Transcript Particle Physics Option

• Why we believe there’s a strong force.
• Why Colour?
– Why not something with no inappropriate
mental imagery
• Probing the Colour Force
– The study of simple massive quark bound states
• So Far…not speculated what holds the
proton together.
• Also Serious Mystery: W- (sss)
– W- is J = 3/2
• So the spin wave function can be: |1  2  3>
• c(spin)f(flavour) = |s1s2s3> |  1  2  3>
• COMPLETELY SYMETRIC!!
Slides available at: www-pnp.physics.ox.ac.uk/~huffman/
• Must have another part to wave function
– Y = c(spin)f(flavour)j(space) = symetric
– Not allowed for Fermions!
• Not only is colour needed, we know it
must have an antisymetric wave function
and it must be in a singlet state with zero
net colour.
J Y  c c 
One of the more interesting things is the ‘width’ in the mass of the J/y.
On a mass resonance:
Breit-Wigner:
 E  
4 2 j  1
2

2
ee
4
2S1  12S2  1E  ER 
2

 2 4
hc 200 MeV fm


 0.133 fm
pc
1500MeV
e+
Solve Hydrogen atom but with a reduced mass
of me/2:
e-
Get first bound state of -6.8eV.
Can do the same for the strong force.
4  s hc
V (r )  
 F0 r
3 r
Use this potential for the quarks
and fit to S and F0.
Discover: S = 0.3 and F0 is 16 tons!
Why so narrow?
2  ee 
3  
  
2
1). Gluons are spin 1 and
massless like photons
2). Gluons have parity -1
  91keV
One gluon forbidden:
c cbar is colour singlet;
gluons have colour charge
Spectroscopic notation:
n2s+1LJ
J/y  13S1 cc0  13P0 or 23P0
ln is true only for 1/r potential
Two gluons: P and C problem
Three gluons allowed but it is
now suppressed by a factor of
s6
J/y has JPC = 1-- like the photon
What is the Isospin of the J/y?
We already know that I3=0 because of the Quark composition.
Because I3 ≤ I, we know that I = 0,1,2,3…. Integer not ½ integer.
Look at the decays of the J/y to I-spin eigenstates:
J/y  r+-, r00, r-+ in almost equal proportions
Both rho and pion have Isospin I = 1
So the J/y could have I = 0,1, or 2 only.
Use the 1x1 Clebsh-Gordon table to settle the matter.
I,I3
r+-
0,0 
1
1,1 1,1 
3
1,0 
1
1,1 1,1
2
2,0 
1
1,1 1,1 
6
r-+
r00
1
1,0 1,0 
3
1
1,1 1,1
3
yes

1
1,1 1,1
2
no
2
1,0 1,0 
3
1
1,1 1,1
6
no
How the Beit-Wigner width is measured:
2  ee 
3  
  
In Theory
Conservation of
Probability

  E dE

0
exp
2

exp
   EBreit dEBreit
0
3
 
  2 2  ee      0.09 MeV
2
  
2
In Fact
DEbeam
3
3.15
E
Why This Shape?

Would like to motivate this with a
relativistic wave equation.
To do this try using E2=p2c2+m2c4
Operator equivalents of E and P:

E  i
t

p  i 
So the QM equivalent of
Energy and momentum
conservation is:
2 2
2
E  p c  m2c 4 
2

f x, t 
2
2 2 2
2 4






c

f
x
,
t

m
c f x, t 
2
t
This is called the Klein-Gordon equation.
Solutions to
K-G equation:
 
E
p x
i t i



f ( x, t )  Ne

With: E   c 2 p 2  m 2 c

1
4 2
For now, ignore negative E solutions. Throw in a
potential (But can we stay fully relativistic??)
2

f
2
2 2 2
2 4




c

f

m
c f  Vf
2
t

Naïve Solution Becomes: f ( x, t )  N e
 
E
p x
i t i


E  c 2 p 2  m2 c 4  V 
1
2
If we assume V is a real function you can multiply
the K-G equation by f* and multiply the complex
conjugate of the K-G equation by f. The potential,
V, drops out.
You then get a continuity equation that you can
interpret as some sort of conservation of probability

 
i   f
f     p  
f
f
     f f  ff
t  t
t 
i

r  
 j 0
t
  0

2
2




 2E N
 N 

   p
0
t   
  
Suppose that V is a purely imaginary constant
V = i
 
t 2 2 2 4
p x
2
i p c  m c  i  i




f ( x , t )  Ne
 Ne
 
p x
i


1

e
t
i p 2 c 2  m 2 c 4


1

i
2  1
2 2
2 4

p
c

m
c


Assume  is << (p2c2+m2c4) and expand
the second term:
 
p x
i


f ( x, t )  Ne
e

t 2 2 2 4
i p c  m c

Note that we no longer have Probability
conservation!!!

1
2 t

2





1
2
In the Rest Frame of the particle p=0 and we
t

are left with:
i mc 2 t
f (t )  Ne

2
Transform into ‘Energy’
space:
E
f E    e f (t )dt
i t


 e
t

i  E  mc 2  i 
2 

0
2
f E f E  
*
i

2
( E  mc )  i
2
N
i
dt 
 2 ( E  mc 2 )  i 
2

1
2

( E  mc 2 ) 2 
4
And The Breit-Wigner form emerges.
• Probability per unit time of a transition from
initial state |i> to final state |f> is constant.
• Call it W
2
2
W
M fi r E 

Mfi is the ‘matrix element’, in QM you recognize it as
r(E) is the ‘density of states available at energy E’.
fVi
Understanding the:
dLIPS
How many QM ‘states’ are there in a
volume ‘V’ up to a given momeum ‘P’?
z
y
x
How big is this?
Dx
Dpx
Understanding dLIPS
6 – dimensions are needed in QM to describe a particle.
Three in space x0, y0, z0;
and three in momentum px0, py0, pz0
x0 y0 z0 px0 py 0 pz 0  Volume
Want to increment these values by the smallest amount
possible and be assured I have crossed to a new state:
x0  Dx  y0  Dy z0  Dz  px0  Dpx  p y 0  Dp y  pz 0  Dpz  
x0 y0 z0 px 0 p y 0 pz 0  DxDpx DyDp y DzDpz  ManyCrossTerm s
The smallest term in this group is the DxDp term…
the uncertainty principle tells us its size!
Understanding dLIPS
Smallest distinguishable
volume  one state!
DxDpx DyDp y DzDpz  2 
3
DxDp x DyDp y DzDp z
2 
3
DxDyDz
Dp x Dp y Dp z
2 

d p
dV
 dN
3
2 
3
3
 DN
 DN
Understanding dLIPS

d pi
dVi
 dNi
3
2 
3
Fine for the ith particle.
But suppose we have N total particles in the final state
(and we only need to worry about the final state because
in any experiment we take great pain to put the initial
particles into a single, well-defined state).
Ni+Nj = wrong!! These states are more like dice!
Or calculation of specific heat of a crystal.
How many possible combinations
of two distinguishable pairs are possible?
Understanding dLIPS

d pi
dNT  dN1dN2 dN3 dNn   dVi
3
2 
i 1
n
3
…This isn’t Lorentz invariant.
… V1V2…Vn is annoying
Turns out the Matrix Element has Volumes that
will cancel with these volume elements.

d pi
dNT  
3
i 1 2 Ei 2 
n
3
Is Lorentz Invariant!!!
Understanding dLIPS
Now we need to add up
all of our dN’s to get the
total.

d pi
n
NT  V   
3
i 1 2 E 2 
3
n
Anticipate the 1/Vn terms from the matrix element.
Integration is over all possible values of the ith momentum.

d pi
dNT
d
r E  



dE dE i 1 2Ei 2 3
n
3
But we do not have independent momenta, if all but one
of the momenta is known, the last one is also known!
Understanding dLIPS
3

d pi  1
d


r E  

3

dE i 1  2 Ei 2   2 En
n 1
Needed to keep
Lorentz inv.
And this is perfectly fine…just remember to apply energy
and momentum conservation at the end.
But using properties of the Dirac delta Function we can
re-cast this equation in the following form (and explicitly
include energy and momentum conservation.
Understanding dLIPS
We do have Energy and Momentum conservation.
So, for example, in CM frame with total Energy W.
n




4
3
2  W   Ei    pi 
Note the additional factors
i 1

  i 1 
R
n

n

d pi



4
3
2   W   Ei    pi 
3
2Ei 2 
i 1

  i 1 
 
all i 1
3
