Download SU(3) symmetry and Baryon wave functions

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 q q̄ (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
System of two quarks: I3=I3(1)+I3(2)
u
,
|I(1)-I(2)| ≤ I ≤ |I(1)+I(2)|
d
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
• 8 matrices have detU=1 and form an SU(3) group
• The 8 matrices are:
• In SU(3) flavor, 3 quark states are :
s
• SU(3) uds flavour symmetry contain SU(2) ud flavour symmetry
• Isospin
• Ladder operators
• Same matrices for u
•
s and d
s
λ3 and 2 other diagonal matrices are not independent, so de fine λ 8
as the linear combination:
• Only need 2 axes (quantum numbers) : (I3,Y)
Quarks:
Anti-Quarks:
All other combinations give zero
Combining uds quarks for baryons
• First combine two quarks:
•
a symmetric sextet and anti-symmetric triplet
• Add the third quark
1. Building with sextet:
Symmetric decuplet
Mixed symmetry
octet
2. Building with the triplet:
Mixed symmetry
octet
Totally antisymmetric singlet
• In summary, the combination of three uds quarks decomposes into:
combination of three uds quarks in strangeness, charge
and isospin axes
Octet
Decuplet
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
not exist
• Combination of three quarks
bound state of qq does
• 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
1 2 but not 1 3 , …
- normalized linear combination
symmetric under 1 2, 1 3, 2
are sym. under inter-change of quarks
is totally
3
Baryon decuplet
• The spin 3/2 decuplet of symmetric flavour and symmetric spin wavefunctions
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
Thank you
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:
qu=+2/3
qd,s=-1/3
 magnetic moment of baryon B:
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