Porters 5 Forces of Sainsburys

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Group Theory
For particle physicists Lecture 5
Rachel Dowdall

University of Glasgow, UK

University of Glasgow – p. 1/2

Last Lecture
SU (3) Flavour symmetry Classiﬁcation of mesons and baryons This time: Finish the classiﬁcation 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 Ψ = φFlavour χspin ξcolour ηspace Overall, this should be symmetric under exchange of quarks for mesons and antisymmetric for baryons We considered φFlavour last lecture If we consider ground states (zero ang. momentum L), then ηspace is symmetric. Let’s now consider ξcolour , 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 (conﬁnement) 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.g. mesons qq have 3 ⊗ 3 = 8 ⊕ 1, so must be in the 1 rep 1 ξmeson = √ (rr + gg + bb) 3 Just like for ﬂavour 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 ξbaryon = √ (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 ξcolour is antisymmetric So far, we have ξcolour ηspace is antisymmetric So χspin φﬂavour 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 χspin , we just apply the Clebsch-Gordan decomposition for SU (2) Quarks q can have spin up ↑ or down ↓ 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 ↑↑↑, √ (↑↑↓ + ↑↓↑ + ↓↑↑), √ (↓↓↑ + ↓↑↓ + ↑↓↓), ↓↓↓ 3 3 Mixed symmetry Ms , spin-1/2 −1 1 √ (2 ↓↓↑ − ↑↓↓ − ↓↑↓), √ (2 ↑↑↓ − ↑↓↑ − ↓↑↑) 6 6 Mixed symmetry MA , spin-1/2 1 1 √ (↑↓↓ − ↓↑↓), √ (↑↓↑ − ↓↑↑) 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 ﬂavour wave functions φFlavour χspin The ﬂavour singlet state is not allowed because that would make Ψ symmetric

University of Glasgow – p. 7/2

Baryon decuplet
We can only combine 10 with the symmetric spin-3/2 states, then Ψ 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

i The highest weight tensor is vj1 ...in = N δi1 1 ...δin 1 δj1 1 δjm 1 where 1 ...jm

Note v is symmetric in both the upper and lower indices and j1 i δi1 vj1 ...in = 0 1 ...jm This gives a reducible representation

i, j = 1, 2, 3 and the 1 component is the...