Origin of SU(3) –Why a simple extension of SU(2) is not enough Extending the Graphical method of finding states Application to Baryon and Meson spectrum.

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Presentation transcript:

Origin of SU(3) –Why a simple extension of SU(2) is not enough Extending the Graphical method of finding states Application to Baryon and Meson spectrum

I3I3 0 1/2 -1/2 1 Fundamental SU(2) Representation 0 1/2 -1/2 -2/3 1 I3I3 -1/3 +2/3 Y Fundamental SU(3) Representation Fundamentally Different….see?

Unlike SU(2), antiquarks can’t fit on the graph the same way as quarks 0 1/2 -1/2 -2/3 1 I3I3 -1/3 +2/3 Y 0 1/2 -1/2 -2/3 1 I3I3 -1/3 +2/3 Y Quarks Antiquarks

Use Same Graphical method to combine quark- antiquark pairs into mesons 0 1/2 -1/2 -2/3 1 I3I3 -1/3 +2/3 Y 0 1/2 -1/2 -2/3 1 I3I3 -1/3 +2/3 Y Quarks Antiquarks

Use Same Graphical method to combine quark- antiquark pairs into mesons 1/2 0 -1/2 1 I3I3 +1 Y Caveat: only one of the states at S=I 3 =0 is fully antisymetric.

Baryon Woes We have one decplet, two Octets, and one singlet 10 s  8 ms  8 ma  1 a. –The decuplet is fully symmetric –The first octet is symmetric in the interchange of the first two quarks. –The second octet is symmetric in the interchange of the second two quarks (therefore antisymmetric in the interchange of the first two quarks). –The singlet is fully antisymmetric in the interchange of any two quarks. BUT: Nature only has 10 s  8 s ! WHY????

Baryon Woes: The Wave Function Not a problem in Mesons….a quark and an antiquark are fundamentally distinguishable particles. We have to pay attention to the form of the baryon wave function (antisymmetric, fermion):  =  (space)  (flavour)  (spin)  (colour) –  (colour) is always antisymmetric This is related to the fact that no hadron is observed to have any net ‘colour’ charge. –So  (space)  (flavour)  (spin)  symmetric –Assume, for the moment, that there is no orbital angular momentum between the quarks in the baryon. (l = 0) Then  (space) will be symmetric.

Baryon Woes: The Wave Function  =  (space)  (flavour)  (spin)  (colour) –  (space) l=0 symmetric. –So we need  (flavour)  (spin) to be symmetric also. Look at  (spin) –Combine 3 spin 1/2 objects using SU(2)…the group that correctly handles spins. 2  2  2 = 4 s  2 ms  2 ma –4 s is symmetric, 2 ms is symmetric in the exchange quarks 1  2, 2 ma is antisymmetric in the exchange of quarks 1  2. –Combine with 10 s  8 ms  8 ma  1 a –(10 s  8 ms  8 ma  1 a )  (4 s  2 ms  2 ma )  Keep symmetric ones (10 s  4 s ) and two (8  2) combinations survive.

Hadrons Are Looking Up! We can understand/predict a whole bunch of stuff! –Why the Baryons cluster like they do. –IF the strong force is a ‘colour SU(3) c ’ group Why the l=0 Decuplet MUST have j=3/2 spin. Why the l=0 Octets MUST have j=1/2 spin. Predicts structure of l  0 states as well. (many not found yet). Can get mass splittings and magnetic moments! –Why the Mesons only form Octets + singlets. The symmetry we must invoke for more quarks: SU(4) for Charm…up to SU(6) to get to top quarks. But this makes NO sense because even SU(4) is broken beyond all recognition.