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Baryon Predictions

Baryon Predictions. Wave functions of Baryons. Baryon Magnetic Moments Baryon masses. Need to explain Parity and Charge Conjugation. Hadrons Magnetic moments. m q related to the intrinsic spin S of the quark. m = (q/mc) S and therefore for each spin-up quark:.

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Baryon Predictions

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  1. Baryon Predictions • Wave functions of Baryons. • Baryon Magnetic Moments • Baryon masses. • Need to explain Parity and Charge Conjugation

  2. Hadrons Magnetic moments • mq related to the intrinsic spin S of the quark. • m=(q/mc)S • and therefore for each spin-up quark: Spin down just changes the sign

  3. Hadron Magnetic moments • Need a particles which are long-lived and have some intrinsic spin. Proton! Total Magnetic Moment should equal the vector sum of the magnetic moments of the constituent quarks. Reminder: The order of the spin arrows designates which quark has that spin.

  4. Hadron Magnetic moments Doing the calculation for the first term: So we expect mproton to be:

  5. Hadron Masses • Seems Simple enough • Just add up the masses of the quarks • Mp = Mu + Md = 2*Mu = 620 MeV/c2 • Experimentally  Mp = 139 MeV/c2 • What???? p+ is |u, d-bar>. This is a particle made up of two like-sign charged quarks. Why doesn’t it fly apart? Strong Nuclear Force!

  6. Hadron Masses Electromagnetic Force Hyperfine splitting in hydrogen atom: Caused by the spin of the electron interacting with the spin of the proton Strong Nuclear Force!

  7. Hadron Masses Strong Nuclear Force! Masses are more equal, Force is much more powerful. Fit to some meson masses and find As = 160*(4pmu/h)2 MeV/c2 S1•S2 Meson Calculated Observed p 140 138 r 780 776 K 484 496 K* 896 892

  8. Hadron Masses Amazingly we can take the meson mass formula as the lead for estimating baryon masses: Fit to some baryon masses and find As’ = 50*(4pmu/h)2 MeV/c2 Caution: There are tricks you need in order to calculate those spin dot products. Example: if all masses are equal (proton, neutron): Again see Griffiths, page 182.

  9. More Conserved Stuff • We need to cover some more conserved quantum numbers and explain some notation before moving on. • Parity and Charge Conjugation: • Parity Y(x,y,z)Y(-x,-y,-z) notreflection in a mirror! • Define the parity operator ‘P’ such that: • P | Y(x,y,z)> = | Y(-x,-y,-z)> • |> is an eigenstate of P if P|> = p|> • P2|> = p2|> = |> so p = 1 • Parity is a simple group. Two elements only.

  10. Eigenstates of Parity • Suppose we have a force that only acts radially between two particles. • Then the wave function Y = y(r)yqyqbar • P | yq>  | yq> = -P| yqbar> • Parity is a Multiplicative quantum number, not additive. • Given q1 and q2 • J = S1 + S2 • P = P1*P2

  11. Eigenstates of Parity • For once, Baryons are easy! • For Mesons with no ang. Momtenum • P|Yb>|Ybbar> = -1 |Yb>|Ybbar> • DEFINE: P |Yb> 1 P |Ybbar> -1 • So in general, for baryons with orbital angular momentum between the quarks: • P |Yb> = (-1)l |Yb> • Unfortunately, because baryon number is conserved anyway this relation is essentially useless.

  12. Eigenstates of Parity • y(r) can be separated into the angular part Ylm(,) and a purely radial part so: • y(r) = (r) Ylm(,) space-part of wave function • P Ylm(,) = (-1)l Ylm(,) • And P| Y > = (-1)l pq pqbar| Y > = (-1)l(1)(-1)| Y > • P| Y > =(-1)l+1 | Y > • For MESONS only (since pq=1, pqbar=-1)

  13. Charge Conjugation • Cis an operator which turns all particles into antiparticles: • C|q> = |q-bar> • changes sign of charge, baryon #, flavour quan. Num. • Leaves momentum, spin, position, Energy unchanged. • Most particles are NOT eigenstates of C • C|Y>  a|Y> (where a = number) • eg.

  14. Charge Conjugation • Neutral Mesons are eigenstates of C |Y>=|Y(space)>|Y(spin)>|q,qbar> If we apply C to the diagram on the left we change nothing but the ‘particleness’. This doesn’t effect |q,qbar> but has the same effect on |Y(space)> as if we’d used the parity operator. C |Y(space)> = (-1)l+1 |Y(space)>

  15. Charge Conjugation • Neutral Mesons are eigenstates of C |Y>=|Y(space)>|Y(spin)>|q,qbar> If we apply C|Y(spin)> what do we get? Lets try this on a S=1 or 0 meson |ms> = |0> C |Y(spin)> = (-1)s+1 |Y(spin)> so C |Y> = (-1)l+s |Y> neutral mesons only

  16. Conserved by Strong force: • Isospin, Quark Flavor • (I, I3, U, D, S, C, B, and T) • Parity • Charge Conjugation • Electric Charge • Energy/momentum • Angular Momentum / Spin

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