Announcements
Homework returned now
•
•
Switching to more lecture-style class starting today
• Good luck on me getting powerpoint lectures ready every day
Schedule slowing down a bit
• Monday – reading assignment only
• Wednesday – homework only
9/19
Questions from the Reading Quiz
“Can you explain how to obtain and the meaning of the
Feynman invariant amplitude? ”
iM
•
•
How to obtain it – depends on the theory
• For * theory – read chapter 5
What it means
• I don’t know
• It’s an amplitude – the amplitude that you go into a particular state
• But with some factors taken out to make it simpler
Questions from the Reading Quiz
“I'm also kind of confused on the concept of D, mainly
whether or not is has any physical significance”
•
Definition:
D  
i out
d 3pi
 2 
3
2 Ei
 2    pout  pin  iM
4
4
What is its physical significance?
• It is most of the factors in the probability for a process to occur
• It allows for any number of particles in the final state
• Works for 2 or 3 (or more) particle final states
• It ignores details of the incoming state
• Can work for decay rates or cross-sections
• Mostly, it’s just a calculational tool
2
Questions from the Reading Quiz
“How is it that we just assume using first order
perturbation theory will work?”
• It is an approximation
• In general, for perturbation theory to work, we need a small parameter
• The small parameter in this theory is g
• If g is not small, then this approximation is poor
• In QED, for example, the parameter is e
• e = 0.3, which is kind of small
• As you go to higher order, you get factors of about e2/162
• In QCD, the coupling is significantly larger than one
• Perturbation theory fails
How to calculate everything
Fermi’s Golden Rule
•
The general formula for probability is:
1
1
4
4
P
2

VT



 pout  pin  iM

j in 2 E jV i out 2 EV
i
•
2
For 1-2 particles in the initial state, and 2-3 particles in the final state, these
become:
D

2M

D
4 E2p1  E1p 2
p
D  two  
d  iM
2

16 Ecm
D  three  
1
dE dE

8  2 
5
1
2
2
,
d 1 d12 iM
2
.
Comments on D,  and :
D  two  
•
p
d  iM
2

16 Ecm
2
The quantity D is Lorentz invariant
• Usually calculated in the c.m. frame
• The quantity p is the momentum of either particle in this frame
• The quantity Ecm in any frame is just s
D
D


4 E2p1  E1p 2
2M
• Decay rate is not Lorentz invariant
• Listed value is in comoving (c.m.) frame
• In arbitrary frame, formula becomes D/2E
• Cross section is Lorentz invariant in any frame where particles are colinear
• Head-on collisions or stationary target
• Then magnitude symbols not really necessary
Differential cross-sections and decays:
D

2M
D  two  
p
d  iM
2

16 Ecm
p
  two  
d  iM
2
2 
32 M
2
2
•
Combining these:
•
Sometimes you are asked for the “differential decay rate”
d
p
• That means: don’t do the integral

iM
2
2
d  32 M
•
Similarly for cross-sections
p
D

d  iM

2

64 Ecm E2p1  E1p2
4 E2p1  E1p 2
d
p
2

iM
2
d  64 Ecm E2p1  E1p2
2
2
The general procedure
•
Find the Feynman invariant amplitude
•
Multiply it by its complex conjugate
• Later, this will be harder than it looks now
iM
iM
2
  iM
 iM 
•
Rewrite this quantity in terms of the given or final energies or angles
•
Calculate D using one of the two formulas we have
D  two  
•
p
d  iM
2

16 Ecm
2
, D  three  
Find the decay rate or cross-section
D
D

, 
.
2M
4 E2p1  E1p2
1
8  2 
5
*
 dE1 dE2 d 1 d12 iM
2
.
An easy problem:
Calculate the rate for  decay in the * theory.
iM  ig
•
•
iM
2
  k    p   *  p
 g2
This is a decay, so we use the decay formula
Two particles in final state, so we need that formula too
2
D
p
2
pg

D
d  iM .

2

2M
16 Ecm
32 2 MEcm
2
pg
• Center of mass energy is the mass  
8 M 2
• Final particles have equal and opposite 3-momentum
• Final particles have identical energies
g2

2
2
1
1
E 2M
p  2 M m
16 M 2
M 2  4m 2
 d
A hard problem:
Calculate the rate for muon decay. Treat all final state particles as massless.
   p   e   p1   e  p2     p3 
•
•
iM
2
 64GF2  p  p2   p1  p3 
This is a decay, so we use the decay formula
Three particles in final state, so we need that formula too
D

2m
D
1
dE dE

8  2 
5
1
2
d 1 d12 iM
We need to:
• Write all quantities in terms of the final energies or angles
• Determine limits of integration
• Perform all integrals
2
.
Rewriting the Amplitude


 p   e  p    p    p 


1
e

2
3
iM
2
 64GF2  p  p2   p1  p3 
•
Conservation of momentum
p  p1  p2  p3
p  p2  p1  p3
•
Square this quantity, remembering everything except the muon is massless
2
2
2
m
  2 p  p2  2 p1  p3
p  p  p  p 

•
2
1
3
Working in the rest frame of the muon, so
p  p2  E E2  p 2  p   m E2
p1  p3  12 m2  m E2
iM
2
 32GF2 m2  m E2  2E22 
Limits on integration
D

2m
D
1
8  2 
5
 dE1 dE2 d 1 d12 iM
2
.
iM
2
 32GF2 m2  m E2  2E22 
1
2
2
2

G
dE
dE
d

d

m
E

2
m
E
F
1
2
1
12   2
 2 
5
16
 d 1  4
• Particle 1 can go any direction we want
• The azimuthal angle runs from 0 to 2
d12  2

• The energy integrals are tricky:
• The total momentum is zero: they form a triangle
• The total energy is m: this is the perimeter
• No side can have more than half the total perimeter
E1  12 m , E2  12 m , E3  12 m
E3  m  E1  E2
E1  E2  12 m
p2
p1
p3
Announcements
Homework returned in boxes
•
•
Small error in problem 4.9
Note my solutions are online for past homework problems
(a formula for momentum of the final
particles was found in problem 2.8b)
9/21
Finishing the Problem
Calculate the rate for muon decay. Treat all final state particles as massless.
E1  12 m ,
E2  12 m ,
E1  E2  12 m .
 d
1
 4
 d
12
 2
1
2
2
2

G
dE
dE
d

d

m
E

2
m
E
F
1
2
1
12   2
 2 
5
16
1
 3 GF2  dE1 dE2  m2 E2  2m E22 
2
1m
1m
1
2 
2 2 
2
2
 3 GF   m E2  2m E2  dE2 1
dE1
0
m  E2
2 
2
1m
1
2 2 
 3 GF   m2 E22  2m E23  dE2
0
2

1
2
3
2
F
G

1
3
m E  m E
2
3
2
2
4
4
2

1m
2 
0
1
2 5

G
F m .
3
192
Turning it Into Numbers
Calculate the rate for muon decay. Treat all final state particles as massless.

1
2 5
G
F m
3
192
GF  1.16637 105 GeV 2 ,
m  105.658 MeV  0.105658 GeV
2
1
5
5

1.16637

10
GeV
0.105658
GeV



3 
192
  3.000911019 GeV  3.00911010 eV
6.58211016 eV  s
  
3.00911010 eV
1
  2.1874 106 s  2.1874 s
 exp  2.1970 s
The general procedure
iM
1. Find the Feynman invariant amplitude
2. Multiply it by its complex conjugate
2
*
iM   iM  iM 
• Later, this will be harder than it looks now
3. Rewrite this quantity in terms of the given or final energies or angles
4. Calculate D using one of the two formulas we have
p
1
2
D  two  
d

i
M
,
D
three

dE1 dE2 d 1 d12 iM


5 
2

16 Ecm
8  2 
5. Find the decay rate or cross-section
D
D

, 
.
2M
4 E2p1  E1p2
2
.
Sample Calculation – Step 3
The amplitude for the scattering process
e-(p1) + -(p2)  e-(p3) + -(p4)
is given at right, in the limit where all
masses are negligible. Find the
differential cross-section if they are
colliding head-on, each with energy E.
3. Rewrite this quantity in
terms of the given or final
energies or angles
p1   E , 0, 0, E 
p2   E , 0, 0,  E 
iM
2

2e4
 p1  p3 
2
 p1  p2  p3  p4  


   p1  p4  p2  p3  
p3

p1
p2
p4
p3   E , E sin  cos  , E sin  sin  , E cos  
p4   E ,  E sin  cos  ,  E sin  sin  ,  E cos  
The dot products
iM
2

2e4
 p1  p3 
2
 p1  p2  p3  p4    p1  p4  p2  p3  
p1   E , 0, 0, E 
p3   E , E sin  cos  , E sin  sin  , E cos  
p2   E , 0, 0,  E 
p4   E ,  E sin  cos  ,  E sin  sin  ,  E cos  
p1  p2  E 2  E  E   2E 2
p3  p4  E 2  E 2 sin 2  cos 2   E 2 sin 2  sin 2   E 2 cos 2   2E 2
p2  p3  E 2  E 2 cos 
p1  p4  E  E cos 
2
2
iM
2

2e4
1  cos  
2
 4  1  cos  2 


p1  p3  E 2  E 2 cos 
Sample Calculation – Steps 4 and 5
…Find the differential cross section…
iM
2

2e4
1  cos  
2
 4  1  cos  2 


4. Calculate D using one of the two formulas we have
D  two  
p
d  iM
2

16 Ecm
2

E
 d  iM 
2
16 2 2 E
4
e
d
2 
16
5. Find the decay rate or cross-section
4
e
D
D
 2 

2 2
128

E
8E
4 E2p1  E1p 2
4  1  cos  
 1  cos 
4  1  cos  
d
e

d  128 2 E 2 1  cos  2
4
2
2
2
d
4  1  cos  
1  cos  
2
2
Comments on this problem
e-(p1) + -(p2)  e-(p3) + -(p4)
4  1  cos  
d
e

d  128 2 E 2 1  cos  2
4
2
• Total cross section is infinity
• Caused by  = 0
• Classically, because particles always scatter a little
• Experimentally, small angles cannot be detected
• This expression assumes you are in the c.m. frame
• Since cross-section is invariant, good idea to write in terms of s
2
s  Ecm
  2E   4E 2
2
d
e 4  1  cos  

d  32 2 s 1  cos  2
4
2
• Unfortunately,  is not Lorentz invariant
• If we had already found total cross-section,
this problem would not occur
Sample Calculation – Step 3
The amplitude for H 
is given by the
formula at right, where p and p’ are the momenta
of the final state particles, m is the mass of the
electron, and v is a constant. What is the rate for
this decay? Let M be the mass of the Higgs.
e+e-
3. Rewrite this quantity in
terms of the given or final
energies or angles
iM
2
H  k   e  p   e  p
k  p  p
k 2  p 2  p 2  2 p  p 
M 2  2m2  2 p  p
iM
2
m2
 2  p  p  m 2 
v
2
m2 1 2
m
 2  2 M  m 2  m 2   2  M 2  4m 2 
v
2v
Sample Calculation – steps 4 and 5
H  k   e  p   e  p
iM
2
m2 1 2
m2
2
2
 2  2 M  m  m   2  M 2  4m 2 
v
2v
4. Calculate D using one of the two formulas we have
p
D  two  
d  iM
2

16 Ecm
2
m2
2
2

d

M

4
m

2
2 

16 M
2v
p
pm2
2
2
D
M

4
m

2 
8 Mv
5. Find the decay rate or cross-section
2
pm
D
2
2

M

4
m


2 2 
2 M 16 M v
p
1
4
M 2  m2
m2
2
2 3/2

M  4m 
2 2 
32 M v
Announcements
Homework not yet graded
•
•
Wednesday: Problems 4.9 & 4.10
Friday: Reading quiz I and problems 5.1, 5.2, 5.3, 5.5, and 5.6
9/24
Feynman Diagrams
Where they fit:
1. Find the Feynman invariant amplitude
2. Multiply it by its complex conjugate
• Later, this will be harder than it looks now
3. Rewrite this quantity in terms of the given or final energies or angles
4. Calculate D using one of the two formulas we have
5. Find the decay rate or cross-section
How to draw Feynman Diagrams
• Make a list of initial particles on the left and final particles on the
right
• Label them by their four-momenta (and spin)
• Start drawing the right type of line for each initial and final particle
• Line for  particle
• Arrow right for  particle
• Arrow left for * particle
• Find every possible way to connect everything
together using the allowed couplings
• To tree level only
• You now have a series of Feynman diagrams
• You need to calculate each one
• You need to add their contribution
How to calculate Feynman Diagrams
FOR EACH DIAGRAM
• Conserve four-momentum at each vertex
Multiply the following factors:
• Include one factor of – ig for each vertex
• Include one factor of i/(k2-M2) for each  propagator
• Include one factor of i/(p2-m2) for each  propagator
Add all the diagrams together