Lecture 9: Quarks II Quarks and the Baryon Multiplets Colour and Gluons Confinement & Asymptotic Freedom Quark Flow Diagrams Section 6.2, Section 6.3, Section 7.1 Useful Sections in Martin & Shaw:

Baryons: Spin numbers of 1/2 and 3/2 suggest the superposition of 3 fermions Absence of anti-particles suggests there is not substantial anti-quark content  So try building 3-quark states Start with 2: Building Baryons (note that m(   ) ≠ m(  + ) so they are not anti-particles, and similarly for the  * group)

 So try building 3-quark states Now add a 3rd: The baryon decuplet !! and the   sealed the Nobel prize  The Decuplet ddd ddu duu uuu dds uus dss uss sss uds Baryons: Spin numbers of 1/2 and 3/2 suggest the superposition of 3 fermions Absence of anti-particles suggests there is not substantial anti-quark content (note that m(   ) ≠ m(  + ) so they are not anti-particles, and similarly for the  * group)

1 I3I3 Y p (938 )   (1321)  0 (1315) n (940)   (1197)  0 (1193)  (1116)   (1189) But what about the octet? It must have something to do with spin... (in the decuplet they’re all parallel, here one quark points the other way) We can ''chop off the corners" by artificially demanding that 3 identical quarks must point in the same direction But why 2 states in the middle? Coping with the Octet ddd ddu duu uuu dds uus dss uss sss uds ways of getting spin 1/2:   u d s  u d s  u d s u d s  u d s u d s these ''look" pretty much the same as far as the strong force is concerned (Isospin)  00 J=1/2 J=3/2

ddd ddu duu uuu dds uus dss uss sss uds We can patch this up again by altering the previous artificial criterion to: ''Any pair of similar quarks must be in identical spin states" What happened to the Pauli Exclusion Principle ??? Why are there no groupings suggesting qq, qqq, qqqq, etc. ?? What holds these things together anyway ?? Quark Questions Not so crazy  lowest energy states of simple, 2-particle systems tend to be ''s-wave" (symmetric under exchange) ( ) So having 2 states in the centre isn’t strange... but why there aren’t more states elsewhere ?! J=1/2 J=3/2   u u s  u u s   u u s i.e. why not and ??? 1 I3I3 Y p (938 )   (1321)  0 (1315) n (940)   (1197)  0 (1193)  (1116)   (1189) The lowest energy state has them

Pauli Exclusion Principle Perhaps, like charge, it also helps hold things together ! We see states containing up to 3 similar quarks  this ''charge" needs to have at least 3 values (unlike normal charge!) Call this new charge ''colour," and label the possible values as Red, Green and Blue We need a new mediating boson to carry the force between colours (like the photon mediated the EM force between charges) call these ''gluons" Colour  there must be another quantum number which further distinguishes the quarks (perhaps a sort of ''charge")

In p-n scattering, u and d quarks appear to swap places. But their colours must also swap (via gluon interactions).  This suggests that an exchange-force is involved... But then we run into trouble while trying to conserve charge at an interaction vertex: R G (a quark-screw!)  the only way out is to attribute colour to the gluons as well. For the above case, the gluon would have to carry away RG quantum numbers Gluons

But we currently have no reason to exclude exchanges which do not change the quark color as well! So, for example, a gluon composed of the superpostion RR 1/  2       BB would couple red and blue quarks without changing their colours Non-Colour-Changing Gluons To handle all possible interchanges, we therefore need different gluons with colour quantum numbers RG, RB, GR, GB, BR, BG

To couple to green as well, we just need one more gluon: RR 1/     +     BB GG Allowing these 2 additional gluons results in a higher degrees of symmetry since we are making use of all possible pair combinations of RGB with the anti-colours  Maybe this is a good thing to do... let’s try it and see ! ''Better" Symmetry Since appropriate superpositions of this gluon with will yield the necessary red-green and blue-green couplings 1/  2   RR   BB 

ICIC YCYC B R G ICIC YCYC B G R RGRG BGBG RBRB BRBR GBGB GRGR RRBB GG Another way:...starting to look familiar ?! Central SU(3) states: 1/  2 (  ) 1/  6(  ) 1/  3(  ) RRBB GG RR BB GG In ''SU(3)-speak", the last state is actually a separate (singlet) representation of the group which is not realized in nature, so we end up with 8 gluons. Another Way: SU(3) The reason we get 9 states for the mesons is that the symmetry there is not perfect, so there is mixing. But, for colour, the symmetry is assumed to be perfect.

We could explain only having the quark combinations seen if we only allowed ''colourless" quark states involving either colour-anticolour, all 3 colours (RGB), or all 3 anticolours. If the carriers of the force (the gluons) actually carry colour themselves, the field lines emanating from a single quark will interact: q q q ''flux tube" * formally still just a hypothesis (calculation is highly non-perturbative) Flux Tubes (hence the analogy with ''colour", since white light can be decomposed into either red, green & blue or their opposites - cyan, magenta & yellow)

For this configuration, the field strength (flux of lines passing through a surface) does not fall off as 1/r 2 any more Can be stopped by terminating field line on another colour charge  Ah! So only colourless states have finite energy ! ''Confinement" q q q q q q q qq q PoP ! ''fragmentation" Confinement The field energy will thus scale with the length of the string and so as L   then E    it will remain constant. Clearly we can’t allow this!!

Need Colourless States... So what about qqqqqq states ? Sure  that’s basically the deuteron (np = uuuddd) How about qqqqq ? Pentaquarks

for q > 0.16e, the number of fractionally charged particles is less than 4x10  22 per nucleon (Halyo et al., PRL 2000) Search for Fractional Charge Search for Free Fractional Charges (M. Perl et al.) v x = qE/6  r v z = 2r 2  g/9 

q q q Getting very close to a quark: q RGRG q RBRB q So, on average, the colour is ''smeared" out into a sort of ''fuzzy ball" Thus, the closer you get, the less colour charge you see enclosed within a Gaussian surface. So, on distance scales of ~ 1 fm, quarks move around each other freely  ''Asymptotic Freedom" Asymptotic Freedom

 the opposite of what happens with vacuum polarization in QED!) Note that asymptotic freedom means that the running coupling will decrease with higher momentum transfer This also means that perturbative QCD calculations will work at high energies! High Energy Limit

Where Are The Coupling Constants Running ???

G RBRB B u = qq annihilation R So how do we now interpret pion exchange?? duuduu duddud uduudu dduddu p p n n Pion Exchange Revisited = qq creation B d R GBGB G  (ud)  or (ud)

 K  + K + Quark Flow Diagrams p + p  p + n +  +  ssss suussuus KK+KK+ uduuududuuud uddduuududdduuud pppp npnp udduuudduu udduuudduu +p+p +p+p  ++ p +  +   ++  p +  + Quark Flow Diagrams: