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Lecture 9: Quarks II

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Lecture 9: Quarks II Lecture 9: Quarks II • Quarks and the Baryon Multiplets • Colour and Gluons • Confinement & Asymptotic Freedom • Quark Flow Diagrams Useful Sections in Martin & Shaw: Section 6.2, Section 6.3, Section 7.1 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(Σ-) ≠ m(Σ+) so they are not anti-particles, and similarly for the Σ* group) ⇒ So try building 3-quark states Start with 2: 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(Σ-) ≠ m(Σ+) so they are not anti-particles, and similarly for the Σ* group) ⇒ So try building ddd ddu duu uuu 3-quark states dds uus Now add a 3rd: uds dss uss The baryon decuplet !! and the Ω- sealed the Nobel prize ⇒ sss Y But what about the octet? n 1 p (940) (938) It must have something to do with spin... (in the decuplet they’re all parallel, 0 (1193) Σ- Σ Σ+ I here one quark points the other way) Λ (1116) 3 (1197) (1189) We can ''chop off the corners" by J=1/2 0 artificially demanding that 3 identical Ξ- -1 Ξ quarks must point in the same direction (1321) (1315) But why 2 states in the middle? ddd ddu duu uuu ways of getting spin 1/2: dds uus ↑ ↓ ↑ ↓ ↑ ↑ ↑ ↑ ↓ uds u d s u d s u d s 0 Σ dss uss these ''look" pretty much the same as far as the strong J=3/2 force is concerned (Isospin) Λ sss So having 2 states in the centre isn’t strange... Y but why there aren’t more states elsewhere ?! n 1 p (940) (938) i.e. why not ↑ ↓ ↑ and ↑ ↑ ↓ ??? 0 u u s u u s (1193) + Σ- Σ Σ I Λ (1116) 3 (1197) (1189) We can patch this up again by altering J=1/2 the previous artificial criterion to: 0 Ξ- -1 Ξ The lowest energy state has (1321) (1315) ''Any pair of similar quarks must them be in identical spin states" ddd ddu duu uuu Not so crazy ⇒ lowest energy states of simple, 2-particle systems tend to be dds uus ( ''s-wave" (symmetric under exchange) ) uds What happened to the Pauli Exclusion Principle ??? dss uss Why are there no groupings suggesting J=3/2 qq, qqq, qqqq, etc. ?? sss What holds these things together anyway ?? Pauli Exclusion Principle ⇒ there must be another quantum number which further distinguishes the quarks (perhaps a sort of ''charge") 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" 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 To handle all possible interchanges, we therefore need different gluons with colour quantum numbers RG, RB, GR, GB, BR, BG 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 1/√2 ( ⎜ RR 〉 - ⎜BB 〉 ) would couple red and blue quarks without changing their colours To couple to green as well, we just need one more gluon: 1/√6 ( ⎜ RR 〉 + ⎜ BB 〉 - 2 ⎜ GG 〉 ) Since appropriate superpositions of this gluon with 1/ √ 2 ( ⎜ RR 〉 - ⎜ BB 〉 ) will yield the necessary red-green and blue-green couplings 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 ! Y Another way: C YC G R B I IC C B R G RG BG ...starting to look familiar ?! Central RB RR BB BR GG SU(3) states: 1/√2 ( ⎜RR 〉 - ⎜ BB 〉 ) 1/√6( ⎜RR 〉 + ⎜BB 〉 - 2⎜GG 〉 ) 1/√3( ⎜RR 〉 + ⎜BB 〉 + ⎜GG 〉 ) GB GR 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. 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. (hence the analogy with ''colour", since white light can be decomposed into either red, green & blue or their opposites - cyan, magenta & yellow) If the carriers of the force (the gluons) actually carry colour themselves, the field lines emanating from a single quark will interact: ''flux tube" q q q * formally still just a hypothesis (calculation is highly non-perturbative) For this configuration, the field strength (flux of lines passing through a surface) does not fall off as 1/r2 any more ⇒ it will remain constant. The field energy will thus scale with the length of the string and so as L → ∞ then E → ∞ Clearly we can’t allow this!! Can be stopped by terminating field line on another colour charge ⇒ Ah! So only colourless states have finite energy ! ''Confinement" PoP ! q q q q q q q q q q ''fragmentation" Need Colourless States... So what about qqqqqq states ? Sure ⇒ that’s basically the deuteron (np = uuuddd) How about qqqqq ? Pentaquarks Search for Free Fractional Charges (M. Perl et al.) 2 vz = 2r ρg/9η vx = qE/6πηr for q > 0.16e, the number of fractionally charged particles is less than 4x10-22 per nucleon (Halyo et al., PRL 2000) Getting very close to a quark: RB q q q q q RG So, on average, the colour is ''smeared" out q 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" Note that asymptotic freedom means that the running coupling will decrease with higher momentum transfer (the opposite of what happens with vacuum polarization in QED!) This also means that perturbative QCD calculations will work at high energies! Where Are The Coupling Constants Running ??? So how do we now interpret pion exchange?? d d p u u n u R RB R d B B u d = qq creation - or + = qq annihilation π π (ud) (ud) G GB d G u n d d p u u Quark Flow Diagrams: - + φ → K + K p + p → p + n + π+ s K- u u u p d d n s u d φ s d u K+ + s u π u p u p p + π+ → Δ++ u u d d → p + π+ u u + π+ d π d Δ++ d d p u u p u u .
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