Physics 214 UCSD/225a UCSB
Lecture 10
• Halzen & Martin Chapter 4
– Electron-muon scattering
– Cross section definition
– Decay rate definition
– treatment of identical particles => symmetrizing
– crossing

Electrodynamics of Spinless particles
• We replace p µ with p µ + eA µ in classical EM for a particle of charge -e moving in an EM potential A µ
• In QM, this translates into: i
!
µ " i
!
µ + eA
µ
• And thus to the modified Klein Gordon
Equation:
(
!
µ !
µ
+ m 2
)
" = #
V
"
V
= # ie (
!
µ
A
µ
+
A
µ !
µ
)
# e 2 A 2
V here is the potential energy of the perturbation.

Two-by-two process
Overview
• Start with general discussion of how to relate number of scatters in AB -> CD scattering to
“beam & target independent” cross section in terms of W fi
.
• Calculate W fi
for electron-muon scattering.
• Calculate cross section from that
• Show relationship between cross section and
“invariant amplitude” (or “Matrix Element”).

j
µ = Reminder from last lecture
" µ j
µ
=
0
!
= i
&
'
"
J
= % i
(
"
#
$"
$ t
% " $" t
#
# , " % " , " #
)
*
)
Plane wave solutions are:
!
t
( )
=
Ne
" ip
µ x
µ
4-vector current for the plane wave solutions we find:
!
=
2 E | N | 2 j
=
2 p | N | 2
"
#
$
J
µ =
2 p
µ
| N | 2
The 2|N| 2 is an arbitrary normalization

Cross Section for AB -> CD target
• Basic ideas: scatter beam
# of scatters = (flux of beam) x (# of particles in target) x σ
W fi
Cross section = σ =
(initial flux)
(number of final states)
W fi
= rate per unit time and volume
“Cross section” is independent of characteristics of beam and target !!!

Aside on wave function
Normalization
W fi
Cross section = σ =
(initial flux)
(number of final states)
W fi
∝ N 4
Number of final states/initial flux ∝ N -4
Cross section is thus independent of choice of wave function normalization
(as it should, of course!)
We will see this explicitly as we walk through this now.

Two-Two process AB -> CD
• Normalize plane wave in constant volume
– This is obviously not covariant, so the volume normalization better cancel out before we’re done!
V
" !
dV
=
2 E
#
N
=
1
V
• # of particles per volume = 2E/V = n
• # of particles A crossing area per time = v
A
n
A
• Flux(AB) = v
A
n
A
(2E
B
/V) = v
A
(2E
A
/V) (2E
B
/V)

Aside on covariant flux
• Flux = v
A
(2E
A
/V) (2E
B
/V)
• Now let target (i.e. B) move collinear with beam (i.e.
A): Flux = ( v
A
- v
B
) (2E
A
/V) (2E
B
• Now take v=p/E: Flux = (E
B p
A
/V)
+ E
A
p
B
) 4/V 2
• Now a little relativistic algebra:
( p
A
µ
( E
A p B
µ
E
B
)
) 2
2
!
m 2
A m
B
2 =
(
=
( p 2 + m 2
)
E
A
A
E
B
!
p
A
( p 2 + m 2
) p
B
B
)
2
!
m 2
A m
B
2 p
A
= !
p
B
Putting the pieces together and adding some algebra:
( p
A
µ p B
µ
) 2
Flux
=
4
!
m
A
2 m
B
2
V 2
= ( p
A
E
B
( p
A
µ p B
µ
) 2
+ p
B
E
A
) 2
!
m
A
2 m
B
2
Obviously covariant!
(up to 1/V 2 normalization factor that is arbitrary, and will cancel)

Number of final states/particle
• QM restricts the number of final states that a single particle in a box of volume V can have:
Number of final states
2E particles
=
Vdp 3
(2
!
) 3 2 E
This follows from Exercise 4.1 in H&M that you will do as homework exercise.

Putting the pieces together
W fi
Cross section = σ =
(initial flux)
(number of final states)
W fi
σ = v
A
(2E
A
/V) (2E
B
/V)
Vdp
C
3
(2
!
) 3 2 E
C
Vdp
D
3
(2
!
) 3 2 E
D
Next we calculate W fi

Electron Muon Scattering
• Use what we did last lecture
– Electron scattering in EM field
• With the field being the one generated by the muon as source.
– Use covariant form of maxwell’s equation in
Lorentz Gauge to get V, the perturbation potential.
• Plug it into T fi

e -
In form of diagrams e -
µ e p
A
X field of muon e p
C
Electron-muon scattering
p
B p
D µ -

Electron Muon scattering
 2 A µ = J µ
(2)
Maxwell Equation
Note:  2 e iqx = -q 2 e iqx
J
µ
(2)
= !
eN
B
N
D
( p
D
+ p
B
)
µ e i ( p
D
!
p
B
) x
A
µ
T fi
=
= !
1 q 2
J
µ
(2)
!
"
(1)
µ
!
1 q 2
J
µ
(2) d 4 x
= q
Note the symmetry: (1) <-> (2)
T fi
= !
iN
A
N
B
N
C
N
D
(2
#
) 4 $ (4 ) ( p
D
!
iM
=
( ie ( p
A
+ p
C
)
µ
)
!
ig
µ % q 2
+ p
C
( ie ( p
D
!
p
A
+ p
B
)
%
)
!
p
B
) M
Note the structure: Vertex x propagator x Vertex

Reminder
$
$
$
"
$
#
1
0
!
0
1
0
0
0
0
0 0
!
1 0
0 0 0
!
1 &
'
'
'
'
%
= g
µ (
A
µ
B
µ
= g
µ !
A
µ
B
!

Reminder: T fi
-> W fi
W fi
= lim t
!>"
T fi
2 t
#
T fi
2 tV
Last time we didn’t work in a covariant fashion. This time around, we want to do our integrations across both time and space, i.e. W is a rate per unit time and volume.
T fi
= !
iN
A
N
B
N
C
N
D
(2
"
) 4 # (4 ) ( p
D
+ p
C
!
p
A
!
p
B
) M
As last time, we argue that one δ -function remains after || 2 while the other gives us a tV to cancel the tV in the denominator.

Putting it all together for W fi
N
=
1
V
W fi
=
T fi
2 tV
T fi
= !
i (2
V
"
)
2
4
# (4 ) ( p
D
+ p
C
!
p
A
!
p
B
) M
W fi
=
(2
!
) 4
" (4 ) ( p
D
+ p
C
V 4
# p
A
# p
B
)
M
2

Putting it all together for σ
W fi
σ = v
A
(2E
A
/V) (2E
B
/V)
Vdp
C
3
(2
!
) 3 2 E
C
Vdp
D
3
(2
!
) 3 2 E
D d
!
=
V 2
4 v
A
E
A
E
B
(2
"
) 4
# (4 ) ( p
D
+ p
C
V 4
$ p
A
$ p
B
)
M
2
V 2
(2
"
) dp c
3
6 dp 3
D
4 E c
E
D d
!
=
1
64
" 2
# (4 ) ( p
D
+ p
C
$ p
A v
A
E
A
E
B
$ p
B
)
M
2 dp c
3 dp 3
D
E c
E
D

Aside on outgoing states
• While the incoming states have definite momentum, the outgoing states can have many momenta.
• The cross section is thus a differential cross section in the outgoing momenta.
e e p
A p
C d
!
=
2
M
4 v
A
E
A
E
B
1
16
" 2
# (4 ) ( p
D
+ p
C
$ p
A
$ p
B
) dp c
3 dp
D
3
E c
E
D
µ p
B p
D µ -
(incoming flux is still not covariant)

It is customery to re-express d
!
=
2
M
4 v
A
E
A
E
B
1
16
" 2
# (4 ) ( p
D
+ p
C
$ p
A
$ p
B
) dp c
3 dp
D
3
E c
E
D
As:
F = flux factor: d
!
=
M
2 dQ
F
F
=
4
( p
µ
A p B
µ
)
2 !
m
A
2 m
B
2 =
4 v
A
E
A
E
B dQ = Lorentz invariant phase space: dQ
=
1
16
!
2
" (4 ) ( p
D
+ p
C
# p
A
# p
B
) dp c
3 dp 3
D
E c
E
D

In the center-of-mass frame:
F
=
4 p i
( E
A
+
E
B
) 2 =
4 p i s dQ
=
1
4
!
2 p f
4 s d
" d
!
d
" cm
=
2
M
64
# 2 s p f p i
You get to show this as homework !

Electron-electron scattering
• With identical particles in the final state, we obviously need to allow for two contributions to M.
– Option 1:
• C attaches at vertex with A
• D attaches at vertex with B
– Option 2:
• C attaches at vertex with B
• D attaches at vertex with A e p
A e p
B p
D e p
C e e e p
A p
B p
C e p
D e -
• As we can’t distinguish C and D,the amplitudes add before M is squared.
M
= !
e 2 $
#
"
( p
$
A
+ p
C
( p
D
)
µ
( p
B
!
p
B
)
+ p
D
2
)
µ +
( p
A
+ p
(
D p
C
)
µ
( p
B
!
p
B
) 2
+ p
C
)
µ

e e e e -
Electron-positron and crossing e p
A p
C
( p
A
(
+ p
C
(
!
p
D
)
µ
(
!
p
B
)
!
(
!
p
B
!
p
)
)
2
D
)
µ
-p
B
-p
D e p
A
-p
B e p
C
-p
D e -
( p
A
!
p
B
( p
C
)
µ
(
!
p
D
!
(
!
p
D
+ p
C
)
)
2
)
µ
M
= !
e 2 $
#
"
( p
$
A
(
+ p
C
(
!
p
D
)
µ
(
!
p
B
)
!
(
!
p
B
!
p
D
)
) 2
)
µ +
( p
A
!
p
B
( p
C
)
µ
(
!
p
D
!
(
!
p
D
)
)
+ p
C
2
)
µ

Electron-positron and crossing
M
= !
e 2
"
$
#
$
( p
A
Electron - electron
+ p
C
( p
D
)
µ
( p
B
!
p
B
)
+ p
D
2
)
µ +
( p
A
+ p
D
( p
C
)
µ
( p
B
!
p
B
)
+ p
C
2
)
µ
M
= !
e 2
Electron - positron
"
$
#
$
( p
A
(
+ p
C
(
!
p
D
)
µ
(
!
p
B
)
!
(
!
p
B
!
p
D
)
) 2
)
µ +
( p
A
!
p
B
( p
C
)
µ
(
!
p
D
!
(
!
p
D
)
)
+ p
C
2
)
µ
Only difference is: p
D p
B
→ -p
B
→ -p
D

e e e -
e -
Electron-Electron scattering e p
A p
C p
A p
B p
B e -
Electron-Positron scattering e p
A p
C crossing p
D e e p
D e crossing e e p
A
-p
B p
C e -
-p
B
-p
D p
C
-p
D e e e -
-

E-mu vs e-e vs e-ebar scattering
M
= !
e 2
Electron - muon
"
( p
A
+ p
C
)
µ
( p
B
+ p
D
( p
D
!
p
B
) 2
)
µ
M
= !
e 2
"
$
#
$
( p
A
Electron - electron
+ p
C
( p
D
)
µ
( p
B
!
p
B
)
+ p
D
2
)
µ +
( p
A
+ p
D
( p
C
)
µ
( p
B
!
p
B
)
+ p
C
2
)
µ
M
= !
e 2
Electron - positron
"
$
#
$
( p
A
(
+ p
C
(
!
p
D
)
µ
(
!
p
B
)
!
(
!
p
B
!
p
D
)
) 2
)
µ +
( p
A
!
p
B
( p
C
)
µ
(
!
p
D
!
(
!
p
D
)
)
+ p
C
2
)
µ