ビッグバン元素合成におけるビッグバン元素合成におけるスタウ原子衝突スタウ原子衝突

ビッグバン元素合成におけるビッグバン元素合成におけるスタウ原子衝突スタウ原子衝突

上村正康上村正康 九大理、理研九大理、理研肥山詠美子肥山詠美子 理研理研

初田哲男初田哲男 東大理東大理

浜口幸一浜口幸一 東大理東大理

柳田　勉柳田　勉 東大理東大理

東北大・理・化学木野康志

What is stau ( ) What is stau ( ) ??

What is stau ( ) What is stau ( ) ??Wikipedia

Stau may refer to one of the following:

• In particle physics, stau is a slepton which is the hypoth

etical superpartner of a tau lepton

• An obsolete letter Stigma in the Greek Alphabet

• In German language, Stau is a word meaning 'traffic jam'

2 %τ %τ

e

μτ

Lepton(spin 1/2)

%e%μ%τ

Slepton(spin 0)

Supersymmetricpair

selectron

smuon

stau(scalar tau)

Why the stau is so importantWhy the stau is so important ? ?

• Higgs mechanism• Supersymmetry, supersymmetric particles • Gravitino (dark matter) • etc.

3

Large Hadron Collider@CERNThe first beam (10 Sep. 2008)

Objectives of LHC

Finding evidence for a supersymmetric particle might be possible as early as the year 2009. Finding the Higgs might be possible by 2010.

By A. Seidan (California) at APS meeting 2008

Perspective of LHC

In this In this talk…talk…

4

Long lifetime ≈ 1,000-100,000 s

Heavy mass: ≈ 100-1,000 GeV/c2

Interaction: Coulomb

%τ ⇒ X

The stau can form an exotic atom/molecule

M(stau) ≥ M(Fe)

X(Stau): negatively charged point-like heavy particle

I do not discuss the supersymmetry (and cosmology), but atomic physics.I denote the Long lived scalar lepton as

The X particle has

mX = 100 GeV/c2

mp = 0.94 GeV/c2

Stau atomStau atom

a

n=

h2

μZe2n2

μ-1 =m

N-1 + m

X-1 ≈m

N-1

E

n=−

μZ2e4

2h2

1n2

r (4He)

2=1.68 fm

r (d)2=2.14 fm

Reduced mass

(mN= m

X)

Root mean charge radius of nucleus

(for example; hydrogen like atom)

System m / GeV c–2 E / keV a /f m HX– 0.930 24.7 29.52 DX– 1.830 48.5 15.41 TX– 2.733 72.3 10.53

3HeX– 2.733 290.1 5.75 4HeX– 3.594 370.8 4.61 6LiX– 5.306 1,255.1 1.91 7LiX– 6.134 1,423.2 1.78

Binding energy and Bohr radius (point nuclear charge)

5

-5

-3

-1

1

3

5

0 5 10 15

r

V(r)

-0.15

-0.10

-0.05

0.00

0.05

0.10

0.15

wave function

Gaussian distribution

b = 1.37 fm for 4Heb = 1.75 fm for db = 2.576 fm for 7Beb = 0.7144 fm for p

ρ(r ) = Ze

(πb)3/ 2exp −

r 2

b2

⎛

⎝⎜⎞

⎠⎟

−

Ze2

rerf −

rb

⎛⎝⎜

⎞⎠⎟

Pure Coulombic potential

ρ(r )

−

Ze2

r

φG(r )

φC (r )

rXHe2 =6.84 fm

rXHe2 =6.29 fm

EXHe1s =−337 keV

EXHe1s =−397 keV

X4He : Hydrogen-like ion

wavefunction

Nuclear charge density

Nuclear charge density

6

Lifetime: more than 1,000 s

nucleosynthesis (the first three minutes)

M. Pospelov (PRL 98, 231301, 2007)

6Li production by X-particle atomic collision

7

X4He + D→ 6Li + X + 1.137 MeV

X4He + D→ 6Li + X + 1.137 MeV

catalyzed nuclear fusion

X4He + D→ 6Li + X + 1.137 MeV

X4He + D→ 6Li + X + 1.137 MeV

Cf. muon catalyzed fusion

Nucleosynthesis in Big Bang and the Li probNucleosynthesis in Big Bang and the Li problemlem

7Be

6Li 6Li (observed) 7Li (observed)

7Li (X 1/3)6Li (X 1000)

}

66Li (Li (= 4He + D)) 4He + D→ 6Li + γ

8

Observed value

7Li

Very slow（ E1:forbidden）

Li problemLi problem

“The Economist” (June, 2007)

Dr Pospelov's catalytic mechanism, by contrast, explains both discrepancies in one fell swoop. It also makes predictions about the detailed properties of the supersymmetric partners—and, as luck would have it, suggests that although they cannot be made on Earth at the moment, they should be in range of the Large Hadron Collider, a particle accelerator being built near Geneva, which should open for business later this year. It should not, therefore, take long to find out if his explanation for the lithium problem is correct. If it is, he can claim to have found SUSY's traces before the particle physicists did.

Posperov’s result significantly limited the current understanding of stau.

Particle phys. (U. Tokyo)

Is it true?

Nuclear Phys.Atomic Phys.(Exotic Atom)

Maybe yes, but…

… Pospelov employed a simple scaling, and paid no attention to the nuclear interaction and low energy atomic collision.

Stau could be created at big baStau could be created at big ban,n,

T ? mXc2

Catalyzed reactions

4HeX + d → 6Li + X4HeX + t → 7Li + X4HeX + 3He → 7Be + X6LiX + p → 4He + 3He + X7LiX + p → 4He + 4He + X7BeX + p → 8BX + γpX + 4He → 4HeX + pdX + 4He → 4HeX + dtX + 4He → 4HeX + t7BeX + p → 8BX * → 8BX + γ

Submitted to Prog. Theo. Phys.arXiv:0809.4772v1

6Li production(4HeX)1s + D → 6Li + X +1.137 MeV

7Li destruction

(7BeX)1s + H → (8BX)2p → (8BX)1s + γ

8B → 8Be → 24He

Feshbach-typeresonance

7Be → 7Lielectron capture

53 d 0.77 s 10–16 s

12

Phys. Lett. B 650, 268, 2007

(6Li ≈ 4He + D)

Submitted to Prog. Theo. Phys.

electron capture

Hamiltonian

H =−

h2

2mc

∇r2 −

h2

2Mc

∇R2 +V

XaC (r

1) +V

abC (r

2) +V

XbC (r

3) +V

abN (r

2)

a

bX–

R3 r3

c = 3

a

R2

r2

c = 2bX–

a

R1r1

c = 1bX–

Numerical method

Numerical method

(a, b) = (D, 4He) (7Be, H)

13

Gaussian Expansion Method Gaussian Expansion Method (GEM)(GEM)Gaussian Expansion Method Gaussian Expansion Method (GEM)(GEM) Prog. Part. Nucl. Phys. 51. 223, 2003.

3,4-Nucleons, Nuclei, Hyper Nuclei, 3,5-Quarks Exotic atom/molecule (positronic, muonic, antiprotonic,…)

ΨJ M =φ0Xα (r1)χ J

Xα−d(R1)YJ M(R̂1) + φ0Li(r2 )χ J

Li−X (R2 )YJ M(R̂2 ) +ΨJ Mclosed

Incident channel Reaction channel

X4He + d→ 6Li + X

Wave functionWave function

Molecular and 3-body continuum statesIncl. resonance states

14

ΨJ M=φ

0X 7 Be(r

1)χ

JX 7 Be−p(R

1)Y

J M(R̂

1) +Ψ

J Mclosed

(7BeX)

1s+ p Ä (8BX)

2p

Incident channel

lim

Rc→ ∞Ψ

J M=φ

0c (r

c) u

J(−) (k

cR

c) −S

Ju

J(+) (k

cR

c)⎡

⎣⎤⎦YJ M

(R̂c)

Boundary conditions

Molecular and 3-body continuum states

ΨJ M

closed = bJ vΦJ Mvv=1

vmax

∑

limR→ ∞

ΨJ Mclosed =0

ΦJ Mv H ΦJ M ′v =EJ vδv ′v

ΦJ Mv

Closed channel Closed channel

(7BeX)

1s+ p

Ecm

Resonance state

discretization

Continuum stateResonance state Bound state

Expanded in terms of L2 basis

Diagonarize the 3-body Hamiltonian

15 ΨJ M

closed

6Li + X X4He + d

Φ

J Mv= C

vnNJ lLc

aAlL∑

c=1

3

∑ exp −anr

c2 −A

NR

c2

( ) Yl(r̂

c) ⊗ Y

L(R̂

c)⎡

⎣⎤⎦J M

Rapid convergencemax≈50

Scattering wavefunctionScattering wavefunction

Coupled integro-differential equations

χ JXα−d(R1), χ J

Li−X(R2 )

Vv

c ′c = φ0c (rc )YJ M(R̂c ) H −E ΦJ Mv r̂cR̂c

1Ev −E

ΦJ Mv H −E φ0′c (r ′c )YJ M(R̂ ′c ) r̂ ′c R̂ ′c

φ0

Xα (r1)YJ M(R̂1) H −E ΨJ M r1R̂1

=0

φ0

Xα (r2 )YJ M(R̂2 ) H −E ΨJ M r2R̂2

=0

K1 +U1(R1) −E + ε0Xα( ) χ J

Xα−d(R1) + dR2U12∫ (R1,R2 )χ JLi−X (R2 )

+ d ′R1cVv11∫ (R1, ′R1)χ J

Xα−d( ′R1) + d ′R ′cVv12∫ (R1,R2 )χ J

Li−X (R2 )( )v∑ =0

K2 +U2 (R2 ) −E + ε0Li( ) χ J

Li−X (R2 ) + dR1U21∫ (R1,R2 )χ JXα−d(R1)

+ d ′R1Vv21∫ (R1, ′R1)χ J

Xα−d( ′R1) + d ′R ′cVv22∫ (R1,R2 )χ J

Li−X (R2 )( )v∑ =0

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

Non local potential

Numerical calculation*Variation method*Difference method

Potential between 4He and D; Vdα(r2)Potential between 4He and D; Vdα(r2)

Reproduces the root mean square radius of 6Li (2.54 fm)

the binding energy of 6Li (1.474 MeV)

原子核電荷分布：ガウス型

b = 1.37 fm for 4He, b = 1.75 fm for d

Ze

(πb)3/ 2exp −

r 2

b2

⎛

⎝⎜⎞

⎠⎟

Ze2

rerf −

rb

⎛⎝⎜

⎞⎠⎟

b = 1.37 fm for 4He, b = 1.75 fm for d

Vdα (r ) =500 exp −

r 2

0.92

⎛

⎝⎜⎞

⎠⎟−64.06 exp −

r 2

2.02

⎛

⎝⎜⎞

⎠⎟+2e2

rerf

r

1.372 + 1.752

⎛

⎝⎜

⎞

⎠⎟

Charge form factor S-wave phase shift

Li + e– 4He + D

17

Nuclear potential Vdα(r2)

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1 10 100Internuclear distance 　 r2 (fm)

Vdαr

2)

(MeV

)

Ecm ≈ 0.036 MeV

Coulomb

Coulomb + Nuclear-10 MeV

E1sXHe =0.33 MeV

NuclearNuclearreactionreaction

The nuclear reaction occurs by tunneling effect between the 4He and D.

18

KXHe

Relative energy

Coulomb barrior

VX

Different mechanism from electron screeningDifferent mechanism from electron screening

Reaction cross section

0.000001

0.00001

0.0001

0.001

0.01

0.1

0 20 40 60 80 100

E (keV)

m

b)

Full Coupled

Channel

DWBA

2-channel

CC

(w/o close

d channel)

X4He + d→ 6Li + X

19

0

10

20

30

40

50

0 20 40 60 80 100

Energy (keV)

S-factor (MeV mb)

Gamow peak@36.4 keV (T=10 keV)

DWBA

(1) : K1 +U1(R1) −E + ε0Xα( ) χ̂ J

Xα−d(R1) =0

(2) : K2 +U2 (R2 ) −E + ε0Li( ) χ J

Li−X (R2 ) + dR1U21∫ (R1,R2 )χ̂ JXα−d(R1)

+ d ′R1Vv21∫ (R1, ′R1)χ̂ J

Xα−d(R1)v∑ =0

⎧

⎨

⎪⎪

⎩

⎪⎪

2-channel CC

K1 +U1(R1) −E + ε0Xα( ) χ J

Xα−d(R1) + dR2U12∫ (R1,R2 )χ JLi−X (R2 ) =0

K2 +U2 (R2 ) −E + ε0Li( ) χ J

Li−X (R2 ) + dR1U21∫ (R1,R2 )χ JXα−d(R1) =0

⎧⎨⎪

⎩⎪

Full CC

K1 +U1(R1) −E + ε0Xα( ) χ J

Xα−d(R1) + dR2U12∫ (R1,R2 )χ JLi−X (R2 )

+ d ′R1cVv11∫ (R1, ′R1)χ J

Xα−d( ′R1) + d ′R ′cVv12∫ (R1,R2 )χ J

Li−X (R2 )( )v∑ =0

K2 +U2 (R2 ) −E + ε0Li( ) χ J

Li−X (R2 ) + dR1U21∫ (R1,R2 )χ JXα−d(R1)

+ d ′R1Vv21∫ (R1, ′R1)χ J

Xα−d( ′R1) + d ′R ′cVv22∫ (R1,R2 )χ J

Li−X (R2 )( )v∑ =0

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

Energy levels of the 7BeX + p reaction

7Be + X + p

(7BeX)1s + p

Gamov peak ≈ 114 keV(kT ≈ 35 keV)

Incident channel

8B + X

(8BX)2p

(8BX)1s

γ

E,

-1.86 MeV

0 MeV

-1.33 MeV

-0.1375 MeV

J–

2+

8B[p(3/2–),7Be(3/2–)](2+) Vp-Be = VWS + Vls + VMC

Ein

Reaction rateReaction rate

NA

v =NAh2 2πh2

μkT

⎛

⎝⎜⎞

⎠⎟

3/ 22 J + 1

(2Ip+ 1)(2I

Be+ 1)

p

γ

p+

γJ∑ exp −

Er

kT

⎛

⎝⎜

⎞

⎠⎟

T = 0.3x109 K

NA

v (cm3s−1mol−1)

0.42 ×105 (mX=50 GeV)

0.68 ×105 (mX=100 GeV)

1.0 ×105 (mX=500 GeV)

1.1×105 (mX=∞ GeV)

Strong Strong mmXX dependence dependence!!

22

(mX)

J– → 2+

γ=16πk3

91

2 J + 1(8 BX)

1s2 + Q

1μ(E1) (8 BX)

2pJ −

M ′M μ∑

Q

1μ(E1) = q

iR

iY1μ(R̂

i)

i∑

Er, p

(7BeX)1s + H → (8BX)2p → (8BX)1s + γ

ρMaxwell(E)

J = 1– J = 2–

mX (GeV) Er (keV) p (keV)

γ (eV)

Er (keV) p (keV)

γ (eV)

50 196.5 0.88 9.1 196.8 0.54 9.1

100 185.2 0.82 9.6 185.5 0.51 9.6

500 175.5 0.73 9.9 176.6 0.44 9.9

∞ 173.0 0.73 10.1 173.3 0.44 10.1

Resonance parametersResonance parameters

J– → 2+

γ=16πk3

91

2 J + 1(8BX)

1s2 + Q

1μ(E1) (8BX)

2pJ −

M ′M μ∑

Q

1μ(E1) = q

iR

iY1μ(R̂

i)

i∑

(8BX)2p

(8BX)1s

γ(820 keV)

E, J–

2+

23

SummarySummary† Low energy stau atomic collision is precisely calculated based on full quantum mechanics.

† Three-body closed channel (Stau molecular states HeDX and resonance state 8BX) plays an important role.

† Present calculation supports the Posperov’s pioneering idea, but reduces his reaction rate by one order of magnitude. (The present result is in the current prediction of SUSY.)

† Accurate theoretical predictions of nuclear reaction rate are important in the study of big ban nucleosynthesis.

‡ Stau opens a new page of exotic atom/molecule research, and will provide fruitful results for few-body physics.

X4He + D→ 6Li + X + 1.137 MeV

24

Stau catalyzed big ban nucleosynthesis => supersymmetry

X 7Be + H→ (X 8 B)

2p→ (X 8 B)

1s+ γ

Stau atom

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Electronic atom

Geocentric SystemGeocentric SystemHeliocentric SystemHeliocentric System

He XHe

Stau atomic/molecular system

Nuclei move around the heavy electron (stau)!!

25