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3.67. 3.33. 1.22. 1.11. 0.66. Off by a factor of 3 !!!!!. Could every known quark have a previously un-noticed 3 -fold degeneracy????. Baryon States. State Quark content Mass Spin p uud 938.272 MeV 1/2

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0.66

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  1. 3.67 3.33 1.22 1.11 0.66 Off by a factor of3!!!!! Could every known quark have a previously un-noticed 3-fold degeneracy????

  2. Baryon States StateQuark contentMassSpin p uud 938.272MeV 1/2 n udd 939.565MeV 1/2 uds 1115.683MeV 1/2 +uus 1189.37 MeV 1/2  0uds 1192.632 MeV 1/2  -dds 1197.449 MeV 1/2  0uss 1314.9 MeV 1/2  -dss 1321.32 MeV 1/2 ++uuu1230. MeV 3/2 +uud 1231 MeV 3/2  0 udd 1233 MeV 3/2  -ddd1234 MeV 3/2 *+ uus 1382.8 MeV 3/2 *0 uds 1383.7 MeV 3/2 *- dds 1387.2 MeV 3/2 *0 uss 1531.80 MeV 3/2 *- dss 1535.0 MeV 3/2  -sss1672.45 MeV 3/2

  3. U(1) introduced an interaction term:(qY g m Y )Am a charged current coupled to a vector field Yang-Mills extended this to SU(2)with an interaction term of: (qY gm tY ).Gm 2    · G each is a “vector” on “iso-space” Remember: SU(2) includes ALL POSSIBLE traceless HERMITIAN2x2 matrices the most GENERAL such matrix is a1 + a2 + a3 = a .t 1 0 0 -1 0 1 1 0 0 -i i0 · G • the G = (G1, G2, G3) are three independent • 4-vector fields, but together form an ISO-vector • transform under SU(2) the same way t does: U(· G)U† = U U†· UGU†

  4. 2 Under the SU(2) Iso-spin transformations: 1 2 recall it was from that we determined the correct transformation property of G U(· G)U† G / =RTG-   Not a column vector in iso-space! Not a matrix! the 3 independent fields are isospin components which transform like under a basis transformation G must obey the same COMMUTATOR relations that  does, i.e., they form an adjoint representation of  (a set of functions isomorphic – mathematically equivalent to the  matrices).

  5. Generalizing YANG-MILLS L ~SFi mnFimn= ( ) . ( ) GAUGE FIELDS Fimn = mGin - n Gim - 2ge ijk GjmGkn c The exact form depends on the non-abelian nature of the generators (their commutator rules) because[ti,tj ] = ieijktk More generally Fimn = mGin - n Gim - 2gCijk GjmGkn c “structure constant” of SU(n)

  6. So now, in the face of evidence that quarks carry some 3-fold degeneracy that needs to be explained LET’S TRY: SU(3) This time  must be not only a Dirac spinoru(s)(p) but a 3-column vector in some NEW SPACE as well:

  7. SU(3) will require introducing an interaction term: (g Y g m Y ) .Gm l1, l2, … l8 span the 8-dim space (the 8-independent parameters) of the 3x3 traceless, Hermitian generators of SU(3) Gmthe 8-independent VECTOR fields behave like the lioperators: 0 1 0 0 -i 0 1 0 0 l1= 1 0 0 l2=i0 0 l3= 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 -i 0 0 0 l4= 0 0 0 l5=0 0 0 l6=0 0 1 -1 0 0 i 0 0 0 1 0 0 0 0 1 0 0 l7=0 0 -il8=1/ /30 1 0 0 i0 0 0 -2

  8. Since we have the explicit matrix form of the i, its easy to check any of these! [i, j] = 2iC ijkk 8×8×8 different C ijk but only = 56 distinct combinations 8·7·6 3·2·1 (since exchanges/rearrangements are already defined) C jik = C ikj=-C ijk only 9 of these are non-zero (with only 3 different values!) C 123 = 1 C 147=C 246=C 257 =C 345= C 516= C 637=1/2 C 458 = C 678 = 3/2

  9. SU(3) States If we write the 3COLOR states as 0 b = 0 1 1 g = 0 0 0 r = 1 0 Look at 1r= g changes r g charge 1g= r HEY!!!!! Electrons don’t change their charge when emitting/absorbing photons! but remember our “gluons” carry “charge” 0 1 0 0 -i 0 1 0 0 l1= 1 0 0 l2=i0 0 l3= 0-1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 -i 0 0 0 l4= 0 0 0 l5=0 0 0 l6=0 0 1 -1 0 0 i 0 0 0 1 0 0 0 0 1 0 0 l7=0 0 -il8=1/ /30 1 0 0 i0 0 0-2

  10. SU(3) States If we write the 3COLOR states as 0 b = 0 1 1 g = 0 0 0 r = 1 0 1r= g 2r creates the same color exchange but with different coefficients: a change of phase 1g= r 3, 8: preserve/retain color but may change phase (still exchange/transfer energy & momentum - like s) 4b= 6b= g r 4, 5: 6, 7: 4g= 6r= b b 0 1 0 0 -i 0 1 0 0 l1= 1 0 0 l2=i0 0 l3= 0-1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 -i 0 0 0 l4= 0 0 0 l5=0 0 0 l6=0 0 1 -1 0 0 i 0 0 0 1 0 0 0 0 1 0 0 l7=0 0 -il8=1/ /30 1 0 0 i0 0 0-2

  11. or 6b= r 6, 7: 6r= b How does this work? The gluons form a color OCTET (a multiplet of 8). How can you get an OCTET from a basis of 3?

  12. In SU(3) the 3 fundamental COLOR “charges” (the basis set in the lowest order representation) can be combined in pairs of COLOR/antiCOLOR to give either a SINGLET or an OCTET x x = x xx x x 3 3 = 8 1 x x Must be color/anticolor states like rbrgbrbggrgb While the fundamental process of the “strong” force is q  gq with a primitive vertex: q b r q Gm6or Gm7 Gm1 = (1/2 ) (rg + gr ) Gm2 = (-i/2 )(rg-gr ) rg or gr or inverting

  13. 6of the gluon fields areindependent linear combinationsof the simple gluon fields we enumerated Gm1 = (rg + gr)/2 Gm4 = (bg + gb)/2Gm6 = (rb + br)/2 Gm2=-i(rg - gr)/2 Gm5=-i(bg - gb)/2Gm7=-i(rb + br)/2 TheCOLOR SINGLETwould be1/3 ( rr + gg + bb ) However, just as we argued that (isospin) part of the deuteron’s wavefunction had to be anti-symmetric If the color singlet gluon existed it would be exchanged between color singlet (color-less?) states which as you’ll see momentarily would suggest a long range strong force between neutron and proton the (color) part of any baryon’s wavefunction has to be antisymmetric and that characteristic must not change when color is exchanged! …which simply does not exist …and as a color singlet it would be directly observable as a free particle.

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