Lecture 5: QED
• 
The Bohr Magneton
• 
Off-Shell Electrons
• 
Vacuum Polarization
• 
Divergences, Running Coupling &
Renormalization
• 
Yukawa Scattering and the Propagator
Useful Sections in Martin & Shaw:
Section 5.1, Section 5.2, Section 7.1.2
Bohr Magneton
Bremsstrahlung
Internal electron line
since real electron cannot
emit a real photon and
still conserve energy and
momentum:
consider
+
~ Zα3
Ze
me = meʹ′ + pʹ′e2/2meʹ′ + Eγ
in electron
rest frame
(conservation of energy)
⏐pʹ′e⏐ = ⏐pγ ⏐ = Eγ
⇒
Ze
(conservation of momentum)
me - meʹ′ = Eγ2/2meʹ′ + E γ
So, for a positive energy photon to be produced, the electron has
to effectively ''lose mass" ⇒ virtual electron goes ''off mass shell"
Pair Production
''twisted Bremsstrahlung"
Matrix element will be the
same; Cross-section will only
differ by a kinematic factor
Ze
Ze
Vacuum Polarization
e-
e-
Thus, we never actually ever see a ''bare" charge,
only an effective charge shielded by polarized
virtual electron/positron pairs. A larger charge
(or, equivalently, α) will be seen in interactions
involving a high momemtum transfer as they
probe closer to the central charge.
⇒ ''running coupling constant"
In QED, the bare charge of
an electron is actually infinite !!!
Note: due to the field-energy near an infinite charge,
the bare mass of the electron (E=mc2) is also
infinite, but the effective mass is brought back
into line by the virtual pairs again !!
-q
Another way...
Consider the integral corresponding to the loop diagram above:
At large q the product of the propagators will go as ~ 1/q4
Integration is over the 4-momenta of all internal lines
where d4q = q3 dq dΩ
(just like the 2-D area integral is r dr dθ and
the 3-D volume integral is r2 dr dcosθ dφ )
so the q
integral
goes like:
∞
∫ (1/q4) q3 dq
qmin
∞
=
∫ dq/q
qmin
= ln q
]
∞
qmin
⇒ logarithmically
divergent !!
With some fiddling, these divergences can always be shuffled
around to be associated with a bare charge or mass. But we
only ever measure a ''dressed" (screened) charge or mass, which
is finite. Thus, we ''renormalize" by replacing the bare values
of α0 and m0 by running parameters.
Analogy for Renormalization:
Instead of changing the equations of motion, you could instead
(in principle) find the ''effective" mass of the cannonball by shaking
it back and forth in the water to see how much force it takes to accelerate it.
This ''mass" would no longer be a true constant as it would clearly depend
on how quickly you shake the ball.
Still unhappy? Well, if it’s any comfort, note that the electrostatic
potential of a classical point charge, e2/r, is also infinite as r → 0
(perhaps this all just means that there are really no
true point particles... strings?? something else?? )
Does an electron feel it’s own field ???
''Ven you svet, you smell you’re own stink, Ja? Zo..."
M. Veltman
Steve’s Tips for Becoming a Particle Physicist
1) Be Lazy
2) Start Lying
3) Sweat Freely
eγ
γ
e+
We
νe
Is A Particle?
No: All Are But
Shadows Of The Field
Recap:
Relativity
Subjectivity of position,
time, colour, shape,...
Relativistic Wave Equation
Spin
(Dirac Equation)
Antimatter
Klein-Gordon Equation
Yukawa Potential
Asymmetry?!
(Nobel Prize opportunity missed)
New ''charge" + limited range
''virtual" particles
Strong Nuclear Force
Prediction of pion
Central
Reality of
the FIELD
exchange quanta
Feynman Diagrams
(QED)
6
sheet 2
Consider the scattering of one nucleon
by another via the Yukawa potential: V(r)
=
- g2
4π r
e-Mcr/ħ
In the CM, both nucleons have equal energy and equal &
opposite momenta, p. Take the incoming state of the 1st nucleon
(normalised per unit volume) to be:
ip•r/ħ
Ψo = e
After scattering, the final state will have a new momentum
vector, pʹ′, where |p| = |pʹ′ | :
Ψ = eipʹ′•r/ħ
f
If θ is the angle between p and pʹ′, write an expression
for the momentum transfer, q , in terms of p .
pʹ′
θ
p
q = pʹ′ - p
magnitude of q = 2p sin(θ/2)
Compute the matrix element for the transition.
<Ψf|V(r)|Ψo> =
=
iħ2g2
2q
∫
e-ipʹ′•r/ħ
=
- g2
2
=
- g2
2
=
-iħg2
2q
[
(
- g2 e- Mcr/ħ
4πr
∫
∫
e
∫
(e
e-iq•r/ħ e- Mcr/ħ r dr d(cosθ)
-iqrcosθ/ħ
e- Mcr/ħ r dr d(cosθ)
-(iq+Mc)r/ħ
e-(iq+Mc)r/ħ
iq + Mc
)
eip•r/ħ r2 dr dφ d(cosθ)
+
- e (iq-Mc)r/ħ)
e(iq-Mc)r/ħ
iq-Mc
]
r=∞
r=0
dr
=
iħ2g2
2q
[
e-(iq+Mc)r/ħ
iq + Mc
+
e (iq-Mc)r/ħ
iq-Mc
]
r=∞
r=0
This will go to zero as r → ∞, so we’re left with:
[
<Ψf|V(r)|Ψo> =
- iħ2g2
2q
=
- iħ2g2
2q
=
ħ2g2
(q2+M2c2)
[
1
iq + Mc
+
iq-Mc + iq + Mc
(iq + Mc)( iq-Mc)
~
1
iq-Mc
]
]
1
(natural units)
2
2
q +M
NOTE:
This is all really a semi-relativistic approximation!
Really, we want to consider a 4-momentum transfer
Recall that
so,
P2 = E2 - p2
1
q2+M2
1
q 4 2 - M2
(choosing appropriate
sign convention)
Known as the “propagator” associated with the field
characterised by an exchange particle of mass M
Show that the relation dp/dE = 1/v , where v is the velocity, holds for both
relativistic and non-relativistic limits.
Classically:
Relativistically:
E = p2/2m
dE = (p/m) dp
dp/dE = (m/mv) = 1/v
Also show that, for the present case:
E2 = p2 + m2
dp/dE = E/p = (γm/γmv) = 1/v
p2d(cosθ) = - ½ dq2
q = 2p sin(θ/2)
q2 = 4p2 sin2(θ/2)
cosθ = cos2(θ/2) - sin2(θ/2)
= 1 - 2sin2(θ/2)
sin2(θ/2) = (1 – cosθ)/2
q2 = 2p2 (1 - cosθ)
dq2 = -2p2 dcosθ
p2d(cosθ) = - ½ dq2
Now, from the definition of cross-section in terms of a rate, the relative velocities
of the nucleons in the CM, and using Fermi’s Golden Rule, derive the differential
cross-section dσ
/dΩ .
in our case
Nbeam = Ntarget = 1
vbeam = 2v (relative velocity in CM)
vbeamNbeamNtarget σ
Rate =
Volume
so, dσ =
dσ =
=
=
1
2v
Normalised Volume (= 1)
1
dRate
2v
given by FGR
2π
|<Ψ |V(r)|Ψ >|2
f
o
ħ
1
2v
2π
ħ
π
4v2ħ
(
(
ħ2g2
(q2+M2c2)
2
)
)
ħ2g2
(q2+M2c2)
2
1
(2πħ)3
p2dp dφ dcosθ
dEtot
1
(2πħ)3
1
2) dφ
(
½
dq
2v
dq2dφ
(2πħ)3
dσ
(
ħ2g2
(q2+M2c2)
π
4v2ħ
=
2
)
dq2dφ
(2πħ)3
1
g4
dq2dφ
= (4π)2 2v2 (q2+M2c2)2
Integrate this expression and take the limit as v → c
π
ħ4g4
dq2dφ
= 4v2ħ (q2+M2c2)2
8π3ħ3
σ
=
g4
2v2(4π)2
πg4
= v2(4π)2
∫
[
1
-
2 2 2
(q +M c )
πg4
σ=
Mc2(4π)2 = π
taking v ∼ c
πg4
= v2(4π)2
dq2dφ
(q2+M2c2)2
]
q2 = 0
q2 = ∞
( )( )
g2
4πħc

strong
coupling
constant
(~1)
2
ħ
Mc

∫
dq2
(q2+M2c2)2
since integral was from cosθ = -1 to 1
(where 1 = zero momentum transfer)
2
“range” of
Yukawa
potential
(~1fm)
σ = π × 10-30 m2
~ 30 mb
(within a factor of 2-3)