1
QUANTUM FIELD THEORY II
(PHYS7652) LECTURE NOTES
Lecture notes based on a course given by Maxim Perelstein.
We begin with discussing the path integral formalism in
Quantum Mechanics and move on to it’s use in Quantum Field
Theory. We then study renormalization and running couplings in
abelian and non-abelian gauge theories in detail.
Presented by:
MAXIM PERELSTEIN
LATEX Notes by: JEFF ASAF DROR
2013
CORNELL UNIVERSITY
Contents
1 Preface
1.1 Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 Path Integrals in Quantum Mechanics
2.1 Harmonic Oscillator . . . . . . . . . . .
2.2 Anharmonic Oscillator . . . . . . . . .
2.3 Correlation Functions . . . . . . . . . .
2.A Jacobian of the Fourier Transformation
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3 Quantum Field Theory
3.1 Free Klien Gordan Field . . . . . . . . .
3.2 External Force . . . . . . . . . . . . . . .
3.3 Interacting Theory . . . . . . . . . . . .
3.4 LSZ Reduction . . . . . . . . . . . . . .
3.5 Statistical Mechanics and Path Integrals
4 Quantization of Electromagnetism
4.1 Classical Electromagnetism . . .
4.2 Quantization . . . . . . . . . . .
4.3 Path Integrals in QED . . . . . .
4.4 Correlation Functions . . . . . . .
5 Fermions with Path Integrals
5.1 Grassman Numbers . . . . .
5.2 Free Dirac Field . . . . . . .
5.3 QED . . . . . . . . . . . . .
5.4 Three-point function . . . .
5.5 e+ e− → µ+ µ− . . . . . . . .
5.6 e− µ− → e− µ− . . . . . . .
5.7 Compton Scattering . . . .
5.8 Proof of Ward Identity . . .
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4
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5
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34
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40
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42
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45
47
49
53
55
57
59
CONTENTS
3
6 Loop Diagrams and UV Divergences in QED
6.1 Superficial Divergences . . . . . . . . . . . . .
6.2 Electron Self-Energy . . . . . . . . . . . . . .
6.3 Classical Renormalization and Regulators . . .
6.4 Dimensional Regularization . . . . . . . . . .
6.5 Vacuum Polarization . . . . . . . . . . . . . .
6.6 Vertex Function . . . . . . . . . . . . . . . . .
6.6.1 Physics of F2 . . . . . . . . . . . . . .
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61
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7 Renormalized Perturbation Theory
7.1 φ4 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 QED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80
80
81
8 Running Couplings and Renormalization Group
8.1 “Large Logs” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 QED β-function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84
84
90
9 Non-Abelian Gauge Theories (“Yang-Mills”)
9.1 Looking Closely at Abelian Gauge Theories .
9.2 SU (2) . . . . . . . . . . . . . . . . . . . . . .
9.3 General Recipe . . . . . . . . . . . . . . . . .
9.4 Universal Couplings . . . . . . . . . . . . . . .
9.5 Quantum Chromodynamics . . . . . . . . . .
9.6 Quantization of Yang Mills . . . . . . . . . . .
9.7 Unitary and Ghosts . . . . . . . . . . . . . . .
9.8 Renormalized Perturbation Theory . . . . . .
9.9 RG flows (evolution) . . . . . . . . . . . . . .
10 Broken Symmetries
10.1 SSB of Discrete Global Symmetries . .
10.2 SSB of Continuous Global Symmetry .
10.3 Linear Sigma Model . . . . . . . . . .
10.4 More on the Mexican Hat Potential . .
10.5 SSB of Gauge Symmetries (“The Higg’s
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114
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Mechanism”)
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116
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121
122
123
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Chapter 1
Preface
This set of notes are based on lectures given by Maxim Perelstein in the Quantum Field
Theory II course at Cornell University during Spring 2013. The course uses both Srednicki’s Quantum Field Theory and Peskin and Schroeder’s An Introduction to Quantum
Field Theory as reference texts. I wrote these notes during lectures and as such may contain some small typographical errors and sloppy diagrams. I also make notes for myself
throughout the text of the form, [Q#:...]. They are just to remind to me to go back and
take a look at something and should be ignored. I’ve attempted to proofread and fix as
many of these problems as I can. I hope to continue to revise and add to these notes until
I feel they are complete. If you have any questions or would like a copy of the LATEX file
feel free to let me know at [email protected].
1.1
Conventions
In these notes we use the conventions used by Peskin and Schroeder. We use a “West
coast metric”,


1 0
0
0
 0 −1 0
0 

gµν = 
(1.1)
 0 0 −1 0 
0 0
0 −1
the Weyl basis,
µ
γ =
0 σµ
σ̄ µ 0
(1.2)
where σ µ ≡ (1, σ i ) and σ̄ ≡ (1, −σ i ).
When applying Fourier transform (FT) we put the 2π on the momenta integral such
that:
Z
dp ˜
f (p)e−ipx
(1.3)
f (x) =
2π
4
Chapter 2
Path Integrals in Quantum Mechanics
The primary tool of this course will be what’s known as the path integral. In nonrelativistic quantum mechanics (NRQM) it is an alternative formulation to using the
Schrodinger equation. Consider a particle at xa that evolves with a Hamiltonian
H = H(x, p; t)
(2.1)
What we want to compute is the transition amplitude for this particle to go from its
initial position some other position xb in a time 2T . In QM we denote this amplitude
with a braket:
U = hxb , T |xa , −T i
(2.2)
To simplify this expression consider inserting in a completeness relation N − 1 completeness relations:
Z
=
dx1 dx2 ... dxN −1 hxb , T |xN −1 , T − ∆ti hxN −1 , T − ∆t|xN −2 , T − 2 ∆ti ... (2.3)
hx1 , −T + ∆t|xa , −T i
Now if we have a braket at two nearby times then you can evolve one side to have an
equal time braket1 :
hx, t0 |x0 , t0 + ∆ti = hx, t0 |e−iH(t1 )·(t1 −t0 ) |x0 , t0 i = hx|e−iH(t1 )∆t |x0 i
(2.4)
where in the last step we rewrote the bra and ket without time dependence as the absolute
value of time has no significance.
Z
=
dx1 dx2 ... dxN −1 hxb |e−iH(T −∆t)·∆t |xN −1 i hxN −1 |e−iH(T −2∆t)·∆t |xN −2 i ...
hx1 |e−iH(−T )·∆t |xa i
1
To do this properly with finite ∆t requires Dyson’s formula
5
(2.5)
6
CHAPTER 2. PATH INTEGRALS IN QUANTUM MECHANICS
Now let t1 = −T, t2 = −T + ∆t, ..., tN = T . Thus we now have
Z
=
dx1 dx2 ... dxN −1 hxb |e−iH(tN −1 )·∆t |xN −1 i hxN −1 |e−iH(tN −2 )·∆t |xN −2 i ...
hx1 |e−iH(t1 )·∆t |xa i
U=
=
N
−1 Z
Y
i=1
N
−1 Z
Y
!
dxi
!
dxi
i=1
N
−1
Y
i=0
N
−1
Y
(2.6)
!
hxi+1 |e−iH(ti )∆t |xi i
(2.7)
!
U (xi → xi+1 ; ∆t)
(2.8)
i=0
Note that technically these equations need two less integrals then the number of xi values.
However often people don’t bother to write it like this (since it somewhat ruins the ascetic
beauty in the formula). The point is that you write down each of the brakets with every
position included, but you only integrate over x1 , ..., xN −1 , not over x0 = xa and xN = xb .
Either way they don’t matter much at the end, as long as you know not to integrate over
your end points... Instead this expression is more commonly written as
N Z
Y
U=
!
dxi U (xi → xi+1 ; ∆t)
(2.9)
i=0
Pictorially this integral is show below
t
T
xa
xb
x
..
.
2
1
−T
This is called a path integral. For simplicity we focus on Hamiltonians are of the form
H=
p2
+ V (x, t)
2m
(2.10)
7
Consider the braket above (if we know the value of this term we almost know everything).
We take ∆t to be small.
p2
+ V (x, ti )|xi i
2m
p2
= δ(xi+1 − xi ) − i∆t hxi+1 |
|xi i − i∆tδ(xi+1 − xi )V (xi , ti )
2m
Z
dpi −ipi (xi+1 −xi )
p2
=
e
− i∆t hxi+1 |
|xi i − i∆tV (xi , ti )
2π
2m
Z
dpi −ipi (xi+1 −xi )
×
e
(no sum)
2π
hxi+1 |e−iH∆t |xi i = hxi+1 |xi i − i∆t hxi+1 |
By insertion of complete set of states (note also that hpi |xi i = e−xi pi ) we have,
Z
0
00
2
hxi+1 |p |xi i = dp0 dp00 eixi+1 p e−ixi p p00 2 hp0 |p00 i
Z
dp0 ixi+1 p0 −ixi p0 0 2
=
e
e
p
2π
Z
dpi i(xi+1 −xi )pi 2
2
pi
(no sum)
e
hxi+1 |p |xi i =
2π
(2.11)
(2.12)
(2.13)
(2.14)
(2.15)
(2.16)
Therefore,
−iH(ti )∆t
hxi+1 |e
2
pi
dpi −ipi (xi+1 −xi )
e
+ V (xi , ti )
|xi i =
1 − i∆t
2π
2m
Z
dpi −ipi (xi+1 −xi ) −i∆tH(xi ,pi ,ti )
e
=
e
2π
Z
dpi −ipi ∆t (xi+1 −xi ) −i∆tH(xi ,pi ,ti )
∆t
e
=
e
2π
Z
(2.17)
(2.18)
(2.19)
Note that in the last step we have combined our infinitesimal result into our exponential
(in the end we take ∆t → 0). Notice that we no longer have operators; pi and Hi are
numbers!The expression has only numbers, we have moved away from regular quantum
mechanics. now taking the ∆t → 0 limit we have
Z
dpi −i∆t(pi ẋi −Hi )
−iHi ∆t
e
(2.20)
hxi+1 |e
|xi i =
2π
Thus we have
N
−1 Z
Y
U=
i=1
Z
=
dp0
2π
!
N
−1 Z
Y
!
dpi −i∆t(ẋi pi −Hi )
dxi
e
2π
i=0
!
Z
N
−1 Z
Y
PN −1
dpi
dxi
e−i i=0 ∆t(ẋi pi −Hi )
2π
i=1
(2.21)
(2.22)
8
CHAPTER 2. PATH INTEGRALS IN QUANTUM MECHANICS
Taking the limit we finally have
N
−1 Z
Y
U=
Z
dxi
i=1
dpi
2π
!
e−i
RT
−T
dt(ẋi pi −Hi )
(2.23)
R
where now that we have taken the limit we have just dropped the dp0 integral and the
value of p0 in the exponential. Since we have an infinite number of integrals, adding one
more shouldn’t effect the final result. Only when we discretize our integrals should such
a distinction be of any importance.
For convenience we define
N
−1 Z ∞
Y
Z
Dx = lim
N →∞
i=1
Z
dxi
Dp = lim
;
N →∞
−∞
N
−1 Z ∞
Y
i=1
dpi
(2.24)
−∞
We can now write
Z
U=
DxDp e−i
RT
−T
dt(ẋi pi −Hi )
(2.25)
Note that the momenta integral has nothing to do with the particular dynamics of the
problems and can be done in general since they are just Gaussian integrals. The integral
we need is (we now go back to the discrete limit for a moment, see equation 2.22)
Z
Z
dpi i∆t(pi ẋi − p2i −Vi )
dpi −i ∆t (pi −mẋi )2 i ∆t (mx˙i )2 −i∆tVi
2m
e
=
e 2m
e 2m
(2.26)
2π
2π
Z
dpi −i ∆t (pi )2 i ∆t (mx˙i )2 −i∆tVi
e 2m
(2.27)
=
e 2m
2π
∆t
2
= C(∆t, m)ei 2m (mx˙i ) −i∆tVi
m
2
= C(∆t, m)ei∆t( 2 x˙i −Vi )
(2.28)
= C(∆t, m)ei∆tL[xi ,ti ]
(2.30)
(2.29)
where we have denoted the value of the momenta integral by C(∆t, m). Though not
obvoius, it’s value will turn out to have no impact on physical quantities. Nevertheless
we include for completeness,
r
2πm
C(∆t, m) =
(2.31)
i∆t
Thus we have
!
Z T
Z
N
−1 Z
N
Y
X
L(xj , t) = C Dx exp i
L[x(t), t]dt
U =C
dxi exp i∆t
i=1
Z
=C
iS Dx e j=1
−T
(2.32)
x(−T )=xa ,x(T )=xb
2.1. HARMONIC OSCILLATOR
9
where S is the action. This can be thought of as
X
U=
eS(path)
(2.33)
paths
We worked in natural units. Going back to regular units we have
X
U=
eS(path)/~
(2.34)
paths
A classical system is one that has a large action (alternatively one can take the Bohr
limit of ~ → 0). We are dealing with a rapidly oscillating integrand. The trajectory that
ends up being the dominant contribution is the one that the smallest phase (S(path)/~).
This is the one that has minimum S, as in classical mechanics.
2.1
Harmonic Oscillator
We will now do an example. Consider a simple harmonic oscillator with an external force,
f (t) . The Hamiltonian is given by
H0
p2 ω 2 x2 ω z }| {
+
− −f (t)x
H=
2
2
2}
|
{z
(2.35)
H0
(set m = 1). We consider a force that is turned on and then turned off at some time
later:
f (t)
−t0
t0
t
Initially the system is in the ground state (|0i), what’s the probability to remain in |0i
after t0 ? In standard quantum mechanics we compute probabilities as
P = |h0, t0 |0, −t0 i|2
(2.36)
It’s unclear how this relates to the path integral formulation since so far we’ve only
discussed probabilities of a particle moving from some x eigenstate to another x eigenstate.[Q 1: At the end repeat this problem using standard PT.] Consider (we use Dyson’s
formula discussed above)
U (xa → xa , 2T ) ≡ hxa |e−i
RT
−T
Hdt
|xa i
(2.37)
10
CHAPTER 2. PATH INTEGRALS IN QUANTUM MECHANICS
with T > t0 . We split the braket into 3 parts:
U (xa → xa , 2T ) = hxa |e−2iH0 T e−i
R t0
−t0
= hxa |e−iH0 (T −t0 ) e−i
= hxa |e−iH0 T e−i
R t0
−t0
H 0 dt
R t0
−t0
|xa i
H0
(2.38)
dt −iH0 (t0 −(−T ))
e
H dt −iH0 T
e
|xa i
|xa i
(2.39)
(2.40)
where we have used the fact that H 0 is only applied between −t0 and t0 and in the last
step we were careful to write H 0 in Dysons formula not H0 . We now insert two complete
set of states:
Z
U (xa → xa , 2T ) =
dx+ dx− hxa |e−iH0 T |x+ i hx+ |e−i
R t0
t0
Hdt
|x− i hx− |eiH0 T |xa i
(2.41)
We now try to simplify the hx− |e−iH0 T |xa i term. Consider,
−iH0 T
e
|xa i =
∞
X
e−iH0 T |ni hn|xa i
(2.42)
ψn (xa )e−iEn T |ni
(2.43)
n=0
=
∞
X
n=0
where we have inserted the set of eigenstates of H0 . Now comes a “dirty trick”. Take
H0 → (1 − i)H (at the end we will take → 0).
−iH0 (1−i)T
lim lim e
→0 T →∞
|xa i = lim lim
→0 T →∞
∞
X
ψn (xa )e−iEn T e−T En |ni
(2.44)
n=0
Choose spectrum as En > 0, n > 0 and E0 = 0. If we take T → 0 then the only term
that contributes from the sum is the one that isn’t effected by the damping term, e−T En
(the |0i term).
lim lim e−iH0 (1−i)T |xa i = lim ψ0 (xa ) |0i
→0 T →∞
→0
= ψ0 (xa ) |0i
(2.45)
(2.46)
This key step helps us take position eigenstates and relate them to what we really want
which is energy eigenstates. Define Ū as U with H0 → H0 (1 − i)
Z
R t0
2
dx+ dx− h0|x+ i hx+ |e−i −t0 Hdt |x− i hx− |0i
lim lim Ū (xa → xa , 2T ) = |ψ0 (xa )|
→0 T →∞
(2.47)
= |ψ0 (xa )|2 h0|e−
2
R t0
−t0
Hdt
|0i
= |ψ0 (xa )| h0, t0 |0, −t0 i
Z
(2.48)
(2.49)
∞
dxa lim lim Ū (xa → xa , 2T ) = h0, t0 |0, −t0 i
−∞
→0 T →∞
(2.50)
2.1. HARMONIC OSCILLATOR
11
due to the normalization in the wavefunction. We now go back to our formula for the
path integral. Note that don’t go to our “final” form of our path integral equation since
p2
to simplify the momentum integral we assumed H of the form 2m
+ V (which is not the
case now that we threw in the i). Instead we go to our earlier form (see equation 2.25).
Z ∞
Z
dt(pẋ − (1 − i)H0 + f (t)x)
(2.51)
lim Ū = DxDp exp i
T →∞
−∞
Z
Z Y
dp
(1 − i)
= Dx
exp i∆t
− p2 + 2pẋ (1 + i) − ẋ2 (1 + i)2
2π
2
2
2
2 2
(2.52)
+ ẋ (1 + i) + 2 (1 + i) f (t)x − ω x
where we have used 1+i ≈ (1 − i)−1 , and we ignore the constant term in H0 since it has
no impact on the final result. The term in the exponential can be factored into a constant
term (with respect to dp) and a −(p − ẋ(1 + i))2 term. We define p0 ≡ p − ẋ(1 + i).
We have the integral
r
Z
1 − i 2
1
2π
dp
exp −i∆t
p =
(1 + i)1/2
(2.53)
2π
2
2π i∆t
r
−i
=
(1 + i)
(2.54)
2π∆t
Finally we have
Z ∞
Z
ẋ2
ω 2 x2
dt (1 + i) −
lim Ū = C Dx exp i
(1 − i) + f (t)x
(2.55)
T →∞
2
2
−∞
To do this integral we need a way to put ẋ and x in a similar form. We do this using a
Fourier transform:
Z ∞
dE −iEt
x(t) =
e
x̃(E)
(2.56)
−∞ 2π
Z
dE
ẋ(t) =
(−iE)e−iEt x̃(E)
(2.57)
2π
Z
dE −iEt ˜
f (t) =
e
f (E)
(2.58)
2π
The integration measure can be changed to Dx̃. This is because every variable in x space
has a one-to-one correspondence with the variables in k space and the Jacobian to switch
spaces is one (this is shown in appendix 2.A). We have
Z
Z
Z
dE ˜
dE 0
dt f (t)x =
f (E)
x̃(E 0 )2πδ(E + E 0 )
(2.59)
2π
2π
Z
dE ˜
=
f (E)x̃(−E)
(2.60)
2π
Z
dE ˜
1
˜
f (E)x̃(−E) + f (−E)x̃(E)
(2.61)
=
2
2π
12
CHAPTER 2. PATH INTEGRALS IN QUANTUM MECHANICS
Similarly,
Z
Z
dE
dE 0
dt ẋ =
x̃(E)
x̃(E 0 )2πδ(E + E 0 )(−1)EE 0
2π
2π
Z
dE
=
x̃(E)x̃(−E)E 2
2π
Z
dE 2
1
E x̃(E)x̃(−E) + x̃(−E)x̃(E)
=
2
2π
2
Z
2
Z
lim Ū = C
T →∞
(2.62)
(2.63)
(2.64)
2
E −ω +i
Z
}|
{
1
dE z 2
2
Dx̃ exp i
E (1 + i) − ω (1 − i) x̃(E)x̃(−E)
2
2π
˜
˜
+ f (E)x̃(−E) + f (−E)x̃(E)
(2.65)
The idea is to complete the square for each dx̃. The variable that does that is ỹ(E) =
f˜(E)
x̃(E) + E 2 −ω
2 +i . The arguement in the exponential turns into,
!
!
Z
i
f˜(E)
f˜(−E)
dE
2
2
(...) =
ỹ (−E) − 2
E − ω + i ỹ (E) − 2
2
2π
E − ω 2 + i
E − ω 2 + i
(2.66)
!
!
f˜(−E)
f˜(E)
˜(−E) ỹ(E) −
+ f˜(E) ỹ(−E) − 2
+
f
(2.67)
E − ω 2 + i
E 2 − ω 2 + i
Z
f˜(E)f˜(−E)
dE
i
E 2 − ω 2 + i ỹ(−E)ỹ(E) − 2
(2.68)
=
2
2π
E − ω 2 + i
Thus we can write,
dE f˜(E)f˜(−E)
lim Ū = C
T →∞
2π −E 2 + ω 2 − i
(2.69)
Now consider the special case of f → 0. Then we have no perturbation and the amplitude
to stay in the ground is just 1. So in this case we have (the second exponential is trivial)
Z
Z
1
dE
2
2
1 = C Dỹ exp i
ỹ(E)(E − ω + i)ỹ(−E)
(2.70)
2
2π
Z
Z
1
dE
2
2
Dỹ exp i
ỹ(E)(E − ω + i)ỹ(−E) exp
2
2π
i
2
Z
This is a mathematical relation completely independent of the particular system that we
consider. Thus we can insert it into equation 2.69 to get a final succinct expression for
the amplitude,
!
Z
i
dE f˜(E)f˜(−E)
h0, t0 |0, −t0 i = lim exp
(2.71)
→0
2
2π −E 2 + ω 2 + i
!
2.1. HARMONIC OSCILLATOR
13
The (E 2 − ω 2 )−1 factor is remenicient of the inverse of the simple harmonic oscillator
equation. This hints that this can be rexpressed in terms of the Green’s function for
simple harmonic motion defined by:
∂t2 + ω 2 G(t − t0 ) = δ(t − t0 )
(2.72)
Writing both sides our using a Fourier decomposition,
Z
Z
dE
dE −iE(t−t0 )
2
2
−iE(t−t0 )
−E + ω G̃(E)e
e
=
2π
2π
(2.73)
and so,
G̃ =
E2
1
− ω2
(2.74)
Now taking the inverse Fourier Transform,
Z
0
dE e−iE(t−t )
0
G(t − t) =
2π −E 2 + ω 2
(2.75)
As always there is an ambiguity on the path of integration. We choose the Feynman
decomposition:
Z
0
dE e−iE(t−t )
G = lim
(2.76)
→0
2π −E 2 + ω 2 − i
Using this expression we can rewrite the amplitude in equation 2.71. The argument in
the exponential is
Z
Z
0
dE
eiEt e−iEt
i
0
0
dtdt f (t)f (t ) 2
(2.77)
(...) =
2
2π
E − ω 2 + i
Z
i dE
=
dtdt0 f (t)G(t − t0 )f (t0 )
(2.78)
2 2π
where we have used the symmetric property of the Green’s functions: G(t−t0 ) = G(t0 −t).
Applying this into our equation above we have
Z t0
i
0
0
0
dtdt f (t)G(t − t )f (t )
(2.79)
h0, t0 |0, −t0 i = exp
2 −t0
Alternatively we can express the amplitude using the definition,
Z T
Z
Z
Z
W [f (t)] ≡ lim
Dx exp i
dt(L̄0 + f x) = Dx exp i
T →∞
x(T )=x(−T )
−T
∞
dt(L̄0 + f x)
−∞
(2.80)
where L̄0 is the free Lagrangian with all the inserted. Using Eq. 2.55 we see that we
can write,
h0, t0 |0, −t0 i = lim CW [f ]
(2.81)
→0
14
CHAPTER 2. PATH INTEGRALS IN QUANTUM MECHANICS
if f = 0 then
h0, t0 |0, −t0 i = 1
⇒C=
1
W [0]
(2.82)
So in our new notation the amplitude is given by,
h0, t0 |0, −t0 i =
W [f ]
W [0]
(2.83)
where,
" Z
#
˜
˜
dE f (E)f (−E)
i
W [f ] = W [0] exp
2
2π −E 2 + ω 2 − i
Z
i
0
0
0
= W [0] exp
dtdt f (t)G(t − t )f (t )
2
(2.84)
(2.85)
since W [0] just cancels away in the calculation of a transition amplitude we don’t need
to calculate it explicitly.
2.2
Anharmonic Oscillator
Up until now the Hamiltonian of our system was the simply quadratic free Hamiltonian
with a time dependent perturbation. The eigenstates were left unchanged but the f (t)x
term allowed the possibility of a state transitioning into another state. The fact that we
were able to complete the square and exactly integrate the exponential was instrumental
in finding a simple form for the amplitude. This was a consequence of not having any
terms in the exponential that are more then quadratic in x. We now consider modifying
the Hamiltonian to include such interactions.
Consider the anharmonic oscillator with a driving force:
p 2 ω 2 x2
+
H0 =
2
2
Ha =
(2.86)
λx4
4!
(2.87)
and
H = H0 + Ha − f (t)x
(2.88)
We have some ground state, |Ωi. We add a constant to the Hamiltonian such that Eω = 0.
Our discussion earlier was rather general and we can reuse some of our earlier results
hΩ, t0 |Ω, −t0 i =
where
Z
Wλ [f ] ≡
Z
Dx exp i
∞
−∞
Wλ [f ]
‘
Wλ [0]
λx4
dt(L̄0 − (
)(1 − i) + f (t)x)
4!
(2.89)
(2.90)
2.2. ANHARMONIC OSCILLATOR
15
Due to the quartic term we cannot simply do this integral as before. There is not general
way to integrate this Gaussian. What we can do, is do a small λ expansion (perturbation
theory).
The 1 − i is important in the initial Lagrangian (L̄0 ) to pick out the ground state
and relate W to the amplitude of transition from the ground state. However it turns out
that the 1 − i has no effect on the anharmonic term. [Q 2: Check Peskin.]. Thus we can
take the that’s there to zero. We do a small λ expansion (i.e. Taylor expansion):
Z ∞
Z
Z
−iλ ∞
4
dt(L̄0 − f x) exp
dtx
(2.91)
Wλ [f ] = Dx exp i
4! −∞
−∞
Z ∞
Z
Z
iλ
≈ Dx exp i
dt(L̄0 − f x) 1 −
dtx4 (t)
4!
−∞
2 Z
λ
4
4
+
dt1 dt2 x (t1 )x (t2 ) + ...
(2.92)
(4!)2
We can write this discretely as
Wλ [f ] ≈
N
−1 Z
Y
"
dxi exp i
X
n
i=1
#
iλ X 4
∆t
xj
4!
j
X
λ2
2
4 4
+
(∆t)
xj1 xj2 + ...
(4!)2
j ,j
∆t(L̄0,n − fn xn )
1−
1
(2.93)
2
or compactly as
Z ∞
Z
Z
iλ ∞
4
dt Dx x (t) exp i
dt(L̄0 + f x)
Wλ [f ] = W0 [f ] −
4! −∞
−∞
2 Z
Z
Z
(−i)2 λ
4
4
+
dt1 dt2 Dx x (t1 )x (t2 ) exp i
dt(L̄0 + f x) + ...
2!
4!
(2.94)
First consider W0 [f ]. We already know this function since we calculated it last class (we
called it W [f ] before we had a perturbation):
Z ∞
Z
dt(L̄0 + f x)
(2.95)
W0 [f ] = Dx exp i
−∞
" N
#
N Z
Y
X
=
dxi exp i
∆t(L̄0 j + fj xj )
(2.96)
i=1
j=1
where fj = f (−T + j∆t) and xj = x(−T + j∆t). We know how to do this integral
explicitly (see Eq. 2.85). We hope to rewrite the other terms in terms of W0 [f ]. To do
this we define something called a “functional derivative”:
δW0
1 ∂W0
≡
δf (tj 0 )
∆t ∂fj 0
(2.97)
16
CHAPTER 2. PATH INTEGRALS IN QUANTUM MECHANICS
In our case this is given by
" N
#
N Z
Y
X
δW0
=
dxi ixj 0 exp i
∆t(L̄0,j + fj xj )
δf (tj 0 ) i=1
j=1
Z
Z
= i Dx x(tj 0 ) exp i
dt(L̄0 + f x)
(2.98)
(2.99)
So at least in this case taking the functional derivative of W0 with respect to f at a time
tj 0 does the thing you would expect; it pulls out a factor of ix(j 0 ).
With this in mind we can write
4
Z
1 δ
iλ ∞
dt
W0 [f ]
Wλ [f ] = W0 [f ] −
4! −∞
i δf (t)
2 Z
4 4
(−i)2 λ
1 δ
1 δ
+
W0 [f ] + ...
(2.100)
dt1 dt2
2!
4!
i δf1
i δf2
We begin by considering O(λ):
4
Z
Z
1 δ
i
iλ
0
0
0
exp
dτ dτ f (τ )G(τ − τ )f (τ )
− W0 [0] dt
4!
i δf (t)
2
(2.101)
For brevity we write,
Z
i
0
0
0
exp [...] ≡ exp
dτ dτ G(τ − τ )f (τ )f (τ )
2
We now take the functional derivatives:
"
#
1 δ
1 ∂
iX
exp [...] =
exp
∆t∆tG(τi − τj )fi fj
i δf (t)
i∆t ∂fj 0
2 i,j
X
1 1 i 2 X
∆t (
G(τi − tj 0 )fi +
G(tj 0 − τj )fj ) exp [...]
=
i ∆t 2
i
j
X
X
1
= ∆t(
G(τi − tj 0 )fi +
G(tj 0 − tj )fj ) exp [...]
2
i
j
X
=
∆tG(tj 0 − τj )fj exp [...]
(2.102)
(2.103)
(2.104)
(2.105)
(2.106)
j
Z
=
dτ 0 G(t − τ 0 )f (τ 0 ) exp [...]
(2.107)
So again the functional derivative does what one would expect. It takes the derivative of
the arguement of the exponential and uses the product rule on it. Continuing on gives,
2
Z
1
1 δ
exp ... = G(t − t) exp ... + dτ 0 dτ 00 G(t − τ 0 )G(t − τ 00 )f (τ 0 )f (τ 00 ) exp ...
i δf (t)
i
(2.108)
2.2. ANHARMONIC OSCILLATOR
1 δ
i δf (t)
1 δ
i δf (t)
17
3
Z
Z
1
2
0
0
0
exp ... =
G(t − t) dτ G(t − τ )f (τ ) + G(t − t) dτ 0 G(t − τ 0 )f (τ 0 )
i
i
Z
0
00
000
0
00
000
0
00
000
+ dτ dτ dτ G(t − τ )G(t − τ )G(t − τ )f (τ )f (τ )f (τ ) exp [...] (2.109)
Z
3
=
G(t − t) dτ 0 G(t − τ 0 )f (τ 0 )
i
Z
0
00
000
0
00
000
0
00
000
+ dτ dτ dτ G(t − τ )G(t − τ )G(t − τ )f (τ )f (τ )f (τ ) exp [...] (2.110)
4
Z
3
− 3G (t − t) + G(t − t) dτ 0 dτ 00 G(t − τ 0 )G(t − τ 00 )f (τ 0 )f (τ 00 )
i
Z
3
+ G(t − t) dτ 0 dτ 00 G(t − τ 0 )G(t − τ 00 )f (τ 0 )f (τ 00 )
(2.111)
i
Z
+ dτ 0 dτ 00 dτ 000 dτ 0000 G(t − τ 0 )G(t − τ 00 )G(t − τ 000 )G(t − τ 0000 )
0
00
000
0000
× f (τ )f (τ )f (τ )f (τ ) exp [...]
(2.112)
Z
6
2
= − 3G (t − t) + G(t − t) dτ 0 dτ 00 G(t − τ 0 )G(t − τ 00 )f (τ 0 )f (τ 00 )
i
Z
+ dτ 0 dτ 00 dτ 000 dτ 0000 G(t − τ 0 )G(t − τ 00 )G(t − τ 000 )G(t − τ 0000 )
0
00
000
0000
× f (τ )f (τ )f (τ )f (τ ) exp [...]
(2.113)
exp ... =
2
We now consider Wλ [0] (i.e. the above expression with f = 0). This is the normalization term for every amplitude.
Z ∞
3λ
Wλ [0] = W0 [0] + i W0 [0]
G2 (t − t) dt
(2.114)
4!
−∞
and in summary we have,
Z
iλ ∞
Wλ [f ] = W0 [f ] − W0 [f ]
dt − 3G2 (t − t)
4! −∞
Z
6
+ G(t − t) dτ 0 dτ 00 G(t − τ 0 )G(t − τ 00 )f (τ 0 )f (τ 00 )
i
Z
+ dτ 0 dτ 00 dτ 000 dτ 0000 G(t − τ 0 )G(t − τ 00 )G(t − τ 000 )G(t − τ 0000 )
0
00
000
0000
× f (τ )f (τ )f (τ )f (τ )
(2.115)
18
CHAPTER 2. PATH INTEGRALS IN QUANTUM MECHANICS
Recall that the transitions amplitudes are,
h0, −T |0, T i =
Wλ [f ]
Wλ [0]
(2.116)
So to first order the transition amplitude is,
Z
Z
iλ ∞
6
2
h0, −T |0, T i = 1 −
dt − 3G (t − t) + G(t − t) dτ 0 dτ 00 G(t − τ 0 )G(t − τ 00 )
4! −∞
i
Z
× f (τ 0 )f (τ 00 ) + dτ 0 dτ 00 dτ 000 dτ 0000 G(t − τ 0 )G(t − τ 00 )G(t − τ 000 )G(t − τ 0000 )
0
00
000
0000
× f (τ )f (τ )f (τ )f (τ )
(2.117)
This calculation was quite tedius and was only to first order. Nevertheless there are
two main independent ingredients that we can identify. The order of the interaction
(which is always accompanied by an integration of t) and the Green’s function. You may
think that the factors f (τ )dτ are also independent but that’s not true. The number of
these factors and their arguements are competely determined by the form of the Green’s
functions. Furthemore, as we will see for our purposes we will take these to zero at the
end of the calculations. So we can now identify the pieces that will always be part of the
transition amplitude calculations. These will be our “Feynman Rules”:
→ −iλ
R∞
−∞
dt
→ G(t1 − t2 )
Furthermore, a time will be given to each intersection point in a diagram and a symmetry
factor is also required for each diagram. So for example we can write,
Wλ [0] = W0 [0] 1 +
2.3
+
+
+
+...
Correlation Functions
Now that we have an idea of how to generalize our discussion to more complicated potentials we quickly go back again to consider the simple harmonic oscillator. Consider
the following derivative with t1 < t2 :
1 δ 2 W0 [f ] = W0 [0] G(t1 − t2 )
(2.118)
i2 δf (t1 )δf (t2 ) f →0
2.3. CORRELATION FUNCTIONS
19
Alternatively we can rewrite this expression in terms of the Lagrangian,
Z
Z
1 δ 2 W0 = Dx x(t1 )x(t2 ) exp i
dtL̄0
i2 δf1 δf2 f →0
Z
Z t1
Z T
Z
Z t2
Dx x2 x1
Dx dx1
= lim
Dx dx2
T →∞ t
−T
t1
2
Z t1 Z t2 Z T dtL̄
+
+
× exp i
−T
Z
T
Dx
= lim
T →∞
Z
t2
Z
t2
Z
=
t1
Dx
Dx dx1 x1
−T
Z t1 Z t2 Z T dtL̄
+
+
× exp i
dx2 x2
t1
−T
Z
(2.120)
t2
t1
Z
(2.119)
t1
(2.121)
t2
Z
dx1
dx2 x1 x2 h0, T |x2 , t2 i hx2 , t2 |x1 , t1 i hx1 , t1 |0, −T i W0 [0] (2.122)
where in the second line we have split the integration interval into the different parts.
This is trivial in the exponential. However, on the outside extra spatial integrals over
dx1 and dx2 appear due to the way the Dx are defined; they have the end points fixed
so to split them up you need to have an integral over each midpoint.
We now switch to the Heisenberg picture. We choose our reference time −T . By
definition then we have
|0i ≡ |0, −T i
(2.123)
likewise we define
|xi ≡ |x, −T i .
(2.124)
With this reference point we can move into the Heisenberg interperpotation. The Heisenberg form of the position operator is, x̂(t) = e−iH0 (t+T ) x̂(−T )eiH0 (t+T ) . We want to think
of x1 as an eigenvalue of this operator. In other words
xi |xi i = x̂(ti ) |xi i
This enables us to get rid of the x integrals:
1 δ 2 W = W0 [0] h0|x̂(t2 )x̂(t1 )|0i
i2 δf1 δf2 f →0
(2.125)
(2.126)
This expression holds if t2 > t1 . If t1 > t2 everything is the same by symmetry but
t1 ↔ t2 and so we have,
1 1 δ 2 W = h0|T (x̂(t2 )x̂(t1 )) |0i
(2.127)
W0 [0] i2 δf1 δf2 f →0
where T denotes the time ordering operator which puts the operator with the earliest
time to the right. So we see that correlation functions (key objects in Quantum Field
20
CHAPTER 2. PATH INTEGRALS IN QUANTUM MECHANICS
Theory!) can be related to W by taking derivatives. In the case of the anharmonic
oscillator we have already taken the first derivative of W0 [f ] (see Eq 2.107) we just need
to take a second at a different time,
1 1 δ
h0|T (x̂(t2 )x̂(t1 )) |0i =
W0 [0] i δf1
1
= G(t2 − t1 )
i
Z
dτ G(t2 − τ )f (τ ) exp [...] 0
0
0
(2.128)
f →0
(2.129)
Our discussion did not depend much on the form of the Hamiltonian. If we had Eq
2.113 with or without λ = 0 the results would be the same as long as we label our ground
state by |Ωi (the ground state of the full system (in our case the anharmonic theory),
1 δ 2 Wλ [f ] hΩ|T (x̂(t1 )x̂(t2 )) |Ωi =
Wλ [0] i2 δf1 δf2 f →0
1
1 δ2
=
Wλ [0] i2 δf1 δf2
1
iλ
W0 −
4!
(2.130)
Z
dt
1 δ
i δf
4
!
W0 + ... f →0
(2.131)
We already sketched out Wλ [0]. Calculating
δ 2 Wλ δf1 δf2 f →0
(2.132)
explicitly gives any possible diagram with two external points (two unconnected times, t1
and t2 ). We show this to first order. The expression for Wλ is given in Eq 2.113. Taking
its derivatives gives to first order,
Z
1 δ 2 Wλ iλ
1
=
G(t1 − t2 ) −
dt 3iG2 (0)G(t1 − t2 )
i2 δf1 δf2 f →0
i
4!
+ 12iG(0)G(t − t1 )G(t − t2 ) W0 [0]
(2.133)
The zero’th order contribution is a Green’s function starting at the initial point and
ending at the second (i.e. from t1 to t2 ). The first order contributions are the a double
bubble diagram (due to the G2 (0) factor) with the zero’th order Green’s function and the
second term is a bubble and Green’s function going in and one going out. Now to find the
amplitudes we need to divide by Wλ [0] to get the correlation functions. We calculated
this quantity earlier on and to first order is given by,
Z
3λi
2
Wλ [0] = W0 [0] 1 −
dtG (0)
4!
(2.134)
2.3. CORRELATION FUNCTIONS
21
So to first order we have,
1
G(t1
i
λ
4!
R
dt3G2 (0)G(t1 − t2 ) + 12G(0)G(t − t1 )G(t − t2 )
R
hΩ|T (x̂(t1 )x̂(t2 )) |Ωi =
dtG2 (0)
1 − 3iλ
4!
(2.135)
R
R
1
iλ
iλ
G(t1 − t2 ) + 4! 4G(0)G(t − t1 )G(t − t2 ) 1 − 4! dt3G2 (0)
i
R
=
dtG2 (0)
1 − 3iλ
4!
(2.136)
Z
1
iλ
= G(t1 − t2 ) +
4G(0)G(t − t1 )G(t − t2 )
(2.137)
i
4!
To all orders we have,
+
− t2 ) +
+
=
+
+
1 +
+
+
+
+
+ ...
+ ...
+ ...
We now briefly summarize what we have done. We have the correlation functions for
anharmonic oscillator (x4 ):
X
1 δ 2 Wλ [f ] hΩ|T (x̂(t1 )x̂(t2 )) |Ωi =
=
connected Feynman diagrams
Wλ [0] δf1 δf2
f →0
(2.138)
where Wλ [f ] is the generating functional and fi ≡ f (ti ). We found the propagator (to
lowest order) is given by,
R
e−iE(t1 −t2 )
→ G(t1 − t2 ) = dE
2π −E 2 +ω 2 +i
and the vertex is given by
x
R
→ −iλ dt
R
→ −iλ dt. These rules are the space time Feynman rules. By Fourier transforming both sides we have the momentum space rules:
→
1
−E 2 +ω 2 +i
P
→ −iλδ( i Ei )
22
CHAPTER 2. PATH INTEGRALS IN QUANTUM MECHANICS
2.A
Jacobian of the Fourier Transformation
The discrete Fourier transform is given by
1 X −2πink/N
x̃k = √
e
xn
N n
(2.139)
The Jacobian is then the matrix
Jn,k =
1
∂ x̃k
= √ e−2πink/N
∂xn
N
(2.140)
We want to show that the absolute value of the determinate of this matrix is equal to 1.
Note that the determinate any orthonormal matrix is ±1 since:
1 = det A−1 A = det AT det A = (det A)2
(2.141)
thus if we can show that J is orthonormal then |det J| = 1.
A matrix is orthonormal if its rows and columns are orthonormal. We need to show
that


e−2πij/N
−2πi2j/N 

1 2πik/N 2πi2k/N

 e
(2.142)
hJk |Jj i =
e
,e
, ..., e2πik 

..
N


.
e−2πij
1 2πi(k−j)/N
=
e
+ e2πi2(k−j)/N + ...e2πi(k−j)
(2.143)
N
is equal to 0 for all row and column numbers, k − j 6= 0. Let k − j = m ∈ Z. Then we
have,
N
1 X 2πijm/N
=
e
N j=1

N/2m
1  X 2πijm/N
e
+
=
N
j=1
=
1
N
2m−2
X (`+1)N/2m
X
(2.144)
2N/2m
X
N
X
e2πijm/N + ... +
j=N/2m+1

e2πijm/N 
(2.145)
j=(2m−1)N/2m+1
e2πijm/N
(2.146)
`=0 j=`N/2m+1
shifting the exponential such that each sum starts at j = 1 gives,
(`+1)N/2m
X
j=`N/2m+1
N/2m
2πijm/N
e
`πi
→e
X
j=1
e2πijm/N
(2.147)
2.A. JACOBIAN OF THE FOURIER TRANSFORMATION
23
Thus we finally get that the dot product between an arbitrary row and column is given
by,


!
N/2m
2m−2
X
X
1 
=
e2πijm/N 
e`πi
(2.148)
N
j=1
`=0
|
{z
}
=0
=0
(2.149)
There is one loophole. If k − j = m = 0 then we can’t divide by m. In this case we get,
N
N
1 X 2πijm/N
1 X
e
=
=1
N j=1
N j=1
(2.150)
Thus hJj |Jk i = δjk and the Jacobian is indeed orthonormal and thus
|det J| = 1
as required.
(2.151)
Chapter 3
Quantum Field Theory
3.1
Free Klien Gordan Field
To keep things simple we consider one non-interacting real scalar field (real Klien Gordan
theory). We work in 1 + 1 dimensions.
Since we are working in 1 dimension we have
discretize
space
φ(x, t) −−−−−→ φ = {φ1 (t), φ2 (t), ..., φN (t)}
(3.1)
x
`
z}|{
√
√
where φ` (t) = φ( `∆ , t) ∆ and the ∆ is a normalization factor. The equations of
motion are
φ + m2 φ = 0
(3.2)
To discretize them we need to find the second derivative in terms of finite differences. We
have,
∂
φ(x + h/2) − φ(x − h/2)
∂ 2φ
=
lim
2
∂x
∂x h→0
h
1
= lim 2 (φ(x + h) − 2φ(x) + φ(x − h))
h→0 h
1
→ 2 (φ`+1 − 2φ` + φ`−1 )
∆
So the discrete equations of motion are,
2
1
2
φ̈` +
+
m
φ
−
(φ`+1 + φ`−1 ) = 0
`
∆2
∆2
R
Since we are working in one spatial dimesion we have L = dxL with
1
1
L = (∂µ φ)(∂ µ φ) − m2 φ2
2
2
24
(3.3)
(3.4)
(3.5)
(3.6)
(3.7)
3.1. FREE KLIEN GORDAN FIELD
25
where our metrix is
g
µν
=
1 0
0 −1
(3.8)
Note that this Lagrangian is equivalent to having N of Harmonic oscillators (no sources
or interaction terms) with one at each ` value. Discretizing our Lagrangian gives
!
2
N
X
φ
−
φ
1
`+1
`
L=
φ̇2` −
− m2 φ2`
(3.9)
2
∆
`=1
=
N
X
1
`=1
where
2
φ̇2`

..



V ≡



0
.
N
X
1
−
φ` V`m φm
2
`,m=1
..
− ∆22
.
− m2
0
..
.
0
..
.
0
0
0
... 0
1
0 0
∆2
.. ..
. 0
.
... ...
0
0 0
(3.10)








(3.11)
is a coupling matrix between the fields at different spatial points.
Note the following oscillator correspondence:
Klien-Gordan
t
x
φ`
φ̂` (t)
J (x,t)
NR oscillator
t
label `
x` (t)
x̂` (t)
f` (t)
(for the time being we use hats to denote operators). We work in the Schrodinger picture:
n
o
φ̂ = φ̂1 , ..., φ̂N
(3.12)
√
and limN →∞ φ̂/ ∆ = φ̂(x). A definite field state is given by |φ, ti :
φ̂ |φ0 , ti = φ0 |φ0 , ti
(3.13)
we are interested in a transition between two states hφf , T |φi , −T i. Each state consists
of a direct product of states, one at each point in space. Thus finding the braket between
two field configurations is equivalent to a product of brakets. In other words we can
write,
hφf , T |φi , −T i = hφ01 , T |φ1 , −T i hφ02 , T |φ2 , −T i ... hφ0N , T |φN , −T i
(3.14)
26
CHAPTER 3. QUANTUM FIELD THEORY
with hφ0` , T |φ` , −T i being the product of oscillator states at ∆`. But we already know
what the value of each of these brakets is for a singlet oscillator. Now we are doing the
same calculation but for many oscillators. It is given by (c.f. equation 2.32)
" M −1
#
Z
N
−1 M
−1
X
Y
Y
dφ`,j exp i
∆tL [φn ]
(3.15)
hφf , T |φi , −T i = C
n=1
`=1 j=1 φ =φ
`,0
`i
φ`,M =φ`f
where the label ` refers to spatial points and the j denotes to time slices. Furthermore,
we have introduced the notation dφ`j ≡ dφ` (tj ). We now take the limit of N → ∞ and
M → ∞:
Z
Z
Z
∞
T
hφf , T |φi , −T i = C
Dφ exp i
dt
where
Z
Dφ ≡
N
−1 M
−1
Y
Y
dxL [φ(x, t)] ,
(3.16)
−∞
T
dφi,j .
(3.17)
`=1 j=1
The interpretation is similar to before. The amplitude of going from one field configuration to another is equal to the sum over all possible field configurations. Note the
extension to multidimensional derivation is completely analogous. For 3 spatial dimensions:
Z T
Z
Z ∞
3
hφf , T |φi , −T i = C Dφ exp i
dt
d xL [φ(x, t)]
(3.18)
−T
3.2
−∞
External Force
√
Consider some one dimensional external force, J(x, t) which discretized is given by J` (t) ∆
(one force for each point and since this is just a field it discretizes in the same way as
the field φ). We take the force to turn on at −t0 and end at t0 . We want to compute the
probability for the system to stay in the ground state,|0i. We take the energy of the state
to be 0 (E0 = 0). Everything we did for a single oscillator can be done for N oscillators.
Each of these oscillators has a force which is dependent on time. We denote the force by
(J = {J1 (t), J2 (t), ..., JN −1 (t)}) (here Ji (t) is the f (t) for each oscillator).
Y W [Ji ]
W0 [J(t)]
≡ lim
(3.19)
h0, t0 |0, −t0 i = lim
→0 W0 [0]
→0
W
[0]
0
i
where (c.f. Eq. 2.55)
W0 [J(t)] = lim
T →∞
N
Y
Z
Z
Dφj exp i
T
−T
j=1
φj (−T )=
φj (T )
+
N
X
`=1
X
N
N
X
1 2
1
dt
φ̇` (1 + i) −
(1 − i)φ` V`n φn
2
2
`=1
`,n=1
φ` J` (t)
(3.20)
3.2. EXTERNAL FORCE
27
here we have added the 1 + i and 1 − i factors in an analagous way to our discussion
of Non-relativistic QM. While they are present for notational convenience we will just
ignore them. More typically we will write
Z
Z
N
−1
Y
lim
Dφ` → Dφ
(3.21)
T →∞
j=1
φj (−T )=
φ` (T )
and dt dx → d2 x such that
Z
Z
1 2 2
2 1
µ
µ
µ
W0 [J(t)] = Dφ exp i d x (∂µ φ)(∂ φ) − m φ + J(x )φ(x )
2
2
(3.22)
As before then if we can do the Gaussian integral then we are set. The way you do it is
to diagonalize the argument so you just a product of individual integrals. Since the term
which gives us trouble connects different space points (the derivative) we can diagonalize
it using a Fourier transform,
Z
Z
dk 1 ˙ 2 1 2 2 1 2 2 ˜∗
φ̃ − k φ̃ − m φ̃ + J φ̃
W0 [J(t)] = C Dφ̃ exp i dt
(3.23)
2π 2
2
2
Z
Z
dk 1 ˙ 2 1
2 2
∗
φ̃ − ω(k) φ̃ + J˜ φ̃
= C Dφ̃ exp i dt
(3.24)
2π 2
2
where
Z
dk −ikx
φ(x, t) =
e
φ̃(k, t) ,
2π
ω(k)2 ≡ k 2 + m2
Z
J(x) =
dk −ikx ˜
e
J(k, t)
2π
(3.25)
and we have changed our integration measure using,
N
−1 Z
Y
`=1
Dφ` =
N
−1 M
−1 Z
Y
Y
dφ`m = J
`=1 m=1
N
−1 M
−1 Z
Y
Y
dφ̃j,m
(3.26)
j=1 m=1
where J is a Jacobian which we can forget about since it just adds to the overall constant.
Thus we can write
N
−1 Z
N
−1 Z
Y
Y
Dφ` →
Dφ̃j
(3.27)
`=1
j=1
Now we have (we don’t need to worry about the factor of 2π as it goes to the constant
in front),
Z ∞
N Z
Y
1 ˙ 2 1 2
2
∗ ˜
W0 [J(t)] = C
Dφ exp i
dt
|φ̃k | − ωk |φ̃k | + φ̃k Jk
(3.28)
2
2
−∞
k=1
=C
N
Y
k=1
h
i
W0SHO J˜k (t), ωk
(3.29)
28
CHAPTER 3. QUANTUM FIELD THEORY
where W0SHO is a quantity that we already calculated earlier,
Z
i
SHO
0
0
0
W0
[f, ω] = exp
dt dt f (t)G(t − t )f (t )
2
#
" Z
dE f˜(E)f˜(−E)
i
= exp
2
2π −E 2 + ω 2 − i
(3.30)
(3.31)
We can finally write
"
N
iX
W0 [J] = W0 [0] exp
2 k=1
Z
dE J¯k (E)J¯k (−E)
2π −E 2 + ωk2 − i
#
(3.32)
R
R
P
ikj
−iEt
where J¯k (E) = dtJ˜k (t)e−iEt = dt N
. We have xj = ∆j and we
j=1 e Jj (t)e
define the Fourier conjugate variable to x as p and it must be given by k/∆. Taking the
continuum limit gives
Z
Z
√ −iEt
dx ikj ¯
Jk (E) =
dt
(3.33)
e
J(x, t) ∆ e
∆
Z
1
=√
dt dxJ(x, t)ei∆jp e−iEt
(3.34)
∆Z
1
d2 xJ(x, t)e−i(Et−xp)
(3.35)
=√
∆Z
1
µ
=√
d2 xJ(x, t)e−ipµ x
(3.36)
∆
1 ˜
E)
(3.37)
≡ √ J(p,
∆
R dp
P
˜
˜
now k → ∆ 2π
and (Note: We got J(−p)
by relating J(−p)
to J˜∗ (p).
" Z
#
Z
˜
˜
i
dp
dE
J(p)J(−p)
W0 [J] = W0 [0] exp
(3.38)
2
2
2π
2π −E + p2 + m2 − i
In 3 + 1 dimensions we have
"
i
W0 [J] = W0 [0] exp −
2
Z
˜ J(−p)
˜
d4 p J(p)
(2π)4 p2 − m2 − i
#
we can equally well write this equation as
Z
i
4
4 0
0
0
W0 [J] = W0 [0] exp −
d x d x J(x)G(x − x )J(x )
2
(3.39)
(3.40)
where the Green’s function is defined by (x + m2 − i)G(x − x0 ) = iδ 4 (x − x0 ) ⇒
Z
0
d4 p e−ip·(x−x )
0
G(x − x ) =
(3.41)
(2π)4 p2 − m2 + i
3.3. INTERACTING THEORY
3.3
29
Interacting Theory
Suppose we turn on an interaction term H =
Z
Wλ [J] = Dφ exp i
Z
iλ
= W0 [J] −
4!
Z
iλ
= W0 [J] −
4!
Z
iλ
= W0 [J] −
4!
Z
d3 x 4!λ φ4 (x) →
R
λ 4
` 4! φ` .
P
We obtain
λφ4
d x L̄0 −
+ J(x)φ
4!
Z
Z
4
4
4
d x Dφ φ (x) exp i
d x L̄0 + Jφ + ...
4
Z
Z
1 δ
4
4
d x Dφ
exp i
d x L̄0 + Jφ + ...
i δJ(x)
4
1 δ
4
W0 [J] + ...
dx
i δJ(x)
4
This equation has the exact same form as for a single harmonic oscillator.
We have
Z
iλ
Wλ [0] = W0 [0] −
d4 x − 3G(x − x)2 + ...
4!
(3.42)
(3.43)
(3.44)
(3.45)
(3.46)
As before every term is made up of two types of terms. We have
→ −iλ
x1
R
dx
x2 → G(x1 − x2 )
In Quantum Mechanics we had
δ 2 Wλ [f ] hΩ|T (x̂(t1 )x̂(t2 )) |Ωi =
Wλ [0] δf (t1 )δf (t2 ) f →0
1
(3.47)
In field theory we have [Q 3: show]
δ 2 Wλ [J] hΩ|T (φ̂(x1 )φ̂(x2 ))|Ωi =
Wλ [0] δJ(x1 )δJ(x2 ) J→0
1
(3.48)
(all we did was replace the quantum mechanics source to the field theory source). One
can then develop perturbation theory for fields as we did for the harmonic oscillator. We
get
hΩ|T (φ̂(x1 )φ̂(x2 ))|Ωi =
+
+
+ ...
30
CHAPTER 3. QUANTUM FIELD THEORY
3.4
LSZ Reduction
[Q 4: Maybe insert a proof here instead of just quoting the final result.]
The correlation functions are defined as,
(n)
G (x1 , ..., x2 ) ≡ hΩ|T φ̂(x1 )...φ̂(x2 ) |Ωi
(3.49)
The Fourier transform is
G̃(n) (p1 , ..., pn ) =
Z
n
Y
!
d4 xi eipi xi
G(n) (x1 , ..., xn )
(3.50)
i=1
The LSZ formula says that
(n)
G̃
(p1 , ..., pn ) =
n
Y
j=1
!
√
i Z
(out hp3 ...pn |p1 p2 iin + permutations)
p2j − m2 + i
+ (terms with < n poles)
(3.51)
(3.52)
where the product is the product over the external lines. The bottom line is once you
know that correlators you can then get all the matrix elements. The scattering problem
is reduced to a problem of calculating correlators.
3.5
Statistical Mechanics and Path Integrals
Suppose you have a harmonic oscillator that’s in equilibrium with it’s environment. We
want to treat this oscillator quantum mechanically. If you studied quantum statistical
mechanics then you know that the fundamental object in statistical mechanics is the
partition function which is defined as,
h
i X
−β Ĥ
Z = tr e
=
hn|e−β Ĥ |ni
(3.53)
n
where β = 1/kB T . Let’s define β = it0 (since β is real we know that t0 is imaginary).
With this we know that
X
Z=
hn|e−iĤt0 |ni
(3.54)
n
=
X
U (n, t = 0 → n, t = t0 )
(3.55)
n
Where U (n, t = 0 → n, t = t0 ) is the transition amplitude for a state to go to itself after
time t0 . Using our path integral formalism we can write this as
Z t0
Z
(3.56)
Dx exp i
dtL 0
x(0)=x(t0 )
i
BT
t0 →− k
In other words computing the partition function at some temperature is given by a path
integral with imaginary time.
Chapter 4
Quantization of Electromagnetism
4.1
Classical Electromagnetism
The electromagnetic (EM) field is
Aµ (x) = (φ(x), A(x))
(4.1)
with E = −∇φ − A and B = ∇ × A. The field strength tensor is defined as
Fµν = ∂µ Aν − ∂ν Aµ


0
Ex
Ey
Ez
 −Ex
0
Bz −By 

=
 −Ey −Bz
0
Bx 
−Ez By −Bx
0
(4.2)
(4.3)
The gauge symmetry is the observation that there is some redundancy in our description.
The physics is given by the electric and magnetic fields. One can show that the only
transformation you can do that won’t change the EM fields is
Aµ → Aµ + ∂µ ξ(x)
(4.4)
where ξ(x) is an arbitrary function. Under this gauge symmetry,
Fµν → Fµν
(4.5)
We want to get the equations of motion for this system (Maxwell equations) using the
Lagrangian density.One possibility is to make the Lagrangian out of Fµν . We want to
have a gauge invariant Lagrangian so we use Fµν . To make a Lorentz scalar out of Fµν we
take the product of two Fµν (Alternatively we could take the trace, but Fµν is traceless).
This gives us two possible terms
1
Lgeneral = − Fµν F µν + µναβ F µν F αβ
4
31
(4.6)
32
CHAPTER 4. QUANTIZATION OF ELECTROMAGNETISM
These are the only two non-renormalizable terms. It also turns out that the second term
violates parity. Since we know that electromagnetism does respect parity we don’t have
the second term in EM. Thus
1
L = − Fµν F µν
4
1
= − (∂µ Aν − ∂ν Aµ ) (∂ µ Aν − ∂ ν Aµ )
4
1
= − ∂µ Aν (∂ µ Aν − ∂ ν Aµ )
2
(4.7)
(4.8)
(4.9)
Note that a m2 Aµ Aµ term is not gauge invariant! So we don’t have such a term in our
Lagrangian. The Euler Lagrange equations give
δL
=0
δ∂µ Aν
Aν − ∂ν (∂ µ Aµ ) = 0
∂µ
(4.10)
(4.11)
Note that we have yet to make a gauge choice. We now fix the gauge to our convenience.
Pictorially this is given by
Aµ
Aµ + dξµ
gauge-fixing curve
gauge orbits
The gauge orbits are lines which show all the possible values of Aµ along some gauge
choice. All points along the gauge orbit correspond to physically equivalent systems. All
points along the gauge fixing curve correspond to the different values of Aµ with a single
gauge. One of the simplest gauge choices is the Lorentz gauge: ∂µ Aµ = 0.
The solutions of 4.11 in the Lorentz gauge are plane waves,
Z
Aµ =
d4 k
µ e−ik·x
(2π)4
(4.12)
with k 2 = 0 and µ is the polarization vector (and also the Fourier transform of the vector
field). These are just wave solutions with no mass. To make sure that we are still in the
4.1. CLASSICAL ELECTROMAGNETISM
33
Lorentz gauge we require,
∂ µ Aµ = 0
Z
(4.13)
4
d k µ −i(k·x)
ik µ e
=0
(2π)4
⇒ k·=0
(4.14)
(4.15)
by uniqueness of the Fourier transform. As an example consider a photon moving the z
direction:
kµ = (E, 0, 0, E)
(4.16)
Our when dotted into kµ must give zero. A basis of polarization vectors that obeys the
condition above is,
1
(1) = √ (0, 1, +i, 0)
2
1
(2) = √ (0, 1, −i, 0)
2
1
(3) = √ (1, 0, 0, 1)
2
(4.17)
(4.18)
(4.19)
(4.20)
(1) and (2) correspond to left and right handed circular polarized waves. The third
polarization vector corresponds to a longitudinally polarized wave. However we know
that such a wave does not exist! This extra possible state arose because we didn’t full fix
the gauge of our field. To see this consider some gauge transformed field Āµ . This gauge
transformed field also satisfies the Lorentz gauge if the field connecting the two gauge
fields obeys,
Āµ = Aµ + ∂µ ξ
⇒ξ = 0
(4.21)
(4.22)
To get rid of this extra freedom consider the component of the Fourier transform of ξ
that matches the wave. We have
Z
d4 q ˜
ξ(q)e−iq·x
(4.23)
ξ(x) =
(2π)4
˜
and we consider the particular contribution ξ(k)
with k 2 = 0. We want to know how
adding this ξ to Aν changes the solution. We have,
˜
0ν (k) = ν (k) + kν ξ(k)
(4.24)
0(3)
˜ = −(3)
By choosing ξ(k)
= 0. If one computed the magnetic and electric
ν we can get fields corresponding to this polarization they would turn out to be zero. This is an
unphysical solution. With these two conditions we can fully fix the gauge.
34
4.2
CHAPTER 4. QUANTIZATION OF ELECTROMAGNETISM
Quantization
To zeroth order we treat Aµ as a set of 4 fields. The most naive generalization of our
discussion on scalar fields is
Z
Z
µ
4
W0 [Jµ ] = DA exp i S̄ [A] + Aµ (x)J (x) d x
(4.25)
R 4
where DA = DA0 DA1 DA2 DA3 and S =
d x − 41 Fµν F µν . Note however that we
wrote a term Aµ (x)J µ (x) which does not appear gauge invariant. Applying a gauge
transformation on the source term gives,
Z
Z
Z
µ
4
µ
4
A Jµ d x → A Jµ d x + ∂ µ ξJµ d4 x
(4.26)
Z
Z
µ
4
= A Jµ d x − ξ∂ µ Jµ d4 x
(4.27)
If we restrict our attention to sources with no divergence, i.e. that obey ∂ µ Jµ = 0, then
the extra term is in fact gauge invariant.
We are going to want to evaluate this path integrals explicitly. At first sight this seems
feasible since the kinetic term is quadratic and our integral appears Gaussian. However
it turns out that there are small technical problems which we will discuss.
We take a short detour into how we treat photons with spin s using canonical quantization.
Asµ ∼ sµ e−ip·x
(4.28)
with p2 = 0 or equivalently, p0 = ± |p|. We have a restriction in the Lorentz Gauge on
our given by · p = 0 and 0 = 0. We have an added the superscript for the polarization
vector for spin, s = 1, 2. We found a particular solution. The general solution is a linear
combination of the solutions we found:
!
Z
2
d3 p X
ip·x As (p)sµ e−ip·x + As∗ (p)s∗
(4.29)
Aµ =
µ e
(2π)3 s=1
p0 =|p|
p0 =|p|
This looks identical to the KG field only we now have a spin index s and we have
polarization vectors.
One can go about performing the Canonical Quantization method:
1
As (p) → √ asp
2p0
,
As∗ (p) → √
1 s†
a
2p0 p
(4.30)
where a, a† are creation and annihilation operators respectively. We would then define
the vacuum, |0i such that
asp |0i
(4.31)
for all p, s. There is then a one-photon state given by
as†
p |0i = |p, si
(4.32)
4.3. PATH INTEGRALS IN QED
4.3
35
Path Integrals in QED
Recall that
Z
1
4
µν
µ
W0 [J ] = DA exp i
d x(− Fµν F +Aµ J )
(4.33)
4
|
{z
}
F term
where we restrict our attention to sources such that ∂µ J µ = 0. We can write the exponential argument as
Z
1
4
∂µ Aν (∂ µ Aν − ∂ ν Aµ )
(4.34)
F term =
dx −
2
Z
1
=
d4 x Aν (g µν − ∂ µ ∂ ν ) Aµ
(4.35)
2
Z
µ
where we have integrated by parts to play with the derivatives. Now (here we write the
Fourier Transform of the vector potential as Ã(k))
Z
d4 k ik·x
Aµ (x) =
e õ (k)
(4.36)
(2π)4
Thus we have
d4 k d4 k 0 i(k+k0 )·x
dx
e
Ãν (k 0 )(−k 2 g µν + k µ k ν )õ (k)
4
4
(2π) (2π)
Z
4
1
dk
=−
Ãν (−k)(k 2 g µν − k µ k ν )õ (k)
4
|
{z
}
2
(2π)
1
F term =
2
Z
4
Z
(4.37)
(4.38)
Dµν
The second term is given by
Z
4
µ
d xAµ J =
Z
d4 k
õ (k)J˜µ (−k)
(2π)4
Recall that for a scalar field the source term was,
Z
d4 k
˜
φ̃(k)J(k)
(2π)4
(4.39)
(4.40)
and the kinetic term was
Z
d4 k
φ̃(k)(k 2 − m2 + i)φ̃(−k)
4
(2π)
(4.41)
and we completed the square by changing variables to
−1
˜
φ̃0 (k) = φ̃(k) + k 2 − m2 + i
J(k)
(4.42)
36
CHAPTER 4. QUANTIZATION OF ELECTROMAGNETISM
With this in mind, we define a new à such that
Ã0µ = õ + (D−1 )µν J˜ν
(4.43)
and then our Lagrangian is given by,
1
1
L = Ã0µ Dµν Ã0ν − J˜µ (D−1 )µν J˜ν
2
2
So we have
(4.44)
4
ν d k
µ
−1
˜
˜
W0 [J ] = W0 [0] exp
(4.45)
J (D )µν J
| {z } (2π)4
Green’s function
where we have our Gaussian integral included in our definition of W0 [0]. Everything seems
to have worked flawlessly. The subtlety here is that Dµν is not invertible! To prove this
we study the eigenvalues of the D matrix. One such eigenvector is kν = (k0 , −k) with an
eigenvalue of zero:
k 2 g µν − k µ k ν kν = 0
(4.46)
ν
i
−
2
Z
Since the determinant is the product of the eigenvalues it must be zero, implies that D
isn’t invertible.
This problem arose because while we are integrating over the gauge fixed field we
are also integrating over gauge inequivalent fields. We haven’t done anthing to set these
contributions to zero. We need to find a way such that we integrate only about inequivalent fields. This corresponds to a single gauge-fixing curve (All the Aµ values for a single
gauge choice):
The procedure is called the Faddeev-Popov procedure. Suppose we have some field A0,µ .
We have a gauge transformed field parameterized by α that is given by,
Aα,µ = A0,µ + ∂µ α
(4.47)
We define a gauge transformation function as G [A]. For the Lorentz gauge we have
G [A] = ∂µ Aµ = 0
(4.48)
4.3. PATH INTEGRALS IN QED
37
We will also later use the Generalized Lorentz Gauge given by
G [A] = ∂µ Aµ − ω(x) = 0
(4.49)
In order to ensure we stay on our gauge we introduce a functional delta function (this
can be thought of as a product of delta functions, one at each spacetime point):
Z
Dαδ (G [Aα,µ ])
(4.50)
By inserting this into our path integral the path integral will be zero unless G [A] = 0.
However, we can’t insert this into the path integral unless it’s equal to 1.
Recall that if f (x) has one zero at x0 then,
Z
df (x) δ (f (x)) = 1
(4.51)
dx dx x=x0
We want to generalize this to instead of having f (x) we have, g(a) for vectors of arbitrary
size. To do this consider the Taylor expansion of g around its root (we assume it only
has one root, a0 ):
X ∂gi (aj − a0,j ) + ...
(4.52)
gi (a) = gi (a0 ) +
∂aj a0
j
We want to insert this into a delta function, δ (n) (g(a)). This will only be nonzero near
a = a0 . Thus we have,
Y
δ (g(a)) =
δ (gi (a))
(4.53)
i
!
=
Y
i
δ
X
Jij (aj − a0,j )
where Jij is the Jacobian matrix defined by Jij ≡
∂gi .
∂aj a0
We have,
!
!
δ (g(a)) = δ
X
(4.54)
j
J1j (aj − a0,j ) δ
j
X
J2j (aj − a0,j ) ...
(4.55)
j
We now use the identiy,
δ(α(a − a0 )) =
δ(a − a0 )
|α|
(4.56)
We choose to isolate each delta function in Eq. 4.55 for a different aj :
δ (g(a)) =
δ(a1 − a0,1 ) δ(a2 − a0,2 )
...
|J1,1 |
|J2,2 |
(4.57)
38
CHAPTER 4. QUANTIZATION OF ELECTROMAGNETISM
If we take the Jacobian matrix to be greater then zero then we have the product:
(J1,1 J2,2 ..)−1 =
1
det J
(4.58)
where we have used the fact that the determinant of J is independent of a unitary
transformation. So we finally have,
Z Y !
∂gi
(n)
dai δ (g(a)) det
=1
(4.59)
∂a
j
i
where it is understood that the Jacobian matrix is evaluated at the root of g.
We write the continuum generalization of this equation as,
Z
δG(Aα )
α
=1
Dα(x)δ (G(A )) det
δα
(4.60)
We consider the generalized Lorentz gauge: G(Aα ) = ∂µ Aµ + 1e ∂µ α(x) − ω(x). In this
case it is straight forward to calculate the determinate explicitly. The discretized gauge
transformation is (we work here in 1D just for brevity however, its easy to extend the
results to four dimensions.
Gi (αi ) =
1
(αi+1 − 2αi + αi−1 ) + (∂µ Aµ0 − ω(x))i
∆2 e
|
{z
}
Indepdent of αj
(4.61)
Taking the derivative we have,
1
∂Gi (αi )
δij = 2 (δi+1,j − 2δi,j + δi−1,j ) δij
∂αj
∆e
(4.62)
which doesn’t have any A dependence so it is already evaluated at the root of G. Thus
the determinant has no A or α dependence and we can write:
Z
Z
Z
δG
µ
α
4
µ
W0 [J ] = det
Dα DAδ (G [A ]) exp i S +
d xAµ J
(4.63)
δα
However J, DA, and S are gauge invariant thus we can change our equation to the gauge
shifted Aαµ :
Z
Z
Z
δG
µ
α
α
α
4
α µ
W0 [J ] = det
Dα DA δ (G [A ]) exp i S [A ] +
d xAµ J
(4.64)
δα
We now see that the α integral is trivial, it is just the volume of the α space (we include
the determinant in what we call V ol):
Z
Z
µ
4
µ
d xAµ J
(4.65)
W0 [J ] = V ol × DAδ (G [A]) exp i S [A] +
4.3. PATH INTEGRALS IN QED
39
where we have redefined our integration variable from Aα → A. We use the Generalized
Lorentz Gauge without α as we redefined, Aα → A. (we ignore the constant in front
since as we saw for a scalar field, it cancels away when calculating observables):
Z
Z
µ
µ
4
µ
(4.66)
W0 [J ] = DAδ (∂µ A − ω(x)) exp i S [A] +
d xAµ J
Z
Z
Z
R
2
−i d4 x ω2ξ
µ
4
µ
= N (ξ) Dωe
DAδ (∂µ A − ω(x)) exp i S [A] +
d xAµ J
|
{z
}
1
(4.67)
where ξ is just a constant and we are allowed to through in this integral since we know
our W0 is independent of ω (it is gauge invariant). Thus we have
Z
Z
Z
µ 2
4
µ
µ
4 (∂µ A )
+i
d x (L + Aµ J )
(4.68)
W0 [J ] = N (ξ) DA exp −i
dx
2ξ
where we have performed our ω integral (which is trivial due to the delta function). Thus
our “gauge-fixing procedure” is equivalent to taking L → L − 2ξ1 (∂µ Aµ )2 . Recall that we
had
Z
d4 k 1
S=
õ (−k) k 2 g µν − k µ k ν Ãν (−k)
(4.69)
4
(2π) 2
Adding this extra term we have (note that we can replace 2ξ1 (∂µ Aµ )2 → − 2ξ1 (Aµ ∂µ ∂ν Aν ))
Z
d4 k 1
1
2 µν
µ ν
Ãν (k)
(4.70)
S=
õ (−k) k g − k k 1 −
(2π)4 2
ξ
{z
}
|
Dξµν
We now have det Dξµν 6= 0. We finally have
Z
1
d4 k ˜µ
−1
µ
ν
˜
W0 [J ] = W0 [0] exp −
J (k)(iDξ )µν J (−k)
2
(2π)4
(4.71)
where
i
kµ kν
gµν − (1 − ξ) 2
≡ Gµν = − 2
(4.72)
k + i
k
This is known as the photon propagator. To see that this in indeed the inverse of Dµν
just multiply it out directly:
1
1
kµ kν
−1 µν
2
ν
Dξ
Dξ,νρ = 2 gµν − (1 − ξ) 2
k gνρ − k kρ 1 −
(4.73)
k
k
ξ
1
1
1
2 µ
µ
µ
µ
= 2 k δ ρ − k kρ 1 −
− (1 − ξ)k kρ + 1 −
(1 − ξ) k kρ
k
ξ
ξ
(4.74)
Dξ−1 µν
= δ µρ
(4.75)
40
CHAPTER 4. QUANTIZATION OF ELECTROMAGNETISM
Naively one would say that the propagator result depends on ξ.
How can this be
since we said that ξ is a mathematical convenience? Being more careful, one notices
˜ 2 . However these are zero by our choice of source
that all the ξ terms look like ξ(k · J)
µ
µ
(∂µ J = 0 ⇒ kµ J˜ = 0). Any choice of ξ is valid. There are a few popular choices. One
popular choice is ξ = 1. This is called the Feynman gauge. Another popular gauge is
called the Landau gauge (ξ = 0). Alternatively, you can keep ξ as a parameter and watch
it’s dependence drop at the end.
There is also something interesting in the sign of the photon propagator. The spatial
components of Aµ propagate with the same sign as the scalar field propagator, however
the time-like component has a relative negative sign. This is a consequence of the fact
that A0 doesn’t actually propagate since the momentum conjugate to A0 is zero:
1 ∂
∂L
=−
Fµν F µν
0
∂0 A
4 ∂0 A0
=0
4.4
(4.76)
(4.77)
Correlation Functions
Consider the correlation function:
h0|T µ (x1 )Âν (x2 ) |0i =
1 1
δ 2 W0 [J]
2
µ
ν
W0 [0] i δJ (x1 )δJ (x2 ) J→0
1
= Gµν (x1 − x2 )
i
(4.78)
(4.79)
However we constructed our J µ such that ∂µ J µ = 0. Having a δJ µ implies we can “wiggle”
J µ in any direction we want but in fact we have some restrictions. We now set out to
find what those are
Z
Aµ (x) =
d3 k õ (k)e−ik·x
(4.80)
and
Z
4
µ
Z
d xAµ J →
d4 k õ J˜µ
(4.81)
Now
Z
d4 x1 d4 x2 ... eik1 x1 ... h0|T (Aµ (x1 )...) |0i
δ
δ
∝
... W0 J˜ δ J˜µ (k) δ J˜ν (k)
˜
J→0
G̃(k1 , ...) =
(4.82)
(4.83)
where G̃ is the Fourier transform of the correlator. This can be done for any k. We take
k 2 = 0 with k = (E, 0, 0, E).
3
X
õ (k) =
sµ
(4.84)
s=0
4.4. CORRELATION FUNCTIONS
41
with a basis
1
0µ = √ (1, 0, 0, −1)
2
1
µ = (0, 1, 0, 0)
2µ = (0, 0, 1, 0)
1
3µ = √ (1, 0, 0, 1)
2
Now
J˜µ (k) =
X
J s sµ
(4.85)
and
kµ J˜µ = 0 ⇒ J 0 = 0
so
à · J˜ =
3
X
As J s sµ · sµ = A1 J 1 + A2 J 2
(4.86)
(4.87)
s=1
Thus what we can find out is the correlation function between A1 and A2 . We can only
know
h0|T A1 (k)... |0i
(4.88)
h0|T A2 (k)... |0i
(4.89)
With this we can write down the LSZ reduction formula for photons:
p
2
i Zγ X s
h(k, s) ...i +...
G̃µ (k, ...) = 2
k + i s=1 µ | {z }
S matrix
(4.90)
Chapter 5
Fermions with Path Integrals
Electrons are fermions, thus you can’t have more then one electron in a single state.
That’s why we never thought of electrons as fields until now. On the other hand photons
are bosons and they can all be in the same state. That’s why we had to think of them as
fields right from the start. However the concept of path integrals requires one to think of
fermion fields.
5.1
Grassman Numbers
It turns out the correct way to think about fermion fields is through the concept of
Grassman numbers, θ, η. They are not like ordinary numbers in many ways. However
they still obey rules of addition and multiplication. For Grassman numbers the addition is
typical (commutative). However the multiplication of such number is anti commutative.
i.e.
η+θ =θ+η
ηθ = −θη
(5.1)
(5.2)
θθ = −θθ
⇒θ2 = 0
(5.3)
(5.4)
f (θ) = a + bθ = a + θb
(5.5)
This implies that
Furthermore,
where a and b are ordinary numbers. Note that this is analogous to a Taylor expansion
however we have θ2 = 0 so all higher order terms vanish. These numbers are non-intuitive
and one needs to be careful when working with them.
You can define derivatives of Grassman numbers such as
df
=b
dθ
42
(5.6)
5.1. GRASSMAN NUMBERS
43
and integration
Z
Z
dθf (θ) =
dθf (θ + η)
(5.7)
The definition of the integral implies that (Exercise: use equation 5.5 to derive the
first equation. Maxim isn’t sure how to derive the second equation)
Z
dθ = 0
(5.8)
Z
dθθ = 1
(5.9)
This means that we have
Z
df
dθ
One can generalize this discussion to complex Grassman Numbers,
dθf (θ) = b
1
θ∗ = √ (θ1 + iθ2 ),
2
and
1
θ∗ = √ (θ1 − iθ2 )
2
Z
Z
dθθ = 1 =
θ∗ θ∗
As an example consider integration over a Gaussian
Z
Z
∗
−iθ∗ bθ
dθ dθe
= dθ∗ dθ (1 − θ∗ bθ)
Z
= dθ∗ dθ (θθ∗ b)
=b
This should be compared to the result for real numbers,
r
Z
2π
− 12 bx2
dxe
=
b
(5.10)
(5.11)
(5.12)
(5.13)
(5.14)
(5.15)
(5.16)
We now move on to multidimensional integrals. First consider the real variable integral:
!
Z
1X
dx1 dx2 ... exp −
xi Aij xj
(5.17)
2 i,j
Suppose that A can be diagonalized by some orthogonal transformation,
A = OT AD O
(5.18)
xT Ax = (Ox)T AD (Ox)
(5.19)
= x0T AD x0
(5.20)
which implies that,
44
CHAPTER 5. FERMIONS WITH PATH INTEGRALS
where we have defined, Ox ≡ x0 . The Jacobian for this transformation is,
Jij ≡
∂x0i
= Oij
∂xj
(5.21)
Thus we have,
Z
1X 0
= det O
dx1 ; dx2 ; ... exp −
xi AD,ii x0j
2 i
Z
Y
1
= det O
dxi exp − xi AD,ii xi
2
Y
N/2
−1
= det O(2π) ( AD,ii )
!
(5.22)
(5.23)
(5.24)
i
= (2π)N/2
det O
det A
(5.25)
We now want to find the analogous result for the Grassman integral,
Z
!
dn θdn θ∗ exp −
X
θi∗ Bij θj
(5.26)
i,j
We again want to tranform the integration variable and diagonalize B, however we need
to be a bit careful as we are working with Grassman numbers. A set of N Grassman
variables obey the relations:
1
i ...i θi θi ...θiN
N! 1 N 1 2
i1 i2 ...iN θ1 θ2 ...θN = θi1 θi2 ...θiN
θ1 θ2 ...θN =
(5.27)
(5.28)
These relations are a consequence of the anticommutating nature of the Grassman variables and are easy to show for N = 3 though I can’t figure out a general proof yet.
If we consider a unitary transformation on the Grassman variables: θi0 = Uij θj then,
Y
θi0 =
i
1 i1 i2 ...iN 0 0
θi1 θi2 ...θi0N
N!
1 i1 i2 ...iN
Ui1 ,i01 θi01 Ui2 ,i02 θi002 ...UiN ,IN0 θi0N
N!
!
Y
1 i1 i2 ...iN
=
Ui1 ,i01 Ui2 ,i02 UiN ,IN0 i01 ,i02 ,...,i0N
θj
N!
j
Y
Y
θi0 = det U
θj
=
i
j
(5.29)
(5.30)
(5.31)
(5.32)
5.2. FREE DIRAC FIELD
45
Thus when changing variables we pick up a determinate, analogous to the earlier situation
with the real variables. With this we have,
Y
dθi∗
i
Y
dθj = (det U )∗ det U
Y
j
dθi∗
Y
i
=
Y
dθi∗
Y
i
dθj
(5.33)
j
dθj
(5.34)
j
and so the integral is,
Z
!
dn θdn θ∗ exp −
X
θi∗ Bij θj
Z
=
!
dn θ0 dn θ0∗ exp −
i,j
X
0∗
θi BD,ij θj0
(5.35)
i,j
Z
=
!
dn θdn θ∗ exp −
X
θi∗ BD,ii θj
(5.36)
i
=
Y
BD,ii
(5.37)
i
= det B
5.2
(5.38)
Free Dirac Field
The Dirac field is given as


ψ1 (x)
 ψ2 (x) 

ψ(x) = 
 ψ3 (x) 
ψ4 (x)
(5.39)
To have Fermi Dirac statistics we make ψa (x) into a complex Grassman number valued
field.
The free Dirac field is given by
LD = ψ̄(i∂/ − m)ψ
(5.40)
with ψ̄ = ψ † γ 0 and ∂/ = ∂µ γ µ .
We want to produce our generating functional. We need to introduce sources. We
want something of the form of source multiplied by the Dirac field. We want to add a
term for the source that takes the form of a source multiplied by a scalar. A simplest
thing one can imagine is


η1 (x)
 η2 (x) 

η=
(5.41)
 η3 (x) 
η4 (x)
46
CHAPTER 5. FERMIONS WITH PATH INTEGRALS
where this term transforms in the same way as the Dirac spinor. The η source is also a
complex Grassman number valued field. Since we also want a Hermitian Lagrangian our
source takes the form
η̄ψ + ψ̄η
(5.42)
One can treat η and η̄ as independent as well as ψ, ψ̄ as independent. Our generating
functional is
Z
Z
4
W0 [η, η̄] = DψDψ̄ exp i
d x L̄D + η̄ψ + ψ̄η
(5.43)
Q
where Dψ ≡ 4a=1 Dψa = ψ1 ψ2 ψ3 ψ4 . The procedure is just as before. We now complete
the square in an analogous way to what we did earlier.
We have,
Z
Z
Z
d4 k d4 k 0 ikx −ik0 x
4
4
ψ̄(k) (−iγµ (ik µ ) − m) ψ(k)
e e
d x (...) = d x
(2π)4 (2π)4
+ η̄(k)ψ(k) + h.c.
(5.44)
Z
d4 k
=
ψ̄(k) (k/ − m) ψ(k) + η̄(k)ψ(k) + h.c.
(5.45)
(2π)4
Now we shift the fermion fields:
(k/ + m)η
− m2 + i
η̄(k/ + m)
ψ̄ → ψ̄ − 2
k − m2 + i
ψ→ψ−
k2
(5.46)
(5.47)
With this we have,
η̄(k/ + m)(k/ − m)(k/ + m)η
(k 2 − m2 )2
η̄(k/ + m)η
= ψ̄(k/ − m)ψ + 2
(k − m2 )
ψ̄(k/ − m)ψ + η̄ψ + h.c. → ψ̄(k/ − m)ψ +
So our path integral becomes,
Z
Z
η̄(k/ + m)η
4
W0 [η, η̄] = DψDψ̄ exp i
d k ψ̄(k/ − m)ψ + 2
k − m2
Z
η̄(k/ + m)η
= W0 [0] exp i
d4 k 2
k − m2
(5.48)
(5.49)
(5.50)
(5.51)
In real space,
Z
4
4
W0 [η, η̄] = W0 [0] exp − d x d y η̄(x)S(x − y)η(y)
(5.52)
5.3. QED
47
where S is the Green’s function, given by
(i∂/x − m14×4 )S(x − y) = iδ 4 (x − y)14×4
(5.53)
where 14×4 is the four by four unit matrix. Note that this is really four equations. Solving
this equation via Fourier transforms we get
Z
d4 k i(k/ + m)e−ik·(x−y)
S=
(5.54)
(2π)4 k 2 − m2 + i
We now move on to functional differentiation. One can show that
1 δ
−1 δ
1
ˆ
...
... W0 (5.55)
h0|T ψ̂(x1 )...ψ̄ (y1 )... |0i =
W0 [0] i δ η̄(x1 ) i δη(y1 )
η,η̄→0
where δδη̄ ... that we have one functional derivative for each fermionic field being time
ordered. The set of fields can be both Grassman fields and scalar fields. We have
δ
δ
† c 0
b
exp
i
ψ̄η
=
1
+
i(ψ
)
(γ
)
η
(5.56)
cb
δη a
δη a
= −i(ψ † )c (γ 0 )cb δ ba
(5.57)
= −iψ̄a
(5.58)
ˆ 2 ))|0i is antisymmetric since the η’s are Grassman
Further note that h0|T (ψ̂(x1 )ψ(x
numbers.
5.3
QED
Quantum Electrodynamics(QED) describes both spin 21 objects,“matter” (electrons, protons, muons, ...), and a spin-1 object “light” (photons). The most general Lagrangian is
given by
1
L = ψ̄(i∂/ − m)ψ − Fµν F µν + Lint
(5.59)
4
The interaction Lagrangian must be Lorentz invariant, and renormalizable. Renormalizability requires that each term must have dimensionality less than or equal to 4. By our
Lagrangian we know that [ψ] = 3/2 and [Aµ ] = 1. There are very few allowed terms.
We cannot have an odd number of ψ terms since then we won’t have a Lorentz invariant
Lagrangian. The only way to couple them to A0 s in a renormalizable way is to have 2 ψ
and 1 A. There are two potential bilinears we can use:
Aµ ψ̄γ µ ψ
Aµ ψ̄γ µ γ 5 ψ
(5.60)
(5.61)
The second term violates parity. Since we know that electromagnetism conserves Parity
we have,
Lint = eAµ ψ̄γ µ ψ
(5.62)
48
CHAPTER 5. FERMIONS WITH PATH INTEGRALS
where e is a dimensionless number. It will turn out that e is the electric charge of the
quantized ψ fields. One may wonder why the charge has no units. That is a result of
using ~ = 1. This can be seen since
α≈
1
e2 Natural Units e2
=
−−−−−−−−−−→
137
4π~c
4π
(5.63)
√
and hence e = 4πα ≈ 0.3.
We now derive the same result in a deeper way. The Dirac Lagrangian, LD has a
“global U (1)” symmetry:
ψ → eiα ψ
(5.64)
i.e. LD → LD . There is a Noether current associated with this symmetry given by
j µ = ψ̄γ µ ψ
(5.65)
with ∂µ j µ = 0. Promoting this global symmetry to a local symmetry, “local U (1)”:
ψ → eiα(x) ψ
(5.66)
∂µ ψ → ∂µ (eiα ψ) = eiα ∂µ ψ + i∂µ αeiα ψ
(5.67)
LD → LD + i(∂µ α)ψ̄γ µ ψ
(5.68)
This is no longer a symmetry since
and
and our Lagrangian is not invariant. The procedure to “fix” this is to introduce a gauge
field that has the following transformation:
1
local U (1)
Aµ −−−−−−→ Aµ + ∂µ α
e
We now introduce another term known as the covarient derivative:
(5.69)
Dµ = ∂µ − ieAµ
(5.70)
local U (1)
Dµ ψ −−−−−−→ eiα Dµ ψ
(5.71)
and we end with
local U (1)
ψ̄Dµ ψ −−−−−−→ ψ̄Dµ ψ
(invariant under local U (1))
(5.72)
Our Dirac Lagrangian becomes
/ − m)ψ = ψ̄(i∂/ − m)ψ + eAµ ψ̄γ µ ψ
L = ψ̄(iD
(5.73)
Note that if we had Parity violating U (1) then we would have:
ψ → eiγ5 α ψ,
ψ̄ → ψ̄eiγ5 α
(5.74)
term we could have added a slightly different covarient derivative and gotten the γ 5 term
in our final Lagrangian. [Q 5: Show this!] Our final QED Lagrangian is given by
1
L = ψ̄(i∂/ − m)ψ − Fµν F µν + eAµ ψ̄γ µ ψ
4
(5.75)
5.4. THREE-POINT FUNCTION
5.4
49
Three-point function
We get our Feynman rules by looking at different correlation functions. We begin by
looking at a 3-pt function (|Ωi is the messy QFT ground state).
1 1
δ
1 δ
1
δ
µ hΩ|T ψ̄a (x1 )ψb (x2 )Aµ (x3 ) |Ωi =
W
[η,
η̄,
J
]
e
We [0] i δηa (x1 ) i δ η̄b (x2 ) i δJ µ (x3 )
η→0,J→0
(5.76)
where
− 14 F 2
Z
Z
↑
µ
µ
µ
4
We [η, η̄, J ] = DψDψ̄DA exp i
d x L̄D + L̄g + eAµ ψ̄γ ψ + η̄ψ + ψ̄η + J Aµ
(5.77)
We almost know how to calculate this. Without the interaction term we could do it
explicitly. We expand in powers of e:
Z
1 δ
1 δ
1 δ
µ
4
µ
We [η, η̄, J ] = W0 + ie
dx
−
(γ )cd
W0 + O e2
µ
i δJ (x)
i δηc (x)
i δ η̄d (x)
(5.78)
where we define
Z
Z
1
4
4
µ
ν
4
4
d w d zJ (w)Gµν (w − z)J (z) exp − d w d z η̄(w)S(w − z)η(z)
W0 = exp −
2
(5.79)
We now calculate this explicitly:
Z
1 δ
exp [...] = i
d4 zSdb (x − z)ηb (z) exp [...]
(5.80)
i δ η̄d (x)
−1 δ 1 δ
exp [...] = −Sdc (0) exp [...]
i δηc (x) i δ η̄d (x)
Z
Z
4
− d zSdb (x − z)ηb (z) · (−1) d4 wη̄a (w)Sac (w − x) exp [...]
and
1 δ
exp [...] = i
i δJ µ (x)
Z
d4 yJ µ (y)Gµν (x − y) exp [...]
(5.81)
(5.82)
which gives,
Z
4
4
µ
We [η, ¯, J ] = 1 − ie d xd yJ (y)Gµν (x − y) Sdc (0)
Z
4
4
ν
− d zd wSdb (w − z)ηb (z)η̄a (x)Sac (w − x) γcd
W0
µ
(5.83)
50
CHAPTER 5. FERMIONS WITH PATH INTEGRALS
To calculate the correlation function to leading order we need to take three derivatives
of We . This is straight forward and the answer is [Q 6: Do this calculation!]
Z
4
ν
Sda (x − x1 )
hΩ|T ψ̄a (x1 )ψb (x2 )Aµ (x3 ) |Ωi = ie d xGµν (x3 − x) Sbc (x2 − x)γcd
ν
+ Sba (x2 − x1 )(γ )cd Sdc (x − x)
(5.84)
We develop the position space Feynman rules given by
ψa (x)
ψ̄b (y)
Aµ (x)
Aν (y)
→ Sab (x − y)
→ Gµν (x − y)
d
Z
→ −ie
µ
d4 xγcd
c
In principle we can now work out every correlation function using these Feynman rules.
The momentum space Feynman rules are
d
a
b
µ
ν
i(p/ + m)ab
p2 − m2 + i
i
kµ kν
→− 2
gµν − (1 − ξ) 2
k + i
k
→
µ
→ −ieγab
momentum
+ conservation
c
Note that when the momentum flows opposite to the direction in the fermion propagator
we get a negative sign in front of p.
We can now calculate the correlation function,
a, p1
G̃abµ (p1 , p2 , p3 ) =
b, p2
p3
+unconnected
diagrams
5.4. THREE-POINT FUNCTION
51
we ignore unconnected diagrams as they don’t contribute to the S matrix. This diagram
is
i p/2 + m
i p/1 + m
(−i) gνµ − (1 − ξ) p3,µp2p3,ν
3
da
bc
(−ie) (γ µ )cd 2
G̃abµ (p1 , p2 , p3 ) = 2
p2 − m2 + i
p1 − m2 + i
p23 + i
(5.85)
We know from our experience with scalars that in order to get the S matrix elements
through LSZ reduction we need to know the poles of this expression.
Note that if you try to solve the three equations:
p1 + p2 = p 3
p21 = p22 = m2
p23 = 0
(5.86)
(5.87)
(5.88)
you get a contradiction since squaring the first equation gives
p1 · p2 = 0
(5.89)
and if we work in the center of mass frame we have p1 = (m, 0), p2 = (E, p) which would
imply
mE = 0
(5.90)
which is impossible since m > 0. Hence it’s impossible to have a 3 point process that
conserves energy. The first real S matrix process is a four point process:
γ
γ
e
e
e
However in order to see the structure of the S matrix it is sufficient to consider the 3
point function.
p→
Consider an incoming fermion
.
Fp
G̃ → (...) i
p/ + m
p2 − m2 + i
(5.91)
LSZ reductions for Fermions tells us that
2
X
i
h...|...(p, s)i ūs (p) + terms with pole in p2
G̃ = 2
2
p − m + i s=1
(5.92)
here the p2 poles are not important. [Q 7: Isn’t this a p2 pole...?] G̃ is of this form
because,
|(p, s)i = as†
(5.93)
p |0i
52
CHAPTER 5. FERMIONS WITH PATH INTEGRALS
and
X
d3 k
s† s
ik·x
s s
−ik·x
√
ψ=
ak u (k)e
+ bk v (k)e
(2π)3 2Ek s
Z
X s†
d3 k
−ik·x
s
ik·x
s s
√
ψ̄ =
v̄
(k)e
a
ū
(k)e
+
b
k
k
(2π)3 2Ek s
Z
(5.94)
(5.95)
So we cannot directly
P2 get the S matrix from the G̃, but we can use trace technology to
simplify our result s=1 us (p)ūs (p) = p/ + m. Using this result we have,
G̃ = (...)
i
P2
s=1 us (p)ūs (p)
p2 − m + i
(5.96)
Comparing with the LSZ formula we can write,
h...|...(p, s)i = (...) us (p)
(5.97)
So for an incoming fermion we see we have a factor of u(p). The derivation for the other
an incoming antifermion as well as for the outgoing fermions is similar.
Now consider the full photon terms:
(−i)(gµν − (1 − ξ)pµ pν /p2 ) ν
N
p2 + i
(5.98)
where N ν is just the rest of the object which has no µ dependence. Using LSZ reduction
we have for an incoming photon,
2
X
i
2
h...|...(p, s)i s∗
G̃µ = 2
µ + (no pole at p = 0)
p + i s=1
(5.99)
where we have used,
Z
Aµ =
X
d3 k
−ik·x
√
as sµ eik·x + as† s∗
µ e
(2π)3 2Ek s
(5.100)
Using the Ward identity (we discuss this identity more later) we know that
pµ N ν = 0
(5.101)
This means that the (1 − ξ)pµ pν /p2 term in the incoming and outgoing photon contributions go to zero and we simply have,
X
−igµν ν
i
N
=
−
s ∗ (p) h...|...(p, s)i
p2 + i
p2 + i s µ
(5.102)
In the fermion case we rearrange the numerator of the propagator using a spin sum. We
would like to do the same thing here with gµν and write it in terms of µ ’s. To do this
5.5. e+ e− → µ+ µ−
53
we use a trick that we will prove later on which says that as long as gµν is going to be
multiplied by an amplitude we can make the replacement,
gµν → −
X
sµ s∗
ν
(5.103)
s
Since the LSZ formula is true for any s,
N ν sµ (p) = h...|...(p, s)i
(5.104)
Hence we can replace an incoming photon by sµ .
We summarize all the rules for incoming and outgoing particles below
incoming fermion
incoming antifermion
outgoing fermion
outgoing antifermion
incoming photon
outgoing photon
5.5
Fp
p
pF
p
gp
gp
p→
= us (p)
p→
= v̄ s (p)
µ p→
µ p→
p→
= ūs (p)
p→
= v s (p)
= sµ (p)
= s∗
µ (p)
e+e− → µ+µ−
Now that we have all our Feynman rules we can finally do some calculations. As an
example lets consider two electrons turning into two muons. The interaction part of the
Lagrangian is
Lint = −eAν (ψ̄e γ ν ψ̄e + ψ̄µ γ ν ψµ )
(5.105)
The leading order connected diagram is
e, p2
µ, p3
γ
k = p 1 + p2
e, p1
iM
s1 ,s2 ,s3 ,s4
=
µ, p4
(−i)
kα kβ
gαβ − (1 − ξ) 2
k 2 + i
k
s3
× ūµ (p3 )(−ieγ β )v̄µs4 (p4 )
[v̄es2 (p2 )(−ieγ α )use1 (p1 )]
(5.106)
54
CHAPTER 5. FERMIONS WITH PATH INTEGRALS
(note that here µ, e are not variables just labels so we know what mass each u and v
corresponds to. Consider the term
kα
[v̄(p2 )γ α u(p1 )] = (−ie)v̄(p2 )k/u(p1 )
k2
= (−ie)v̄(p2 )(p/1 + p/2 )u(p1 )
= (−ie) (v̄(p2 )me u(p1 ) + v̄(p2 )(−me )u(p1 ))
=0
(5.107)
(5.108)
(5.109)
(5.110)
since (p/ − m)u = 0, (p/ + m)v = 0. Thus the amplitude is indendent of the kα kβ
contribution and hence independent of gauge parameter ξ. The amplitude simplifies to
iMs1 s2 s3 s4 = i
e2
[v̄2 γ α u1 ] [ū3 γα u4 ]
s
(5.111)
where s ≡ (p1 + p2 )2 . Typical experiments have polarized beams. In this case we can
average over incoming particle spins, and sum over outgoing particle spins.
2 X
1
|Ms1 ,s2 ,s3 ,s4 |2 ≡ |M|2
(5.112)
2 s ,s ,s ,s
1
2
3
4
In our case we have
i
h
i
1 e4 h
α
β
tr
(
p
−
m
)γ
(
p
+
m
)γ
×
tr
p
+
m
γ
(
p
−
m
)γ
|M|2 =
/
/
/
/
e
e
µ
α
µ
β
2
1
3
4
4 s2
4
1e
=
(4) pα2 pβ1 + pα1 pβ2 − g αβ p1 · p2 + m2e (4)
2
4s
α β
α β
αβ
2
× p4 p 3 + p3 p4 − g
p3 · p4 + mµ
(5.113)
(5.114)
In the high energy regime (E me , mµ ) we have
|M|2 =
8e4
((p1 · p4 )(p2 · pe ) + (p1 · p3 )(p2 · p4 ))
s2
(5.115)
but,
p1 + p2 = p3 + p 4
(5.116)
(p1 − p4 )2 = (p3 − p2 )
p1 · p4 ≈ p 3 · p2
(5.117)
(5.118)
and
since we have in the high energy regime. So after switching the Mandlestam variables
the amplitude squared takes the form,
|M|2 =
If we work in the COM frame:
2e4 2
2
t
+
u
s2
(5.119)
5.6. e− µ− → e− µ−
55
e
e
θ
then
p1
p2
p3
p4
= (E, 0, 0, E)
= (E, 0, 0, −E)
= (E, E sin θ, 0, E cos θ)
= (E, −E sin θ, 0, −E cos θ)
where s = 4E 2 , t = −2E 2 (1 − cos θ), u = −2E 2 (1 + cos θ). We have the simple result
|M|2 = e4 (1 + cos2 θ)
(5.120)
and
dσ
|M|2
e4
=
=
(1 + cos2 θ)
dΩ
64π 2 s
32πs
The functional dependence is as follows,
(5.121)
# of Events
cos θ
5.6
e−µ− → e−µ−
We have found the amplitude and cross sections for e+ e− → µ+ µ− . We now consider
elastic e− µ− scattering but now in the non-relativistic regime,
p1 →
p4 →
q = p1 − p4
p2 →
p3 →
56
CHAPTER 5. FERMIONS WITH PATH INTEGRALS
We can do a similar calculation to last time:
iM =
ie2
(ū4 γ µ u1 )(ū3 γµ u2 )
t
(5.122)
In the non-relativistic limit we have
p1
p2
p3
p4
= (me + O(p2 ), p)
= (mµ , 0)
= (mµ + O(p2 ), q)
= (me + O(p2 ), p0 )
where q = p − p0 . Hence we have
t = (p1 − p4 )2 = −q2 + O(p4 )
The Dirac spinor in the non-relativistic limit is
s √
ξ
s
+ O(p)
u (p) = m
ξs
1
0
1
2
where ξ =
,ξ =
. The scalars are:
0
1
√
ū4 γ 0 u1 = m(ξ s4 † ξ s1 + ξ s4 † ξ s1 ) = 2mξ s4 † ξ s1 = 2mδs4 s1
i
ū4 γ u1 = O(p)
(5.123)
(5.124)
(5.125)
(5.126)
Thus to leading order we have
iM ≈ −ie2 (2me )(2mµ )
δs4 s1 δs3 s2
q2
(5.127)
The cross-section is given by,
dσ
1
Ee 1 2pCM
e
=
|M|2
2
dΩ
2Ee 2mµ pe 32π ECM
(5.128)
We work in the limit of a stationary muon (though the cross-section is Lorentz invariant).
In this case pcm
≈ pe and Ecm ≈ mµ . We have,
e
dσ
1
=
|M|2
2
2
dΩ
64π mµ
(5.129)
4α2 2
m
q4 e
(5.130)
=
where the spin averaging is trivial due to the spin delta functions.
In non-relativistic quantum mechanics we’d expect the electron to be moving slowly
and the muon to be at rest acting as a source of charge through the potential, V (r) = αr .
5.7. COMPTON SCATTERING
57
0
pe
→
pe →
µ
In this case the Born approximation tells us that
dσ
m2
= e2 Ṽ 2 (q)
dΩ
4π
0
where q = p − p. Comparing the two expressions we identify,
α2 16π 2
q4
4πα
⇒ Ṽ (q) = 2
q
Ṽ 2 (q) =
We have
(5.131)
(5.132)
(5.133)
1
d3 q −iq·r 4πα
e
(5.134)
(2π)3
q2
Z
α
=
d cos θdqeiqr cos θ
(5.135)
π
Z
1
2α
dq sin(qr)
(5.136)
=−
π
q
α
=−
(5.137)
r
where the integral was done by inserting a regulator and taking the regulator to 0 at
the end. We see that in the non-relativistic limit QED repreduces classical electromagnetism as expected. Furthermore, this is the first confirmation that the coupilng in the
Lagrangian, e, really represents an electric charge. Up to now we just assumed the most
natural interpretation and we finally see that this is indeed the case.
Z
V (r) =
5.7
Compton Scattering
Consider e− γ → e− γ. There are two second order vertices:
+
1
Where there is no ambiguity with four-vectors we write, r ≡ |r| and q ≡ |q|.
58
CHAPTER 5. FERMIONS WITH PATH INTEGRALS
We have the amplitude
Ms = sµ (p)Mµ
(5.138)
To get the cross-section we need to average over spins:
X
X
ν∗
|Ms |2 =
sµ s∗
ν M
s
(5.139)
s
The claim is that we can make the replacement:
X
sµ ∗s
ν → −gµν
(5.140)
s
To see why this is consider the photon with the momenta, p = E(1, 0, 0, 1). The Ward
identity tells us that pµ Mµ = 0. Hence
M0 = M3
(5.141)
Since we can choose any basis we wish for the polarization lets consider linear polarization,
1 = (0, 1, 0, 0)
2 = (0, 0, 1, 0)
Explicitly we can now calculate

X
sµ s∗
ν
s
0
 0
=
 0
0
0
1
0
0
0
0
1
0

0
0 

0 
0
(5.142)

0
1
0
0
0
0
1
0

0
0 

0 
1
(5.143)
This is not quite
− gµν
−1
 0
=
 0
0
P
But s sµ sν ∗ and gµν never appear on their own. Instead they come as as a product of
amplitudes. We have,
2 1 2 2 2 3
2 1 2 2 2
0
M
+ M + M + M
= M + M − gµν Mµ Mν∗ = −
(5.144)
X
2
2
µ
ν∗
µ s∗
= M1 + M2 (5.145)
ν M M
s
Hence due to the Ward Identity we can always make the substituion,
X
sµ sν ∗ → −gµν
s
(5.146)
5.8. PROOF OF WARD IDENTITY
5.8
59
Proof of Ward Identity
We finally prove the Ward identity and then discuss the intuition behind it. The defining
equation for the vector potential without choosing a gauge is given by
∂ν ∂ ν Aµ (x) = 0
(5.147)
This equation is solved by the an satz,
Z
Aµ (x) =
d4 p
µ (p)e−ip·x
(2π)4
(5.148)
The gauge transformation of the vector potential under which physics is invarient is,
A0µ = Aµ + ∂µ α
(5.149)
R d4 p −ip·x
Plugging in the potential as well as the Fourier transform of α = (2π)
α̃(p) we get
4e
Z
Z
−ip·x
d4 p
d4 p
0
(−ipµ )e−ip·x α̃
(5.150)
−
e
=
µ
(2π)4 µ
(2π)4
0
Now multiplying by e−ip ·x and integrating over d4 x we have
Z
Z
d4 p
d4 p
0
0
pµ α̃δ(p − p0 )
−
δ(p
−
p
)
=
−i
µ
(2π)3 µ
(2π)4
0µ (p) = µ (p) − iα̃(p)pµ
(5.151)
(5.152)
This is the effect of a gauge transformation on the polarization vector.
Now in general any amplitude with N gauge bosons will take the form,
µ1 µ2 ...µN Mµ1 µ2 ...µN
(5.153)
If a theory is gauge invariant then this amplitude must be invariant under any gauge
transformation. Applying equation 5.152 to some polarization vector i we have
µ1 µ2 ...µN Mµ1 µ2 ...µN → µ1 µ2 ...µN Mµ1 µ2 ...µN − iα̃ pµi Mµ1 ,...,µN
(5.154)
In order for the amplitude to be gauge invariant we must have Ward Identity,
pµi Mµ1 ,...,µN = 0
which is just the Ward Identity.
We now try to explore how this relationship holds. Consider the following
p1
p→
p2
(5.155)
60
CHAPTER 5. FERMIONS WITH PATH INTEGRALS
where we have p is off-shell since otherwise it would violate energy conservation. The
amplitude is,
M = ν ū(p1 )γ ν v(p2 )(−ie)
(5.156)
|
{z
}
Mν
ν
Then the sum, pν M , is zero since,
pν Mν = ū(p1 )p/v(p2 )(−ie)
(5.157)
= ū(p1 )(p/1 + p/2 )v(p2 )(−ie)
(5.158)
= (−ie)ū1 v2 (me − me )ū1 v2
=0
(5.159)
(5.160)
Now if we are considering a more complicated diagram,
×
+
×
where the “ ×” represents some combination of external lines. Then the Ward identity
means that these contributions must cancel. See Peskin chapter 7.4 for details. [Q
8: Expand on this section]
Chapter 6
Loop Diagrams and UV Divergences in
QED
6.1
Superficial Divergences
The first thing one thinks about when hearing about loop diagrams is estimating how
divergent diagrams are. We use what’s known an “naive power-counting” to find the
superficial degree of divergence. For D greater then or equal to zero we expected ∼ ΛD
and we expect logarithmic divergence if D = 0. If D < R0 then our diagrams are not
d4 p
superficially divergent. Each loop contributions a factor of (2π)
4 (4 factors of momenta).
Fermion propagators have dimension −1 and photon propagators have dimensions −2
hence (Pi the number of i propagators)
D = 4L − Pf − 2Pγ
(6.1)
One can relate this expression to the number of external lines by (Exercise)
3
D = 4 − Nf − Nγ
2
61
(6.2)
62
CHAPTER 6. LOOP DIAGRAMS AND UV DIVERGENCES IN QED
In reality the power counting is not precise and diagrams may or may not diverge independently of the superficial degree of freedom. Consider the D ≥ 0 diagrams in QED:
p
pg
gpg
vpg
u
gpg
g
g
FpF
Dpg
E
Diagram
D
D=4
D=3
D=2
Expected # of CT’s
Name
vacuum energy → unphysical
zero!(HW)
3 counterterms(
vacuum polarization
D=1
zero!(HW)
D=0
1 counterterm
D=1
2 counterterms
electron self-energy
D=0
1 counterterm
vertex correction
Our Lagrangian is
1
L = ψ̄(i∂/ − m0 )ψ − Fµν F µν + e0 Aµ ψ̄γ µ ψ
(6.3)
4
√
√
We want to introduce renormalized fields ψ = Z2 ψr and Aµ = Z3 Arµ . We then have
p
1 r µν
L = Z2 ψ̄r (i∂/ − m0 )ψr − Z3 Fµν
Fr + e0 Z2 Z3 Arµ ψ̄r γ µ ψr
(6.4)
4
We introduce a physical mass and we have δm = m − m0 :
p
1 r µν
L = Z2 ψ̄r (i∂/ − m)ψr − Z3 Fµν
(6.5)
Fr + e0 Z2 Z3 Arµ ψ̄r γ µ ψr + δmZ2 ψ̄r ψr
4
We still need to renormalize the coupling constant. We define the new coupling constant
as
√
e0 Z2 Z3
e=
(6.6)
Z1
How we define a coupling is not physical since it’s not a number like mass which has a
definition on it’s own. We just need to require that it won’t be divergent. This form
turns out to be convenient. With this we have
1 r µν
L =ψ̄r (i∂/ − m)ψr − Fµν
Fr + eArµ ψ̄r γ µ ψr +
4
1 r µν
(Z1 − 1)Arµ ψ̄r γ µ ψr + (Z2 − 1)ψ̄r (i∂/ − m)ψr − (Z3 − 1) Fµν
Fr + δmZ2 ψ̄r ψr (6.7)
4
We define
δ1 ≡ Z1 − 1
δ2 ≡ Z2 − 1
δ3 ≡ Z3 − 1
δm ≡ (Z2 − 1)(m) − δmZ2
(6.8)
6.2. ELECTRON SELF-ENERGY
63
Our counterterms are
LCT = δ1 Arµ ψ̄r γ µ ψr + ψ̄r (iδ2 − δm )ψr −
δ3 r µν
F F
4 µν r
(6.9)
Note that from here on we drop the r to denote renormalized fields. Naively we expect
the number of counter terms to be given by the sum of all the divergences → 3 + 1 + 2 + 1.
However we can only possibly write down 4 terms! We know this theory is renormalizable
so the superficial degree of divergence must be overestimating the divergence of this
theory. Consider our renormalized propagator
gpg
µ
ν
q→
q→
= Πµν (q) = A(q 2 )gµν + B(q 2 )qµ qν
(6.10)
The Ward identity tells us that q µ Πµν = 0 ⇒
Aq ν + Bq 2 q ν = 0
(6.11)
since this must be true for all q µ we have
A = −Bq 2 e
gpg
and hence
µ
ν
q→
q→
= (gµν q 2 − q µ q ν )B(q 2 )
(6.12)
(6.13)
The Ward identity provides a constraint and shows that we only need 1 CT for the
propagator. Similarly one can show in the same way that the four photon diagram is in
fact finite and does not need a counterterm. So only need the 4 CT’s.
6.2
Electron Self-Energy
Z
G̃ab (p) =
d4 xe−ix·p hΩ|T ψ̄a (x)ψb (0) |Ωi
(6.14)
where in general we have ψb (y) but we just choose y = 0. Using Feynman diagrams we
can write
G̃ =
F FFyFF FFyFyFF
FFyFFFyFF
+
+|
+
{z
} +...
not 1PI
i
i
i
=
+
(−iΣ(p))
p/ − m0 + i p/ − m0 + i
p/ − m0 + i
(6.15)
(6.16)
where Σ(p) is a matrix that’s equal to the sum of all the possible diagrams with two
external lines and p is the incoming momentum.
64
CHAPTER 6. LOOP DIAGRAMS AND UV DIVERGENCES IN QED
We define 1 particle irreducible (1PI) graphs to be diagrams that you cannot cut into
2 by cutting a single line. We define
Σ̂ =
sum of 1PI graphs only, =
without external lines
FyF FyFyF
+
+ ...
(6.17)
we have (for brevity we initially suppress the +i in the denominator)
i
i
i
i
i
i
+
−iΣ̂
+
−iΣ̂
−iΣ̂
+ ...
G̃ =
p/ − m0 p/ − m0
p/ − m0 p/ − m0
p/ − m0
p/ − m0
(6.18)
=
i
p/ − m0
∞ X
n=0

Σ̂
1
p/ − m0

n
(6.19)
i
1


p/ − m0 1 − Σ̂
p
/−m0
!
i p/ − m0 + i
1
=
p/ − m0 + i
p/ − m0 + i − Σ̂(p)
=
=
i
(6.20)
(6.21)
(6.22)
p/ − m0 − Σ̂(p) + i
where we have used a geometric series for matrices,
∞
X
Ak = (1 − A)−1
(6.23)
k=0
for any matrix A with eigenvalues less than 1.
To better see what this means, we switch to the “proper form” of the propagator (this
is straightforward to check by multiplying the p/ − m + i with its inverse),
(p/ − m − Σ(p) + i)−1 =
p/ + m + Σ(p)
p2 − (m + Σ(p))2 + i
(6.24)
Hence the physical mass is shifted to the point where we still have a pole in the propagator,
p2 = (m0 + Σ̂(p))2
(6.25)
or m ≡ m0 + Σ̂(p2 = m2 ). Thus Σ̂ represents the change in mass from the “bare” mass
calculated to all orders in perturbation theory. We can also write from 6.22
1
G̃−1 = G̃−1
0 − Σ̂(p)
i
(6.26)
so the two-point function contains, apart from the inverse bare propagator, only 1PI
contributions.
6.2. ELECTRON SELF-ENERGY
65
Now we have
G̃ = hΩ|ψ̄r (x)ψr (y)|Ωi
= Z2 hΩ|ψ̄(x)ψ(y)|Ωi
iZ2
=
p/ − m0 + i
(6.27)
(6.28)
(6.29)
where Z2 is the wave function renormalization factor (this is just the old unrenormalized
propagator). By comparing this expression with equation 6.22 we can write
p/ − m0 + i
− p/ + m0 − i
(6.30)
Σ̂ = −
Z2
dΣ̂
= −Z2−1 + 1
dp/
!−1
dΣ̂
⇒ Z2 = 1 −
dp/
(6.31)
(6.32)
Thus if we know how to calculate Σ̂ we can calculate both mass renormalization and
wavefunction renormalization. Note that we have been a little careless above with our
notation. Formally we can write
Σ̂αβ = −(pµ γ µ − m0 + i)αβ (Z2−1 − 1)
(6.33)
∂ Σ̂αβ
= −(Z2−1 − 1)(γ µ )αβ
∂pµ
(6.34)
1 ∂ Σ̂
= (Z2 − 1)γ µ
i ∂pµ
(6.35)
so
or
This is what we really mean when we write
∂
.
∂p
/
We now explicitly calculate Σ̂ to second order in e.
Σ̂(p) =
p→
FyF
`→
p→
(6.36)
4
d ` µ i(/̀ + m0 ) ν
−igµν
γ 2
γ
2
4
(2π)
` − m0 + i (` − p)2 + i
Z
d4 ` µ (`α γ α + m0 ) ν
gµν
= (−ie)2
γ
γ
2
(2π)4 `2 − m0 + i (` − p)2 + i
= (−ie)2
Z
(6.37)
(6.38)
Now using (see for example [Griffith(2008)]) γ µ γ α γ µ = −2γ α and γ µ γµ = 4 it’s easy to
see that
Z
i(−2/̀ + 4m0 )
d4 `
2
(6.39)
Σ̂(p) = −e
4
2
(2π) (` − m20 + i) ((` − p)2 + i)
66
CHAPTER 6. LOOP DIAGRAMS AND UV DIVERGENCES IN QED
Note that this expression has a superficial degree of divergence of order 1. The true
divergence turns out to be of zeroth order as we will see later due to some terms being
odd. To solve this integral and incorporate the divergence we introduce an unphysical
mass µ that will regulate the integral. Note that this is not a unique way to regulate.
We could have used for example dim-reg.
Z
d4 `
(−2i/̀ + 4m0 )
2
(6.40)
Σ̂(p) → −e
2
4
2
(2π) (` − m0 + i) ((` − p)2 − µ2 + i)
The steps are analogous to our discussion in the Fall of scalar loops diagrams. We
introduce Feynman parameters (where A = ((` − p)2 − µ2 + i) and B = `2 − m20 + i)
Z 1
dx
1
=
(6.41)
2
AB
0 (Ax + B(1 − x))
Our expression becomes
Z
Z
(−2i/̀ + 4m0 )
d4 `
2
(6.42)
Σ̂(p) = −e
dx
(2π)4 (((` − p)2 − µ2 + i) x + (`2 − m20 + i)(1 − x))2
Z
Z
d4 `
(−2i/̀ + 4m0 )
2
= −e
dx
(2π)4 (`2 − 2`px + p2 x2 − p2 x2 + (p2 − µ2 ) x + (−m20 )(1 − x) + i)2
(6.43)
Z
Z
4
(−2i/̀ + 4m0 )
d`
(6.44)
= −e2 dx
4
(2π) ((` − px)2 − ∆ + i)2
(6.45)
where ∆ ≡ −p2 x(1 − x) − (µ2 − m20 )x + µ2 . Now define k = ` − px:
Z
Z
d4 k (−2i(k/ + p/x) + 4m0 )
2
Σ̂ = −e
dx
(2π)4
(k 2 − ∆ + i)2
(6.46)
(6.47)
Applying a Wick rotation such that k0 → ik0E and k → kE . We now get (dk0 = idk0E and
k0 → ik0E )
Z
Z 4
d kE (2(ik/E − p/x) + 4m0 )
2
Σ̂ = −e
dx
(6.48)
(2π)4 (kE2 − ∆ + i)2
The denominator is even. Thus we can get rid of the top contribution from k/E 1 :
Z
Z 4
d kE −2(p/x) + 4m0
2
Σ̂ = −e
dx
(6.49)
(2π)4 (kE2 − ∆ + i)2
(6.50)
1
this is where the reduction in the divergence of the diagram occurs
6.2. ELECTRON SELF-ENERGY
67
First since our integral is Wick rotated we are no longer near a pole and we can drop the
i. Now taking the integral is a 4D spherical space we have (S3 = 2π 2 )
Z
Z
dkE kE3 −2(p/x) + 4m0
2
Σ̂ = −e
dx
(6.51)
2
8π 2
(kE2 − ∆)
Z
Z
−e2
kE3
(6.52)
=
dx
(−
p
x)
+
2m
dk
/
0
E
2
4π 2
(kE2 − ∆)
Z
1
−e2
−Λ2
Λ2 − ∆
=
(6.53)
dx (−p/x) + 2m0
+ log
4π 2
2 Λ2 − ∆2
∆
where in the last step we introduced a cutoff. The integral is easy enough to do in
Mathematica. For simplicity we now assume that Λ2 ∆ then we get the somewhat
simpler result,
2
Z
α
Λ
(6.54)
Σ̂ = −
dx p/x − 2m0 log
π
∆
Consider the limit that µ → 0. In this case ∆ → m2 (1 − x) and we have to integrate
the logarithm from 0 to 1. This is divergent as logs aren’t defined at 0. This causes
the electron normalization to diverge. However, the electron normalization itself isn’t
physical. No observables will depend on µ, however we introduce it for convenience so all
the terms in our calculations will be well defined. The physical mass is shifted by
m = m0 + Σ̂(m)
(6.55)
We can use m or m0 in evaluating the Σ̂ since the correction will be of order α2 . To see
this consider the structure of Σ̂:
Σ̂(p/) = p/f (p/2 ) + g(p/2 )
(6.56)
Setting p/ = m and expanding f, p, m we have:
Σ̂(p/ = m) = m0 + δm(1) + ... f (0) + αf (1) + ... + g (0) + αg (1) + ... = 0
(6.57)
The zero’th order result is just the propagator evaluated at p/ = m0 :
f (0) =
p2
1
,
− m20 + i
g (0) = −
p2
m0
− m20 + i
(6.58)
and trivially satisfies the condition. To first order in α we have,
δm(1) f (0) + m0 f (1) + g (1) = 0
(6.59)
Inside f (1) and g (1) there are contributions proportional to m and m2 . However, when to
this order we drop any additional factors of α which arise from inserting m0 + αδm(1) .
Thus we can just use m ≈ m0 to first order.
68
CHAPTER 6. LOOP DIAGRAMS AND UV DIVERGENCES IN QED
We then have the rather simple result,
Z
α
Λ2
Σ̂ ≈ m0 dx (2 − x) log
π
−m20 x2
(6.60)
This integral can be evaluated in Mathematica to be
Σ̂ ≈
Λ2
3α
log 2
4π
m0
(6.61)
So we finally have the mass shift,
m = m0 + Σ̂(m0 ) = m0
Λ2
3α
log 2
1+
4π
m0
(6.62)
There are two approaches towards renormalization. One interpretation is that it is
just mathematical trickery that is used in order to get finite answers. This point of view
is fine and it certainly works however there is a more modern interpretation which says
that there is some physics in renormalization.
Suppose we are evaluating some loop diagram. Our Feynman rules say to integrate
over some loop momentum. If we work in position space we know the relation between
momenta and wavelength. Large momenta correspond to short wavelengths. At very
high loop momenta it corresponds to fluctuations of the field with short wavelengths.
With every theory there is some scale in which we know the theory works. For example
for all we know the SM works well until about the energies that we have gotten to so far.
However when we do loop integrals we are told to integrate all the way to infinity (all
possible virtual particle momenta are allowed). That means that we are extrapolating to
energy scales further then we know that the theory works. With this point of view, there
probably means that this extrapolation has to fail at some energy. This is analogous to
blackbody radiation problem in quantum mechanics. Physicists tried to extrapolate their
understanding of blackbody radiation to regimes where they didn’t know it held.
The suspicion about these divergences in QFT is that they are there because we are
missing some essential pieces of physics that occur at some shorter length scale that we
are unable to probe. Putting in a cutoff scale, Λ, is just some crude model for what
happens at some short length scale. Another way to model how a theory behaves at
small length scales is to switch from field theory to string theory. Depending on the
particle high energy theory you must use a different regulator. Different regulators are
essentially different toy models for how to model the effects at higher energy scale. The
freedom of choice of regulator is a representation of our ignorance of high energy physics.
In particular since we know that gravity cannot be explained by QFTs we believe that
the highest Λ can be is at the Planck scale (though it is likely much earlier). Hence Λ is
some scale that is of the Planck scale or less. Renormalizability tells us that the details of
the high energy physics don’t effect the effects at the low energy regime. By definition it
means that the dependence of the cutoff can be absorbed into the CT’s and won’t effect
physics quantities. The cutoff scale is taken is some sense to be a real physical energy
scale. The interpretation of a formula such as 6.62 is an observer at a low energy will see
m ≈ m0 however at a large energy there are going to be corrections.
6.3. CLASSICAL RENORMALIZATION AND REGULATORS
6.3
69
Classical Renormalization and Regulators
Consider an electron. It has some electric field
e
Einstein told us that energy is equivalent to mass hence the electric field must change
the mass of the particle. This correction is what we calculated using equation 6.62.
Classically one can go ahead to calculate this same result.
|E|2
δm = d r
8π
Z ∞
e 2
4π
2
r dr 2
=
8π 0
r
Z ∞
dr
∝ e2
r2
0
→∞
Z
3
(6.63)
(6.64)
(6.65)
(6.66)
Classically this energy doesn’t converge!The problem occurred since we extrapolated classical EM to a length scale in which it was no longer valid. Classical EM is a low energy
effective theory. While it does a good job at describing the physics at low energies (large
length scales) it fails at high energies. Alternatively one can introduce a length cutoff
(analogous to the energy cutoff, Λ that we are used to in QFT:
2
Z
∞
δm =∝ e
r0
∝
e2
r0
dr
r2
(6.67)
(6.68)
This quantity is no longer infinite. One could ask why this length scale was required
since to our current understanding the electron truly is a point charge. The problem is
QM. In QFT the vacuum is not static, but electron positron pairs pop in and out of the
vacuum. The vacuum of QED is filled with these pairs:
70
CHAPTER 6. LOOP DIAGRAMS AND UV DIVERGENCES IN QED
+
-
+
+ +
+ +
-
These pairs can appear as long as they disappear in a timescale as determined by the
uncertainty principle. Normally we don’t care about them, since they average to zero
(equally likely to be oriented in all direction). However if you have a real electron then
they are more likely to orient themselves along the electric field and screen the field.
As you approach the electron of order the inverse electron mass these effects become
important. The wavelength that such effects will occur is the Compton wavelength
λ=
1
≈ rc
me
(6.69)
is known as the Compton wavelength. We identify r0 with the Compton wavelength and
we get
δm ∝ e2 me
(6.70)
This is starting to look very similar to our equation that we derived with QFT except
with the Logarithmic divergence. Knowing the higher energy theory of QED told us the
exact form of the correction. In essence our QED calculation is just an estimate that will
be fixed when more physics is added. One can go ahead and renormalize classical EM if
one wanted as well.
1
, m0 ≈ 511keV , Λ ≤ MP lanck ∼
Consider our correction term in equation 6.62. α = 137
19
10 GeV . This makes the correction
3α
Λ2
log 2 ≈ 0.2
4π
m0
(6.71)
So even with having this huge cutoff scale the correction is still not large. Note such
an argument doesn’t work using the φ4 theory. In the end we don’t really care since we
renormalize our theories anyways.
When picking a regulator it’s important to have it respect the symmetry of your
problem, since its possible that the high energy theory does respect the theory (of course
it is also possible that it won’t break the symmetry but you don’t want to remove the
other possibility). As an example recall our derivation of the photon terms. We were not
allowed to write down a mass term:
1 2
m Aµ Aµ
2
since we wanted gauge invariance. When finding the photon propagator:
q→
q→
∝ gµν q 2 − q µ q ν Π(q 2 )
gpg
µ
ν
(6.72)
(6.73)
6.4. DIMENSIONAL REGULARIZATION
71
we would in general renormalize the mass and create a mass. However experimentally
we know the photon is massless to a high precision. We want to somehow ensure that
the photon will remain massless at this high energy scale. We want to introduce a
regulator that respects gauge invariance. The cutoff is not a gauge invariant cutoff (this
is not obvious at this point). The preferred regulator in QED is through dimensional
regularization or “dim reg”.
6.4
Dimensional Regularization
Using dim reg we now consider the integral we found earlier with a hard cutoff
Z
γ µ (/̀ + m0 )γµ
dd `
2
Σ̂(p) = e
(2π)d (`2 − m20 + i) ((` − p)2 − m20 + i)
(6.74)
Note that we have changed the dimension of each object in this expression from 4 → d.
It’s not clear what the γ matrices even mean in more then 4 dimensions. A discussion in
Peskin and Schroeder tells us that γ µ γµ = d, γ µ γ ν γν = −(d − 2)γ ν . These expression are
analytic continuations of the expressions of integer d.
Z
Z 1
dd kE − d2 − 1 xp/ + d2 m0
2
dx
Σ̂(p) = 2e
(6.75)
d(2π)d
(kE2 + ∆)2
0
2−d/2
Z 1 1
2e2
d
d
Γ(2
−
d/2)
dx
m
−
−
1
x
p
=
(6.76)
/
0
(4π)d/2
2
2
∆
0
In 4 dimensions this expression is infinite because of the Γ term. However in 4 − dimensions we have
Z 1
Z 1
2
e2
Σ̂(p) = 2
dx xp/ − 2m0
− γ + log 4π − log ∆ +
dx xp/ − m0 + O()
8π
0
0
(6.77)
where γ is the Euler constant and equal to 0.566....
6.5
Vacuum Polarization
Consider
G̃µν (q) =
=
gpg
g gcg gcgcg
q→
q→
+
+
+ ...
qµ qν
i
= − 2 gµν − 2
q
q
i
qµ qα
i
qβ qν
+ − 2 gµα − 2
(iΠαβ ) − 2 gβν − 2
+ ...
q
q
q
q
(6.78)
(6.79)
(6.80)
72
CHAPTER 6. LOOP DIAGRAMS AND UV DIVERGENCES IN QED
where we use the Landau gauge for the propagator. We define
Π=
c cgc
+
+ ...
(6.81)
and
Π̂ =
X
1PI graphs
(6.82)
Note that this is completely analogous to Σ and Σ̂ from earlier. Then
G̃µν
i qµ q ν
i
qµ qα
qβ qν
αβ
gµν − 2
+ − 2 gµα − 2
iΠ̂
− 2 gβν − 2
q
q
q
q
q
αβ
σρ
+ (...)µα iΠ̂
(...)βσ iΠ̂
(...)ρν + ...
(6.83)
i
=− 2
q
The Ward identity says that (see equation 6.13)
Π̂αβ
qα q β
= q gαβ − 2
Π(q 2 )
q
2
(6.84)
which implies
q µ qα
qµ qα
qαqβ
αβ
2
αβ
Π̂ = q gµα − 2
gµα − 2
g − 2
Π(q 2 )
q
q
q
qµ q β
1
2
β
β
β
= q gµ − 2 qµ q + qµ q + 2
Π(q 2 )
q
q
β
qµ q
Π(q 2 )
= q 2 gµβ − 2
q
(6.85)
(6.86)
(6.87)
and hence
G̃µν
qµ qν
qµ q β
qβ qν
β
gµν − 2
+ gµ − 2
gβν − 2
(−i)(i)Π(q 2 )
q
q
q
2
+ (...) (...) (...) −iΠ(q 2 ) + ...
(6.88)
i
=− 2
q
We define Aαβ ≡ gαβ −
G̃µν
qα qβ
.
q2
With this we can write as
o
−i n
2
αβ
β
2 2
= 2 Aµν + Aµ Aβν Π(q ) + Aµα A Aβν Π(q ) + ...
q
(6.89)
but
qµ q β
qβ qν
qµ qν
β
gµ − 2
gβν − 2
= gµν − 2
q
q
q
(6.90)
6.5. VACUUM POLARIZATION
73
or equivalently Aµβ Aβν = Aµν . This gives
G̃µν
−i
Π(q 2 )
2 2
= 2 Aµν + Aµν 2 + Aµν Π(q ) + ...
q
q
−1
−iAµν 2
1
−
Π(q
))
=
q2
−i
qµ qν
1
= 2 gµν − 2
q
q
1 − Π(q 2 )
(6.91)
(6.92)
(6.93)
(6.94)
So we have our old propagator with an extra factor, since the photon propagator is
qµ qν
−i
gµν − 2
(6.95)
q2
q
Note that here the Π̂ doesn’t come additive to the q 2 value as it did for the massive fields.
The pole is still at q 2 = 0 ⇒ mγ = 0. One can trace the reason that this happened to
the Ward identity (and hence gauge invariance). The only renormalization comes about
in the wavefunction renormalization:
q µ qν
i
(6.96)
G̃µν = − 2 gµν − 2 Z3
q
q
or equivalently in Feynman gauge:
G̃µν = −
igµν
Z3
q2
(6.97)
The amplitude we want is,
←`
−iΠ̂µν =
q→
q→
`+q →
µν
− iΠ̂ (q) = (−1) (−ie)
2
Z
d4 ` Tr γ µ −/̀ + m γ ν /̀ + /q + m
(2π)4
(`2 − m2 ) (` + q)2 − m2
(6.98)
The most divergent part is (using a hard cutoff)
Z
d4 ` (a`2 gµν + b`µ `ν )
∼ gµν Λ2
`4
(6.99)
74
CHAPTER 6. LOOP DIAGRAMS AND UV DIVERGENCES IN QED
However this object doesn’t obey q µ Π̂µν = 0. Hence this is not true at O(Λ2 )! To solve
this integral in a gauge invariant way we use dim reg. We can greatly simplify the amount
of work that we need to do by identifying that the final answer must be in the form,
Π̂µν = (q 2 g µν − q µ q ν )Π(q 2 )
(6.100)
Thus we just need to calculate either the q µ q ν or the g µν part and just infer the other
part. As we will see it will be easier to calculate the q µ q ν contribution. The denomenator
is:
Z
1
1
1
(6.101)
=
dx 2
2
2
2
2
2
` − m + i (` + q) − m + i
(` + qx) − ∆ + i
where
∆ ≡ −q 2 x (1 − x) + m2
We now shift the integration variable and conside the trace:
Tr [...] → Tr −/q + /̀ + /qx + m γν −/̀ − /qx + m γµ
= −Tr /̀ + /q (x − 1) γν /̀ + /qx γµ + m2 Tr [γµ γν ]
= −Tr /̀γν /̀γµ − Tr /q (x − 1) γν /qxγµ + m2 Tr [γµ γν ] + linear in `
= 4 2`µ `ν − `2 g µν + 4x (1 − x) 2q µ q ν − q 2 g µν
(6.102)
(6.103)
(6.104)
(6.105)
(6.106)
∆ only depends on q 2 so there is only one term that has a q µ q ν contribution. We work
out this term and infer the value of the rest of the terms. We need the integral,
Z
i
2
1
d4 `
=
(1 + log 4π)
− γ (1 − log ∆)
(6.107)
(2π)4 [`2 − ∆]2
(4π)2
i
∆
2
=
− log
−γ
(6.108)
16π 2 4π
Thus
Z
e2
2
∆
Π̂µν = −q q − 2 dx2x (1 − x)
− log
−γ
+ (...) g µν
4π
4π
Z
2α
2
∆
Π(q 2 ) = −
dx (1 − x)
− log
−γ
π
4π
µ ν
⇒
(6.109)
(6.110)
Recall that earlier we showed that for an electron scattering off a muon in the nonrelativistic limit given by
the amplitude went as iM ∼
e2
.
q2
The Born approximation tells us that
hp|iT |pi ∼ Ṽ (p − p0 )
(6.111)
6.5. VACUUM POLARIZATION
75
We had
V =
e2
e2
→ Ṽ ∼
r
|q|2
(6.112)
what we really need to do is
e2 Z
0 3
(we need to have the renormalized propagator for the photon). This results in iM ∼ |q|
2
√
using instead of the bare charge we have e = e0 Z3 . At higher energies we need to
include the following correction as well:
!
e20 Z3
1 − Π̂(0)
iM ∼
(6.113)
|q|2 1 − Π̂(−q2 )
Going through the calculation one can show
α
α e−2mr
+ ...
V (r) =
1+ √
r
4 π (mr)3/2
(6.114)
++-
The interaction gets stronger then the Coulomb expectation as the charges get close.
This goes back to why this effect is called vacuum polarization. Imagine the muon:
++-
++-
++-
Hence at low energy what you see is the screened potential. This is the effect of having
e → e20 Z3 . As you start throwing more energetic electrons toward the muon, then you
get the electron very close the muon. You can get rid of some of the screening effect and
the stronger the field will be. You’re interaction with the muon will not be determined
by e2 but more by something that’s more close to the bare charge. Eventually when you
are very close the center of the muon you will be able to measure Z3 . This is the concept
behind the running coupling constant. This effect is observed and well tested.
2
76
CHAPTER 6. LOOP DIAGRAMS AND UV DIVERGENCES IN QED
6.6
Vertex Function
Consider the vertex function which represents the following diagram
p0
q ≡ p0 − p
0
G̃µ (p, p ) =
µ
p
in the limit of p2 → m2e , p02 → m2e but q has to stay off shell by energy momentum
conservation. Now recall that
P
i(p/0 + m)
i 2s=1 us (p0 )ūs (p0 )
i
= 02
=
(6.115)
p/0 − m + i
p − m2 + i
p02 − m2 + i
so we have a
P s0
0
us (p0 )ūs (p0 ) ν
u (p)ūs (p)
0
G̃µ = 02
Γ (p, p ) 2
Dµν (q)
p − m2 + i
p − m2 + i
P
(6.116)
The interesting object is Γν . To leading order we have just a single vertex:
→ Γν = −ieγ ν
(in the terminology we will introduce shortly, F1 = 1, F2 = 0). Next to leading order we
have the diagram:
q−
p
→
-`−q
q→
`−p↓
p
→
`%
ν
3
Γ = (−ie)
Z
/̀ + m
(−igρσ)
dd ` ρ i(/̀ + /q + m)
ν i
γ
γ 2
γσ
d
2
2
2
(2π)
(` + q) − m + i ` − m + i (` − p)2 + i
(6.117)
6.6. VERTEX FUNCTION
77
We won’t solve this explicitly however we consider the Lorentz structure. The only
nontrivial scalar in this problem is q 2 . From the fact it’s a four vector we already know
that we will have something of the form,
Γν = γ ν A(q 2 ) + (p + p0 )ν B(q 2 ) + (p − p0 )ν C(q 2 )
(6.118)
In general B and C can be matrices as well however since mu(p) = p/u(p) we can write
the expressions in terms of ordinary numbers. From the Ward identity we know that
q µ G̃µ = 0
(6.119)
and so,
ū(p0 )qν Γν u(p) = 0
(6.120)
z }| {
0
2
0
2
ū(p )
/q A(q ) + q · (p + p ) B(q ) + q · (p − p ) C(q ) u(p) = 0
| {z }
|{z}
(6.121)
=p2 −p02 =0
0
2
q 2 6=0
=ū(p)(p
/ −p
/0 )u(p0 )=0
where we have used the fact that q ≡ p0 − p and hence we have C = 0. Now we use the
Gordon identity:
0
(p + p)µ iσ µν qν
0 µ
0
ū(p )γ u(p) = ū(p )
+
u(p)
(6.122)
2m
2m
where σ µν = 2i [γ µ , γ ν ].
We quickly prove this identity below. Consider
1 µ ν
(γ γ − γ ν γ µ ) pν
2
1
= −pµ −
2γ µ p/ − 2g µν pν
2
µ
= −p − γ µ p/ − pµ
−pµ + iσ µν pν = −pµ −
µ
= −γ p/
(6.123)
(6.124)
(6.125)
(6.126)
Furthermore,
1
p0µ + iσ µν p0ν = p0µ + p0ν (γ ν γ µ − γ µ γ ν )
2
1
= p0µ +
2p/0 γ µ − 2p0µ
2
0 µ
= p/ γ
(6.127)
(6.128)
(6.129)
Subtracting these two equations gives
µ
(p0 + p) + iσ µν (p0 − p)ν = p/0 γ µ + γ µ p/
0
µ
ūp0 (p + p) + iσ µν (p0 − p)ν up = ūp0 mγ µ up + ūp0 γ µ mup
0
0
µ
(p + p)µ
0 µ
µν (p − p)
ū(p )γ u(p) = −ūp0
up
+ iσ
2m
2m
(6.130)
(6.131)
(6.132)
78
CHAPTER 6. LOOP DIAGRAMS AND UV DIVERGENCES IN QED
Using this identity we see that
iσ µν qµ
Γ (p, p ) = (−ie) γ F1 (q ) +
F2 (q 2 )
2m
0
ν
ν
2
(6.133)
where the Fi (q 2 ) are known as form factors. To lowest order we have no loop and F1 (q 2 ) =
1 and F2 (q 2 ) = 0. To second order they are given by [Q 9: Show this]
Z 1
e2
Γ(2 − d/2) (2 − )2
F1 =
dxdydzδ(x
+
y
+
z
−
1)
+
(4π)d/2 0
∆2−d/2
2
Γ(3 − d/2) 2
2
2
2
q [2(1 − x)(1 − y) − xy] + m 2(1 − 4z + z ) − (1 − z)
∆3−d/2
(6.134)
where ∆ = (1 + z)2 m2 − xyq 2 + zµ2 and = 4 − d as well as
α
F2 =
2π
Z
0
1
2m2 z(1 − z) dxdydzδ(x + y + z − 1)
2
∆
µ →0
(6.135)
α
where F2 (q 2 → 0) = 2π
. Note that the UV divergences are completely hidden inside F1
[Q 10: Is this true to all orders?]
6.6.1
Physics of F2
p0
→
Consider the semi classical picture of a fermion in an EM field
p→
EM field
Lint = eĀµ ψ̄γ µ ψ
(6.136)
To leading order we have
0
hp0 , s0 |iT |p, si = ūs (p0 )(−ieγ µ )us (p0 )õ (p − p0 )
and the Feynman rule is,
q
×
→ −ieγ µ õ (q)
(6.137)
6.6. VERTEX FUNCTION
79
In non relativistic electron scattering we have (for a scalar
Aµ = (φ((x)), 0) ⇒
potential,
s
√
ξ
õ = (φ̃(q), 0)). Recall that in the Weyl basis us = 2m
ξs
hp0 , s0 |iT |p, si = −ie ū(p0 )γ 0 u(p)φ̃(q) = −ie(2m)δss0 φ̃(q)
(6.138)
If we have a vector potential Aµ = (0, A) with B = ∇ × A then
hp0 , s0 |iT |p, si = −ieūs0 (p0 )γ i us (p)Ãi (q)
e s0† i s = −i(2m)
ξ σ ξ B̃i (q)
m
0 σi
i
[Q 11: I don’t see how we got from A to B.] (recall that γ =
).
σi 0
Now in the quantum mechanics Born approxiamtation we have,
hp0 , s0 |iT |p, si ∼ Ṽ (q)
(6.139)
(6.140)
(6.141)
This equation shows that the potential is given by V = me Ŝ · B = µ̂ · B. From none
relativistic QM we know that µ = g 2m
S, where g is the Landé g factor. Hence to leading
order it was found that g = 2. See Peskin and Schroeder pg 187-188 for more details.
Now consider the next to leading order expression
iσ µν qν
µ
= −ieū(p ) γ F1 (0) +
F2 (0) u(p)õ (q)
2m
0
×
The scalar field is V = eF1 (0) · φ(x). With the vector field one can show that you get the
same structure we found before, V ∝ S · B with a different g factor
g = 2 + 2F2 (0) = 2 +
α
π
(6.142)
This was the first radiative correction that was predicted and confirmed by experiment.
Chapter 7
Renormalized Perturbation Theory
7.1
φ4 Theory
Recall φ4 theory:
1
1
1
L = (∂µ φ)2 − m20 φ2 − λ0 φ4
2
2
4!
(7.1)
where
G̃(p) =
p2
iZ
+ (no poles)
− m2 + i
Define new renormalized fields as φ̃r = Z −1/2 φ which implies that G̃r (p) =
and we have
L=
(7.2)
i
p2 −m2 +i
Z
Z
Z2
(∂µ φr )2 − m20 φ2r −
λ0 φ4r
2
2
4!
+ ...
(7.3)
Experimentally we want to measure the mass, m and the coupling, λ. λ can be measured
from φφ → φφ at p → 0. More precisely we can measure
iM4 (s = 4m2 , u = t = 0) = λ
(7.4)
Notice that when we do this measurement we don’t know about what’s a loop diagram,
what’s tree level. This term includes all diagrams. We want to rewrite our Lagrangian
in terms of these physical parameters:
1
1
1
1
1
L = (∂µ φr )2 − m2 φ2r − λφ4r + δz (∂µ φ2 )2 − δµ φ2r − δλ φ4r
2
2
4!
2
2
4!
where δz = Z − 1, δm = Zm20 − m2 , δλ = Z 2 λ0 − λ. Our new Feynman rules are
80
(7.5)
7.2. QED
81
=
i
p2 − m2 + i
= −iλ
×
= i(p2 δz − δm)
×
= −iδλ
we impose our renormalization conditions:
−1 G̃2r (p2 )
2 2=0
p =m
d 2 2 −1 G̃r (p )
2 2 = −i
dp2
(7.6)
(7.7)
p =m
At first order these conditions are trivially satisfied. At higher orders these conditions
give us our values of δz , δm .
7.2
QED
The QED Lagrangian in terms of the renormalized parameters is,
1
L = ψ̄r (i∂/ − m)ψr − eψ̄r γ µ ψr Ar,µ − (Fr,µν )2
4
1
+ ψ̄r iδ2 ∂/ − δm ψr − eδ1 ψ̄1 γ µ ψr Ar,µ − δ3 (Fr,µν )2
4
(7.8)
The original theory was gauge invariant so we better maintain that gauge invariance. The
counter-terms must also be gauge invariant. The only way that we happen is if δ1 = δ2 .
To see this consider part of QED Lagrangian (the rest is trivialy gauge invariant on its
own),
∆L = δ2 ψ̄i∂/ψ + ieδ1 ψ̄1 γ µ ψAµ
(7.9)
Now consider a gauge shift Aµ → Aµ − 1e ∂µ α(x), ψ → e−iα :
δL = −δ2 ψ̄iγ µ i∂µ αψ − iδ1 ψ̄1 γ µ ψ∂µ α
= (δ2 − δ1 )ψ̄γ µ ψ∂µ α
(7.10)
(7.11)
82
CHAPTER 7. RENORMALIZED PERTURBATION THEORY
In order for this to be gauge invariant we require δ1 = δ2 . This highly non-trivial result
implies that the renormalization of the QED coupling is equal to the renormalization of
the fermion wavefunction renormalization! This implies that Z1 = Z2 and the renormalization couplings is simplify given by (see Eq. 6.6),
e = e0
p
Z3
(7.12)
We have the Feynman rules
→
i
p
/−m+i
→
−igµν
q 2 +i
×
→ i(p/δ2 − δm )
×
−i(g µν q 2 − q µ q ν )δ3
−ieγ µ
The renormalization conditions are
d
dp
/
d
dq 2
−1 =0
p
/=m
−1 −1 = −i
p
/=m
= −igµν
q 2 =0
= −ieγµ
The first two conditions provide us with δz , δm , the third conditions gives δ3 and the last
condition gives δ1 .
We now consider the two point function at 1-loop. We have already derived this
expression (the amputated version):
= i g µν q 2 − q µ q ν Π(q 2 )
(7.13)
gcg
7.2. QED
83
We have
gpg g gcg gxg
=
+
+
+ ... = −
igµν
1
ˆ 2)
+ i 1 − Π(q
q2
(7.14)
P
where Π̂ =
1PI diagrams. Our condition is that Π̂(q 2 = 0) = 0. We have (the
amputated version)
+
= i g µν q 2 − q µ q ν Πloop (q 2 ) + δ3
(7.15)
gcg gxg
and hence we know that
δ3 = −Πloop (q 2 = 0)
(calculated last lecture)
(7.16)
Chapter 8
Running Couplings and
Renormalization Group
8.1
“Large Logs”
Consider λφ4 theory. The phenomena happens in almost all QFT’s however it is convenient and easier to√work with φ4 theory. We consider 2 → 2 scattering. We will work at
a COM energy of s = µ m. We will basically be computing Green’s function like
this:
s channel
+
+
+
u channel
+
=
t channel
+ ... +
The first diagram is trivial and given by λ. Based on dimensional analysis we know that
the second diagram is
λ2
Λ
non-log
Γ=c
log √
+
+ δλ
(8.1)
Λ-ind terms
16π 2
s
Λ
λ2
δλ is given when Γ = 0. When s = 4m2 ⇒ δλ = −c 16π
log m
+ ... . Hence Γ is given by
2
λ2
Λ
Λ
non-log
log √ − log
Γ=c
+
(8.2)
Λ-ind terms
16π 2
m
s
λ2
m
non-log
=c
log √
+
(8.3)
Λ-ind terms
16π 2
s
(8.4)
Going to higher order terms we have higher order powers of these logs:
2
λ2
m
λ3
m
0
non-log
Γ=c
log √ +
+c
log √
+ ...
16π 2
(16π 2 )2
s Λ-ind terms
s
84
(8.5)
8.1. “LARGE LOGS”
85
√
Our perturbation theory is not a series in λ but it’s a series in λ log ms . This is an issue
since you lose control of your calculation at sufficiently high energies. This issue can be
avoided by organizing your perturbation theory in a better way.
We define
√
λ̃ ≡ i
s=µ
We impose
a different
renormalization condition given by δλ : Γ = 0 when s = µ2 ⇒ δλ =
λ2
log Λµ and we have
−c 16π
2
λ2
Γ=c
16π 2
µ
log √ + non-log Λ-ind terms
s
(8.6)
Instead of on shell renormalization condition you can impose your renormalization condition at any scale you want. In the end your results do not depend on the scale that you
renormalize however that’s only true for the sum of the series. Since we always truncate
our series, the error made by that truncation depends on the scale that you chose for
your renormalization scheme. Our λ̃(µ) is called the running coupling.
Type of PT
Bare PT
Renormalized
PT (on-shell)
Off-shell Renormalized PT
Coupling
λ0 → “bare coupling”
λ → “physical
coupling”
λ̃(µ) → “running
coupling”
Notes
doesn’t include
screening (scat√
tering at s ≈ Λ)
includes
screening (scattering at
√
s ≈ 2m )
√
partial screening 2m < s < Λ
o
screening
The “Renormalization Group evolution curve” is shown below
86
CHAPTER 8. RUNNING COUPLINGS AND RENORMALIZATION GROUP
λ̃(µ)
λ0
λ
µ
Λ
Now consider the Lagrangian:
L=
1
1
1
1
1
1
(∂µ φ)2 − m̃(µ)2 φ2 − λ̃(µ)φ4 + δZ (∂µ φ)2 − δm φ2 − δλ φ4
2
2
4!
2
2
4!
(8.7)
where as before we have
δZ = Z − 1, δm = Zm20 − m̃2 , δλ = Z 2 λ0 − λ̃
Before we had a certain scale and that was m. now our scale is µ instead. Before we go
ahead and start calculating. We try to think what our δ ’s will depend on :
(8.8)
δi = δi m̃, λ̃, Λ, µ
We can invert our formulas for δi to get
λ̃ = λ̃ (m0 , λ0 , µ, Λ)
m̃ = m̃ (m0 , λ0 , µ, Λ)
(8.9)
(8.10)
Z̃ = Z(m0 , λ0 , µ, Λ)
(8.11)
We fix m0 , λ0 , Λ (UV physics). We define the Beta functions as
∂ λ̃ ,
β≡µ ∂µ m0 ,λ0 ,Λ fixed
∂ m̃
βm ≡ µ
,
∂µ
1 ∂Z γ≡
µ
2Z ∂µ m0 ,λ0 ,Λ fixed
where for historical reasons the last expression has a different name (the γ function).
Note that if you use a different regulator you would keep that fixed instead of Λ.
To 1 loop in φ4 theory one can show that (assuming µ m)
β(λ̃) ≡ µ
∂ λ̃
3λ̃2
=
d log µ
16π 2
(8.12)
8.1. “LARGE LOGS”
87
This is called the RG equation (RGE). λ̃(Λ) ≡ λ0 ⇒
λ0
Z Λ
dλ̃
3
=
d log µ
2
16π 2 µ=µ
λ̃(µ) λ̃
3
1
1
Λ
=
−
log
2
16π
µ
λ̃(µ) λ0
Z
(8.13)
(8.14)
3λ0
Λ
1 + 16π
2 log µ
1
=
λ0
λ̃(µ)
λ0
λ̃(µ) =
3λ0
Λ
1 + 16π
2 log µ
(8.15)
(8.16)
(8.17)
This is the plot we showed earlier.
We don’t need to use λ0 we could use the non-relativistic coupling to estimate the
coupling at a different scale: λ̃(µ) = λN R ⇒ λ̃(µ) = 3λNλNRR µ . The large log will no
1+
16π 2
log
µ0
longer be in the PT. Note as µ → 0, λ → 0. Hence at arbitrarily low energies the particle
is free. These kinds of theories are called IR − f ree theories. It turns out that in practice
the running coupling does not run to zero. Instead when the coupling gets close to m, it
begins to level off. We will not show this. This turns out to be due to particles having a
nonzero mass.
Note further that the coupling blows up at
log
16π 2
Λ
=−
µ
3λ0
16π 2
ΛL = µ exp −
3λ(λ0 )e
(8.18)
(8.19)
This is called the “Landau pole”. The coupling becomes large at some scale. At this point
perturbation theory no longer makes sense. The theory is strongly coupled.
We now develop the formalism required to calculate these β functions. Consider the
following
via CT
z}|{
hΩ|T (φr (x1 )...φr (xn )) |Ωi = G(n)
(x
,
...,
x
;
λ̃,
m̃,
Λ, µ)
1
n
r
(8.20)
we also have
(n)
hΩ|T (φ(x1 )...φ(xn )) |Ωi = Gb (x1 , ..., xn ; λ0 , m0 , Λ)
(8.21)
88
CHAPTER 8. RUNNING COUPLINGS AND RENORMALIZATION GROUP
(n)
(n)
(n)
φr = Z −1/2 φ , Gb = Z n/2 Gr . But Gb is clearly independent of µ (the unrenormalized
propagator doesn’t depend on what renormalization scale you use). Hence we know that
d (n)
G =0
dµ b
d
µ
Z n/2 G(n)
=0
r
dµ
!
∂ λ̃ ∂
∂ m̃ ∂
∂
n n/2 ∂Z
µ Z n/2
+ Z n/2
+ Z n/2
+
Z
G(n)
r = 0
∂µ ∂ λ̃
∂µ ∂ m̃
∂µ 2Z
∂µ
∂
∂
∂
βλ
+ βm
+
+ nγ G(n)
r = 0
∂
m̃
∂µ
∂ λ̃
µ
(8.22)
(8.23)
(8.24)
(8.25)
(8.26)
where in the last step we used the definitions of β and γ functions as well as divided by
Z n/2 .
This is called the Callan-Symanzik equation. It applies very generally and not just to
4
λφ . One can show that to one loop one doesn’t need to worry about the βm . Conceptually
it is the same as the other functions however it is more convenient to ignore this term.
In other words we take m̃ → 0.
(2)
(4)
We consider 1-loop calculations: Gr and Gr .
G(2)
r
f
p→
=
+
c
fff
p→
p→
|
0,
fxf
p→
+
δZ
{z
by construction
p→
+ O(λ2 )
(8.27)
}
Our one loop diagram gives the CT. To first order we have,
Z
1
d4 `
2
Π(p ) = λ
= λN Λ2
4
2
(2π) ` + i
(8.28)
where N is a number. Note that this integral is independent of p. Our renormalization
condition is Π(µ2 ) + δZ = 0, which implies that δZ = −Π(µ2 ). To first order we have
G(2)
r =
p2
i
+ i
(8.29)
which is independent of λ̃, m̃, an µ. Thus the Callan-Symanzik equation says that in λφ4
the Gamma function is zero to second order. i.e. γ = 0 + O(λ2 ). Recall that our four
point function was given by
s channel
+
+
+
u channel
+
=
t channel
+ ... +
G̃(4)
r =
=
(8.30)
4
Y
i=1
i
2
pi + i
!
2
−iλ̃ + −iλ̃ (V (s) + V (t) + V (u)) − iδZ + O(λ̃3 )
(8.31)
8.1. “LARGE LOGS”
89
Consider the second loop:
p2
p4
p3
p1
It contribution to the vertex
Z
1
V (s) =
2
Z
1
=
2
Z
1
=
2
is
dd ` 1
1 (2π)d `2 (` + p)2 p=p1 +p2
Z
dd `
1
dx
d
2
(2π)
(x(` + 2p` + p2 ) + (1 − x)`2 )2
Z
dd `
1
dx
(2π)d
(`2 + 2`px + p2 x2 + ∆)2
(8.32)
(8.33)
(8.34)
where ∆ ≡ −p2 x2 + p2 x.
Z
Z
1
dd `
1
=
dx
d
2
(2π)
((` + px)2 + ∆)2
Z
Z
1
dd `
1
=
dx
2
(2π)d (`2 + ∆)2
Using the known integral we have
Z
Γ(2 − d/2)
1 1
dx i
=
(−∆)d/2−2
d/2
2 0
(4π) Γ(2)
We now take d = 4 − :
Z
1
Γ()
= i dx
(−∆)−
2
2
(4π) Γ(2)
Z
i
1
1
2
2
2 2
=
dx
−
γ
+
(6γ
+
π
)
1
−
log
p
(x
−
x)
E
E
32π 2
12
Z
i
1
2 2
dx
− log(p (x − x)) − γE
=
32π 2
1
i
2
=
+ 2 − log(p ) − γE
32π 2 or using the Mandelstam variable s ≡ p1 + p2 = p we can finally write
i
1
iM =
− log s + 2 − γE
32π 2 (8.35)
(8.36)
(8.37)
(8.38)
(8.39)
(8.40)
(8.41)
(8.42)
90
CHAPTER 8. RUNNING COUPLINGS AND RENORMALIZATION GROUP
or (the two constant are equivalent...)
i
1
=−
− log s + γE log 4π
32π 2 (8.43)
We set the renormalization condition given by s = µ2 ⇒ (∵ we have massless particles in
2
2
the CM frame) t = µ2 (1 − cos θ), u = µ2 (1 + cos θ). Our renormalization condition gives
us (using crossing symmetry)
δZ = −iλ̃2 (V (s) + V (t) + V (u))
3
λ̃2
2
− 3 log µ + 3γE log 4π
=
32π 2 We have
∂
∂
µ
+β
+ nγ G̃(4)
r = 0
∂µ
∂ λ̃
(8.44)
(8.45)
(8.46)
At order λ̃ we have 0 = 0. At O(λ̃2 ) we have µ dependence:
µ
∂ (4)
∂
3iλ̃2
G̃r = µ
(−iδλ ) =
∂µ
∂µ
16π 2
and we have
(8.47)
(4)
∂ G̃r
β
= −iβ + O(λ̃)
(8.48)
∂ λ̃
To order λ̃2 the gamma term still doesn’t contribute since O(γ) = λ̃, O(Gr ) = λ̃2 . Using
our equation relating the β functions we have
β=
8.2
3λ̃2
16π 2
(8.49)
QED β-function
We begin by considering the two photon Green’s function
Z
(2γ)
Gµν (p) =
d4 xe−ix·p hΩ|T (Aµ (x)Aν (0))|Ωi
=
g gFFg gxg
+
= i gµν
pµ p ν
− 2
p
+ O(e4 )
convention
z}|{
2
1 + e f (p, ) − δ3 (e, , µ)
+
p2
1
+ i
(8.50)
(8.51)
(8.52)
Here we also set me = 0 (or equivalently µ me ). Thus we have βm = 0. The CallanSymanzik (CS) equation is given by (this is exactly the same as before with renaming
the variables)
∂
∂
(8.53)
µ
+β
+ 2γA G(2γ)
µν = 0
∂µ
∂e
8.2. QED β-FUNCTION
91
We need to compute δ3 in order to know our µ dependence.
pµ pν
i gµν − p2
q→
1
=
2
q
1 − Π̂(q 2 )
gpg
(8.54)
Our on-shell renormalization condition was Π̂(q 2 = 0) = 0. We now abandon this condition and have the condition Π̂(q 2 = µ2 ) = 0. At one loop we have
FF
q→
q→
+
x
= Πloop (q 2 ) − δ3
(8.55)
and hence we have δ3 = Πloop (µ2 ). We found earlier that
Z 1
−8e2
Γ(2 − d/2)
2
dx(1
−
x)x
Πloop (q ) =
(4π)d/2 0
∆2−d/2
|
{z
}
(8.56)
1/6
0
z}|{
where ∆ = m2e −x(1 − x)q 2 . Expanding the denominator we have
4 e2
1
2
2
µ-independent
Πloop (q ) = −
− log µ +
constants
3 16π 2 (8.57)
where we have used
∆ = elog ∆ = e log ∆ = 1 + log ∆ + ...
(8.58)
3
3
Now β = O(e )(we’ll see why soon). If we are consistent to O(e ) we have the CS
equation gives:
∂
(−Pµν δ3 ) + 2γA Pµν = 0
∂µ
1
= i gµν − pµp2pν p2 +i
. Hence we have
µ
where we have defined Pµν
(8.59)
e2
1 ∂
γA =
δ3 =
2 ∂µ
12π 2
(8.60)
Hence
1 ∂
e2
hΩ|ψ̄(x)ψ(0)|Ωi ⇒ γψ = µ δ2 =
(8.61)
2 ∂µ
16π 2
This function is somewhat abstract and doesn’t have a physical meaning. However we
now go on to find the β function which is physical:
+
hΩ|T ψ̄(x)ψ(y)Aµ (0) |Ωi =
+
+
+
x
x
+
+
x
+
x
+
(8.62)
92
CHAPTER 8. RUNNING COUPLINGS AND RENORMALIZATION GROUP
The CS equation is given by
∂
∂
µ
+β
+ γA + 2γψ G(3) = 0
∂µ
∂e
(8.63)
To lowest order we have
The counterterms are
+
}ED
x
=
i
ν i
0
0 (−ieγ ) Dµν (q) ≡ (−ie)Fµ (p, p )
p/
p/
+
+
x
G(3) =
x
(8.64)
x
1
= −ieFµ + (ip/δ2 )Fµ + ...
p/
= −ie(δ1 − 2δ2 − δ3 )Fµ
(8.65)
(8.66)
so we have
∂
0 = (−ie)µ (δ1 − 2δ2 − δ3 )Fµ + β(−i)Fµ + (2γψ + γA )(−ieFµ )
∂µ
∂
β = e −µ (δ1 − 2δ2 − δ3 ) − 2γψ − γA
∂µ
1 ∂
β = eµ
−δ1 + δ2 + δ3
∂µ | {z } 2
(8.67)
(8.68)
(8.69)
=0
where δ1 = δ2 is zero by gauge invariance.
Thus we have
eµ ∂
e3
β=
δ δ3 =
2 ∂µ
12π 2
Recall that
α(µ) =
α(µ0 )
1−
2α(µ0 )
3π
log µµ0
The coupling goes as
(
)
1/137
me
L
(8.70)
(8.71)
8.2. QED β-FUNCTION
where ΛL = me exp
3π
2αN R
93
mP lanck
Chapter 9
Non-Abelian Gauge Theories
(“Yang-Mills”)
9.1
Looking Closely at Abelian Gauge Theories
QED is an “Abelian Gauge Theory”. You start with a Dirac Lagrangian:
L = iψ̄γ µ ∂µ ψ − mψ̄ψ
(9.1)
This is the only thing you can write down that is renormalizable, Lorentz invariant, and
only has a fermion field. Once you impose these requirements you see an accidental
“global U (1)” symmetry:
ψ → eiα ψ, ψ̄ → e−iα ψ̄
and hence L → L. What if we promote the global symmetry to a local symmetry? Then
we have “local U (1)” (this is also known as “gauge U (1)”).
ψ → eiα(x) ψ,
ψ̄ → e−iα(x) ψ̄
⇒
ψµ ψ → eiα ψ + i∂µ αeiα ∂µ ψ
(9.2)
and hence L0 6= L. We can fix this by letting ∂µ → Dµ where the covarient derivative
obeys
Dµ ψ → eiα Dµ ψ
(9.3)
This holds if
Dµ = ∂µ − ieAµ
(9.4)
1
∂ α.
e µ
where Aµ → Aµ +
This procedure is unique. To turn this into a real field we need
to add kinetic terms for A. These terms should be quadratic in ∂ µ Aµ and require gauge
invariance. The only way this can be done is by taking Fµν F µν or µναβ F µν F αβ . The
second term violates parity so we are left with the first term. We get our Lagrangian and
adding in a normalization we have
1
L = iψ̄γ µ Dµ ψ − mψ̄ψ − Fµν F µν
4
94
(9.5)
9.2. SU (2)
95
Here it was easy to find Fµν . However in different gauge theories it would be harder to
construct. We now consider a useful way to build such terms. Note that phase factors
commute with the covarient derivative (by construction). In other words
Dµ (eiα ...) = eiα Dµ (...)
(9.6)
with that in mind consider the commutator:
[Dµ , Dν ] ψ → eiα [Dµ , Dν ] ψ
iα
(9.7)
{(∂µ − ieAµ )(∂ν − ieAν ) − (∂ν − ieAν )(∂µ − ieAµ )} ψ → e [Dµ , Dν ] ψ
(9.8)
−ie(∂µ Aν − ∂ν Aµ )ψ → eiα [Dµ , Dν ] ψ
(9.9)
iα
−ieFµν ψ → e [Dµ , Dν ] ψ
(9.10)
(9.11)
but Fµν doesn’t transform under gauge symmetry. So we know that Fµν ψ → eiα Fµν ψ.
Thus we have
−eiα ieFµν ψ = eiα [Dµ , Dν ] ψ
(9.12)
[Dµ , Dν ] ψ = −ieFµν ψ
(9.13)
Thus we can finally write
9.2
SU (2)
We now generalize U (1) → SU (2). We put two fields with the same mass in a “doublet”:
ψ1 (x)
Ψ(x) ≡
(9.14)
ψ2 (x)
L=
2
X
iψ̄a γ µ ∂µ ψa − mψ̄a ψ
(9.15)
a=1
= iΨ̄γ µ ∂µ Ψ − mΨ̄Ψ
(9.16)
ψ̄1
where Ψ̄ ≡
. This Lagrangian has an accidental global symmetry given by rotation
ψ̄2
between the two doublets,
Ψ→VΨ
(9.17)
where V is a 2 × 2 unitary matrix. The possible V matrices form a group known as U (2).
We can write V as
!
3
X
σk
(9.18)
V = exp iα0 + i
αk
2
k=1
96
CHAPTER 9. NON-ABELIAN GAUGE THEORIES (“YANG-MILLS”)
where 1, σ2k are the generators of U (2) algebra. We can split this symmetry into two
parts: U (2) = U (1) × SU (2). SU (2) is defined in terms of its algebra:
hσ σ i
σk
j
i
,
= iijk
(9.19)
2 2
2
where ijk are the structure constants of SU (2).
We first consider the first identity generator. If we demand a local symmetry out of
this generator we would get back QED for two different fields. Now consider the case
of det V = 1 (an SU (2) symmetry). The identity operator does not obey this condition
since it is not traceless. Thus we only have the Pauli matrices as our generators and we
have
!
3
X
σk
(9.20)
V = exp i
αk
2
k=1
We promote this symmetry to a local (gauge) SU (2) symmetry.
σk V (x) = exp iαk (x)
2
(9.21)
where
Ψ→VΨ ,
Ψ̄ → V † Ψ̄
but
∂µ Ψ → V ∂µ Ψ + (∂µ V )Ψ
(9.22)
and hence L is not invariant under this transformation. We need to introduce a covarient
derivative. We define
Dµ Ψ → V Dµ Ψ
(9.23)
We have
12×2 ∂µ − ig õ
(9.24)
(∂µ − ig õ )Ψ → V ∂µ Ψ + (∂µ V )Ψ − ig Ã0µ V Ψ
(9.25)
where õ is some matrix.
To find our covarient derivative. We enforce the requirement in equation 9.23. In other
words we set this equation to V (∂µ − ig õ )Ψ:
i
(9.26)
Ã0 V = V Ã − (∂µ V )
g
i
Ã0 = V ÃV † − (∂µ V )V †
(9.27)
g
P
For linear order in α we can write V = 1 + i 3k=1 αk σ2k . It is sufficient for us to restrict
ourselves to Hermitian, traceless 2 × 2 matrices. In this case we can write
õ =
3
X
a=1
Aaµ
σa
2
(9.28)
9.2. SU (2)
97
With this one can show that our requirement for the covarient derivatives finally says
1
a
a
abc b c
Aaµ → Aa0
Aµ α
µ = Aµ + ∂µ α + g
(9.29)
Note that our vector boson transformation looks identical to the U (1) transformation
with an extra part. All of this can be applied equally well to U (1) however since the
group is so simple it is fairly trivial.
We write our Lagrangian as
L = iΨ̄γ µ Dµ Ψ − mΨ̄Ψ
(9.30)
= LD [ψ1 ] + LD [ψ2 ] + g
3
X
Aaµ
2
X
a=1
ψ̄i
!
σ a
2
i,j
ij
γ µ ψj
(9.31)
where LD [ψ] = iψ(∂/ − m)ψ. We can try to guess our Feynman rules. This is a bit
premature since we haven’t quantized our gauge fields. However let’s give it a shot. We
have the interactions
gE
µ,a
j
→
−igγ µ
σ a
2
ij
i
0 1
we have different fermion interaction.
1 0
We have yet to make these gauge fields dynamical. We need to add a kinetic term for
these fields. To do this we use the trick we learned from QED:
a
a
a
a
aσ
aσ
aσ
aσ
[Dµ , Dν ] Ψ =
∂µ − igAµ
∂ν − igAν
− ∂ν − igAν
∂µ − igAµ
Ψ
2
2
2
2
For example for σ1 =
(9.32)
a b σ
a
a
a b σ
= −ig ∂µ Aν − ∂ν Aµ − igAµ Aν
,
Ψ
2 2
σa
Ψ
= −ig ∂µ Aaν − ∂ν Aaµ + gabc gAbµ Acν
{z
} 2
|
(9.33)
(9.34)
a
Fµν
(9.35)
a
a
a
We call this Fµν
our field strength. This object is antisymmetric, Fµν
= −Fνµ
. We define
k
k σ
−ig F̃µν Ψ ≡ [Dµ , Dν ] Ψ or F̃µν = Fµν 2 . We have
− ig F̃µν Ψ → −ig F̃µν V Ψ
(9.36)
F̃µν → V F̃µν V †
(9.37)
and
98
CHAPTER 9. NON-ABELIAN GAUGE THEORIES (“YANG-MILLS”)
We have two options of terms that conserve parity to add into our Lagrangian F̃µν F̃ µν
or F̃µµ F̃ νν . However F̃µν = 0. We’re not done since F̃µν is still a matrix. The simplest
thing we can do is take the trace of this (we can take the determinate but that’s a mess):
tr F̃µν F̃ µν
(9.38)
The trace is gauge invariant:
tr F̃ F̃ → tr V F̃ V † V F̃ V †
†
= tr V V F̃ F̃
= tr F̃ F̃
(9.39)
(9.40)
(9.41)
Lastly we need to normalize this term. In terms of the non-matrix fields we have
1 (9.42)
L = − tr F̃µν F̃ µν
2
1 k,j
δ
z 2}| {
σk σ i
1 X k j,µν
Fµν F
tr
=−
2 k,j
2 2
(9.43)
3
1 X k k,µν
=−
F F
4 k=1 µν
(9.44)
so this normalization gives the same normalization we had in QED. This is the kinetic
term of 3 “photons” + cubic and quartic terms in A (due to the extra gijk Ajµ Akν addition).
These are self interactions of the gauge field!In QED these vertices did not exist at tree
level. However this is a property of non-Abelian gauge fields. We have diagrams such as
vg
vu
u
uv
and
even without any other particles. What’s important to note is that
the gijk Ajµ Akν where not optional, they were required by gauge invariance.
As an example, consider ψ ψ̄ → V V where V is a gauge boson. This an analogue to
e+ e− → γγ. This corresponds to
fg
fg
b
photon
+ crossing
+
deguv
Note that the last diagram is fundamental to non-Abelian gauge theories and is required
in order for the Ward Identity to work.
9.3
General Recipe
We now discuss gauge theories more generally. To construct the theory you need to
9.3. GENERAL RECIPE
99
• Choose a symmetry group (any Lie group, G). In our example this was SU (2).
• Choose a representation r(G). In our example we chose the fundamental representation (∼ , in Young Tableaux notation).
• Introduce n ≡ dim (r(G)) fields - “n-tuplet”. In our example we had n = 2ψ fields.
These fields are called “matter fields” (as opposed to gauge fields). These n fields
must be in the same representation of the Lorentz group. For Yang-Mills theories
these have to be spin 0 or spin 1/2. In our example we have spin 1/2.
• Consider a global symmetry
ψ →Vψ
(9.45)
where V ∈ r(G). We build the most general Lagrangian that is invariant under this
transformation:
L = iψ̄γ µ ∂µ ψ − ψ̄ψ
(9.46)
• Promote the global symmetry to local. In other words we rotate our ψ such that
ψ → V (x)ψ with
P Ng
V (x) = ei i αi (x)Ti
(9.47)
where Ng is the number of generators and the Ti are generators of the algebra
G in the representation r(G). The number of generators is independent of the
representation chosen. The number, Ng is equal to the dimension of the ad-joint
representation (dim(Adj(G))). Where for the ad-joint representation we have
(T i )jk = fijk
(9.48)
The generators are given in terms of the structure constants by [Ti , Tj ] = fijk Tk ,
where fijk are the structure constants of G. These Ti ’s are n × n matrices.
To promote to a local symmetry we need to require
∂µ → Dµ = ∂µ − ig õ
(9.49)
where õ always has the same form and given by
õ =
Ng
X
Aiµ (x)T i
(9.50)
i=1
and the T i are the generators in the representation r(G). We introduce one new
number that is our gauge coupling. Note that in order to enforce gauge invariance
you are forced to have the same coupling for each Aµ . This is the case for simple
groups (groups that do not have any generators that commute with all other generators). For non-simple groups we are allowed to have such coupling differences.
g
z }| {
For example you can split the couplings for U (2) through U (2) = U (1) × SU (2).
| {z }
g0
100
CHAPTER 9. NON-ABELIAN GAUGE THEORIES (“YANG-MILLS”)
We have the gauge field transformation
i
õ → V õ V † − (∂µ V )V †
g
(9.51)
It’s a simple exercise to show that
1
Aiµ → Aiµ + ∂µ αi + f ijk Ajµ αk
g
(9.52)
Note that there is no dependence on r(G). Further note that you can have any
multiple of matter fields. However there are a single set of gauge fields that are
fixed by G.
One last comment is often people say “gauge fields belong in the adjoint representation, Adj(G)” (The more correct way to say it is transforms as the adjoint
representation). For global transformations (α = const) we have
Aiµ = Aiµ + f ijk Ajµ αk
= Aiµ + (T i )jk Ajµ αk
(9.53)
(9.54)
so Aiµ transforms through adjoint representation. However this is only for a global
transformation and is strictly wrong in general due to the ∂α αi term.
• The Field Strength Tensor is defined as
ig F̃µν ψ [Dµ , Dν ] ψ
(9.55)
where Dµ are n×n matrices where n is the dimension of the representation. Further
F̃µν =
Ng
X
i
Fµν
Ti
(9.56)
i=1
i
where Fµν
= ∂µ Aiν − ∂ν Aiµ + gf ijk Ajµ Akν . This is independent of r(G). The transformation law is given by
F̃µν → V F̃µν V †
(9.57)
• The Gauge Field Lagrangian is given by
1 L = − tr F̃µν F µν
2
assuming that the generators are normalized as tr (T i T j ) = 12 δ ij .
(9.58)
9.4. UNIVERSAL COUPLINGS
101
• The general Feynman rules are as follows given below (the first result will be derived
soon):
k→
−igµν δba gauge depend.
= 2
+
µ,a
term
q + i
ν,b
p→
iδij
=
p/ − m + i
i
j
ggg
fFf
Dg
j
= −igγ µ (T a )ij
µ,a
i
vug
vu
uv
µ,a
k&
←q
p%
ν,b
= gf abc (g µν (k − p)ρ + g νρ (p − q)µ + g ρµ (q − k)ν )
ρ,c
= −ig 2 f abe f cde (g µρ g νσ − g µσ g νρ + (b ↔ c, ν ↔ ρ) + (b ↔ d, ν ↔ σ))
where i, j = 1, ..., dim(r(G)) and a, b = 1, ..., Ng . We can motivate the cubic and
quartic vertices by looking at the kinetic term for the gauge bosons:
Ng
X
a=1
a
Fµν
F a,µν =
X
∂µ Aaν − ∂ν Aaµ + gf abc Abµ Acν
∂ µ Aa,ν − ∂ ν Aa,µ + gf ade Ad,µ Ae,ν
a
(9.59)
9.4
Universal Couplings
An important property of gauge theories is that every particle charge under that the
gauge group will have the same coupling constant unless the group is Abelian. We now
prove this statement below.
Suppose you have some Lie group with abstract generators τa , then the generators
obey,
[τa , τb ] = ifabc τc
(9.60)
and a representation of the group with generators Ta obeys the same algebra,
[Ta , Tb ] = ifabc Tc
(9.61)
Now if the theory has two fields ψA , ψB (we consider fermions though one could just as
easily discuss bosons) in the same reprsentation of a gauge group then by the definition
of a gauge group they must transform the same way:
a
ψA → eiTa θ ψA
θa
ψB → eiTa ψB
(9.62)
(9.63)
102
CHAPTER 9. NON-ABELIAN GAUGE THEORIES (“YANG-MILLS”)
Now the Lagrangian of these fields is of the form,
L = ψ̄A DµA γψA + ψ̄B iDµB γ µ ψ + ...
(9.64)
where the covariant derivatives are,
DµA = ∂µ − gA Wµa Ta
(9.65)
DµB = ∂µ − gB Wµa Ta
(9.66)
where here we assume different gauge couplings. But the first derivative implies that the
Wµa field transforms as,
Wµa → Wµ + f abc Wµ,b θc +
1
∂µ θa
gA
(9.67)
1
∂µ θa
gB
(9.68)
and the second derivative implies that
Wµa → Wµ + f abc Wµ,b θc +
However Wµa are the same fields in either case. Thus we must have gA = gB .
In the above discussion we assumed that the fields were in the same representation
of the gauge group. This discussion fails if the fields are in different representations.
In an Abelian group since there are no commutation relations, every field can be in a
different representation of the Lie group. Thus every Abelian field can have different
gauge couplings.
9.5
Quantum Chromodynamics
Quantum Chromodynamics (QCD) is Yang-Mills theory with an SU (3) gauge group. For
SU (N ) the number of generators is Ng = N 2 − 1. In this case since N is 3 we have 8
generators which in turn gives 8 gauge fields Aaµ (a = 1, 2, ..., 8). We call these fields
gluons. There is one coupling constant g. They are given by structure constants f abc .
We want to write the matter fields that are charged under this gauge group. The
fundamental matter fields are the quarks. We only consider QED so we call these spin 1
Dirac fermions (in reality due to the weak interaction we should split these into Weyl
2
fermions). The quarks are qf = (u, d, s, c, b, t) (f = 1, ..., 6) we call this the “flavor” of
the quarks. The spin of the particles are the same. The electric charges split into 3 sets
of doublets. Their could be a symmetry in between these sets. However it would not
exact since they have different masses. We take one of the quarks for simplicity (they
all have the same interactions). These transform as the fundamental representation of
SU (3). The transformations act on vectors given by U . dim() = N (Young Tableaux
notation).


U1
U =  U2 
(9.69)
U3
9.6. QUANTIZATION OF YANG MILLS
103
where 1, 2, 3 → r, g, b (“colour”). The Lagrangian is given by
9
L=
X
iq¯f γ µ Dµ qf − mq¯f qf −
f
1 X a a,µν F F
4 a=1 µν
(9.70)
P
where Dµ = ∂µ − ig 8a=1 Aaµ T a and T a are the generators of SU (3) in the fundamental
representation, the “Gell-Mann matrices”.
9.6
Quantization of Yang Mills
We have
Z
W0 =
" Z
!#
Ng
X
1
DA exp i
d4 x −
F a F a,µν
4 a=1 µν
(9.71)
Q g Q3 Q
a,µ
where DA = N
a=1
µ=0
x dAx . We tried this in QED and we ran into a problem. The
procedure was to go to Fourier space and we got a Green’s function which we needed to
invert and it was not invertible. We traced that problem back to the fact that this integral
is redundant (we were dealing with all the gauge fields). We required the Faddeev-Popov
procedure. The idea is the same here. The gauge freedom gives
1
α(x)
Aaµ −−→ Aaµ + ∂µ αa + f abc Abµ αc
g
(9.72)
There is a huge amount of redundancy in this integral which we need to eliminate.
The procedure is as follows. You choose some functional of the gauge field and you
choose this to be zero, G [A] = 0. Geometrically:
A
Int space G[A]=0
Orbits
We have a mathematical identity:
Z
δG [Aα ]
α
1 = Dαδ (G [A ]) det
δα
(9.73)
We have
Z
W0 =
Z
δG [Aα ]
4
α
α
DA Dα exp i
d x (L [A ]) δ (G [A ]) det
δα
α
(9.74)
104
CHAPTER 9. NON-ABELIAN GAUGE THEORIES (“YANG-MILLS”)
where we shifted our integration measure to Aα . Furthermore, the Lagrangian is by
definition gauge invariant so we have L [Aα ] = L [A]. We now relabel our integration
variable.
Z
Z
δG [Aα ] 4
W0 = DADα exp i
d x (L [A]) δ (G [A]) det
(9.75)
α
δα
A →A
R
Now it’s clear that Dα = N is just a constant. So we can forget about it
Z
Z
δG [Aα ] 4
W0 = DA exp i
d x (L [A]) δ (G [A]) det
(9.76)
α
δα
A →A
For U (1) we had G [A] = ∂µ Aµ − ω(x) and G [Aα ] = ∂µ Aµ + 1e ∂ µ α − ω(x). Clearly
δG
= const (A independent).
δα
For non-Abelian gauge theories we have
G [A] =
Ng
Y
(∂µ Aa,µ − ω a (x))
(9.77)
a=1
and
Y 1 µ a
a,µ
abc b,µ
G [A ] =
∂µ A + ∂ α + f A αc − ωa (x)
g
a
(9.78)
δG
δca
=
∂µ ∂ µ + f abc ∂µ Ab,µ
δαc
g
(9.79)
α
So we have
The definition of the first term will become clear soon. In order to make sense
expression we go back to the discrete version of our integrals. x → xi , α(x)
Aaµ (x) → Aaµ,i .
1 α
a
α
abc b,µ c
a,µ
G [A (x)] = ∂µ A + ∂ αa + f A α − ω a (x)
g
a,µ
a,µ
b,µ
µ
A − Ai
1 1
a
a
abc Ai+1 − Ai c
→ i+1
+
α
+
α
−
2α
+
f
α
i+1
i−1
i
∆
g ∆2
∆
αc − αic
+ f abc Ab,µ i+1
− ωia
∆
of this
→ αi ,
(9.80)
(9.81)
So we have
δGa
dGai
→
δαc
dαjc
1
=
(δca )(δi+1,j + δi−1,j − 2δi,j ) + ...
2
g∆
a,i
≡ Mc,j
(9.82)
(9.83)
(9.84)
9.6. QUANTIZATION OF YANG MILLS
105
The next step was to convert everything in the path integral into a term in the Lagrangian.
The delta function becomes a gauge fixing term. In QED we had an extra
Z
R 4 2
Dωe− d xω /2ξ
(9.85)
and we have
1 X
Lgauge fix. =
(∂µ Aa,µ )2
2ξ a
(9.86)
which gives
Z
W =
Z
1 a a,µν
1
4
a,µ 2
DA exp i
d x − Fµν F
+ (∂µ A )
4
2ξ
(9.87)
The last question is how to convert the determinant into a term into the Lagrangian.
This is done by something which are known as ghosts. These in some sense behave like
ordinary fields. They take part into the Feynman rules but just cannot propagate.
Ghosts are essentially just way to exponentiate the determinate term so we will be able
i,a
Mj,b
}|
{
z
to perform a Taylor expansion. We want to rewrite the determinate: det ∂µ ∂ µ δ ab − gf abc ∂µ Ac,µ
in an exponential form. Faddeev and Popov showed that you can write
det ∂µ ∂ µ δ ab − gf abc ∂µ Ac,µ =
Z
Z
b
4
a
µ ab
abc
c,µ
DcDc̄ exp i
d xc̄ ∂µ ∂ δ − gf ∂µ A
c
(9.88)
The new ghost fields, c, c̄ are Grassman variables. Recall that for two Grassman variables:
Z
dθdθ̄eθ̄Aθ = A
(9.89)
Earlier we showed that,
Z
!
Y
dθi dθ̄i
P
e
i,j
θ̄i Ai,j θj
= det A
(9.90)
i
So any determinate you wish can be written as a product of integrals over Grassman
numbers. The c(x) are “ghost fields”. They are anti-commuting however unlike Dirac
fields they don’t have an extra index and are spin-0 particles. Anti-commuting spin-0
fields violate spin-statistics! On the other hand spin-statistics is only for real asymptotic
particles. So far we have not introduced a source for these ghost fields and adding
this source will violate spin-statistics theorem. In other words these cannot be external
particles (there are no S matrix elements for these particles).
106
CHAPTER 9. NON-ABELIAN GAUGE THEORIES (“YANG-MILLS”)
Adding in these ghost fields we have
Z
Z
1 a a,µν
1
4
a,µ 2
a
µ ab
W = DADcDc̄ exp i
d x − Fµν F
+
(∂µ A ) + c̄ ∂µ (D ) cb
(9.91)
4
2ξ
Z
Z
1 a a,µν
1
4
a,µ 2
µ ab
= DADcDc̄ exp i
d x − Fµν F
+
(∂µ A ) − (∂µ c̄a ) (D ) cb
4
2ξ
(9.92)
where we have just integrated by parts to which the derivative term. Furthermore Dµ is
just the just regular covarient derivative since we have
(9.93)
∂µ ∂ µ δ ab − gf abc ∂µ Ac,µ = ∂µ ∂ µ δ ab − gf abc Ac,µ
ab c,µ
c
= ∂µ ∂ µ δ ab − g TAdj
A
(9.94)
The Feynman rules for these ghosts are,
k→
µ, a
ν, b
k→
µ, a
=
δ ab
−i
2
k + i
kµ kν
gµν − (1 − ξ) 2
k
−i
2
k + i
←
a
ν, b
=
δ ab
p
abc µ
c, µ = gf p
b
Notice that ghosts can only show up in pairs. With this in mind and the fact that they
cannot show up in external states, they only appear in loop diagrams. Thus in tree level
you can just forget about ghosts.
9.7
Unitary and Ghosts
The fact that ghosts can only show up in loops is not obviously consistent with the
unitarity of the S matrix. This topic is a bit subtle. We will not do many calculations
however we will go over the logic and series of statements that describe the situation.
The statement of unitarity is the statement that (this is the same as saying that
probability is correctly normalized)
S †S = 1
(9.95)
9.7. UNITARY AND GHOSTS
107
gn where S is the S matrix. We split S into the part that’s associated with no collision
and the interested part which is due to the interactions, S = 1 + iT (see for example
[Peskin and Schroeder(1995)] pg. 104).
S = 1 + iT
(9.96)
(1 + iT ) 1 − iT † = 1
1 + i T − T † + T †T = 1
⇒ −i T − T † = T † T
(9.97)
Plugging this into 9.95 we have
This is further discussed in [Peskin and Schroeder(1995)]. What this says is that (for φ4 )
+ ... = (
*
)
d3p/E1 d3p/E2
This is known as the optical theorem. Unitarity becomes more intricate for non-Abelian
theories. We denote V as the following amplitude: (ψi ψ̄j → Va Vb , two fermions into two
gauge bosons). The square of this corresponds to ψi ψ̄ → ψk ψ̄` .
+
+
=Im
+
+
+
This is equivalent to
Z 3
X µν s1
d k1 d3 k2
Vs1 ,s2 (1, 2 → k1 k2 ) Vs∗1 ,s2 (k1 , k2 → 3, 4) =
M µ (k1 )sν2 (k2 )sρ1 ∗ sσ2 ∗ M∗ρσ
E1 E2
s s
1 2
(9.98)
P
Recall that in QED we had 2s=1 sµ (k)s∗
ν (k) = −gµν + Akµ kν where we relied on the
Ward identity to simplify our results.
This was all relatively simple however things get complicated in non-Abelian gauge
theories. It turns out that
external gauge bosons this replacement is not
P for ν2 or(more
(
(
(
allowed (we don’t have (((
µ (
ν → −gµν ). This is because when you have two gauge
bosons you can make one boson unphysical. If you make one boson longitudinal then
Ward identity for the other boson will not work. What always holds is
3
X
s=0
sµ ∗,s
ν = −gµν
(9.99)
108
CHAPTER 9. NON-ABELIAN GAUGE THEORIES (“YANG-MILLS”)
however this requires summing over all the polarizations. You can work this out and we’ll
basically be working this out on the homework.
This now sounds like a problem. Since on the left hand side of the unitarity relation
we have ’s (we have external gauge bosons) and on the right hand side we don’t have
any such terms. The term with ghosts fixes this problem. This is quite neat. The ghosts
are in a sense there to preserve unitarity. The formal proof of this is known as the BRST
theorem.
We now go back to tree level. We just said that at tree level we don’t worry about
ghosts. However consider ψi ψ̄j → Va Vb . We have
M = Mµν sµ1 (p1 )sν2
(9.100)
In QED we would have
QED
X
2
z}|{
|M| = gµα gνβ Mµν M∗,αβ
(9.101)
s1 ,s2
However in non-Abelian theories we can’t do that. We have two options
1. We can have explicit ’s. If p = (E, 0, 0, E) then = √12 (0, 1, ±i, 0). You can just
P
compute s1 ,s2 |Ms1 ,s2 |2 without using the spin sum formula. We will do this on
the homework. This always works and you never need to mention ghosts.
2. There is a second way to deal with this that is more
but a little bit
P3 convenient
s ∗,s
= −gµν (notice our
weird. This is to remember that we still
s=0 µ ν
P have
summing limits). What you can do is use µ ∗ν → −gµν but then you always need
to include an extra diagram:
2
In other words our tactic is to add an extra diagram with the external gauge bosons
are replaced with external ghosts.
9.8
Renormalized Perturbation Theory
This just means that we have counter terms. We start with the Yang-Mills Lagrangian.
We assume that the fermion is massless. This makes the calculations for the β functions
a lot easier but this is not necessary.
1 a a,µν
1
/ i,j ψ j − Fµν
L = ψ̄ i iD
F
+ (∂µ Aa,µ )2 − (∂µ c̄a )(Dµ )ab cb + LCT
4
2ξ
(9.102)
9.8. RENORMALIZED PERTURBATION THEORY
109
where
1
LCT = δ2 ψ̄i∂/ψ − δ1 g ψ̄γ µ Aaµ T a ψ − δ3 (∂µ Aaν − ∂ν Aaµ )2 + (4 more CT’s)
4
(9.103)
In the case of QED we did not have the last 4 terms since ghosts didn’t exist. In this
case gauge invariance no longer implies that δ1 = δ2 . This is because now for example
(∂µ Aaν − ∂ν Aaµ ) is not even gauge invariant since the gauge invariance is only saved by the
ghost CT’s.
What we are really interested in is the β function. To do this we need to use the CS
equations. We need to compute
gave
hAAi −−→ γA
gave
ψ̄ψ −−→ γψ
gave
∂e(M )
ψ̄ψA −−→ β = M
∂M
The formula we had in QED for the β function was
∂
1
β = eM
−δ1 + δ2 + δ3
∂M
2
(9.104)
If you look back at that procedure. Nothing in it used the Feynman rules of the theory
in anyway. We just used the structure of counterterms. In non-Abelian Yang Mills we
can do the exact same thing. Just like in QED the three point function will be
×
+
+
×
+
×
+
+
×
+
+
i
j
This calculation is historically significant and lead to a Nobel prize. The counterterms
have the following conditions
δ2 : Σ2 (k/ = M ) = 0
We evaluate the following
(9.105)
110
CHAPTER 9. NON-ABELIAN GAUGE THEORIES (“YANG-MILLS”)
+
k→
k→
×
The first diagram gives,
Feynman gauge
z }| {
d
(−igµν )
d
`
iδ
jk
ν
b
µ
a
iM = (ig)2
γ
(T
)
δab
γ
(T
)
k`
ij
/̀ + k/ + i
(2π)d
k 2 + i
Z
X
dd ` µ 1
1
a a
2
(T T )i`
γµ 2
= (ig)
γ
d
/̀ + k/ k
(2π)
a
{z
}
|
{z
}|
same as QED
C2 (rf )δi`
Z
ig 2 Γ(2 − d/2)
=
k/C2 (rf )δi`
(4π)d/2 (k 2 )2−d/2
(9.106)
(9.107)
(9.108)
This gives (using our renormalization condition for Σ2 )
δ2 = −
g 2 Γ(2 − d/2)
C2 (rf )
(4π)d/2 (M 2 )2−d/2
(9.109)
In this calculation we got lucky. We were able to factor our answer into a QED contribution and a group theory contribution. This will not happen again. We now calculate
δ1 . For this we use
→ igδ1 γ µ (T a )ij
Our renormalization condition is to set F1 (q 2 = M 2 ) = 0. We need to calculate
= (T b T a T b )ij ΓµQED (q, ...)
i
j
we have the matrix coefficient:
T b T b T a + T b T a , T b = C2 (rf )T a + T b f abc T c
(9.110)
9.8. RENORMALIZED PERTURBATION THEORY
111
Structure constants are antisymmetric so we can write this as
1
T b T b T a + T b T a , T b = C2 (rf )T a + f abc T b T c + T b T c
2
1
a
= C2 (rf )T + f abc T b T c − T b T c
2
1
a
= C2 (rf )T + f abc T b , T c
2
1
a
= C2 (rf )T + f abc f bce T e
2
1
a
f bac f bce T e
= C2 (rf )T −
|
{z }
2
(9.111)
(9.112)
(9.113)
(9.114)
(9.115)
b ) (T b )
(TAdj
ac Adj ce
1 b b
= C2 (rf )T a − (TAdj
TAdj )ae T e
2
1
a
= C2 (rf )T − C2 (Adj)δae T e
2
(9.116)
(9.117)
(9.118)
We now discuss the group structure of another diagram contribution:
a
j
i
This has the group theory contribution of
1 abc bce e
1
(T b T c )ij f abc = f abc T b , T c =
f f
T
2
2 | {z }
(9.119)
−C2 (Adj)δae
The full calculation gives
δ1 =
−g 2 Γ(2 − d/2)
(C2 (rf ) + C2 (Adj))
(4π)d/2 (M 2 )2−d/2
There is one more set of diagrams to calculate which is
+
+
×
+
(9.120)
112
CHAPTER 9. NON-ABELIAN GAUGE THEORIES (“YANG-MILLS”)
We now calculate the group factors. The fermionic loop has a group theory factor of,
(T a )ij (T b )ji = tr T a T b = C(rf )δ ab
(9.121)
while for the gluon loop we have,
f acd f bcd = C2 (Ajd)δab
(9.122)
A full calculation of the last counterterm gives


2
X
Γ(2 − d/2)  5
4
g
C2 (Adj) −
C(rf )
δ3 =
d/2
2
2−d/2
(4π) (M )
3
3 r
(9.123)
f
2
We have (M 2 ) = e log M = 1 + log M 2 + ... (and a similar expansion for 4π). Using
these expressions we have
1 Γ(2 − d/2) 1
1
2
=
+ γE − log M − log(4π) + O()
(9.124)
(4π)d/2 (M 2 )2−d/2 d=4− 16π 2 ∂
Using M ∂M
log M 2 = 2, Our expression for the β function is
#
"
g3
4X
11
β(g) = −
C2 (Adj) −
C(rf )
16π 2 3
3 f
(9.125)
We now interpret this result. The first interesting thing to notice is that this is non
zero if there are no fermions at all. What’s even more interesting is that β < 0. This
is completely different from the φ4 case and the QED case where the β function was
positive. This is interesting since we know that β controls how the coupling changes with
scale that we are studying the system at.
We had the graph in QED of
g
M
9.8. RENORMALIZED PERTURBATION THEORY
113
The reason this worked in this way was because the β function was positive. However in
this case we will have the opposite effect. Instead we have
g
1
M
This is called an asymptotically free theorem. However at low energies this theory becomes non-perturbative. You cannot take a zero momentum scattering (or equivalently
take the quantum mechanics limit). QCD of course has fermions however it doesn’t have
enough fermions to change the sign of the β function. In QCD the energy that we have
non perturbative results is about 1GeV or below.
In QED we had screening:
-+
-+
+
- +
- +
-+
+
-
+
In non-Abelian Yang Mills we have “anti-screening” due to gauge fields. To understand
where this anti-screening comes from we consider a SU (2) gauge theory. We have
1 a a,µν
L = − Fµν
F
+ gAaµ j a,µ + Lmatter
4
(9.126)
where j a,µ is the current (it’s easy to show that our earlier form of the Lagrangian is
a
equivalent to this with the appropriate conserved current) and Fµν
= ∂µ Aaν − ∂ν Aaµ +
gabc Abµ Acν .
We consider the equations of motion for the gauge field,
δL
δL
= ∂ν
a
δAµ
δ∂ν Aaµ
(9.127)
114
CHAPTER 9. NON-ABELIAN GAUGE THEORIES (“YANG-MILLS”)
Recall that in QED we had Ei = F0i . We can extend this idea to a non-Abelian gauge
theories and write, Eia = F0ia . This is often called the “chromo-electric field”. Note this
is a bit of a misnomer in this case since we are dealing with SU (2) but people often use
this term for all non-Abelian gauge theories. The equations of motion are
∂i Eia = gρa − gabc Abi Eic
(9.128)
where ρa ≡ j a,0 is called (in analogy with QED) the charge density. This is the analogue
to Gauss’s law.
Consider a point-like charge, ρa = δ 3 (x)δ a,1 . Note that we needed to specify a direction
in the group space for the charge. Notice that Gauss’s law is a non-linear equation. The
g
2
QED classical solution is E1 = 4πr
= E3 = 0. You can choose a gauge such that
2 r̂, E
a
a
A = 0 and A0 6= 0. If we do this then (since aab = 0) the difficult term disappears. So
we can recycle the QED solution.
This was a classical computation. However the β function arose from a quantum
solution (it came from loop corrections which necessarily imply the use of quantum mechanics). These non-linear terms inevitably come in at the quantum level. In quantum
mechanics we can’t just set A = 0. In other words, δA 6= 0. We have
->
A2
+1
+1
+3
-1
->
E1
anti-screening!
Consider δA2 . The sources are δE3 . So we have
∇ · E3 = gδA2 · E⊥
(9.129)
∇ · δE1 = −igδA2 · δE3
(9.130)
So we have
This is demonstrated above.
9.9
RG flows (evolution)
So far to display our running couplings we have shown plots of g as a function of the
renormalization scale. There is an alternative way to represent this.
9.9. RG FLOWS (EVOLUTION)
115
IR free
g
g( )
Such theories are called IR free. These include λφ4 , QED , Yang-Mills when β > 0.
Alternatively we can have UV or asymptotically free theories:
g
g( )
Asymptotically Free
This includes theories such as Yang-Mills with no/little matter (β < 0). The alternative,
which we didn’t see in our calculation but is possible is called “fixed-point”. This occurs
when β = 0, but λ 6= 0. You have an interacting theory but the coupling doesn’t run.
For example
*
*
IR-stable fixed point
The theory is “attracted” to some stable coupling. We have
β ∼ aλ + bλ2 + ...
(9.131)
Another thing that’s possible is so called Conformal Field Theories. These have β = 0
for any λ. For this to be the case you require some symmetry to make it this way. This
is also known as scale invariance.
Chapter 10
Broken Symmetries
10.1
SSB of Discrete Global Symmetries
Classical Analysis
The symmetries we have discussed so far have been exact. There are two types of symmetry breaking. Explicit symmetry breaking is fairly trivial and less interesting. Spontaneous symmetry breaking (SSB) plays a key role and nature and can be very subtle.
A field transformation is φ → Λφ (where φ is a scalar or a fermion, Λ ∈ Rep(G), and
G can be a Lie or discrete group. Λ is x-independent, global or a function of x and local.
It is a symmetry if
L → L + ∂µ J µ
(10.1)
As an example consider a local discrete symmetry. We take the Lagrangian to be given
by
1
L = (∂µ φ)2 − V (φ)
(10.2)
2
2
with V (φ) = µ2 φ2 + λ4 φ4 . This theory is invariant under φ → φ0 = −φ. This is a Z2
symmetry. This theory cannot produce an odd number of external legs. This is clear
diagrammatically:
An odd number of external legs is forbidden due to the Z2 symmetry. This symmetry
can be broken in two different ways.
116
10.1. SSB OF DISCRETE GLOBAL SYMMETRIES
117
1. This can be done explicitly by adding the term ∆V = φ3 to the Lagrangian. You
can then do perturbation theory in . To order 0 Z2 is a good symmetry.
2. Alternatively this can be broken spontaneously. We denoted the quadratic term
2
as µ2 φ2 . However we don’t really know the sign of µ2 . The way to tell the sign
of terms is by the requirement that the Hamiltonian ( 21 (φ̇2 + (∇φ)2 ) + V (φ)) is
bounded from below. To be bounded from below we must have λ > 0. However we
can have µ2 > 0 or µ2 < 0. In either case we have
2
V
2
>0
V
-v
<0
v
The first case is what we’ve done before. Consider µ2 < 0. Classically our system
2
will settle down in the lowest energy state. δV
= 0 ⇒ v ≡ −µ
. You don’t know
δφ
λ
which point the system will fall to. If we lived in a universe where this was present
and the field had enough time to evolve and settle at a lowest energy state. By
measuring this field we could find whether out field will be in either v or −v. So
even though the Lagrangian is symmetric about this symmetry, to an observer it
will appear the system is not symmetric. Once you specify which configuration your
system is in you have spontaneously broken the symmetry.
Quantization
In the normal case we quantize the field by finding the classical equations of motion:
φ + µ2 φ = −λφ3
(10.3)
We first solve the system in the case of λ = 0:
φ ∼ αk e−ik·x + αk∗ eik·x
(10.4)
where k 2 = µ2 . We then replace αk → âk , α̂k∗ = â†k . The vacuum is given by âk |0i = 0
for all k. Hence we have
h0|φ̂|0i = 0
(10.5)
It turns out that this approach doesn’t really work when µ2 < 0. The reason is very
simple. If you try to just solve this classically with µ2 < 0 then formally the solution still
works but now k 2 < 0 which means that at least something in k must be imaginary. In
particular it means that k 0 is imaginary, k 0 = ik̄. Hence the solutions aren’t oscillator
by grow in time. i.e.. φ ∝ ekt . Physically this is because we are looking at an unstable
system (when λ = 0) since the potential goes to negative infinity. We can’t just throw
118
CHAPTER 10. BROKEN SYMMETRIES
away λ because it’s what stabilizes the potential. We cannot take it to be zero. The
lesson is that we cannot just ignore this λ in the beginning.
The obvious thing to do is instead of ignoring λ which was basically equivalent to
expansions around small perturbations around zero, what we need to do is expand about
the actual minimum of the theory. This is the same thing you would do in classical
mechanics. If you want to find the oscillations of a ball in such a potential well then you
should expand about the minima.
Around the minima we have (from the Euler Lagrange equations)
φ +
δV
=0
δφ
(10.6)
φ0 = +v or φ0 = −v. We have
δV 1 δ 3 V δ 2 V φ +
(φ − φ0 ) +
+
(φ − φ0 )2 + ... = 0
δφ φ0 δφ2 φ0
2! δφ3 φ0
{z
}
| {z }
|
≡m2
interactions
(10.7)
It’s easy to show that all the interactions are proportional to λ. We define a new field
ξ(x) = φ(x) − φ0 . So we have
ξ + m2ξ ξ = 0
(10.8)
up to O(λ). Our new field is ξ and it has solutions of
ξ ∼ αe−ik·x + α∗ eik·x
(10.9)
with k 2 = m2ξ where m2ξ = −2µ2 . The quantization is trivial. We do the same replacement
we had before
ξ ∼ âk e−ik·x + â†k eik·x
(10.10)
We have the vacuum given by âk |0i = 0 for all k and then h0|x̂|0i = 0. However what is
the condition based on our initial field? Recall that φ̂ = φ0 + ξˆ and thus
h0|φ̂|0i = φ0
(10.11)
This is called the vacuum expectation value (VEV) of the system. This is a huge difference
in the physics!If for example the electric field (and hence Aµ ) had a VEV there would be
a background field everywhere which would mean if you had a charged particle it would
be accelerated in some direction in space.
We can now create particles
â†k |0i = |ki
(10.12)
These are massive particles that obey E 2 = k2 + m2ξ .
10.2. SSB OF CONTINUOUS GLOBAL SYMMETRY
119
You can insert the interactions in two different ways. One way is to rewrite the
potential in terms of ξ:
µ2 2 λ 4
φ + φ
2
4
1 2 2
= mξ ξ + λ3 ξ 3 + λ4 ξ 4
|{z}
|{z}
2
V (φ) =
∝λv
(10.13)
λ
Note that we have gained a cubic interaction out of a Lagrangian invariant under Z2 .
10.2
SSB of Continuous Global Symmetry
Classical Analysis
+iφ2
Consider V (Φ) = µ2 |Φ|2 + λ |Φ|4 where Φ = φ1√
. The symmetry of this Lagrangian is
2
iα
2
Φ → e Φ where α is a constant. If µ > 0 then the VEV is at Φ = 0. If µ2 < 0 then we
have
V
1
Vacuum Manifold
2
2
2
−µ
δV
and δV
= δΦ
. Note that before we had two minima but now
∗ = 0 which gives |Φ| =
δΦ
λ
we have an infinite number of vacua. All the different vacua are known as the vacuum
manifold. We can get all the vacuums by applying the transformations on the VEV:
r
−µ2
1
= eiα Φ0
(10.14)
Φα = eiα √
λ
2
This is a very general property of spontaneous breaking. We have spontaneous breaking
of U (1).
Quantization
Our equation of motion are
δV
=0
δφ1
δV
φ2 +
=0
δφ2
φ1 +
(10.15)
(10.16)
120
CHAPTER 10. BROKEN SYMMETRIES
Due to the symmetry the physical results will not depend on the vacuum we choose. We
choose to expand about φ1 = v and φ2 = 0. The equations of motion give
δV v
δ 2 V δ 2 V = 0 (10.17)
interaction
√
φ1 +
(φ
−
)
+
+
+
1
δφ1 δφ21 (v,0)
δφ1 δφ2 (v,0) terms
2
(φ1 ,φ2 )=(v,0)
δ 2 V δ 2 V ξ+
φ2 = 0 (10.18)
ξ + 2 δφ
δφ1 δφ2 1 (v,0)
in terms of ξ ≡ φ1 −
√v .
2
(v,0)
We also have a second equation:
δ 2 V δ 2 V φ2 = 0
(10.19)
φ2 +
ξ+ 2
δφ2 δφ1 (v,0)
δφ2 (v,0)
2
V We define a mass matrix: m2ij ≡ δφδi δφ
. We can put our equations together:
j
(v,0)
For our example we have V =
µ2
2
ξ
φ2
+M
(φ21 + φ22 ) +
λ
4
ξ
φ2
=0
(10.20)
2
(φ21 + φ22 ) . We can take the derivatives:
δ2V
= µ2 + 3λv 2 = −2µ2
2
δφ1
(10.21)
δ2V
(φ1 ,φ2 )=(v,0)
= 2λφ1 φ2 −−−−−−−→ 0
δφ1 δφ2
(10.22)
δ2V
λ
(φ1 ,φ2 )=(v,0)
= µ2 + + λφ21 + 3λφ22 −−−−−−−→ 0
2
δφ2
2
(10.23)
Lastly
Hence not only is the mass matrix diagonal but one of the eigenvalues is zero. In the
2
language of quantum field theory the δφVi become masses. We have two particles with
masses m2ξ = −2µ2 and mφ2 = 0. The massless particles are known as Goldstone bosons.
This mass also turns out to be included in the quantum corrections.
The existence of the particle is easy to understand classically. Consider a mechanical
system at the point near (v, 0) (see diagram of potential above). It’s clear that there is
one direction along φ1 that there is a restoring force. However we can also move along
φ2 along the vacuum manifold and not change the energy and hence there is no restoring
force. In the language of oscillator that means you have an oscillator with zero frequency.
The fact that the potential is flat was a consequence of the theory. This just follows from
the symmetry of the theory. Since the symmetry holds to all orders quantum corrections
do not change the mass of the Goldstone boson.
10.3. LINEAR SIGMA MODEL
10.3
121
Linear Sigma Model
Now that we know how this works in the U (1) case we will generalize these ideas in what’s
known as the linear sigma model. We start with N real fields, φi :
L=
N
X
1
i=1
2
(∂µ φi )2 + V (φi )
(10.24)
We consider a potential that is invariant under rotation of the fields about themselves:
φi = Ri,j φi . Where R is orthogonal: RT = R−1 . The only way this can occur is if the
potential only depends on the magnitude of the vector


φ1
 φ2 


φ =  .. 
 . 
φN
We take
V (φ) = µ2 φ2 + λφ4
(10.25)
!2
= µ2
X
i
φ2i + λ
X
φ2i
(10.26)
i
2
. The manifold is
As before we take µ2 < 0. The vacuum manifold is given by φ2 = −µ
λ
one dimension lower then the number of degrees of freedom in the system SN −1 . We pick
a vacuum
 
0
 .. 
 
φ= . 
 0 
v
We define σ(x) = φN (x) − v and we get
2
cubic and
1
2
−2µ σ +
V (φ) =
quartic
2
terms
(10.27)
The key point is that we don’t have any mass terms except for σ so we have N − 1
Goldstone bosons. We could have predicted how many Goldstone bosons will appear by
just thinking about the symmetry of the theory. We can express the symmetry of the
Lagrangian as R = eiαi Ti where Ti are the generators of O(N ). The dimensionality of the
O(N ) group is N (N2−1) . This comes from the condition that Ti ’s need to be antisymmetric.
Consider
Rφ0 = (1 + iαi Ti + ...)φ0
(10.28)
122
CHAPTER 10. BROKEN SYMMETRIES
We have two possibilities,
Ti φ0 = 0 Unbroken gen.
Ti φ0 6= 0 Broken gen.
The Ti matrices take the form



Ti = 


0
..
.



...
1 
−1 0
(10.29)
The number of broken generators is equal to the number of Goldstone bosons. This is
called Goldstone theorem. We did two examples but you can do any group you want.
In nature we don’t think anything like this is happening exactly. But in QCD there
are a lot of combinations of quarks. The pions for example are much lighter then the
other particles. You can discuss the physics of pions as Goldstone bosons. You can
add explicit breaking to give them some mass to correct that they are not massless. In
many beyond the SM theories people speculate that different particles are approximate
Goldstone bosons.
10.4
More on the Mexican Hat Potential
We now go back to the original Mexican Hat potential, V (Φ) = µ2 |Φ|2 + λ |Φ|4 , with
+iφ2
. If this were a mechanics problem then it would be clear
µ2 = −λv 2 < 0 and Φ ≡ φ1√
2
that it is more convenient to use fields that obey the symmetry of the problem. In this
case this would be a degree of freedom that represents the angle. There is nothing that
stops you from doing this in Quantum Field Theory.
With this in mind we introduce a non-linear field redefinition
1
Φ = eiπ(x)/v √ (v + σ(x))
2
(10.30)
Our potential becomes
µ2
λ
(v + σ)2 + (v + σ)4
2
4
1 2 2
= mσ σ + λ3 σ 3 + λ4 σ 4
2
V (Φ) =
(10.31)
where we drop the unimportant constant term and the linear term disappeared since
10.5. SSB OF GAUGE SYMMETRIES (“THE HIGG’S MECHANISM”)
δV
δφ
123
= 0. The kinetic term is given by
2
1 σ
+ ∂µ σ |∂µ Φ| = i∂µ π 1 +
2
v
1
1
1
1
= (∂µ σ)2 + (∂µ π)2 + (∂µ π)2 σ + 2 (∂µ π)2 σ 2
2v
|2
{z 2
} |v
{z
}
interactions
can. normalization
2
Notice that the interaction terms have couplings with dimensions of inverse mass!These
interactions appear non-renormalizable. However that is very naive and power counting is
not sufficient to test Renormalizability. This theory is still renormalizable (it is completely
equivalent to the original φ4 theory).
Further notice that π only enters the Lagrangian in terms of it’s derivative. Our U (1)
symmetry acts on π and σ in a very simple way:
π → π + α “shift”
σ→σ
(10.32)
(10.33)
To have an invariant Lagrangian we can’t have any dependence on π, since π is not
invariant. For example if we had a aπ 2 term we could apply a shift and get a(π + α)2
which differs from the first Lagrangian. Hence we can only have (as we see above),
L [σ, ∂µ π].
In the original theory renormalizability was clear and here the symmetry is clear.
Further we have
hΩ|T (π̃(p)...)|Ωi ∝ pµ · (...)
(10.34)
10.5
SSB of Gauge Symmetries (“The Higg’s Mechanism”)
Abelian Case
Today we will consider the Abelian example of SSB. Later we will the more intricate SM
SU (2) × U (1) → U (1) mechanism. We have the same potential as before only now we
consider a gauge symmetry:
1
L = |Dµ φ|2 − V (Φ) − Fµν F µν
4
(10.35)
where V (Φ) = −µ2 |Φ|2 + λ |Φ|4 (we make the minus sign on µ2 explicit such that now
µ2 > 0) and Dµ = ∂µ + ieAµ . Our VEV is
v2 =
and our vacuum is Φα = eiα √v2 and Aµ = 0.
µ2
λ
(10.36)
124
CHAPTER 10. BROKEN SYMMETRIES
We’ll start by solving this problem heuristically and follow up with a more rigorous
derivation. We fix Φ to it’s vacuum value, Φ0 = √v2 . This gives
2
v
e2 v 2
2
Aµ Aµ
LΦ kin = |Dµ Φ| = ieAµ √ =
(10.37)
2
| 2 {z }
mass term
We got a mass term for the gauge field, mA = ev. When we first began talking about QED
we rejected these terms because they were not gauge invariant (Aµ → Aµ − 1e ∂µ α). We
started with a theory that’s completely gauge invariant by shifting the Φ field, Φ → eiα(x) Φ
spontaneously broke gauge invariance. Picking the vacuum.
We now attack this problem more carefully. We begin by shifting our field:
1
Φ = eiπ/v √ (v + σ(x))
2
(10.38)
This shift gives
2
1 σ
+ ∂µ σ + ieAµ (v + σ)
∂µ + π 1 +
2
v
1
1
σ 2
= (∂µ σ)2 + (∂µ π + evAµ )2 1 +
2
2
v
This is the same kinetic term as before except for the extra gauge field terms:
|Dµ φ|2 =
(10.39)
1
ev∂µ πAµ + e2 v 2 Aµ Aµ
(10.40)
2
Note that 10.39 is still manifestly gauge invariant since σ → σ, π → π + α and Aµ →
Aµ − 1e ∂α .
The second mass term we knew would be there based on our earlier heuristic arguments. However there is a new interesting term, ev∂µ πAµ . This term is very strange.
This leads to propagators starting as one field and turning into another. This means that
we are probably not using a good basis. There is probably more then one way to solve
this problem and understand what the physics is. We will use the standard solution. The
idea is instead of trying to redefine fields, we have another approach. Recall that in QED
Lgauge fix = − 2ξ1 (∂µ Aµ )2 . This allowed us to compute a propagator. This is a remnant
of the choice we made that said that G [A] = ∂µ Aµ = 0. This was a good choice but it
was not unique.
Having said that, with theories with spontaneously broken symmetries we use what’s
known as Rξ gauge. Our gauge fixing Lagrangian is
1
Lgauge fix = − (∂µ Aµ − ivξπ)2
2ξ
1
e2 v 2 ξ 2
(∂µ Aµ )2 + evπ(∂µ Aµ ) −
π
2ξ
2
1
e2 v 2 ξ 2
= − (∂µ Aµ )2 − ev(∂µ π)Aµ −
π
2ξ
2
=−
(10.41)
10.5. SSB OF GAUGE SYMMETRIES (“THE HIGG’S MECHANISM”)
125
where we have integrated by parts. The middle term cancels the term we didn’t like
ev∂µ πAµ in equation 10.40. However there is a new strange mass term for the π field.
Based on physical arguments we there should not be any mass for π.
We have three propagators:
k→
−i
kµ kν
µ
ν
= 2
gµν − 2
(1 − ξ)
k − m2A
k − ξmA
1
p→
π
= 2
p − m2A
1
p→
σ
= 2
, m2σ = 2µ2
2
p − mσ
ggg
hhh
hhh
Note that we have two poles in the gauge boson propagator. One makes sense since it
corresponds to that we have a massive gauge boson. However we have a second gauge
dependent pole. This gauge dependent pole is cancelled by the π field. The propagators
always come together:
deged dehed
A
π
+
→ gauge independent
π cannot be real fields. We cannot have diagrams with external π fields.
To understand our system better we introduce Unitary gauge, ξ → ∞. In this limit
π drops out of the theory completely. The photon propagator becomes
k→
kµ kν
−i
µ
ν
gµν − 2
(10.42)
→ 2
k − m2A
mA
ggg
To understand the particle we consider the case of the particle on shell:
ggg
µ
k→
ν
2
→
− k →
m2A
3
X
−i
s (p)∗s
ν (p)
k 2 − m2A s=1 µ
(10.43)
We started with the theory of the photon that has two degrees of freedom and a
complex scalar field with two degrees of freedom. The overall number of degrees of
freedom was 2 + 2 = 4. After the symmetry is broken the picture changes, the gauge
boson gets a mass and has three degrees of freedom while the σ is a real scalar field and
has 1 degree of freedom. So we still have 3 + 1 = 4 degrees of freedom. The gauge boson
“ate” the π field and became massive.
Non-Abelian Case
Consider a scalar field, Φ, which is charged under some non-Abelian gauge symmetry, G.
In other words that
Φ → Φ0 = ΛΦ
(10.44)
126
CHAPTER 10. BROKEN SYMMETRIES
a
where Λ† Λ = 1 and Λ = eiαa (x)T (a = 1...Ng ) with Ng = dim (ajd(G)). Our Lagrangian
becomes
1 a a,µν
F
+ |Dµ Φ|2 − V (Φ)
(10.45)
L = − Fµν
4
where Dµ = ∂µ − igAaµ T a . The vacuum state (Φ0 ) is where
δV =0
(10.46)
δΦi Φ0
a
a
If Φ0 is a solution then Φα = eiα T Φ0 is also a solution. There are two possibilities. Either
we have T b Φ0 6= 0 or T a Φ0 = 0. The first case corresponds to the broken generators while
the second corresponds to unbroken generators. We Nb such that b = 1, ..., Nb then the
vacuum manifold is an Nb -dimensional space. Each broken generator gives a Goldstone
boson.
Just as we did last time parametrize our field non-linearly as
Φ = eiΠ
b T b /v
Φ0 [(σ − fields)]
(10.47)
The gauge transformation gives π b (x) → π b + αb (x), and σ → σ. Expanding about the
vacuum, Φ0 gives a term
|Dµ Φ|2 → g 2
Nb
X
Aaµ Ab,µ (T a Φ0 )† (T b Φ0 )
(10.48)
a,b=1
Note that the sum is over broken generators since the unbroken ones gives 0. We now
examine this term more closely. Claim: T a Φ0 (where
a runs over unbroken generators)
2V
2
with zero eigenvalue. Proof: Suppose
are eigenvectors of the mass matrix Mi,j
≡ δΦδi δΦ
j
Φ0
that Φ → eiαa Ta Φ is a symmetry. Then if we consider infinitesimal α such that
Φ → Φ + δΦ
δΦi = αa (T a Φ)
Since V must be invariant under this transformation (by assumption) then
δV =
δV
δΦi = 0
δΦi
which implies that
αa (T a Φ)i
δV
=0
δΦi
Taking the derivative of both sides gives:
αa
δ
δV
δ2V
(T a Φ)i
+ αa (T a Φ)i
=0
δΦj
δΦi
δΦi δΦj
10.5. SSB OF GAUGE SYMMETRIES (“THE HIGG’S MECHANISM”)
Plugging in Φ =
δV Φ0 δΦ
i
127
gives
Φ0
δ 2 V (T Φ0 )i
δΦi δΦj Φ0
a
(10.49)
as required.
The Mass matrix, M i,j is Hermitian, so eigenvectors can be chosen to be orthonormal:
(10.50)
(T a Φ0 )† T b Φ0 ∝ v 2 δ ab
P
which implies that g 2 v 2 b Abµ Aa,µ is a mass term for the gauge bosons. The number of
massive gauge bosons is equal to the number of broken generators. We don’t have any
Goldstone bosons left (all have been “eaten”). The masses are m2 ∝ |T a Φ0 |2 .
As an examplewe consider
SU (2). We define Φ is a fundamental “spin-1/2” rep
2
Φ1
resentation, Φ =
. The potential is given by V = −µ2 Φ† Φ + λ Φ† Φ where
Φ2
2
2
†
Φ Φ = |Φ1 | + |Φ2 | . The Lagrangian is invariant under Φ → ΛΦ where Λα = e−iα·σ/2 .
The vacuum is given by
0√
Φα = Λα
(10.51)
v/ 2
2
where v 2 = µλ .
We have
σ a Φα 6= 0
(10.52)
for a = 1, 2, 3 so we have three broken generators and hence 3 Goldstone bosons. Our
three gauge bosons are
2
2
σa
0
=v
√
(10.53)
2
v/ 2 8
2 2
for all a = 1, 2, 3 which implies that m2A g 4v .
This is very similar to the situation in the Standard Model, except in that case we
have an extra U (1) symmetry. In the Standard Model we see weak interactions which
give masses of mW ∼ 80GeV, mZ = 90GeV and only left handed fermions feel the weak
interaction. The Lagrangian is given by
L = iψL† ∂µ − igWµa ta σ̄ µ ψL + iψR† ∂µ σ µ ψR
(10.54)
ψL,1
where ψL =
.
ψL,2
†
The naive guess for the mass terms (the Dirac mass terms) is m ψL,a
ψR + ψR† ψL,a .
However this is not gauge
invariant!We need the Higg’s mechanism to fix this problem.
∗
H
We have H =
. The vacuum is given by
H0
v
hH 0 i = √
2
(10.55)
128
CHAPTER 10. BROKEN SYMMETRIES
which implies that mW =
gv
√
.
2
The Yukawa couplings are given by
λv †
ψR + ψR† ψL,1 + ...
LY uk = λ ψL† HψR + ψR† H † ψL = √ ψL,1
2
|{z}
(10.56)
mF
so but adding the Yukawa couplings to the Higgs field and give it a VEV we immediately
get fermion masses. Note
that to derive this formula we require the transformation
0
a a
H = eiπ t /v √v
where σ is the physical Higg’s boson.
+σ
2
Bibliography
[Griffith(2008)] D.J. Griffith. Introduction to Elementary Particles. Wiley-VCH, 2nd
edition, 2008.
[Peskin and Schroeder(1995)] M. Peskin and D. Schroeder. An Introduction to Quantum
Field Theory. West View Press, 1995.
129