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Wave functions of Baryons. Baryon Magnetic Moments Baryon masses. Need to explain Parity and Charge Conjugation
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q related to the intrinsic spin S of the quark. = (q/mc)S and therefore for each spin-up quark: Hadrons Magnetic moments Spin down just changes the sign
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Need a particles which are long-lived and have some intrinsic spin. Proton! Hadron Magnetic moments 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.
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Hadron Magnetic moments Doing the calculation for the first term: So we expect proton to be:
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Hadron Masses Seems Simple enough –Just add up the masses of the quarks M = M u + M d = 2*M u = 620 MeV/c 2 Experimentally M = 139 MeV/c 2 –What???? + is |u, d-bar>. This is a particle made up of two like-sign charged quarks. Why doesn’t it fly apart?
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Hadron Masses Hyperfine splitting in hydrogen atom: Caused by the spin of the electron interacting with the spin of the proton
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Hadron Masses Masses are more equal, Force is much more powerful. Fit to some meson masses and find A s = 160*(4 m u /h) 2 MeV/c 2 S 1S 2 MesonCalculatedObserved 140 138 780 776 K 484 496 K* 896 892
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Hadron Masses Fit to some baryon masses and find A s ’ = 50*(4 m u /h) 2 MeV/c 2 Amazingly we can take the meson mass formula as the lead for estimating baryon masses: 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.
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More Conserved Stuff We need to cover some more conserved quantum numbers and explain some notation before moving on. Parity and Charge Conjugation: –Parity (x,y,z) (-x,-y,-z) not reflection in a mirror! –Define the parity operator ‘P’ such that: P | (x,y,z)> = | (-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.
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Eigenstates of Parity Suppose we have a force that only acts radially between two particles. –Then the wave function = (r) q qbar P | q > | q > = -P| qbar > Parity is a Multiplicative quantum number, not additive. –Given q1 and q2 –J = S 1 + S 2 –P = P 1 *P 2
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Eigenstates of Parity For once, Baryons are easy! –For Mesons with no ang. Momtenum P| b >| bbar > = -1 | b >| bbar > –DEFINE: P | b > 1 P | bbar > -1 –So in general, for baryons with orbital angular momentum between the quarks: P | b > = (-1) l | b > Unfortunately, because baryon number is conserved anyway this relation is essentially useless.
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Eigenstates of Parity (r) can be separated into the angular part Y l m ( , ) and a purely radial part so: – (r) = (r) Y l m ( , ) space-part of wave function –P Y l m ( , ) = (-1) l Y l m ( , ) –And P| > = (-1) l p q p qbar | > = (-1) l (1)(-1)| > – P| > = (-1) l+1 | > For MESONS only (since p q =1, p qbar =-1)
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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 | > | > (where = number) –eg.
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Charge Conjugation Neutral Mesons are eigenstates of C | >=| (space)>| (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 | (space)> as if we’d used the parity operator. C | (space)> = (-1) l+1 | (space)>
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If we apply C| (spin)> what do we get? Lets try this on a S=1 or 0 meson m s = |0> C | (spin)> = (-1) s+1 | (spin)> so C | > = (-1) l+s | > neutral mesons only Charge Conjugation Neutral Mesons are eigenstates of C | >=| (space)>| (spin)>|q,qbar>
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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
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