Axion as a Dark matter 20195513 Minsang Yu 1 Contents 1. - PowerPoint PPT Presentation

Axion as a Dark matter 20195513 Minsang Yu 1

Contents 1. Spontaneous Symmetry Breaking (SSB) U 1 A problem 2. 3. Strong CP problem 4. Axion Search 5. Summary 2

1. Spontaneous Symmetry Breaking (SSB) • Consider a simple Lagrangian for a (real) scalar field: ℒ = 1 2 𝜖 𝜈 𝜚𝜖 𝜈 𝜚 − 𝑊 𝜚 • For a ”symmetric” potential: “ Symmetric ” potential means: 𝑊 −𝜚 = 𝑊 𝜚 2 𝜈 2 𝜚 2 + 𝜇 4 𝑊 𝜚 = − 1 4 𝜚 4 • Our goal: Take a look at the ground states! 3

1. Spontaneous Symmetry Breaking (SSB) • Here, It is known that : “ constant configuration of scalar field such that minimizes the potential term also minimizes the kinetic term at the same time. ” 𝜖𝑊 𝜖𝐿 𝜚 = 𝑑𝑝𝑜𝑡𝑢𝑏𝑜𝑢 , 𝜖𝜚 = 0 𝜖𝜚 = 0 4

1. Spontaneous Symmetry Breaking (SSB) • For simplicity, let ’ s consider only two potentials: 𝑊 𝜚 = − 𝟐 𝟑 𝜚 2 + 1 𝑊 𝜚 = + 𝟐 𝟑 𝜚 2 + 1 4 𝜚 4 4 𝜚 4 5

1. Spontaneous Symmetry Breaking (SSB) 𝜖𝑊 • constant field solutions for 𝜖𝜚 = 0 𝑊 𝜚 = − 𝟐 𝟑 𝜚 2 + 1 4 𝜚 4 1. ϕ = 0 • Question: Does the solution enjoys the same symmetry as the potential? Yes! 1. ϕ = 0 = (−0) 𝜚 = 0 6

1. Spontaneous Symmetry Breaking (SSB) 𝜖𝑊 • constant field solutions for 𝜖𝜚 = 0 𝑊 𝜚 = + 𝟐 𝟑 𝜚 2 + 1 4 𝜚 4 1. ϕ = 0 2. ϕ = +1 3. ϕ = −1 • Question: Does the solution enjoys the same symmetry as the potential? 𝜚 = 0 Yes! 1. ϕ = 0 = (−0) No … 2. ϕ = +1 ≠ −(+1) 𝜚 = 1 𝜚 = −1 No … 3. ϕ = −1 ≠ −(−1) Two solutions do not follow the symmetry of the potential 7

1. Spontaneous Symmetry Breaking (SSB) • For more intuition, let’s consider some perturbations: 𝜚 𝑦 = 𝜚 0 + 𝜀𝜚(𝑦) derivative of constant is zero. ℒ 𝜚 = 1 𝜖 𝜈 𝜚 0 + 𝜀𝜚 𝑦 2 𝜖 𝜈 𝜚 0 + 𝜀𝜚 𝑦 − V 𝜚 0 + 𝜀𝜚 𝑦 → ℒ 𝜚 = 1 2 𝜖 𝜈 𝜀𝜚𝜖 𝜈 𝜀𝜚 + 1 2 − 1 4 2 𝜚 0 + 𝜀𝜚 𝑦 4 𝜚 0 + 𝜀𝜚 𝑦 8

1. Spontaneous Symmetry Breaking (SSB) • For more intuition, let ’ s consider some perturbations: 𝜚 𝑦 = 𝜚 0 + 𝜀𝜚(𝑦) 2 − 1 4 1 1 For 𝜚 = 0 𝜀𝜚 → −𝜀𝜚 2 𝜖 𝜈 𝜀𝜚𝜖 𝜈 𝜀𝜚 + → ℒ 0 = 2 𝜀𝜚 𝑦 4 𝜀𝜚 𝑦 2 − 1 4 For 𝜚 = ±1 → ℒ ±1 = 1 1 𝜀𝜚 → −𝜀𝜚 2 𝜖 𝜈 𝜀𝜚𝜖 𝜈 𝜀𝜚 + 2 ±1 + 𝜀𝜚 𝑦 4 ±1 + 𝜀𝜚 𝑦 • What governs the underlying physics is potential, and it has a symmetry. • What we see in our experience is small perturbations, and the symmetry is “ broken ” . (or, “ hidden ” . There IS always a symmetry, but we just don ’ t see it explicitly. ) 9

1. Spontaneous Symmetry Breaking (SSB) OK. We got SSB with a real scalar field. How about a complex scalar field? 10

1. Spontaneous Symmetry Breaking (SSB) Real scalar field Complex scalar field ℒ = 1 ℒ = 1 2 𝜖 𝜈 𝜚 ∗ 𝜖 𝜈 𝜚 − 𝑊 𝜚 2 𝜖 𝜈 𝜚𝜖 𝜈 𝜚 − 𝑊 𝜚 2 𝜈 2 𝜚 2 + 𝜇 4 2 𝜈 2 𝜚 2 + 𝜇 4 𝑊 𝜚 = − 1 𝑊 𝜚 = − 1 4 𝜚 4 4 𝜚 4 𝜚 → 𝜚𝑓 𝑗𝜄 𝜚 → −𝜚 11

1. Spontaneous Symmetry Breaking (SSB) 4 𝜚 4 ( 𝜈 2 < 0 ) 𝜇 4 1 2 𝜈 2 𝜚 2 + 𝑊 𝜚 = − Symmetric solution: Re 𝜚 = 0, Im 𝜚 = 0 Non-Symmetric solution: Re 𝜚 = 𝜈 𝜇 , Im 𝜚 = 0 Re 𝜚 Q: If you also consider some perturbations here: Im 𝜚 A1: sym. solution falls to non-sym. solution. A2: non-sym. solution climbs the valley, or spins around. A degree of freedom is equivalent to a new particle. It’s called a Goldstone boson . 12

ҧ 1. Spontaneous Symmetry Breaking (SSB) • In strong interaction, ത 𝑒𝑒 ≠ 0 , breaking the axial symmetry U 1 𝐵 . 𝑣𝑣 , • Therefore, there should be a Goldstone boson related to the SSB. • There were two possibilities: 1. pion “ eats ” the goldstone boson and gets a polarization. 2. or, another particle lighter than pion should exist. • … and none of them was observed. This is so called ‘ 𝐕 𝟐 𝑩 problem ’ . 13

2. U 1 A problem • OK. Something is wrong. But where did we miss? “ Maybe we used wrong boundary conditions on QCD vacuum. ” Gerard ‘t Hooft 14

2. U 1 A problem classical Quantum level • Baryonic current: 𝑕 2 𝑏𝜈𝜉 = 𝜖 𝜈 𝐿 𝜈 (from Ward-Takahashi identity) 𝜈 = 𝜈 = 0 𝜈𝜉 𝐺 𝜖 𝜈 𝐾 𝐶 32𝜌 2 𝐺 𝜖 𝜈 𝐾 𝐶 𝑏 (Follows least action principle in average, but there are some offsets.) 𝑕 2 • Quantum corrections on action: 32𝜌 2 න 𝑒 4 𝑦𝜖 𝜈 𝐿 𝜈 𝜀𝑋 = 𝜈 = 0 • Naïve boundary condition: 𝐵 𝑏 න 𝑒 4 𝑦𝜖 𝜈 𝐿 𝜈 = 0 𝜀𝑋 = 0 (at spatial infinity) • ‘ t Hooft boundary condition: 0 𝜈 = ቊ න 𝑒 4 𝑦𝜖 𝜈 𝐿 𝜈 ≠ 0 𝐵 𝑏 𝑕𝑏𝑣𝑕𝑓 𝑢𝑠𝑏𝑜𝑡𝑔𝑝𝑠𝑛 𝑝𝑔 0 𝜀𝑋 ≠ 0 (at spatial infinity) 15

2. U 1 A problem • Actually, using ‘t Hooft B.C. : න 𝑒 4 𝑦𝜖 𝜈 𝐿 𝜈 ∝ 𝑂 𝑗𝑜𝑢𝑓𝑕𝑓𝑠 • It means that there are n-vacua with the same energy (=degenerated): (𝑜 = 0) ⇒ corresponding set 𝐵 𝜈 0 n-vacua (𝑜 = 1) ⇒ corresponding set 𝐵 𝜈 1 {|𝑂 = 𝑜⟩} … • Vacuum transition • Classical: no transition btw different vacuum: 𝑜 𝑛 = 0 • Quantum: transition is OK: 𝑜 𝑛 ≠ 0 16

2. U 1 A problem • Well, to bypass the “transition - able” vacuum, How about… 𝜄 𝜄 ′ = 0. 𝑓 −𝑗𝑜𝜄 |𝑜⟩ 𝜄 = ෍ 𝜄: 𝑔𝑠𝑓𝑓 𝑞𝑏𝑠𝑏𝑛𝑓𝑢𝑓𝑠 𝑢ℎ𝑓𝑜 𝑜 • OK. Now let’s see what happens to action 𝑇 in order to make 𝜄 as the eigenstate of the ℒ : 𝑇 𝑓𝑔𝑔 = 𝑇 𝑅𝐷𝐸 + 𝜄 𝑕 2 𝜈𝜉 ത 32𝜌 2 න 𝑒 4 𝑦𝐺 𝐺 𝑏𝜈𝜉 𝑏 17

2. U 1 A problem • One last thing: • If we assume quark mass as real, we should get some proper coordinate. • Such coordinate can be obtained through chiral transformation: ҧ 𝜄 = 𝜄 + arg det 𝑁 𝜄 𝑕 2 𝜈𝜉 ത 32𝜌 2 න 𝑒 4 𝑦𝐺 𝑇 𝑓𝑔𝑔 = 𝑇 𝑅𝐷𝐸 + ҧ 𝐺 𝑏𝜈𝜉 𝑏 a CP-violating term 18

2. U 1 A problem 𝜄 𝑕 2 𝜈𝜉 ത 32𝜌 2 න 𝑒 4 𝑦𝐺 𝑇 𝑓𝑔𝑔 = 𝑇 𝑅𝐷𝐸 + ҧ 𝐺 𝑏𝜈𝜉 𝑏 a CP-violating term “ OK. Now we know why U 1 A problem happens. ” “ It ’ s because of the CP violating term. Problem solved! ” “ Now let ’ s go get ҧ 𝜄 ! ” (As ҧ 𝜄 is a free parameter, it lies somewhere on 0, 2𝜌 .) 19

2. U 1 A problem 𝜄 ~10 −10 (extremely small) 𝑒 𝑜 ~5 ∙ 10 −16 ഥ ഥ 𝜄 𝑓 ∙ 𝑑𝑛 ⇒ 20

2. U 1 A problem • 𝜄 is a free parameter in 0, 2𝜌 . • So it ’ s totally OK for ҧ 𝜄 being any value. ANY value. • But why ZERO?? • … this is the famous strong CP problem : “ Why there ’ s no sign of strong interaction violating CP? ” 21

ҧ 3. Strong CP problem • Let’s do reverse engineering for ҧ 𝜄 being zero: 𝜄 𝑕 2 𝜈𝜉 ത 32𝜌 2 න 𝑒 4 𝑦𝐺 𝑇 𝑓𝑔𝑔 = 𝑇 𝑅𝐷𝐸 + ҧ 𝐺 𝑏𝜈𝜉 𝑏 𝜄 = 𝜄 + arg det 𝑁 chiral transformation (for real quark mass) 𝜄 22

ҧ ҧ 3. Strong CP problem • Let ’ s do reverse engineering for ҧ 𝜄 being zero: “ Maybe there is something that cancels ഥ 𝜾 at the ground state. ” 𝑕 2 𝜈𝜉 ത 32𝜌 2 න 𝑒 4 𝑦𝐺 𝑇 𝑓𝑔𝑔 = 𝑇 𝑅𝐷𝐸 + 𝜄 + 𝑡𝑝𝑛𝑓𝑢ℎ𝑗𝑜𝑕 𝐺 𝑏𝜈𝜉 𝑏 𝜄 + 𝑡𝑝𝑛𝑓𝑢ℎ𝑗𝑜𝑕 = 𝜄 + arg det 𝑁 + 𝑡𝑝𝑛𝑓𝑢ℎ𝑗𝑜𝑕 chiral transformation (for real quark mass) 𝜄 23

ҧ ҧ 3. Strong CP problem • Let ’ s do reverse engineering for ҧ 𝜄 being zero: “ Maybe there is something that cancels ഥ 𝜾 at the ground state. ” 𝑕 2 𝜄 + 𝑏 𝑦 𝜈𝜉 ത 32𝜌 2 න 𝑒 4 𝑦𝐺 𝑇 𝑓𝑔𝑔 = 𝑇 𝑅𝐷𝐸 + 𝐺 𝑏𝜈𝜉 𝑏 𝑔 𝑏 𝜄 + 𝑏 𝑦 = 𝜄 + arg det 𝑁 + 𝑏 𝑦 𝑔 𝑔 𝑏 𝑏 chiral transformation (for real quark mass) Peccei and Quinn: “ something is the axion field a(x) ” 𝜄 U 1 PQ : 𝑏 𝑦 ⟶ 𝑏 𝑦 + 𝛽𝑔 𝑏 24

ҧ ҧ 3. Strong CP problem • What happens when U 1 PQ is imposed? 1. A particle corresponding to U 1 PQ is assumed: 𝑅 = 𝜍𝑓 𝑗𝑏 𝑦 /𝑔 𝑏 𝑕 2 𝜄 + 𝑏 𝑦 2. Such particle ’ s mass has a form: 𝜈𝜉 ത 32𝜌 2 න 𝑒 4 𝑦𝐺 𝑇 𝑓𝑔𝑔 = 𝑇 𝑅𝐷𝐸 + 𝐺 𝑏 𝑏𝜈𝜉 𝑛 𝑏 ~𝑓 𝑗𝑏 𝑦 /𝑔 𝑔 𝑏 𝑏 𝜄 + 𝑏 𝑦 = 𝜄 + arg det 𝑁 + 𝑏 𝑦 3. Such particle also experiences chiral transformation 𝑔 𝑔 (or, the mass term is now added to M) 𝑏 𝑏 chiral transformation (for real quark mass) 𝜄 25

3. Strong CP problem • Now, let ’ s do what we did on SSB section: 𝜖𝑊 𝑓𝑔𝑔 • Question: “ find 𝑏 that = 0 ” 𝜖𝑏 • Answer: “ The minima happens at 𝑏 = − ҧ 𝑏 ” 𝜄𝑔 𝑏 • Then: ҧ 𝑏 = 0 at the ground state! → CP violation might not be observed! 𝜄 + 𝑔 26