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

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Originally form Brian Meadows, U. Cincinnati Bound States

originally from B. 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, …

Brian Meadows, U. Cincinnati Gell-Mann-Nishijima Relationship Applies to all hadrons  Define hypercharge Y = B + S + C + B’ + T  Then electric charge is Q = I 3 + Y / 2 Relatively recently added Third component of I-spin Bayon #

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. I 3 Use your book to find the masses of the  ’s and the K’s K-K- K 0 (497) --  0 (135) ++  ’ (548/960) K   KK I3I3       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 (J P = 1 - ) K*-K*- K * 0 (890) --  0 (775) ++  0 /  (783)/(1020) K   K  I3I3       Y Vector mesons: All mesons here have Spin J = 1 and Parity P = -1

Brian Meadows, U. Cincinnati  Also works for baryons with same J P “Eight-Fold Way” (Baryons) n (935) p --  0 (1197) ++  0 (1115) --  0 (1323) I3I3       Y {8} J P = 1/2 + Centroid is at origin Elect. charge Q = Y + I 3 /2

Brian Meadows, U. Cincinnati  Also find {10} for baryons with same J P “Eight-Fold Way” (Baryons) --  0 (1385) ++ --  0 (1532)  - (1679) ???  ++ ++  0 (1238) -- Y I3I3       {10} J P = 3/2 + G-M predicted This to exist Centroid is at origin

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

 3 quark flavors [uds] calls for a group of type SU(3)  SU(2): N=2 eigenvalues(J2,Jz), N 2 -1=4-1=3 generators (J x,J y,J z )  SU(3): N=3 eigen values (uds), N 2 -1=9-1=8 generators (8 Gell-Man mat.) or smarter: SU(3) Flavor I3I3       Y d u s V +/- T +/- U +/- Y I 3 T +, T - U +, U - V +, V -

 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 Physics 841, U. Cincinnati, Fall, 2009Brian Meadows, U. Cincinnati SU(3) Flavor I3I3      Y d u s {3}

Physics 841, U. Cincinnati, Fall, 2009Brian 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

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:  + = |1,+1>= -ud  0 = |1, 0>= (uu-dd)/sqrt(2)  - = |1,-1>= +ud  An iso-singlet (e.g.  or  ’) would be  = |0,0>= (uu+dd)/sqrt(2)

 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  1 and octet  8. Brian Meadows, U. Cincinnati Mesons – Flavour Wave-functions

Brian Meadows, U. Cincinnati Mesons – Mixing (of I =Y=0 Members)  In practice, neither  1 nor  8 corresponds to a physical particle. We observe ortho-linear combinations in the J P =0 - (pseudo-scalar) mesons:  =  8 cos  +  1 sin  ¼ ss  ’ = -  8 sin  +  1 cos  ¼ (uu+dd)/sqrt 2  Similarly, for the vector mesons:  = (uu+dd)/sqrt 2  = ss  What is the difference between  and  ’ (or  and , or K 0 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 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” (m 1 and m 2 ) for the quarks are: m u =m d =310 MeV/c 2 and m s =483 MeV/c 2.  The operator produces (S=1) or for (S=0) Determine empirically

Brian Meadows, U. Cincinnati Mesons – Masses in MeV/c 2 L=0 q q q q J P = 0 - S 1. S 2 = -3/4 h 2 J P = 1 - S 1. S 2 = + 1/4 h 2 What is our best guess for the value of A?See page 180

 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) Baryons L l x x S = ½ or 3/2

Brian Meadows, U. Cincinnati  We find {8} and {10} for baryons Ground State Baryons --  0 (1385) ++ --  0 (1532)  - (1679) ???  ++ ++  0 (1238) -- Y I3I3       {10} J P = 3/2 + n (935) p --  0 (1197) ++  0 (1115) --  0 (1323) I3I3       Y {8} J P = 1/2 + 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 {8 12 } and {8 23 }  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: Color Wave-functions {1} = [(u(ds-ds) + d(su-us) + s(ud-du)] / sqrt(6)

Brian Meadows, U. Cincinnati 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 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