Elementary Particle Physics - SS 04 Chapter 8

Contents
Elementary Particle Physics - SS 04
Chapter 8
Ian C. Brock
30th July 2004 – 17 : 26
8 From Dirac to Renormalisation 1
8.1 The Dirac Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
8.2 Helicity and Helicity Conservation . . . . . . . . . . . . . . . . . . . . . . . 4
8.3 Adjoint Spinors and Current Conservation . . . . . . . . . . . . . . . . . . 5
8.4 Decay Rates and Cross-Sections . . . . . . . . . . . . . . . . . . . . . . . . 7
8.4.1 Decays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
8.4.2 Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.5 Feynman Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
8.6 Higher Orders and Renormalisation . . . . . . . . . . . . . . . . . . . . . . 14
8.7 (g − 2) of Electron and Muon . . . . . . . . . . . . . . . . . . . . . . . . . 16
8.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
8 From Dirac to Renormalisation
The Dirac equation was the real starting point of relativistic quantum mechanics. It is the
equation that provides the wavefunctions of spin 1/2 particles and leads to the introduction
of antiparticles as equivalent partners to particles.
Using the wavefunctions which are the solutions of the Dirac equation one can calculate
scattering processes of fundamental particles. In the first part of the chapter I will introduce
the Dirac equation as well as “Fermi’s Golden Rule”, which one needs when one wants to go
from the transition amplitude or matrix elements, which one can calculate from Feynman
graphs to decay rates or cross-sections.
I have already introduced Feynman graphs as pictures which represent these calculations
in perturbation theory. What I have not mentioned so far is that when one starts to
actually calculate higher order graphs one very often gets answers which are infinite! This
held up the development of Quantum electrodynamics (the quantum field theory of the
electromagnetic interaction) for about 20 years! In the second half of this chapter I will
discuss the problem and the solution.
Lecture 16, 23/06/2004

8.1: The Dirac Equation 2
8.1 The Dirac Equation
The Klein-Gordon equation was introduced in Chapter 1, when I discussed antiparticles.
A particle in free space is described by the de Broglie wave function:
Ψ(x, t) = N exp[i(p · x − Et)] = Ne −ipx
Starting from the relativistic energy-momentum relationship and substituting operators
gives the Klein-Gordon equation:
∂ 2 Ψ
∂ 2 t = (∇2 − m 2 )Ψ
The equation is suitable for describing spinless bosons, as no spin variable has been introduced.
The Klein-Gordon equation is second order in the derivatives, which is the reason
for the negative energy solutions. The Klein-Gordon equation also has another problem,
in that it does not guarantee a positive-definite probability density for position, due to
the 2nd order time derivative. Although we now know the implications of negative energy
solutions, it provided the motivation for Dirac to try to find an equation that was linear
in the derivatives and Lorentz invariant, in the hope that the probability problem and the
one of of negative energy solutions could be solved.
The simplest form that can be written down is that for massless particles, in the form of
the Weyl equations:
∂Ψ
∂t
∂Ψ ∂Ψ ∂Ψ
= ± σ1 + σ2 + σ3
∂x ∂y ∂z
If this equation is valid it should also satisfy the Klein-Gordon equation. We square and
compare coefficients to find:
σ 2 1 = σ 2 2 = σ 2 3 = 1
σ1σ2 + σ2σ1 = 0 etc.
m = 0
These results hold for both signs on the right-hand side of the Weyl equation and so must
both be considered. The σ’s cannot be numbers, since they do not commute. In fact
the equations define the 2 × 2 Pauli matrices which you should already know from atomic
physics:
σ1 =
0 1
1 0
0 −i
, σ2 =
i 0
1 0
, σ3 =
0 −1
The equations can also be written in terms of the energy and momentum operators:
Eχ = −σ · pχ
Eφ = +σ · pφ
From Dirac to Renormalisation Lecture 16, 23/06/2004
.

8.1: The Dirac Equation 3
where E and p are the operators. χ and φ are two-component wavefunctions called spinors.
They are separate solutions of the two Weyl equations. As I will discuss in a few minutes
the equations have in total four solutions corresponding to particle and antiparticle states
with two spin substates of each.
If the fermion mass is included we need to enlarge the equations by including a mass term,
which leads to the Dirac equation:
EΨ = (α · p + βm)Ψ
The matrices α and β are 4×4 matrices operating on four-component spinor wavefunctions.
They are given by:
0
α =
σ
σ
,
0
I
β =
0
0
,
−I
where each element denotes a 2 × 2 matrix and ‘1’ denotes the unit 2 × 2 matrix and σ
are the Pauli matrices. The matrix α has three components. This representation is the
so-called Dirac-Pauli representation of the matrices – other representations are possible.
A more common way to write the Dirac equation is in its covariant form, replacing the
operators by derivatives:
µ ∂
iγ − m Ψ = 0
∂x µ
(iγ µ ∂µ − m) Ψ = 0
where the γµ (with µ = 0, 1, 2, 3) are related to the above matrices via:
γ µ ≡ (β, βα)
Note that Ψ is a four-element column matrix. However, it is not a 4-vector. I will return
to combinations that are 4-vectors a bit later. Writing out the matrices explicitly we have:
γ 0
I 0
0 σ
=
, γ =
, γ
0 −I
−σ 0
5
0 I
= .
I 0
It is useful to define the product matrix:
γ 5 ≡ iσ 0 σ 1 σ 2 σ 3 =
0 I
I 0
It is also very common to write down the Dirac equation in momentum space. Before and
after an interaction with particle propagate as plane waves in free space. The fermion field
can then be written as:
Ψ(x) = u(p)e −ipµxµ
From Dirac to Renormalisation Lecture 16, 23/06/2004

8.2: Helicity and Helicity Conservation 4
where u now only depends on energy and momentum and not on the position of the particle.
u is then a solution of:
(γ µ pµ − m)u(pµ) = 0
Although this equation does not look very different it is. The Dirac equation contains
derivatives. The above equation is a pure matrix equation and as such is much easier to
solve. One can construct four independent solutions to the Dirac equation – two of them
have positive energy and two of them have negative energy/ The positive energy four-spinor
solutions are:
u (1,2) = √
E + m
χ (1,2)
p·σ
E+m χ(1,2)
χ (1) =
1
0
Rewriting the negative energy solutions such that E > 0 we have:
v (1) (p) = u (4) (−p) = √ E + m
p·σ
E+m χ(2)
, χ (2)
0
=
1
χ (2)
v (2) (p) = −u (3) (−p) = √ p·σ
E + m E+mχ(1) χ (1)
where E is the energy of the particle or antiparticle defined by E = p 2 + m 2 .
The normalisation I have adopted is that from Halzen and Martin of 2E particles per unit
volume.
Using the definition of v given above, they satisfy the Dirac equation with the sign of pµ
reversed:
(γ µ pµ + m)v(pµ) = 0
8.2 Helicity and Helicity Conservation
For a massless particle with positive energy we have E = |p|. Inserting this in the first
equation for the spinors we have:
The quantity
σ · p
χ = −χ
|p|
H ≡
σ · p
|p|
= −1
is called the helicity or handedness of a particle. It measures the sign of the component of
the spin of the particle, jz = ± 1 in the direction of motion (z-direction). The z-component
2
of spin and the momentum vector p together define a screw sense. H = +1 corresponds
to a right-handed screw while H = −1 corresponds to a left-handed screw. The solution
From Dirac to Renormalisation Lecture 16, 23/06/2004

8.3: Adjoint Spinors and Current Conservation 5
of the equation given above represents a left-handed, positive energy particle. It can also
represent a particle with negative energy −E and momentum −p. In this case:
−Eχ = −σ · (−p)χ
H =
σ · (−p)
= +1
|p|
Thus the equation represents either a left-handed particle or a right-handed antiparticle
state.
From the definition you can see that helicity is a well-defined Lorentz-invariant quantity for
a massless particle. Such particles travel with the velocity of light. If one makes a Lorentz
transformation to another reference frame with velocity v < c, it is therefore impossible to
reverse the helicity. Neutrinos have a very small mass and for all practical purposes are
well described by one of the two Weyl equations. Solutions of the Dirac equation are not
pure helicity eigenstates, due to the mass term. Rather they are an admixture of the LH
and RH functions. However, at high energies, when the particles are highly relativistic,
the Weyl equations provide a good description also for massive particles.
An important consequence of this is that for interactions involving vector or axial vector
fields (i.e. those mediated by vector or axial vector bosons), helicity is conserved in the
relativistic limit. As the mediators of the strong, electromagnetic and weak interactions
are vector bosons, whose coupling is either vector or a mixture of vector and axial vector this
constraint applies to all fundamental interactions. The reason for helicity conservation is
that such interactions do not mix the LH and RH solutions of the Weyl equations. Writing
these as ΨL ≡ χ and ΨR ≡ φ, consider as an illustration the momentum operator U = p
operating on a mixture of LH and RH helicity states, i.e. Ψ ∗ UΨ, where Ψ = aΨL + bΨR,
a and b are constants and Ψ ∗ is the complex transpose of the two-component spinor Ψ. I
have omitted the space-time dependence e −ipx , as it cancels in the product. Making use of
the Weyl equations one finds that the cross terms are zero since Ψ ∗ L ΨR − Ψ ∗ R ΨL = 0. If U
were a scalar operator the cross terms would be proportional to Ψ ∗ L ΨR + Ψ ∗ R ΨL = 0 and
would not disappear.
In scattering processes at high energy a LH particle remains LH and RH remains RH.
This fact together with the conservation of angular momentum determines the angular
distributions in many interactions. It also implies that if one makes a beam of RH electrons
these will only interact with LH positrons and not with RH positrons. For studies which
one wants to make with polarised particles this is a great help, as one only needs to polarise
one of the beams.
8.3 Adjoint Spinors and Current Conservation
The adjoint spinor is defined as:
¯Ψ = Ψ † γ 0 = (Ψ ∗ 1, Ψ ∗ 2, −Ψ ∗ 3, −Ψ ∗ 4)
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From Dirac to Renormalisation Lecture 16, 23/06/2004

8.3: Adjoint Spinors and Current Conservation 6
From the conservation of probability, the rate of decrease of the number of particles in a
given volume is equal to the total flux of particles out of that volume. Thus:
− ∂
ρ dV = j ·
∂t
ˆn dS = ∇ · j dV
V
S
where I have used Green’s (or Gauss’) theorem. The probability and flux density are
therefore related by the “continuity” equation:
∂ρ
∂t + ∇ · j = 0
Expressing this in terms of the 4-vector j µ = (ρ,j) we have
∂µj µ = 0
I have not shown that the quantity so defined is a 4-vector, but one can.
Returning to the Dirac equation:
(iγ µ ∂µ − m)Ψ = 0
we can also write it in terms of the adjoint spinor as:
i∂µ ¯ Ψγ µ + m ¯ Ψ = 0
Multiplying the Dirac equation from the left by ¯ Ψ and the adjoint equation from the right
by Ψ and adding we obtain:
¯Ψγ µ ∂µΨ + ∂µ ¯ Ψ γ µ
Ψ = ∂µ
¯Ψγ µ
Ψ = 0
Thus
j µ = ¯ Ψγ µ Ψ
satisfies the continuity equation and suggests that we should identify j µ with the probability
and flux densities. In order to avoid problems with negative probability densities Pauli and
Weisskopf inserted the charge of the electron −e into j µ , so that ρ is the charge density
instead of the probability density:
j µ = −e ¯ Ψγ µ Ψ
j µ is the electron four-vector density, often called the electron current. To complete the
identification one must show that j µ transforms as a 4-vector, which again can be done.
********************* End skip *********************
From Dirac to Renormalisation Lecture 16, 23/06/2004
V

8.4: Decay Rates and Cross-Sections 7
The quantity
j µ = ¯ Ψγ µ Ψ
is the electron four-vector density, often called the electron current. It can be shown to
transform as a 4-vector.
For future reference it is convenient to write down the different forms of currents consistent
with Lorentz covariance. Ψ and ¯ Ψ are not 4-vectors, but one can form the following bilinear
quantities:
¯ΨΨ Scalar +under P
¯Ψγ µ Ψ Vector −under P (space components)
¯Ψγ 5 Ψ Pseudoscalar −under P
¯Ψγ 5 γ µ Ψ Axial Vector +under P (space components)
I have left out the tensor, as it is very rarely needed.
8.4 Decay Rates and Cross-Sections
In order to calculate the decay rate or scattering cross-section one needs 2 ingredients:
• The amplitude (M) or matrix element for the process and
• The phase space or density of final states available.
The amplitude contains all the dynamical information and we calculate it by evaluating the
relevant Feynman diagrams using the Feynman rules appropriate to the interaction we are
considering. The phase space factor contains only kinematical information; it depends on
the masses, energies and momenta of the participants and reflects the fact that a process
is more likely to occur if there is a lot of “room to manoeuvre” in the final state. For
example the decay of a heavy particle into many light secondaries has a large phase space
factor, as there are many different ways to apportion the available energy. In the β-decay
of the neutron there is almost no extra mass to spare and the phase space factor is very
small.
The transition rate for a given process is determined using Fermi’s “Golden Rule” given
by:
transition rate = 2π
¯h |M|2 × (phase space)
A non-relativistic derivation of the Golden Rule can be found most books on quantum
mechanics. For a relativistic version you have to consult a quantum field theory book. End of
Lecture
16
From Dirac to Renormalisation Lecture 17, 25/06/2004

8.4: Decay Rates and Cross-Sections 8
8.4.1 Decays
Suppose particle 1 decays into several other particles 2, 3, 4, ..., n. The decay rate is given
by the formula:
3
2 S c d p2
dΓ = |M|
2¯hm1 (2π) 3 3 c d p3
2E2 (2π) 3 3 c d pn
· · ·
2E3 (2π) 3
2En
×(2π) 4 δ 4 (p1 − p2 − p3 · · · − pn)
where pi = (Ei/c, pi) is the 4-momentum of the ith particle. The delta function enforces
the conservation of energy and momentum. We assume that the decaying particle is at
rest: p1 = (m1c,0). S is the product of statistical factors: 1/j! for each group of j identical
particles in the final state. The above formula gives the differential rate for a decay where
the momentum of particle 2 lies in the range d3p2 about the value p2, etc. Normally we are
not interested in individual momenta and so we integrate over the outgoing momenta to
get the total decay rate Γ for the mode we are considering. If there are only two particles
in the final state the result is:
Γ = S
c
2 2 1 |M|
δ
¯hm1 4π 2 E2E3
4 (p1 − p2 − p3) d 3 p2 d 3 p3
If we have a 2-body decay into massless secondaries π 0 → γγ then one can show that
(exercise?):
Γ =
S
|M|
16π¯hm1
2
i.e. the integration can be carried out independently of the form of |M| 2 (strictly true
in the case of spin 0 particles and also if we want the spin averaged amplitude). M is
evaluated at the momenta dictated by the conservation laws: p3 = −p2 and |p2| = mc/2.
One can consider the more general case of a 2-body decay in which the outgoing particles
have masses m2 and m3. In this case we have:
Γ =
S|p|
8π¯hm 2 1c |M|2
where p is the magnitude of either outgoing momentum. It is given in terms of the 3 masses
involved:
c
|p| = m
2m1
4 1 + m4 2 + m4 3 − 2m2 1m2 2 − 2m2 1m2 3 − 2m2 2m2 3
If both decay particles are massless then |p| = m1c/2, as we had above. The spin factor is
S = 1
2 for π0 → γγ, whereas it is S = 1 for π 0 → ν¯ν The final formula for 2-body decay is
very simple and general. If there are three or more particles in the final state the integrals
cannot be done until the form of M is known.
From Dirac to Renormalisation Lecture 17, 25/06/2004

8.5: Feynman Rules 9
8.4.2 Scattering
Suppose particles 1 and 2 collide, producing particles 3, 4, ..., n. The cross-section is given
by:
dσ = |M| 2 ¯h 2 S
4 (p1 · p2) 2 − (m1m2c2 ) 2
2 c d p3
(2π) 3 2 c d p4
2E3 (2π) 3 2 c d pn
· · ·
2E4 (2π) 3
2En
×(2π) 4 δ 4 (p1 + p2 − p3 − p4 · · · − pn)
This equation determines the cross-section for a process in which the momentum of particle
3 lies in the range d 3 p3 about the value p3, that of particle 4 lies in the range d 3 p4 about
the value p4 and so on. We often want to consider the case where we look at the angle
at which particle 3 emerges. We then integrate over all the other momenta and over the
magnitude of p3; what’s left gives us dσ/dΩ, the differential cross-section for the scattering
of particle 3 into solid angle dΩ.
By far the most common process that we will encounter is:
1 + 2 → 3 + 4
We start in the CM frame. In this frame p2 = −p1. Hence p1 · p2 = E1E2/c 2 + p 2 1. One can
then show that:
(p1 · p2) 2 − (m1m2c 2 ) 2 = (E1 + E2)|p1|/c
Thus
dσ =
2 ¯hc S|M|
8π
2c (E1 + E2)|p1|
d 3 p3 d 3 p4
E3E4
δ 4 (p1 + p2 − p3 − p4)
Again one can carry out the integration without knowing the explicit form of M and we
obtain:
dσ
dΩ =
2 ¯hc S|M|
8π
2
(E1 + E2) 2
|pf|
|pi|
where pf is the magnitude of either outgoing momentum and pi is the magnitude of either
incoming momentum. This equation applies to almost all scattering processes that we will
consider.
8.5 Feynman Rules
If we are given a matrix element we now know how to calculate a cross-section or a decay
rate. How do we calculate the matrix element? To start with I will discuss QED and restrict
myself to electrons, positrons and photons. Extending to the other leptons is trivial (they
are just a repeat of electrons and positrons). Extension to quarks is also straightforward
– one just has to take into account the different charge of the quarks and remember that
they come in 3 colours.
Free electrons and positrons are represented by the wave functions:
From Dirac to Renormalisation Lecture 17, 25/06/2004

8.5: Feynman Rules 10
ψ(x) = ae−ip·xu (s) (p) ψ(x) = aeip·xv (s) (p)
with s = 1, 2 for the 2 spin states
The spinors satisfy the Dirac equations in momentum space
(γ µ pµ − m)u = 0 (γ µ pµ + m)v = 0
and their adjoints satisfy
ū(γ µ pµ − m) = 0 ¯v(γ µ pµ + m) = 0
The solutions are orthogonal
ū (1) u (2) = 0 ¯v (1) v (2) = 0
normalised
ūu = 2m ¯vv = −2m
and
complete
s=1,2 u(s) ū (s) = (γ µ pµ + m)
s=1,2 v(s) ¯v (s) = (γ µ pµ − m)
I gave a set of solutions in the last lecture.
⇒ Transparency A complete set of spinors (spinor.tex)
We will normally (but not always) average over the electron and positron spins, and in
that case it does not matter that the solutions are not pure spin up and spin down. End of
Lecture
********************* Begin skip *********************
17
A free photon is represented by the wave function:
A µ (x) = ae −ip·x ɛ µ
(s)
where s = 1, 2 for the two spin states (or polarisations) of the photon. The polarisation
vectors satisfy the momentum space Lorentz condition The Lorentz condition in the
Maxwell equations is given by ∂µA µ = 0. The Maxwell equations in terms of the 4-vector
A µ = (V, A) are:
[∂µ (∂ µ A ν − ∂ ν A µ )] = 4π ν
J
c
The equation is invariant under a gauge transformation A µ → A µ + ∂ µ λ, where λ is an
arbitrary function of space and time. This gauge freedom can be exploited to impose an
extra constraint on the potential ∂µA µ = 0. One then has:
[✷A ν ≡ ∂µ∂ µ A ν ] = 4π ν
J
c
The potential still has a further gauge freedom provided that ✷λ = 0. I will not discuss
the details here. Most often the Coulomb gauge will be used, where A 0 = 0, which implies
that ∇ · A = 0. Thus the polarisation vectors satisfy:
ɛ µ pµ = 0
From Dirac to Renormalisation Lecture 18, 01/07/2004

8.5: Feynman Rules 11
The solutions are orthogonal:
and normalised:
In the Coulomb gauge we have:
µ ∗
ɛ (1) ɛµ (2) = 0
ɛ µ ∗ ɛµ = 1
ɛ 0 = 0 ɛ · p = 0
one can then show that the polarisation vectors obey the completeness relation:
∗
ɛ(s) ɛ i (s) j = δij − ˆpiˆpj
s=1,2
A convenient pair (in the Coulomb gauge, where ɛ · p = 0 and p points in the z direction)
is:
ɛ(1) = (1, 0, 0) ɛ(2) = (0, 1, 0)
A massless particle has only two possible spin states.
********************* End skip *********************
The procedure to calculate the amplitude M for a particular Feynman graph is sketched
in the transparency and given in somewhat more detail below:
⇒ Transparency Feynman rules (chapter08 figs.sxi)
1. Notation. Label the incoming and outgoing 4-momenta p1, p2, ..., pn and the corresponding
spins s1, s2, ..., sn; label the internal 4-momenta q1, q2, ..., qn. Assign arrows
to the lines: arrows on external fermion lines indicate whether it is an electron or a
positron; arrows on internal fermion lines are assigned so that the direction of flow
is preserved (i.e. every vertex must have one arrow entering and one arrow leaving).
2. External lines. External lines contribute the following factors:
• Incoming electrons: u
• Outgoing electrons: ū
• Incoming positrons: ¯v
• Outgoing positrons: v
• Incoming photons: ɛ µ
µ ∗
• Outgoing photons: ɛ
3. Vertex factors. Each vertex contributes a factor
igeγ µ
The dimensionless coupling constant ge is related to the charge of the positron via
ge = e = √ 4πα. For up-type quarks we must use 2
√
4πα and for down-type quarks
√ 3
4πα.
we use 1
3
From Dirac to Renormalisation Lecture 18, 01/07/2004

8.5: Feynman Rules 12
4. Propagators. Each internal line contributes a factor:
Electrons and positrons:
Photons:
i(γ µ qµ+m)
q2−m2 −igµν
q2 5. Conservation of energy and momentum. For each vertex write a delta function of
the form:
(2π) 4 δ 4 (k1 + k2 + k3)
where the k’s are the 4-momenta coming into the vertex. If the arrow for a particle
leads out of the vertex, then k is minus the 4-momentum of the line. If the arrow for
an antiparticle comes into the vertex (i.e. we have an outgoing antiparticle) then k
is also minus the 4-momentum of the line.
6. Integrate over internal momenta. For each internal momentum q write a factor
and integrate.
d 4 q
(2π) 4
7. Cancel the delta function. The result at the end will include a factor
(2π) 4 δ 4 (p1 + p2 + ... − pn)
corresponding to overall energy-momentum conservation. Cancel this factor and
what remains is −iM.
The procedure is then to write down all diagrams contributing to the process in question,
calculate the amplitude for each and add them up to get the total amplitude, which is then
inserted into the appropriate formula for the cross-section or lifetime.
One extra factor must not be forgotten. Fermions have antisymmetric wavefunctions.
Therefore we have to insert a minus sign in combining amplitudes that differ only in the
interchange of two identical external fermions. As the total amplitude will eventually be
squared, it does not matter which diagram you associate the minus sign with.
8. Antisymmetrisation. Include a minus sign between diagrams that differ only in the
interchange of two incoming (or outgoing) electrons (or positrons), or an incoming
electron with an outgoing positron (or vice versa).
Fermion loops need some special attention which I will say a bit about later.
The propagator has a form consistent with what one may naively expect. The exchange of
particles of high mass is suppressed by a factor 1/m 2 , as are large 4-momentum transfers.
I will not say any more in these lectures on how one can extract the explicit form.
********************* Begin skip *********************
From Dirac to Renormalisation Lecture 18, 01/07/2004

8.5: Feynman Rules 13
The explicit form of the propagator can be extracted from the Lagrangian. In general the
Lagrangian density contains a part Lfree for each participating field and an interaction term
Lint. As discussed at the beginning of the lectures, the form of the interaction term is given
by imposing local gauge invariance. The propagator is obtained by applying the Euler-
Lagrange equation to the spin-specific free Lagrangian. This yields the free field equation
for the particle, which is then written in momentum space using the usual prescription.
The propagator is i multiplied by the inverse of the field corresponding to the motion
operator.
• Spin 0 is the Klein-Gordon operator. In Hilbert space this is:
∂µ∂ µ + m 2 φ = 0
In momentum space using i∂µ → pµ this becomes:
p 2 − m 2 φ = 0
Hence the propagator is given by:
DF (spin 0; p) =
i
p 2 − m 2
• Spin 1/2 is the Dirac operator. In Hilbert space this is:
In momentum space we have
[iγ µ ∂µ − m] ψ = 0
[p − m] ψ = 0
where p = pµγ µ . Hence the propagator is given by:
DF (spin 1/2; p) =
i
p − m
= i p + m
p 2 − m 2
• Spin 1 with mass is the Proca operator. In Hilbert space we have:
∂µ (∂ µ A ν − ∂ ν A µ ) + m 2 A ν = 0
In momentum space this is given by:
2 2
−p + m ν
gµν + pµpν A = 0
where the metric gµν is given by
gµν =
The propagator is then given by:
⎛
⎜
⎝
1 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 −1
⎞
⎟
⎠
DF (Spin 1; m = 0; p) = −i gµν − pµpν/m 2
p 2 − m 2
From Dirac to Renormalisation Lecture 18, 01/07/2004

8.6: Higher Orders and Renormalisation 14
• Massless spin 1 is the Maxwell operator. In Hilbert space we have:
[∂µ (∂ µ A ν − ∂ ν A µ )] = 0
With the Lorentz condition ∂µA µ = 0, we have in momentum space:
2 ν
−p gµν A = 0
which implies that the propagator is as I gave above for the photon:
DF (Spin 1; m = 0; p) = −i gµν
p 2
Note that one cannot simply take the limit of the Proca operator as m → 0 to
obtain the propagator for a massless particle. Current conservation ensures that
pµpν/m 2 → 0 for massless particles.
********************* End skip *********************
The propagators as defined above are only valid for stable particles. For the Z and W ±
boson one has to include the Breit-Wigner form of the resonance.
8.6 Higher Orders and Renormalisation
While the basics for QED were laid down in the 30’s it took until the 50’s before it was
really established as a proper theory. The reason is that if you consider higher order
Feynman graphs and try to calculate them you typically get infinity! For example if you
consider electron-muon scattering, one diagram that one must consider is where the virtual
photon splits into an electron-positron pair for a short time (called vacuum polarisation).
⇒ Transparency Electron-muon scattering higher order (emuscat.mnf)
Bubble looks fairly harmless. However, when you calculate the graph the momentum of
the electron or positron is not defined and so you have to integrate over all possible values.
When you do so you find that if k is the 4-momentum of the electron (and q − k is the
momentum of the positron), the integral diverges as ln |k|!
Such infinities appear very many loop diagrams. How to get out of the problem? For large
values of q2 , i.e. hard scattering the integral has a term that goes as:
I(q 2 ) = α
3π ln
2 M
where M 2 is a large cut-off value rather than the true infinite value. I said that at each
electromagnetic vertex one has to write the factor √ 4πα = e. If, however, instead of e you
write:
eR ≡ e
−q 2
1 − e2
2 M
ln
12π2 m2 From Dirac to Renormalisation Lecture 18, 01/07/2004
1/2

8.6: Higher Orders and Renormalisation 15
and if eR is finite then remarkably all such graphs give you a finite correction. Now you
have to think what is the e that one inserts in such graphs? The charge you measure is
defined in terms of electron scattering at long range. Thus the charge you measure also
includes the effect of such higher order graphs. If this is in fact eR and this is finite, then
I should calculate the graphs with e (which may well be infinite), but at the end I should
write the result in terms of eR that subtracts an infinite cut-off term and leaves me with a
finite answer! Or instead I just use eR everywhere and as such have automatically included
such higher order corrections (be careful there are a number of technical details in the
calculations that have to clarified when you do this).
One also finds that masses have to be corrected in a similar way. The measured mass is
given by an (infinite) naked mass minus an (infinite) correction. In order for renormalisation
to work it is necessary that all the divergent terms can all be absorbed into the masses
and coupling strengths.
If you think this is strange, I don’t blame you. The proof of the pudding is first of all
whether it works! In fact it works extremely well. As you may of heard the Lamb shift
between the S and P states in hydrogen probes exactly such effects as you are looking
at different distances between the electron and the nucleus. Close to the nucleus there is
a cloud of e + e − pairs that are polarised due to the charge of the nucleus. This cloud is
exactly the sort of correction graph that I just drew.
An intuitive explanation of this effect can be seen by considering the screening of a charge
q by a dielectric medium. The negative end of the dipole is attracted towards q and the
positive end is repelled. The “halo” of negative charge partially cancels the field due to q.
⇒ Transparency Charge screening (screen1.jpg)
⇒ Transparency Vacuum fluctuations and coupling strengths vs. distance (screen2.jpg)
In QCD, i.e. due to the self-interaction of the gluons, this in fact works the other way round!
While the vacuum polarisation diagrams that contain fermion-antifermion pairs result in a
reduction of the effective colour charge, those with gluon loops lead to an increase at large
distances. Which of the 2 wins is given by the factor:
a ≡ 2f − 11n
where f = 6 is the number of quark flavours and n = 3 is the number of colours. So in
our world, the QCD coupling decreases at short distances, which leads to the effect known
as asymptotic freedom, i.e. even though quarks interact strongly with each other, at short
distances we can regard them as free particles and as such measure their properties and
even “observe” them.
As a result of the renormalisation the coupling strengths are no longer constants, but vary
with q 2 . For the electromagnetic coupling strength αQED, while the value at q 2 = 0 is
≈ 1/137 at MZ = 91 GeV its value is ≈ 1/128. As we have already seen, the variation of
the strong coupling strength is stronger:
From Dirac to Renormalisation Lecture 18, 01/07/2004

8.7: (g − 2) of Electron and Muon 16
⇒ Transparency αs vs. q 2 (cernlep8 5-04.jpg)
The plot shows the same trend as the one in the previous chapter, which was from the
PDG. They show that LEP and PETRA data are very consistent when treated in a similar
way. In addition the variation of αS with q 2 within a single experiment is clearly seen. A
similar trend is seen in the measurements made at HERA:
⇒ Transparency αs vs. q 2 (2004 LLouise Poeschl6.jpg)
In most of these measurements the energy of the measured jets sets the scale, i.e. the
value of αS that one measures. These are not all the same (as they are in e + e − ), as the
proton is a composite object and the struck quark can therefore carry a variable fraction
of the proton momentum. In addition, ep interactions are a scattering process and not an
annihilation process; hence the q 2 varies from one event to the next. In e + e − q 2 is (mostly)
just given by the c.m. energy √ s.
The masses of particles also depend on the q 2 of the process that measures them. First
evidence for such a variation of the b quark mass has been shown (albeit model dependent)
by the LEP experiments.
⇒ Transparency Running b quark mass (cernlep10 5-04.jpg)
8.7 (g − 2) of Electron and Muon
Turning back to QED, another effect of the vacuum polarisation is that it leads to a
slight modification of the magnetic dipole momentum of elementary fermions. One of the
great triumphs of the Dirac equation is that it explained why the magnetic moment of the
electron is:
with S = 1σ
and the gyromagnetic ratio
2
µ = −g e
2m S
g = 2
There is however a small correction to this number, which can be both precisely calculated
and measured. The calculations have been made to several orders of perturbation theory:
⇒ Transparency Some leading diagrams contributing to the magnetic moment (g-2-feyn.jpg)
The theoretical expectation is:
g − 2
2
theory
e
= 0.5
α
− 0.32848
π
α
2 + 1.19
π
= (115 965 230 ± 10) × 10 −11
α
3 + · · ·
π
From Dirac to Renormalisation Lecture 18, 01/07/2004

8.7: (g − 2) of Electron and Muon 17
while the measured value is:
g − 2
2
exp
e
= (115 965 218 59 ± 38) × 10 −11
where the uncertainty in the theory actually comes mainly from the uncertainty in the
value of α. You see a remarkable agreement between experiment and data.
Measurements have also been made of (g − 2) for the muon. Over the past few years a
group at Brookhaven has been working extremely hard to improve the accuracy of the
measurements. The idea of the experiment is illustrated in the next transparency:
⇒ Transparency Sketch of the main idea behind a g − 2 experiment (g-2.jpg)
In the experiment at BNL, polarised muons are injected into a superconducting storage
ring. The ring provides a very uniform static dipole field, B. If the g factor of a muon
were exactly equal to two, the spins and momenta of the muons would stay parallel during
storage. This would mean that the cyclotron frequency, ωc, of the muon is equal to its spin
precession frequency, ωa. However, due to the anomaly on the magnetic moment, the spin
precesses faster than the momentum ωs = ωc(1 + aµγ).
From the theoretical side, an idea of the Feynman graphs that have to be calculated and
the size of the various contributions is given in the transparency:
⇒ Transparency Prediction for muon g − 2 (pw eps03 talk.pdf, P.42)
The results for µ + and µ − are summarised in the following transparency:
⇒ Transparency Muon g − 2 measurements and predictions (g2results2.eps)
For the record the numbers are:
aµ + = 11 659 203(8) × 10−10
aµ − = 11 659 214(9) × 10−10
aµ = 11 659 208(6) × 10 −10
a e+ e −
µ = 11 659 181(8) × 10 −10
a τ µ = 11 659 196(7) × 10 −10
(0.7 ppm)
(0.7 ppm)
(0.5 ppm)
(0.7 ppm)
(0.6 ppm)
The average measured value from aµ differs from the predictions made using the e + e − and
the τ decay data by 2.7σ and 1.4σ respectively. The 2 different theoretical predictions
come from using e + e − → hadrons data at lower energies or using τ decay data for this
energy range. There appear to be some incompatible measurements between 850 MeV and
1 GeV. Discrepancies at lower energies have now disappeared after a reevaluation of the
e + e − data.
It is unlikely that one will see much improvement in these measurements (the predictions
may well improve), as the proposal to continue the experiment at BNL with the aim of
From Dirac to Renormalisation Lecture 18, 01/07/2004

8.8: Summary 18
achieving an accuracy of 0.4 ppm, which should be sensitive to electroweak corrections was
unfortunately not approved.
The difference between the 2 values is somewhat under 3 standard deviations. While
this cannot yet be seen as overwhelming evidence that there is some physics beyond the
standard model it does provide a hint. It may indicate that there are some new, as yet
unobserved, particles that can also contribute to the vacuum polarisation. One favourite
explanation is supersymmetry, that introduces integer spin partners of all fermions and
half-integer partners of all bosons. There are good theoretical arguments for the existence
of supersymmetry, but no direct experimental evidence has yet been seen.
8.8 Summary
Infinity - infinity give wonderful agreement between experiment and theory!
In this chapter I gave a very brief overview of the Dirac equation. I showed in principle how
one can calculate cross-sections and decay rates using the Fermi Golden Rule and Feynman
graphs. I discussed what happens when one tries to calculate higher orders. Although the
expected contributions are small, in fact the graphs often given infinite contributions.
These are resolved using the procedure called renormalisation, which absorbs these infinite
corrections into a redefinition of the masses and charges of the fundamental particles. While
this procedure may seem to be very arbitrary, the proof of the pudding, is that it works
wonderfully. I have illustrated the level of agreement between theory and experiment with
the (g − 2) measurements. I will show more comparisons in Chapter 10. End of
Lecture
18
From Dirac to Renormalisation Lecture 18, 01/07/2004