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) 2. U 1 A problem 3. Strong CP problem 4. Axion Search 5. Summary

2

• Consider a simple Lagrangian for a (real) scalar field:
• For a ”symmetric” potential:
• Our goal: Take a look at the ground states!
• 1. Spontaneous Symmetry Breaking (SSB)

ℒ = 1 2 𝜖𝜈𝜚𝜖𝜈𝜚 − 𝑊 𝜚 𝑊 𝜚 = − 1 2 𝜈2𝜚2 + 𝜇4 4 𝜚4

“Symmetric” potential means: 𝑊 −𝜚 = 𝑊 𝜚

3

• 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.”

• 1. Spontaneous Symmetry Breaking (SSB)

𝜚 = 𝑑𝑝𝑜𝑡𝑢𝑏𝑜𝑢,

𝜖𝑊 𝜖𝜚 = 0 𝜖𝐿 𝜖𝜚 = 0

4

• For simplicity, let’s consider only two potentials:
• 1. Spontaneous Symmetry Breaking (SSB)

𝑊 𝜚 = − 𝟐 𝟑 𝜚2 + 1 4 𝜚4 𝑊 𝜚 = + 𝟐 𝟑 𝜚2 + 1 4 𝜚4

5

• constant field solutions for

𝜖𝑊 𝜖𝜚 = 0

1. ϕ = 0

• Question: Does the solution enjoys the same

symmetry as the potential? 1. ϕ = 0 = (−0) Yes!

• 1. Spontaneous Symmetry Breaking (SSB)

𝑊 𝜚 = − 𝟐 𝟑 𝜚2 + 1 4 𝜚4 𝜚 = 0

6

𝑊 𝜚 = + 𝟐 𝟑 𝜚2 + 1 4 𝜚4

• constant field solutions for

𝜖𝑊 𝜖𝜚 = 0

1. ϕ = 0 2. ϕ = +1 3. ϕ = −1

• Question: Does the solution enjoys the same

symmetry as the potential? 1. ϕ = 0 = (−0) Yes! 2. ϕ = +1 ≠ −(+1) No… 3. ϕ = −1 ≠ −(−1) No…

• 1. Spontaneous Symmetry Breaking (SSB)

𝜚 = 0 𝜚 = 1 𝜚 = −1

Two solutions do not follow the symmetry of the potential

7

• For more intuition, let’s consider some perturbations: 𝜚 𝑦 = 𝜚0 + 𝜀𝜚(𝑦)

ℒ 𝜚 = 1 2 𝜖𝜈 𝜚0 + 𝜀𝜚 𝑦 𝜖𝜈 𝜚0 + 𝜀𝜚 𝑦 − V 𝜚0 + 𝜀𝜚 𝑦 → ℒ 𝜚 = 1 2 𝜖𝜈𝜀𝜚𝜖𝜈𝜀𝜚 + 1 2 𝜚0 + 𝜀𝜚 𝑦

2 − 1

4 𝜚0 + 𝜀𝜚 𝑦

4

• 1. Spontaneous Symmetry Breaking (SSB)

derivative of constant is zero.

8

• For more intuition, let’s consider some perturbations: 𝜚 𝑦 = 𝜚0 + 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.)

• 1. Spontaneous Symmetry Breaking (SSB)

For 𝜚 = 0 → ℒ0 =

1 2 𝜖𝜈𝜀𝜚𝜖𝜈𝜀𝜚 + 1 2 𝜀𝜚 𝑦 2 − 1 4 𝜀𝜚 𝑦 4

For 𝜚 = ±1 → ℒ±1= 1

2 𝜖𝜈𝜀𝜚𝜖𝜈𝜀𝜚 + 1 2 ±1 + 𝜀𝜚 𝑦 2 − 1 4 ±1 + 𝜀𝜚 𝑦 4

𝜀𝜚 → −𝜀𝜚 𝜀𝜚 → −𝜀𝜚

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)

ℒ = 1 2 𝜖𝜈𝜚𝜖𝜈𝜚 − 𝑊 𝜚 𝑊 𝜚 = − 1 2 𝜈2𝜚2 + 𝜇4 4 𝜚4

11

Real scalar field

ℒ = 1 2 𝜖𝜈𝜚∗𝜖𝜈𝜚 − 𝑊 𝜚 𝑊 𝜚 = − 1 2 𝜈2 𝜚 2 + 𝜇4 4 𝜚 4

Complex scalar field

𝜚 → −𝜚 𝜚 → 𝜚𝑓𝑗𝜄

• 1. Spontaneous Symmetry Breaking (SSB)

12

𝑊 𝜚 = −

1 2 𝜈2 𝜚 2 + 𝜇4 4 𝜚 4 (𝜈2 < 0)

Symmetric solution: Re 𝜚 = 0, Im 𝜚 = 0 Non-Symmetric solution: Re 𝜚 = 𝜈

𝜇 , Im 𝜚 = 0

Re 𝜚 Im 𝜚

Q: If you also consider some perturbations here: 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.

• 1. Spontaneous Symmetry Breaking (SSB)
• In strong interaction, ത

𝑣𝑣 , ҧ 𝑒𝑒 ≠ 0, breaking the axial symmetry U 1 𝐵.

• 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.

• r, 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?

14

“Maybe we used wrong boundary conditions on QCD vacuum.”

Gerard ‘t Hooft

• 2. U 1 A problem
• Baryonic current:
• Quantum corrections on action:
• Naïve boundary condition:
• ‘t Hooft boundary condition:

15

𝜖𝜈 𝐾𝐶

𝜈 = 0

classical

𝜖𝜈 𝐾𝐶

𝜈 = 𝑕2 32𝜌2 𝐺 𝑏 𝜈𝜉𝐺 𝑏𝜈𝜉 = 𝜖𝜈𝐿𝜈 (from Ward-Takahashi identity)

(Follows least action principle in average, but there are some offsets.)

Quantum level

𝜀𝑋 = 𝑕2 32𝜌2 න 𝑒4𝑦𝜖𝜈𝐿𝜈 𝐵𝑏

𝜈 = 0

(at spatial infinity)

න 𝑒4𝑦𝜖𝜈𝐿𝜈 = 0 𝜀𝑋 = 0

𝐵𝑏

𝜈 = ቊ

𝑕𝑏𝑣𝑕𝑓 𝑢𝑠𝑏𝑜𝑡𝑔𝑝𝑠𝑛 𝑝𝑔 0 (at spatial infinity)

න 𝑒4𝑦𝜖𝜈𝐿𝜈 ≠ 0 𝜀𝑋 ≠ 0

• 2. U 1 A problem
• Actually, using ‘t Hooft B.C. :
• It means that there are n-vacua with the same energy (=degenerated):
• Vacuum transition
• Classical: no transition btw different vacuum: 𝑜 𝑛 = 0
• Quantum: transition is OK: 𝑜 𝑛 ≠ 0

16

න 𝑒4𝑦𝜖𝜈𝐿𝜈 ∝ 𝑂 𝑗𝑜𝑢𝑓𝑕𝑓𝑠

n-vacua {|𝑂 = 𝑜⟩} (𝑜 = 0) ⇒ corresponding set 𝐵𝜈 0 (𝑜 = 1) ⇒ corresponding set 𝐵𝜈 1 …

• 2. U 1 A problem
• Well, to bypass the “transition-able” vacuum, How about…
• OK. Now let’s see what happens to action 𝑇 in order to make 𝜄 as the

eigenstate of the ℒ:

17

𝜄 = ෍

𝑜

𝑓−𝑗𝑜𝜄|𝑜⟩ 𝜄: 𝑔𝑠𝑓𝑓 𝑞𝑏𝑠𝑏𝑛𝑓𝑢𝑓𝑠 𝜄 𝜄′ = 0. 𝑢ℎ𝑓𝑜 𝑇𝑓𝑔𝑔 = 𝑇𝑅𝐷𝐸 + 𝜄 𝑕2 32𝜌2 න 𝑒4 𝑦𝐺

𝑏 𝜈𝜉 ത

𝐺

𝑏𝜈𝜉

• 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

18

𝑇𝑓𝑔𝑔 = 𝑇𝑅𝐷𝐸 + ҧ 𝜄 𝑕2 32𝜌2 න 𝑒4 𝑦𝐺

𝑏 𝜈𝜉 ത

𝐺

𝑏𝜈𝜉

a CP-violating term

• 2. U 1 A problem

19

“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 .)

𝑇𝑓𝑔𝑔 = 𝑇𝑅𝐷𝐸 + ҧ 𝜄 𝑕2 32𝜌2 න 𝑒4 𝑦𝐺

𝑏 𝜈𝜉 ത

𝐺

𝑏𝜈𝜉

a CP-violating term

• 2. U 1 A problem

20

𝑒𝑜 ~5 ∙ 10−16 ഥ 𝜄 𝑓 ∙ 𝑑𝑛 ⇒ ഥ 𝜄 ~10−10 (extremely small)

• 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:

21

“Why there’s no sign of strong interaction violating CP?”

• 3. Strong CP problem
• Let’s do reverse engineering for ҧ

𝜄 being zero:

22

𝑇𝑓𝑔𝑔 = 𝑇𝑅𝐷𝐸 + ҧ 𝜄 𝑕2 32𝜌2 න 𝑒4 𝑦𝐺

𝑏 𝜈𝜉 ത

𝐺

𝑏𝜈𝜉

ҧ 𝜄 = 𝜄 + arg det 𝑁 𝜄

chiral transformation

(for real quark mass)

• 3. Strong CP problem
• Let’s do reverse engineering for ҧ

𝜄 being zero:

23

𝑇𝑓𝑔𝑔 = 𝑇𝑅𝐷𝐸 + ҧ 𝜄 + 𝑡𝑝𝑛𝑓𝑢ℎ𝑗𝑜𝑕 𝑕2 32𝜌2 න 𝑒4 𝑦𝐺

𝑏 𝜈𝜉 ത

𝐺

𝑏𝜈𝜉

“Maybe there is something that cancels ഥ 𝜾 at the ground state.” ҧ 𝜄 + 𝑡𝑝𝑛𝑓𝑢ℎ𝑗𝑜𝑕 = 𝜄 + arg det 𝑁 + 𝑡𝑝𝑛𝑓𝑢ℎ𝑗𝑜𝑕 𝜄

chiral transformation

(for real quark mass)

• 3. Strong CP problem
• Let’s do reverse engineering for ҧ

𝜄 being zero:

24

𝑇𝑓𝑔𝑔 = 𝑇𝑅𝐷𝐸 + ҧ 𝜄 + 𝑏 𝑦 𝑔

𝑏

𝑕2 32𝜌2 න 𝑒4 𝑦𝐺

𝑏 𝜈𝜉 ത

𝐺

𝑏𝜈𝜉

“Maybe there is something that cancels ഥ 𝜾 at the ground state.” ҧ 𝜄 + 𝑏 𝑦 𝑔

𝑏

= 𝜄 + arg det 𝑁 + 𝑏 𝑦 𝑔

𝑏

𝜄

chiral transformation

(for real quark mass)

Peccei and Quinn: “something is the axion field a(x)” U 1 PQ: 𝑏 𝑦 ⟶ 𝑏 𝑦 + 𝛽𝑔

𝑏

• 3. Strong CP problem
• What happens when U 1 PQ is imposed?

25

𝑇𝑓𝑔𝑔 = 𝑇𝑅𝐷𝐸 + ҧ 𝜄 + 𝑏 𝑦 𝑔

𝑏

𝑕2 32𝜌2 න 𝑒4 𝑦𝐺

𝑏 𝜈𝜉 ത

𝐺

𝑏𝜈𝜉

ҧ 𝜄 + 𝑏 𝑦 𝑔

𝑏

= 𝜄 + arg det 𝑁 + 𝑏 𝑦 𝑔

𝑏

𝜄

chiral transformation

(for real quark mass)

• 1. A particle corresponding to U 1 PQ is assumed:

𝑅 = 𝜍𝑓𝑗𝑏 𝑦 /𝑔

𝑏

• 2. Such particle’s mass has a form:

𝑛𝑏~𝑓𝑗𝑏 𝑦 /𝑔

𝑏

• 3. Such particle also experiences chiral transformation

(or, the mass term is now added to M)

• 3. Strong CP problem
• Now, let’s do what we did on SSB section:
• Question: “find 𝑏 that

𝜖𝑊𝑓𝑔𝑔 𝜖𝑏

= 0”

• Answer: “The minima happens at 𝑏 = − ҧ

𝜄𝑔

𝑏”

• Then: ҧ

𝜄 +

𝑏 𝑔

𝑏 = 0 at the ground state! → CP violation might not be observed! 26

• 4. Axion Search
• A rough approximation of axion mass: 𝑛𝑏~

Λ𝑅𝐷𝐸

2

𝑔

𝑏 ~6𝜈𝑓𝑊

1012𝐻𝑓𝑊 𝑔

𝑏

• How to detect?
• wait for the axion spontaneously decay into photons.
• 𝑢𝑚𝑗𝑔𝑓𝑢𝑗𝑛𝑓 ≫ 𝑢𝑣𝑜𝑗𝑤𝑓𝑠𝑡𝑓
• or, use inverse Primakoff effect.
• 𝑄 ∝ 𝐶0

2 ∙ 𝑊 ∙ 𝑅

• 𝐶0: magnetic field
• 𝑊: volume of the cavity
• 𝑅: Quality factor of the cavity

27

Inverse Primakoff effect Spontaneous decay

• 4. Axion Search

28

• 5. Summary
• 1. SSB in the strong interaction proposes a goldstone boson
• but nothing found (U 1 A problem).
• 2. Using ‘t Hooft’s boundary condition, a CP violating term gives the answer

for U 1 A problem

• but no sign of CP violation in the strong interaction is found (the strong CP problem).
• 3. Proposing a new symmetry U 1 PQ imposes a SSB particle axion, and its

field cancels out the CP violation term at the ground state.

29

Thank you!

30