Lecture 21: Apr 10th 2012
Reading Griffiths chapter 7
HW due Thursday: 7.37, 7.39
Calculate the e+e-  +- CM frame matrix element. In the ultra-relativistic limit show
that it obeys the s,t,u crossing relationship with e--  e-0) Feynman rules for iM
a) Draw a diagram labeling all the incoming and outgoing momentum and spins. Use
arrows to indicate particles (with direction of time) and antiparticles (opposite direction
of time) and internal lines (preserving continuity of the direction of flow). Note down
which spins are known and which will be summed over (final or internal state spins) or
averaged over (initial state spins if unknown).
b) Write down spinors for particles and antiparticles. u for incoming electrons and u for
outgoing electrons. u for incoming positrons and  for outgoing positrons. e m for
incoming photons and e m * for outgoing photons. For instance for an incoming electron,
s3
1, we write us1 (p1 ) and for an outgoing electron, 3, u ( p3 )
c) Vertex factors of ige, use a different index for each vertex term.
Each vertex represents a separate interaction and has its own index
As an example with particles a vertex factor is written between the outgoing (first) and
incoming (last) particle to form a current term.
s3
b and c form the current terms, example u ( p3 ) us1 (p1 ), which will reduce to factors
proportional to momentum just as in our scalar particle transition rate. This is the basic
unit for a particle in transition and is called a particle current, i.e. the J we derived.
Where there is an initial or final state photon at the vertex the vertex factor will be
summed over  with the photon if there is one. e m = e/ .
d) Propagators for the photons or electrons/positrons: -ig/q2,
i(q + m)/(q2 - m2)
Note in the photon propagator case the g matrix converts  to  so it is summed with
 of the first current. The propegator is what relates the two interactions in this case.
In the case of the electron propagator each current has a photon which is summed with
the  vertex and the propagator has it’s own sum over q, so no further sums have to be
done to get a scalar result for the matrix element probability. q is often written as q/
e) Add an integral over the internal four momentum d4q/(2)4 and a four dimensional
delta function for each vertex and use one to integrate over the internal four momentum
and cancel the other since it corresponds to overall four momentum conservation.
f) antisymmetrization. Crossed diagrams will exist for identical sets of particles in the
final state. The second diagram will get a negative sign (Fermi statistics).
Diagrams can have their electrons moved from the initial to the final state where they
occur as a positron producing an equivalent in probability diagram ( a type of crossing
symmetry). To have a consistent convention you have to be careful about the signs in
those cases.
1) Non relativistic scattering of unpolarized identical particles
Before tackling the more “complex” relativistic case where we must evaluate all the
matrix and vector multiplications it is instructive to consider a non relativistic case where
the matrix math simplifies.
æu ö
u( p)1 = N ç A ÷,
è uB ø
N=(|E|+m)
æ 1ö
uA = ç ÷,
è 0ø
uB =
æ1ö
1
1 æ pz ö
( p × s )ç ÷ =
ç
÷,
E+m
è0ø E + m è px + ipy ø
æu ö
u( p) 2 = N ç A ÷,
è uB ø
æ 0ö
uA = ç ÷,
è 1ø
uB =
æ0ö
1
1 æ px - ipy ö
( p × s )ç ÷ =
ç
÷
E+m
è1ø E + m è - pz ø
If p<<m then all the p terms in the spinors go to 0 because they are divided by m. Then
u s1g m us3 = 2mds1,s3 using u iuj = 2mij,= 1,2 spins and the fact that only the 0 matrix will
contribute since the other are anti-block diagonal.
Then using the result
M=
2
-ge2 é
m
ùéëu(4)g m u(2)ùû + ge éu(4)g m u(1)ùéëu(3)g m u(2)ùû
u(3)
g
u(1)
ë
û
ë
û
t
u
Now consider all the various spin combinations so we can later average over the initial
states and sum over the final states.
æ1 1 1 1ö
æ -1 -1 -1 -1 ö
æ1 1 ö
M (1, 2 ® 3, 4) M ç , ® , ÷ = M ç , ® , ÷ = -4ge2 me2 ç - ÷
è2 2 2 2ø
è2 2
èt uø
2 2ø
identical particles in the end state so both the t and u channel diagrams contribute
and
Note that the summation condition over and u does not allow this interaction to flip the
spins. This is expected at first order for slow moving electrons, since they will only
interact via the electric field, which we know from the atom does not continually flip
spins. However, at higher energy spin flips are possible because the moving electrons
are starting to represent a strong current and can interact via the magnetic part of the
field. Also for the first set of spins we have true identical particles we must include
crossed diagrams in M. In the second two matrix elements we also do not flip the spins
since that would create a sum over
and u that is 0, but the crossed u channel diagram
leaves particle 4 with the spin of particle 1 so that matrix multiplication is not 0.
To evaluate the full matrix element for unpolarized we have to average over the number
of initial spin states and sum over the final spin states.
M =
2
1
2
Mi
å
(2s1 + 1)(2s2 + 1) i
The probability will be proportional to the matrix element times it’s complex conjugate.
Each of the three processes above is independent so we will square each and then add
them. Averaging over initial spin states introduces two factors or ½. Summing over the
final states each matrix element there are three distinct possibilities that occur twice each.
2
é
1
1ö
1 1ù
2 2 2 æ1
M = 2( 4me ge ) êç - ÷ + 2 + 2 ú
4
u û
ëè t u ø t
substituting in our expressions for t and u and simplifying
2
M = 8(m g
2
)
2 2 2
e e
é
ù
ú
2 ê 1
1
1
+
ú
4ê
16 p ê sin 4 q cos4 q sin 2 q cos2 q ú
ë
2
2
2
2û
and putting this in:
d/d = (1/8)2 |M|2 pi/pf 1/E2
where in the non relativistic limit E = 2me
This type of result will be typical even in the full treatment in that we will end up with a
number of s, t and u type scattering or annihilation terms.
3) Unpolarized relativistic electron-electron or electron-muon scattering
For results where we will sum over spins the terms of the matrix element squared can
also be substantially simplified.
The matrix element is:
M=
2
-ge2 é
m
ùéëu(4)g m u(2)ùû + ge éu(4)g m u(1)ùéëu(3)g m u(2)ùû
u(3)
g
u(1)
ë
û
ë
û
t
u
consider just the first term. The first term also is relevant for electron-muon scattering
M1 =
-ge2
( p1 - p3 )
2
[u (3)g
m
u(1)][ u (4)g m u(2)]
Each term is a scalar and the gamma matrix indices indicate any necessary sums
This allows us to group terms with the same particles together.
M1 =
2
ge4
( p1 - p3 )
4
[u (3)g
m
u(1)][ u (3)g n u(1)] [ u (4)g m u(2)][ u (4)g n u(2)]
*
Consider terms such as [ u (3)g m u(1)][ u (3)g n u(1)]
*
*
[u(3)g
n
u(1)] = [ u (3)g n u(1)] = [ u+ (1)g n +g 0+ u(3)] = [ u+ (1)g 0g 0g n +g 0u(3)] = [ u (1)g 0g n +g 0u(3)] = [ u(1)g n u(3)]
[u(3)g
m
u(1)][ u (3)g n u(1)] = [ u (3)g m u(1)][ u (1)g n u(3)]
+
*
*
where the adjoint gamma matrix
g n = g 0g n +g 0 = g n
Evaluate the sum over the spins of particle 1 and then write indices of the matrix
multiplication explicitly to understand how to contract the sum over the spins of particle
3
[u(3)g
m
[
]
u(1)][ u (1)g n u(3)] = [ u (3)g m ( p/ 1 + m)g n u(3)] = u (3)i (g m ( p/ 1 + m)g n ) ij u(3) j = (g m ( p/ 1 + m)g n ) ij u(3) j u (3) i
Contracting the row and column matrices of particle three will force i=j. Move the ajoint
operator for particle 3 to the opposite end and then sum over the spins of particle 3.
[u(3)g
m
u(1)][ u (1)g n u(3)] = (g m ( p/ 1 + m)g n ) ij ( p/ 3 + m) ji = Tr[g m ( p/ 1 + m)g n ( p/ 3 + m)]
Thus in general
[u(a)g
m
u(b)][ u (a)g n u(b)] = Tr[g m ( p/ b + mb )g n ( p/ a + ma )]
*
where all the traces of gamma matrices and momentum vectors have been worked out.
Thus in our example
*
ge4
*
u (3)g m u(1)][ u (3)g n u(1)] [ u (4)g m u(2)][ u (4)g n u(2)]
2 [
t
g4
2
M1 = e2 Tr[g m ( p/ 1 + me )g n ( p/ 3 + me )]Tr g m ( p/ 2 + mm )g n ( p/ 4 + mm )
4t
4 is from the averaging over initial spins
M1 =
2
[
]
Tr éëg m ( p/ 1 + me ) g n ( p/ 3 + me )ùû = Tr éëg m p/ 1g n p/ 3 ùû + meTr éëg m p/ 1g n ùû + meTr éëg mg n p/ 3 ùû + me2Tr éëg mg n ùû
Using trace identities
10) Trace of an odd number of gamma matrices is 0
mn
12) Tr g m g n = 4g
[
]
17) Tr éëg p/ 1g n p/ 3 ùû = 4éë p1m p3n + p3m p1n - gmn ( p1 × p3 )ùû
m
[
]
Tr[g m ( p/ 1 + me )g n ( p/ 3 + me )] = 4 p1m pn3 + p3m p1n + gmn ( me2 - p1 × p3 )
[
][
]
M1 =
ge4
4 p1m pn3 + p3m p1n + g mn ( me2 - p1 × p3 ) p2m p4 n + p4 m p2n + gmn ( mm2 - p2 × p4 )
t2
M1 =
ge4
8[( p1 × p2 )( p3 × p4 ) + ( p1 × p4 )( p2 × p3 ) - ( p1 × p3 ) mm2 - ( p2 × p4 ) me2 + 2me2 mm2 ]
t2
2
2
Often we evaluate these matrix elements in relativistic limits.
In the ultra relativistic limit the Madelstam variables become:
s = ( p1 + p2 ) = 2p1 × p2,
2
t = ( p1 - p3 ) = -2 p1 × p3,
2
u = ( p1 - p4 ) = -2 p1 × p4,
2
and
M1 =
2
2ge4 2
s + u2 ]
2 [
t
note t = ( p1 - p3 ) = -2 p2 (1- cosq ) = -4 p 2 sin2
2
q
2
which is valid if the lepton masses can be taken as the same.
This cross section includes and infinity at theta = 0 which corresponds to infinite range
scattering.
4) Crossing symmetry and e-e+  -+
Starting from e--  e-- you can get to e-e+  -+ by moving the final state electron
electron (p3) to the initial state as a positron (-p2) and by moving the initial state muon
(p2) to become a final antimuon (-p3).
Since you are just switching p3 with –p2 or p2 with –p3 then this is equivalent to
changing s and t
t 2 + u2
M = 2g
s2
2
4
e
Which is quite interesting considering one of the interactions is a scattering interaction
and the other annihilation! This only works at high energy. At low energy there is not
enough energy to form the muon pair and there is no s channel process. Also notice that
the annihilation interaction is not divergent. This makes sense since you can’t have
annihilation at infinite range. In fact we can calculate the total cross section.
In CM d/d = (1/8)2 |M|2 pi/pf 1/E2
using the exact expression for s, t and u in CM
s = ( p1 + p2 ) = 4( p2 + m 2 ),
2
q
t = ( p1 - p3 ) = -2 p2 (1- cosq ) = -4 p 2 sin2 , and
2
2
2
2
2q
u = ( p1 - p4 ) = -2 p (1+ cosq ) = -4 p cos ,
2
2
which can be integrated!
a2
8 4pa 2
ds =
ò (1+ cos q)df sinqdq = 4s 2p 3 = 3s
4s
2
a2
The cross section is finite and goes down at higher energy, which is what is observed.
The t channel scattering cross section is still divergent at =0. The point =0 corresponds
to the point where b=. That the total cross section is infinite just represents the fact that
a 1/r2 potential still has a small effect no matter how far away you get. However, we are
interested in computing scattering cross section or probabilities that for a particle between
b and b+db to scatter to angle theta. As you get to very large b you are probing very
small scattering angles between ~0 and +d. The cross section and the probability to
go to any of the angles will still be finite for any finite b.