space-time that
ˆ
ˆ
0
L0 d4 xµ = S 0 ,
4 µ
Ld x =
S=
R0
R
and the action integral is shown to be Lorentz invariant.
Variation of the eld
The variation of the action integral by taking the deformation of the correct eld
component
φν
φ̃ν;α = φν + αξν ,
to be
where
ξ
is an arbitrary eld that vanishes at
the start-point and end-point in space-time. We have that
φ̃ν → φν
for
α → 0.
ˆ
!
ˆ
dφ̃ν;α µ
dS d
dφν µ
4 µ
4 µ
δS = δ
L φν , µ , x d x = 0 ⇔
=
L
φ̃
,
,
x
d
x
ν;α
dx
dα α=0
dxµ
R
R dα
α=0
ˆ
R
∂L ∂ φ̃ν;α
∂L
+
d
φ̃
∂ φ̃ν;α ∂α
∂ dxν;α
µ
4 µ
d x ∂α
dφ̃
∂ dxν;α
µ
= 0.
α=0
We are actually taking functional partial derivatives here, not very rigoursly... We
1
can do some manipulation of the last part of the integral by integration by parts
for each of the
µ
integrals:
ˆ
∂L
R ∂ dφ̃ν;α
dxµ
1
´
uv 0 dx = uv −
´
∂
∂α
dφ̃ν;α
dxµ
!
d4 xµ = 0
u0 vdx
6 of 88
=0
3
ˆ
End

=
∂L
dφ̃
∂ dxν;α
µ


d  ∂L  ∂
φ̃ν;α d4 xµ
µ
dφ̃ν;α
∂α
R dx
∂ dxµ
Start


ˆ
d  ∂L  ∂ φ̃ν;α 4 µ
=0−
d x .
µ
dφ̃
∂α
R dx
∂ ν;α
∂
φ̃ν;α 
∂α
−
dxµ
ˆ

=−
R

d  ∂L 
ξν d4 xµ .
dxµ ∂ dφ̃ν;α
dxµ
Where the last part follows since
∂
∂α φν;α
∂
∂α (φν
=
+ αξν ) = ξν
which must vanish
at the start- and endpoint in spacetime. Hence we nd
ˆ
d  ∂L 
∂L
4 µ
ξν − µ
ξν d x dφ̃
dx
R ∂ φ̃ν;α
∂ dxν;α
µ
α=0



ˆ
∂L
d  ∂L 
4 µ

=
ξν d x − µ
dφ̃
dx
R ∂ φ̃ν;α
∂ dxν;α
µ
α=0
"
!#
ˆ
∂L
d
∂L
=
ξν d4 xµ = 0 .
− µ
dx
∂ dφµν
R ∂φν
dS =
dα α=0


dx
If this is to be zero, we must have that the integrand is zero, since the variation
ξν
is arbitrary. Hence we nd the eld version of the Euler-Lagrange equations to be
∂L
d
− µ
∂φν
dx
∂L
Euler-Lagrange eqs
!
dφν
∂ dx
µ
= 0.
We can dene the proper eld theoretic generalization of the canonical momenta
to be the
Canonical
momentum
density
∂L
∂L
Π =
=
,
∂ (∂0 φν )
∂ φ̇ν
ν
2
which is a eld in its own right, and not the same as the four-momentum at all .
The interpretation is more like the it is the intrinsic momentum that the eld carries,
not of any particle.
One can go from the Lagrangian formulation to the Hamiltonian formulation of
the elds in the same way as in regular analytical mechanics by performing a Legendre transformation, which with the proper generalization gives os the Hamiltonian
density
H
H = Πν φ̇ν − L .
The canonical momentum density is a contraction of the tensor eld Pνµ = ∂ (∂∂Lφ ) , which
µ ν
enters in the denition of the stress-energy tensor T µν .
2
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Hamiltonian density
3
Example 2
(Derive the Lagrangian density for a classical eld of harmonic oscilla-
tors, and nd the eld equations)
.
Consider rst the
the classical simple harmonic oscillators with mass
atated by a distance
where
ηi
a.
Let the
i'th
m
and spring constant
oscillator have the position
is the deviation from the position
(ηi+1 − ηi ),
x-direction and say that we have
ia.
k
seper-
xi = ia + ηk , i ∈ Z,
We have each spring is compressed
and the Lagrangian for each oscillator is then
1
1
1
1
Li = Ti − Vi = mẋ2i − k(ηi+1 − ηi )2 = mη̇i2 − k(ηi+1 − ηi )2 .
2
2
2
2
And for all oscillators we have
L=
X1
X
1
1
1
mη̇i2 − k(ηi+1 − ηi )2 = a
a−1 mη̇i2 − a−1 k(ηi+1 − ηi )2 .
2
2
2
2
i
i
We have that
ma−1 = `
is the linear mass density, and
ka = Y ,
where
Y
is Youngs
modulus. Hence we have
(ηi+1 − ηi )2
1 X 2
`η̇i − Y
.
L= a
2
a2
We have
For
a→0
we
(ηi+1 − ηi )2
ηi+1 − ηi 2
∂η(x, t)
=(
) →
for a → 0 .
2
a
a
∂x
´
P
dx, and η becomes continous in t
have that a
i →
and
x
and we
have
ˆ
1
L=
2
`
∂η(x, t)
∂t
Hence, dropping the constant factor
sions is
L = `
∂η
∂t
2
−Y
∂η
∂x
2
2
−Y
∂η(x, t)
∂x
2
dx .
1
2 the Lagrangian density in one spatial dimen-
It is straightforward to see that in three spatial
dimensions we have
L = `
If we have
∂η
∂t
2
−Y
` = Y = 1,
∂η
∂x
2
+
∂η
∂y
2
+
∂η
∂z
2 !
= `∂ 0 η∂0 η + Y ∂ i η∂i η
we have
L = ∂ 0 η∂0 η + ∂ i η∂i η = ∂ µ η∂µ η .
Putting into the Euler-Lagrange equation, we have
∂ (∂ µ η∂µ η)
∂L
∂L
∂ (∂ ν η∂ν η)
0 =
− ∂µ
=
− ∂µ
∂η
∂ (∂µ η)
∂η
∂ (∂µ η)
ν
∂ (∂ η∂ν η)
= ∂µ
= ∂µ (δνµ ∂ ν η) = ∂µ (∂ µ η) = ∂µ ∂ µ η ⇔ ∂ 2 η = 2 η = 0.
∂ (∂µ η)
Hence, the governing equation is the wave equation, and the solutions are given by
plane waves
exp(ikµ xµ ).
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4
LORENTZ TRANSFORMATIONS
4 Lorentz transformations
4.1 Lorentz transformation of coordinates and derivatives
The laws of physics must be the same in all inertial systems, and the equations
of motion must have the same form in any inertial system.
The class of linear
transformations of coordinates that leaves the 4-norm invariant is called the Lorentz
Lorentz transforms
transformations, and is of the form
xµ = Λµρ xρ ,
This transformation must of course be bijective, and hence we must be able to
perform Lorentz transformations back and forth between dierent inertial system,
so there must always exist an inverse Lorentz transformation.
The metric tensor
gµν
The metric
is used to take inner products, raise and lower indicies.
We have that the metric is symmetric;
gµν = gνµ ,
and that the inverse metric
has the same elements as the metric, but with raised indicies and hence we have
gµα
g αν
=
g αν g
µα
=
Inverse metric
δµν .
We further have some restrictions on the possible transformation.
First, the
metric tensor must be unchanged under a Lorentz transformation, that is
gµν Λµρ Λνσ = gρσ .
We must have
det Λ = ±1
to leave the spacetime volume invariant.
This also
comes from the fact that Lorentz transformations just are rotation in space-time. The
Lorentz transformations constitutes a group, since a composition of any two Lorentz
transformations is again a Lorentz transformation, and every Lorentz transformation
has an inverse, and the Lorentz group can be represented in dierent ways.
Innitesimal transform
A innitesimal Lorentz transformation is given by
Λµρ = δ µρ + δω µρ ,
, and the inverse transformation is given by
Λµρ = δ µρ − δω µρ ,
Since we have
1.
(1 + δω)(1 − δω) = 11 − 1δω + δω1 − δω 2 = 1 + O(δω 2 ) =
If we restrict ourselves to the class of Lorentz transformations with
0
(the proper subgroup) and even further take Λ 0
≥ 1,
det Λ = 1
that corresponds to keep the
direction of time, theese form the orthochronous subgroup. This subgroup is what
Orthochronous
is normally referred to very slobby as the Lorentz group, and is actually also a Lie
group
group (manifold and a group), since all of the transformations are continuous, and
Lie group
hence we can reach all transformations from the identity transformation by doing a
series of innitesimal transformation. Also, we can develope the Lie algebra of the
Lorentz group, which alow us to investigate the structure of the transformations even
further.
9 of 88
Lie algebra
sub-
4
Example 3 (Show that for an innitesimal Lorentz transformation Λµρ = δµρ + δωµρ ,
δω µρ is an antisymmetric tensor (problem 2.1)). Let Λµρ = δ µρ +δω µρ be an innitesimal
Lorentz transformation (that is to rst order in
δω ).
To let the length of a 4-vector
be invariant, we must for any Lorentz transformation have that
gµν Λµρ Λνσ = gρσ .
Hence, by inserting
Λµρ = δ µρ + δω µρ
and
Λνσ = δ νσ + δω νσ ,
expanding parentheses and
contracting indicies, we have that
gµν (δ µρ + δω µρ )(δ νσ + δω νσ ) = gµν (δ µρ δ νσ + δ µρ δω νσ + δω µρ δ νσ + δω µρ δω νσ )
= gµν (δ µρ δ νσ + δ µρ δω νσ + δ νσ δω µρ + O(δω 2 ))
= gρσ + gµν (δ µρ δω νσ + δ νσ δω µρ )
Where we have neglected
µ ν
for gµν Λ ρ Λ σ
=
δω µρ δω νσ
since
δω µρ δω νσ = O(δω 2 )
gρσ to hold, we must have gµν (δ µρ δω νσ
in the last line. Hence,
+ δ νσ δω µρ )
= 0.
By contraction,
we nd
0 = gµν (δ µρ δω νσ + δ νσ δω µρ )
= gµν δ µρ δω νσ + gµν δ νσ δω µρ
= gρν δω νσ + gµσ δω µρ
= δωρσ + δωσρ ⇒
δωρσ = −δωσρ .
Hence
δω µρ
is an antisymmetric tensor. Notice, that we also have
δω κτ = −δω τ κ
by
simply applying the inverse metric tensor two times to raise indicies, i.e. appling the
tensor
g κσ g τ ρ
to both sides of the equation.
4.2 Unitary transformations in QM
Sym.transforms in QM
In quantum mechanics a continuous symmetry transformation is given by an element
of the unitary group, which is called an unitary operator.
An element
U
in the
Unitary group
(continuous) unitary group takes the general form
U = exp (iαa T a ) ,
where
Ta
is the generators of the unitary group, and
αa
are a set of real numbers,
that must have same dimension as the number of generators of the group.
a
generators T must be hermitian,
(T a )†
=
The
T a , since
U † = exp (iαa T a )† = exp −iαa (T a )† = exp (−iαa T a ) .
From this we see that we must have
U †U = I ,
so the inverse operator is the
10 of 88
Inverse
4
adjoint,
U −1 = U † .
For the group of continuous symmetry transformation, that is
3 by innitesimal
transformations that are connected to the identity transformation
transformations, and hence theese are the only ones we need to consider:
U = I + iαa T a .
A, the operator transforms as
α with eigenvalue a transforms as α 0 = U −1 α
If we do an unitary transformation of an operator
A0
=
U −1 AU , and an eigenvector
0
and must have a
Operator
transforma-
tion
= a, where the last equalty is required to leave the physics invariant
under the transformation. We see that
0
A0 α
= U −1 AU U −1 α
= U −1 Aα = U −1 aα
0
= aU −1 α = aα .
Hence the eigenvalue of
A0
remains the same as for
A
for the transformed eigen-
vector.
We must have that any Lorentz transformation in QM depends on the coordinate
transformation
Λ,
transformation
Λ = 1 + δω ,
so the unitary operator is
U (Λ).
As an innitesimal Lorentz
we can dene the generators of the Lorentz group by
M µν
i
U (1 + δω) ≡ I + δωµν M µν ,
2
since there can be maximally
42 = 16 generators for the group, but we have some
restrictions that limits the number of independent generators to
M µν antisymmetric since
M µν
δωνµ M νµ , and
δωµν
=
µν
also M
must be hermitian operators. Since
because
U (Λ)
δωµν
Λ y U (Λ)
6:
We must have
is antisymmetric, and
must be an isomorphism
represents an unique transformation, the group product (composition
of Lorentz transformations) must be preserved, and hence we have
U (Λ0 )U (Λ) = U (Λ0 Λ) ,
U (Λ)−1 U (Λ0 )U (Λ) = U (Λ−1 Λ0 Λ) .
As opposed to transformation that involve some kind of discrete transformation as represented
by antiunitary operators, U −1 = −U † or some group product of a continuous group and a discrete
group.
3
11 of 88
Isomorphism
4
Example 4 (Show that for an Lorentz transformation Λ, the unitary representation
−1 M µν U (Λ) = Λµ Λν M ρσ (problem 2.2)). We have that the representaobeys U (Λ)
ρ σ
tion map is a homomorphism, so we must have that
for any Lorentz transformations
0
Lorentz transformation as Λ
Λ, Λ0 .
U (Λ)−1 U (Λ0 )U (Λ) = U (Λ−1 Λ0 Λ)
We can use that we can write an innitesimal
= 1 + δω ,
and do the following calculation on
Λ−1 Λ0 Λ:
Λ−1 Λ0 Λ = Λ−1 (1 + δω)Λ
= Λ−1 1Λ + Λ−1 δωΛ
= 1 + Λ−1 δωΛ .
And since we have for an innitesimal transformation that
i
µν and
2~ δωµν M
1+
U (1 + δω) = I +
Λ−1 δωΛ also is an innitesimal transformation, we have that
U (Λ)−1 U (1 + δω)U (Λ)
i
U (Λ)−1 (I + δωµν M µν )U (Λ)
2~
i
−1
U (Λ) (I + δωµν M µν )U (Λ)
2~
i
I + δωµν U (Λ)−1 M µν U (Λ)
2~
= U (1 + Λ−1 δωΛ)
i
= I+
Λ−1 δωΛ ρσ M ρσ
2~
i
= I + δωµν Λµρ Λνσ M ρσ
2~
i
= I + δωµν Λµρ Λνσ M ρσ ,
2~
δωµν is an arbitrary innitesimal Lorentz transformation, we must have
−1
that U (Λ)
M µν U (Λ) = Λµρ Λνσ M ρσ , which was what we wanted.
and since
4.2.1 Derivative operators
The derivative operator
∂µ
Derivative operator
transforms like
∂ µ = Λ−1
ρ
µ
∂ρ = Λµρ ∂ρ ,
µ
∂ = Λµρ ∂ ρ ,
So this is also a linear transformation.
The operator
∂2
is easily shown to be
invariant.
One can also show that given two elds
φ, ϕ,
we have
∂µ φ∂ µ ϕ = ∂ µ φ∂µ ϕ .
Example 5 (Show that ∂µ φ∂ µ ϕ = ∂ µ φ∂µ ϕ).
fullles
We use that the metric and its inverse
δνσ = gνξ g ξσ :
∂µ φ∂ µ ϕ = δνσ ∂σ φ∂ ν ϕ = gνξ g ξσ ∂σ φ∂ ν ϕ = ∂ ξ φ∂ξ ϕ = ∂ µ φ∂µ ϕ .
As we wanted to show.
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4
Example 6 (Show that ∂ µ = Λµρ ∂ρ ).
∂µ ≡
By the chain rule we have
∂Λνρ xρ ∂
∂
∂xν ∂
∂
=
= Λνµ ν ≡ Λνµ ∂ ν .
=
ν
ν
µ
µ
µ
∂x
∂x ∂x
∂x ∂x
∂x
Hence, the inverse yields the wanted relation:
Λ−1
µ
∂ = Λ−1
ρ µ
µ
ρ
Λνµ ∂ ν = δ νρ ∂ ν = ∂ ρ .
As we wanted to show.
13 of 88
4
4.3 Lie algebra of the Lorentz group
Example 7
(Find the commutator relations of the generators of the Lorentz'
group (problem 2.3))
.
We have that
innitesimal transformation
U (Λ)−1
=
U (Λ−1 )
=I−
U (Λ)−1 M αβ U (Λ) = Λαρ Λβσ M ρσ .
Λ = 1+
δω , we have Λ−1
= 1 − δω ,
Using the
and hence
i
µν
2 δωµν M . We nd
i
i
αβ
ργ
ργ
I − δωργ M
I + δωργ M
M
= δ αµ + δω αµ δ βν + δω βν M µν
2
2
i
i
i
i
M αβ I + M αβ δωργ M ργ − δωργ M ργ M αβ I − δωργ M ργ M αβ
δωξθ M ξθ
2
2
2
2
= δ αµ δ βν + δω βν + δω αµ δ βν + δω βν M µν
Removing all terms that are
O(δω 2 ),
we nd
i
M αβ + δωργ M αβ M ργ − M ργ M αβ = M αβ + δ αµ δω βν M ρσ + δω αµ δ βν M µν ⇒
2
i
h
i
δωργ M αβ , M ργ = δω αµ M µβ + δω βν M αν .
2
We now factor out
and
1
2 δωργ on the RHS by using the antisymmetry of
M µβ = −M βµ and
δω βν = −δωνβ
changing the dummy index, to see that we can take out the
metric
=
=
=
=
=
=
i
h
i
δωργ M αβ , M ργ
2
1
2δω αµ M µβ + 2δω βν M αν
2
1
− δω αµ M βµ − δωµα M βµ − δω βν M αν + δωνβ M αν
2
1
− δω αγ M βγ − δωρα M βρ − δω βγ M αγ + δωρβ M αρ
2
1
− g αρ δωργ M βγ − g αρ δωργ M βρ − g βρ δωργ M αγ + g βρ δωργ M αρ
2
1
αρ
βγ
βρ
αγ
αγ
βρ
βγ
αρ
− g M −g M
− δωργ g M − g M
2
1
− δωργ g αρ M βγ − (α ↔ β) − (ρ ↔ γ) .
2
Hence, we nally have the commutator relations
h
i
M αβ , M ργ = i g αρ M βγ − (α ↔ β) − (ρ ↔ γ) .
Example 4 shows that the generators transforms as we expect any second-order
tensor to do, so everything looks good. We can go even further by taking
Λ = 1 + δω
and derive some commutation relations for the generators of the Lorentz group to
Commutation
tions
14 of 88
rela-
4
nd the Lie algebra of the Lorentz group as done in Example 7, so we get the
commutation relations
h
i
M αβ , M ργ = i g αρ M βγ − (α ↔ β) − (ρ ↔ γ) .
This analysis shows that there are 6 generators for the Lorentz group; 3 boosts
and 3 rotations (which gives 6 spacetime rotations), and the generators are the
momentum, angular momentum and so forth.
Theese commutator relations for the generator can be rewritten, if we take the
denitions
1
Ji ≡ ijk M jk , Ki ≡ M i0 ,
2
where
Ji
is the angular momentum operator (at least, it has the same form and
Angular
Ki is the boost operaνρ
tor, which is the time part of M . With theese denitions, we get the following
operator
propities as the regular angular momentum operator), and
momentum
Boost operator
equivalent commutation relations
[Ji , Jj ] = iijk Jk ,
[Ji , Kj ] = iijk Kk ,
[Ki , Kj ] = −iijk Jk .
Lie algebra
This is the Lie algebra of the Lorentz group, and the physics behind this is of
course that it does matter in which way one does rotations/boosts.
algebra, since it is self-dening.
It is a closed
From this it is also somewhat easy to see that
there are no nite-dimensional hermitian (matrix) representation, since we will get
a contradiction from the fact that the trace of the operators is non-zero, but should
be zero for nite hermitian matrices.
4.4 Lorentz transformation of scalar elds
A classical eld
φ(x) = φ(t, x)
must be a Lorentz scalar and hence transform under
0
a Lorentz transformation of the coordinates x
have
φ(x) =
= Λx
0
and the eld φ
= Λ (φ)
so we
φ0 (x0 ).
Similary a quantum eld, where
φ̂(x)
now is an operator, we must have that the
transformation is given by unitary operators, and transforms according to
φ̂(x) → φ̂0 (x) = U (Λ)−1 φ̂(x)U (Λ) = φ̂(Λ−1 x) .
Derivatives of the eld transform as
ρ
U (Λ)−1 ∂ µ φ̂(x)U (Λ) = Λµρ ∂ φ̂(x̄) ,
15 of 88
Lorentz scalar:
φ0 (x0 )
= φ(x)
φ(x) 7→
4
2
U (Λ)−1 ∂ 2 φ̂(x)U (Λ) = ∂ φ̂(x) ,
Hence
∂ 2 φ̂(x)
is a lorentz scalar, and
∂ µ φ̂(x)
is a Lorentz vector.
4.5 Generalization of the Lorentz transformation
Say that we have a physical object
♦A (x)
that has some generic index
A
that runs
over some discrete set. For this quantity to be Lorentz invariant, we must have that
under a Lorentz transformation
Λ
we have
U (Λ)−1 ♦A (x)U (Λ) = DAB (Λ)♦B (Λ−1 x) ,
where
DAB (Λ)
is some tensor/matrix that depends on the transformation
Λ.
To
satisfy the group properties of the Lorentz group, we must have that
DAB (Λ)DBC (Λ0 ) = DAB (ΛΛ0 ) .
We call any matrices/tensors that fullls this for a representation of the Lorentz
Representation
group, since we will have the same commutation relation for a set of some generating
matrices
S αβ
B
A
(set running over the double index
A, B )
which may or may not
be hermitian depending on what we choose
S
αβ
B
A
, (S
ργ
)AB
=i
g
αρ
S
βγ
B
A
− (α ↔ β) − (ρ ↔ γ) ,
and a innitesimal transformation is given by
i
DAB (1 + δω) = IδAB + δωµν (S µν )AB .
2
Similarly one can go to tensor objects of arbitrary many indicies
♦A1 ...AN (x)by
requiring that they are Lorentz invariant, that is transforms as
U (Λ)−1 ♦A1 ...AN (x)U (Λ) = DAB11 · · · DABNN (Λ)♦B1 ...BN (Λ−1 x) .
4.6 Representation of the Lorentz group
Denition 8.
A representation of a Lie algebra
homomorphism
ρ
vector space
x∈L
over some Lie group
(product/Lie bracket preserving) from
V , gl(V ),
we have
Representation
L
where the Lie bracket of
ρx ∈ gl(V )
and
V
L
L
is a
to endomorphisms of a
is the commutator.
Hence for
ρ[x,y] = ρx ρy − ρy ρx .
Informally, if we have some operators or matrices that has the same algebra as
the Lorentz group, we say that it is a representation.
Raising/lowering oper-
One can dene the non-hermitian raising/lowering operators
Ji+ =
for which we have
ators
1
1
(Ji + iKi ) , Ji− = (Ji − iKi ) ,
2
2
†
Ji+
=
Commutation
Ji− , and the commutation relations
tions
16 of 88
rela-
4
h
Ji+ ,
Jj−
i
= 0,
h
i
Ji+ , Jj+ = iijk Jk+ ,
h
i
Ji− , Jj− = iijk Jk− ,
which can be shown to be equivalent to the regular Lie algebra of the Lorentz
group.
Now, we have two closed rotation algebras for each
Ji+ , Jj−
that doesn't
talk to each other, so that must mean that we can represent the Lorentz group by
a group product of two smaller identical groups, that are isomorphic to the Lorentz
group. There are no nite dimensional hermitian representation of the algebra, but
Finite
if we allow the matrices of the representation to be non-hermitian, we can nd nite
matrices
non-hermitian
dimensional representations.
The commutation relations for
Ji+ , Jj−
is just the standard angular momentum
algebra from quantum mechanics, so there exists a number
n±
such that
2n± + 1
n± =
3
1
2 , 1, 2 , . . . is exactly the spin number from quantum mechanics. The representa-
denes the dimension of the matrices in the representation, and this number
0,
4
tions are irreducible, which means that the only subrepresentations for a given
n±
are given by matrices of full rank or of zero-rank. We also know that for each
n±
Spin number
Irreducible reps.
the matrices have a set of (orthogonal) eigenvectors of full dimension that has the
eigenvalues
− |n± | , − |n± | + 1, . . . , n± − 1, n±
sentations by a set of numbers
. Thus we can label dierent repre-
(2n+ + 1, 2n− + 1)
for the corresponding representations matrices of
sion of the quantum eld
ϕ̂
is
which is the set of dimensions
Ji+ , Jj− ,
(2n+ + 1, 2n− + 1).
and hence the dimen-
Now since
Ji+
†
= Ji− ,
we
can go from one representation to another just by taking the hermitian conjugate
(2n+ + 1, 2n− + 1) y (2n− + 1, 2n+ + 1).
Ji =
+
Ji together as
Now since we have the regular (hermitian) angular momentum operator is
Ji−
+
Ji+ , the spin of the eld is reduced to adding the spin of
Ji− ,
Spin of eld
in regular quantum mechanics. So we can in this way deduce that dierent types of
quantum elds, since they must be Lorentz invariant, have dierent spin, and the
representation of the Lorentz group we use corresponds to elds of particles with
5
dierent spin . We then know that the eigenvalues of
1, . . . , n+ + n−
Ji
is
|n+ − n− | , |n+ − n− | +
when the two parts of eld spin is coupled. Depending
n±
we have
that the coupling gives rise to some decomposition of the representations in terms of
higher/lower dimensional representations as given by the ClebschGordan expansion.
The representation as written before gives that the quantum eld
sion
(2n+ + 1, 2n− + 1),
ϕ̂ haves dimen-
and the four rst representations are the most common:
4
Denition. A subspace W of V that is invariant under the group action is called a subrepresentation.
One can say that the very existence of the Lorentz group leads to the fact that there must exist
spin.
5
17 of 88
Eigenvalues of
Ji
5
CONTINUOUS SYMMETRIES AND NOETHER'S THEOREM
(2n+ + 1, 2n− + 1)
Name of representation/eld
Spin of corresponding eld
(1, 1)
Scalar
0
(2, 1)
Left-handed spinor
(1, 2)
Right-handed spinor
1
2
1
2
(2, 2)
Vector
1
5 Continuous symmetries and Noether's theorem6
Theorem 9 (Formal version). To every dierentiable symmetry generated by local
actions, there corresponds a conserved current.
Theorem 10 (Informal version). If a system has a continuous symmetry property,
then there are corresponding quantities whose values are conserved in time.
´
Theorem 11 (Useful eld theoretic version). Suppose the action integral S = Ld4 xµ
is invariant under a transformation of the eld φ (in the Lagrangian, which may be
both classical or quantum) given by φ → φ + αf , where α is a scalar and f is a eld.
Then there is exists a quantity that is conserved, namely a current j µ (the Noether
´
current), and some charge Q = d3 xi j 0 (x) (the Noether charge).
Noether's theorem
Noether's theorem was initially formulated for discrete systems and yields result
as momentum and energy conservation for theese systems. However, it is quite easy
to generalize it to eld theories, where it is very powerful and useful, and with no
modications can also be applied to quantum eld theories.
Example 12
(Various conservation laws)
.
If the Hamiltonian (Lagrangian) is in-
dependent of time, Noether's theorem gives that the total energy of the system is
conserved.
If the Lagrangian has an cyclic coordinate, the corresponding generalized momentum
conjugate to the cyclic coordinate is conserved, since
d
dt
∂L
∂ q̇i
∂L
− ∂q
=
i
d
dt
(pi ) = 0 ⇒
pi = constant.
All of the discussion below applies also to elds with more components, but we
will just formulate it for a scalar eld. If we take
and
f
φ → φ + αf ,
where
α
is a scalar
is some eld that this leaves the action invariant (that is given, we assume).
Then when taking
α
to be a eld also,
that the action is necessarily invariant!
0
Lagrangian as L
= Lorginal + δL,
where
α → α(x),
then it is no longer the case
Assuming that we can write/rewrite the
Lorginal
the part of the Lagrangian that depends on
is the original Lagrangian and
αf , we can isolate δα =
δL
is
∂µ αdxµ to write
´ 4
0
δL =
d xδL0 =
µ α(x), then the variation of the action of the form δS =
´ 4
´ 4
d xδL = d x [j µ (x)∂µ α(x)], then j µ (x) will be the conserved Noether current,
´ 4
when applying the equations of motion of the elds since δS =
d xδLorginal = 0.
j µ (x)∂
When we do a variation in the action with respect to
α, δS 0 /δα,
and we want this
to be zero when the equations of motions are imposed for it to be a true symmetry.
Hence we must have
6
http://en.wikipedia.org/wiki/Noether's_theorem
18 of 88
Noether current
5
ˆ
ˆ
d x [j (x)δα] = d4 x [j µ (x)∂µ α(x)]
ˆ
= − d4 x [∂µ j µ (x)] α(x)
0 =
4
µ
Where the last line follows from integration by parts. Hence we must have
∂µ j µ (x) = 0 ,
so
j µ (x)
is conserved. We can write this as a continuity equation
∂j 0 (x)
∂ρ(x)
+ ∇ · j(x) = 0 ⇔
+ ∇ · j(x) = 0 ,
def
∂t
∂t
and so we have the Noether charge
Noether charge
ˆ
Q(t) =
d3 xi j 0 (x) ,
is conserved, which can easily be shown from the continuity equation.
This can be formulated very neatly in Pouls method below.
Pouls method for applying Noether's theorem:
1. Identify
2. Take
3.
α
φ → φ + αf .
as it was a eld,
S is no longer invariant.
that terms
∝ α(x)
α → α(x)
Factor
(making it local).
δS 0 =
most vanish for
´
d4 xδL =
δS 0 = 0
4. Apply equations of motion to check that
since
δS 0 = 0
´
d4 x [j µ (x)∂µ α(x)] by noting
α
is actually a constant.
to identify conserved current
j µ (x).
5. Identify conserved charge.
19 of 88
5
Example 13 (Symmetry of the complex eld Lagrangian L = −∂µ φ∗ ∂ µ φ − m2 |φ|2 ).
For the given Lagrangian, we can treat
φ
and
φ∗
as independent elds, since we can
express any of them as a sum of the real and imaginary part of
dependent.
We see that
φ∗ → φ∗ e−iα ),
L
is invariant under the transformation
φ,
which are in-
φ → φeiα
since we will obtain the same Lagrangian. Hence, since
α
(and
is a con-
tinuous parameter, we have a continuous symmetry, and will now look for the corresponding conserved current of the theory. We will look at the innitesimal symmetry
transformation, that is
By inserting in
L
φ → φ+iφα (and φ∗ → φ∗ −iφ∗ α), and we will take α → α(x).
we nd
L0 = −∂µ (φ∗ − iφ∗ α) ∂ µ (φ + iφα) − m2 |φ|2
= Lorginal − i (∂µ φ∗ ∂ µ (φα) − ∂µ (αφ∗ ) ∂ µ φ) + O(α2 ) ⇒
δL = i (∂µ (αφ∗ ) ∂ µ φ − ∂µ φ∗ ∂ µ (φα))
= i ((∂µ α) φ∗ ∂ µ φ + α (∂µ φ∗ ) ∂ µ φ − ∂µ φ∗ (∂ µ φ) α − ∂µ φ∗ φ (∂ µ α))
= i ((∂µ α) φ∗ (∂ µ φ) − (∂µ φ∗ ) φ (∂ µ α))
= i (φ∗ ∂ µ φ − ∂ µ φ∗ φ) (∂µ α) .
Hence, the candidate for the conserved current is
j µ = i (φ∗ ∂ µ φ − φ∂ µ φ∗ ),
for which
we now must apply the equations of motion, which is easily seen to be the KleinGordon equation for each of the elds:
∂L
−∂µ
∂φ
∂L
∂ (∂µ φ)
2 ∗
2 ∗
=0⇒∂ φ =m φ
∂L
,
−∂µ
∂φ∗
∂L
∂ (∂µ φ∗ )
= 0 ⇒ ∂ 2 φ = m2 φ .
We then have
∂µ j µ = i∂µ (φ∗ ∂ µ φ − φ∂ µ φ∗ )
= i φ∗ ∂ 2 φ − φ∂ 2 φ∗
= i φ∗ m2 φ − φm2 φ∗ = 0 ,
so we see that
jµ
is a conserved current.
20 of 88
5
5.1 Energy-momentum tensor
Figure 1: Components of the Energy-momentum tensor.
For any physical theory, we must have that the equations of motion must be invariant
under a spacetime translation by a xed vector
the Lagrangian
L
aµ ,
xµ → xµ + aµ .
that is
Hence
Spacetime translation
µ
cannot depend on x but only dierences in coordinates that are
invariant under Lorentz transformations.
Now, to use Noethers theorem we assume that
7
by a Taylor expansion (for small values of
aµ
L
aµ ) that
and
φ
xµ ,
we have
aµ ∂ µ L(x) and
δφ(x) =
depends of
δL(x) =
∂ µ φ(x)
∂L
∂L
δφ +
δ(∂µ φ)
∂φ
∂(∂µ φ)
∂L
∂L
∂L
δφ + ∂µ
=
− ∂µ
δφ
∂φ
∂(∂µ φ)
∂(∂µ φ)
|
{z
}
δL =
=0 for eqs. of motion
∂L
∂L
δφ = ∂µ
aν ∂ ν φ(x) ⇒
∂(∂µ φ)
∂(∂µ φ)
∂L
aν ∂ ν L(x) = aν ∂µ
∂ν φ ⇒
∂(∂µ φ)
∂L
ν
ν
0 = aν ∂µ
∂ φ − δµ L
∂(∂ µ φ)
≡ aν Tµν ,
= ∂µ
hence the quantity
Tµν
is conserved, and
T
is just what we will call the energy-
momentum tensor:
tensor
Tµν =
∂L
∂ ν φ − δµν L .
∂(∂ µ φ)
We can also write
T µν = g µα Tαν = g µα
7
Energy-momentum
∂L
∂L
∂ ν φ − g µα δαν L =
∂ ν φ − δ µν L .
α
∂(∂ φ)
∂(∂µ φ)
f (x + a) = f (x) + aµ ∂ µ f (x) + O(a2 ) ⇔ δf = aµ ∂ µ f (x)
21 of 88
6
FREE QUANTUM FIELDS
The energy-momentum tensor contains a lot of informations, and its components
interprets as given in gure 1. Especially we see
T00 =
so this is just the
H,
∂L
∂ 0 φ − L = Πφ̇ − L = H ,
∂(∂ 0 φ)
the hamiltonian density, which
Tµν
tells us is conserved.
6 Free quantum elds
6.1 Canonical quantization
The standard procedure for obtaining a quantum eld theory is to perform a canonical
Canonical
quantization which comes to replacing classical elds and densities by operators, and
tion
quantiza-
imposing commutation relations given below (putting 'hats' on everything). Hence
we substitute
φ(x) 7→ φ̂(x)
which is now the quantum eld (which we would like
to nd; there may be more quantum elds
φ̂ν
is a generalization of the coordinates, and
Π(x) 7→ Π̂(x),
which each gives an operator) which
and impose the following
canonical commutation relations (all given at the same time
h
Quantum eld
t)
Canonical c.r.
i
φ̂(x, t), φ̂(x0 , t) = 0 ,
±
h
i
Π̂(x, t), Π̂(x0 , t) = 0 ,
±
h
i
φ̂(x, t), Π̂(x0 , t) = iδ 3 (x − x0 ) ,
±
where we must choose the commutator for bosonic elds, and anti-commutator
for fermionic elds. Strictly, these commutation relations are only valid for free elds
(to be dened below), and we don't really know or care about interacting elds, since
these can just be considered perturbations of the free eld.
6.2 The Klein-Gordon equation and free elds
The Klein-Gordon equation is is
Klein-Gordon eqs
−∂ 2 + m2 f (x) = 0 ,
where
f (x) = f (t, x, y, z)
and
m
is some real parameter.
The Klein-Gordon
appears in many dierent contexts, and most importantly when one tries to square
the Hamiltonian of a free relativistic particle
H2
=
P2
+
m2
=
∇2
+
H =
√
P 2 c2 + m2 c4 =
where
ψ
P 2 + m2 ⇒
m2 and inserts it in the square Schrödinger equation, which
then yields that the equation of motion for a free particle is
0,
√
is the wave-function.
−∂ 2
+
m2
ψ(x) =
Looking at it this way however leads to having
wavefunctions with negative energy - and even worse; normalization is not retained.
The Klein-Gordon equation has plane-wave solutions of the type
e
i(~k·~
x−ωk t)
q
µ
, ωk ≡ ~k 2 + m2 , ~k ∈ R3 ⇔ eikµ x ,
22 of 88
Single relativistic particle
7
FREE SPIN-0 FIELDS
which can put together in an superposition to form any real solution
f (x)
(by
Fourier transformation) as
f (x) =
=
where
a(~k)
ˆ
h
i
1
3~
~k)ei(~k·~x−ωk t) + a∗ (~k)e−i(~k·~x−ωk t)
d
k
a(
(2π)3
ˆ
h
i
1
3~
~k)eikµ xµ + a∗ (~k)e−ikµ xµ ,
d
k
a(
(2π)3
is the Fourier coecients of
f.
A free eld theory is a theory where the particles doesn't interact.
In mathe-
matical terms, a free eld theory is that has linear equations of motion, that is, the
Lagrangian contains terms that is at most quadric. Hence, a free eld must have a
equation of motion given by the Klein-Gordon equation
−∂ 2 + m2 φ̂(x) = 0 ,
which has the Lagrangian
1
1
L = − ∂ µ φ̂∂µ φ − m2 φ̂2 + Ω0 .
2
2
Perhaps with more indicies, several eld components and so on.
7 Free spin-0 elds
When canonical quantization is performed, it yields the free bosonic eld when using
commutators. In principle we should also be able to use anticommutators for all we
know now, but we would quickly see that for the free eld we would get nonsense,
which implies that spin-0 particles are bosons.
For a classical eld the general solution
ˆ
φ(x) =
φ
is given by
General solution
h
i ˆ
h
i
d3~k
ikµ xµ
∗ ~ −ikµ xµ
~
˜ a(~k)eikµ xµ + a∗ (~k)e−ikµ xµ ,
a(
k)e
+
a
(
k)e
=
dk
(2π)3 2ωk
where the measure
˜ ≡
dk
d3~k
is Lorentz invariant as can be seen below.
(2π)3 2ωk
23 of 88
7
Example 14
showing
´
=
.
d3~k
is Lorentz invariant) This is equivalent to
(2π)3 2ωk
3
~
d k
2
2
0
dk 0 (2π)
3 δ(k + m )θ(k ), which is Lorentz invariant since it is
(Show that
d3~k
3
FREE SPIN-0 FIELDS
˜ =
dk
(2π) 2ωk
4
a d measure. We have
ˆ
ˆ
d3~k
δ(k 2 + m2 )θ(k 0 ) =
dk
(2π)3
ˆ
d3~k X θ(k 0 )
=
= dk 0
(2π)3 k =±ω 2k0
dk 0
0
0
since
k 2 + m2 = 0
The factor
θ(k 0 )
~k
X
k0 =±ωk
only has a zero at
k02 = m2 + ~k 2 = ωk2 ⇒ k0 = ±ωk .
arises since we only want orthosynchronous transformations.
Going to quantum eld theory, we want to 'put hats' on
φ,
we see that the only factors that could become operators are
and
a∗ (~k) → â† (~k).
θ(k 0 )
d(k 2 + m2 )/dk 0
d3~k
˜ ,
= dk
(2π)3 2ωk
k
for xed
d3~k
(2π)3
and trying to do this
a(~k),
so
a(~k) → â(~k)
Hence, a general quantum eld is given by
ˆ
φ̂(x) =
General free quantum
eld
h
i
˜ â(~k)eikµ xµ + â† (~k)e−ikµ xµ .
dk
Now, we can nd the momentum density eld operator, which for the Klein-
Momentum density
Gordon eld is given as
ˆ
h
i
˜ â(~k)∂0 eikµ xµ + â† (~k)∂0 e−ikµ xµ
Π̂(x) = ∂0 φ̂(x) = dk
ˆ
h
i
˜ k â(~k)eikµ xµ − â† (~k)e−ikµ xµ .
= −i dkω
Applying the commutation relations of
φ̂(x)
tation relations for the annihilation operator
and
â(~k)
Π̂(x)
can then give us commu-
and the creation operator
Applying the formulas and collecting in commutators of for
â(~k)
and
â† (~k),
â† (~k).
we nd
â(~k)
that
and
â† (~k)
h
i
â(~k), â(~q) = 0 ,
h
i
â† (~k), â† (~q) = 0 ,
h
i
â(~k), â† (~q) = (2π)3 2ωk δ 3 (~k − ~q) ,
which is the same commutation relations as for the usual creation/annihilation
operators. Hence we should expect the same physical interpretation, in the matter
of creating or annihilating states.
We can rewrite the Hamiltonian in terms of the creation/annihilation operators
by doing a lot of tedious algebra and integration:
24 of 88
Hamiltonian
c.r.
7
ˆ
Ĥ(t) =
=
=
=
=
ˆ
ˆ
FREE SPIN-0 FIELDS
2 1
1
d xH = d x Πφ̇ − L = d x Π̂2 + ∇φ̂ + φ̂2
2
2
ˆ
1
˜ k â† (~k)â(~k) + â(~k)â† (~k)
dkω
2
ˆ
h
i
˜ k â† (~k)â(~k) + 1 â(~k), â† (~k)
dkω
2
ˆ
˜ k â† (~k)â(~k) − π 3 δ 3 (~k − ~k)
dkω
ˆ
˜ k â† (~k)â(~k) ,
dkω
3
where the term
´
3
h
˜ k π 3 δ 3 (~k − ~k)
dkω
i
3
is very much innity, but we can allow
ourself to ignore it since it is a 'constant shift', and theese has no inuence on the
expectation values.
There exists a state of the Hamiltonian
Ĥ
that has energy zero, since
bounded from below. We will call this state the vacuum state
â(~k) |0i = 0
† k)
(and equivivalently h0| â (~
terminology that
â† (~k) |0i
= 0).
is an eigenstate of
|0i,
Ĥ
is
and it must fulll
We will now see as expected by
Ĥ 8 :
ˆ
†
˜ 0 ωk0 â† (~k 0 )â(~k 0 )â† (~k) |0i
dk
ˆ
h
i
˜ 0 ωk0 â† (~k 0 ) â(~k 0 ), â† (~k) + â† (~k)â(~k 0 ) |0i
=
dk
ˆ
h
i
˜ 0 ωk0 â† (~k 0 ) â(~k 0 ), â† (~k) |0i
=
dk
ˆ
˜ 0 ωk0 â† (~k 0 )2π 3 2ωk0 δ 3 (~k 0 − ~k) |0i
=
dk
ˆ
=
d3~k 0 ωk0 â† (~k 0 )δ 3 (~k 0 − ~k) |0i
Ĥâ (~k) |0i =
= ωk â† (~k) |0i ,
p
ωk = ~k 2 + m2 . This we can
† k) created a particle with momentum ~
interpret as â (~
k and mass m. The same way
QN
†
†
†
† ~
~
~
~
we can show that â (k1 )â (k2 ) · · · â (kM ) |0i =
n=1 â (kn ) |0i ≡ |k1 k2 · · · kN i is an
PN
eigenstate of Ĥ with eigenvalue ωk1 + ωk2 + . . . ωkN =
n=1 ωkn :
so
8
â† (~k) |0i
h
is really an eigenstate with eigenvalue
i
Using â(~k0 ), â† (~k) = â(~k0 )â† (~k) − â† (~k)â(~k0 ) = 2π 3 δ 3 (~k0 − ~k) ⇔ â(~k0 )â† (~k) = â† (~k)â(~k0 ) +
2π 3 δ 3 (~k0 − ~k)
25 of 88
Vacuum state
7
FREE SPIN-0 FIELDS
ˆ
Ĥ |k1 k2 · · · kN i =
=
=
=
=
=
˜ 0 ωk0 â† (~k 0 )â(~k 0 )â† (~k1 )â† (~k2 ) · · · â† (~kM ) |0i
dk
ˆ
n
o
˜ 0 ωk0 â† (~k 0 ) â(~k 0 )â† (~k1 ) â† (~k2 ) · · · â† (~kM ) |0i
dk
ˆ
h
i
˜ 0 ωk0 â† (~k 0 ) â(~k 0 ), â† (~k1 ) + â† (~k1 )â(~k 0 ) â† (~k2 ) · · · â† (~kM ) |0i
dk
ˆ
˜ 0 ωk0 â† (~k 0 ) 2π 3 2ωk0 δ 3 (~k 0 − ~k1 ) + â† (~k1 )â(~k 0 ) â† (~k2 ) · · · â† (~kM ) |0i
dk
ˆ
d3~k 0 ωk0 δ 3 (~k 0 − ~k1 ) â† (~k 0 ) · · · â† (~kM ) |0i
ˆ
n
o
˜ 0 ωk0 â† (~k 0 )â† (~k1 ) â(~k 0 )â† (~k2 ) · · · â† (~kM ) |0i
+ dk
ˆ
h
i
˜ 0 ωk0 â† (~k 0 )â† (~k1 ) â(~k 0 ), â† (~k2 ) + â† (~k2 )â(~k 0 ) · · · â† (~kM ) |0i
ωk1 |k1 k2 · · · kN i + dk
.
.
.
=
N
X
!
ωkn
|k1 k2 · · · kN i ,
n=1
and hence
|k1 k2 · · · kN i
is an eigenstate of
Ĥ .
Notice that the calculation
above implies that the particles obey Bose-Einstein statistics, since
â† (~kj )â† (~ki ) by the commutation rules, so we have that
â† (~ki )â† (~kj ) =
|k1 · · · ki · · · kj · · · kN i =
Bose-Einstein
tics
|k1 · · · kj · · · ki · · · kN i for arbitrary i, j , so the states are symmetric under exchange
of particles.
7.1 Lorentz transformation of elds and operators
We know from the discussion of section 4.4 that we must have
φ̂(Λ−1 x).
U (Λ)−1 φ̂(x)U (Λ) =
In terms of the creation/annihilation operators, we have
ˆ
h
i
˜ â(~k)eikµ xµ + â† (~k)e−ikµ xµ U (Λ)
φ̂(x)U (Λ) = U (Λ)
dk
ˆ
h
i
˜ U (Λ)−1 â(~k)U (Λ)eikµ xµ + U (Λ)−1 â† (~k)U (Λ)e−ikµ xµ =
=
dk
ˆ
h
i
−1
˜ â(~k)eikΛ−1 x + â† (~k)e−ikΛ−1 x
(φ̂(Λ x) ⇒) =
dk
ˆ
h
i
0
˜ 0 â(Λ−1~k 0 )eiΛ−1 k0 Λ−1 x + â† (Λ−1~k 0 )e−iΛ−1 k0 Λ−1 x
(changing variables to k = Λk) =
dk
ˆ
h
i
˜ â(Λ−1~k)eiΛ−1 kΛ−1 x + â† (Λ−1~k)e−iΛ−1 kΛ−1 x
=
dk
ˆ
h
i
˜ â(Λ−1~k)eikµ xµ + â† (Λ−1~k)eikµ xµ ,
=
dk
U (Λ)
−1
−1
This gives us that the creation/annihilation operators transforms as they should,
that is
U (Λ)−1 â(~k)U (Λ) = â(Λ−1~k) , U (Λ)−1 â† (~k)U (Λ) = â† (Λ−1~k) ,
26 of 88
Transformation
statis-
7
and we have that as a consequence of this that
a state
|k1 k2 · · · kN i
FREE SPIN-0 FIELDS
â† (~k) = U (Λ)−1 â† (Λ~k)U (Λ),
so
transforms as
N
−1
Y
U (Λ) |k1 k2 · · · kN i =
i=1
N
−1
Y
=
i=1
N
Y
!
†
U (Λ)â (~ki )U (Λ)
−1
U (Λ)â† (~kN ) |0i
!
U (Λ)â† (~ki )U (Λ)−1
!
†
−1
U (Λ)â (~ki )U (Λ)
=
U (Λ)â† (~kN )U (Λ)−1 |0i
|0i
i=1
= |Λk1 Λk2 · · · ΛkN i
where we in line two have inserted an
|0i → U (Λ)−1 |0i,
since we must have
|0i = U (Λ)−1 |0i for the vacuum state since nothing happens there.
|k1 k2 · · · kN i
Hence have that
transforms as any vector is suppose to.
7.2 Simple inner products and expectation values of states
For the vacuum state we have that it is normalized, so
h0|0i = 1 ,
and it is easy to see (using
â(~k) |0i = 0
and
h0| â† (~k) = 0)
that
D
E
hk1 · · · kN |0i = k1 · · · kN |â(~kM ) · · · â(~k1 )|0 = 0 unless k1 = k2 = . . . = kN = 0 ,
and the normalization
0
k|k = (2π)2 2ωδ 3 (k − k0 ) .
We also have that
D
D
ˆ
E
h
i µ
µ
ik
x
†
−ik
x
µ
µ
˜ â(~k)e
0|φ̂|0
=
0| dk
+ â (~k)e
|0
ˆ
h
i
˜ h0| â(~k) |0i eikµ xµ + h0| â† (~k) |0i e−ikµ xµ
=
dk
ˆ
h
i
˜ 0eikµ xµ + 0e−ikµ xµ = 0.
=
dk
k|φ̂|0
E
ˆ
h
i ˜ 0 â(~k 0 )eikµ0 xµ + â† (~k 0 )e−ikµ0 xµ |0
k| dk
ˆ
i h
0 xµ
0 xµ
0
0 ikµ
† ~ 0 −ikµ
˜
~
~
~
=
0| dk â(k)â(k )e
+ â(k)â (k )e
|0
=
= ...
= eikx .
27 of 88
8
INTERACTION AND SCATTING IN SPIN-0 THEORIES
8 Interaction and scatting in spin-0 theories
Figure 2: Sketch that shows why we can treat interactions as pertubations; elds are
free most of the time anyway.
As the gure above indicates, particles are most of the time free (or can at least
be treated so), and the physics (that we are interested in) happens on very short
timescales and spatial regions (but may be 'wild'), so we can just treat interaction
between particles as a pertubation to the previous derived free particle Hamiltonian.
To develope this, it is neat to realize that quantum mechanics can take place in
dierent 'pictures', that is dierent views on how the system evolves in time an
place, but which of course yields same physics (expectation values).
Pictures
We have the
three most important: The Schrödinger picture (time evolution of the wavefunction),
the Heisenberg picture (time evolution of operators), and the interaction picture (a
mix between the two former).
The Schrödinger picture
Schrödinger picture
In the Schrödinger picture, one looks at the operators are constant, but the states
evolve in time. To make translations in time, we need to dene the time evolution
operator
U (t, t0 ) ≡ exp −iĤ(t − t0 ) ,
which of course is unitary and have the property that
|ψ(t)i = U (t, t0 ) |ψ(t0 )i .
When taking expectation values of an operator
Q̂
we get an expression of the
kind
D
E
E
D
E D
E D
Q̂(t) = ψ(t)|Q̂|ψ(t) = ψ(t)|U † (t, t0 )Q̂U (t, t0 )|ψ(t) = ψ(t)|eiĤ(t−t0 ) Q̂e−iĤ(t−t0 ) |ψ(t) .
S
28 of 88
8
The Heisenberg picture
Heisenberg picture
In the Heisenberg picture, the states are constant in time, but the operators evolves
in time. Time evolution of the operators are given by
Q̂(t) = U † (t, t0 )Q̂(t0 )U (t, t0 ) = eiĤ(t−t0 ) Q̂(t0 )e−iĤ(t−t0 ) ,
so when we are taking expectation values we have
D
E
Q̂(t)
H
D
E D
E D
E
,
= ψ|Q̂(t)|ψ = ψ(t0 )|U † (t, t0 )Q̂U (t, t0 )|ψ(t0 ) = Q̂(t)
S
so the two pictures are equvalent when it comes to the physics.
8.1 The interaction picture, Dyson expansion
Interaction picture
In the interaction picture we will have a mix of the two, since this will suit us best
when we want to do pertubation theory in quantum eld theory. We assume that
the Hamiltonian (of the Schrödinger picture) can be written as
Ĥ = Ĥ0 + ĤI ,
where
Ĥ0
is the 'unpertubed' Hamiltonian (which we assume we can solve exactly), and
ĤI
is the interaction Hamiltonian.
For an operator
Q̂,
we will do a time evolution by
Ĥ0
only, so
Q̂I (t) = eiĤ0 (t−t0 ) Q̂(t0 )e−iĤ0 (t−t0 ) ,
and we must compensate for not taking the full Hamiltonian for time evolution,
so the states must be pushed forward in time by
|ψ(t)i = eiĤ0 (t−t0 ) e−iĤ(t−t0 ) |ψ(t0 )i ,
since we wil now have
D
E
Q̂(t)
I
D
E
ψ(t)|Q̂I (t)|ψ(t)
D
E
=
ψ(t0 )|eiĤ(t−t0 ) e−iĤ0 (t−t0 ) eiĤ0 (t−t0 ) Q̂(t0 )e−iĤ0 (t−t0 ) eiĤ0 (t−t0 ) e−iĤ(t−t0 ) |ψ(t0 )
D
E D
E
D
E
=
ψ(t0 )|eiĤ(t−t0 ) Q̂(t0 )e−iĤ(t−t0 ) |ψ(t0 ) = Q̂(t)
= Q̂(t)
,
=
H
S
so the physics is the same. The advantage comes now when dening UI (t, t0 ) ≡
i
Ĥ
0
e (t−t0 ) e−iĤ(t−t0 ) , and the job is now to evaluate this in a good way. We see that
fullles the dierential equation
i
∂
UI (t, t0 ) = eiĤ0 (t−t0 ) Ĥ − Ĥ0 e−iĤ(t−t0 )
∂t
= eiĤ0 (t−t0 ) ĤI (t0 )e−iĤ(t−t0 )
= ĤI (t)U (t, t0 ) ,
which is to say that it is the interaction that drives the time evolution of the
29 of 88
8
state. This equation can we solve for
get an approximation to
UI (t, t0 )
UI (t, t0 )
in an iterative way, so that we can
as good as we want. We have then the following
integral equation:
ˆ
t
∂
UI (t0 , t0 )dt0 = I − i
∂t
UI (t, t0 ) = I +
t0
ˆ
t
ĤI (t0 )U (t0 , t0 )dt0 .
t0
To a zero'th order approximation we simply have
UI0 (t, t0 ) = I ,
and then to a
rst order approximation we get
ˆ
UI1 (t, t0 )
t
=I −i
ĤI (t0 )dt0 ,
t0
and to a second order approximation we have
ˆ
ˆ tˆ
t
UI2 (t, t0 ) = I − i
t0
To
n'th
t2
ĤI (t1 )dt1 −
dt2 dt1 ĤI (t2 )ĤI (t1 ) ,
t0
t0
order we have the Dyson series
UIn (t, t0 )
ˆ
n
X
=I+
(−i)j
j=1
Dyson series
ˆ
···
dtn · · · dt1 ĤI (tn ) · · · ĤI (t1 )
t0 <t1 <···<tn <t
This formal solution will eventually diverge when taking many terms, and so is
an asymptotic series (diverges in the innity, but converges at high nite order of
Asymptotic series
expansion, which yields a good approximation).
If we look at the expansion, we see that it is actually time-ordered with the latest
times at the left and the earliest times at the right. We can dene the time-ordering
symbol
T
operators
to simplify the expression quite a bit. The time-ordering symbol of two
A
and
B
is given as

A(t )B(t )
1
2
TA(t1 )B(t2 ) =
B(t )A(t )
2
1
if t1 > t2
if t2 > t1
= θ(t1 − t2 )A(t1 )B(t2 ) + θ(t2 − t1 )B(t2 )A(t1 ) ,
and so forth to many operators.
30 of 88
Time ordering
Time ordering symbol
8
Figure
3:
Illustration
that
shows
what
the
time
ordering
does
(Source:
http://bolvan.ph.utexas.edu/~vadim/Classes/2008f.homeworks/dyson.pdf )
If we take a look at gure 3, we see that for the case with the intergral over two
operators, we because of the symmetry that we have two identical operators that are
beeing time orderet that the two intergrals in the second line are identical and hence
we have
ˆ
t
T
0
ĤI (t )dt
0
ˆ t ˆ
2
t
dt2 dt1 ĤI (t2 )ĤI (t1 )
= T
t0
ˆ
t0
tˆ t
≡
t0
dt2 dt1 TĤI (t2 )ĤI (t1 )
t0 t0
ˆ t ˆ t1
ˆ tˆ
=
t0
t0
ˆ tˆ
t2
t0
and so on for larger exponents than
T
0
0
ĤI (t )dt
t0
t0
dt2 dt1 ĤI (t2 )ĤI (t1 ) ,
t0
t
dt1 dt2 ĤI (t2 )ĤI (t1 )
t0
= 2
ˆ
t2
dt2 dt1 ĤI (t2 )ĤI (t1 ) +
2
we have:
ˆ
n
ˆ
···
= n!
dtn · · · dt1 ĤI (tn ) · · · ĤI (t1 ) .
t0 <t1 <···<tn <t
We then see that the formal solution is
ˆ t
ˆ t
n
X
j 1
= I+
(−i)
···
dtn · · · dt1 TĤI (tn ) · · · ĤI (t1 ) ⇒
j! t0
t0
j=1
ˆ t
∞
0
0
UI (t, t0 ) = UI (t, t0 ) = T exp −i
ĤI (t )dt .
UIn (t, t0 )
t0
This result we can just expand at any order we like to get the required accuracy.
31 of 88
8
8.2 Interaction picture in spin-0 quantum eld theories
Now comes the real deal; going to quantum eld theory in the interaction-picture.
The time-evolution of the eld is just by the free eld (Heisenberg-evolution), so we
have for some initial eld conguration
evolution
φ̂I (x)
φ̂(t0 , ~x)
that the interaction picture time-
is:
φ̂I (x) = φ̂I (t, ~x) = eiĤ0 (t−t0 ) φ̂(t0 , ~x)e−iĤ0 (t−t0 ) .
All of the perturbation happens in the states.
We must take the elds as time ordered since, we are going to expand the interaction Hamiltonian according to the time-ordered Dyson expansion, so we dene the
time ordering of the eld operators is given by
Time ordering
Tφ̂(x)φ̂(x0 ) = θ(t − t0 )φ̂(x)φ̂(x0 ) + θ(t0 − t)φ̂(x0 )φ̂(x) .
Say we want to nd the vacuum expectation value of this two-point correlation
function
Two-point
correlation
function
0
0
F(x, x ) = Tφ̂(x)φ̂(x ) ,
which could be interpret as nding the amplitude of particles of the eld interacting
with as
φ2 .
It is useful to make sure that
Ĥ |0i = 0
(can by done by an arbitrary
constant shift in Hamiltonian) and if the vacuum state has the lowest energy, that is,
that
|0i is the ground state of Ĥ .
expres the true ground state of
If we don't have that this is fullled, then we must
Ĥ ,
called
|Ωi,
in terms of
|0i
in some way to make
True ground state
sure that we can take vacuum expectation values, since we are treating everything
above the free eld as a pertubation. The formula relating the two is
|Ωi =
where
e−iĤT |0i
,
T →∞(1−i) e−iE0 T hΩ|0i
lim
D
E
E0 = Ω|Ĥ|Ω
is the true ground state energy. When taking expectation
|Ωi
of some time-ordered product of eld operators, we have
values with repect to
the the relation to the vacuum state as
True-vacuum
state relation
D
D
n
o E
Ω|T φ̂(x1 ) · · · φ̂(xN ) |Ω =
lim
T →∞(1−i)
so we can always just evaluate
n
o E
´T
0|T φ̂(x1 ) · · · φ̂(xN )e−i T dtHI (t) |0
D
E
,
´T
0|Te−i T dtHI (t) |0
D
n
o E
´T
0|T φ̂(x1 ) · · · φ̂(xN )e−i T dtHI (t) |0
and nor-
malize it properly afterward.
If we now want to evaluate the vacuum expectation value of the two-point correlation function, we have that
D
E
D
E
D
E
0|Tφ̂(x)φ̂(x0 )|0 = θ(t − t0 ) 0|φ̂(x)φ̂(x0 )|0 + θ(t0 − t) 0|φ̂(x0 )φ̂(x)|0 ,
32 of 88
ground
8
and thus we only need to calculate
D
E
0|φ̂(x)φ̂(x0 )|0
to get the full time-ordered
product, since we can just swap the names of the variables. We have
D
0|φ̂(x)φ̂(x0 )|0
E
ˆ ˆ
h
ih
i ˜ dk
˜ 0 â(~k)eikµ xµ + â† (~k)e−ikµ xµ â(~k 0 )eikµ0 x0µ + â† (~k 0 )e−ikµ0 x0µ |0
0|
dk
ˆ ˆ
D h
i E
˜ dk
˜ 0 0| â(~k), â† (~k 0 ) |0 eikµ xµ eikµ0 x0µ
=
dk
=
ˆ
=
d3~k
0
eik(x−x ) .
3
(2π) 2ωk
8.2.1 The Feynman propagator
Figure
4:
Feynman
outside
Plot
that
propagator,
of
the
shows
and
light-cone,
the
one
but
amplitude
can
falls
and
phase
see
that
it
is
of
very
quickly
of
the
non-zero
(Source:
http://golem.ph.utexas.edu/category/2009/03/the_algebra_of_grand_unied_t_1.html).
The last integral can be evaluated by Cauchy's residue theorem in terms of Bessel
functions, and totally when adding the step-functions
θ
Feynman solved it in terms of the Feynman propagator
makes it complicated, but
∆(x − x0 )
to nd
Feynman propagator
33 of 88
8
D
ˆ
0
E 1
eik(x−x )
1
d4 k
0|Tφ̂(x)φ̂(x0 )|0 = ∆(x − x0 ) ≡
, > 0.
i
i
(2π)4 k 2 + m2 − i
One can say that the Feynman propagator propagates the eld interaction (in
terms of some virtual particles, that mediates the interaction and only exists for a
Virtual particles
short while). It is of course Lorentz invariant, but is not exactly causal in the usual
sense. It is non-zero outside of the lightcone, but only noticeable around
x − x0 = 0
which one can interprets as the virtual particles may break causality and travel faster
than light, but as the name suggest the virtual particles must disappear before we can
ever peform measurements that shows their existence. We can't actually construct
such experiments, since this would destroy the quantum states upon measuring.
In momentum space, the Feynman propagator is just the Fourier transform of
1
i ∆(x
− x0 ),
so we have
1˜
−i
∆(p) = 2
i
p + m2 − i
Mathematical details: The poles of the Feynman propagator is
4
(o the real axis' in C ), and the small
k 0 = ± (ωk − i)
Poles
> 0 is just to make integration simple so we
won't integrate over some singularities in
R4 ,
and in the end we will just take
→0
if we ever need it.
8.2.2
n-point
correlation function and Wick's theorem
When we have a correlation function with
n
eld terms,
n
o
F(x1 , . . . , xn ) = T φ̂(x1 ) · · · φ̂(xn ) ,
and we would like to evaluate the time-ordered vacuum expectation value of this,
we can start with
Tφ̂(x)φ̂(x0 )
and do induction from there.
Dene:
ˆ
φ̂+
I (x)
≡
˜ ~k)eikµ xµ , φ̂− (x) ≡
dkâ(
I
We then have the original eld given by
φ̂(x)φ̂(x0 ) (for
t>
ˆ
˜ † (~k)e−ikµ xµ .
dkâ
−
φ̂I (x) =h φ̂+
I (x) + φ̂I (x)
i.
t0 ) in terms of the commutator
φ̂+
I (x),
0
φ̂−
I (x )
We can rewrite
and some other
terms. We nd that
+
−
− 0
+ 0
0
0
φ̂I (x)φ̂I (x0 ) = φ̂+
(x)φ̂−
φ̂− (x)φ̂+
I (x)φ̂I (x ) + φ̂
I (x ) +
I (x ) + φ̂I (x)φ̂I (x )
hI
i I
+ 0
+
− 0
− 0 +
−
+ 0
−
− 0
= φ̂+
I (x)φ̂I (x ) + φ̂I (x)φ̂I (x ) + φ̂I (x )φ̂I (x) + φ̂I (x)φ̂I (x ) + φ̂I (x)φ̂I (x )
h
i
+
− 0
−
= φ̂I (x), φ̂I (x ) + N (φ̂+
I , φ̂I ) ,
where
−
+
+ 0
− 0 +
−
+ 0
−
− 0
N (φ̂+
I , φ̂I ) = φ̂I (x)φ̂I (x ) + φ̂I (x )φ̂I (x) + φ̂I (x)φ̂I (x ) + φ̂I (x)φ̂I (x )
is a normal ordering of
−
φ̂+
I , φ̂I ,
that is, all of the
â
operators to the right and all of
†
the â 's to the left. For any normal ordering of elds, we have that
34 of 88
Normal ordering
8
h0|N |0i = 0 .
We have that
h
φ̂+
I (x),
i
0
φ̂−
I (x )
ˆ ˆ
h
i
˜ 0 â(~k)â† (~k 0 )eikx−k0 x0 − â† (~k 0 )â(~k)eikx−k0 x0
˜ dk
dk
ˆ ˆ
h
i
˜ 0 â(~k), â† (~k 0 ) eikx−k0 x0
˜ dk
=
dk
ˆ
d3 k
0
eik(x−x ) ,
=
3
(2π) 2πωk
=
Which is a part of the Feynman propagator written in sums of step-functions. If
we evaluate the whole time-ordered product, we will then nd that
to its contraction
φ(x)φ(x0 ) (now dropping the subscript I
φ̂(x)φ̂(x0 ) is equal
since it is understood that
Contraction
we are working the interaction picture, and the hats) plus some normal-ordered terms
N,
where the contraction is dened as:
h
i
 φ̂+ (x), φ̂− (x0 )
I
I
i
φ(x)φ(x0 ) ≡ h
 φ̂+ (x0 ), φ̂− (x)
I
I
We can absorb
φ(x)φ(x0 )
, t > t0
1
= ∆(x − x0 ) .
i
, t0 > t
into the normal ordering since it is just a number and
thus we can write
n
o
T φ̂(x)φ̂(x0 ) = N φ̂(x)φ̂(x0 ) + φ̂(x)φ̂(x0 ) ,
and to be explicit in this notation
D
n
o E
D
E
0|T φ̂(x)φ̂(x0 ) |0
=
0|N φ̂(x)φ̂(x0 ) + φ̂(x)φ̂(x0 ) |0
D
E
=
0φ̂(x)φ̂(x0 )|0 = φ̂(x)φ̂(x0 ) h0|0i
=
1
∆(x − x0 ) .
i
Wick's theorem is very useful, since it states that any time-ordered products of
Wick's theorem
elds can be written as contractions plus normal-ordered products. We have explicit:
n
o
T φ̂(x1 ) · · · φ̂(xn ) = N φ̂(x1 ) · · · φ̂(xn ) + all possible contractions .
As a consequence of this, the only terms that will count when calculating the
vacuum expectation values is the fully contracted terms (that is with contraction
symbol connecting every eld), because the uncontracted part of the term can be
normal-ordered to give zero. Further because of this we have that any time-ordered
product of uneven number of eld operators has expectation value 0, because for the
the terms that are contracted as much as possible we still have it
D
E
∝ 0|φ̂|0 = 0.
35 of 88
Fully contracted
8
Hence
n
o
T φ̂(x1 ) · · · φ̂(xn ) is equal to sums of products of Feynman propagators,
that reduces the workload quite a bit when evaluating the pertubation series.
It is now a matter of combinatorics to determine the the number of distinct ways
to do this, since the Feynman propagator is symmetric in the argument,
1
1
∆(x − x0 ) = ∆(x0 − x) ,
i
i
∆(x − x0 ) = ∆(x0 − x)
so we have many terms that are the same when evaluating later on.
Example 15
(Find the vacuum expectation value of
.
terms of Feynman propagators)
n
o
T φ̂(x1 )φ̂(x2 )φ̂(x3 )φ̂(x4 )
in
We have
T {φ(x1 )φ(x2 )φ(x3 )φ(x4 )} = N (φ(x1 )φ(x2 )φ(x3 )φ(x4 ) + all possible contractions)
= N φ(x1 )φ(x2 )φ(x3 )φ(x4 ) + φ(x1 )φ(x2 )φ(x3 )φ(x4 )
+φ(x1 )φ(x2 )φ(x3 )φ(x4 ) + φ(x1 )φ(x2 )φ(x3 )φ(x4 )
+ φ(x1 )φ(x2 )φ(x3 )φ(x4 ) + φ(x1 )φ(x2 )φ(x3 )φ(x4 ) .
Hence the vacuum expectation value is only the last three terms.
1
1
∆(x1 − x2 ) ∆(x3 − x4 )
i
i
1
1
+ ∆(x1 − x3 ) ∆(x2 − x4 )
i
i
1
1
+ ∆(x1 − x4 ) ∆(x2 − x3 )
i
i
= −∆(x1 − x2 )∆(x3 − x4 )
h0|T {φ(x1 )φ(x2 )φ(x3 )φ(x4 )} |0i =
−∆(x1 − x3 )∆(x2 − x4 )
−∆(x1 − x4 )∆(x2 − x3 ) .
We can generalize the notion of contraction to also contain external contractions
External contraction
by some state that is not the vacuum state. We have
φ(x) |pi = e−ipx |0i ,
hp| φ(x) = h0| eipx .
Again Wick's theorem applies, so the calculation of
D
E D
E
f |Ŝ|i = k1 k2 . . . km |Ŝ|p1 p2 . . . pk
is simplied a great deal when doing this. Each particle should be drawn as a external
line, that then might go into an interaction vertex.
D
i k1 k2 . . . km |T̂ |p1 p2 . . . pk
E
, since
D
Actually one
E
ˆ
k1 k2 . . . km |I|p1 p2 . . . pk = 0
calculates just
unless it is the
same number of particles, and they have the same momentum.
36 of 88
External line
8
8.2.3 Feynman diagrams and rules in spin-0 theories
The previous use of Wicks theorem can be represented as Feynman diagrams, which
Feynman diagrams
are spacetime diagrams with a 'time axis' and a 'space axis'. The rules of computing
numbers from these diagrams can be derived from the contractions, and does depend
on what the perturbation looks like. Each Feynman propagator is an internal line,
that is drawn as a line between the two spacetime points with verticies on the ends.
At each vertex we write
−ig
(in momentum space), where
g
is the coupling constant
of the interaction. At every vertex momentum is conserved, which can also be shown
by explicit calculations, so this must also be implied on the diagrams; the total
momentum of each vertex must be zero.
All of the combinatorial factors from the propagator-expansion, is exactly canceled when drawing all of the topologically inequivalent Feynman diagrams. Finally,
Topologically inequiv-
to obtain the expectation value (to the given order), everything is multiplied together.
alent
Example 16
tonian
rst
HI
order
(Find
=
we
λ
D
´
n
o E
´∞
0|T φ̂(x1 )φ̂(x2 )e−i ∞ dtHI (t) |0
d3 zφ(z)2
−i
e
have
´∞
∞
using
dtHI (t)
Feynman
=
gives h0|T {φ(x1 )φ(x2 )} |0i
=
´ 3
−iλ 0|T φ(x1 )φ(x2 ) d zφ(z)φ(z) |0
just
I − iλ
1
i ∆(x1
for the interaction Hamil-
diagrams
´
to
d3 zφ(z)2 .
− x2 ).
.
rst
order)
The
The
rst
second
and by example 14 we have by
To
term
term
is
z 7→ x3 , x4
ˆ
−iλ 0|T
d3 zφ(x1 )φ(x2 )φ(z)φ(z) |0
ˆ
= iλ d3 z (∆(x1 − x2 )∆(z − z)
∆(x1 − z)∆(x2 − z) + ∆(x1 − z)∆(x2 − z))
ˆ
= iλ d3 z (∆(x1 − x2 )∆(z − z) + 2∆(x1 − z)∆(x2 − z))
This is equal to the sum of the Feynman diagrams below, where the last two terms
is the second order terms
Often it is easier to go to momentum space, since the Feynman propagators are
easier, and if we just have a few particles that we know the momentum of at some
earlier and later times, they will enter the equations very simple.
´
If we have a bubble as in example 16, where the bubble is
d3 z∆(0),
´
d3 z∆(z − z) =
we have this is divergent since
ˆ
∆(0) =
which diverges for large
k.
transforming the expression for
d4 k
1
4 k 2 + m2 − i ,
(2π)
We should then introduce a ultraviolet cuto
∆(0)
Λ
and
so
37 of 88
Ultraviolet cuto
9
∆(0) =
SCATTERING THEORY
i
Λ2 .
16π 2
This bubble can be thought of as a loop-correction to the Feynman propagator,
and it takes another course to gure out how to renormalize this.
Feynman rules
The Feynman rules for spin-0 elds are
Figure 5: Components of Feynman diagrams for spin-0 elds in momentum space.
9 Scattering theory
In the interaction picture we can dene the time evolution operator take from some
very early time
−∞
to some very late time
∞,
and assume that the particles are
free to begin with and end up free. Then we call the time-evolution operator for
Ŝ ,
the S-operator or the S-matrix, which then glues the interaction together with the
free states to obtain some amplitude for a given proces to happen. Hence
ˆ ∞
0
0
Ŝ = UI (∞, −∞) = T exp −i
ĤI (t )dt
−∞


ˆ


= T exp −i
d4 xHI (x)
all spacetime
We have that
Ŝ Ŝ † = I ,
38 of 88
S-operator
9
which implies that probability is conserved, and
Ŝ
SCATTERING THEORY
is of course unitary (by de-
nition).
It is useful to dene another operator
T̂
such that
Ŝ ≡ Iˆ + iT̂ .
T̂
is is really the pertubation part of
Ŝ .
Further we can dene an operator
T
T
operator
such that


X
T̂ ≡ (2π)4 δ (4) 
X
pi −
incoming
kj  T̂ .
outgoing
When evaluating the perturbations we will always get the factor
9 δ (4)
P
incoming pi
which is just energy conservation, so this we take out so the expressions for
T
−
P
outgoing
kj
,
is more
compactly written.
If we want to evaluate the probability amplitude for a initial state
|ii = |Φ(t = −∞)i
with some number of particles and their momenta to end up ind a nal state
|f i = |Φ0 (t = ∞)i.
We want to join these states, so since we are in the inter-
action picture, we have have that
we are evaluating
|Φ(t)i = UI (t, −∞) |ii
and
|Φ0 (t)i = UI (t, ∞) |f i
hΦ0 (t)|Φ(t)i
Φ0 (t)|Φ(t) ≡ hf |ii = hf | UI† (t, ∞)UI (t, −∞) |ii
= hf | UI (∞, t)UI (t, −∞) |ii = hf | UI (∞, ∞) |ii
D
E
=
f |Ŝ|i .
So when we want to evaluate these scattering amplitudes, we can get a good
physical quantities (when absolute squared to get the probability, at least).
Now,
physically all particles are given as a wavepacket in momentum-space, but one just
look at the states of pure momentum to make life easier, but still remember that
it is quite nonphysically, and some integrals over all spacetime will diverge, which
we need to take care of by assuming that the universe has volume
integral is
T,
so the total spacetime volume is
|ii = |p1 p2 . . . pk i
and
|f i = |k1 k2 . . . km i.
V T.
We then have

X
incoming
where
T ≡ hf | T̂ |ii
and the time
Hence we will assume that

hf |ii = (2π)4 δ (4) 
V
pi −
X
kj  T ,
outgoing
is the scattering matrix element, for cases where there is an
actual scattering, that is
hk1 k2 . . . km |p1 p2 . . . pk i = 0.
Scattering matrix element
Feynman diagrams in
When using Feynman diagrams, we nd, by construction, that
scattering
iT =
X
All Feynman diagrams (in momentum space) .
For pure states of denite momentum only! Real, physically wavepackets doesn't have the
explicit factor.
9
39 of 88
9
Example 17
for
the
HI = g
a† (p
A
(Evaluate the transition amplitude for the proces
scalar
´
A, B, C
particles
.
d3 zφA (z)φB (z)φC (z))
)a† (p
under
the
hf | = h0| a(kA )a(kB ).
A+B → A+B
interaction
Given that the initial state
B ) |0i of the elds and the nal state
giving the ket
SCATTERING THEORY
hamiltonian
|ii = |pA pB i =
|f i = |kA kB i = a† (kA )a† (kB )D|0iE,
We then have that we can evaluate
iTf i = i T̂
to rst order
iTf i
ˆ
D E D
E 3
ˆ
= i T̂ = f |Ŝ − I|i = f |gT d zφA (z)φB (z)φC (z)|i
ˆ
3
†
†
= g 0| a(kA )a(kB )T d zφA (z)φB (z)φC (z) a (pA )a (pB )|0
= 0,
since when inserting
a† (~k), a(kA ), a(kB )
φ(z) =
´
commute
h
i
˜ a(~k)eikµ z µ + a† (~k)e−ikµ z µ , since the operators
dk
and so does a(~
k), a† (pA ), a† (pB ) since the particles
are of dierent species, and the canonical commutation relations only apply to particles of the same kind. Going to second order, we perform the calculation by doing
external contractions and noting that only full contractions contribute
iTf i
ˆ ˆ
=
f |gT
d z1 d z2 φA (z1 )φB (z1 )φC (z1 )φA (z2 )φB (z2 )φC (z2 )|i
ˆ ˆ
2
3
3
= g kA kB |T
d z1 d z2 φA (z1 )φB (z1 )φC (z1 )φA (z2 )φB (z2 )φC (z2 )|pA pB
ˆ
= g 2 d6 zi hkA kB |TφA (z1 )φB (z1 )φC (z1 )φA (z2 )φB (z2 )φC (z2 )|pA pB i .
3
3
There are in total X (26?) ways of doing this, so this is much easier evaluated using
Feynman diagrams to nd
T,
since there are only two topologically inequivalent
diagrams that contribute. Theese are
Hence we have by multiplying all of the factors according to the Feynman rules for
each term (ignoring
iT
in the Feynman propagator)
−i
−i
+ (ig)2 · 1 · 1 · 1 · 1 ·
2
(kb − pa ) + m2c
(pb + pa )2 + m2c
ig 2
ig 2
+
.
(kb − pa )2 + m2c
(pb + pa )2 + m2c
= (ig)2 · 1 · 1 · 1 · 1 ·
=
40 of 88
9
SCATTERING THEORY
9.1 Cross sections
Cross section
σ is a measurable quantity, that can be computed from the scatD
E2
amplitudes f |Ŝ|i . To derive a formula for it, we have by assuming
The cross section
tering
the universe is nite in space and time, so the total spacetime volume is
(2π)n δ (n) (0)
=
Ln , where
L
VT
and
is a combination of space and/or time:


2
D
E2
X
X
4
pi −
kj  |T |2
f |Ŝ|i = (2π) δ (4) 

incoming
outgoing
X
X


= (2π)4 δ (4) 
pi −
incoming
kj 
d4 xei(
incoming
pi −
P
outgoing
kj ) 

X
= (2π)4 V T δ (4) 
X
pi −
incoming
h0|0i = 1.

P
outgoing

Recall that
ˆ
kj  |T |2 .
outgoing
We have that
hp|pi
D
E
0|a(p)a† (p)|0
hp|pi =
= (2π)3 2ωp δ (3) (p − p) h0|0i
= 2ωp V .
Now, we consider
2→N
particle scattering in the following, with the two in-
coming particles called 1 and 2.
We have the norm squared of the states given
by
hi|ii = hp1 p2 |p1 p2 i = hp1 |p1 i hp2 |p2 i = 4ωp1 ωp2 V 2 ,
hf |f i =
N
Y
2ωkj V
j=1
We need the scattering probability
P =
P
to be normalized properly, so hence
|hf |ii|2
,
hf |f i hi|ii
and the probability pr. unit time for the scattering proces we get from dividing
by the timespan of the universe
Ṗ
=
=
T
to get
|hf |ii|2
T hf |f i hi|ii
P
P
2
(2π)4 V δ (4)
p
−
k
incoming i
outgoing j |T |
.
Q
4ωp1 ωp2 V 2 N
j=1 2ωkj V
Since each of the particles are in a nite box of volume
V,
momenta is actually
41 of 88
|T |2
9
SCATTERING THEORY
quantized (free particle in a box) and we must sum over them, but in the limit of
large, this sum tends to an integral. We should then correct
Ṗ
V
V
by the factor
d3 kj
(2π)3
so
P
2 N
p
−
k
incoming i
outgoing j |T | Y V
3
QN
3 d kj
2
(2π)
4ωp1 ωp2 V
j=1 2ωkj V
j=1
P
P
2 N
(2π)4 δ (4)
incoming pi −
outgoing kj |T | Y
˜j .
dk
4ωp1 ωp2 V
(2π)4 V δ (4)
Ṗ
=
=
P
j=1
Rewriting the expressions using the Mandelstam variables we can show that
4 k~v1 k ωp1 ωp1 = 4
q
q
(p1 · p2 ) − m21 m22 = 2 s2 − 2 m21 + m22 + (m1 − m2 )2 ,
|
{z
}
Lorentz invariant
dσ
is a Lorentz invariant quantity, and hence
is Lorentz invariant as we would like
it to be.
Hence, if we now choose a frame where particle 2 is xed (Fixed Target system) so
p~2 = ~0
that
and particle 1 moves with speed
s ≡ − (p1 + p2 )
2
k~v1 kF T ,
we have from Mandelstam variables
is the square of the center-of-mass energy.
k~
p1 kF T
1
=
2m2
Fixed Target
CM energy
q
2
s2 − 2 m21 + m22 s + m21 − m22 ,
and we have, which can be shown
√
m2 k~
p1 kF T =
We divide
Ṗ
by
s k~
p1 kCM
k~v1 k /V = k~
p1 k /V ωp1 , which is the incoming ux of prticle 1, to
get a quantity that say how much likelihood there is for scattering to occur, which
we will call the dierential cross section
10
dσ :
Dierential
tion
(2π)4 δ (4)
P
dσ =
incoming pi
−
P
outgoing
N
kj |T |2 Y
4 k~v1 kF T ωp1 ωp2
˜j
dk
j=1
2
where
≡
|T |
dLIPSN (p1 + p2 )
4 k~v1 k ωp1 ωp2
=
|T |2
dLIPSN (p1 + p2 )
4 k~
p1 kF T m2
=
|T |2
√ dLIPSN (p1 + p2 ) ,
4 k~
p1 kCM s
dLIPSN (p1 +p2 ) ≡ (2π)4 δ (4)
P
incoming pi −
Lorentz Invariant Phase Space measure for
used that
k~v1 kF T ωp1 = kp1 kF T
and
N
ωp2 = m2
P
outgoing kj
Q
N
˜
j=1 dkj is the
outgoing particles. In line three we
since particle 2 is not moving, and
10
It is implicitly understood that we must integrate over all unknown variables in the expression
of dσ to get rid of the δ -function.
42 of 88
cross
sec-
9
m2 k~
p 1 kF T =
from line 3 to 4 we used that
for
√
s k~
p1 kCM
SCATTERING THEORY
to obtain a simple expression
dσ .
For a proces of the kind
frame where
B
A + B → C + D,
is xed and
A
is shoot at
assuming we are in the Fixed Target
B.
invariant, we can calculate it in the CM frame,
dLIPS2 (p1 + p2 ) is Lorentz
where p
~A + p~B = ~0 (but the particles
Since
may still move in dierent directions with dierent speeds, since their mass is not
the same). In CM we have
p0total = ωA+B = ωA + ωB =
√
s,
and we have
dLIPS2 (ωA+B ) = dLIPS2 (pA + pB ))
˜C dk˜D
= (2π)4 δ (4) (pA + pB − kC − kD ) dk
√ 1
=
δ ωC + ωD − s δ (3) (~kC + ~kD )d3 k~C d3 k~D .
2
4 (2π) ωC ωD
We can integrate over
d3 k~D
to get rid of
δ (3) -function,
from which we now get
that that the energies of the particles are xed to
r r ~ 2
2
ωC = kC + m2C , ωD = k~C + m2D ,
and we then have
dLIPS2 (ωA+B )
dLIPS2 (ωA+B ) =
=
1
2
given as
δ ωC + ωD −
√ 3
s d k~C
4 (2π) ωC ωD
√ 1
~ 2 ~ s
k
δ
ω
+
ω
−
C d kC dΩCM ,
C
D
4 (2π)2 ωC ωD
where we in the last line switched to spherical coordinates and have the dierential
solid angle
dΩCM = sin θdθdφ
. Integrating over
~ kC ,
Dierential solid angle
we get by some calculation
that after some algebra and delta-function manipulation gives us
~ kC √ dΩCM .
dLIPS2 (ωA+B ) =
16π 2 s
Hence we will get the dierential cross section to be
~ kC |T |
√
√ dΩCM ⇒
4 k~
p1 kCM s 16π 2 s
~ |T |2 kC .
64π 2 s k~
p1 kCM
2
dσ =
dσ
dΩCM
=
The dierential decay rate of the proces
dΓCM
A → 1, 2
Decay rate
is given by
~ 1 k1 =
|T |2 dΩCM
32π 2 m2A
43 of 88
9
SCATTERING THEORY
9.2 Mandelstam variables
Figure
6:
The
three
channels
of
2-2
particles
scattering
(Source:
http://en.wikipedia.org/wiki/Mandelstam_variables).
To simplify
1+2 → 3+4
particle scattering, one can dene the Lorentz invariant
s ≡ − (p1 + p2 )2 = − (p3 + p4 )2
t ≡ − (p1 − p3 )2 = − (p2 − p4 )2
u ≡ − (p1 − p4 )2 = − (p2 − p3 )2
then the expression for the scattering process becomes much easier.
We can
view scattering in the Feynman diagrams as going along certain channels (since each
incoming particle can scatter with the three others, we have three diagrams). We
further have the property that
s + t + u = m21 + m22 + m23 + m24 ≡ M 2 ,
which puts a constraint on the variables.
Hence we have just two degrees of
freedom in our equations. Also we have some inner product relations:
p1 · p2 =
1
m21 + m22 − s
2
p1 · p3 =
1
t − m21 − m23
2
p1 · p4 =
1
u − m21 − m24
2
p3 · p4 =
1
m23 + m24 − s
2
p2 · p3 =
1
u − m22 − m23
2
Inner products
44 of 88
FREE SPIN-½ FIELDS
10
p2 · p4 =
1
t − m22 − m24
2
See appendix for more relations.
Example 18 (Show the inner product relations).
We have
−s = (p1 + p2 )2 = p21 + p22 + 2p1 · p2 = −m21 − m22 + 2p1 · p2 ⇒
p1 · p2 =
1
m21 + m22 − s
2
and
−t = (p1 − p3 )2 = p21 + p23 − 2p1 · p3 = −m21 − m23 − 2p1 · p3 ⇒
p1 · p3 =
1
t − m21 − m23
2
and
−u = (p1 − p4 )2 = p21 + p24 − 2p1 · p4 = −m21 − m24 − 2p1 · p4 ⇒
p1 · p4 =
Example 19
1
u − m21 − m24
2
A + B → A + B for the scalar
´ 3
particles A, B, C under the interaction Hamiltonian HI = g
d zφA (z)φB (z)φC (z)).
To
1
3! g
second
´
(Write the scattering amplitude for
order
d3 zφ(x
in
the
perturbation
expansion
1 )φ(x2 )φ(z)φ(z), an expression for
iT = g 2
iT
of
φ3 -theory
(with
ĤI
=
then takes the form
1
1
1
+
+
.
m2 − s m2 − t m2 − u
10 Free spin-½ elds
10.1 Spinor representations of the Lorentz group
After the scalar elds with
tation is
½
(2, 1)
respectively
(2n+ + 1, 2n− + 1) = (1, 1)
(1, 2),
the next simplest represen-
that gives rise to quantum elds describes spin-
particles, are called left-handed (Weyl) spinors respectively right-handed (Weyl)
spinors, which corresponds to spin-up and spin-down for the eld quanta. The
corresponding elds are two dimensional and given as
(2, 1) :
(1, 2) :
ψ̂a (x) =
ψ̂ȧ† (x) =
ψ̂1 (x)
ψ̂2 (x)
!
,
ψ̂1̇† (x) ψ̂2̇† (x)
,
45 of 88
Weyl spinors
FREE SPIN-½ FIELDS
10
where we have introduced the dot-notation, where we use undotted indicies for
Dot notation
the left-handed spinors and dotted indicies (and dagger on the eld) for the right-
Ji+
handed spinors. Notice that because of
h
†
= Ji−
we have that
i†
ψ̂a (x) = ψ̂ȧ† (x)
Under a Lorentz transformation we must then have for the left-handed spinor
elds that
Lorentz
transforma-
tion
U (Λ)−1 ψ̂a (x)U (Λ) = Lab (Λ)ψ̂b (Λ−1 x) ,
where
Lab (Λ) are 2×2-matrices that depends on the transformation Λ.
To satisfy
the group properties of the Lorentz group, we must have that
Lab (Λ)Lbc (Λ0 ) = Lab (ΛΛ0 ) .
We call any matrices/tensors that fullls this for a representation of the Lorentz
Representation
group, since we will have the same commutation relation for a set of some generating
matrices
SLαβ
b
SLαβ
a
which may or may not be hermitian depending on what we choose,
b
a
,
b
SLργ a
=i
g
αρ
SLβγ
b
a
− (α ↔ β) − (ρ ↔ γ)
,
and a innitesimal transformation is given by
b
i
Lab (1 + δω) = Iδab + δωµν SLµν a ,
2
and we must have that
SLαβ
are antisymmetric matrices (because of
δω
being
antisymmetric), so
SLµν
b
a
= − SLνµ
b
a
.
One can show that the Pauli matrices and the identity matrix are generators for
Pauli matrices
this representation via the relations
SLij
b
a
b
1
= ijk σ k
2
a
,
SLk0
b
a
=
i k b
.
σ
2
a
Likewise we for the right-handed spinor elds have that
U (Λ)−1 ψ̂ȧ† (x)U (Λ) = Rȧḃ (Λ)ψ̂ḃ† (Λ−1 x) ,
and
where the
i
µν ḃ
Rȧḃ (1 + δω) = Iδȧḃ + δωµν SR
,
ȧ
2
µν ḃ
generators SR
again of course are 2 × 2-matrices.
ȧ
We have by the
intimate relation between the left/right-handed elds that
Intimate relation
46 of 88
10
µν ḃ
SR
ȧ
=−
i∗
µν b
SL a
h
FREE SPIN-½ FIELDS
.
We can then always just choose one handedness and then it is easy to translate
it to the other handedness by some of these relations above.
Example 20 (Show that
µν
SR
ḃ
ȧ
=−
h
SLµν
b i∗
a
.
)
We have
h
i†
†
= [U (Λ)]† ψ̂a† (x) U (Λ)−1
U (Λ)−1 ψ̂a (x)U (Λ)
= U (Λ)−1 ψ̂ȧ† (x)U (Λ)
= Rȧḃ (Λ)ψ̂ḃ† (Λ−1 x)
h
i†
= Lab (Λ)ψ̂b (Λ−1 x)
h
i†
= Lab (Λ) ψ̂b† (Λ−1 x) ,
and hence must have
†
Rȧḃ = Lab .
Hence we have by using the innitesimal trans-
formation that
i
µν ḃ
ȧ
2
†
i
µν b
b
= Iδa + δωµν SL a
2
h
b i†
i
= Iδab − δωνµ SLµν a
2
h
b i∗
i
= Iδab − δωνµ SLνµ a
2
h
b i∗
i
b
= Iδa − δωµν SLµν a ,
2
and by using the antisymmetry of
δω
any innitesimal transformation that
SLνµ , we nd that since
h
i∗
µν ḃ
µν b
SR
=
−
S
.
L a
ȧ
and
Now, say that we have a eld of two spinor undotted indicies
it must hold for
Cab (x).
Under a
Two indicies
Lorentz transformation it must transform as
U (Λ)−1 Cab (x)U (Λ) = Lac (Λ)Lbd (Λ)Ccd (Λ−1 x) .
The question is now whether we can the (four) components of
Cab (x) into smaller
sets that does not mix under a Lorentz transformation. Now by coupling of angular
momentum, we know from regular quantum mechanics when coupling two spin-½
particles (same algebra, but not same particle-interpretation here) that the statespace is spanned by the Clebsch-Gordan expansion triplets (s
(s
= 0).
= 1)
and singlets
This leads to the interpretation that the quantum eld can be in spin-0
(in which case it is a scalar eld, antisymmetric) or in spin-1 (in which case it is a
4-vector eld, symmetric). This can be formulated as the group-theoretic relation
(2, 1) ⊗ (2, 1) = (1, 1)A ⊕ (3, 1)S .
Decomposition of
47 of 88
Cab
FREE SPIN-½ FIELDS
10
This gives that we can always write
Cab (x)
as
Cab (x) = εab D(x) + Gab (x) ,
D(x) ∈ (1, 1)A
where
is an scalar eld, and
εab = −εba
is a
2×2
antisymmetric
εab
matrix of constants, which makes normalization holds, so we can choose
!
0 −1
εab =
1
= −iσ 2 ,
0
and will show up to have function as some kind of spinor-metric that can be used
to raise indicies, and
Gab (x)
is symmetric. Now, since
D(x)
is a scalar eld, it must
transform as one, and hence we have by linearity
U (Λ)−1 εab D(x)U (Λ) = εab U (Λ)−1 D(x)U (Λ) = Lac (Λ)Lbd (Λ)εcd D(x)(Λ−1 x) ⇒
εab = Lac (Λ)Lbd (Λ)εcd ,
so
εab
is an invariant symbol of the Lorentz group. Like the metric, we can dene
an inverse
εab
that turns out to be
ε
ab
0
=
!
1
= iσ 2 ,
−1 0
so we thus have
εab εbc = δac , εab εbc = δ ac .
The same relations are of course true for dotted indicies. Mixed symbols have no
meaning, since the dierent spinor elds are independent.
Example 21 (Check that εab εbc = δac ).
bc
εab ε =
0 −1
1
0
!
We have by matrix multiplication that
0
1
!
−1 0
=
1 0
0 1
!
= δac .
Thus we can dene raised indicies of elds and other quantities using the spinor
ab
metric ε ;
♦a (x) ≡ εab ♦b (x) .
We further, by antisymmetry, have that equality relations
♦a (x) = εab ♦b (x) = ♦b (x)εab = −εba ♦b (x) .
Likewise, if we contract indicies, we have
♦a a = εab ♦a a = −εba ♦a a = −♦a a .
We have that for a eld with two mixed indicies,
Caȧ (x),
we have the group
48 of 88
Spinor metric
FREE SPIN-½ FIELDS
10
relation
(2, 1) ⊗ (1, 2) = (2, 2) ,
Caȧ (x) is actually a 4-vector
µ
11
12
invariant
symbol
σaȧ such that
which implies that
C µ (x).
Hence we can nd an
σaµȧ
Caȧ (x) = σaµȧ Cµ (x) ,
where we nd that we actually have
σaµȧ = (I, ~σ )aȧ = I, σ 1 , σ 2 , σ 3
. We can
aȧ
µḃb , and we
also nd the inverse relation, for which the invariant symbol is called σ
σ µḃb
have
1
C µ (x) = − σ µḃb Abḃ (x) .
2
In this case we have
σ µḃb ≡ εab εȧḃ σaµȧ = (I, −~σ )bḃ .
10.1.1 Other useful relations
We have some metric relations
σaµȧ σµaȧ = −2εab εȧḃ ,
εab εȧḃ σaµȧ σaνȧ = −2g µν .
Note that we also have
h
For the generators
σaµḃ
S µν ,
SLµν
i†
h
i†
= σbµȧ , σ µȧb = σ µḃa .
we have that
ab
= SLµν
ba
µν
⇔ SR
ȧḃ
µν
= SR
ḃȧ
.
10.1.2 Index free notation
We can develop an index free notation for the spinors, so it is easier to keep track
of the products.
The rule is that left-handed (undotted) elds are summed with
indicies going
and right-handed (dotted) elds are summed as
,
.
Since we are
working with fermions, observable objects (elds and so on) must anticommute, so
ab = −ba,
which this index free notation neatly takes care of and makes consistent.
We have
♦ ≡ ♦a a = εab εac ♦b c = −εba εac ♦b c = −♦a a = a ♦a = ♦ .
The same for the the dotted indicies:
11
12
Index free notation
It is invariant since it consists of invariant objects; Pauli matrices and the spinor metric.
The existence of this symbol is guaranteed by the group product relations.
49 of 88
Anti-commutation
FREE SPIN-½ FIELDS
10
♦† † ≡ ♦†ȧ ȧ† = εȧḃ εȧċ ♦ċ† †ḃ = −εḃȧ εȧċ ♦ċ† †ḃ = −♦ȧ† †ȧ = †ȧ ♦ȧ† = † ♦† .
Likewise, for hermitian conjugation, we have
[♦]† = [♦a a ]† = †ȧ ♦ȧ† = † ♦† ,
which agrees with normal behavior of the objects.
10.2 Bispinors, the Majorana and Dirac eld
We can build the Lagrangian for a free spin-½ eld by trail-and-error and a little
insight. We want the equations of motion to be linear and the elds
ψ̂a (x) and ψ̂ȧ† (x),
so the Lagrangian can be no more that quartic in these terms, and the Lagrangian
must be hermitian.
ψψ +
The terms with no derivatives can only be proportional to
ψ † ψ † (which doesn't vanish because of the index free notation, even though
the elds anti-commute). The term with derivatives cannot be
∂ µ ψ∂µ ψ + ∂ µ ψ † ∂µ ψ † ,
since this gives a Lagrangian that is unbounded below. The only other choice is the
term
iψ † σ µ ∂µ ψ ,
which is almost hermitian except for some boundary terms from
partial integration, that vanishes anyway when doing suitable integration later.
Thus our free Lagrangian is of the form
1
1
L = iψ † σ µ ∂µ ψ − mψψ − m∗ ψ † ψ † ,
2
2
where
m is a complex parameter, which with no loss of generality can be assumed
to be real since the Lagrangian is independent of the phase of
m.
Hence we have
Lagrangian
1 L = iψ † σ µ ∂µ ψ − m ψψ + ψ † ψ † ,
2
and will later interpret
m
as the mass of the eld. The Euler-Lagrangian equa-
tions then gives the equations of motion
∂L
d
−
∂ψ † dxµ
∂L
∂ [∂µ ψ † ]
∂L
d
− µ
∂ψ dx
∂L
∂ [∂µ ψ]
Equations of motion
= 0 ⇒ −iσ µ ∂µ ψ + mψ † = 0 ,
= 0 ⇒ −iσ µ ∂µ ψ † + mψ = 0 .
In dotted/undotted notation we then have
−iσ µȧc ∂µ ψc + mψ ȧ† = 0 , −iσaµċ ∂µ ψ ċ† + mψa .
We can write these two equation as one system of four equations
mδac
−iσaµċ ∂µ
−iσ µȧc ∂µ
mδ ȧċ
Here we have that the object
ψc , ψ ċ†
!
ψc
ψ ċ†
!
=
~0
~0
!
.
Bispinors
is called a bispinor, that as we will see
constitutes a new representation of Lorentz group as the
(1, 2)⊕(2, 1) representation,
50 of 88
10
FREE SPIN-½ FIELDS
which we can now develop to some really neat and powerful notation, since we now
can describe both spin-up and spin-down particles (and antiparticles) in one
eld structure.
We can go even further and introduce the
4×4
gamma matrices
Gamma matrices
γ 0 , γ 1 , γ 2 , γ 3 , that is dened as
γµ ≡
0
σaµċ
σ µȧc
0
!
,
and the gamma matrices has the dening property that they fulll the anticommutation relations
A.C.R and C.R of the
gamma matrices
{γ µ , γ ν } = −2g µν I4 .
Example 22 (Show that γ µ γµ = 4I4 from the dening properties).
γ µ γµ = gµν γ µ γ ν =
=
We have
1
1
(gµν γ µ γ ν + gνµ γ µ γ ν ) = (gµν γ µ γ ν + gµν γ ν γ µ )
2
2
1
gµν {γ µ , γ ν } = −gµν g µν I4 = −4I4 .
2
If we introduce the Majorana eld
Ψ≡
Majorana eld
ψc
ψ ċ†
!
,
which is a bispinor, then the equations of motion reduce to the Majorana equation
Majorana equation
(−iγ µ ∂µ + m) Ψ = 0 .
Here we interpret
½
Ψ
as a real eld, which will correspond to a eld of spin-
particles that are their own antiparticles. We can also do a complex eld theory,
where this is not the case by combining two Majorana elds. If we dene the bispinor
ψa ≡
ψ ψ†
,
a = 1, 2,
which contains both a left- and right-handed eld, we
can then dene a Lagrangian that is a sum of left-handed and right-handed spinor
elds as
1 L = iψa† σ µ ∂µ ψa − m ψa ψa + ψa† ψa† ,
2
we can take the complex linear-combination of the elds
1
1
χ ≡ √ (ψ1 + iψ2 ) , ξ ≡ √ (ψ1 − iψ2 ) ,
2
2
and rewrite the Lagrangian in terms of these complex spinor elds. We then nd
1 L = iχ† σ µ ∂µ χ + iξ † σ µ ∂µ ξ − m χξ + ξ † χ† ,
2
which is hermitian. This Lagrangian has a
χ → e−iα χ
and
ξ → e+iα ξ
U (1) symmetry, so the transformation U (1)
leaves the Lagrangian invariant, and we have a corre-
sponding conserved Noether current and charge. The equations of motion can easily
51 of 88
symmetry
FREE SPIN-½ FIELDS
10
be derived
∂L
d
− µ
∂χ dx
∂L
∂ [∂µ χ]
= 0 ⇒ −i∂µ χ† σ µ + iξ † σ µ ∂µ ξ + mξ ,
and so forth, and this we can rewrite as a system of equations by
mδac
−iσaµċ ∂µ
−iσ µȧc ∂µ
mδ ȧċ
!
!
χc
~0
~0
=
ξ ċ†
!
.
By dening the complex Dirac eld
Dirac eld
χc
Ψ≡
!
,
ξ ċ†
the equations of motion can be written as the Dirac equation
Dirac equation
(−iγ µ ∂µ + m) Ψ = 0 .
Using Feynman's slash-notation by dening
A ≡ γ µ Aµ ,
we have
Feynman
notation
(i
∂ − m) Ψ = 0 .
Note that both the Dirac and Majorana equation (having same form, but dierent interpretation) obey the Klein-Gordon equation (or at least, four Klein-Gordon
equations), since we by multiplying the operator
(i
∂ − m)
by
(−i
∂ − m)
nd that it
is exactly the Klein-Gordon equation(s).
Example 23
(Show that
a a = −a2 ).
We have by the dening anti-commutation
relation that
1
a a = aµ γ µ aν γ ν = aµ aν ({γ µ , γ ν } + [γ µ , γ ν ])
2
1
1
1
=
aµ aν ({γ µ , γ ν } − i4S µν ) = aµ aν ({γ µ , γ ν } − i4S νµ ) = aµ aν {γ µ , γ ν }
2
2
2
= −aµ aν g µν I4 = −aµ aµ I4 = −a2 I4 .
Now, the problem with the Dirac eld is that it is not quite hermitian, so this
has to be taken care of. We have
Ψ† =
If we dene the matrix
β
χ†ċ ξ c
.
β
as
β≡
we can dene the barred
0
δ ȧċ
δac
0
!
= γ0 ,
Ψ, which is called the Dirac adjoint to simplify notation
as
Ψ ≡ Ψ† β =
matrix
ξ c χ†ċ
.
52 of 88
Dirac adjoint
slash-
FREE SPIN-½ FIELDS
10
We then see that
1×4
Ψ
is some kind of generalization of the hermicity and is a
Ψ
vector eld, and is dened so to make sure that
transforms as a 4-vector
under Lorentz transformations, which is not the case for just
representation is not unitary.
dene for any matrix
Ψ† ,
since the Lorentz
This barred notation can be generalized and so we
A
A = βA† β
We have by a simple calculation that
ΨΨ = χξ + ξ † χ† ,
and by some algebra and partial integration
Ψγ µ ∂µ Ψ = χ† σ µ ∂µ χ + ξ † σ µ ∂µ ξ + ∂µ ξσ µ ξ † ,
but since the last term is a divergence term, and the equation of motions have
a gauge freedom, we can just forget about it. Hence we nd by comparison that we
have the equivalent Dirac Lagrangian
Dirac Lagrangian
L = iΨγ µ ∂µ Ψ − mΨΨ = Ψ (i
∂ − m) Ψ .
In this notation we also have that the Lagrangian has an
itly the transformation
Ψ→
e−iα Ψ and
Ψ→
U (1)
symmetry, explic-
e+iα Ψ leaves the Lagrangian invariant.
We can project to the bi-spinors back onto the right- or left-handed eld by a proper
Projection
projection matrix. It is easy to see that we have
and that
P∓ =
ΨL =
I4 − γ 5
Ψ,
2
ΨR =
I4 + γ 5
Ψ,
2
I4 ∓γ 5
2
is a genuine projection operator (P∓
2
= P∓ ).
For the Majorana eld, we can rewrite the Lagrangian and eld as such that we
will get the same Lagrangian
L = iΨγ µ ∂µ Ψ − mΨΨ,
but with
Ψ = Ψt C ,
where
C
is
the charge conjugation matrix as dened below, since all we have to do is formally
Dirac→Majorana:
ψ → χ
ψ → χ, ξ
and
ψ → ξ.
Thus, since notation, Lagrangian, and equation of motion is
the same, we can treat them very much alike, besides a few but crucial points: The
Majorana eld has no
for this symmetry.
U (1)
symmetry and hence no corresponding conservation law
For the Majorana eld, we can explicitly write the simplied
Lagrangian with charge conjugation if we want to:
L = iΨt Cγ µ ∂µ Ψ − mΨt CΨ .
10.2.1 Charge conjugation
Now, since
L
is hermitian and swapping
χc ↔ ξ ċ†
in
Ψ,
gives
†
Ψ ↔ Ψ ,,
the La-
grangian is invariant under such a discrete transformation since
53 of 88
No
U (1)
symmetry for
Majorana elds
FREE SPIN-½ FIELDS
10
†
iΨγ µ ∂µ Ψ − mΨΨ
L† =
†
†
= −iΨ† [γ µ ]† ∂µ Ψ − mΨ† Ψ
†
†
= iΨ† γ µ ∂µ Ψ − mΨ† Ψ
= iΨγ µ ∂µ Ψ − mΨΨ = L ,
even though the spinor elds
ξ, χ
swap places, it doesn't change the result and
the hermicity of the Lagrangian.This transformation is called charge conjugation,
and the corresponding unitary matrix
C
that implements this is the
4×4
charge
conjugation matrix
C=
!
εac
0
0
εȧċ
,
with
C −1 χC = ξ , C −1 ξC = χ ,
and
C −1 LC = L .
Going to the Dirac eld we have that
t
C
Ψ ≡ CΨ =
which is just
Ψ
with
ξ ↔ χ,
ξc
!
χ†ċ
,
so one can say that the charge conjugate is just the
Dirac eld with the 2-spinors' charges interchanged. For a Majorana eld we have
Ψ = ΨC ,
which is to say that the spinors have the same charge.
10.2.2 Lorentz transformation of the Dirac and Majorana elds
We know from the representation theory of the Lorentz group for spin-½ particles,
that we have innitesimal transformations given by
b
i
Lab (1 + δω) = Iδab + δωµν SLµν a ,
2
i
µν ḃ
,
ȧ
2
h
i∗
µν ḃ
µν b
and that SR
=
−
S
. This we can collect into a more unied manner
L a
ȧ
b
ij
that ts our bispinor notation. One can as mentioned before show that SL
=
a
b
b
b
1 ijk
σ k a and SLk0 a = 2i σ k a , and similarly for the right-handed part of the
2
eld. This implies that we can write
SLµν
b
a
i
= + (σ µ σ ν − σ ν σ µ )ab ,
4
54 of 88
Charge conjugation
10
µν
SR
ȧ
ḃ
FREE SPIN-½ FIELDS
i
= − (σ µ σ ν − σ ν σ µ )ȧḃ ,
4
and we can thus see that that when going to bispinor representation as the Dirac
or Majorana eld, that we can collect the left- and right-handed parts in a matrix
so that we can dene a set of 16
4×4
matrices named
S µν
that allows us to treat
S µν
the Lorentz transformation of the spinor elds in a unied manner
+ SLµν
S µν ≡
b
a
−
0
where the denition of
S µν
0
!
=
µν ȧ
SR
ḃ
i µ ν
[γ , γ ] ,
4
. We explicitly have
U (Λ)−1 Ψ(x)U (Λ) = D(Λ)Ψ(Λ−1 x) ,
where we for an innitesimal transformation have
i
D(1 + δω) = I + δωµν S µν ,
2
so in this notation
S µν =
i
4
[γ µ , γ ν ]
is again the generators of the Lorentz group.
10.3 Canonical quantization of the Dirac eld
Since the Dirac eld describes fermions, we must use anti-commutators, since the
elds must be antisymmetric under exchange.
Imposing the standard canonical
quantization relations thus gives us for the Weyl spinors:
A.C.R.
for
Weyl
spinors
n
o
n
o
ψ̂a (x, t), ψ̂b (x0 , t) = 0 , π̂ a (x, t), π̂ b (x0 , t) = 0 ,
n
o
ψ̂a (x, t), π̂ b (x0 , t) = iδab δ 3 (x − x0 ) .
Using that we have the canonical momentum
π̂ b (x) ≡
π̂ b
density given as
∂L
= iψḃ† (x)σ 0bḃ ,
∂ [∂0 ψb (x)]
we nd by substituting that
n
o
n
o
ψ̂a (x, t), ψḃ† (x0 , t) σ 0bḃ = iδab δ 3 (x−x0 ) ⇔ ψ̂a (x, t), ψḃ† (x0 , t) = iσa0ḃ δ 3 (x−x0 ) .
Similarly we can nd relations for the right-handed eld. For the bispinor representation as the Dirac eld we nd that the equivalent anti-commutation relations
is given by
Ψα (x, t), Ψβ (x0 , t) = 0 , Ψα (x, t), Ψβ (x0 , t) = i γ 0 αβ δ 3 (x − x0 ) .
55 of 88
A.C.R. for Dirac eld
FREE SPIN-½ FIELDS
10
Example 24 (Show the anti-commutation relations for Ψ).
Ψ(x, t), Ψ(x0 , t) =
We have
Ψ(x, t)Ψ(x0 , t) + Ψ(x0 , t)Ψ(x, t)
!
!
!
!
χc
χb 0
χc
χb0
+
0
0
ξ ċ†
ξ ḃ †
ξ ċ†
ξ ḃ †
!
!
χc χb0 − χc χb0
χc χb0 + χb0 χc
= 0.
=
0
0
0
0
ξ ċ† ξ ḃ † − ξ ċ† ξ ḃ †
ξ ċ† ξ ḃ † + ξ ḃ † ξ ċ†
=
=
The Dirac LagrangianL
0.
= Ψ (i
∂ − m) Ψ, gives the Dirac equation (i
∂ − m) Ψ =
equation (−i
∂ − m) Ψ = 0, they together implies that the
Together with the
Klein-Gordon equation is fullled (our 4 copies of it at least), since we by multiplying
have
(i
∂ − m) (−i
∂ − m) = ∂
∂ − i
∂ m + mi
∂ + m2
= ∂
∂ + m2
= −∂ 2 + m2 ,
which is exactly the Klein-Gordon equation. Hence there exists plane wave solutions, and they must fulll either
(i
∂ − m) Ψ = 0
or
(−i
∂ − m) Ψ = 0.
(i
∂ − m) Ψ = 0
for the physical solutions, that fullls the Dirac equation
We look
and hence
have a total of four independent plane-wave solutions which can be grouped as
us (p)eipx , vs (p)e−ipx ,
where
us (p)
and
vs (p)
are bispinors and with
or spin down, that solves the Dirac equation.
be function of the 4-momentum
p
s=±
corresponding to spin up
When solving,
us and vs
seems to
, but since they must satisfy the Klein-Gordon
equation also, we have the constraint
p
p0 = ωp = p2 + m2 .
Restriction of
p0 = ωp
The general solution is
thus
ˆ
˜
dp
Ψ(x, t) =
Xh
i
bs (p)us (p)eipx + d†s (p)vs (p)e−ipx ,
s
where
bs (p)
and
d†s (p)
are a set of (scalar) operators, and we have
ˆ
Ψ(x, t) = Ψ† β =
˜
dp
i
Xh
b†s (p)us (p)e−ipx + ds (p)v s (p)eipx .
s
We are to interpret
momentum
p
b†s (p) as the creation operator of a particle of type b with three
and spin
s,
and
(the antiparticle of particle
b)
d†s (p)
as the creation operator of a particle of type
with three momentum
pretation goes for the annihilation operators
Ψ(x, t)
p
and spin
bs (p), ds (p).
s.
d
The same inter-
Now the interpretation of
says that it is the same to annihilate a particle as to create an antiparticle.
56 of 88
Creation/annihilation
FREE SPIN-½ FIELDS
10
10.3.1 Solution of us (p), vs (p) and properties of the bi-spinors
We can solve the Dirac equation for each of the 4 independent plan wave solutions,
where we have that the Dirac equation reduces to
(i
∂ − m) us (p)eipx = (iγ µ ∂µ − m) us (p)eipx = (iγ µ ipµ − m) us (p)eipx ⇒
(p + m) us (p) = 0 ,
and
(i
∂ − m) vs (p)e−ipx = (iγ µ ∂µ − m) vs (p)e−ipx = (−iγ µ ipµ − m) vs (p)e−ipx ⇒
(−p + m) vs (p) = 0 .
The mass
m
is understood to be
mI4 ,
so we have two times 4 equations to solve
to nd the bispinor motion. This is easy to solve in the rest frame, but horrible in
a general frame, but luckily the bispinors transforms as four-vectors (since they are
four-vectors), so we can just boost them to a general frame afterward. In the rest
frame we have that the spatial components of the four-momentum is zero,
p =
γ0p
0
=
−γ 0 m − mI4 us (0) =
0
p = 0,
so
−γ 0 m, so the equations reduces to
γ m − mI4 vs (0) =
−mI2
0
0
mI2
mI2
0
0
−mI2
!
−
!
−
mI2
0
0
mI2
mI2
0
0
mI2
!!
us (0) ,
!!
vs (0) .
This gives 4 unique solutions up to a normalization constant, and we nd that
we can normalize the solutions as

1


 

 0 
√ 


u+ (0) = m   , u− (0) = m 

 1 

0



0



√  1 
√ 



v+ (0) = m 
 , v− (0) = m 
 0 

−1
√
The barred bispinors
Ψ ≡ Ψ† β
0


1 
,
0 

1

−1

0 
.
1 

0
are given as row vectors, and we nd that
57 of 88
10
FREE SPIN-½ FIELDS
us (p) (p + m) = 0 , v s (p) (−p + m) = 0 ,
u+ (0) =
v + (0) =
√ √ m 1 0 1 0
, u− (0) = m 0 1 0 1 ,
√
m
0 −1 0 1
, v − (0) =
√ m 1 0 −1 0 .
The boost to a general frame is given by exponentiation of the innitesimal
transformation
Λik = 0
for
1 + δω
i 6= k ,
, and since the boost contains no rotations, we have that
and so the only non-vanishing components is in the rst column
and rst row (pure boosts). Hence we have
D(Λ) = exp (iΛµν S µν ) = exp iΛj0 S 0j = exp (iη p̂ · K) ,
where we in the last equation have rewritten the transformation in terms of the
rapidity
η ≡ arsinh(kpk /m)
and
boost. The formal solution for the
p̂
is the unit 3-momentum; the direction of the
u's
and
v 's
Rapidity
is thus
us (p) = exp (iη p̂ · K) us (0) , vs (p) = exp (iη p̂ · K) vs (0) .
Not very useful, but we can derive some useful consequences from the solutions.
The barred spinors is given as
us (p) = us (0) exp (−iη p̂ · K) , v s (p) = v s (0) exp (−iη p̂ · K) ,
where we use that the gamma-matrices are invariant under barring, and so
must
K
be in the chosen representation of
S µν .
We thus see that we have the following relations for products of the kind
w s ws 0 ,
which are all scalar products
us (p)us0 (p) = us (0)e−iηp̂·K eiηp̂·K us (0) = us (0)us (0) = 2mδss0 , v s (p)vs0 (p) = −2mδss0 ,
us (p)vs0 (p) = 0 , v s (p)us0 (p) = 0 .
58 of 88
10
Example 25
FREE SPIN-½ FIELDS
.
(Show the relations for the product of the spinors)
We have by a
direct calculation for two of the cases:
u+ (p)u+ (p) = us (0)e−iηp̂·K eiηp̂·K us (0)

1
√ 
 0
√ = us (0)us (0) = m 1 0 1 0
m
 1

0



 = 2mδss0 ,


and

u+ (p)u− (p) =
√
m
1 0 1 0
√
0

 
 1 

m
 0  = m0 = 0 .
 
1
We can also calculate more general products with gamma matrices and dierent
momenta, given as the Gordon identities:
2mus0 (p0 )γ µ us (p) = us0 (p0 )
−2mv s0 (p0 )γ µ vs (p) = v s0 (p0 )
p0 + p
Gordon identities
µ
− 2iS µν p0 − p
ν
us (p) ,
µ
p0 + p − 2iS µν p0 − p ν vs (p) ,
Now, if we instead take products of the kind
ws 0 w s ,
we get
4×4
matrices. We
can calculate the spin averaged sums of products (which is the only physical relevant
Spin averaged sums
thing we need anyway), and we nd
X
s
X
s
us (p)us (p) = −p + m ,
vs (p)v s (p) = −p − m .
These are called the completeness relation for the spinors, and gives a simple
result for spin sums. The relations are easy to prove in the rest frame, since we can
just use that we have

1


1 0 1 0



 
 0 
 0 0 0 0 



u+ (0)u+ (0) = m 
 1  1 0 1 0 = m 1 0 1 0  ,


 
0 0 0 0
0

0


0 0 0 0

 


 1 
 0 1 0 1 



,
u− (0)u− (0) = m   0 1 0 1 = m 

 0 
 0 0 0 0 
1
0 1 0 1
59 of 88
Completeness relation
10

1 0 1 0


FREE SPIN-½ FIELDS

0 0 0 0

1 0 1 0







 0 1 0 1 
 0 1 0 1 
 0 0 0 0 
0






,
us (0)us (0) = m 
+m  0 0 0 0  = m  1 0 1 0  = mγ +mI4 = −p+m
1
0
1
0






0 1 0 1
0 1 0 1
0 0 0 0
X
s
which is proven likewise for the other spinors in the rest frame. Going to prove
that the completeness relation holds in a general frame is more complicated, but can
be done by showing that
mation, that is
m γ 0 + I4
exp (iη p̂ · K) m
γ0
is invariant under a general Lorentz transfor-
+ I4 exp (−iη p̂ · K) = m γ 0 + I4 .
10.4 Quantization in terms of b, d operators
We can rewrite the general solution for the Dirac equation
b, d
Ψ
and
Ψ
to isolate the
Dirac equation
operators. We nd after some clever algebra that
ˆ
ˆ
d x e−ipx us (p)γ 0 Ψ(x)
3
bs (p) =
,
b†s (p)
,
d†s (p)
ˆ
ds (p) =
d3 x eipx Ψ(x)γ 0 us (p) ,
=
ˆ
3
−ipx
d x e
0
Ψ(x)γ vs (p)
=
d3 x eipx v s (p)γ 0 Ψ(x) .
From these, we can from the canonical commutation relations work out the algebra for the
b, d
operators. We nd that
3·2
of the 9 unique relations anti-commute
Algebra for
b, d
and is trivially zero, namely
o
n
bs (p), bs0 (p0 ) = 0 , ds (p), ds0 (p0 ) = 0 , bs (p), d†s0 (p0 ) = 0 ,
and likewise for the hermitian conjugates:
o
n
o
n
o
n
b†s (p), b†s0 (p0 ) = 0 , d†s (p), d†s0 (p0 ) = 0 , b†s (p), ds0 (p0 ) = 0 .
The only non-trivial commutation relations is
and
{bs (p), ds0 (p0 )}.
n
o
n
o
bs (p), b†s0 (p0 ) and ds (p), d†s0 (p0 )
We nd for these that
n
o
bs (p), b†s0 (p0 ) = (2π)3 2ωp δ (3) (p − p0 )δss0 ,
n
o
ds (p), d†s0 (p0 ) = (2π)3 2ωp δ (3) (p − p0 )δss0 ,
This shows that the
bs (p), ds0 (p0 ) = 0 .
b's and d's constitutes a set of creation/annihilation algebras,
60 of 88
10
FREE SPIN-½ FIELDS
that doesn't talk together for the dierent particles, nor for dierent spins.
†
think of bs (p) as creating a particle of spin
an antiparticle. Likewise we think of
s
bs (p)
We
†
and momentum p, and ds (p) as creating
as annihilating an particle and of
as annihilating an antiparticle. Note the way that
Ψ
is written in terms of
b
ds (p)
and
d† ,
that this implies that creating an antiparticle is equivalent to destroying a particle.
Likewise from
Ψ
we interpret that creating a particle is equivalent to destroying
Hence the quanta of the Dirac eld is spin-½ particles and the
an antiparticle.
corresponding spin-½ antiparticles.
Majorana eld
Writing the algebra for the Majorana is done the same way, but we now have
only one set of operators since the condition
t
Ψ = CΨ
implies that
bs (p) = ds (p).
This interprets as the quanta of the Majorana eld is spin-½ particles, that is also its
own antiparticle.
10.5 The free Dirac eld and its Hamiltonian
We can write a Hamiltonian for the free Dirac eld, which is given by the Legendre
Hamiltonian
transform of the Lagrangian. We nd
Dirac eld
H=
Xˆ
for
h
i
˜ p b† (p)bs (p) + d† (p)ds (p) .
dpω
s
s
s
Now, this Hamiltonian is bounded from below, so there exists a vacuum state
|0i
Vacuum state
such that
bs (p) |0i = ds (p) |0i = 0 .
Acting on the vacuum state with the creation operators is an eigenstate of the
Eigenstate
†
Hamiltonian, as one can see by the calculation for bs (p) |0i:
Hb†s (p) |0i
=
Xˆ
i
h
˜ 0 ωp0 b†0 (p0 )bs0 (p0 ) + d†0 (p0 )ds (p0 ) b† (p) |0i
dp
s
s
s
s0
=
Xˆ
i
h
˜ 0 ωp0 b†0 (p0 )bs0 (p0 )b† (p) + d†0 (p0 )ds (p0 )b† (p) |0i
dp
s
s
s
s
s0
=
Xˆ
ˆs
i
h
n
o
˜ 0 ωp0 b†0 (p0 ) bs0 (p0 ), b† (p) + (−1)2 b† (p)d†0 (p0 )ds (p0 ) |0i
dp
s
s
s
s
0
h
n
o
i
˜ 0 ωp0 b† (p0 ) bs (p0 ), b† (p) + b† (p)d† (p0 )ds (p0 ) |0i
dp
s
s
s
s
ˆ
h
i
˜ 0 ωp0 b† (p0 ) (2π)3 2ωp0 δ (3) (p − p0 ) |0i
=
dp
s
=
= ωp b†s (p) |0i ,
ωp . Likewise one can do the
†
†
†
same for ds (p) |0i to show it has eigenenergy ωp also. Now, consider bs (p)bs (p) |0i.
†
†
†
†
Since we by the commutation rules have bs (p)bs (p) = −bs (p)bs (p), we must have
b†s (p)b†s (p) |0i = 0, which implies that there can maximally be one particle in each
Eigenenergy
state, which we already know from the Pauli principle and the way we constructed
Pauli principle
the Dirac theory.
One-particle state
so hence it is an eigenstate with eigenenergy
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free
10
FREE SPIN-½ FIELDS
Now we can write the one-particle (with plus)/antiparticle (with minus) states
as
b†s (p) |0i = |p, s, +i
, d†s0 (k) |0i = k, s0 , − .
Multi-particle states can be written as
b†s (p)d†s0 (k) |0i = |p, s, + k, s, −i = −d†s0 (k)b†s (p) |0i = − |k, s, − p, s, +i ,
b†s1 (p1 ) · · · b†sn (p)d†r1 (k1 ) · · · d†rn (kn ) |0i = |p1 , s1 , + · · · pn , sn , + k1 , r1 , − · · · kn , rn , −i .
We have the inner product of two states (which easy generalizes to multi-particle
states) by the anti-commutation-relations is given by
p, s, q|p0 , s0 , q 0 = (2π)3 2ωp δ (3) p − p0 δss0 δqq0 .
We also nd
h0|Ψ(x)|0i = 0 ,
hp, s, +|Ψ(x)|0i = 0 ,
hp, s, −|Ψ(x)|0i = vs (p)e−ipx ,
p, s, +|Ψ(x)|0 = us (p)e−ipx ,
p, s, −|Ψ(x)|0 = 0 .
10.6 Conserved charges
We have a
and
U (1)
Ψ → Ψe−iα .
symmetry of the Dirac eld under the transformation
Ψ → Ψeiα
The conserved Noether current is given by
j µ (x) = Ψγ µ Ψ ,
which can be shown by using Poul's method as below. The corresponding conserved charge is
ˆ
Q=
3
0
d xj (x) =
Xˆ
h
i
˜ b† (p)bs (p) + d† (p)ds (p) ,
dp
s
s
s
which is actually electrical charge.
We see from this that the electron and its
antiparticle have opposite charge.
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11
INTERACTION AND SCATTERING IN SPIN-½ THEORIES
Example 26 (Show the conserved current is j µ (x) = Ψγ µ Ψ).
The transformation
Ψ → eiα Ψ ⇔ Ψ → e−iα Ψ, α ∈ R,
is a
U (1)
continuous symme-
try, as we have that the Lagrangian is invariant:
L0 = e−iα Ψ (i
∂ − m) eiα Ψ = eiα e−iα Ψ (i
∂ − m) Ψ
= Ψ (i
∂ − m) Ψ = L .
By Noether's theorem, we thus have a corresponding conserved current and charge.
We derive an expression for the current by Poul's method by assuming that
eld,
α = α(x)
formation
α
is a
(making the transformation local), and take the innitesimal trans-
Ψ → (1 + iα) Ψ.
We then have the transformation becomes
L0 = (1 − iα) Ψ (i
∂ − m) (1 + iα) Ψ
∂ − m) Ψ − iα(x)Ψ (i
∂ − m) Ψ + Ψ (i
∂ − m) [iα(x)Ψ]
= Ψ (i
−iα(x)Ψ (i
∂ − m) [iα(x)Ψ]
= L − iα(x)Ψ (i
∂ − m) Ψ + Ψ (i
∂ − m) iα(x)Ψ + O(α(x)2 )
Hence, the change in the Lagrangian is
∂ − m) Ψ + Ψ (i
∂ − m) [iα(x)Ψ]
δL = −iα(x)Ψ (i
= α(x)Ψ
∂ Ψ − Ψ
∂ [α(x)Ψ]
= α(x)Ψγ µ ∂µ Ψ − Ψγ µ [∂µ α(x)] Ψ − Ψγ µ α(x) [∂µ Ψ]
= −Ψγ µ Ψ [∂µ α(x)]
For the action to be invariant we must thus have, since a constant factor doesn't
matter, that the conserved current is
j µ (x) = Ψγ µ Ψ .
11 Interaction and scattering in spin-½ theories
Going to interaction Hamiltonians of the spin-½ elds, we can reuse much of the
developed stu from the bosonic spin-0 theory and just generalize it properly.
11.1 Interaction picture in spin-0 quantum eld theories
The time-evolution of the bispinor eld is just by the free Dirac eld (Heisenbergevolution), so we have for some initial eld conguration
Ψα (t0 , x) that the interaction
α
picture time-evolution ΨI (x) is:
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Interaction picture
11
ΨαI (x) = eiĤ0 (t−t0 ) Ψα (t0 , x)e−iĤ0 (t−t0 ) .
We must take the elds as time ordered since we would like to calculate the spin-½
∆(x − x0 ),
equivalent of the Feynman propagator
which we can expand the time-
ordered Dyson expansion in terms of. Now, since the eld is fermionic, the operators
anti-commute, and thus we dene the time ordering of the eld operators by
Time ordering
TΨα (x)Ψβ (x0 ) ≡ θ(t − t0 )Ψα (x)Ψβ (x0 ) − θ(t0 − t)Ψβ (x0 )Ψα (x) .
x0 > x00 , we have that we can calculate 0|Ψ(x)Ψ(x0 )|0
0|Ψ(x)Ψ(x0 )|0 αβ = 0|Ψα (x)Ψβ (x0 )|0 ) given as
Assuming
ˆ ˆ
0|Ψα (x)Ψβ (x0 )|0 =
˜0
˜ dp
dp
ˆ
=
ˆ
=
Assuming
x00 > x0
(a matrix,
D
Ei
Xh
0
eipx e−ipx us (p)α us0 (p)β 0|bs (p)b†s0 (p0 )|0
s,s0
i
Xh
0
˜
eip(x−x ) us (p)α us (p)β
dp
s
˜ ip(x−x0 ) (−p + m) .
dpe
αβ
we likewise nd that
ˆ ˆ
0|Ψβ (x0 )Ψα (x)|0 =
˜0
˜ dp
dp
s,s0
ˆ
˜
dp
=
ˆ
=
D
Ei
Xh
0
eipx e−ipx vs (p)α v s0 (p)β 0|ds (p)d†s0 (p0 )|0
Xh
0
eip(x −x) vs (p)α v s (p)β
i
s
˜ −ip(x−x0 ) (−p − m) .
dpe
αβ
If we then take the expectation value of the time-ordered product with respect
to the vacuum state we
|0i,
we then nd
0|TΨα (x)Ψβ (x0 )|0 = 0|θ(t − t0 )Ψα (x)Ψβ (x0 ) − θ(t0 − t)Ψβ (x0 )Ψα (x)|0
ˆ
1
d4 p (−p + m)αβ ip(x−x0 )
e
=
i
(2π)4 p2 + m2 − i
1
≡
S(x − x0 )αβ ,
i
where
S(x − x0 )αβ
is the fermionic Feynman propagator. In momentum space we
thus nd that the fermionic Feynman propagator is given as
S̃(p)αβ =
(−p + m)αβ
p2 + m2 − i
Fermionic
propagator
.
We nd by easy use of anti-commutation relations for the creation/annihilation
operators that
64 of 88
Feynman
11
0|TΨα (x)Ψβ (x0 )|0 = 0|TΨα (x)Ψβ (x0 )|0 = 0 ,
which also makes good sense because of Pauli's principle.
11.1.1
n-point
correlation function and Wick's theorem
When we have a correlation function with
n
eld terms of barred and unbarred eld
components (some other ordering than thus perhaps),
Fα1 ...αn α1 ···αn (x1 , . . . , xn , x1 , . . . , xn ) = T Ψα1 (x1 ) · · · Ψαn (x1 )Ψα1 (x1 ) · · · Ψαn (xn ) ⇔
Fα1 ...αn (x1 , . . . , xn ) = T {Υα1 · · · Υαn } ,
and we would like to evaluate the time-ordered vacuum expectation value of expressions like this one. We thus dene normal ordering for fermionic eld analogously
Normal ordering
to the bosonic spin-0 case, but with the generalization that all daggers (creation
operators
b† (p), p† (q)) goes to the left, and all non-daggers
b(p), p(q))
(annihilation operators
goes to the right. Hence by the properties of the operators, these terms
all vanish and we have for any normal ordering of eld components
We can dene a contraction of two eld components
Υα , Ωβ
h0|N |0i = 0 .
(either barred or
Contraction
unbarred elds) as

{Υ (x), Ω (x0 )}
α
β
Υα (x)Ωβ (x0 ) ≡
− {Ω (x0 ), Υ (x)}
β
α
, t > t0
, t0 > t
Using the result we found for the Feynman propagators, we have that

 Ψ (x)Ψ (x0 )
α
β
Ψα (x)Ψβ (x0 ) ≡
− Ψ (x0 )Ψ (x)
α
β
, t > t0
,
t0
1
= S(x − x0 )αβ ,
i
>t
Ψα (x)Ψβ (x0 ) = 0 ,
Ψα (x)Ψβ (x0 ) = 0 .
We can absorb the contractions into the normal ordering since it is just a number.Wick's theorem generalized in this case and is very useful, since it states that any
Wick's theorem
time-ordered products of elds can be written as contractions plus normal-ordered
products. We have explicit:
T {Υα · · · Ωβ } = N (Υα (x1 ) · · · Ωβ (xn ) + all possible contractions) .
As a consequence of this, the only terms that will count when calculating the vacuum expectation values is the fully contracted terms (that is with contraction symbol
connecting every eld) of equally many barred and unbarred eld components, be-
65 of 88
Fully contracted
11
cause the uncontracted part of the term can be normal-ordered to give zero. Further
because of this we have that any time-ordered product of uneven number of eld
operators has expectation value 0, because for the the terms that are contracted as
much as possible we still have it
∝ h0|Υα |0i = 0.
External contraction
There are here four possible ways of contracting with external lines, since we
have four bispinors in the general solution. We have for example that contraction of
the eld with the incoming particle state
b†s (p) |0i = |p, s, +i,
is
us (p) |0i
Ψα (x) |p, s, +i = eipx us (p) |0i ,
The contraction with an external outgoing state
hp, s, +|
is given by
hp, s, +| Ψα (x) = h0| us (p)e−ipx .
Generally, the contractions with external antiparticle states is the same as with
particle states, which is contained in the Feynman rules.
11.1.2 Feynman diagrams and rules
The previous use of Wicks theorem can be represented as Feynman diagrams, which
are spacetime diagrams with the 'time-axis' and 'space-axis'. Each Feynman propagator is drawn as a line between the two spacetime points with dots on the ends. The
rules of computing numbers from these diagrams can be derived from the contractions, and does depend on what the perturbation looks like. All of the combinatorial
factors from the propagator-expansion, is exactly canceled when drawing all of the
topologically in-equivalent Feynman diagrams, and must be added up, to a relative
minus sign. The Feynman rules for calculating/drawing the expectation values.
Figure 7: Components of Feynman diagrams for spin-½ elds
How, given a Feynman diagram, we can compute the amplitude is by going
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Feynman diagrams
11
against the arrows on the Feynman diagram and multiply factors together from
left to right.
Relative sign rule
The sign rule says that, if we write the diagrams in standard form (so they look
like
e− e− → e− e−
scattering with ow to from left to right,
→).
Then, taking one
diagram (they will all look the same) as the zero-diagram, we get a relative minus
sign every time every time a line is exchanged, when comparing to the zero-diagram.
Hence, odd number of exchanged lines, gives a relative minus sign.
Benny Lautrups rule
Benny Lautrup has a easier rule, that simply says ever time an antiparticle goes
all the way through the diagram, we get a relative minus sign.
11.2 Scattering in spin-½ theory
To calculate scattering amplitudes in spin-½ theories, we do as always; read o the
Feynman diagrams. We have
iT =
X
All Feynman diagrams (in momentum space, with signs) .
The situation is more complicated than with pure scalar elds, since the elds
now have several components, and the incoming states also can be of several dierent
type, since we have both particles and antiparticles with spin up and down (4 in
total), and the same number of outgoing states.
It is so that it is much easier to calculate
|T |2
directly than just calculating
iT ,
since we have
|T |2 = T T ,
which allows for some simplications. Especially, if we don't care about the spin
of the incoming particles/antiparticles, we can average over the incoming spin, and
likewise we can average for the outgoing states, so
1
|T |2 ≡
n
X
|T |2 .
all spins
(nspinors)
Then this can be converted into a trace of some product of matrices (Casimir's
trick), and we can apply the completeness relations at most times. Now, everything
is reduced to calculating traces of gamma-matrices (see the appendix)
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Casimir's trick
11
Example 27
e−
proces
(Compute
µ−
+
the
e−
→
+
spin-averaged
µ−
which
scattering
interacts
amplitude
under
a
for
scalar
the
pion)
.
We have that the Feynman diagram is given as above with the denition of momenta
and so on. We then have that using the Feynman rules gives
iT =
us (k) (ig) (−i) ur (p) us0 (k0 ) (ig 0 ) ur0 (p0 )
2
(p − k) + m2π
T =
= igg 0
us (k)ur (p) us0 (k0 )ur0 (p0 )
,
m2π − t
gg 0
gg 0
us (k)ur (p) us0 (k0 )ur0 (p0 ) = 2
ur0 (p0 )us0 (k0 ) ur (p)us (k) .
2
mπ − t
mπ − t
And thus
|T |2
=
=
g 2 g 02
2 us (k)ur
(m2π − t)
g 2 g 02
(p) us0 (k0 )ur0 (p0 ) ur0 (p0 )us0 (k0 ) ur (p)us (k)
us (k)ur (p)us0 (k0 )ur0 (p0 ) ur0 (p0 )us0 (k0 )
2 ur (p)us (k)
{z
}|
{z
}
(m2π − t) |
scalar
=
g 2 g 02
2 Tr [ur
(m2π − t)
scalar
(p) ur (p)us (k) us (k)] Tr [us0 (k0 ) us0 (k0 )ur0 (p0 ) ur0 (p0 )]
In second line we moved the scalar
ur (p)us (k) from right to left so we can take traces.
We now spin-average over incoming and outgoing states and use the completeness
relations (and divide by 4)
2
|T |
=
=
=
g 2 g 02
4 (m2π −
2 02
0
0
(−p
+ me ) −k + me Tr −k + me (−p
+ me )
2 Tr
0 0
2
2
p
k + me Tr k p
+ me
t)
g g
4 (m2π −
2 02
2 Tr
t)
4g g
2
(m2π − t)
p · k + m2e
k 0 · p0 + m2e .
68 of 88
12
PHOTON FIELDS AND QUANTUM ELECTRODYNAMICS
12 Photon elds and Quantum Electrodynamics
12.1 Maxwell's equations and Lagrangian formulation
As always, taking our starting point in classical physics, and then impose the canonical quantization relations. Maxwell's equations (with
c = 1 and in Heaviside-Lorentz
Maxwell's equations
units) is
∇ · E = ρ , ∇ · B = 0 , ∇ × E = −Ḃ , ∇ × B = Ė + J ,
with the usual meaning. We can always nd a four-potential
E
and
B
Aµ
so we can write
as
E = −∇A0 + Ȧ , B = ∇ × A .
Maxwell's equations are actually Lorentz invariant, so
and not just some smart labeling. Now,
the same (physical) elds
E
and
B
Aµ is
Aµ is genuine four-potential
not uniquely determined: we will get
for all transformations of the type
Aµ → Aµ − ∂ µ Γ ,
Γ = Γ(x)
is an arbitrary dierentiable function. All transformations that leaves
13 transformations, and
the elds unchanged is called gauge
E
and
B
is said to be
Gauge transformation
gauge invariant. One must thus choose a gauge when working with the four-potential.
Maxwell's equations can be rewritten in terms of the antisymmetric (2,0) Faraday
Faraday tensor
µν dened as
tensor (or eld strength tensor) F
F µν ≡ ∂ µ Aν − ∂ ν Aµ .
F µν
is of course gauge invariant, as can easily be shown.
µ
(Lorentz invariant) four-current J
≡ (ρ, J),
When dening the
we nd that Maxwell's equations can
be written as
∂ν F µν = J µ , µνρσ ∂ ρ F µν = 0 .
The equation
µνρσ ∂ ρ F µν = 0
µν as we dene it. Since
for F
is redundant since it is automatically fullled
F µν is antisymmetric (and the derivative operators
commute), we have that
∂µ ∂ν F µν = ∂µ J µ = 0 .
Hence the four-current
Jµ
is a conserved Noether current (which we also knew
from the continuity equation in electrodynamics).
In total, we have that the equations of motion for the electromagnetic eld is
given by
∂ν F µν = J µ .
13
Means: to measure, ne-tune.
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Four-current
12
The Lagrangian that gives these equations of motion is
Electrodynamics
Lagrangian
1
L = − F µν Fµν + J µ Aµ ,
4
which can be shown to gauge invariant. Writing it in terms of the four-potential,
we nd that
1
Aµ g µν ∂ 2 − ∂ µ ∂ ν Aν + J µ
2
1
1
= − ∂ µ Aν ∂µ Aν + ∂ µ Aν ∂ν Aµ + J µ ,
2
2
L =
which is still gauge invariant under the transformation
Aµ → Aµ − ∂ µ Γ.
But
µ
now, to have no ambiguity in A , we must now explicitly choose a gauge. From the
Euler-Lagrange equations, we have that the equations of motion is
Equations of motion
g µν ∂ 2 − ∂ µ ∂ ν Aν + J µ = 0 .
12.2 The free photon eld and its quantization
We say that the eld is free when
J µ = 0.
We will call
Aµ
for the photon eld,
since the quanta of the eld is spin-1 particles, that we call photons.
Free photon eld
Hence, the
Lagrangian reduces to
1
L = − F µν Fµν .
4
∂µ Aµ = 0,
If we choose to work in the Lorentz gauge, where we choose
the
Lorentz gauge
Lagrangian can be written as
1
1
L = − F µν Fµν − (∂µ Aµ )2
4
2
and the Maxwell equations gives us that
∂ν F µν = ∂ν ∂ µ Aν − ∂ν ∂ ν Aµ = ∂ µ ∂ν Aν − ∂ 2 Aµ = −∂ 2 Aµ = 0 ,
which shows that
Aµ
fullls the massless Klein-Gordon equation, and hence has
14 If one looks at
plane wave solutions.
needs to be expanded in basis of four
to solve.
∂ 2 Aµ = 0
with no restriction, it apparently
4-vectors, since there actually are four equations
However, we have two conditions imposed Lorentz gauge
brings down the degrees of freedom from
4
to
2.
∂ µ Aµ = 0
and
This is true for any gauge, but
it is just easy to see for the Lorentz gauge, and is actually a consequence of the
photon being massless. This also implies that the photon as a spin-1 particle only
has the spin states
±1.
Hence the plane wave solutions is of the kind
µλ (p)e−ipx , where µ∗
λ (p),
ipx
µ∗
λ (p)e
µλ (p) are a 4-vectors, that by the condition that
be real gives that they must be each others conjugate and
λ = ±.
and
Aµ must
We see that the
applying the Lorentz gauge gives
14
Plane wave solutions
A great source: http://courses.washington.edu/phys55x/Physics%20557_lec13.htm
70 of 88
Degrees
4→2
of
freedom
12
ipx
ipx
∂µ µ∗
= ipµ µ∗
= 0 ⇒ pµ µ∗
λ (p)e
λ (p)e
λ (p) = 0 ,
and similarly
pµ µλ (p) = 0 ,
which is to say that 4-vectors
µ
and
pµ
are orthogonal, and would classically
be interpret as the electromagnetic waves are transverse (here goes one degree of
µ
µ
freedom). We will call the for polarization vectors, and specic + (p) for the
µ
right-handed polarization vector and − (p) for the left-handed polarization vector.
µ
µ
µ
The Lorentz gauge is not unique though. Since we have ∂µ (A + ∂ Γ) = ∂µ A =
0 if ∂ 2 Γ = 0, it is not unique, and we can hence choose an Γ. If we take Γ(x) =
Transverse
Polarization vectors
−iaeipx , then we have15 ∂ 2 Γ = p2 iaeipx = 0 and ∂ µ Γ(x) = −ia∂ µ eipx = apµ eipx , and
by inserting the plane wave solution into the Lorentz gauge condition, we nd that
ipx
ipx
µ
∂µ µ∗
+ ∂ µ Γ = ∂µ µ∗
e =0⇒
λ (p)e
λ (p) + ap
µ
µ∗
λ (p) + ap = 0 .
From this, we can always choose
spatial part of
p.
a
so
0λ (p) = 0,
since it only depends on the
Hence we have nally brought the degrees of freedom down to 2
as we argued we could. The general solution for
Aµ
is thus, when we use the adjoint
instead of the conjugate on the Fourier coecient as we are going to quantize soon:
ˆ
0
A (x) = 0 , A(x) =
˜
dp
General solution
i
Xh
∗λ (p)aλ (p)eipx + λ (p)a†λ (p)e−ipx
λ=±
We can solve for the polarization vectors, and choose the basis as the so-called
circular polarization basis for propagation along the

µ+ (p)
0

z -direction,

0
where we here nd
Circular
polarization
basis



 
1 
1 
1 
1 
µ


.
=√ 
, + (p) = √ 


2  −i 
2 i 

0
0
We have in a general direction of polarization a orthonormalization condition for
the polarization vectors:
∗λ (p) · λ0 (p) = δλλ0 ,
pµ µλ (p) = 0 .
Furthermore, we have a completeness relation
15
2.
Hence we see that if the photon had mass, we would have 3 degrees of freedom instead of just
71 of 88
12
X
ν
µν
µ∗
,
λ (p)λ0 (p) = g
λ=±
which makes it much easier to calculate scattering amplitudes, when we average
over the photon polarizations. For just the spatial components we have
X
∗iλ (p)jλ0 (p) = δij −
λ=±
pi pj
.
p2
The way to prove the completeness relation is to argue that we must have the
most general relation given by
X
∗iλ (p)jλ0 (p) = Aδij − Bpi pj ,
λ=±
since the polarization vectors must be at most linear in
∗
constraints λ (p)
· λ0 (p) = δλλ0
and
p,
and then use the
p · λ0 (p) = 0.
Since the photon is a spin-1 particle, it must be a boson, and we must use commutators when imposing the canonical quantization relations, that with the generalized
canonical momentum
Π0 =
∂L
∂L
= ∂ν Aν = 0 , Πi =
= ∂ i A0 − ∂ 0 Ai = −∂ 0 Ai = Ȧi .
∂ Ȧ0
∂ Ȧi
must be
µ
A (x, t), Aν (x0 , t) = 0 ,
µ
Π (x, t), Πν (x0 , t) = 0 ,
Now, however,
Aµ (x, t), Πν (x0 , t) = ig µν δ (3) (x − x0 ) .
[Aµ (x, t), Πν (x0 , t)] = ig µν δ (3) (x − x0 )
implies that there are four
polarization vectors, which becomes polarizations in time and in the longitudinal
direction, which we won't mind, since our circular (transverse) polarization is still
two of the 4 polarization vectors.
After expressing
aλ (p)
and
a†λ (p)
in terms of
A(x),
we have that commutation
relations is the creation/annihilation algebra:
algebra
[aλ (p), aλ0 (q)] = 0 ,
h
i
a†λ (p), a†λ0 (q) = 0 ,
h
i
aλ (p), a†λ0 (q) = (2π)3 2ωp δ 3 (~k − ~q)δλλ0 .
Finally, the Hamiltonian becomes
72 of 88
12
H=
Xˆ
h
i
†
˜
dp ωp aλ (p)aλ (p) + 2E0 V .
λ
12.2.1 Feynman rules and diagrams
The Feynman propagator (or photon propagator) becomes
˜ µν (k) =
∆
g µν
.
k 2 − i
The Feynman rules for photons is straight out easy. The factors is as given below
in gure 8, and when going through the diagram (in the way the rest of the Feynman
rules for the interaction, for example spinors, where we must go against the arrows
along a complete fermion line), we must contract indicies at each vertex where they
join up by a metric in the right indicies, but otherwise the polarization vectors (just
numbers) don't need to be written in any special way.
Again all of the combinatorial factors from the propagator-expansion, is exactly
canceled when drawing all of the topologically inequivalent Feynman diagrams, and
the sum of the diagram is exactly
iT .
Figure 8: Components of Feynman diagrams for photon elds (in momentum space).
73 of 88
12
12.3 Interactions, spinor QED
Figure 9: Components of Feynman diagrams for QED (in momentum space).
We can include interactions with spinor elds (electrons, positrons) very naturally.
Remember that we for the Dirac Lagrangian have a conserved current
Ψγ µ Ψ,
and
that the particles are charged. Hence
Conserved current
J µ (x) = eΨγ µ Ψ ,
must be a 4-current suitable for electrodynamics, since we as we saw in the Dirac
theory has a conserved charge that has all of the same properties as electric charge.
With this, the combined Lagrangian becomes
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12
L = LDirac + Lphoton + Linteraction
1
= Ψ (i
∂ − m) Ψ − F µν Fµν + eΨγ µ Aµ Ψ
4
1 µν
Ψ − mΨΨ ,
= − F Fµν + iΨ
D
4
where we have introduced the gauge covariant derivative
Covariant derivative
≡
D
∂ − ieA .
We have a symmetry for this Lagrangian; namely a local transformation in Dirac
and photon eld simultaneously:
Aµ → Aµ − ∂ µ Γ ,
Ψ → e−ieΓ Ψ ,
Ψ → eieΓ Ψ .
This is a local
U (1)
16 , since
gauge transformation
Γ
may be any (dierentiable)
Gauged symmetry
function. The gauge covariant derivative transforms as
→ e−ieΓ
eieΓ .
D
D
F µν
of
as
D
is invariant in itself under this transformation. We can express
F µν
in terms
a commutator relation:
F µν =
i µ
[D , Dν ] .
e
We must also have
∂µ F µν = eΨγ ν Ψ .
There are also Feynman rules for QED, which is just a combination of the rules
for photons and electrons/positrons, together with a vertex rule for the interaction
derived from
Linteraction .
We have them given in gure 9.
12.4 Massive gauge bosons
As an extension of the workings of the photon eld, we could imagine spin-1 bosons
that aren't mass-less. In this case we from the gauge invariance (choosing the Lorentz
gauge) get the constraint
pµ µλ (p) = 0,
and thus still have three degrees of freedom.
Hence the polarization vectors in a given basis must consist of three vectors, corresponding to the polarizations
16
λ = 0, ±, which is of course also the allowed spin states
Or we may say that we have 'gauged the U (1) symmetry'.
75 of 88
Degrees of freedom
12
of the particle. Otherwise, we have same Feynman rules as for massless photons, but
must use the completeness relation:
X
ν
µν
εµ∗
+
λ (k) ελ (k) = g
λ=0,±
kµ kν
M2
12.5 The Higgs mechanism for massive gauge bosons
17 For a scalar eld
φ
we have the scalar electrodynamics Lagrangian
1
1
1 2 2
µν
†
L = − [D φ] [Dµ φ] − V (φ) − Fµν F
, V (φ) ≡ λ φ φ − v
,
4
4
2
µ
†
Dµ ≡ ∂µ − igAµ
is the covariant derivative and,
v
is some real number, and
λ, g
are coupling constants. The Lagrangian looks like it is massless This Lagrangian is
invariant under a local gauge transformation:
Aµ → Aµ − ∂ µ Γ ,
φ → e−ieΓ φ .
φ(x)
has a vacuum expectation value
to a eld
φ(x) =
ρ
√1
2
h0|φ|0i =
√1 v , but we can shift the eld
2
that has a vacuum expectation value that is zero
(v + ρ(x)),
h0|ρ|0i = 0
by setting
and a general gauge transformation gives a phase factor that
results in
1
φ(x) = √ (v + ρ(x)) e−iχ(x)/v ,
2
where
ρ(x) is a real (uctuation eld) and then we have Higgs potential given by
1
1
2
2
3
4
V (φ) = λ v ρ(x) + vρ(x) + ρ(x)
.
4
4
Now, under the given gauge transformation, but further in the unitary gauge
where
χ(x) = 0
we have
[Dµ φ]† [Dµ φ] = |Dµ φ|2 =
=
=
=
1
|(∂µ − ig (Aµ + 0)) (v + ρ(x))|2
2
1
|(∂µ ρ(x) − igAµ (v + ρ(x)))|2
2
1 µ
1
∂ ρ(x)∂µ ρ(x) + g 2 (v + ρ(x))2 Aµ (x)Aµ (x)
2
2
1
1 µ
∂ ρ(x)∂µ ρ(x) + g 2 ρ(x)2 + 2vρ(x) Aµ (x)Aµ (x) + g 2 v 2 Aµ (x)Aµ (x)
2
2
Substituting back into the Lagrangian we nd
17
Unitary gauge
[Srednicki]
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13
L=−
NON-ABELIAN GAUGE THEORIES
1 µ
1
1
∂ ρ(x)∂µ ρ(x) + g 2 ρ(x)2 + 2vρ(x) Aµ (x)Aµ (x) −V (φ)− Fµν F µν − g 2 v 2 Aµ (x)Aµ (x) .
2
4
2
If we identify
M = gv , the gauge eld Aµ (x) has acquired a mass term, and hence
we interpret it as the interaction with Higgs potential has given mass to the boson.
This is the Higgs mechanism, which is generalized to non-abelian gauge theories in
a similar manner; we will get something
∝ Aµ (x)Aµ (x)
that we interpret as a mass
4
term. We will actually for any ρ potential get the Higgs mechanism, which can be
argued for on physical grounds, since it is related to the 4 space-time dimensions.
13 Non-abelian gauge theories
13.1 Extending the symmetry groups to non-abelian
So far we have only considered theories (Lagrangians) with corresponding abelian
symmetry groups, that is where the symmetry transformations
A, B fullls AB = BA
(regular complex number multiplication as the group product of
U (1)
for example).
We could also have theories with non-abelian symmetry groups, where the symmetry
transformations doesn't commute so
Abelian symmetry
Non-abelian symmetry
AB 6= BA.
Example 28. We have that U (1) and SO(2) are isomorphic (clearly), U (1) w SO(2),
and since
U (1)
hence so must
is clearly an abelian symmetry group, it has only 1 generator, and
SO(2),
and thus is also abelian.
If we have a symmetry group
transformations
A, B
that
AB
erty. We must also have that
have that
tors
[A, B]
{T a }a∈I
S,
then we must have that for two symmetry
is also a symmetry transformation by the group prop-
A−B
is a symmetry transformation, and hence we
is a symmetry transformation. Given that we have a set of genera-
for the symmetry group, where
numel(I) = numgen = dim (S )
is the
number of generators of the group, this implies that
h
i
T a , T b = if abc T c ,
where
f abc
Structure constants
is called the structure constants of the group. From this we see that
the symmetry group is abelian if and only if
f abc = 0
for all
a, b, c ∈ I .
If this is
not the case, the group is non-abelian. Even further, if the symmetry group is also
a dierentiable manifold (we have some continuous parameters to label the group
elements), then the symmetry group is also a Lie group and we have a Lie algebra
with the commutator as the bracket product.
Lie group and algebra
Thus the Jacobi identity must be
satised for the generators and hence for the structure constants, so we must have
hh
i
i hh
i
i h
i
T a , T b , T c + T b , T c , T a + [T c , T a ] , T b = 0 ⇔ f abd f dce +f bcd f dae +f cad f dbe = 0 .
Hence we can do ne with just looking at innitesimal transformations when we
want to extract information about the group. The general innitesimal transforma-
77 of 88
13
tion is given by
U = I + ωaT a .
Normally we work with a representation of the symmetry, and so most of the
times we only have
SU (N )
and
SO(N )
symmetry groups (perhaps a group product
U (1) to get U (N ) or O(N )), and so we the generators are some appropriate matrices
SU (2)
that span the group.
The simplest non-abelian symmetry (Lie) group is
erators (and thus is a Lie group of dimension 3,
SU (2),
which has three gen-
dim (SU (2)) = 3),
that doesn't
commute and hence is non-abelian, since the generators can be chosen as the Pauli
SO(N )
matrices.
The group
generators is
SO(N )
is a non-abelian Lie group for
N ≥ 3,
and the number of
dim (SO(N )) = N (N − 1) /2.
The more general group
number of generators is
SU (N )
SU (N )
is a non-abelian Lie group for
dim (SU (N )) =
N ≥ 2,
and the
N 2 − 1. We have that a general innitesimal
transformation in this case can be written as
Ukl = δkl − igθa (T a )kl ,
where the generators
Ta
can all be taken to be traceless hermitian matrices.
Non-abelian gauge theories is extensions of QED to a non-abelian symmetry
group and such a theory is also called Yang-Mills theory.
Yang-Mills theory
13.2 Quantum Chromodynamics
QCD
Quantum Chromodynamics QCD is constructed by take the Lagrangian from QED
as a starting point and then take enough copies of the spinors and photon elds so we
can get a
SU (3) symmetry group.
Thus we have, that since
dim (SU (3)) = 32 −1 = 8
that we must have 8 copies of the QED elds. The generators can be taken as the
3×3
18
Gell-Mann matrices . So for
(color index) and
from
a, b, c = 1 . . . 8
(gluon index) and
i, j = 1 . . . 3
I = 1 . . . 6 (avor index; theory brings down the degrees of freedom
dim (SU (3)) = 8
to 6) we have the elds
ΨiI , ΨiI
(the quark elds) and
Aµa
(the gluon elds) and the QCD Lagrangian is given as
1
a
)ij ΨjI − mI ΨiI ΨiI − F aµν Fµν
LQCD = iΨiI (
D
4
1
= (QED like terms) − gf abc Aaµ Abν ∂ µ Aνc − g 2 f abe f cde Aaµ Abν Acµ Adν .
4
6
X
1
a
k − mk ) Ψk
= − F aµν Fµν
+
iΨk (i
D
4
k=1
This Lagrangian is invariant under a
have the covariant derivative
SU (3)
transformation by construction. We
)ij ≡ (
D
∂ δij − ig Aa Tija .
Now the electric charge is
replaced by the charges of the quarks (two kinds of charges,
18
2
3 |e| and
− 13 |e|).
The
http://en.wikipedia.org/wiki/Gell-Mann_matrices
78 of 88
Gell-Mann matrices
13
Lagrangian gives only the that there are three possible vertices: QED vertices, a
3-gluon vertex, and a 4-gluon vertex.
13.3 Electroweak interaction
The Lagrangian for the electroweak interaction is given by
1
1
LEW = − F~µν · F~ µν − Bµν B µν − V (φ) − [Dµ φ]† [Dµ φ] ,
4
4
where we have
~ ν − ∂ν A
~ µ + g2 A
~µ × A
~ν ,
F~µν ≡ ∂µ A
Bµν ≡ ∂µ Bν − ∂ν Bµ ,
~
σ
~ µ · + g1 Bµ Y ,
Dµ ≡ ∂µ − i g2 A
2
g1 , g2
are coupling constants,
I2 ), V (φ) ≡ 41 λ φ† φ − 12 v
2 2
Y = − 12 I2
is the weak hypercharge (2
×2
matrix
φ
is a set
is the Higgs potential. It is understood that
of two complex scalar elds and we write them in terms of four real elds
φ1 + iφ2
φ(x) ≡
!
.
φ3 + iφ4
This Lagrangian is invariant under the
SU (2) × U (1)
gauge transformation:
φ(x) → e−i~α(x)·~σ/2−iβ(x) φ(x) ,
~ µ (x) ,
~ µ (x) → A
~ µ (x) − 1 ∂µ α
~ (x) + α
~ (x) × A
A
g2
Bµ (x) → Bµ (x) −
The invariance is easy for the
1
∂µ β(x) .
g1
F~µν · F~ µν , Bµν B µν
terms, that is just as in QED.
The diculty is with the covariant derivative, but using
~σ (x)
~σ (x)
~σ (x)
~ µ (x) × α
~
Aµ (x) ·
,α
~ (x) ·
= A
~ (x) ·
,
2
2
2
it is done by a direct calculation in the where we take
19
and further take
φ2 = φ3 = φ4 = 0
and expand
φ1
α
~ (x) = ~0
around the
β(x) = 0
√
minimum at v/ 2
and
so we have
φ(x) =
φ1
0
!
1
=√
2
v + H(x)
0
!
,
This is a gauge choice also, since we can just translate φ by a constant to the vacuum expectation
value and insist on it to be real so that the unwanted terms cancel.
19
79 of 88
Unitary gauge
13
where
H(x)
is the Higgs eld. By calculation we then nd
Higgs eld
2
1 − [Dµ φ]† [Dµ φ] = − v 2 g2 A3µ g1 Bµ + g22 A1µ − iA2µ A1µ + iA2µ .
8
If we now dene the elds
Wµ± ≡ A1µ ∓ iA2µ ,
1
g2 A3µ g1 Bµ ,
Zµ ≡ p 2
g2 + g12
Wµ±
is complex elds for two bosons of same mass (as it will show soon) and
Zµ Wµ± , Zµ
bosons
is a real eld boson. We see that we can write the Lagrangian as
!
v 2 g22 + g12
g22 v 2 + −
LEW
Zµ +
Wµ Wµ ,
4
4
1
1
1
2
MZ2 Zµ + MW
Wµ+ Wµ−
= − F~µν · F~ µν − Bµν B µν − V (φ) −
4
4
2
p
1
where we in the last line have dened MZ ≡ v
g22 + g12 and MW ≡ 21 vg2 . We
2
1
1
1
= − F~µν · F~ µν − Bµν B µν − V (φ) −
4
4
2
here see that we now get the appropriate mass terms for a real and complex eld,
which we interpret as
Wµ± , Zµ
bosons has gained mass.
We can dene
cos θW ≡
where
θW
MW
g2
=p 2
MZ
g1 + g22
g1
, sin θW ≡ p 2
,
g1 + g22
Weak mixing angle
is the weak mixing angle. We then nd the relation
Aµ
Zµ
!
=
cos θW
sin θW
− sin θW
cos θW
We also from this see that the photon eld
!
Bµ
!
A3µ
Aµ haven't gained any mass, consistent
with experimental ndings. Again it is the Higgs mechanism at play.
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13
13.4 The Standard Model
Figure 10: Vertex rules for Feynman diagrams for the standard model.
The Standard Model is the combination of the electroweak theory and quantum
chromodynamics, so we have a combined theory for these interactions. We have a
QCD sector, a electroweak sector and a Higgs sector, where the Higgs boson again
gives mass terms to the right particles (perhaps some discrepancy with the neutrino).
The symmetry group is
SU (3) × SU (2) × U (1)
| {z } |
{z
}
strong
electroweak
81 of 88
A
NICE-2-KNOW RELATIONS
A Nice-2-know relations
A.1 Delta functions
ˆ
eipx dx
δ(x) =
R
ˆ
1
f (0)
|c|
f (cx)dx =
R
ˆ
δ(g(x)dx =
R
X
zeros xi
1
|g 0 (xi )|
ˆ
X
f (x)δ(g(x)dx =
R
zeros xi
f (xi )
|g 0 (xi )|
A.2 Dirac adjoint/barred identities
The denition of the Dirac adjoint for a scalar
Barred spinors
a
is
a = a∗
For some spinor
Ψ
we have
Ψ ≡ Ψ† β
A combination of
γ -matrices A
Ψ ≡ βA† β
We have that
β = γ0
and
β2 = I
From [Srednicki] (38.15) we have the identities
γµ = γµ
iγ 5 = iγ 5
γµγ5 = γµγ5
S µν = S µν
iγ 5 S µν = iγ 5 S µν
For the bi-spinors we have
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γ -MATRICES
B
us (p) ≡ u†s (p)β , us (p) ≡ βus (p) .
A.3 Traces of γ -matrices
γ -matrix
We have a bunch of useful rules when calculating traces of
γ5
2
γ -matrices:
=I
µ 5
γ , γ = γ5γµ + γ5γµ = 0
Tr [odd
# of
Tr γ 5 (odd
γ -matrices] = 0
# of
γ -matrices) = 0
Tr [ab] = −4aµ bµ ≡ −4 (ab)
Tr γ 5 ab = 0
Tr [abcd ] = 4 [(ad) (bc) − (ac) (bd) + (ab) (cd)]
20 .
More can be found in [Srednicki] page 294-297, and on Wikipedia
B γ -matrices
The gamma matrices is a set of four
4×4
matrices that is a representation of the
generators of the Dirac algebra, which is to say the anti-commutation relations
{γ µ , γ ν } = −2g µν I4 .
We can from thus further deduce the property that
γ0
2
= I , γi
2
= −I .
Also, they obey the following commutation relations:
i µ ν
[γ , γ ] ≡ S µν ,
4
where
S µν
is the generator of the Lorentz group.
The Dirac algebra have many representations, but the representation used in
[Srednicki] is the Weyl or chiral representation is given in terms of the Pauli matrices:
20
http://en.wikipedia.org/wiki/Gamma_matrices
83 of 88
technology
C
µ
γ ≡
0
σµ
σµ
0
QUANTUM MECHANICS
!
,
Or:

γ ≡
1
γ ≡
γ2 ≡
γ3 ≡
!
0 I
0
I
0 0 0 1
0
0
σ1
−σ 1
0
0
σ2
−σ 2
0
0
σ3
−σ 3
0

 0 0
=
 0 1

1 0

!  0
 0
=
 0

−1

!  0
 0
=
 0

−i

!  0
 0
=
 −1

0


1 0 
,
0 0 

0 0
0
0 1


1 0 
,
−1 0 0 

0 0 0

0 0 −i

0 i 0 
,
i 0 0 

0 0 0

0 1 0

0 0 −1 
.
0 0 0 

1 0 0
0
Also, we have the fth gamma matrix:

5
0 1 2 3
γ ≡ iγ γ γ γ =
−I
0
0
I
!
−1

 0
=
 0

0
0
0 0


−1 0 0 
,
0 1 0 

0 0 1
that obeys
γ5
2
=I,
and
γ µ , γ 5 = 0 ⇔ γ µ γ 5 = −γ 5 γ µ
C Quantum Mechanics
C.1 Basic equations
Time dependent Schrödinger equation
i
d S = Ĥ(t)S
dt
Uncertainty relation
84 of 88
C
1 σÂ σB̂ ≥ Â, B̂ ,
2i
QUANTUM MECHANICS
σx̂ σp̂ ≥
1
2
C.2 Commutator relations
General relations:
i
h
Q̂, Û
h
i
Q̂, Q̂
i
h
Q̂ + P̂ , Û
h
i
Q̂P̂ , Û
∀Q̂, Û , P̂ ∈ L (H) , ∀λ ∈ C :
i
h
= − Û , Q̂
h
i
= 0 , Q̂, λ = 0
i
i h
h
= Q̂, Û + P̂ , Û
h
i h
i
= Q̂ P̂ , Û + Q̂, Û P̂
Canonical commutation relation:
[x̂, p̂] = i
Angular momentum commutation relations:
Li , Lj = iijk Lk
i j
L , x = iijk xk
i j
L , p = iijk pk
Or explicitly:
h
i
h
i
h
i
L̂x , L̂y = iL̂z , L̂y , L̂z = iL̂x , L̂z , L̂x = iL̂y
h
i
L̂, L̂2 = 0
h
i
h
i
h
i
L̂x , r̂ = (0, iẑ, −iŷ) , L̂y , r̂ = (−iẑ, 0, ix̂) , L̂z , r̂ = (iŷ, −ix̂, 0)
h
i
L̂x , p̂ = (0, ipˆz , −i~pˆy ) ,
h
i
h
i
L̂y , p̂ = (−ipˆz , 0, ipˆx ) , L̂z , p̂ = (ipˆy , −ipˆx , 0)
h
i
L̂, p̂2 = 0 ,
h
i
L̂, r̂2 = 0
85 of 88
D
COLLECTION OF FEYNMAN RULES
D Collection of Feynman rules
D.1 Scalar elds
D.2 Spinor elds
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D
COLLECTION OF FEYNMAN RULES
D.3 Photon elds
D.4 QED
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E
MANDELSTAM VARIABLES
E Mandelstam variables
To simplify
1 + 2 → 3 + 4 particle scattering, one can dene the Lorentz invariant
s ≡ − (p1 + p2 )2 = − (p3 + p4 )2
t ≡ − (p1 − p3 )2 = − (p2 − p4 )2
u ≡ − (p1 − p4 )2 = − (p2 − p3 )2
s + t + u = m21 + m22 + m23 + m24 ≡ M 2 ,
p1 · p2 =
1
m21 + m22 − s
2
p3 · p4 =
1
m23 + m24 − s
2
p1 · p3 =
1
t − m21 − m23
2
p2 · p3 =
1
u − m22 − m23
2
p1 · p4 =
1
u − m21 − m24
2
p2 · p4 =
1
t − m22 − m24
2
In the center of mass frame we further have that
1
kp1 k = kp2 k = √
2 s
q
2
s2 − 2 m21 + m22 s + M12 − m22
1
kp3 k = kp3 k = √
2 s
q
2
s2 − 2 m23 + m24 s + M32 − m24
and
√
s = p01 + p02 = p03 + p04
√
√
t = p01 − p03 = p02 − p04
u = p01 − p04 = p02 − p03
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