Wave functions of Baryons Baryon Magnetic Moments Baryon

• • 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 • and therefore for each spin-up quark: Spin down just changes the sign

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.

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

Hadron Masses • Seems Simple enough – Just add up the masses of the quarks • Mp = Mu + Md = 2*Mu = 620 Me. V/c 2 • Experimentally Mp = 139 Me. V/c 2 – 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

Hadron Masses are more equal, Force is much more powerful. Fit to some meson masses and find As = 160*(4 pmu/h)2 Me. V/c 2 S 1 • S 2 Meson p r K K* 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: Fit to some baryon masses and find As’ = 50*(4 pmu/h)2 Me. V/c 2 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.

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| > • P 2| > = p 2| > = | > 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 = y(r)yqyqbar • P | yq> = -P| yqbar> • Parity is a Multiplicative quantum number, not additive. – Given q 1 and q 2 – J = S 1 + S 2 – P = P 1*P 2

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 • 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)

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)

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)>

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

Conserved by Strong force: • Isospin, Quark Flavor – (I, I 3, U, D, S, C, B, and T) • Parity • Charge Conjugation • Electric Charge • Energy/momentum • Angular Momentum / Spin