1/25 - SLAC Indico (Indico)€¦ · U leptonic mixing matrix (lives in generation space; rotates...

1 / 25

U

leptonic mixing matrix (lives in generation space ; rotates from charged lepton αto neutrino i) with three angles (index order reversed wrt quarks) :

Uαi =

1 0 00 c23 s230 −s23 c23

c13 s13e−iδ

0 1 0−s13e

iδ 0 c13

c12 s12 0−s12 c12 0

0 0 1

P

=

c12c13 c13s12 s13e−iδ

−c23s12 − c12s13s23eiδ c12c23 − s12s13s23e

iδ c13s23s23s12 − c12c23s13e

iδ −c12s23 − c23s12s13eiδ c13c23

P

P = diage−iφ1/2, e−iφ2/2, 1 for Majorana, identity for Dirac

2 / 25

U

leptonic mixing matrix (lives in generation space ; rotates from charged lepton αto neutrino i) with three angles (index order reversed wrt quarks) :

Uαi =

1 0 00 c23 s230 −s23 c23

c13 s13e−iδ

0 1 0−s13e

iδ 0 c13

c12 s12 0−s12 c12 0

0 0 1

P

=

c12c13 c13s12 s13e−iδ

−c23s12 − c12s13s23eiδ c12c23 − s12s13s23e

iδ c13s23s23s12 − c12c23s13e

iδ −c12s23 − c23s12s13eiδ c13c23

P

P = diage−iφ1/2, e−iφ2/2, 1 for Majorana, identity for Dirac

θ23 ≃ π/4 θ12 ≃ π/6 θ13 ≃ 0.15, 8o δ ∼ 1.4π(global fits of www.nu-fit.org)

for comparaison, in CKM :

θ23 ≃ Vcb ≃ 0.04 θ12 ≃ Vus ≃ 0.225 θ13 ≃ Vub ≃ 0.004

2 / 25

The drunken Unitarity triangle

Not hear much about “leptonic unitarity triangle”1.not measure elements at tree in CC2. Also, it drinks.

3 / 25


Not hear much about “leptonic unitarity triangle”1.not measure elements at tree in CC2. Also, it drinks.Amplitude to oscillate from flavour α to β over distance L :

Aαβ(L) = Uα1U∗β1 + Uα2U

∗β2e

−i(m22−m2

1)L/(2E) + Uα3U∗β3e

−i(m23−m2

1)L/(2E)

3 / 25




∗β2e

−i(m22−m2

1)L/(2E) + Uα3U∗β3e

−i(m23−m2

1)L/(2E)

at L = 0 unitarity : ⇒ Aαβ = 1 for α = βAαβ = 0 for α 6= β

⇔ unitarity triangle(in complex plane)

Uµ1U∗

e1

Uµ2U∗

e2 Uµ3U∗

e3

3 / 25




∗β2e

−i(m22−m2

1)L/(2E) + Uα3U∗β3e

−i(m23−m2

1)L/(2E)

at L = 0 unitarity : ⇒ Aαβ = 1 for α = βAαβ = 0 for α 6= β

⇔ unitarity triangle(in complex plane)

Uµ1U∗

e1

Uµ2U∗

e2 Uµ3U∗

e3

At L = t 6= 0, two of the vectors rotate in the complex plane, with frequencies(m2

j −m21)/2E

oscillations ↔ time-dependent non-unitarity“drunken unitarity” ⇒ intuition when can use 2-flavour approx

3 / 25

About two- flavour analyses : atm/LBL νµ disappearance

Amplitude to oscillate from flavour µ to τ over distance L :

Aµτ (L) = Uµ1U∗τ1 + Uµ2U

∗τ2e

−i(m22−m2

1)L/(2E) + Uµ3U∗τ3e

−i(m23−m2

1)L/(2E)

4 / 25




∗τ2e

−i(m22−m2

1)L/(2E) + Uµ3U∗τ3e

−i(m23−m2

1)L/(2E)

Uµ1U∗

τ1

Uµ2U∗

τ2Uµ3U

∗

τ3

At L ∼ (m23 −m2

1)/E , vector “3” rotates,atfrequency (m2

3 −m21)/2E

4 / 25




∗τ2e

−i(m22−m2

1)L/(2E) + Uµ3U∗τ3e

−i(m23−m2

1)L/(2E)

Uµ1U∗

τ1

Uµ2U∗

τ2Uµ3U

∗

τ3

At L ∼ (m23 −m2

1)/E , vector “3” rotates,atfrequency (m2

3 −m21)/2E

⇒ “Atmospheric” neutrinos, also LBL(νµ disappearance via ∆m2

31 oscillations) :

Aµτ (L) ≃ Uµ1U∗τ1 + Uµ2U

∗τ2 + Uµ3U

∗τ3e

−i(m23−m2

1)L/(2E)

Uµ3U∗τ3 oscillates on timescale t = L ∼ (m2

3 −m21)/E

Uµ2U∗τ2 ∼ stationary, measure θ23

4 / 25

Three flavour probability required for CPV ∝ triangle area

Pαβ(L) = δαβ − 4∑

i<j

ReUαiU∗βiU

∗αjUβj sin2 xji

2

+2∑

i<j

ImUαiU∗βiU

∗αjUβj sin xji

Last term violates CP ⇔ opposite sign for να → νβ .(Also matter effect opposite sign forν).

∝ area of triangle ∝ J = 8c213s13c23s23c12s12 :

Suppose triangle base ∈ Re = Uµ1U∗τ1.

Then base*height ∝ ImUµ1U∗τ1U

∗µjUτ j

∝ J sin δ

Uµ1U∗

τ1

Uµ2U∗

τ2Uµ3U

∗

τ3

5 / 25

Leptogenesis

a class of recipes, that use (majorana) neutrino mass models to generate thematter excess

what matter excess ?

required ingredients ?

a simple seesaw model

how it works...

6 / 25

Preambule

1. about “What the stars (and us) are made of” (5% of U)

≈ H ≈ baryons

7 / 25

Preambule


≈ H ≈ baryonsnot worry about lepton asymmetry : is (undetected) Cosmic NeutrinoBackground ...so how to measure asym? ? ?

7 / 25

Preambule



2. I am made of baryons(defn) ... observation... all matter we see is made ofbaryons (not anti-baryons)

3. quantify as (s0 ≃ 7nγ,0)

YB ≡nB − nB

s

∣∣∣∣0

= 3.86 × 10−9ΩBh2 ≃ (8.53 ± 0.11)× 10−11

PLANCK

7 / 25

Preambule



2. I am made of baryons(defn) ... observation... all matter we see is made ofbaryons (not anti-baryons)

3. quantify as (s0 ≃ 7nγ,0)

YB ≡nB − nB

s

∣∣∣∣0

= 3.86 × 10−9ΩBh2 ≃ (8.53 ± 0.11)× 10−11

PLANCK

⇒ Question : where did that excess come from ?

7 / 25

Where did the matter excess come from ?

1. the U(niverse) is matter-anti-matter symmetric ?= islands of particles and anti-particlesX no ! not see γs from annihilation

8 / 25



2. U was born that way...

X no ! After birth of U, there was “inflation” (only theory explaining coherent temperature fluctuations in microwave

background that arrive from causally disconnected regions today...) “60 e-folds” inflation ≡ VU →> 1090

VU

(nB − nB) → 10−90(nB − n

B), s from ρ of inflation...

8 / 25



2. U was born that way...

X no ! After birth of U, there was “inflation” (only theory explaining coherent temperature fluctuations in microwave

background that arrive from causally disconnected regions today...) “60 e-folds” inflation ≡ VU →> 1090

VU

(nB − nB) → 10−90(nB − n

B), s from ρ of inflation...

3. created/generated/cooked after inflation...

8 / 25

Three ingredients to prepare in the early U (old russian recipe)

Sakharov

1. B violation : if Universe starts in state of nB − nB = 0, need B to evolve tonB − nB 6= 0

9 / 25


Sakharov


2. C and CP violation : ...particles need to behave differently fromanti-particles.Present in the SM quarks, observed in Kaons and Bs, searched for inleptons (...T2K,future expts)

9 / 25


Sakharov



3. out-of-thermal-equilibrium ...equilibrium = static. “generation” =dynamical processNo asym.s in un-conserved quantum #s in equilibrium

9 / 25


Sakharov



3. out-of-thermal-equilibrium ...equilibrium = static. “generation” =dynamical processNo asym.s in un-conserved quantum #s in equilibriumFrom end inflation → BBN, Universe is an expanding, cooling thermalbath, so non-equilibrium from :

slow interactions : τint ≫ τU = age of Universe (Γint ≪ H) phase transitions :

9 / 25

ingredient 1 : Does the SM conserve B ?

B, L are global symmetries of the SM Lagrangian (q, ℓ doublets, e, u, d singlets)

LSM ⊃ qD/ q , ℓD/ ℓ , ℓHe , qHu , qHd

so, classically, there are conserved currents, and B and L are conserved. (SoB + L and B − L are conserved.)

10 / 25





Good—proton appears stable :τp >∼ 1033 yrs (τU ∼ 1010 yrs).

10 / 25





Good—proton appears stable :τp >∼ 1033 yrs (τU ∼ 1010 yrs).

But the SM does not conserve B + L...In QFT, there is the axial anomaly......anomalously, the fermion current associated to a classical symmetry is notconserved.

see Polyakov,“Gauge Fields + Strings,”

6.3=qualitative effects of instantons

10 / 25

ingredient 1 : the SM does not conserve B + L

B + L is anomalous. Formally, for one generation(α colour ) :

∑

SU(2)singlets

∂µ(ψγµψ) + ∂µ(ℓγµℓ) + ∂µ(qαγµqα) ∝1

64π2W A

µνWµνA.

where integrating the RHS over space-time counts “winding number” of theSU(2) gauge field configuration.⇒ Field configurations of non-zero winding number are sources of a doubletlepton and three (for colour) doublet quarks for each generation.

11 / 25

ingredient 1 : the SM does not conserve B + L

B + L is anomalous. Formally, for one generation(α colour ) :

∑

SU(2)singlets

∂µ(ψγµψ) + ∂µ(ℓγµℓ) + ∂µ(qαγµqα) ∝1

64π2W A

µνWµνA.

where integrating the RHS over space-time counts “winding number” of theSU(2) gauge field configuration.⇒ Field configurations of non-zero winding number are sources of a doubletlepton and three (for colour) doublet quarks for each generation.

E

t

Left-handed fermions

thanks to V Rubakov

11 / 25

SM B+L violation : rates

’t HooftKuzmin Rubakov+

Shaposhnikov

At T = 0 is tunneling process (from winding # to next, “instanton”) : Γ ∝ e−8π/g2

At 0 < T < mW , can climb over the barrier : ΓB+L ∼e−mW /T T < mW

α5T T > mW

⇒ fast SM B+L at T > mW

ΓB+L > H for mW < T < 1012 GeV

SM B+L called “sphalerons”⇒ if produce a lepton asym, “sphalerons” partially transform to a baryonasym. ! !⋆ ⋆ ⋆ SM B+L is ∆B = ∆L = 3 (= Nf ). No proton decay ! ⋆ ⋆ ⋆

12 / 25

Summary of preliminaries : A Baryon excess today :

• Want to make a baryon excess ≡ YB after inflation, that corresponds today to∼ 1 baryon per 1010 γs.• Three required ingredients : B , CP , TE .Present in SM, but hard to combine to give big enough asym YB

Cold EW baryogen ? ? Tranberg et al...

⇒ evidence for physics Beyond the Standard Model (BSM)

13 / 25

Summary of preliminaries : A Baryon excess today :

• Want to make a baryon excess ≡ YB after inflation, that corresponds today to∼ 1 baryon per 1010 γs.• Three required ingredients : B , CP , TE .Present in SM, but hard to combine to give big enough asym YB

Cold EW baryogen ? ? Tranberg et al...

⇒ evidence for physics Beyond the Standard Model (BSM)

One observation to fit, many new parameters...

⇒ prefer BSM motivated by other data ⇔ mν ⇔ seesaw ! (uses non-pert. SM

B+L )

13 / 25

The type I seesaw Minkowski, YanagidaGell-Mann Ramond Slansky

• add 3 singlet N to the SM in charged lepton and N mass bases, at scale > Mi :

add 18 parameters :M1,M2,M3

18 - 3 (ℓ phases) in λL = LSM + λαJNJℓα · φ− 1

2NJMJNcJ

MI unknown (∝/ v = 〈φ0〉), and Majorana (L ). CP in λαJ ∈ C .

14 / 25

The type I seesawMinkowski, Yanagida

Gell-Mann Ramond Slansky




2NJMJNcJ


• at low scale, for M ≫ mD = λv , light ν mass matrix

νLα νLβ

NA

MA

Xx xvλαA vλβA

9 parameters :m1,m2,m3

6 in UMNS

[mν ] = λM−1λTv2

forλ ∼ ht , M ∼ 1015 GeV

λ ∼ 10−7, M ∼ 10 GeV∼ .05 eV

“natural” mν ≪ mf : mν ∝ λ2, and M > v allowed.14 / 25

The type I seesawMinkowski, Yanagida

Gell-Mann Ramond Slansky




2NJMJNcJ


• at low scale, Higgs mass contribution

φ φNA

ν

λαA λβA

δm2φ ≃ −

∑

I

[λ†λ]II8π2

M2I ∼

mνM3I

8π2v4v2

for M >∼ 107 GeV > v2 tuning problem

( ? adding particles to cancel 1 loop...but higher loop ? Need symmetry to cancel ≥ 2

loop ?)

⇒ do seesaw with MI<∼ 108 GeV ?

(NB, in this talk, φ = Higgs, H = Hubble)14 / 25

Leptogenesis in the type 1 seesaw : usually a Fairy TaleFukugita Yanagida

Buchmuller et alCovi et al

Branco et alGiudice et al

...

Once upon a time, a Universe was born.

15 / 25




...Once upon a time, a Universe was born.

15 / 25




...

Once upon a time, a Universe was born.At the christening of the Universe,the fairies give the Standard Model and theSeesaw (heavy sterile Nj with L masses and CP interactions) to the Universe.

15 / 25




...

Once upon a time, a Universe was born.At the christening of the Universe,the fairies give the Standard Model and theSeesaw (heavy sterile Nj with L masses and CP interactions) to the Universe.The adventure begins after inflationary expansion of the Universe :

1 Assuming its hot enough, a population of Ns appear—they like the heat.

2 As the temperature drops below M , the N population decays away.

3 In the CP ,L interactions of the N , an asym. in SM leptons is created.

15 / 25




...

If this asymmetry can escape the big bad wolf of thermal equilibrium...

15 / 25




...Once upon a time, a Universe was born.At the christening of the Universe,the fairies give the Standard Model and theSeesaw (heavy sterile Nj with L masses and CP interactions) to the Universe.The adventure begins after inflationary expansion of the Universe :

1 Assuming its hot enough, a population of Ns appear—they like the heat.

2 As the temperature drops below M , the N population decays away.

3 In the CP ,L interactions of the N , an asym. in SM leptons is created.

4 If this asymmetry can escape the big bad wolf of thermal equilibrium...

5 the lepton asym gets partially reprocessed to a baryon asym bynon-perturbative B + L -violating SM processes (“sphalerons”)

And the Universe lived happily ever after, containing many photons. And forevery 1010 photons, there were 6 extra baryons (wrt anti-baryons).

15 / 25

Estimate something : TE + dynamicsSuppose M1 ≪ M2,3,Treheat > M1 ∼ 109GeV

Recipe : calculate suppression factor for each Sakharov condition, multiplytogether to get YB :

nB − nBs

∼1

3g∗ǫL,CPηTE ∼ 10−3ǫη (want 10−10)

s ∼ g∗nγ , ǫ = lepton asym in decay, η ∼ TE process

16 / 25


1 produce a population of N1s, via e.g. (qℓα → NtR)Get thermal density nN ≃ nγ if M1

<∼ T , and τprod < τU :

16 / 25




Γprod ∼ σvn ∼h2t λ

2

T 2

T 3

π2∼

h2t λ

2

π2T > H ≃

10T 2

mpl

, ⇒λ2

π2>

10T

mpl

∣∣∣∣T=M1

Suppose satisfied...

16 / 25




Γprod ∼ σvn ∼h

2

t λ2

T 2

T3

π2∼

h2

t λ2

π2T > H ≃

10T 2

mpl

, ⇒λ2

π2>

10T

mpl

∣

∣

∣

∣

T=M1

Suppose satisfied...2 Lepton asym is produced in N int. (eg decays) if there is CP ; can surviveafter inverse processes(eg decays) =“washout” become rare enough

ΓID(φℓ → N) ≃ Γdecay e−M1/T =

[λλ†]11M1

8πe−M1/T <

10T 2

mpl

16 / 25





2

t λ2

T 2

T3

π2∼

h2

t λ2

π2T > H ≃

10T 2

mpl

, ⇒λ2

π2>

10T

mpl

∣

∣

∣

∣

T=M1


ΓID(φℓ → N) ≃ Γdecaye−M1/T =

[λλ†]11M1

8πe−M1/T <

10T 2

mpl

At temperature Tα when Inverse Decays turn off,

nN

nγ(Tα) ≃ e−M1/Tα ≃

H

Γ(N → ℓαφ)can calculate this

16 / 25





2

t λ2

T 2

T3

π2∼

h2

t λ2

π2T > H ≃

10T 2

mpl

, ⇒λ2

π2>

10T

mpl

∣

∣

∣

∣

T=M1


ΓID(φℓ → N) ≃ Γdecaye−M1/T =

[λλ†]11M1

8πe−M1/T <

10T 2

mpl

At temperature Tα when Inverse Decays turn off,

nN

nγ(Tα) ≃ e−M1/Tα ≃

H

Γ(N → ℓαφ)can calculate this

so (1/3 is from SM B+L , s ∼ g∗nγ , ǫα is lepton asym in decay)

nB − nBs

∼1

3

∑

α

ǫαnN(Tα)

g∗nγ∼ 10−3ǫ

H

Γ(want 10−10)

16 / 25

Estimate ǫ, the CP and L asymmetry in decays

Recall (in S-matrix) CP : 〈φℓ|S |N〉 → 〈φℓ|S |N〉 = 〈φℓ|S |N〉, ( η =anti-η)

17 / 25



In leptogenesis, need CP ,L interactions of NI ...for instance :

finite temp :Beneke etal 10

ǫαI =Γ(NI →φℓα)− Γ(NI → φℓα)

Γ(NI →φℓ)+Γ(NI → φℓ)(recall NI = NI )

∼ fraction N decays producing excess lepton

17 / 25



In leptogenesis, need CP ,L interactions of NI ...for instance :

finite temp :Beneke etal 10

ǫαI =Γ(NI →φℓα)− Γ(NI → φℓα)

Γ(NI →φℓ)+Γ(NI → φℓ)(recall NI = NI )

∼ fraction N decays producing excess lepton

X λNI ℓα

φ

×λ∗ λ

λNJXNI ℓα

φ

φ

ℓ

+λ∗ λ λ

NJ

XNI ℓα

φ

φ

ℓ

Just try to calculate ǫ1 ?• no asym at tree• asym at tree × loop, if CP from complex cpling and on-shell particles in theloop (divergences cancel in diff, need Im part of Feynman param integrtn)

17 / 25

CP , complex couplings, loops unitarity and all that...(estimate ǫ, no loop caln)

Kolb+Wolfram,NPB ’80, Appendix1 the S-matrix S ≡ 1 + iT is CPT invariant

〈φℓ|S |N〉 = 〈N |S |φℓ〉 (= 〈φℓ|S†|N〉∗)

18 / 25



〈φℓ|S |N〉 = 〈N |S |φℓ〉 (= 〈φℓ|S†|N〉∗)

and unitary : SS† = 1 = (1 + iT )(1 − iT †)

⇒ iT − iT † + TT† = 0

18 / 25



〈φℓ|S |N〉 = 〈N |S |φℓ〉 (= 〈φℓ|S†|N〉∗)


⇒ iT − iT † + TT† = 0

⇒ i〈φℓ|T |N〉 − i〈φℓ|T †|N〉+ 〈φℓ|TT†|N〉 = 0

18 / 25



〈φℓ|S |N〉 = 〈N |S |φℓ〉 (= 〈φℓ|S†|N〉∗)


⇒ iT − iT † + TT† = 0


|〈φℓ|T |N〉|2 = |〈φℓ|T †|N〉|2 − i〈φℓ|T †|N〉〈N |TT†|φℓ〉

+i〈N |T |φℓ〉〈φℓ|TT†|N〉+ ...

18 / 25



〈φℓ|S |N〉 = 〈N |S |φℓ〉 (= 〈φℓ|S†|N〉∗)


⇒ iT − iT † + TT† = 0


|〈φℓ|T |N〉|2 = |〈φℓ|T †|N〉|2 − i〈φℓ|T †|N〉〈N |TT†|φℓ〉

+i〈N |T |φℓ〉〈φℓ|TT†|N〉+ ...

2 We are interested in a CP asymmetry :

ǫ ∝∫dΠ

(|〈φℓ|T |N〉|2 − 〈φℓ|T |N〉|2

)

so (this formula exact, if I kept 2s and sums ;∫

dΠ = phase space )

ǫ ∝∫dΠIm

〈φℓ|T †|N〉〈N |TT

†|φℓ〉

⇒ need complex cplings, and on-shell particles in a loop18 / 25

Estimating ǫ for hierarchical NI

Consider simple case : M1 ≪ M2,3. Suppose lepton asym generated in CP ,Ldecays of N1 :

ǫα1 =Γ(N1→φℓα)− Γ(N1→ φℓα)

Γ(N1→φℓ)+Γ(NI → φℓ)(recall N1 = N1)

(NB, no intermediate N1 because cplg combo real)

X λNI ℓα

φ

×λ∗ λ

λNJXNI ℓα

φ

φ

ℓ

+λ∗ λ λ

NJ

XNI ℓα

φ

φ

ℓ

19 / 25





[κ]αβ ∼[mν ]αβ

v2

Xλ

N1 ℓα

φ

×λ∗

κN1 ℓα

φ

φ

ℓ+

λ∗κN1 ℓα

φ

φ

ℓ

19 / 25






v2

Xλ

N1 ℓα

φ

×λ∗

κN1 ℓα

φ

φ

ℓ+

λ∗κN1 ℓα

φ

φ

ℓ

Recall

Γǫ ∼ Im

〈φℓ|T |N〉∗〈N |T |φℓ〉〈φℓ|T †|φℓ〉

19 / 25





Xλ

N1 ℓα

φ

×λ∗

κN1 ℓα

φ

φ

ℓ+

λ∗κN1 ℓα

φ

φ

ℓ

Recall

Γǫ ∼ Im


Γǫ ∝

∫dΠIm

M∗(N → φℓ)M(N → φℓ)M(φℓ→ φℓ)

ǫ1 ∼ MImλλκ∗

8π|λ|2

19 / 25






v2

Xλ

N1 ℓα

φ

×λ∗

κN1 ℓα

φ

φ

ℓ+

λ∗κN1 ℓα

φ

φ

ℓ

Recall

Γǫ ∼ Im


Γǫ ∝

∫dΠIm

M∗(N → φℓ)M(N → φℓ)M(φℓ→ φℓ)

ǫ1 ∼ MImλλκ∗

8π|λ|2<

3

8π

mmaxν M1

v2∼ 10−6 M1

109GeV>∼ 10−6

so for M1 ≪ M2,3, need M1>∼ 109 GeV to obtain sufficient ǫ

19 / 25

Maybe want MK < 109 GeV ? Leptogenesis with lighter NI ...

1. MI ∼ MJ ⇔ resonantly enhance ǫ ... up to ǫ <∼ 1.

2. at lower T , age of Universe is longer, takes more care to getout-of-equilibrium...

3. need to generate L asym before Electroweak PT (to profit fromsphalerons)...

⇒ leptogenesis via NI decay/scattering works for degen NI down to MI ∼ TeV

For even lighter MI , can make asym by coherent oscillations and scatterings ofNI ...

20 / 25

νMSM : type 1 seesaw below 100 GeV gives BAU and DMAsaka + Shaposhnikov

...thesis Canetti

ingredients : SM +

N2,3 : 100 MeV <∼ M2,3

<∼ 10 GeV, ∆M <

∼

10−6 eV YB ,ΩDM

keV YB ,NOT ΩDM

Yukawas ∋ give 2 light SM neutrinos via seesawN1 : M1 ∼ keV. WDM candidate.

feebly coupled (negligeable contribution mν,SM )

scenario :Population of N2,3 produced via Yukawas before EPTProduce ∆L → YB , before EPT, via oscillations of N2,3 coherent with Yukawascattering to νSMProduce ∆L >

∼ 10−5 via osc. and decay of N2,3 after EPT

Can produce sufficient distribution of N1 via osc.

tests :N2,3 : beam dump, SHIPN1 as DM : X -rays from DM decay, WDM bounds (depend on momentum distribution)

21 / 25

How does asym generation work ? (very simplified !)

1 at T <∼ TeV (recall λ <

∼ 10−7) , produce N2,N3 via Yukawa interaction λNℓ · φ

22 / 25



∼ 10−7) , produce N2,N3 via Yukawa interaction λNℓ · φ2 N2,N3 oscillate (almost degenerate)3 back to νL via λ

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∼ 10−7) , produce N2,N3 via Yukawa interaction λNℓ · φ2 N2,N3 oscillate (almost degenerate)3 back to νL via λat τU ∼ τosc , 1,2,3 are coherent, so CPV from λ-∆M2-λ gives flavour asyms inνLα (to small)

*lepton number in ℓL + NR is conserved* (actually, LSM+ helicity of NI )

from τosc → τEWPT , asyms in νLα seed asyms in N −→ asyms in νLα (enough asym)

...works also in detailed calculations with all available technology...(eg also include lepton number violating interactions)

Teresi HambyeEijima + Shaposhnikov

Ghiglieri+ Laine

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U2 = Tr[λM−2λ†]

arXiv :1208.460723 / 25

Summary

Leptogenesis is a class of recipes, that use (majorana) neutrino mass models togenerate the matter excess. The model generates a lepton asymmetry (beforethe Electroweak Phase Transition), and the non-perturbative SM B+L violnreprocesses it to a baryon excess.⋆ efficient, to use the BSM for mν to generate the Baryon Asym.⋆ using SM B+L violn (∆B = ∆L = 3) avoids proton lifetime bound

⋆ it works ...rather well, for a wide range of parameters

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more CPV in U ,mν if Majorana

• suppose that all parameters in L that can be complex (U and mν i), are complex

• 3 angles and 6 phases in generic unitary matrix U (18 real parameters in arbitrary

3 × 3 complex matrix. Unitarity UU† = 1 reduces this to 9.)

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• five relative phases between the fields eL, µL, τL, ν1, ν2, ν3 ...so can choose the5 relative phases among LH fermions, to remove all but one phase in the mixingmatrix.

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• five relative phases between the fields eL, µL, τL, ν1, ν2, ν3 ...so can choose the5 relative phases among LH fermions, to remove all but one phase in the mixingmatrix.• now check if can make the masses real : if dirac masses, absorb phase of masswith νRI . If νL3 has Majorana mass, between self and anti-self, choose absolutephase of νL3 to make the mass real. Now all LH fermion phases are fixed, andcannot remove phases from mν1,mν2.⇒ extra CPV in processes where Majorana mass appears linearly (not as mm∗,not in kinematics = not in oscillations)

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