1. 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

2. The Bohr Magneton
Bohr Magneton

3. Bremsstrahlung & Pair ProductionZe Ze +
Internal electron line
since real electron cannot
emit a real photon and
still conserve energy and
momentum:
~ Z3
in electron
rest frame
consider
me = me + pe2/2me + E (conservation of energy)
pe = p = E(conservation of momentum)
me  me = E2/2me + E  
So, for a positive energy photon to be produced, the electron has
to effectively ''lose mass"
 virtual electron goes ''off mass shell"
Pair Production Ze
''twisted Bremsstrahlung"
Matrix element will be the
same; Cross-section will only
differ by a kinematic factor

4. Vacuum Polarization e e In QED, the bare charge of
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, ) will be seen in interactions
involving a high momemtum transfer as they
probe closer to the central charge.
q
 ''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 !!

5.  (1/q4) q3 dq Another way... = ln q ] =  dq/q
Renormalization
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 d
(just like the 2-D area integral is r dr d and
the 3-D volume integral is r2 dr dcos d )
so the q
integral
goes like: ∞
qmin
= ln q ]
 (1/q4) q3 dq ∞
qmin ∞
qmin
=  dq/q
 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 0 and m0 by running parameters.

6. The Underwater Cannonball
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.

7. Point Charges and Sweat
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

8. Steve’s Tips for Becoming a Particle Physicist
1) Be Lazy
2) Start Lying
3) Sweat Freely

9. e - e+  e-
W - e

10. Is A Particle?

11. No: All Are But
Shadows Of The Field

12. Relativistic Wave Equation
The Story So Far
Recap:
Relativity
Subjectivity of position,
time, colour, shape,...
Relativistic Wave Equation
Spin
(Dirac Equation)
Antimatter
Klein-Gordon Equation
Asymmetry?!
Yukawa Potential
(Nobel Prize opportunity missed)
New ''charge" + limited range
''virtual" particles
exchange quanta
Strong Nuclear Force
Central
Reality of
the FIELD
Feynman Diagrams
(QED)
Prediction of pion

13. 6 V(r) = e-Mcr/ħ Yo = eip•r/ħ Yf = eip¢•r/ħ
sheet 2
Consider the scattering of one nucleon
by another via the Yukawa potential:
V(r) = e-Mcr/ħ
- g2
4pr
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:
Yo = eip•r/ħ
After scattering, the final state will have a new momentum
vector, p¢, where |p| = |p¢ | :
Yf = eip¢•r/ħ
If q is the angle between p and p¢, write an expression
for the momentum transfer, q , in terms of p . p p¢ q
q = p¢ - p
magnitude of q = 2p sin(q/2) 13

14. ∫ ∫ ∫ ∫ [ + ] ( ) ( - ) <Yf|V(r)|Yo> = e-iq•r/ħ e- Mcr/ħ
Compute the matrix element for the transition.
<Yf|V(r)|Yo> = ∫
e-ip¢•r/ħ
eip•r/ħ
e- Mcr/ħ
- g2
4pr
( )
r2 dr df d(cosq) ∫
e-iq•r/ħ
e- Mcr/ħ
r dr d(cosq)
- g2 2 = ∫
e-iqrcosq/ħ
e- Mcr/ħ
r dr d(cosq)
- g2 2 = ∫
e(iq+Mc)r/ħ dr
-iħg2 2q =
( )
e (iq-Mc)r/ħ
e-(iq+Mc)r/ħ
iħ2g2 2q =
[ ]
e(iq-Mc)r/ħ
iq  Mc
iq-Mc
r = ∞
r = 0 14

15. [ + ] [ + ] [ ] e-(iq+Mc)r/ħ = e (iq-Mc)r/ħ iq  Mc iq-Mc
iħ2g2 2q =
[ ]
e (iq-Mc)r/ħ
iq  Mc
iq-Mc
r = ∞
r = 0
This will go to zero as r ® ∞, so we’re left with:
<Yf|V(r)|Yo> = 1
- iħ2g2 2q
[ ]
iq  Mc
iq-Mc
iq-Mc
- iħ2g2 2q
[ ]
+ iq  Mc
(iq  Mc)(
iq-Mc) =
ħ2g2
(q2+M2c2) = 1
q2+M2 ~
(natural units)

16. NOTE: This is all really a semi-relativistic approximation!
Really, we want to consider a 4-momentum transfer
Recall that P2 = E2  p2 1
q2+M2
so, 1
q42 M2
(choosing appropriate
sign convention)
Known as the “propagator” associated with the field
characterised by an exchange particle of mass M

17. dp/dE = E/p = (gm/gmv) = 1/v dp/dE = (m/mv) = 1/v
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
E2 = p2 + m2
dE = (p/m) dp
dp/dE = E/p = (gm/gmv) = 1/v
dp/dE = (m/mv) = 1/v
Also show that, for the present case: p2d(cosq) = - ½ dq2
q = 2p sin(q/2)
cosq = cos2(q/2) - sin2(q/2)
= 1 - 2sin2(q/2)
q2 = 4p2 sin2(q/2)
sin2(q/2) = (1 – cosq)/2
q2 = 2p2 (1 - cosq)
dq2 = -2p2 dcosq
p2d(cosq) = - ½ dq2 17

18. ( ) ( ) in our case Nbeam = Ntarget = 1
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 ds /dW .
in our case
Nbeam = Ntarget = 1
vbeam = 2v (relative velocity in CM)
Normalised Volume (= 1)
Rate =
vbeamNbeamNtarget s
Volume
so, ds = dRate 1 2v
given by FGR
ds = 1 2v 2p ħ
|<Yf|V(r)|Yo>|2
(2pħ)3
p2dp df dcosq
dEtot 2 =
( ) 1
(2pħ)3 df
ħ2g2
(q2+M2c2) 2v
(- ½ dq2) 2p ħ =
( )
dq2df
(2pħ)3
ħ2g2
(q2+M2c2) 2 p
4v2ħ 18

19. ∫ ∫ ( ) [ ] ( ) ( ) = dq2df (2pħ)3 ħ2g2 (q2+M2c2) p 4v2ħ ds = dq2df
( )
dq2df
(2pħ)3
ħ2g2
(q2+M2c2) 2 p
4v2ħ ds =
dq2df
8pħ3
ħ4g4
(q2+M2c2)2 p
4v2ħ =
dq2df
(q2+M2c2)2 1
(4p)2 g4
2v2
Integrate this expression and take the limit as v ® c s =
dq2df
(q2+M2c2)2 g4
2v2(4p)2 ∫ =
dq2
(q2+M2c2)2
pg4
v2(4p)2 ∫ = 1
(q2+M2c2)
pg4
v2(4p)2 -
[ ]
q2 = 0
q2 = ∞
since integral was from cosq = -1 to 1
(where 1 = zero momentum transfer) ħ Mc
= p g2
4pħc
( ) ( )
taking v ~ c
s =
pg4
Mc2(4p)2
s = p × m2
~ 30 mb
strong
coupling
constant
(~1)
“range” of
Yukawa
potential
(~1fm)
 
(within a factor of 2-3)