More about Unparticles

More about Unparticles

Bohdan GRZADKOWSKI University of Warsaw

• Deconstruction and spontaneous symmetry breaking • UnCosmology:

– The equation of state - hand-waving arguments – Freeze-out and thaw-in

Deconstruction and spontaneous symmetry breaking

h0|O_U(x)OU(0)|0i =

Z d4p (2π)2e −ipx ρU(p2) • Scaling: O_U(x) → O_U0 (x0) = s−dU_O U(x) for x → x0 = sx

• The spectral density: ρU(p2) =

Z

d4x eipxh0|O_U(x)OU(0)|0i ⇒ ρU(p2) = Ad_Uθ(p0)θ(p2)(p2)dU−2

• The phase space:

dΦU(pU) = Ad_Uθ(p0)θ(p2_U)(p2_U)dU−2 d4pU (2π)4 with AdU = 16π5/2 (2π)2dU Γ(dU + 1₂) Γ(dU − 1)Γ(2dU)

• Unparticles behave as a collection of dU massless particles ⇒ continuous spectrum

Deconstruction of unparticles

K¨allen-Lehman representation of the Feynman propagator:

i∆U_F(p2) = Z

d4xeipxh0|T {O_U(x)OU(0)}|0i =

Z ∞ 0 dm2 2π ρ(m 2 ) i p2 − m2 _{+ iε}

with ρU(m2) = Ad_Uθ(m2)(m2)dU−2. Deconstruction (Stephanov’07):

OU → ∞ X n=0 Fnϕn with m2_n = ∆2n Then i∆U_F(p2) = Z

So, the undeconstructed result has been confirmed. Now, let’s focus on the non-trivial phase: Im ( _∞ X n=0 F_n2 p2 − m2 n + iε ) = −X n F_n2πδ(p2 − m2_n) → ∆→0 − Ad_U 2 θ(p 2 )(p2)dU−2

So, each peak becomes lower as F_n2 ∼ ∆2 → 0, but their density increases.

• Each mode ϕn breaks the scale invariance.

• In the limit lim N →∞ N X n=0

The deconstruction for t → uOU decay i λ Λd_UU uγ¯ µ(1 − γ5)t ∂ µ_O U −→ i λ Λd_UU uγ¯ µ(1 − γ5)t ∞ X n=0 F_n∂µϕ_n ⇓ Γ(t → uϕn) = λ2 Λ2d_U U mtE_u2 2π F 2 n with Eu = m2_t − m2_n 2mt and F_n2 = AdU 2π ∆ 2 (m2_n)dU−2

Number of states |ϕ_ni in the interval (Eu, Eu + dEu): dN = dEu2m_∆2t

⇓ dΓ dEu = 2mt ∆2 Γ(t → u + ϕn) = λ2 Λ2d_U UAdU m2_t 2π2 E 2 u(m 2 t − 2mtEu) (d_U−2) θ(mt − 2Eu)

Spontaneous symmetry breaking with unparticles and Higgs boson physics

(Delgado, Espinosa, Quiros’07)

• The IR divergence!

• A possible regularization δ V 0 = ζ|H|2 P ϕ2_n is not scale invariant.

Since the scaling invariance is anyway violated by the vacuum expectation value 6= 0

through |H|2OU so we adopt

δ V 0 = ζ|H|2 X

n

ϕ2_n

as the IR regulator. Then

vn = hϕni = −

κUv2

2(m2

n + ζv2)

Fn

The minimization for H reads:

m2 + λv2 + κU X n Fnvn + ζ X n v_n2 = 0

Inserting vn one gets in the continuum limit (∆ → 0):

for λU ≡ dU 4 ζ d_U−2 Γ(dU − 1)Γ(2 − dU) and (µ2_U)2−dU ≡ κ2_U Ad_U 2π Veff = m2|H|2− 2dU−1 dU λU(µ2_U)2−dU|H|2dU + λ|H|4

Even if m2 = 0 one can get the vacuum expectation value 6= 0 (ΛU provides the

The equation of state - hand-waving arguments

(in progress with Jose Wudka)

The trace anomaly of the energy momentum tensor for a gauge theory with massless fermions:

θ_µµ = β

2gN [F

µν

a Fa µν] (1)

where β denotes the beta function and N stands for the normal product.

Non-trivial IR fixed point at g = g?, so in the IR we assume

β = γ(g − g?), γ > 0

in which case the running coupling reads

g(µ) = g? + cµγ; β[g(µ)] = γcµγ

where c is an integration constant and µ is the renormalization scale.

From the thermal average of (1) choosing the renormalization scale µ = T and

using hθ_µµi = ρ_U − 3p_U, we get

ρU − 3pU =

β 2g?

ρU − 3pU = AT4+γ ⇓ ρU = σT4 + A  1 + 3 γ  T4+γ and pU = σ T4 3 + A γ T 4+γ

where σ is an integration constant.

⇓ pU = 1₃ρU  1 − Bργ/4_U  for B ≡ A σ1+γ/4

One can expect that A ∝ Λ−γ_U , therefore we obtain

ρNP = π 2 30T 4 _× ( gIR + f  T Λ_U γ for T <_{∼ Λ}U gBZ for T >_{∼ Λ}U

where gBZ = 2(n2_c − 1 + 7₈ncnf) for SU (nc) with nf flavours in the BZ sector.

• From the continuity at T = ΛU, the constant f could be determined: f = gBZ−gIR.

ρNP = π 2 30T 4 _× ( gIR + f  T Λ_U γ for T <_{∼ Λ}U gBZ for T >_{∼ Λ}U Deconstruction (Stephanov’07): OU → ∞ X n=0 Fnϕn with m2_n = ∆2n

The above result fits the following guess for the effective number of degrees of freedom: gU(T ) ∝ R T2 0 dM 2_ρ(M2_)θ(Λ2 U − M2) R Λ2_U 0 dM 2_ρ(M2₎ where ρ(M2) ∝ (M2)(dU−2)_{. Then} gU(T ) ∝  T ΛU 2(dU−1)

Freeze-out and thaw-in

♣ Brief history of the Universe in the presence of unparticles (no mass-gap).

• T  MU: the BZ sector in form of massless particles (no unparticles yet), thermal

equilibrium with the SM is maintained (assumption), so T = TBZ = TSM

• T <_{∼ M}U:

– The BZ sector starts to decouple, as the average energy is no longer sufficient

to create mediators.

– However, the thermal equilibrium may still be maintained (T = TBZ = TSM)

depending on the strength of effective couplings between the SM and the extra

sector (which at higher temperature, T >_{∼ Λ}U, is made of the BZ matter, while

below ΛU of unparticles).

Let’s denote by T_f the decoupling temperature at which

Γ(SM ↔ N P ) ' H

where H is the Hubble parameter

H2 = 8π

There are 2 interesting cases: • MU > Tf > ΛU:

– Tf is determined by the condition

Γ(SM ↔ BZ) ' H

– For T > Tf the SM and the BZ sectors evolve in thermal equilibrium, but even

for T < Tf their temperatures remain equal (T = TBZ = TSM) since ΛU > v.

• ΛU > Tf:

– Till T = ΛU the SM and unparticles still have the same temperature.

– For ΛU >_{∼ T >∼ T}f still the equilibrium is maintained (assumption, in general this

depends on dU). The decoupling temperature Tf must be now determined by

Γ(SM ↔ OU) ' H

– Till T ∼ v temperatures of SM and unparticles remain equal, at T ∼ v they

split.

♣ The Banks-Zaks phase. L_BZ = 1 MU φ†φ
(¯qBZqBZ) Then ΓBZ ∝ T3 M_U2 and H ∝ T2 MP l =⇒ decoupling for T <_∼ Tf −BZ

♣ The unparticle phase.

L_U = cU

Λd_UBZ−dU

M_Uk OUOSM for k = dSM + dBZ − 4

The most relevant operators for scalar unparticles are

ΓU

H ∝ T

2d_U−5

=⇒



dU > 5₂ decoupling for T < Tf −U freeze-out

dU < 5₂ decoupling for T > Tf −U thaw-in

BBN constraints

Big-Bang Nucleosynthesis =⇒ ∆ρrad

ρtot    T =T_BBN < 7% (95% CL) ⇓ ∆ρU ρtot     T =T_BBN < 7% Assume ΛU > Tf-_U > v = 246 GeV. • d(s)_U > 5₂ decoupling for _{T < Tf-U} ρU = π2 30 gIR T 4  gγ gSM gν + gγe gγe 4/3 | {z } 1.2·10−2 and ρSM = π2 30 gγν T 4 ⇓ ∆ρU ρtot     T =T_BBN < 7% =⇒ gIR <∼ 20

To be compared with e.g. gBZ = 2(n2_c − 1 + 7₈ncnf), for nc = 3 and nf = 10,

• d(s)_U < 5₂ decoupling for T > Tf −U

An operator responsible for keeping the equilibrium down (below T ∼ mH where

L_s becomes irrelevant) to TBBN is needed:

L_v = c(v)_U Λ 3−d_U M_U3 (BµνB µν ) OU with d (v) U < 1 2

– Note that L_v could be generated radiatively through O_U − H mixing (from L_s).

Assuming the equilibrium down to the BBN temperature TBBN ∼ 0.1 MeV we

Summary

• Rough arguments for the equation of state for unparticles: pU = 1₃ρU

h

1 − Bρδ/4_U i

• Rough arguments for the energy density for unparticles ”derived”:

ρNP = π2 30T 4 _×     gIR + (gBZ − gIR)  T Λ_U δ for T <_{∼ Λ}U gBZ for T >_{∼ Λ}U

• Unparticles in equilibrium: freeze-out and thaw-in.

• BNN bounds on the number of degrees of freedom for unparticles. Things to be done: