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## PowerPoint Slideshow about 'SU(3) symmetry and Baryon wave functions' - kadeem

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Introduction

- Fundamental symmetries of our universe
- Symmetry to the quark model:
- Hadron wave functions
- Existence (mesons) and qqq (baryons)
- Idea: extend isospin symmetry to three flavors (Gell-Mann, Ne’eman 1961)
- SU(3) flavour and color symmetry groups

Unitary Transformation

- Invariant under the transformation
- Normalization:

U is unitary

- Prediction to be unchanged:

Commutation U & Hamiltonian

- Define infinitesimal transformation

(G is called the generator of the transformation)

Symmetry and conservation

- Because U is unitary

G is Hermitian, corresponds to an observable

- In addition:

G is conserve

Symmetry conservation law

For each symmetry of nature there is an observable conserved quantity

- Infinitesimal spatial translation: ,

Generator px is conserved

- Finite transformation

Isospin

- Heisenberg (1932) proposed : (if “switch off” electric charge of proton )

There would be no way to distinguish between a proton and neutron (symmetry)

- p and n have very similar masses
- The nuclear force is charge-independent
- Proposed n and p should be considered as two states of a single entity (nucleon):

Analogous to the spin-up/down states of a spin-1/2 particle

Isospin: n and p form an isospin doublet (total isospin I=1/2 , 3rd component I3=±1/2)

Flavour symmetry of strong interaction

- Extend this idea to quarks: strong interaction treats all quark flavours equally
- Because mu≈md (approximate flavour symmetry)
- In strong interaction nothing changes if all u quarks are replaced by d quarks and vs.
- Invariance of strong int. under u d in isospin space (isospin in conserved)
- In the language of group theory the four matrices form the U(2) group
- one corresponds to multiplying by a phase factor (no flavour transformation)
- Remaining three form an SU(2) group (special unitary) with det U=1 Tr(G)=0
- A linearly independent choice for G are the Pauli spin matrices

The flavour symmetry of the strong interaction has the same transformation properties as spin.

- Define isospin: ,
- Isospin has the exactly the same properties as spin (same mathematics)
- Three correspond observables can not know them simultaneously
- Label states in terms of total isospin I and the third component of isospin I3

: generally

d u u d

System of two quarks: I3=I3(1)+I3(2) , |I(1)-I(2)| ≤ I ≤ |I(1)+I(2)|

Combining three ud quarks

- First combine two quarks, then combine the third
- Fermion wave functions are anti-symmetric
- Two quarks, we have 4 possible combinations:

(a triplet of isospin 1 states and a singlet isospin 0 state )

- Add an additional u or d quark

Grouped into an isospin quadruplet and two isospin doublets

- Mixed symmetry states have no definite symmetry under interchange of quarks 1 3 or 2 3

Combining three quark spin for baryons

- Same mathematics

SU(3) flavour

- Include the strange quark
- ms>mu/md do not have exact symmetry u d s
- 8 matrices have detU=1 and form an SU(3) group
- The 8 matrices are:
- In SU(3) flavor, 3 quark states are :

SU(3) uds flavour symmetry contain SU(2) ud flavour symmetry

- Isospin
- Ladder operators
- Same matrices for u s and d s
- and 2 other diagonal matrices are not independent, so de fine as the linear combination:

Combining uds quarks for baryons

- First combine two quarks:
- a symmetric sextet and anti-symmetric triplet
- Add the third quark

1. Building with sextet:

2. Building with the triplet:

- In summary, the combination of three uds quarks decomposes into:

Mixed symmetry octet

Symmetric decuplet

Totally anti-symmetric singlet

Mixed symmetry octet

combination of three uds quarks in strangeness, charge and isospin axes

OctetDecuplet

Charge: Q=I3+1/2 Y

Hypercharge: Y=B+S (B: baryon no.=1/3 for all quarks

S: strange no.)

SU(3) colour

- In QCD quarks carry colour charge r, g, b
- In QCD, the strong interaction is invariant under rotations in colour space SU(3) colour symmetry
- This is an exact symmetry, unlike the approximate uds flavor symmetry
- r, g, b SU(3) colour states:

(exactly analogous to

u,d,s flavour states)

- Colour states labelled by two quantum numbers: I3c(colour isospin), Yc(colour hypercharge)

Quarks:

Anti-Quarks:

Colour confinement

- All observed free particles are colourless
- Colour confinement hypothesis:

only colour singlet states can exist as free particles

- All hadrons must be colourless (singlet)
- Colour wave functions in SU(3) colour same as SU(3) flavour
- Colour singlet or colouerless conditions:
- They have zero colour quantum numbers I3c=0, Yc=0
- Invariant under SU(3) colour transformation
- Ladder operators are yield zero

Baryon colour wave-function

- Combination of two quarks
- No qq colour singlet state Colour confinement bound state of qq does not exist
- Combination of three quarks
- The anti-symmetric singlet colour wave-function qqq bound states exist

Baryon wave functions

- Quarks are fermions and have anti-symmetric total wave-functions
- The colour wave-function for all bound qqq states is anti-symmetric
- For the ground state baryons (L=0) the spatial wave-function is symmetric (-1)L
- Two ways to form a totally symmetric wave-function from spin and isospin states:

1. combine totally symmetric spin and isospin wave-function

2. combine mixed symmetry spin and mixed symmetry isospin states

- both and are sym. under inter-change of quarks

1 2 but not 1 3 , …

- normalized linear combination is totally

symmetric under 1 2, 1 3, 2 3

Baryon decuplet

- The spin 3/2 decuplet of symmetric flavour and symmetric spin wave-functions

Baryon decuplet (L=0, S=3/2, J=3/2, P=+1)

- If SU(3) flavour were an exact symmetry all masses would be the same (broken symmetry)

Baryon octet

- The spin 1/2 octet is formed from mixed symmetry flavor and mixed symmetry spin wave-functions

Baryon octet (L=0, S=1/2, J=1/2, P=+1)

- We can not form a totally symmetric wave-function based on the anti-symmetric flavour singlet as there no totally anti-symmetric spin wave –function for 3 quarks

Baryons magnetic moments

- Magnetic moment of ground state baryons (L = 0) within the constituent quark model: μl =0 , μs ≠0
- Magnetic moment of spin 1/2 point particle:
- for constituent quarks:
- magnetic moment of baryon B:

qu=+2/3

qd,s=-1/3

Baryons magnetic moments

- magnetic moment of the proton:
- further terms are permutations of the first three terms

Baryons: magnetic moments

- result with quark masses:
- Nuclear magneton

Thank you

Reference: University of Cambridge, Prof. Mark Thomson’s lectures 7 & 8, part III major option, Particle Physics 2006

WWW.hep.phy.cam.ac.uk/~thomson/lectures/lectures.html

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