Theory For particle physicists Lecture 5 Rachel Dowdall University of Glasgow, UK University of Glasgow – p. 1/2 Last Lecture SU (3) Flavour symmetry Classi? cation of mesons and baryons This time: Finish the classi? cation of baryons Colour SU (3) symmetry Young’s tableaux University of Glasgow – p. 2/2 Wave functions The wave function for a particle is the product ? = ? Flavour ? spin ? colour ? space Overall, this should be symmetric under exchange of quarks for mesons and antisymmetric for baryons We considered ? Flavour last lecture If we consider ground states (zero ang.

momentum L), then ? pace is symmetric. Let’s now consider ? colour , this will turn out to be antisymmetric University of Glasgow – p. 3/2 SU (3) Colour symmetry Colour was proposed as an additional quantum number to explain how states that looked the same could co-exist Quarks can have three “colours” r, g, b, antiquarks have r, g, b Processes do not depend on the colour, i. e. invariant under SU (3)C We don’t observe colour in nature (con? nement) so we think that bound states of quarks must be in a colour singlet, or colourless state – Not proven theoretically why this is true e.

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g. mesons qq have 3 ? 3 = 8 ? , so must be in the 1 rep 1 ? meson = v (rr + gg + bb) 3 Just like for ? avour SU (3) except that SU (3)C is an exact symmetry Baryons qqq have 3 ? 3 ? 3 = 10 ? 8 ? 8 ? 1 and the colourless state is 1 ? baryon = v (rgb ? rbg + gbr ? grb + brg ? bgr) 6 Exercise:

Check this is antisymmetric University of Glasgow – p. 4/2 Colour We are not allowed to have qq or qqq as these do not contain a singlet Can have exotics qqqq, qqqqq but these have not been found Since particles are in the singlet state the wave function ? colour is antisymmetric So far, we have ? colour ? space is antisymmetric So ? spin ?? vour must be symmetric for baryons Aside: We say the standard model has SU (3) ? SU (2) ? U (1) symmetry – Colour is the SU (3) part The Lagrangian for QCD is an SU (3) gauge theory, see QFT and standard model courses University of Glasgow – p. 5/2 Spin wave function For ? spin , we just apply the Clebsch-Gordan decomposition for SU (2) Quarks q can have spin up ^ or down v For mesons we want the decomposition 2 ? 2 = 3 ? 1 Symmetric combinations, S, spin-3/2 For baryons 2 ? 2 ? 2 = 2 ? (3 ? 1) = 4 ? 2 ? 2 and we have: 1 1 ^^^, v (^^v + ^v^ + v^^), v (vv^ + v^v + ^vv), vvv 3 3 Mixed symmetry Ms , spin-1/2 ? 1 v (2 vv^ ? ^vv ? v^v), v (2 ^^v ? ^v^ ? v^^) 6 6 Mixed symmetry MA , spin-1/2 1 1 v (^vv ? v^v), v (^v^ ? v^^) 2 2 (1) (2) (3) University of Glasgow – p. 6/2 Symmetric combinations Recall decomposition of u ? d ? s gave 3 ? 3 ? 3 = 10 ? 8 ? 8 ? 1 10 is the symmetric decuplet, 8 are mixed symmetry, 1 is completely antisymmetric Combine the spin and ? avour wave functions ? Flavour ? spin The ? avour singlet state is not allowed because that would make ? symmetric University of Glasgow – p. 7/2 Baryon decuplet We can only combine 10 with the symmetric spin-3/2 states, then ? is antisymmetric and we have

University of Glasgow – p. 8/2 Baryon octet The 8 states have mixed symmetry and when they are combined with the mixed symmetry MS , MA spin states, the total wavefunction is antisymmetric University of Glasgow – p. 9/2 Tensors We can also write the decomposition into irreps as SU (3) tensors Start with a basis |i of the fundamental, 3 rep and |j of the conjugate 3 A general vector |v in 3 is given by |v = v i |i Make a tensor product of n 3 states and m 3 states 1 … jm |j1 … in = |i1 … |in |j1 … |jm i Since this is a basis, an arbitrary state can be written 1 … in 1 … jm |T = Tji1 … jm |j1 … in i The highest weight tensor is vj1 … in = N ? i1 1 …? in 1 ? j1 1 ? jm 1 where 1 … jm Note v is symmetric in both the upper and lower indices and j1 i ? i1 vj1 … in = 0 1 … jm This gives a reducible representation i, j = 1, 2, 3 and the 1 component is the highest weight of |i or |j University of Glasgow – p. 10/2 Tensors We can obtain other states by applying the lowering operators to v j1 i This preserves the symmetry and ? i1 vj1 … in = 0 1 … jm 3 and 3 are not equivalent irreps so cannot map directly between them But, we can use o? = Ui l Uj m Ukn olmn = det(U )oijk = oijk ijk to write v a = oabc u[bc] i. e. ap 3 to an antisymmetric tensor in 3 ? 3 Note oijk and oijk are the same since the eqn is invariant under conjugation – we can map the other way as well So to do the tensor product decomposition, we want to split into symmetric, traceless pieces and antisymmetric pieces University of Glasgow – p. 11/2 Example Consider 3 ? 3, one up one down index ui vj = (ui vj ? 1 i 1 i k ? j u vk ) + ? j uk vk 3 3 the ? rst piece is symmetric and traceless and 8-dimensional, the second is invariant, so 3? 3=8? 1 Consider 3 ? 3, this only has upper indices ui v j = (ui v j + uj v i ) + 1 ijk o oklm ul v m 2 oklm ul v m is a 3 tensor, and the ? st part is symmetric dimension six 3? 3=6? 3 This is still rather a lot of work, so we use Young’s tableaux to represent these decompositions University of Glasgow – p. 12/2 Dimensions of tensor products We have a new way of labelling an SU (3) representation: (n, m) Because of the conditions on the tensors the dimension of that representation is dim(m, n) = 1 (n + 1)(m + 1)(n + m + 2) 2 So the fundamental rep. can be called (1, 0) or 3 The conjugate rep. is (0, 1) or 3 The irrep (2, 0) is dimension 6 but its conjugate (0, 2) is denoted 6 Similarly (2, 1) is dimension 15 but its conjugate (1, 2) is denoted 15

University of Glasgow – p. 13/2 Young’s tableaux/diagrams Graphical method for decomposing SU (N ) tensors into irreducible pieces A tensor in the fundamental representation 3 is denoted vi = For a 3 tensor uj , we raise the indices with oklj uj and represent it as oklj uj = for SU (N ) this would have N ? 1 boxes University of Glasgow – p. 14/2 Young’s tableaux A symmetric tensor w(abc) is denoted w(abc) = and for n upper indices w(i1 … in ) = i1 … in and for antisymmetric lower indices, we get ok1 l1 j1 … okm lm jm uj1 … ujm = So a general tensor is 1 … in Tji1 … jm = k1 … km l1 … lm k1 … m i1 … in l1 … lm University of Glasgow – p. 15/2 Young’s Tableaux Some rules Each row must be shorter than the row above, i. e. not: A diagram cannot have more than N rows since the maximally antisymmetric tensor is, e. g. for SU (3) oijk = University of Glasgow – p. 16/2 Dimension of a diagram Since each diagram is a representation, it has a dimension. To ? nd it: 1. Write the diagram as a quotient dim = / 2. For the numerator, write N in the ? rst box, then count up along the top row and down each column e. g. , N = 3 3 4 5 2 3 1 =1? 2? 3? 3? 4? 5 University of Glasgow – p. 17/2 Dimension of a diagram

For the denominator, count number of boxes to the left in the same row, and below in the same column, then add one 5 3 1 3 1 1 =1? 1? 1? 3? 3? 5 To evaluate the quotient, multiply all the numbers in the numerator together then divide by all the numbers in the denominator The denominator only depends on the shape of the diagram, the numerator depends on the group as well University of Glasgow – p. 18/2 Example In SU (3) the dimension of is given by 3 4 / 2 The dimension of 3 4 5 3 1 1 = 2? 3? 4 =8 3? 1? 1 for SU (3) is given by / 3 2 1 = 3? 4? 5 = 10 3? 2? 1 but in SU (6), the same diagram has dimension 6 7 8 / 3 2 1 = 56

University of Glasgow – p. 19/2 Tensor products Draw the two diagrams you wish to decompose, In the 2nd diagram, write as in the ? rst row, bs in the 2nd, etc. e. g. a b ? Add the boxes with as to the left/below the other diagram making all possible allowed new diagrams with no two as in the same column. Repeat for bs, cs etc a b ? > a ? a a b b a > a b ? ? ? a b (4) University of Glasgow – p. 20/2 Tensor products Cancel additional columns of N boxes For each box, de? ne number of as above and to the right as na , then only keep diagrams with na ? nb ? nc … Using the last rule, the ? rst and third diagrams are excluded ? a b ? a b (5) Then compute the dimensions, we obtain 3? 3=8? 1 University of Glasgow – p. 21/2 SU (2) Young’s diagrams Young’s diagrams are particularly simple for N = 2 Both 2 and 2 are represented by 2 ? 2 = 1 ? 3 is given by ? and 3 ? 2 = 4 ? 2 is given by = ? which re? ects the fact that 2 is equivalent to 2 and the antisymmetric tensor only has two indices ? = ? University of Glasgow – p. 22/2 Homework Using Young’s tableaux show that for SU (3) 6 ? 8 = (0, 2) ? (1, 1) = 24 ? 15 ? 6 ? 3 To get marks you will need to: – Express 6, 8 as diagrams – Draw all of the diagrams obtained by moving the a, b labels – Say (brie? ) why certain diagrams are disallowed – Compute the dimension of each remaining diagram and correctly identify its irrep. If you wish to draw them in LaTeX, then include the line \usepackage{young} Use & for each box on a line and \cr to start a new line, e. g. \begin{Young} 3 & 4 \cr 2 \cr \end{Young} will draw 3 4 2 University of Glasgow – p. 23/2 Next time The Lorentz group The Poincare group Their representations Useful references: 1. H. Georgi, Lie Algebras in Particle Physics 2. Particle data group reivews 3. Greiner & Muller, Quantum mechanics symmetries University of Glasgow – p. 24/2