Baryogenesis and Particle—Antiparticle Oscillations
Seyda Ipek UC Irvine
SI, John March-Russell, arXiv:1604.00009
Baryogenesis and ParticleAntiparticle Oscillations Seyda Ipek UC - - PowerPoint PPT Presentation
Baryogenesis and ParticleAntiparticle Oscillations Seyda Ipek UC Irvine SI, John March-Russell, arXiv:1604.00009 Sneak peek There is more matter than antimatter - baryogenesis SM cannot explain this There is baryon number
SI, John March-Russell, arXiv:1604.00009
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ΩΛ ∼ 0.69 ΩDM ∼ 0.27 ΩB ∼ 0.04
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number of baryons: η = nB n ¯
B
nγ ' 6 ⇥ 10−10
PDG
CMB
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Need to produce 1 extra quark for every 10 billion antiquarks!
Sakharov, JETP Lett. 5, 24 (1967)
Three conditions must be satisfied: 1) Baryon number (B) must be violated
can’t have a baryon asymmetry w/o violating baryon number!
2) C and CP must be violated
a way to differentiate matter from antimatter
3) B and CP violating processes must happen out of equilibrium
equilibrium destroys the produced baryon number
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2HDM, MSSM, NMSSM, …
Heavy right-handed neutrinos,…
asymmetric dark matter
+ Affleck-Dine
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Let’s re-visit SM CP violation
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We see SM CP violation through neutral meson mixing Bd − Bd K − K Bs − Bs D − D
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−Lmass = M ψ ψ + m 2 (ψc ψ + ψ ψc) Take a Dirac fermion with an approximately broken U(1) charge
Dirac mass Majorana mass
−Lint = g1 ψ X Y + g2 ψc X Y + h.c. with interactions ψ : pseudo-Dirac fermion
We will want the final state XY to carry either baryon or lepton number
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H = M − i 2Γ
M = ✓ M m m M ◆
Hamiltonian: eigenvalues: |ψH,Li = p|ψi ± q|ψci
Γ ' Γ ✓ 1 2 r eiφΓ 2 r e−iφΓ 1 ◆
r = |g2| |g1| ⌧ 1
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1/∆m
1/Γ
important parameter:
x ≡ ∆m Γ ∆m = MH ML ' 2m x ⌧ 1 x 1 x ∼ 1 Too fast Too slow Just right Goldilocks principle for oscillations
0.0 0.5 1.0 1.5 2.0 2.5 3.0 0.0 0.2 0.4 0.6 0.8 1.0 time Probability
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✏ = Z ∞ dt Γ( / c → f) − Γ( / c → ¯ f) Γ( / c → f) + Γ( / c → ¯ f) ✏ ' 2x r sin Γ 1 + x2
r = |g2| |g1| ⌧ 1
For
10-4 10-2 1 100 104 10-5 10-4 10-3 10-2 10-1 x = 2m / Γ ϵ
r = 0.1 sin φΓ = 0.5
exact approximation
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Say the final state f has baryon number +1
e.g. RPV SUSY ˜ g u d d
Baryon asymmetry is produced due to oscillations and decays:
✏ ' 2x r sin Γ 1 + x2
B = ✏ nψ
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Big Bang?
Time
M ∼ 300 GeV
H u b b l e r a t e
1 10 100 10-6 10-4 10-2 1 z = M/T Rates (eV)
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Big Bang?
Time
M ∼ 300 GeV
H u b b l e r a t e Decay rate
1 10 100 10-6 10-4 10-2 1 z = M/T Rates (eV)
Γ . H(T ∼ M) Out-of-equilibrium decay:
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Big Bang?
Time
M ∼ 300 GeV
Γ . H(T ∼ M) Out-of-equilibrium decay:
Oscillations start when
ωosc > H
H u b b l e r a t e Decay rate ωosc = 2 m
1 10 100 10-6 10-4 10-2 1 z = M/T Rates (eV)
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−Lscat = 1 Λ2 ¯ ψ Γaψ ¯ f Γaf
Particles/antiparticles are in a hot/dense plasma with interactions
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ωosc > Γann, Γscat
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Big Bang?
Time
further delayed
M ∼ 300 GeV
H u b b l e r a t e Decay rate ωosc = 2 m Elastic scattering rate A n n i h i l a t i
r a t e
1 10 100 10-6 10-4 10-2 1 z = M/T Rates (eV)
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O± = diag(1, ±1) H : Hamiltonian Y: Density matrix
zH dY dz = i
Γ± 2 [O±, [O±, Y]] shσvi± ✓1 2{Y, O± ¯ YO±} Y 2
eq
◆
Oscillations Vanishes if scatterings are flavor blind Annihilations
z = M/T
not redshift!
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m = 5 x 10-6 eV, x = 10 m = 2 x 10-6 eV, x = 4 m = 0.25 x 10-6 eV, x = 0.5
M = 300 GeV, Γ = 10−6 eV
Symmetric initial conditions: ∆(0) = 0 Oscillations are delayed for smaller m Smaller asymmetry
: particle asymmetry
10 20 30 40 50
0.00001 0.000012 0.000014 z Δ(z)
r = 0.1 sin φΓ = 0.5
∆(z) ≡ Yψ − Yψc
∆(z) = ✏ Yeq(1) exp ✓ − Γ 2H(z) ◆ sin2 ✓ m 2H(z) ◆
z = M/T
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: (massless) fermion
f Λ: interaction scale
flavor-sensitive
flavor-blind
e.g. scalar e.g. vector
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d2∆(y) dy2 + 2 ξ ω0 d∆(y) dy + ω2
0 ∆(y) = 0
ω0 ≡ m yH , ξ ≡ ΓS
ann/ΓV scat
2m
Ignoring decays, particle asymmetry is given by
ξ 1 ξ < 1
underdamped, system oscillates y = z2 z = M/T
∆(z) ≡ Yψ − Yψc
10-6 10-4 10-2 1 1 10 50 100 m (eV) zosc
σ0 = 1 ab σ0 = 1 fb
mass difference (eV)
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ωosc = Γann(zosc) flavor-blind interactions ωosc = Γscat(zosc) flavor-sensitive interactions
z = M/T
when
start
zosc
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: total number of particles
η
1 10 100 10-12 10-9 10-6 10-3 z Σ(z)
No annihilation <σv> = 10-2 ab <σv> = 1 ab
M = 300 GeV m − 2 × 10−6 eV Γ = 10−6 eV sin φΓ = 0.5 r = 0.1 − 0.3
Σ(z) ≡ Yψ + Yψc
z = M/T
flavor-sensitive
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Oscillations are delayed Smaller asymmetry
1 10 100 10-12 10-9 10-6 10-3 z Δ(z)
No annihilation <σv> = 10-2 ab <σv> = 1 ab
M = 300 GeV m − 2 × 10−6 eV Γ = 10−6 eV sin φΓ = 0.5 r = 0.1 − 0.3
∆(z) ' ✏ Yeq(zosc) exp ✓
2H(z) ◆ sin2 ✓ m 2H(z) ◆
z = M/T
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Annihilations freeze-out
H u b b l e r a t e Decay rate ωosc A n n i h i l a t i
r a t e
1 5 10 50 100 10-8 10-6 10-4 10-2 1 z = M/T Rates (eV)
zf ' 20
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Annihilations freeze-out
H u b b l e r a t e Decay rate ωosc A n n i h i l a t i
r a t e Elastic scattering rate
1 5 10 50 100 10-8 10-6 10-4 10-2 1 z = M/T Rates (eV)
Oscillations start
zosc ' 60 zf ' 20
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Annihilations freeze-out
H u b b l e r a t e Decay rate ωosc A n n i h i l a t i
r a t e Elastic scattering rate
1 5 10 50 100 10-8 10-6 10-4 10-2 1 z = M/T Rates (eV)
Oscillations start
zosc ' 60 Γann ⌧ Γ
Annihilations are Boltzmann suppressed
Oscillate a few times Produce the baryon asymmetry ⌘ ' ✏ Σ(zosc) ⇠ 10−10 zf ' 20
1 10 100 10-12 10-9 10-6 10-3 z Abundancies
σ0 = 1 ab σ0 = 10−2 ab
Σ(z) ∆(z) ∆B(z)
Flavor-sensitive
M = 300 GeV m = 2 × 10−6 eV Γ = 10−6 eV φΓ = π/6
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∆B = YB − YB : baryon asymmetry
η d∆B(z) dz ' ✏ Γ zH Σ(z)
For z > zosc baryon asymmetry is given by:
z = M/T
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Inflation Temperature Time
1014 K 1 ns 10 ns 13.7 billion years hot 1013 K 1 s 1010 K
p n
300,000 years 3,000 K
Big Bang
Equal number
antiparticles Antiparticles turn into particles Particles decay into protons and neutrons galaxies… are formed n n n n n n p p p p p n p Atoms, stars,
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global symmetries are broken by gravity
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Hall, Randall, Nuc.Phys.B-352.2 1991 Kribs, Poppitz, Weiner, arXiv: 0712.2039 Frugiuele, Gregoire, arXiv:1107.4634
Pilar Coloma, SI, arXiv:1606.06372
SI, John March-Russell, arXiv:1604.00009 SI, McKeen, Nelson, arXiv: 1407.8193
Dirac gauginos are awesome - less tuning for heavier stops Solves SUSY CP and flavor problems, … Has Dirac gauginos
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Fields SU(3)c SU(2)L U(1)Y U(1)R Q = ˜ q + θ q 3 2 1/6 1 ¯ U = ˜ ¯ u + θ ¯ u ¯ 3 1
1 ¯ D = ˜ ¯ d + θ ¯ d ¯ 3 1 1/3 1 Φ ¯
D = φ ¯ D + θ ψ ¯ D
¯ 3 1 1/3 1 ΦD = φD + θ ψD 3 1
1 W ˜
B,α ⊃ ˜
Bα 1 1 1 ΦS = φs + θ S 1 1
Bino has +1 R-charge Singlino (S) has -1 R-charge (pseudo)-Dirac gauginos! Sfermions have +1 R-charge
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Fields SU(3)c SU(2)L U(1)Y U(1)R Q = ˜ q + θ q 3 2 1/6 1 ¯ U = ˜ ¯ u + θ ¯ u ¯ 3 1
1 ¯ D = ˜ ¯ d + θ ¯ d ¯ 3 1 1/3 1 Φ ¯
D = φ ¯ D + θ ψ ¯ D
¯ 3 1 1/3 1 ΦD = φD + θ ψD 3 1
1 W ˜
B,α ⊃ ˜
Bα 1 1 1 ΦS = φs + θ S 1 1
New superfields non-gauge couplings for Dirac partners
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Take the bino and the singlino: −Lmass ⊃ MD ˜ BS + M ∗
D ˜
B†S† ˜ B ≡ (1, 1, 0)+1 S ≡ (1, 1, 0)−1
…because (anomaly mediation) (Small) Majorana mass for the bino m ˜
B = β(g)
g Fφ
some conformal parameter
m3
3/2
16π2M 2
Pl
. |Fφ| . m3/2
Arkani-Hamed, et al, hep-ph/0409232 Fox, Nelson, Weiner, hep-ph/0206096
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ACME, Science 343 (2014)
e e 𝛿
EW loop
What SUSY has:
eL eR ˜ eR ˜ eL
χ χ
𝛿
╳ ╳
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eL
eR ˜ eR ˜ eL χ
χ
𝛿
╳ ╳
Due to the UR(1) symmetry:
We can have large CP violating parameters w/o affecting EDMs
(similar story for the flavour problem)
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−Lmass → MDBS + 1 2 ⇣ m ˜
B ˜
B ˜ B + mSSS ⌘ + h.c.
Let’s also consider R-parity violation Mass terms: −Leff = G ˜
B ˜
B ¯ u ¯ d ¯ d + GSS ¯ u ¯ d ¯ d + h.c. −Lmass = M ψ ψ + m 2 (ψc ψ + ψ ψc)
antibino
−Lint = g1 ψ X Y + g2 ψc X Y + h.c. with interactions
bino
Remember from before: u d d u d d
GS ∼ gS λ00 m2
φ
GB ∼ gY λ00 m2
sf
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H = ✓ MD m m MD ◆ − i 2Γ ✓ 1 2reiφΓ 2re−iφΓ 1 ◆
Γ ' M 5 (32π)3 |G ˜
B|2
r = |GS| |G ˜
B| ⌧ 1
−Lscat = g2
Y
m2
sf
¯ ψγµPLψ ¯ Fγµ(gV + gAγ5)F with annihilations + elastic scatterings:
gV,A = Y 2
R ± Y 2 L
2 F = ✓ fL f †
R
◆
flavor sensitive, oscillations are delayed, etc etc
Σ(z) Δ(z) ΔB(z)
1 10 100 10-12 10-9 10-6 10-3 z Abundancies
Σ(z) ∆(z) ∆B(z)
msf = 10 TeV
msf = 20 TeV M = 300 GeV m − 2 × 10−6 eV Γ = 10−6 eV φΓ = π/6, r = 0.1
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z = M/T
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(no lighter than ~ 3 TeV)
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|ψ(t)i = g+(t)|ψi q pg−(t)|ψci, |ψc(t)i = g+(t)|ψci p q g−(t)|ψi g±(t) = 1 2 ⇣ e−imHt− 1
2 ΓHt ± e−imLt− 1 2 ΓLt⌘
Time dependent oscillations
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d2∆(y) dy2 + 2 ⇠ !0 d∆(y) dy + !2
0 ∆(y) = −✏ !2 0 Σ(y)
∆(z) ' A ✏Yeq(1) exp ✓
2H(z) ◆ sin2 ✓ m 2H(z) + ◆ Σ(z) = 2 Yeq(1) exp ✓ − Γ 2H(z) ◆ for z > 1 For: Solution is
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Γ = 10−7 eV Γ = 10−6 eV Γ = 10−5 eV Γ = 10−4 eV
M = 300 GeV φΓ = π/6, r = 0.1 σ0 = 1 ab
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zosc ∼ 6 r 2 × 10−6 eV m ✓ M 300 GeV ◆ zosc ∼ ln " 107 ✓ M 300 GeV ◆3 ✓2 × 10−6 eV m ◆ ⇣ σ0 1 fb ⌘# zosc ' 80 ✓ M 300 GeV ◆3/5 ✓2 ⇥ 10−6 eV m ◆1/5 ⇣ σ0 1 fb ⌘1/5 Hubble Flavor-blind Flavor-sensitive
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Rates (eV) z = M/T
Big Bang?
Time
M ∼ 300 GeV
1 5 10 50 100 500 10-8 10-6 10-4 10-2 1
Hubble rate Decay rate ωosc = 2m annihilation rate s c a t t e r i n g r a t e
hσvi = 1 fb