What does μ-τsymmetry imply about leptonic CP violation?*)

Teppei BabaTokai University

In collaboration with Masaki Yasue

*) based on hep-ph/0612034 to be published in Physical Reviews D (March. 2007)

Content1. Introduction 2. What’s μ-τsymmetry ?3. μ-τsymmetry-breakings 　 and CP ph

ase4. Summary

1.Introductionee e e

e

e

M M MM M M M

M M M

We don’t know which masses give Dirac CP phase

(*) Charged lepton masses are diagonalized

Dirac CP phase?

??

However, there is an ambiguity, where phases of Mij (ij=e,,) are not uniquely determined because of the redefinition of phases of the neutrinos.

Observed quantities such as the mixing angles and the Dirac phase are independent of this ambiguity.

We can give the Dirac phase in terms of phases Mij (ij=e,,).

2 2 2 213 23 12

2.3 0.41 0.18sin 0.9 10 sin 0.44 1 sin 0.314 1

0.9 0.22 0.15

Experimental data give useful constraints on Mij.

Constraints on Mij Constraints on δ⇒

We study general property of leptonic CP violation without referring to specific relations among Mij.

13 23 12, , ,

functions of , , ,

ee e eij ij ij ij

e

ije

M M M M M M MM M M M

M i j eM M M

The mixing angles and Dirac CP phase δ are to be given as functions of Mij.

Problemμ-τ symmetry gives consistent results with experimental data. But, It can not give Dirac CP Violation. Why?

2.What’s μ-τsymmetry ?μ-τsymmetry gives a constraint that Lagrangian is invariant under transformation of νμ→σντ ,

ντ→σνμ (σ=±1)

(*) sign is just our convention.

μ-τ Symmetric Part+

μ-τSymmery Breaking Part

μ-τsymmetry

13

23 23

sin 01cos sin 12

ie

CP Violation ()

can not be

obtained

We clarify which flavor neutrino mass determines as general as possible.

extended to experiment:13

23 23

sin 01cos sin2

Why doesμ-τsymmetry give no Dirac CP violation?

We need μ-τSymmery Breaking Part

Definition of mass matrix

0

0

0

ee e e e e

sym b e e

e e

M M M M M

M M M M M M M M

M M M M M

sym b

0+

A B B B BB D E B D iEB E D B iE D

M M M

†

A B CWith M M B D E

C E F

M μ-τ symmetric partμ-τsymmetry brea

king part

2 2 2 2e e e e

e e

M M M M M M M MM M M M

We can fomally divide M into: 13 23 231sin 0, cos sin 12

ie

sym , , : real

i i

i

i

A e B e Be B D E A D Ee B E D

M

2 cos 2 sin 01 sin cos2 cos 1sin

i

isym

i

eU e

e

diagonalized by Usym

2 2

2sin2 2

XX X

22

2cos22

XXX

22 82

A D E A D E BX

B

μ-τsymmetric partThe phase iB e B

23 13 12This gives = , =0 and no Dirac CP violation. We calculate :

4

Usym gives UPMNS

12 13 12 13 13

12 23 23 12 13 23 12 23 12 13 23 13 13 13 13

23 12 23 12 13 23 12 23 12 13 23 13

cos , etc sin sin

i i

ii iPMNS CP

ii i

c c e s c s e

U s c e s c s e c c s s s e s c c J

s s e c c s e s c c s s e c c

12 13 23 231, ; 0, 2

s c s 13sin sin 0CPJ

3.μ-τsymmetry-breaking and CP phase

We estimate Dirac CP violation induced by –symmetry breakings

1.First, we use perturbation with Mb treated as a perturbative part to estimate .

2.Next, we perform exact estimation of that gives the perturbative results.

12 122

12 12

2

12 12

2

2 2 2 sin 2 2 cos 22

2 sin 2 cos 2 213 122

2 sin 2 cos2 21

2

i

atm

i

atm

i

atm

B R D iE e Bm

R B e D iE D iE

m

R B e D iE D iE

m

23 2

1

2 1s

23 2

1

2 1c

The phase structure of |3> suggests Δ and γ :

* *

1 112 12 12 1213 232 2 2 2

1 3 2 3

2

22 2

, , 1

i i

atm

c B s D i E e s B e c D i Ea a

m m m mmRm

12

112 13

12

2 0112 1

i

i

c

s e a

s e

12

112 23

12

2 0122 1

is ec ac

and

13

23

23

3

i

i

i

s e

s e

c e

132

1 12

1

is e

ii

3-1. Perturbation with b

0 B B

B D iE

B iE D

M

iB e B

1,

These δ 、 Δ and γ consistently describe |1> and |2>

12 1213 2

12 12

2

12 12

2

2 2 cos 2 sin 22

2 sin 2 Re cos 2 2

2 2

2 sin 2 Im cos 2 2

2 2

ii

atm

i

atm

i

atm

R B R D iE es e

m

R B e R D

m

R B e R E

m

12

12 12 13

12 12 13

211 12

1

i i

i i

c

s i e c s e

s i e c s e

12

12 12 13

12 12 13

212 12

1

i

i

i

s e

c i s s e

c i s s e

Δ and γ can be calculated

132

1 12

1

3

is e

i

i

3-1. Perturbation with b

0 B B

B D iE

B iE D

M

Suggested UPMNS

1 2 3

13 13 12 12

23 23 12 12

23 23 13 13

, ,

1 0 0 1 0 0 0 00 0 0 0 1 0 00 0 0 0 0 0 1

, ,

i i

i iPMNS

i i

K

c s e c s ee c s s e c

e s c s e cU

1212 13

12 12 13 12 12 13

12 12 13 12 12 13

1 , 2 , 3

22 21 1 1 12

11 1

i i

ii i

i i i

s ec s e

s i e c s e c i s s e i

is i e c s e c i s s e

U

We guess the appropriate form of the PMNS matrix

γ <<1<<1

This expression gives perturbative result

If

iB e B

3-2. Exact results

2, , , ,0i i iPMNSB e B C e C E e E U

21

† 22

23

0 00 00 0

, , , ,PMNS PMNS

mm

mU U

M

2 213 13 3

1 2 213 13

2 22 23 23 23 23

2 23 23 23 23 23

2 Re '

2 Re '

c A sc s

c D s F s c E

s D c F s c E

We have used redefined masses to control phase-ambiguities of γ:

which gives the following formula:

23 2312 12

2 1 13

13 23 23 133

23 23

23 2313 23

13

2tan 2 ,

2tan 2 ,

Re cos 2 sin 2 Im

,

i

i

i

c B s Cec

e s B c CA

E D i E

c B s Cs ec

Another redundant phase ρcan also be removed by the redefinition of masses. But we keep ρ to see its trace in CP violation.

Exact result for a

23 23

23 23

arg

arg

i i

i i

s e B c e C

c e B s e C

receives main contribution from &

arg , argB B

B B

, γ <<1 <<1

B B BC B B

23 231 1* ,

2 2c s

Exact result for a 23 23 13Re cos 2 sin 2 Im 'iE D i E s e X

23 - 23 13Re ' cos 2 sin 2 - cos 'E D s X x Re part :

2 2 2 2 2 2sin sin

Re Rei i

Dx

e E D e E D

23 4 2

Maximal atmospheric mixing x=0(s⇒ 13cos’=0) & D_=0(M=M)

2 with iE e E

⇒ Maximal CP violation if M=M

23

If the textures are approximately – symmetric

Which masses give which phases

δ depends on B －ρ depends on B+

Δ depends on D －γ depends on E －

( ＊ ) δ+ρ is Dirac CP Violating phase

†

0AM M D

B BBB

BE i

E D iD E

B D

B

EBM

13 2

12 2

2

2

2 2sin 2

2 2sin 2

i

atm

i

atm

atm

Bem

Be

mDm

Em

d

r

q s

q

sg

- -

+

-

-

» D

» D

D » - D

» D

e

・・ We can determine the phase of δ and ρ, and θ23

・・ Maximal atmospheric mixing conditions are given by

1313

sin 0sin cos 0 and

cos 0 Maximal CP violation

M M　

23 23 23 23arg , argi i i is e B c e C c e B s e C

4.Summary

223 13, with sin sin cos , sin

4 2

m M M

2, ,i i ie e e ee e e

M M M M M M

・・ Redefined flavor masses given byRedefined flavor masses given by

・・ give the weak-base independence of the Jarlskog invariant:give the weak-base independence of the Jarlskog invariant:

2 2 2 2 2 2

12 23 31 12 23 31

Im Ime e e eCP m m m m m mJ

M M M M M M

・・ The phases of Mν are so constrained to give δ and ρ via B+ and B － .

ee e e

e

e

M M MM M M M

M M M

・・ We can determine which masses provide which phases.

†

0AM M D

B BBB

BE i

E D iD E

B D

B

EBM

† † †

† † †

ee e e e

ee e e e

B B M M M M M M

B B M M M M M M

The work to discuss phases of Mν is in progress.

・ δ depends on B －・ ρ depends on B+

・ Δ depends on D －・ γ depends on E －

END

Three versions of M and UPMNS

12 13 12 13 13

212 23 23 12 13 23 12 23 12 13 23 13

223 12 23 12 13 23 12 23 12 13 23 13

i i i

i i ii PDGPMNS

i i ii

A Be Ce c c s c s e

B e D Ee U s c s c s e c c s s s e s c

C e E e F s s c c s e s c c s s e c c

M

12 13 12 13 13

12 23 23 12 13 23 12 23 12 13 23 13

23 12 23 12 13 23 12 23 12 13 23 13

1 0 00 0 0 0

i i

ii i iPMNS

i ii i

c c e s c s eA B CB D E U e s c e s c s e c c s s s e s cC E F e s s e c c s e s c c s s e c c

M

12 13 12 13 13

212 23 23 12 13 23 12 23 12 13 23 13

223 12 23 12 13 23 12 23 12 13 23 13

i ii i

ii i i iPMNS

i i ii i

c c e s c s eA Be CeB e D Ee U s c e s c s e c c s s s e s cC e E e F s s e c c s e s c c s s e c c

M

There are other two versions

For the redefined masses, we have the PDG version of UPMNS:

2) The intermediate one (γ is excluded from UPMNS):

1) The original one:

2, ,i i ie e e ee e e

M M M M M M・・ Redefined flavor masses given byRedefined flavor masses given by

reassure the weak-base independence of the Jarlskog reassure the weak-base independence of the Jarlskog invariant:invariant:

2 2 2 2 2 2

12 23 31 12 23 31

Im Ime e e eCP m m m m m mJ

M M M M M M

Now, we study which masses of Mν give which phases.

( )

( )

23 2312 2 2

13

2 2

2sin 2 2

2 2 2 2122

i ii c e B s e CXe

m c m

B i B Bm m

B B B

C B B

g grq

g

s

-

+ - +

+ -

+ -

-= =D D+ D+» »D D

æ ö= + ÷ç ÷ç ÷ç ÷÷ç =- -çè ø

e e

e e

・・ We can determine θ23, and the phase of ρand δ

・・ Maximal atmospheric mixing conditions are given by

・・ We can determine which masses provide which phases.

・ δ depends on B －

・ ρ depends on B+ ・ γ depends on E －

2and

M M

23 23 23 23arg , argi i i is e B c e C c e B s e C

5.Summary a 2

23 13, with sin sin cos , sin4 2

m M M

2, ,i i ie e e ee e e

M M M M M M

・・ Redefined flavor masses given byRedefined flavor masses given by

・・ reassure the weak-base independence of the Jarlskog invariant:reassure the weak-base independence of the Jarlskog invariant:

2 2 2 2 2 2

12 23 31 12 23 31

Im Ime e e eCP m m m m m mJ

M M M M M M

†

0AM M D

B BBB

BE i

E D iD E

B D

B

EBM