The Oldest Unsolved Problem:the mystery of three generations

C.S. Lam

McGill and UBC, Canada

arXiv:1002.4176 (phys. Letts. B)arXiv:1003.0498(talk, summary)

Who Ordered That?

Carl Anderson

(1905—1991)

1928: Dirac Eq1932: positron

e+

e-

1936: muon (Anderson)

Millikan Blackett

Isidor Issac Rabi

(1898—1988)

1933: Nobel Prize (Dirac)1936: Nobel Prize (Anderson)

oldest mystery in particle physics

趙忠堯 (1930)

the plot thickens• now there are three generations of quarks and

leptons. What tells them apart?

• Maybe a new set of (generation, horizontal) quantum numbers to tell them apart?

Heisenberg invented

isotopic spin in 1932

to distinguish the newly discovered neutron from the proton

a horizontal symmetry?

• if so, must be (spontaneously) badly broken to account for the mass differences and mixing

a horizontal symmetry?

• if so, must be (spontaneously) broken to account for mass differences and mixing

• is there any trace of symmetry left?

yes for leptons, very little for quarks

a horizontal symmetry?

• if so, must be (spontaneously) broken to account for mass differences and mixing

• is there any trace of symmetry left?

yes for leptons, very little for quarks• what is the unbroken symmetry for leptons?

a horizontal symmetry?

• if so, must be (spontaneously) broken to account for mass differences and mixing

• is there any trace of symmetry left?

yes for leptons, very little for quarks• what is the unbroken symmetry for leptons?• is there a direct experimental test to decide

whether there is a horizontal symmetry?

residual leptonic symmetry(regularity in neutrino mixing)

2 2 01

1 2 36

1 2 3

U

1 2 3

e

tri-bimaximal mixing

how is this regularity related to the unbroken horizontal symmetry, and what symmetry might that be?

beta decay

quarks

two .. math slides• horizontal symmetry, symmetry breaking, and mass

matrices (L, R complications)

• mass matrices, masses, mixings, `residual’ symmetry of the mass matrices and regularity of the mixing matrix

?

( , , , , ) .aL R L RH e e N H G

horizontal symmetry

symmetry breaking( , , , , ) .aL R L RH e e N H G

1

2. . ......L R L R R Re Ne e N N NM M h cM

LR, RR, mass matrices

LL mass matrices

†ee eM M M 1

NTM MMM

,L e L L Le M e M

( , , , )a aeff eff L LH H e

effective LL Hamiltonian

math slide I3

eM

LL mass matrices

2 2 2( , , )ediag m m m

TU U agM di

2 2 01

1 2 36

1 2 3

U

1 2 3

e

(special unitary) residual symmetry operators

[ , ] 0eF F iaM d g

2

1 2 3

[ , ] 0, 1

, ,

G G

G G

M

G

1 2 21

2 2 13

2 1 21G

1 2 212 1 2

32 2

21

G

1 0 0

0 0 1

0 13

0

G

G eigenvalues: +1, -1, -1

It is reversible if F is non-degenerate

diagonalization

same diagonalization

math slide II

two.. math slides• horizontal symmetry, symmetry breaking, and mass

matrices (L, R complications)

• mass matrices, masses, mixings, `residual’ symmetry of the mass matrices and regularity of the mixing matrix

• relation between regularity of mixing and unbroken horizontal symmetry (2.5 criteria)

• the case of tri-bimaximal mixing?

2.5 criteria

1. horizontal symmetry contains residual symmetry of mass matrices

2. vev must be determined to obey 1.

,F GG

vev( , , , , ) .aL R L RH e e N H G

aa aF

aa aG

( , , , )a aeff eff L LH H e ,F G

( , , , )a aeff eff L LH H e

2.5 criteria (summary)

1. horizontal symmetry contains residual symmetry of mass matrices

2. vev must be determined to obey 1.

3. residual symmetry should determine the mixing matrix (reversibility)

,F GG

aa aF aa aG

F must be non-degenerate

tri-bimaximal mixing & variations

2

1

F

,F GG

3 1

4 2,A F G

3 3,S F G

4 1 1 2 3{ , } , , ,S F G F G G G

1 2 3, ,

2 2 01

1 2 36

1 2 3

G GU G

1 2 3 4, , ,NGF G G G S

An experimental test ofhorizontal symmetry

• Fermion masses in the SM come from the coupling with a single Higgs

• In horizontal symmetry models, they come from couplings with several Higgs.

• Therefore the coupling of any of these Higgs is no longer proportional to the fermion mass. Instead, they are proportional to the Clebsch-Gordan coefficients of the horizontal symmetry group.

• When a Higgs is found, the deviation of its fermion-pair decay rates from the SM predictions may indicate the presence of a horizontal symmetry.

L Rq q L Re e

illustration: `SM Higgs’ is a horizontal singlet

1. the SM Higgs does not decay to a fermion pair whose L and R belong to different (horizontal) irreducible representations (IR)

2. its decay rate to every member of an IR is the same, instead of proportional to the square mass of the member

3. if the top quark (or the tau) belongs to an irreducible triplet, then the Higgs decay rate is 1/3 to 1/9 of the SM rate. This makes the Higgs detection more difficult

4. if the top quark (or the tau) belongs to an irreducible doublet, then the Higgs decay rate is 1/2 to 1/4 of the SM rate

5. if all decay rates are the same as the SM rates, then horizontal symmetry either does not exist, or all fermions belong to a 1-dimensional IR

6. otherwise we know which IR each fermion belongs to, which greatly constrains possible horizontal groups and models.

L Rq q L Re e

Conclusion

• The regularity of neutrino mixing suggests the presence of a horizontal symmetry or groups containing for leptons, but we do not know theoretically why these groups

• No simple regularity seems to be present in quark mixing. That makes the idea of horizontal symmetry somewhat of an enigma

• An experimental test for the presence of horizontal symmetry is suggested: measure the SM Higgs decay rates to fermion pairs and compared them with the SM rates.

• The experimental results should give much better insight into this oldest unsolved puzzle in particle physics: `who order that?’

4S4 3 4, , ,A S S

M

The coupling of the SM Higgs to fermion pairs is of the form

In the presence of a horizontal symmetry,

the mass matrix is then

The SM Higgs is a singlet, with . .Its coupling to the ith lepton pair is

For irreducible representations,

hence

In SM, the Higgs to ith fermion pair decay rate is proportional to

With horizontal symmetry, if i belongs to an irreducible triplet, the it is proportional to

where R is the contribution from non-singlet Higgs. Its value is model dependent, but for those models in which the vev can be chosen for M to be diagonal, then

A similar formula holds if i belongs to an irreducible doublet.

1. Horizontal symmetry?

2. An experimental test.