Physics 214 UCSD/225a UCSB
Lecture 12
• Halzen & Martin Chapter 6
– Spin averaged e-mu scattering
• Introduce traces, and allude to trace theorems.
– Use crossing to derive spin averaged
e+e- -> mu+mu– Detailed discussion of the result.
Spinless
( )
! t, x = Ne
vs
Tfi = !i % J µfi Aµ d 4 x + O(e 2 )
( )
! t, x = u( p)e
"ip µ x µ
J µ = !ie(" # ($µ " ) ! ($ µ " # )"
Spin 1/2
)
"ip µ x µ
J µ = !e"# µ"
Tfi = !i $ J Aµ d x + O(e )
µ
fi
4
We basically make a substitution:
(p
f
+ pi ) ! u f " µ ui
µ
And all else in calculating |M|2 remains the same.
2
Example: e- e- scattering
For Spinless (i.e. bosons) we showed:
µ
" ( p + p )µ ( p + p )
%
(
p
+
p
)
(
p
+
p
)
A
C
B
D
µ
A
D
B
C
µ
''
M = !e 2 $$
+
2
2
( pA ! pC )
( pA ! pD )
#
&
For Spin 1/2 we thus get:
# (u " µ u )(u " u ) (u " µ u )(u " u ) &
M = !e 2 %% c A D 2µ B ! D A C 2µ B ((
( pA ! pC )
( pA ! pD )
$
'
Minus sign comes from fermion exchange !!!
Spin Averaging
• The M from previous page includes spinors in
initial and final state.
• In many experimental situations, in particular
in hadron collissions, you neither fix initial nor
final state spins.
• We thus need to form a spin averaged
amplitude squared before we can compare
with experiment:
1
1
2
2
M =
M = !M
!
4 spin
(2sA + 1)(2sB + 1) spin
A
C
B
D
2
Last time we did the
nonrelativistic case.
This time we do the complete
derivation.
Note: This is quite possibly the most
painful derivation we do this quarter.
“Easiest”: e- mu- -> e- mu• Easiest because it has only one diagram !!!
ek for electron momenta.
p for muon momenta.
prime for outgoing momenta.
µ-
ek
k’
p
p’
µ-
µ
$
&
$
&
u
(
k
)
#
u(k)
u
(
p
)
#
u(
p)
"
"
µ
%
'
%
'
2
M = !e
t
t = q2 = (k’-k)2 = (p’-p)2 = scalar product of 4-momenta
Spin averaged |M|2 (1)
M =
2
1
!
( 2sA + 1)( 2sB + 1) spin
M =
2
1
M
!
4 spin
2
+
e4
µ
$
%
&
'
)
M = 2 ! u ( k " )# u(k) gµ$ u ( p" )# u( p) ( u ( k " )# u(k) g%& u ( p" )# u( p) *
4t spin
2
(
) (
)(
) (
This is a Scalar product of 4-vectors,
multiplied by its complex conjugate,
to get a positive definite number.
Unfortunately, the scalar product mixes e and mu currents.
To form the spin average, we separate e and mu,
to execute the spin average on e and mu independently
)
Spin averaged |M|2 (2)
*
e4
µ
#
$
%
M = 2 + u ( k ! )" u(k) gµ# u ( p! )" u( p) &' u ( k ! )" u(k) g$% u ( p! )" u( p) ()
4t spin
(
2
) (
)(
) (
)
4
e µ!
"#
M = 2 Lelectron Lmuon gµ" g!#
t
2
Where we defined:
µ!
electron
L
(
)(
1
µ
!
= $ u ( k " )# u(k) u ( k " )# u(k)
2 spin
)
*
We can now focus on doing the sum over spins for this tensor!
Summary of where we are
$
&
$
&
u
(
k
)
#
u(k)
u
(
p
)
#
u(
p)
"
"
µ
%
'
%
'
M = !e2
t
4
e
2
M = 2 Lelectron µ( Lmuon µ(
t
*
1
µ(
µ
(
Lelectron = ) $%u ( k " )# u(k) &' $%u ( k " )# u(k) &'
2 e! spins
µ
Lmuon
µ(
*
1
µ
(
$%u ( p" )# u( p) &' $%u ( p" )# u( p) &'
=
)
2 muon ! spins
All that’s left to do is the sum over spins.
Note the structure of this expression:
$
.
*
#
$%u ( p! )" u( p) &' = .(
.
.
%
(
*
)( 4x4 ) *
*
*)
#
+&
-/
#
/
- = (1x1)
-/
-, /
'
The complex conjugate of this object is thus
identical to the hermitian conjugate of this !!!
We can use the latter in order to rearrange terms,
while ignoring the lorentz index for the moment.
#
$%u ( k ! )" u(k) &'
T*
#
= $%u ( k ! )" " u(k) &'
T*
0
T*
= $%u T * (k)" #T *" 0u( k ! ) &' = $%u (k)" # u( k ! ) &'
Where in the last step , we used the commutation properties.
To be explicit:
0 " T*
#$! ! %& = ! "T *! 0 = '! " ! 0 = ! 0! "
#$! 0 %&
"
#$! %&
T*
T*
=!0
for ν=1,2,3
= '! "
At this point we get the electron tensor:
Lelectron
µ!
1
= ( $%u ( k " )# µ u(k) &' $%u (k)# ! u( k " ) &'
2 s, s "
Next, we are going to make all summations explicit, by writing
out the gamma-matrices and spinor-vectors as components.
Lelectron
µ!
1
= ( $%u ( k " )# µ u(k) &' $%u (k)# ! u( k " ) &'
2 s, s "
1
! s"
= ( ( $%uis " ( k " )# ijµ u sj (k) &' $%uls (k)# lm
um ( k " ) &'
2 s, s " ijlm
1
!
s"
s"
s
s
$
&
$
&
= ( # ijµ# lm
u
(
k
)u
(
k
)
u
(k)u
"
"
(
(
i
m
l
j (k) '
%
'
%
2 ijlm
s"
s
Here we now apply the completeness relations:
! u ( p)u ( p) = p "
s
s
µ
µ
+ m = ( 4 x4 )
s
See H&M exercise 5.9 for more detail on completeness relation.
Lelectron
µ!
1
!
s$
s$
s
s
%
'
%
'
= # " ijµ" lm
u
(
k
)u
(
k
)
u
(k)u
$
$
#
#
i
m
l
j (k) (
&
(
&
2 ijlm
s$
s
Now apply to this the relationship:
! u ( p)u ( p) = p "
s
s
µ
µ
+ m = ( 4 x4 )
s
And you get as a result:
Lelectron
µ!
1
#
µ
#
!
= ) %& k#" $ + m '( mi $ ij %& k# $ + m '( jl $ lm
2 ijlm
Mathematical aside:
Let A,B,C,D be 4 matrices.
#A B
ij
jl
Clm Dmi ! Tr( A " B " C " D)
ijlm
This weird sum is thus nothing more than
the trace of the product of matrices !!!
I won’t prove this here, but please feel free to convince yourself.
Lelectron
µ!
Lelectron
µ!
1
#
µ
#
!
= ) %& k#" $ + m '( mi $ ij %& k# $ + m '( jl $ lm
2 ijlm
1
#
µ
%
!
&
= Tr ' k#" $ + m $ k% $ + m $ ()
2
(
) (
)
Lelectron
Lmuon
[
1
= Tr ( k#" $ # + m)$ µ ( k % $ % + m)$ !
2
1
= Tr ( p#" $ # + m)$ µ ( p% $ % + m)$ !
2
µ!
µ!
[
]
]
electron
kA
electron
k’C
pB
muon
p’D
4
e
µ!
M = 2 Lelectron Lmuon µ!
t
2
muon
“All” that’s left to do is apply trace theorems.
(There’s a whole bunch of them in H&M p.123)
Lelectron
µ!
[
1
= Tr ( k#" $ # + m)$ µ ( k % $ % + m)$ !
2
]
1
= Tr[ k#" $ # $ µ k% $ % $ ! + k#" $ # $ µ $ ! m + m$ µ k % $ % $ ! + m 2$ µ $ ! ]
2
Trace of product of any 3 gamma matrices is zero!
Lelectron
2
1
m
= Tr[ k#" $ # $ µ k % $ % $ ! ] +
Tr[$ µ $ ! ]
2
2
µ!
Next, define unit vectors a,b for µ and ν coordinate:
a!" ! # " µ
$
b$ " # "
%
This will allow us to use trace theorems
to evaluate the remaining traces.
Aside
• What does this mean?
a!"
!
!
a!" # "
$
µ
b$ " # "
%
Is a scalar product of 4-vectors.
As al is a unit vector, it projects out a component of γl .
The components of γl are themselves 4x4 matrices.
We introduce this to be able to use this trace theorem:
Tr [ abcd ] = 4 #$( a ! b ) ( c ! d ) " ( a ! c ) ( b ! d ) + ( a ! d ) ( b ! c ) %&
2 Trace Theorems to use:
(2)
(1)
(3)
(4)
Tr[( k"! # )(a$# )(k% # )(b& # )] =
"
$
(1x2) (3x4)
%
&
- (1x3) (2x4)
+ (1x4) (2x3)
= 4 [( k ! ' a)(k ' b) ( ( k ! ' k)(a ' b) + ( k ! ' b)(k ' a)]
µ )
µ)
µ )
!
!
= 4 [ k k ( ( k ' k)g + k k ! ]
Tr[(a$# )(b& # )] = 4a ' b = 4g
$
&
µ)
Now put it all together …
Electron - Muon scattering
µ!
electron
L
= 2[ k " k + (m # k " $ k)g
µ !
2
µ!
µ
!
+ k k"
2
"
Lmuon
=
2
p
p
+
(M
# p" $ p)gµ! + pµ p!" ]
[ µ!
µ!
]
4
e
µ! muon
M = 2 Lelectron L µ!
t
4
8e
= 2 [( k "p")( kp) + ( k "p)( kp") # m 2 pp" # M 2 kk " + 2M 2 m 2 ]
t
2
This is the “exact” form. Next look at relativistic approx.
Aside:
gµ! g µ! = 4
Relativistic approx. for e-mu scattering:
8e 4
M = 2 [( k !p!)( kp) + ( k !p)( kp!)]
t
2
Let’s use the invariant variables:
s = (k+p)2 ~ 2kp ~ 2k’p’
u = (k - p’)2 ~ -2kp’ ~ -2k’p
2
M = 2e
4
(s
2
+u
t
2
)
2
Next look at ee -> mumu, and get it via crossing.
emu -> emu
=> ee -> mumu
via crossing
k’C
electron
kA
muon
pB
s = (k+p)2
t = (k’ - k)2
electron
k’C
muon
p’D
- pB
electron
kA
electron
k’C
s
t
muon
pB
muon
p’D
“s” = (k’ - k)2 = t
“t” = (k+p)2 = s
e-mu- -> e-mu- => e+e- -> mu+mu2
M = 2e
4
(s
2
+u
t
2
)
2
2
M = 2e
4
(t
2
+u
s
2
)
2
2
Recall exercise 4.2 from H&M:
M pf
d!
=
2
d" cm 64 # s pi
We will now use this for relativistic e+e- -> mu+mu- scattering.
2
pf ~ p i
=>
(because we can neglect masses)
M
d!
#
2
d" cm 64 $ s
t = -2 k2 (1 - cosθ)
u = -2 k2 (1 + cosθ)
s ~ 4k2
α = e2 /4π
Relativistic
limit
(see next slide)
(
M = e4 1 + cos 2 !
=>
2
d!
$
2
#
1+ cos % )
(
d" cm 4s
2
2
4
($
+ &
+ &
! (e e ' µ µ ) #
3s
)
Aside on Algebra
t = -2 k2 (1 - cosθ)
u = -2 k2 (1 + cosθ)
s ~ 4k2
(t
(
2
2
) = 4k
2
)=(
+u
2
s
2
2
t +u
s
2
M = 2e
4
(t
2
+u
s
2
)
2
#$1 ! 2 cos" + cos " + 1 + 2 cos" + cos " %&
4
16k
2
1 + cos "
4
2
2
)
2
(
M = e 1 + cos !
2
4
2
)
Result worthy of discussion
1. σ ∝ 1/s must be so on dimensional grounds
2. σ ∝ α2 two vertices!
3. At higher energies, Z-propagator also
contributes:
More discussion
4. Calculation of e+e- -> q qbar is identical as
long as sqrt(s) >> Mass of quark.
+ "
! (e e # qq ) = 3$
# of flavors
% e ! (e e
2
q
q" flavor
+ "
+
"
#µ µ )
Charge of quark
Sum over quark flavors
Measurement of this cross section was very important !!!
Measurement of R
! (e +e" # qq )
2
R=
= 3$ % eq
+ "
+ "
! (e e # µ µ )
q" flavor
Below charm threshold: R = 3 [ (2/3)2 + (1/3)2 + (1/3)2 ] =2
Between charm and bottom: R = 2 + 3(4/9) = 10/3
Above bottom: R = 10/3 + 3(1/9) = 11/3
Measurement of R was crucial for:
a. Confirm that quarks have 3 colors
b. Search for additional quarks
c. Search for additional leptons
Experimental Result
Ever more discussion
5. dσ/dΩ ∝ (1 + cos2θ)
5.1 θ is defined as the angle between e+ and
mu+ in com. cos2θ means that the outgoing
muons have no memory of the direction of
incoming particle vs antiparticle.
Probably as expected as the e+e- annihilate
before the mu+mu- is created.
5.2 Recall, phase space is flat in cosθ. cos2θ
dependence thus implies that the initial state
axis matters to the outgoing particles. Why?
Helicity Conservation in relativistic limit
• You showed as homework that uL and uR are
helicity eigenstates in the relativistic limit, and
thus:
µ
µ
u ! u = ( uL + uR )! ( uL + uR )
• We’ll now show that the cross terms are zero,
and helicity is thus conserved at each vertex.
• We then show how angular momentum
conservation leads to the cross section we
calculated.
Let’ do one cross product explicitly:
u ! µ u = ( uL + uR )! µ ( uL + uR )
1
1
5
0
5
uL = u ! = u
1"
!
!
=
u
1+
!
(
)
(
)
2
2
1
uR = (1+ ! 5 ) u
2
1
µ
5
µ
5
µ 1
5
5
uL ! uR = u (1+ ! )! (1+ ! ) u = u !
1"
!
1+
!
u=0
(
)
(
)
4
4
T* 0
L
T*
Here we have used:
Helicity conservation holds for all
vector and axialvector currents as E>>m.
! 5! µ = "! µ ! 5
! 5 = ! 5T *
! 5! 5 = 1
eL- eR+ -> muL- muR+
Jz +1 -> +1
eL- eR+ -> muR- muL+
Jz +1 -> -1
eR- eL+ -> muL- muR+
Jz -1 -> +1
eR- eL+ -> muR- muL+
Jz -1 -> -1
Next look at the rotation matrices:
1
#u
1
d11 (! ) = (1+ cos ! ) "
Cross products cancel in
•
•
•
•
•
2
s
1
#u
1
d#1#1 (! ) = (1+ cos! ) "
2
s
1
#t
1
d#11 (! ) = (1# cos! ) "
2
s
1
#t
1
d1#1 (! ) = (1# cos ! ) "
2
s
Spin average:
M ! (1+ cos 2 " )
2
dJ1-1
Initial Jz
final Jz
Conclusion on relativistic limit
• Dependence on scattering angle is given
entirely by angular momentum conservation !!!
• This is a generic feature for any vector or
axialvector current.
• We will thus see the exact same thing also for
V-A coupling of Electroweak interactions.