Lecture 29: May 8th 2012
Homework due Thursday
Homework: Derive the weak neutral current cross section and the neutral charged current
interference term(i.e. calculate the traces for the cross term) in the cross section for
scattering. Find the total cross section including all terms.
Consider an energy range where the momentums are small compared to the mass of the
W and Z boson but still large compared to the mass of the electron.
Note the charged current version was derived in lecture 25.26.
Reading 9.6-9.7
1) Neutral weak interactions: Z boson
-igZ m
g ( cV - c A g 5 )
2
-i( gmn - qm qn / M Z2 )
Weak Z vertex factor:
Weak Z propagator:
q 2 - M Z2
For small q this propagator reduces to:
GN = GF =
igmn
gz2
and
the
constants
or
M Z2
4M Z2
2gZ2
,
8M Z2
2) Consider electron-neutrino electron scattering:
M=
GN é
u(3)g m ( cV - cAg 5 ) u(1)ùûéëu(4)g m (1- g 5 ) u(2)ùû
ë
2
{
M =16GN2 E 4 4 ( cV + cA ) + ( cV - cA ) (1+ cosq )2
2
2
2
ds GN2 E 2
2
2
=
4 ( cV + cA ) + ( cV - cA ) (1+ cosq )2
2
dW 16p
s=
ò
}
{
}
{
}
GN2 2
2
2
E 4 ( cV + cA ) + ( cV - cA ) (1+ cosq )2 sinq dq
8p
GN2 2 ì
8
2
2ü
s=
E í8 ( cV + cA ) + ( cV - cA ) ý
î
þ
8p
3
{
s=
GN2 2
2
2
E 3(cV + c A ) + (cV - c A )
3p
s=
4GN2 2 2
E {cV + cV c A + c A2 }
3p
}
However this interaction can also happen via the charged W and there can be interference
between the weak charged and neutral currents!
3) Electron-neutrino electron scattering: Z/W interference
However this interaction can also happen via the charged W and there can be interference
between the weak charged and neutral currents!
Consider the matrix elements again.
G
M = N u (3)g m c A - cV g 5 u(1) u (4)g m 1- g 5 u(2)
2
[
M=
(
[
) ][
(
][
]
)
]
GF
u (3)g m (1- g 5 ) u(1) u (4)g m (1- g 5 ) u(2)
2
where GN = GF
Note in this expression of the two matrix elements particle 3 and 4 for the W refer to the
neutrino and the electron rather than vice-versa. However, if we are neglecting the
electron and neutrino masses the sums over 3 and 4 spinors will be the same.
You can then just add the matrix elements
G
M = F u (3)g m (c A + 1) - (cV + 1)g 5 u(1) u (4)g m 1- g 5 u(2)
2
[
(
) ][
(
)
]
and write down the total cross section using the by substituting
Comparing:
4
4G 2
s W = GF2 E 2, s Z = F E 2 {cV2 + cV c A + c A2 }
p
3p
4GF2 2 2
4G 2
4
E {cV + cV cA + cA2 } + F E 2 {cA + cV } + GF2 E 2 ,
3p
p
p
one can see the interference term.
s=
Note the W term was isotropic, the Z term had an angular distribution and we don’t yet
know in the case of the interference term. Let’s evaluate the cross term to verify this
result and derive the angular dependence:
Must be careful to label the particles the same in this case!
G
M Z = N éëu(3)g m cA - cV g 5 u(1)ùûéëu(4)g m 1- g 5 u(2)ùû
2
(
MW =
)
(
)
GF é
u(4)g m (1- g 5 ) u(1)ùûéëu(3)g m (1- g 5 ) u(2)ùû
ë
2
[
]
GF2
M =
u (3)g m (cV - c A g 5 ) u(1)u (1)g n (1- g 5 ) u(4)u (4)g m (1- g 5 ) u(2)u (2)g n (1- g 5 ) u (3)
2
2
M =
2
GF2 é m
Tr ëg ( cA - cVg 5 ) p/ 1g n (1- g 5 ) p/ 4g m (1- g 5 ) p/ 2gn (1- g 5 ) p/ 3 ùû
4
Looks difficult!
n
m
5
n 5
m
n 5
m
5
n
é m
ù
GF2 ê(g cV p/ 1g p/ 4 + g cAg p/ 1g g p/ 4 - g cV p/ 1g g p/ 4 - g cAg p/ 1g p/ 4 )ú
M =
Tr
ú
4 ê(g m p/ 2g n p/ 3 + g mg 5 p/ 2g n g 5 p/ 3 - g mg 5 p/ 2g n p/ 3 - g m p/ 2g ng 5 p/ 3 )
ë
û
2
m
n
m 5
n
é
ù
GF2 ê2 ( cV + cA ) (g cV p/ 1g p/ 4 - g g p/ 1g p/ 4 )ú
M =
Tr
ú
4 ê(g m p/ 2g n p/ 3 - g mg 5 p/ 2g n p/ 3 )
ë
û
2
[
]
Terms with an odd # of gamma 5’s will contract to Tr g 5g b g d = 0. Trace 15.
M =
2
GF2 é
Tr ë2 ( cV + cA ) (g m cV p/ 1g n p/ 4g m p/ 2gn p/ 3 + g mg 5 p/ 1g n p/ 4g mg 5 p/ 2g n p/ 3 )ùû
4
GF2
M =
4 ( cV + cA ) Tr éëg m cV p/ 1g n p/ 4g m p/ 2g n p/ 3 ùû
4
2
See HW 7.37
M = 64GF2 (c A + cV )( p1×p2 )( p3 ×p4 )
M = 256GF2 ( cA + cV ) E 4
Same exact form for the W and the Z/W term except multiplied by (c A + cV )
4) Given these results one could compare various scattering cross sections.
Comparing:
4 2 2
4GF2 2 2
4GF2 2
2
s W = GF E , s Z =
E {cV + cV c A + c A }, s WZ =
E {c A + cV }
p
3p
p
s W +Z =
4GF2 2 2
4GF2 2
4
E {cV + cV c A + c A2 } +
E {c A + cV } + GF2 E 2,
3p
p
p
by only Z
by only W
by both W and Z
The Z only process has an angular dependence:
ds GF2 2
2
2
=
E 4 ( cV + cA ) + ( cV - cA ) (1+ cosq )2 sinq
dq 8p
{
}
Given this pattern of results one can propose a set of simple experiments to extract cV and
cA for the charged leptons.
Use a beam of medium energy neutrinos so that the approximations q<<MW,Z and
q>>me, can be applied. Muon neutrinos so that scattering interactions that result in final
state electrons only involve the Z and process.
From the angular dependent part of the
(cV - c A ) = s a
cross section:
p
3GF2 E 2
From the isotropic part of the process.
p
(cV + c A ) = s i 2
GF E 2
By taking the ratio of these processes to the
dependence on GF and energy.
s Z /sW =
1 2
cV + cV cA + cA2 }
{
3
via the W you can eliminate the
5) Fermion scattering at the Z pole: Here we have to use the full Z propagator
-i( gmn - qm qn / M Z2 )
q 2 - M Z2
é
ù -i ( gmn - qm qn / M Z ) é
ù
ig
-ig
-iM = êu(2) Z g m ( cVe - cVeg 5 ) u(1)ú
u(3) Z g n ( cV - cAg 5 ) u(4)ú
ê
2
2
ë
û
ë
û
2
q - MZ
2
2
Take this to be at high energy where the fermions can be approximated as massless.
Then the dirac equation for massless particles can be used to eliminate the qm terms
(contracted with the proper gamma matrix).
qn = p3n + p4n contracts with g n to to give
éu(3) ( p/ + p/ ) ( c - c g 5 ) u(4)ù
3
4
V
A
ë
û
Then by the dirac equation ignoring the lepton and quark masses we can compute take:
u(3) p/ 3 = 0, p/ 4u(4) = 0
Note this approximation is not as good for the b quark and in fact the probability Z
interactions to b quarks is slightly different than the other down type quarks. Similarly
between the c quarks and the other up quark. However the differences are at the 1%
percent level.
then
M=
[
][
]
gZ2
n (2)g m (c Ae - c Ae g 5 ) u(1) u (3)g n (cV - c A g 5 )n (4)
2
2
4 (q - M Z )
é
ù
g2
2
M = ê 2 Z 2 ú n (2)g m (cVe - c Ae g 5 ) u(1)u (1)g n (cVe - c Ae g 5 )n (2) u (3)g m (cV - c A g 5 )n (4)n (4)g n (cV - c A g 5 )
êë 4 (q - M Z ) úû
2
[
][
é
ù
g2
M = ê 2 Z 2 ú Tr g m (cVe - c Ae g 5 ) p/ 1g n (cVe - c Ae g 5 ) p/ 2 Tr g m (cV - c A g 5 ) p/ 4 g n (cV - c A g 5 ) p/ 3
êë 4 (q - M Z ) úû
2
2
[
] [
]
]