Why we believe there’s a strong force. Why Colour? –Why not something with no inappropriate...
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Transcript of Why we believe there’s a strong force. Why Colour? –Why not something with no inappropriate...
• 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: - (sss)– - is J = 3/2
• So the spin wave function can be: |1 2 3>
• (spin)(flavour) = |s1s2s3> | 1 2 3>
• COMPLETELY SYMETRIC!!
Slides available at: www-pnp.physics.ox.ac.uk/~huffman/
• Must have another part to wave function– = (spin)(flavour)(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.
ccJ One of the more interesting things is the ‘width’ in the mass of the J/.
On a mass resonance:
fm
MeV
fmMeV
pc
ch
EESS
jE
R
ee
133.01500
200
412124124
2221
22
Breit-Wigner:
e+
e-
Solve Hydrogen atom but with a reduced mass of me/2:
Get first bound state of -6.8eV.Can do the same for the strong force.
rFr
chrV s
03
4)(
Use this potential for the quarksand fit to S and F0.Discover: S = 0.3 and F0 is 16 tons!
Why so narrow?2
23
ee
keV91
Spectroscopic notation:n2s+1LJ
J/13S1 c 13P0 or 23P0
ln is true only for 1/r potential
1). Gluons are spin 1 and massless like photons
2). Gluons have parity -1
One gluon forbidden:c cbar is colour singlet; gluons have colour charge
Two gluons: P and C problem
Three gluons allowed but it isnow suppressed by a factor of s
6
J/ has JPC = 1-- like the photon
What is the Isospin of the J/?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/to I-spin eigenstates: J/ +-, 00, -+ in almost equal proportions
Both rho and pion have Isospin I = 1So the J/could have I = 0,1, or 2 only.Use the 1x1 Clebsh-Gordon table to settle the matter.
1,11,16
10,10,1
3
21,11,1
6
10,2
1,11,12
11,11,1
2
10,1
1,11,13
10,10,1
3
11,11,1
3
10,0
I,I3 +- 00 -+
yes
no
no
How the Beit-Wigner width is measured:2
23
ee
E
Ebeam
In Theory
In Fact
Conservation ofProbability
3 3.15
MeV
dEEdEE
ee
BreitBreit
09.02
32
22
0exp0 exp
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:
ipt
iE
So the QM equivalent of Energy and momentum conservation is:
txcmtxct
tx
cmcpE
,,, 42222
2
22
42222
This is called the Klein-Gordon equation.
With:
xpit
Ei
Netx
),(
Solutions to K-G equation:
For now, ignore negative E solutions. Throw in a potential (But can we stay fully relativistic??)
Naïve Solution Becomes:
2
14222 cmpcE
Vcmc
t
422222
22
xpit
Ei
eNtx
),(
2
14222 VcmpcE
If we assume V is a real function you can multiply the K-G equation by * and multiply the complex conjugate of the K-G equation by . The potential, V, drops out.You then get a continuity equation that you can interpret as some sort of conservation of probability
Suppose that V is a purely imaginary constantV = i
02
22
N
pNE
t0
jt
0
i
p
ttt
i
2
1
42222
14222
21
4222
1
),(
cmcp
icmcp
tixp
i
xpiicmcp
ti
eNe
Netx
Assume is << (p2c2+m2c4) and expand the second term:
22
14222
),(
tcmcp
ti
xpi
eNetx
Note that we no longer have Probability conservation!!!
In the Rest Frame of the particle p=0 and we are left with:
22
)(
tmct
i
NetTransform into ‘Energy’ space:
2)(2
)(
20
22
imcE
iNdte
dtteEt
imcEi
tE
i
4)(
1
2)(
222
2
2
*
mcEimcE
iEE
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 EMW fi 22
Mfi is the ‘matrix element’, in QM you recognize it as
(E) is the ‘density of states available at energy E’.
iVf
Non-relativistic!
Want to increment these values by the smallest amount possible and be assured I have crossed to a new state:
6 – dimensions are needed in QM to describe a particle.
Three in space x0, y0, z0; and three in momentum px0, py0, pz0
Understanding dLIPS
Volumepppzyx zyx 000000
ermsManyCrossTpzpypxpppzyx
ppppppzzyyxx
zyxzyx
zzyyxx
000000
000000
The smallest term in this group is the xp term…the uncertainty principle tells us its size!
6 dimensio
ns!!
Understanding dLIPS
32 zyx pzpypxSmallest distinguishablevolume one state!
dN
pddV
Nppp
zyx
Npzpypx
zyx
zyx
3
3
3
3
2
2
2
Understanding dLIPS
Fine for the ith particle. ii
i dNpd
dV 3
3
2
But suppose we have N total particles in the final state(and we only need to worry about the final state becausein 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
n
i
iinT
pddVdNdNdNdNdN
13
3
3212
…This isn’t Lorentz invariant.
… V1V2…Vn is annoying
Turns out the Matrix Element has Volumes that will cancel with these volume elements.
n
i i
iT
E
pddN
13
3
22
Is Lorentz Invariant!!!
Understanding dLIPSNow we need to add upall of our dN’s to get thetotal.
n
i
inT
E
pdVN
13
3
22
Anticipate the 1/Vn terms from the matrix element.Integration is over all possible values of the ith momentum.
n
i i
iT
E
pd
dE
d
dE
dNE
13
3
22
But we do not have independent momenta, if all but one of the momenta is known, the last one is also known!
Understanding dLIPS
And this is perfectly fine…just remember to apply energyand 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.
1
13
3
2
1
22
n
i ni
i
EE
pd
dE
dE
Needed to keepLorentz inv.