SU(3) symmetry and Baryon wave functions

Sedigheh Jowzaee

PhD seminar- FZ Juelich, Feb 2013

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

q q̄

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)

• 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

Symmetry and conservation

• 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)

Isospin

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: 3λ λ8

• Only need 2 axes (quantum numbers) : (I3,Y)

All other combinations give zero

Quarks: Anti-Quarks:

• First combine two quarks:

• a symmetric sextet and anti-symmetric triplet

• Add the third quark

Combining uds quarks for baryons

1. Building with sextet:

2. Building with the triplet:

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

Symmetric decupletMixed symmetry

octet

Mixed symmetry octet

Totally anti-symmetric singlet

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

Octet Decuplet

Charge: Q=I3+1/2 YHypercharge: 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 I3

c=0, Yc=0

– Invariant under SU(3) colour transformation– Ladder operators are yield zero

• 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 colour wave-function

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 2006WWW.hep.phy.cam.ac.uk/~thomson/lectures/lectures.html