Lecture 26: April 26th 2012
Reading 9.1-9.5
Reading for next time 9.5-9.6
Homework
1) 9.5, 9.29
When Comparing to the experimental result, remember that the decay to a muon/electron
is only part of the total decay width of the tau. Is solving the problem using the
assumption muon mass is small supported by the experimental data?
2) What quarks or anti-quarks does n e not interact with.
3) 9.16
1) Consider two scattering interactions for electrons at medium energy:
t channel neutrino electron scattering: e-n e ® n ee-,12 ® 34
M=
[
][
]
GF
u (3)g m (1- g 5 ) u(1) u (4)g m (1- g 5 ) u(2)
2
M = 64GF2 ( p1 × p2 )( p3 × p4 )
2
In CM frame
æ æ
2 ö1/ 2 ö
m
e
( p1 × p2 ) = ( p4 × p3 ) = E e Ene + (E - m ) Ene = Ene E e çç1+ ç1- 2 ÷ ÷÷ = 2E 2
è è Ee ø ø
2
M = 256GF2 E 4
2
e
2 1/ 2
e
We can express this this in terms of s, t and u
s = ( p1 + p2 ) = 2( p1 × p2 ) = 4E 2
2
M =16GF2 s2
2
s channel anti-neutrino electron scattering via annihilation: e-n e ® n ee-,12 ® 34
M=
GF é
n (2)g m (1- g 5 ) u(1)ùûéëu(4)g m (1- g 5 ) n (3)ùû
ë
2
M = 64GF2 ( p1 × p3 )( p2 × p4 )
2
In CM frame
( p1 × p3 ) = ( p2 × p4 ) = E 2 (1- cosq)
M = 64 E 4 (1- cosq)
2
2
t = ( p1 - p3 ) = -2( p1 × p3 )
2
M =16GF2 t 2
2
In electron (anti)neutrino scattering s channel has the angular dependence and t is
isotropic.
Also the s channel diagram can be obtained from the t channel diagram by an s to t
crossing.
2) Scattering cross sections
using d/d = (1/8)2 |M|2 S |pf|/|pi| 1/E2, where E is the total energy of both particles
t channel neutrino electron scattering: e-n e ® n ee-,12 ® 34
M = 256GF2 E 4
2
s=
4
p
GF2 E 2
s channel anti-neutrino electron scattering: e-n e ® n e-,12 ® 34
s=
1 2 2
2
1 2 2
GF E ò (1- cosq ) sinq =
GF E ò (sinq - 2cosq sinq + cos2 q sinq ) dq
2p
2p
s=
ö
1 2 2æ
1
GF E ç-cosq + cos2 q - cos3 q ÷
è
ø
2p
3
s=
4 2 2
GF E
3p
There is also a difference in the total cross section!
Why is anti-neutrino scattering one third as likely? We can make a helicity agreement.
The backward scattering interaction for the anti-neutrino basically violates helicity
conservation for the antineutrino. This relationship would have been different if the W+
didn’t interact with neutrinos in a pure V-A way.
Helicity conservation is an important factor in how neutrinos interact that has to be
considered when detecting neutrinos and anti-neutrinos.
3) Pion decay
Since the anti u and d quarks in the pion interact via the strong force it is incorrect to
consider this as an annihilation diagram. However, it is instructive.
,
é
ù ig é
ù
-ig
-ig
-iM = ên (2) w g m (1- g 5 ) u(1)ú mn2 êu (3) w g m (1- g 5 )n (4)ú
ë
ûM ë
û
2 2
2 2
2
G
2
M = F n (2)g m (1- g 5 ) u(1)u (1)g n n (2) u (3)g m (1- g 5 )n (4)n (4)(1- g 5 ) u(3)
2
[
][
[
]
][
]
GF2
M =
Tr g m (1- g 5 )( p/ 1 + md )g n (1- g 5 )( p/ 2 - mu ) g m (1- g 5 ) p/ 4 g n (1- g 5 ) p/ 3
2
The terms to first order in the quark mass are will cancel our since they have an odd
number of gamma matrices. The term to second order in the quark masses turns out to
be zero uses eq 7.66 g m g 5 = -g 5g m and g 5g 5 = I and chapter 7 trace 15.
2
M = 64GF2 ( p1 × p4 )( p2 × p3 )
2
In CM the quark momentums will be approximately 0 and we will gets terms with the
quark mass times the energy of the outgoing particles.
However, this is not quite a correct treatment. The distribution of momentums of the
quarks, and thus the possible angular distributions of terms like ( p1 × p4 ), will be
constrained by the strong force and we should treat them in a more generalized way.
[
]
GF m
F u (3)g m (1- g 5 )n (4)
2
Where F will be a function of the pion or quark momentums pm = p1m + p2m times a scalar
function of the initial momentum vector p 2, f p p2 = f p ( mp ).
M=
( )
[
]
GF2 f p2 m n
M =
p p u (3)g m (1- g 5 )n (4)n (4)g n (1- g 5 ) u(3)
2
GF2 f p2 m n
2
M =
p p Tr g m (1- g 5 ) p/ 4 g n (1- g 5 ) p/ 3
2
GF2 f p2 m n
2
M =
p p {8( p4m pn3 + p3m pn4 - g mn ( p3 × p4 )) + 8ip1a p3b eanbm }
2
The last is anti-symmetric under interchange of p1 and p3, which are equal/opposite.
2
M = 4GF2 f p2 [2( p × p3 )( p × p4 ) - p2 ( p3 × p4 )]
2
[
]
[
]
using
p = p1 + p2 = p3 + p4
( p× p3 ) = (( p3 + p4 ) × p3 ) = me2 + ( p3 × p4 )
( p × p4 ) = (( p3 + p4 ) × p4 ) = ( p3 × p4 )
p2 = ( p3 + p4 ) = p32 + p42 + 2( p3 × p4 ) = me2 + 2( p3 × p4 ) = mp2
1
1
( p3 × p4 ) = mp2 - me2
2
2
2
éæ
æ1
1
1 öæ 1
1 ö
1 öù
2
M = 4GF2 f p2 ê2ç me2 + mp2 - me2 ÷ç mp2 - me2 ÷ - mp2 ç mp2 - me2 ÷ú
è2
2
2 øè 2
2 ø
2 øû
ëè
2
2 2
2
2
2
2
2
2
2
M = 2GF f p mp + me mp - me - mp mp - me
[(
)(
M = 2G f p m ( mp - m
2
2
F
2
2
e
2
The decay width is: G =
2
e
)
)
pf
M
(
2
8pmp
Where using homework 3.16
(mA,0,0,0) = (EB + EC,0,0,0)
(mA,0,0,0) = (EB + (pB2 + mC2)1/2,0,0,0)
(mA,0,0,0) = (EB + (EB2 - mB2 + mC2)1/2,0,0,0)
m A = EB + (EB2 - mB2 + mC2)1/2
(m A - EB)2 = EB2 - mB2 + mC2
m A2 -2 m A EB + EB2 = EB2 - mB2 + mC2
EB = (m A2 + mB2 - mC2)/2 m A
pf =
1
mp2 - me2 )
(
2mp
2
)]
G=
2
1
GF2 f p2 me2 2
2
2
2 2 2
2
2
m
m
2G
f
m
m
m
=
m - me2 )
e)
F p
e( p
e)
3 ( p
3 ( p
16pmp
8pmp
If we compare the decay width to muons and electrons
Again we can make a helicity argument. The electron or muon is forced to be in the
wrong helicity state which can happen more often the muon.
2) Quark generations
Consider the decay:
The Width should be: G =
GF2 f K2 me2 2
(mK - me2 )
8pmK3
Since for f p and f K involve the same type of strong interactions between quarks by
isospin symmetry we don’t expect them to be different.
However is we compare
G ( K - ® e-n e )
and
we find:
3
2
2
fK2 mp ( mK - me ) mK
=
=
¹ 0.126
2
mp
G ( p - ® e-n m ) fp2 m3 ( mp2 - me2 )
K
2
We expected the kaon decay width to be larger!
To solution to this is to have a suppression factor when decaying across generations in
the matrix element. For the standard d to u decay there is a factor of cosqc and for the
suppressed s to u decay there is a factor of sin 2 q c and qc =13.1.
It was also observed that s to d transitions only occurred at extremely small rates. There
had to be a mechanism that almost completely suppressed them.
The solution was the introduction of weak quark eigenstates as the states that couple to
the weak force and express the strong eigenstates as superposition’s of the weak
eigenstates and viceversa.
Weak states expressed as super-positions of strong states.
d ' = d cosq c + ssin qc
s' = -d sin q c + scosq c
and strong states expressed as super-positions of weak states.
d = d ' cosq c - s' sin qc
s = d ' sinq c + s' cos qc
or represented as a rotation matrix
æ d ' ö æ cosq c sin qc öæ dö
ç '÷=ç
÷ç ÷
è s ø è-sin q c cosq c øè s ø
æ d ö æcos qc
ç ÷=ç
è s ø è sin q c
-sin q c öæ d ' ö
÷ç ÷
cosqc øè s' ø
The weak force will only relate quarks of the same weak generation
Then for weak quark decay.
For the interaction within the generation d  u:
é
ù
-igw m
g 1- g 5 u(1,d)ú
êëu (3,u)
û
2 2
é
ù
-igw m
g 1- g 5 u(1,d ' )cosqc - u(1,s' )sin qc ú
êëu (3,u)
û
2 2
(
)
(
)(
)
Then only the contribution within the weak generation contributes.
é
ù
-igw cosqc m
g 1- g 5 u(1,d' )ú
êëu (3,u)
û
2 2
(
)
The amplitude will have a factor of cos2 qc
For interactions across the generation s  u:
é
ù
-igw m
g 1- g 5 u(1,s)ú
êëu (3,u)
û
2 2
é
ù
-igw m
g 1- g 5 u(1,d ' )sinqc + u(1,s' )cosqc ú
êëu (3,u)
û
2 2
(
)
(
)(
)
Only allow interactions within the weak generation contributes
é
ù
-igw sin q c m
g 1- g 5 u(1,d' )ú
êëu (3,u)
û
2 2
The amplitude will have a factor of sin2 q c
(
)