• • • • Wave functions of Baryons. Baryon Magnetic Moments Baryon masses. Need to explain Parity and Charge Conjugation Hadrons Magnetic moments mq related to the intrinsic spin S of the quark. • m = (q/mc)S q q h m iz m i S z Sz mc mc 2 • and therefore for each spin-up quark: 2 eh mu 3 2mu c 1 eh md 3 2md c 1 eh ms 3 2 ms c Spin down just changes the sign Hadron Magnetic moments • Need a particles which are long-lived and have some intrinsic spin. Proton! pspin up 1 2u u d u u d u u d 3 2 permutations Total Magnetic Moment should equal the vector sum of the magnetic moments of the constituent quarks. pspin up m1 m 2 m 3 z pspin up pspin up m u Suz m u Suz m d S dz pspin up Reminder: The order of the spin arrows designates which quark has that spin. Hadron Magnetic moments pspin up 1 2u u d u u d u u d 3 2 permutations Doing the calculation for the first term: 2 2 u u d m1 m 2 m 3 z u u d 3 2 2 h 2 m u m d 9 2 So we expect mproton to be: 1 h m proton (4m u m d ) 3 2 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? Hadron Masses Hyperfine splitting in hydrogen atom: Caused by the spin of the electron interacting with the spin of the proton Ehf 8p p e 2 3c 2 Aem 0 Se S p me m p 2 Se S p me m p ~ 6.0 10 6 eV S1 S 2 M (meson) m1 m2 As m1m2 Hadron Masses 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 3 h p 4 1 h r 4 3 h K 4 1 h K* 4 2 2 2 2 Calculated 140 780 484 896 Observed 138 776 496 892 Hadron Masses Amazingly we can take the meson mass formula as the lead for estimating baryon masses: S1 S2 S3 S2 S1 S3 M (baryon) m1 m2 m3 As m3m2 m1m3 m1m2 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): J 2 S1 S 2 S3 S1 S 2 S3 2S1 S 2 S3 S 2 S1 S3 2 h2 2 9 j j 1 4 2 2 2 Again see Griffiths, page 182. 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) not reflection 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. Eigenstates of Parity • Suppose we have a force that only acts radially between two particles. – Then the wave function Y = (r)qqbar • P | q> | q> = -P| qbar> • Parity is a Multiplicative quantum number, not additive. – Given q1 and q2 – J = S 1 + S2 – P = P1*P2 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 Eigenstates of Parity (r) can be separated into the angular part Ylm(,) and a purely radial part so: (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) Charge Conjugation • C is 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> – eg. (where a = number) C ud u d ud Charge Conjugation • Neutral Mesons are eigenstates of C qr1 q r2 |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)> Charge Conjugation • Neutral Mesons are eigenstates of C qr1 |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> q r2 C 0 C q q q q q q q q 2 2 1 C |Y(spin)> = (-1)s+1 |Y(spin)> so C |Y> = (-1)l+s |Y> neutral mesons only s 1 0 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