Brian Meadows, U. Cincinnati

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Brian Meadows, U. Cincinnati Bound States Originally form Brian Meadows, U. Cincinnati

What is a Bound State? Imagine a system of two bodies that interact. They can have relative movement. If this movement has sufficient energy, they will scatter and will eventually move far apart where their interaction will be negligible. If their interaction is repulsive, they will also scatter and move far apart to where their interaction is negligible. If the energy is small enough, and their interaction is attractive, they can become bound together in a “bound state”. In a bound state, the constituents still have relative movement, in general. If the interaction between constituents is repulsive, then they cannot form a bound state. Examples of bound states include: Atoms, molecules, positr-onium, prot-onium, quark-onium, mesons, baryons, … originally from B. Meadows, U. Cincinnati

Gell-Mann-Nishijima Relationship Applies to all hadrons Define hypercharge Y = B + S + C + B’ + T Then electric charge is Q = I3 + Y / 2 Relatively recently added Bayon # Third component of I-spin Brian Meadows, U. Cincinnati

“Eight-Fold Way” (Mesons) M. Gell-Mann noticed in 1961 that known particles can be arranged in plots of Y vs. I3 Use your book to find the masses of the p’s and the K’s K- K0(497) p- p0(135) p+ /’ (548/960) K+ I3 +1 -1 Y Pseudo-scalar mesons: All mesons here have Spin J = 0 and Parity P = -1 Centroid is at origin Brian Meadows, U. Cincinnati

“Eight-Fold Way” (Meson Resonances) Also works for all the vector mesons (JP = 1-) K*- K*0(890) - 0(775) + 0/ (783)/(1020) K*+ I3 +1 -1 Y Vector mesons: All mesons here have Spin J = 1 and Parity P = -1 Brian Meadows, U. Cincinnati

“Eight-Fold Way” (Baryons) Also works for baryons with same JP JP = 1/2+ n (935) p S- S0(1197) S+ L0(1115) X- X0(1323) I3 +1 -1 Y {8} Elect. charge Q = Y + I3/2 Centroid is at origin Brian Meadows, U. Cincinnati

“Eight-Fold Way” (Baryons) Also find {10} for baryons with same JP JP = 3/2+ Y D++ D+ D0(1238) D- +1 {10} S- S0(1385) S+ Centroid is at origin -1 X- X0(1532) W-(1679) ??? I3 -1 +1 G-M predicted This to exist Brian Meadows, U. Cincinnati

Discovery of the W- Hyperon Brian Meadows, U. Cincinnati

SU(3) Flavor 3 quark flavors [uds] calls for a group of type SU(3) Y SU(2): N=2 eigenvalues(J2,Jz) , N2-1=4-1=3 generators (Jx,Jy,Jz) SU(3): N=3 eigen values (uds), N2-1=9-1=8 generators (8 Gell-Man mat.) or smarter: Y I3 T+, T- U+, U- V+, V- I3 +1 -1 +1/2 -1/2 Y d u s V+/- T+/- U+/-

Physics 841, U. Cincinnati, Fall, 2009 Brian Meadows, U. Cincinnati SU(3) Flavor At first, all we needed were three quarks in an SU(3) {3}: SU(3) multiplets expected from quarks: Mesons {3} x {3} = {1} + {8} Baryons {3} x {3} x {3} = {1} + {8} + {8} + {10} Later, new flavors were needed (C, B, T ) so more quarks needed too I3 +1 -1 +1/2 -1/2 Y d u s {3} Physics 841, U. Cincinnati, Fall, 2009 Brian Meadows, U. Cincinnati

Physics 841, U. Cincinnati, Fall, 2009 Brian Meadows, U. Cincinnati Add Charm (C) SU(3)  SU(4) Need to add b and t too !  Many more states to find ! Some surprises to come Physics 841, U. Cincinnati, Fall, 2009 Brian Meadows, U. Cincinnati

Brian Meadows, U. Cincinnati Hadron and Meson Wavefunctions Brian Meadows, U. Cincinnati

Mesons – Isospin Wave-functions Iso-spin wave-functions for the quarks: u = | ½, ½ > d = | ½, -½ > u = | ½, -½ > d = - | ½, +½ > (NOTE the “-” convention ONLY for anti-”d”) So, for I=1 particles, (e.g. pions) we have: p+ = |1,+1>= -ud p0 = |1, 0> = (uu-dd)/sqrt(2) p- = |1,-1> = +ud An iso-singlet (e.g. h or h’) would be h = |0,0> = (uu+dd)/sqrt(2) Brian Meadows, U. Cincinnati

Mesons – Flavour Wave-functions They form SU(3) flavor multiplets. In group theory: {3} + {3}bar = {8} X{1} Flavor wave-functions are (without proof!): NOTE the form for singlet h1 and octet h8. Brian Meadows, U. Cincinnati

Mesons – Mixing (of I=Y=0 Members) In practice, neither h1 nor h8 corresponds to a physical particle. We observe ortho-linear combinations in the JP=0- (pseudo-scalar) mesons: h = h8 cosq + h1 sinq ¼ ss h’ = - h8 sinq + h1 cosq ¼ (uu+dd)/sqrt 2 Similarly, for the vector mesons: w = (uu+dd)/sqrt 2 f = ss What is the difference between w and h’ (or f and h, or K0 and K*0(890), etc.)? The 0- mesons are made from qq with L=0 and spins opposite  J=0 The 1- mesons are made from qq with L=0 and spins parallel  J=1 Brian Meadows, U. Cincinnati

Determine empirically Mesons – Masses In the hydrogen atom, the hyperfine splitting is: For the mesons we expect a similar behavior so the masses should be given by: “Constituent masses” (m1 and m2) for the quarks are: mu=md=310 MeV/c2 and ms=483 MeV/c2. The operator produces (S=1) or for (S=0) Determine empirically Brian Meadows, U. Cincinnati

Mesons – Masses in MeV/c2 JP = 1- S1.S2 = +1/4 h2 JP = 0 - S1.S2 = -3/4 h2 q q L=0 L=0 q q What is our best guess for the value of A? See page 180 Brian Meadows, U. Cincinnati

Brian Meadows, U. Cincinnati

Brian Meadows, U. Cincinnati

Brian Meadows, U. Cincinnati Baryons L l x Baryons are more complicated Two angular momenta (L,l) Three spins Wave-functions must be anti-symmetric (baryons are Fermions) Wave-functions are product of spatial(r) x spin x flavor x color For ground state baryons, L = l = 0 so that spatial(r) is symmetric Product spin x flavor x color must therefore be anti-symmetric w.r.t. interchange of any two quarks (also Fermions) Since L = l = 0, then J = S (= ½ or 3/2) S = ½ or 3/2 Brian Meadows, U. Cincinnati

Brian Meadows, U. Cincinnati Ground State Baryons We find {8} and {10} for baryons n (935) p S- S0(1197) S+ L0(1115) X- X0(1323) I3 +1 -1 Y {8} JP = 1/2+ S- S0(1385) S+ X- X0(1532) W-(1679) ??? D++ D+ D0(1238) D- Y I3 +1 -1 {10} JP = 3/2+ L = l = 0, S = ½ L = l = 0, S = 3/2 Brian Meadows, U. Cincinnati

Flavor Wave-functions {10} Completely symmetric wrt interchange of any two quarks Brian Meadows, U. Cincinnati

Flavor Wave-functions {812} and {823} Two possibilities: Anti-Symmetric wrt interchange of 1 and 2: Anti-Symmetric wrt interchange of 2 and 3: Another combination 13 = 12+23 is not independent of these Brian Meadows, U. Cincinnati

Flavor Wave-functions {1} Just ONE possibility: All baryons (mesons too) must be color-less. SU (3)color implies that the color wave-function is, therefore, also a singlet: color is ALWAYS anti-symmetric wrt any pair: color = [R(GB – BG) + G(BR – RB) + B(RG – GR)] / sqrt(6) Anti-symmetric wrt interchange of any pair: {1} = [(u(ds-ds) + d(su-us) + s(ud-du)] / sqrt(6) Color Wave-functions Brian Meadows, U. Cincinnati

Another combination 13 = 12+23 is not independent of these Spin Wave-functions Clearly symmetric wrt interchange of any pair of quarks Clearly anti-symmetric wrt interchange of quarks 1 & 2 Clearly anti-symmetric wrt interchange quarks 2 & 3 Another combination 13 = 12+23 is not independent of these Brian Meadows, U. Cincinnati

Baryons – Need for Color The flavor wave-functions for ++ (uuu), - (ddd) and - (sss) are manifestly symmetric (as are all decuplet flavor wave-functions) Their spatial wave-functions are also symmetric So are their spin wave-functions! Without color, their total wave-functions would be too!! This was the original motivation for introducing color in the first place. Brian Meadows, U. Cincinnati

Brian Meadows, U. Cincinnati Example Write the wave-functions for + in the spin-state |3/2,+1/2> For {8} we need to pair the (12) and (13) parts of the spin and flavor wave-functions: Neutron, spin down: Brian Meadows, U. Cincinnati

Magnetic Moments of Ground State Baryons Brian Meadows, U. Cincinnati

Masses of Ground State Baryons Brian Meadows, U. Cincinnati