what does μ-τsymmetry imply about leptonic cp violation? *)
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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
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