Mass bounds for thermal baryogenesis from particle decays · Mass bounds for thermal baryogenesis...

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Transcript of Mass bounds for thermal baryogenesis from particle decays · Mass bounds for thermal baryogenesis...

Mass bounds for thermal baryogenesis from particledecays

Juan Racker

Instituto de Fı́sica corpuscular (IFIC), Universidad de Valencia-CSIC

COSMO 2013

Mass bounds for thermal baryogenesis from particle decays – p.1/34

Outline of the talk:

� Brief Introduction

� Problems for having baryogenesis at low energy

� Some solutions and bounds

– p.2/34

The matter-antimatter asymmetry of the UniverseObservations:

(a) The Universe is globally asymmetric: the amount of antimatter isnegligible with respect to the amount of matter.� Cosmic rays from the sun.� Planetary probes.� Galactic cosmic rays.

� BESS-Polar experiment −→ HeHe

< 1 × 10−7.� Absence of strong γ−ray flux from nucleon-antinucleon

annihilations in clusters of Galaxies (like Virgo cluster).

=⇒ Matter and antimatter domains should be larger than 20 Mpc.[Steigman, 1976]

Actually they must be larger than ∼ the visible Universe ( cosmicdiffuse γ-ray background ) . [Cohen, De Rújula, Glashow, 1998]

– p.3/34

(b) Baryon density� Big Bang Nucleosynthesis.

The abundances of the light elements D, 3He, 4He, and 7Lidepend mainly on one parameter, nB/nγ.

� CMB anisotropies.

YB ≡ nB − nB̄s

=nBs

≃ 8,5 × 10−11

– p.4/34

The annihilation catastrophe

Nucleons and antinucleons remain in chemical equilibrium untilΓann < H, which occurs at

Tfo ∼ 22 MeV

If the Universe was locally-baryon-symmetric, then

YB fo ∼ 7 × 10−20 !!!

Conclusion: There was a baryon asymmetry at T ∼ O (102) MeV.

Origin?↓

initial conditions��� or dynamic generation

– p.5/34

Sakharov’s conditionsBasic requirements to dynamically generate a baryon asymmetry:

� Baryonic number ( B) violation

� C and CP Violation

� Departure from thermal equilibrium

In thermal baryogenesis from the decay of a particle with mass M :

H(T = M)

Interaction rates∝ f(Mi/M, couplings)

M

MPl

– p.6/34

Is baryogenesis possible in the SM?� B violation: Yes → sphalerons (violate B + L but conserveB − L).

� C violation: Yes

� CP Violation: Not enough → JCP/T12c ∼ 10−18

� Departure from thermal equilibrium: No → mH > 114GeV

implies that the EW phase transition is not strongly first order.

Conclusion: physics beyond the SM is needed to explain the origin ofthe cosmic asymmetry.

– p.7/34

Problems for thermal Baryogenesis at low energy

Example: Type I Leptogenesis

The singlet Majorana neutrinos of the type I seesaw can generate a

lepton asymmetry when decaying in the primitive Universe.

Y fB = −κ ǫ η (constant ǫ)

� κ =28

79Y eqN (T ≫M1) ∼ 10−3

� ǫ =∑

α

ǫα =∑

α

γ(N1 → Hℓα) − γ(N1 → H̄ℓ̄α)∑

β γ(N1 → Hℓβ) + γ(N1 → H̄ℓ̄β)

� η = efficiency ∼ 1

strength of interactions

– p.8/34

CP violation in decays

Ni

h

ℓα

(a) Tree

Ni

ℓ̄β

Nj

h

ℓα

(b) Vertex

�Ni

ℓβ, ℓ̄β

h, h̄

Nj

h

ℓα

(c) Wave

– p.9/34

– p.10/34

Two problems to lower the energy scale

ǫ ∼ 3

16π

λ2α2

M2M1 (hierarchical)

� Connection with light neutrino masses:Type I seesaw: ǫ ∼ 3

16πmiv2M1 (type I seesaw)

|ǫ| ≤ ǫDImax =

3

16π

M1

v2(m3 −m1) =⇒ M1 & 109 GeV (η ≤ 1)

Some alternatives: Inverse seesaw, radiative seesaws, ...

� Even with no connection to neutrino masses:Washout processes inherent to the existence of CP violation

washouts ∝(

λ2α2

M2

)2

large ǫ → large λα2 → too much washout at LE → How low?

– p.11/34

Ways to have Baryogenesis at low energy scales

� Mass degeneracy:

when M2 −M1 ∼ΓN2

2, |ǫ| ∼ 1

2

Im[

(λ†λ)221

]

(λ†λ)11(λ†λ)22≤ 1

2

Note: However in the type I seesaw the mixing between active and sterile neutrinos is:

mixing ∼mD

M∼

r

M≪ 1.

� Three body decays: It’s more easy to satisfy the o.e.c.

� Hierarchy of couplings:� Take λα1 as small as necessary.

E.g. λα1 ∼ 10−7 to have Γ ∼ H(T = M1) for M1 = 1 TeV.� Take λα2 much larger to have enough CP violation.

[T. Hambye, 2002]

– p.12/34

Other ways and bounds

L-violating CP asymmetry

ǫ ∝ λ2α2

M1

M2, washouts ∝

[

λ2α2

M1

M2

]2

50

500

100

1000

3 5 30 50 10

Bou

nd o

n M

1 [T

eV]

M2/M1

[JR, arXiv:1308.1840]– p.13/34

L-conserving CP asymmetry

ǫα ∝ λ2β2

(

M1

M2

)2

, washouts ∝[

λ2β2

(

M1

M2

)2]2

.

Inverse seesawParticle content: SM + νRi , sLi (singlet fermions).The mass matrix of the neutral sector in the basis νL, νcR, sL is

M =

0 mD 0

mTD 0 M

0 MT µ

mν = mDMT−1

µM−1mTD ∼ mD

( µ

M

) (mD

M

)

(mD, µ≪M)

νRi , sLi combine to form quasi-Dirac fermions with mass ∼M .

mixing ∼ mD

M∼

µ – p.14/34

1e+06

1e+07

1e+08

1e+09

2.5 3 5 30 10

Bou

nd o

n M

1 [G

eV]

M2/M1

— µ2 ≫ ΓN2— µ2 ≪ ΓN2

[JR, M. Peña, N. Rius, 2012]

Note: This is for 2 flavors. The bound can be up to a factor ∼ 4 smaller for 3 flavors.

– p.15/34

100000

1e+06

1e+07

1e+08

2.5 3 5 30 10

Trh

[GeV

]

M2/M1

— µ2 ≫ ΓN2— µ2 ≪ ΓN2

Note: The Upper bound on Trh from gravitino overproduction can be satisfied

– p.16/34

Massive decay productsIn baryogenesis from annihilations, χχ→ ψu, it is possible to takemψ > mχ =⇒ Boltzmann suppression ∝ e−mψ/T of the washoutswithout reducing the CP asymmetry.[Y. Cui, L. Randall, B. Shuve, 2012],

[N. Bernal, S. Colucci, F-X. Josse-Michaux, JR, L. Ubaldi, 2013]

In decays, e.g. taking a massive H2 in N1 → H2ℓ, like in the inertdoublet model, there are two opposite effects:

- Boltzmann suppression of the washouts (but not as much as forannihilations, since mH2

< M1).

- Phase space suppression of the CP asymmetry

⇓ SM + H2 + Ni, with H2 and Ni odd under a Z2

– p.17/34

2

345

20

50

1

10

100

0.98 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Bou

nd o

n M

1 [T

eV]

Mh/M1

Tsfo=140

Tsfo=80

[JR, arXiv:1308.1840]

– p.18/34

Initial thermal densityIf the N1 are produced at T ≫M1 by a process different from theYukawa interactions, then λα1 can be chosen small enough to havethe N1 decay at T ≪M2.⇒ It is possible to have large λα2 and consequently a big ǫ, but atthe same time small washouts at the moment the N1 start to decayand produce the BAU.

M1min ∼ 2500 (2000) GeV for Tsfo = 140 (80) GeV.

In this way the small neutrino masses, DM, and BAU can besimultaneously explained at the TeV scale in the inert doublet model,without degenerate heavy neutrinos or fine tuning among phases.

Note: The interaction that creates the N1 must decouple before they decay.

– p.19/34

Almost degenerate neutrinos

1e-08

1e-07

1e-06

1e-05

0.0001

0.001

0.01

250 1000 10000

∆ M

/M1

M1 [GeV]

r > 10-5

r > 10-3

r > 10-1

δ ≡ M2−M1

M1

, r = smallest Yukawa couplinglargest Yukawa coupling

δ × r . 10−8 , for 250 GeV . M1 . 4 TeV and

δ × r . 3 × 10−9 , for 250 GeV . M1 . 1 TeV .– p.20/34

Additional slides ...

– p.21/34

Relevant processes for N1–Leptogenesis

�N1

ℓj

H

�ℓjH

N1

(a) Decay and inverse decay (production) of N1.

ΓN1=

1

8π(h†h)11M1 .

– p.22/34

�Hℓ

ℓ̄

N1,2,3 �Hℓ

ℓ̄

N1,2,3

�ℓℓ

N1,2,3

(b) ∆L = 2 scatterings mediated by N1,2,3.

ℓ t̄

N1 H̄ Q3

Q3

N1

t

H

�N1

Q̄3t̄

H

(c) ∆L = 1 scatterings mediated by the Higgs.

– p.23/34

ǫNiℓα = ǫNiℓα (vertex) + ǫNiℓα (wave)

ǫNiℓα (vertex) =1

j

f(yj)Im

[

λ∗αjλαi(λ†λ)ji

]

(λ†λ)ii

ǫNiℓα (wave) = − 1

j 6=i

Mi

M 2j −M 2

i

Im[(

Mj(λ†λ)ji +Mi(λ

†λ)ij)

λ∗αjλαi]

(λ†λ)ii

with yj ≡M 2j /M

2i and f(x) =

√x(1 − (1 + x) ln[(1 + x)/x]).

[Covi, Roulet, Vissani, 1996]

– p.24/34

Boltzmann equationsSimple unflavored version:

dYNdz

= − 1

zHs

(

YNY eqN

− 1

)

γD

dYLdz

=1

zHs

{

ǫ

(

YNY eqN

− 1

)

γD − YLY eqL

γD2

}

with Yx ≡ nxs

and z ≡ M1

T.

�dYNdz

= −K(z)

z(YN − Y eq

N ) with K(z) ∼ ratesH

.

�dYLdz

= source - washouts

Source = CP violation × L violation × departure from eq.

Washouts = asymmetries (YL) × rates (γ).

– p.25/34

The role of m̃1

It determines the amount of departure from eq. and the intensity ofthe washouts.Reference value given by the equilibrium mass m∗ :

ΓN1

H(T = M1)=m̃1

m∗

,

with m∗ ≃ 1,08 × 10−3 eV .

� m̃1 ≫ m∗ → strong washout regime:

• Independence from initial conditions.

• η ∝ m̃−11 (YL ∼ source/wo ∼ ( ǫ dY eq

N /dz)/wo ) .

� m̃1 ≪ m∗ → weak washout regime:

• Very dependent on initial conditions.

• If Y iN = 0 → η ∝ m̃1

1� m̃21 .

– p.26/34

Is leptogenesis possible with ǫ = 0?

Flavor effects

N1 → ℓd H

� T & 1012 GeV: The Yukawa interactions of the charged leptonsare out of equilibrium→ ℓd is the only relevant “direction” in flavor space.

� T . 1012 GeV: The Yukawa interactions of the τ (and eventuallythe µ) are in equilibrium→ they project ℓd into the flavor eigenstates (ℓτ , ℓµ, ℓe) →decoherence

Note: similarly for the antileptons, with N1 → ℓ̄′d H̄

– p.27/34

Boltzmann equations

Define Y∆α≡ 1

3YB − YLα (B/3 − Lα is conserved by sphalerons)

dY∆α

dz≈ f(z)ǫα − Y∆α

Kαw(z) (α = e, µ, τ),

with z ≡M1/T , Kα ≡ |〈ℓα|ℓd〉|2

The asymmetries Y∆αevolve (approximately) independently.

– p.28/34

Two types of CP violation

Γ(N1→ℓµh)Γ(N1→ℓ̄µh̄)

Γ(N1→ℓτh)

Γ(N1→ℓ̄τ h̄)

Γ(N1→ℓdh)

Γ(N1→ℓ̄′dh̄)

-�

6

?

��

���3

���+

(a) ℓ′d = ℓd, ǫ 6= 0

Γ(N1→ℓµh)Γ(N1→ℓ̄µh̄)

Γ(N1→ℓτh)

Γ(N1→ℓ̄τ h̄)

Γ(N1→ℓdh)

Γ(N1→ℓ̄′dh̄)

-�

6

?

��

���3

(b) ǫ = 0, ℓ′d 6= ℓd, ǫα 6= 0

– p.29/34

1e-05

0.0001

0.001

0.01

0.1 1 10

|Y∆ i

|/|ε|

, |

YB

-L|/|

ε|

Z = M1/T

— |Y∆τ/ǫτ | —

∣Y∆µ/ǫµ

∣ — |YB−L/ǫµ|ǫτ = −ǫµ Kτ = 0,1 Kµ = 0,9 m̃1 = 0,01 eV

– p.30/34

The relevant set of BE for the case µ2 ≫ ΓN2is

dYN1

dz=

−1

sHz

(

YN1

Y eqN1

− 1

)

γD1,

dY∆α

dz=

−1

sHz

{

(

YN1

Y eqN1

− 1

)

ǫα1 γD1−

i

γNiℓαhyℓα

−∑

β 6=α

(

γℓβh

ℓαh+ γ

ℓβ h̄

ℓαh̄+ γhh̄ℓαℓ̄β

)

[yℓα − yℓβ ]

}

,

where z ≡M1/T , YX ≡ nX/s, yX ≡ (YX − YX̄)/Y eqX , and

Y∆α≡ YB/3 − YLα.

– p.31/34

Instead, for µ2 ≪ ΓN2

dYN1

dz=

−1

sHz

(

YN1

Y eqN1

− 1

)

γD1,

dYN2−N̄2

dz=

−1

sHz

α

γN2

ℓαh[yN2

− yℓα] ,

dY∆α

dz=

−1

sHz

{(

YN1

Y eqN1

− 1

)

ǫα1 γD1− γN1

ℓαhyℓα + γN2

ℓαh[yN2

− yℓα]

−∑

β 6=α

(

γℓβh

ℓαh+ γ

ℓβ h̄

ℓαh̄+ γhh̄ℓαℓ̄β

)

[yℓα − yℓβ ]

}

.

– p.32/34

0.01

1

100

10000

1e+06

1e+08

1e+10

10000 100000 1e+06 1e+07 1e+08 1e+09

γ τ /

γ N2 ,

γ

τ / |

γ Σ F

CI|

T [GeV]

M2 = 107 GeV , (λ✝ λ)22 = 10-4

— γτ/γN2— γτ/ |γΣ FCI|

– p.33/34

Light neutrino masses:

mi ∼λ2

21v2

M1

+ µ2λ2

22v2

M 22

+ λ′22λ22v

2/M2 .

Taking mi ∼ matm ∼ 0,05 eV, we get

λα1 ∼ 10−5 − 10−4, µ2/M2 ∼ 10−8 − 10−6, λ′α2 ∼ 10−8 − 10−7 .

Moreover,

ΓN2/M2 ∼ 5 × (10−4 − 10−2) ⇒ (typically) µ2 ≪ ΓN2

Note: For M1 & 5 × 106 GeV, and still not considering large finetunings related to phase cancellations, it is also possible to haveµ2 & ΓN2

.

– p.34/34