Lecture 24: April 19th 2012
Reading for next time: Griffiths 9.1,9.2
Homework due next Thursday:
Homework due Thursday: 9.3, 9.10
1) Briefly on hadron structure
M =
2
[
][
]
[
*
ge4 mn
mn
m
n
m n
m n
mn
L Fmn , L = u (3)g u(1) u (3)g u(1) = 2 p1 p3 + p3 p1 - g ( p1 × p3 )
2
t
]
The second part is unknown. However, we know that it has to have a form of a second
rank tensor(two independent indices) and must involve the initial, p2, final, p4, and
propagator q four momentums where only two are independent. Writing this in terms of
the initial momentum p=p2 and q the most general form is:
F mn = -F1gmn + F2
p m pn
q m qn
qm pn + p m qn
+
F
+
F
3
4
p2
p2
p2
Next if we consider this in lab from or even CM frame where the nucleon is very heavy p
squared is just a mass squared.
Consider in lab frame where nucleon is very heavy p squared is just a mass squared.
All the F terms turn out not to be independent. Writing down the independent terms
(redefining F)
æ qm qn
ö F æ
1 öæ
1 ö
F mn = F1ç 2 - g mn ÷ + 22 ç p m + qm ÷ç pn + qn ÷
2 øè
2 ø
è q
ø M è
then
M =
é æ qm qn
ö F2 æ m 1 m öæ n 1 n öù
ge4
mn
L
F
g
÷ + 2 ç p + q ÷ç p + q ÷ú
mn ê 1ç
2
t2
2 øè
2 øû
ø M è
ë è q
M =
é æ q m qn
ö F æ
2ge4 m n
1 öæ
1 öù
m n
mn
p
p
+
p
p
g
p
×
p
( 1 3 )]êF1ç 2 - gmn ÷ + 22 ç pm + qm ÷ç pn + qn ÷ú
3 1
2 [ 1 3
t
2 øè
2 øû
ø M è
ë è q
2
2
q is just p1-p3 so this will reduce to terms with ( p1 × p3 )(F1 only), ( p1 × p),
only). Absorbing any constants into the F terms
(F2
M =
2
ù
4ge4 é
F
F p × p3 ) + 22 (( p1 × p)( p3 × p))ú
2 ê 1( 1
û
t ë
M
The first term (t type term) will gives sin 2
M =
2
q
q
and the second (u type term) a cos2 .
2
2
é
q
qù
F1 sin 2 + F2 cos 2 ú
ê
q
2
2û
4 p 2 sin 4 ë
2
ge4
The F terms are known as form factors. They are going to depend on the details of the
charge distribution inside the proton, charge location and velocity. The F terms can be
identified as linear combinations of the Fourier transforms of the charge and the magnetic
moment distribution inside the proton. Redefining this in this way.
M =
2
éæ 2 k 2q2 2 ö 2 q
ù
q2
2
2q
F
cos
F
+
k
F
sin
êç F1 (
)
ú
÷
1
1
2
q
4M2 ø
2 2M 2
2û
4 p 2 sin 4 ëè
2
ge4
In this parameterization the F terms are literally the Fourier transforms of the electric
charge and magnetic moment distributions. Again note that the F functions are functions
of q2. For instance for proton at q=0, a long wavelength electron that can’t probe the
structure of the nucleon, F1=1 and F2=1 and for the neutron F1=0 and F2=1. The proton
has a charge of 1 and a magnetic moment of 1 times kappa while the neutron only has a
magnetic moment.
For a point like particle kappa = 0 and F1=1 and F2=1
é 2q
q2
qù
cos
sin 2 ú
ê
2
q
2 2M
2û
4 p 2 sin 4 ë
2
The formula for electron muon scattering in lab frame.
M =
2
ge4
The determination of these form factors and the equivalent ones for the gluons is what we
call determining the PDFs or parton distribution functions of the proton. In hadron
physics these effect the angular distributions and energies of the produced particles.
The formalism is almost identical if the proton is broken apart except that the form
factors are slightly different to account for the formation of final particles.
2) Renormalization
If quantum electrodynamics we find that there are divergent terms proportional to the
logarithm of the momentum of the propagator particle as you integrate to infinity.
Consider that these diagrams typically have 4 internal propagators that must be integrated
over and 4 vertices that introduce delta functions. One delta function conserves overall
momentum and energy conservation so only three of the propagators integrals can be
performed using the delta functions leaving one that must be integrated. This integral,
which can involve momentums, from zero to infinity leads to the divergent terms.
One way to fix this is to insert a new propagator term into the matrix element
computation.
-M2/(q2-M2), with a large mass M
This term looks somewhat like a propagator term and is equivalent to saying that at large
energy the particles can interact via a new force with a force carrying bosom of mass M.
In fact this is now just what we think happens. We had many problems that the
phenomenology of the standard model did not answer such as dark matter and
unification. Introduction a new set of particles and interactions as can solve most of these
problems. Performing the new integral you find that for finite M you find new terms that
modify the constants but otherwise leave the algebra of the calculations unchanged. Thus
you have the same interaction but with new effective constants.
In low energy reactions the masses and coupling constants we observe are these modified
versions. At higher energies there the additional terms that modify the coupling constants
in an energy dependent way. This “running” of the coupling constants with the energy
of the interactions has been observed in experiment.
Example for EM force
ì
æ M 2 öü
g2
gR = ge í1- e 2 ln ç 2 ÷ý
è m øþ
î 12p
ì
a ( q 2 ) = a ( 0) í1+
î
for small,large x
a ( 0) æ -q 2 öü
fç
÷ý, f ( x ) = x / 5, ln ( x )
3p è m 2 øþ
One can even calculate the mass M you need to prevent problems anywhere in the model
which you will find to be about 1TeV, the scale at which we expect to find new physics
at the LHC. All the quantum field theories are renormalizable in this way and all are
consistent assuming we find new particles and interactions at a higher scale. However,
now the problem has just been pushed off to a new energy scale since these new
interactions will have their own divergences. You can either postulate another higher
energy scale where still other interactions take over or an theory with a mechanism for
exact cancelation of the divergent terms (for instance Super symmetry where an exact
symmetry between fermions and super symmetric bosons (and vice-versa) leads to
exactly canceling divergent terms.
3) Introduction to the weak force.
The Feynman rules for the weak force are similar to the electromagnetic force except in
that we need a different vertex factor and propagator.
Vertex factor: The vertex factor should be constructed to give a bilinear covariant when
sandwiched between the initial and final state particle (bilinear covariant terms were are
the Lorentz invariant constructions). Also the vertex factor should represent the parity
violating nature of the weak force. We know that a combination of a  and a 5 term
would violate parity since one term is parity even and the other parity odd. From our
study of the neutrino solution to the Dirac equation we know that a (1-5) will only
interact with the left handed neutrino. Without the evidence of the neutrino we would
have to use a general superposition of these terms (next weeks homework). In fact for
the Z we will have to be more general.
Weak vertex factor:
-igw m
g (1- g 5 )
2 2
Properties: Bilinear covariant, parity violating, only interacts with left(right) handed
(anti)neutrinos
Propagator:
-i( gmn - qm qn / M 2 )
q2 - M 2
Properties: Denominator for massive force carrier and numerator from sum over massive
boson wave functions.
For small q (for instance the energy of nuclear decays) this propagator reduces to:
igmn
M2
In this limit if we consider the constants in the vertex factors and the propagator for a tree
g2
level process we have: W 2 .
8M
Considering that we expect a unification between the EM and weak forces then
we can expect gW and ge to have similar values.
2gW2
gW2
1
-5
2
In fact:
,
where
.
a
=
=
=
G
=
1.166x10
/GeV
W
F
2
8M
4p 29.5
2) Aside on the weak coupling constant given running coupling constants.
The coupling constant for the electromagnetic has been shown experimentally to run, get
larger, with energy. This can be shown explicitly in QED by calculating all the higher
order diagrams, which include e+e- pairs forming around a charged particle and changing
it’s effective potential or coupling strength. Effectively this can be taken as a correction
to the coupling strength giving a renormalized coupling strength.
ì a ( 0) æ -q 2 öü
a ( q ) = a ( 0) í1+
fç
÷ý, f ( x ) = x / 5, ln ( x ) for small,large x
3p è m 2 øþ
î
The electromagnetic and weak forces have similar strength at high energy. If the weak
force runs differently then it might have a different strength at lower energy. In fact we
would expect the weak force to run differently since the higher order diagrams will
involve different particles and different propagators. The relationship between ge as gW is
not necessarily energy independent. Actually the primary modification is going to the
weak coupling constant is going to come from how the weak boson interact with the
particle that gives them mass, the Higgs boson and otherwise the couplings will run in
similar ways. They will be closer at higher energy.
2
3) Charged W weak decays and scattering.
For decays and scattering at low energy, q, and the matrix elements will be quite simple
since there are no possible alternative diagrams in different channels. For instance, for
muon decay to an electron and neutrinos.
é
ù ig é
ù
-ig
-ig
-iM = êu (3) w g m (1- g 5 ) u(1)ú mn2 êu (4) w g n (1- g 5 )n (2)ú
ë
ûM ë
û
2 2
2 2
where u (3) is the final state spin ½ muon neutrino (same notation as for a massive spin
½ particle since they are both spin ½ solutions of the dirac equation), u(1) is the initial
state muon, u (4) is the final state electron, and n (2) is the final state anti-electron
neutrino.
m(1) ® n m (3)e(4)n e (2)
I’ve numbered these to be consistent with scattering next
or for particle scattering
é
ù ig é
ù
-ig
-ig
-iM = êu (3) w g m (1- g 5 ) u(1)ú mn2 êu (4) w g n (1- g 5 ) u(2)ú
ë
ûM ë
û
2 2
2 2
m(1)n e (2) ® n m (3)e(4)
where instead you have an initial state electron neutrino. Note that matrix element will
be no different since particle and antiparticle terms introduce ( p/ ± m) factors, but the
neutrino is effectively massless.
For the scattering:
M=
[
][
]
gW2
u (3)g m (1- g 5 ) u(1) u (4)g m (1- g 5 ) u(2)
8MW2
2gw2
The terms in front can be identified with the Fermi coupling constant GF =
8M W2
M=
[
][
]
GF
u (3)g m (1- g 5 ) u(1) u (4)g m (1- g 5 ) u(2)
2
[
][
] [u(4)g (1- g )u(2)][u(4)g (1- g )u(2)]
M =
GF2
u (3)g m (1- g 5 ) u(1) u (3)g n (1- g 5 ) u(1)
2
M =
GF2
u (3)g m (1- g 5 ) u(1)u (1)g n (1- g 5 ) u(3) u (4)g m (1- g 5 ) u(2)u (2)g n (1- g 5 ) u (4)
2
2
2
[
*
5
m
n
][
5
*
]
Perform the sums and averages over spins. Introduce a factor of ½ for the average over
muon spins.
2
G2
M = F Tr éëg m 1- g 5 ( p/ 1 + mm ) g n 1- g 5 p/ 3 ùûTr éëg m 1- g 5 p/ 2g n 1- g 5 ( p/ 4 + me )ùû
4
(
)
(
)
(
)
(
)
Similar traces to those we found for the EM force except that we have the gamma 5
terms. Mass terms drop out since they have an odd number of gamma matrices. g 5 is
four gamma matrices. Evaluating this trace:
M = 64GF2 ( p1 × p2 )( p3 × p4 )
2
Note that when we average over the initial muon states there are two possibilities while
when we sum over the final electron states there are two possibilities for the electron but
only one possible spin state for the neutrinos.
Note that the convention of s, t and u factors is not as useful here since the mass of the W
dominates in the propagator, where for the electromagnetic force the mass of the photon
was zero and the electron mass was typically negligible and q was the dominant term in
the denominator.
The matrix element for decay is the same. There was no crossing type exchange of two
particles. You have an electron antineutrino in the final state instead of a neutrino in the
initial state. This would introduce a negative sign in front of the mass term from the sum
over electron antineutrino states. However, the mass terms are certainly negligible for
neutrinos.