CP Violation in tau -> nu_tau + 3 pi

12 Pages·0.14 MB·English

by  Yung Su Tsai

CP VIOLATION IN τ ν +3π
τ
→
YUNGSU TSAI
8 Stanford Linear Accelerator Center
9
Stanford University, Stanford, California 94309
9
e-mail: [email protected]
1

n We discuss the ways to find CP violation in the decay of τ± → ντ +3π from
a unpolarizedaswellaspolarizedτ±.
J

2
1 1 Introduction

1 IntheStandardModel1,2 noCPviolationcanoccurindecayprocessesinvolv-
v ing leptons either as a parent or a daughter because these processes involve
4
one W exchangeand thus, evenif the CP violating complex coupling exists in
7
these decays, its effect will not show up when the amplitude is squared. We
2
1 need two diagramsto interfere with eachother to see the CP violating effects.
0 The CP violation is caused by the existence of a complex coupling constant
8 whose phase changes sign as one goes from particle to antiparticle as required
9
by the hermiticity of Lagrangian that results in the CPT theorem. Let A be
/
h the complex coupling constant for τ− decay A = Aeiδt, then for τ+ decay
p this coupling constant must be (CP)A(CP)−1 = |A∗| = Ae−iδt in order to
p- preserve the hermiticity of the Hamiltonian3,4. | |
e In quantum mechanics i is changed into i under time reversal in order
h −
to preserve the basic commutation relations such as [x ,p ] = iδ , [J ,J ] =
i j ij i j
:
v iǫ J , etc. Thus under CPT the coupling constant A is invariant
ijk k
i CPT A(CPT)−1 =A.
X
LetuscalltheparticleexchangedinthenewFeynmandiagramthexparti-
r
a cle. Thisparticlecouldbe the chargedHiggsbosonoranentirelynew particle
Nature created in order to have CP violation and the matter-dominated Uni-
verse. In this paper we do not discuss the originof such a particle, we restrict
ourselves to how the existence of such a particle with a complex coupling
constant can be discovered experimentally. The x particle must have spin 0
because if it were spin 1 the x exchange diagram must be proportional to the
W exchange diagram, thus even if they have a relative complex phase, the
imaginary part of the relative phase can never show up3. Let M = cM ,
x W
then M +M 2 = M 2(1+2Rec+ c2). Since CP violation is caused by
W x W
| | | | | |
Imc, there cannot be any CP violationif x is spin 1. Higher spin particles are
excluded because they are unrenormalizable.
In τ± decay CP violation manifests itself in the following way:
1
1. Branchingfractionforsemileptonicdecayofτ− withfinalstatehadronic
interactions is different from the corresponding one for τ+ decay5.
2. The coefficient of p q in the rest frame of hadron decay product of
−→1·−→i
τ− is different from the corresponding one for τ+ lepton, where p is
−→1
the momentum of τ− and q is the momentum of a decay hadron both
−→i
measuredinthe restframeoftotalhadronscm. Since p q isT even,
−→1·−→i
CP violation for this kind of terms can occur only if there is a strong
interaction in the final state because of TCP theorem3,4,6.
3. If τ− is polarized with polarization vector w , we have terms such as
−→
w q ,w (q q ). Under CP transformation we have3,4
−→·−→i −→· −→i×−→j
(CP)(w q )(CP)−1 = w ′ q ′
−→·−→i −−→ ·−→i
and
(CP)(w (q q ))(CP)−1 =w ′ (q ′ q ′) .
−→· −→i×−→j −→ · −→i×−→j
Hence if the coefficients of these terms in τ+ decay do not behave as
above then there is CP violation. Notice that CP changes the sign of
momentum but does not change the sign ofw .
−→
4. For pure leptonic decay, such as τ± ν +ν +µ±, the only term that
τ µ
→
violates CP is w−→τ (P−→µ w−→µ). Since there is no final state interaction
· ×
herewe need T odd productto violate CP because ofthe CPT theorem.
Hence we neednot only τ polarizationbut alsomeasurementof µpolar-
ization. This experiment will show whether a pure leptonic system can
have CP violation7.
Sincethedirectionofpolarizationcanbereversed,itcanbeusedtoisolatethe
coefficients of different spin dependent terms such as w q , w (q q ).
−→·−→i −→· −→i×−→j
Thus the polarization is essential for pure leptonic decay. For semileptonic
decay it is highly desirable especially if the effect is marginal.
When dealing with CP violation it is of utmost importance to be open-
minded. The reason is that experimentally8 there is only K 2π known to
L
→
exhibitCPviolationsofarandthestandardtheoryofKobayashiandMaskawa1
was invented to explain this. It would be a mistake to neglect testing the CP
violation involving leptons just because the CKM theory says there is no CP
violation for τ decay. It is also unconscionable to ignore the possibility of
lighter particlesparticipating inCP violationjust because the Higgs exchange
model in general favors heavy particles for this violation! Fortunately we do
nothavetoassumeanyspecificmodelforourpurpose. Allweneedistoassume
2
theexistenceofaspin0particlecalledxthatiscoupledtoleptonsandquarks
withCPviolatingcomplexcouplingconstants. Ourtaskistodeviseamethod
to test whether the imaginary part of this complex coupling constant is not
zero. I believe that our first priority is to discover the effect in any channel of
decay of τ or semileptonic decay of B or D. After the effect is discovered in
one particular decay mode one can worry about a systematic search for the x
coupling to all quarks and leptons. Only after that can one worry about why
x exists.
This paper is a sequel to my previous work4,5,9,10,3 investigating possible
CP violation involving leptons either as a parent or as a daughter. There are
many issues involved in this vast subject:
(a) What is the best decay mode for the first discovery of the effect?
(b) The longitudinal polarization of e± helps, but can one do without it?
(c) Once the first discovery is made and thus confirming the existence of x
boson, we have to systematically find out the coupling of the x bosonto
all the leptons and quarks. To find out the x coupling to the lepton we
not only require τ to be polarized but also require the measurement of
muon polarization in the decay τ ν +ν +µ.
µ τ
→
(d) In order to answer the first question (a) we have to theoretically inves-
tigate many decay modes. This paper deals with one of them; namely,
how to find CP violation in the decay τ 3π+ν .
τ
→
We are not yet in a position to write a comprehensive summary to answer
question (a). We can give only the intermediate answer to this question. For
τ decay:
1. The decay mode τ± π± +π0 +ν has the largest branching fraction
→
(25%),butCPviolationwillbesuppressedbecauseitneedsinterference
between s and p waves of the 2π produced by x and W bosons. The s
waveissuppressedbytheisospinsymmetry. Thereisafew%breakingof
isosymmetry due to mass difference of π± and π0, thus the suppression
is about the same order of magnitude as the Cabibbo suppression for
decaying into a strange channel3,4.
2. The decay mode τ± e± + ν +ν is next in the size of branching
e τ
→
fraction, but here we need polarization of τ± as well as measurement of
transverse polarization of e± that is an impossibly difficult task.
3
3. The decay mode τ± µ±+ν +ν . This is similar to the above, but
µ τ
→
polarization of µ± can be measured by its decay angular distribution.
µ± can be slow moving if it moves in the direction opposite to τ±5.
4. The decay mode τ± π±π±π∓ν . This reaction was first considered
τ
→
by Choi et al.11. In this paper we deal with the same problem with
polarization of τ taken into account. We also exhibit the phases of a
1
andπ′ resonancesexplicitly,sothatitwillaidthe decisionastohowthe
data should be integrated to exhibit the existence of the CP violating
phase δ .
t
5. InRefs. (3) and(11)the Cabibbo suppressedmodeτ± K±+π0+ν
τ
→
is treated. If x is a charged Higgs boson, this one could be large.
6. More complicated cases such as τ ν + more than 3π, τ ν +K+
τ τ
→ →
more than 2π need to be considered in the future.
7. Semileptonicmodewithoutfinalstateinteractionssuchasτ± π±+ν ,
τ
→
τ± K±+ν cannothaveCPviolationbecausewedonothaveenough
τ
→
kinematic variables to construct T odd triple product and the T even
term is not allowed because there is no final state interaction.
For semileptonic decay of hadrons we have:
1. K± π0+µ±+ν has been considered12,13,14.
µ
→
2. B± π0 + τ± + ν , D± π0 + µ± + ν , D± K0 + µ± + ν ,
τ µ µ
→ → →
B± D0+τ±+ν have been considered15,3.
τ
→
3. t± π0+τ±+ν , K0+τ±+ν , B0+τ±+ν havebeenconsidered16.
τ τ τ
→
2 Calculations
We consider six Feynman diagrams shown in Fig. 1 for the decay τ±
→
π±+π∓+π±+ν . Therearetwoidenticalparticlesdenotedbyq andq . We
τ 1 2
mayarbitrarilychoosewhichoneiscalledq andq ;forexample,q andq are
1 2 1 2
chosensuchthats =(q +q )2 >s =(q +q )2. Inthequarktheorybotha−
1 2 3 2 1 3 1
and π′− consist of ud combination and thus they have the same CP violating
weakphaseeiδw whencoupledtoW− whereasthepossibleCPviolatingphase
inthexcouplingtoπ′ isdenotedbyeiδx. Onlythe relativephaseδt δx δw
≡ −
will show up in the CP violation.
4
p +(q3) p –(q ) p +(q3) p –(q )
2 1
p2 n – r ° p –(q1) p2 n – r ° p –(q2)
W W
+
– –
p t – a1 p t – a1
1 1
Awa
p +(q3) p –(q ) p +(q3) p –(q )
2 1
p2 n – r ° p –(q1) p2 n – r ° p –(q2)
W W
+
t – p ' t – p '
p p
1 1
Awp
p +(q3) p –(q ) p +(q3) p –(q )
2 1
p2 n – r ° p –(q1) p2 n – r ° p –(q2)
X X
+
t – p ' t – p '
p p
1 1
Axp 1–98(cid:13)
8381A1
Figure1: Feynmandiagramsforτ−→π−+π−+π++ντ forCPviolation.
Let
A = c B ei(δa+δw)[eiδ1M (s )+eiδ2M (s )] (1)
wa | wa a1| a 1 a 2
Awp = cwpBπ′ ei(δp+δw)[eiδ1Mp(s1)+eiδ2Mp(s2)] (2)
| |
Axp = cxpBπ′ ei(δp+δx)[eiδ1Mp(s1)+eiδ2Mp(s2)] (3)
| |
where
G
c = F cosθ f f f = c eiδw (4)
wa c a aρπ ρππ wa
2√2 | |
G
cwp = F cosθcfπ′fπ′ρπfρππ = cwp eiδw (5)
2√2 | |
5
G
cxp = x fπ′fπ′ρπfρππ = cxp eiδx (6)
2√2 | |
Ba1, Bπ′ and Br are Breit-Wigner propagators defined by
1 eiδy(q2)
B (q2)= (7)
y m2y−q2−imyΓy(q2) ≡ q(m2y−q2)2+m2yΓ2y(q2)
for y =a , π′, or ρ.
1
Since our purposeis to investigatethe effect ofCP violatingphaseδ and
w
δ we explicitly extract all the final state interaction phases of the problem
x
δ ,δ ,δ , and δ for a , π′, ρ(s ) and ρ(s ) respectively.
a p 1 2 1 1 2
M (s ) =u(p )γ (1 γ )u(p )(2q q q 2q q q )B (s ) , (8)
a 1 2 µ 5 1 2 3µ 3 2µ r 1
− · − · | |
M (s ) =u(p )γ (1 γ )u(p )(2q q q 2q q q )B (s ) , (9)
a 2 2 µ 5 1 1 3µ 3 1µ r 2
− · · − · | |
M (s ) =u(p )(1+γ )u(p )(s s )B (s ) , (10)
p 1 2 5 1 2 3 r 1
− | |
M (s ) =u(p )(1+γ )u(p )(s s )B (s ) . (11)
p 2 1 5 1 1 3 r 2
− | |
The decay rate as well as the decay energy angle distribution of τ− ν +
τ
→
π−+π−+π+ fromapolarizedτ withapolarizationvectorw canbeobtained
−→
from
1 d3p d3q d3q d3q
dΓ = 2 1 2 3 δ4(p p q q q )
2m Z 2E 2w 2w 2w 1− 2− 1− 2− 3
τ 2 1 2 3
(2π)−8 A +A +A 2 . (12)
wa wp xp
× | |
The matrix elements squared can be written as
A +A +A 2 = A 2+ A +A 2+(A+ A +A+ A )
| wa wp xp| | wa| | wp xp| wa wp wp wa
+(A+ A +A+ A ) . (13)
wa xp xp wa
Using Reduce 3.617 we calculate A 2, A +A 2, (A+ A +A+ A )
| wa| | wp xp| wa wp wp wa
and (A+ A +A+ A ) as follows:
wa xp xp wa
A 2 =8c 2 b 2 br 2A (q )+ br 2A (q )
wa wa a 1 1 2 2 1 1
| | | | | | | | | |
(cid:8)
+2br br [cos(δ δ )ReA sin(δ δ )ImA ]
1 2 1 2 2 1 2 2
| | − − − }
+8c 2 b 2m
wa a τ
| | | |
6
w q br 2A (q )+ br 2A (q )+2br br cos(δ δ )A
1 3 2 2 3 1 1 2 1 2 4
{ · | | | | | | −
(cid:2) (cid:3)
+w q 2br 2A (q )+2br br cos(δ δ )A (q )
1 2 3 1 1 2 1 2 5 2
· | | | | −
(cid:2) (cid:3)
+w q 2br 2A (q )+2br br cos(δ δ )A (q )
2 1 5 2 1 2 1 2 5 1
· | | | | −
(cid:2) (cid:3)
+w q br 2A (q )+ br 2A (q )+2br br cos(δ δ )A
3 1 6 2 2 6 1 1 2 1 2 7
· | | | | | | −
(cid:2) (cid:3)
2br br q q sin(δ δ )A , (14)
1 2 3 1 2 8
− | | · − }
where ba = Ba1, bp = Bπ′, br1 = Bρ(s1), br2 = Bρ(s2), δ1 and δ2 are phases
of br and br respectively, and
1 2
A (q )=2[(p q )(q q ) (p q )(q q )]2
1 1 1 1 3 1 3 1
· · − · ·
+(p q m2)[m2(q q )2+m2(q q )3
1· − τ · 1 · 3
2(q q )(q q )(q q )]
1 3 1 3
− · · ·
ReA =(p q)[(q q )(q q )m2 (q q )(q q )(q q )
2 1 1 2 1 3 2 3
· · · − · · ·
(q q )(q q )(q q )+(q q )2(q q )]
2 3 1 3 3 1 2
− · · · · ·
+2.0[(p q )(q q )+(p q )(q q )]
1 1 3 1 3 1
· · · ·
[(p q )(q q ) (p q )(q q )]
1 2 3 1 3 2
× · · − · ·
(q q )m2[(q q )m2 (q q )(q q )]
− · 1 τ · 2 − · 3 2· 3
+(q q )m2[(q q )(q q ) (q q )(q q )]
· 3 τ · 2 · 3 − · 3 1· 2
ImA = (q q )q2Eps(p ,q ,q ,q )
2 3 1 1 2 3
− ·
A (q )=2(q q )(q q )(q q ) [(q q )2+(q q )2]m2
3 1 1 3 1 3 1 3
· · · − · ·
A =2.0 (q q )[(q q )(q q )+(q q )(q q )
4 3 1 2 3 2 1 3
n · · · · ·
(q q )(q q )] (q q )(q q )m2
1 2 3 1 2
− · · − · · o
A (q )=(q q )[ (p q )(q q )+(p q )(q q )]
5 1 3 1 1 3 1 3 1
· − · · · ·
A (q )= 2.0(q q )[(p q )(q q ) (p q )(q q )]
6 1 1 1 1 3 1 3 1
− · · · − · ·
A =(q q )[(p q )(q q )+(p q )(q q )]
7 3 1 1 2 1 2 1
· · · · ·
2(p q )(q q )(q q )
1 3 1 2
− · · ·
A =(q q )Eps(p ,q ,q ,w) (q q )Eps(p ,q ,q ,w)
8 1 2 2 3 2 2 1 3
· − ·
7
(q q )Eps(p ,q ,q ,w)
3 2 1 2
− ·
m =m =0.140 GeV , m =1.777 GeV , q =q +q +q ,
π τ 1 2 3
s =(q +q )2 , s =(q +q )2 , s =(q +q )2 .
1 2 3 2 1 3 3 1 2
A +A 2 = bp2[c 2+ c 2+2c c cos(δ δ )]
wp xp wp xp wp xp x w
| | | | | | | | | | −
M+(s )M (s )+M+(s )M (s )
×h p 1 p 1 p 2 p 2
+2M+(s )M (s )cos(δ δ )
p 1 p 2 2− 1 i
=[2.0(m2 p q) 2m w q]
τ − 1· − τ ·
[c 2+ c 2+2c c cos(δ δ )] bp2
wp xp wp xp x w
×| | | | − | |
[br 2(s s )2+ br 2(s s )2
1 2 3 2 1 3
| | − | | −
+2br br (s s )(s s )cos(δ δ )] (15)
1 2 1 3 2 3 2 1
| | − − −
A+ A +A+ A canbe obtainedfromA+ A +A+ A by letting c
wa wp wp wa wa xp xp wa wp ←
c and δ δ =0:
xp x w
−
A+ W +A+ A = c c babp
wa wp wp wa | wa wp |
2cos(δ δ )Re[M+(s )M (s )]
n p− a a 1 p 1
2sin(δ δ )Im[M+(s )M (s )]
− p− a a 1 p 1
+2cos(δ δ +δ δ )Re[M+(s )M (s )]
p− a 2− 1 a 1 p 2
2sin(δ δ +δ δ )Im[M+(s )M (s)2)]
− p− a 2− 1 a 1 p
+(q q , s s ) . (16)
1 1 1 2
↔ ↔ o
A+ A +A+ A = c c babp
wa xp xp wa | wa xp |
2cos(δ δ +δ δ )Re[M+(s )M (s )]
n p− a x− w a 1 p 1
2sin(δ δ +δ δ )Im[M+(s )M (s )]
− p− a x− w a 1 p 1
+2cos(δ δ +δ δ +δ δ )Re[M+(s )M (s )]
p− a 2− 1 x− w a 1 p 2
2sin(δ δ +δ δ +δ δ )Im[M+(s )M (s )]
− p− a 2− 1 x− w a 1 p 2
8
+(q q , s s ) (17)
1 2 1 2
↔ ↔ o
where
Re(M+(s )M (s )) =4m A (q ) 4w qA (q )
a 1 p 1 τ 9 2 − · 9 2
+4w q (q q )A (s ) 4w q (q q )A (s )(18)
2 3 11 2 3 2 11 2
· · − · ·
Im(M+(s )M (s ))= 4.0(s s )A (q ) (19)
a 1 p 1 − 1− 3 12 1
Re(M+(s )M (s )) =4m A (q ) 4w qA (q )+4w q (q q )A (s )
a 1 p 2 τ 9 2 − · 10 1 · 1 · 3 11 2
+4w q (q q )A (s ) 4w q (q q )A (s ) .(20)
2 3 11 1 3 2 11 1
· · − · ·
ImM+(s )M (s )= 4(s s )A (q ) (21)
a 1 p 2 − 2− 3 12 1
A (q )=(s s )[(p q )(q q ) (p q )(q q )] (22)
9 2 2 3 1 3 2 1 2 3
− · · − · ·
A (q )=(s s )[(p q )(q q ) (p q )(q q )] (23)
10 1 2 3 1 3 1 1 1 3
− · · − · ·
A (s )=(s s )(m2 p q) (24)
11 1 1− 3 τ − 1·
A (q )=q q Eps(p ,q,q ,w) q q Eps(p ,q,q ,w) . (25)
12 1 3 1 2 2 1 3
· − ·
2.1 Observations
1. A and A deal with W exchange, so it deals only with physics con-
wa wp
tained in the Standard Model. The a decay amplitude A is more
1 wa
thoroughly investigated than the π′ decay amplitude A .
wp
2. In order to have CP violation we not only need that the gauge bosons
exchangedmust be different (W and x), but also that the final hadronic
states must be different (a and π′). For example, Eq. (15) shows that
1
the interference between A and A is proportional to cos(δ δ )
wp xp x w
−
for τ− decay whereas for τ+ decay it is cos(δ δ ) which is equal
w x
−
to cos(δ δ ) and thus we cannot have a CP violating effect even if
x w
−
δ δ =0. ThisisbecauseA andA areproportionaltoeachother
w x wp xp
− 6
and thus the imaginary part of the phase difference does not show up.
Equation(16)does notcontainδ orδ , sono CPviolatingeffect. Only
w x
Eq. (17) can contain CP violating effects.
3. When τ’s are not polarized only A (q ) and A (q ), shown in Eq. (22),
9 1 9 2
appearinthe CPviolatingexpression. CPviolationcanthusbechecked
by comparing the coefficients of A (q ) and A (q ) for τ− decay with
9 1 9 2
those of A (q′) and A (q′) for τ+ decay. If they are different, then CP
9 1 9 2
is violated.
9
4. WeareinterestedintheCPviolationinthedecayofτ. TheCPviolation
intheproductionofτ isexpectedtobemuchlessthanα/πthatiscaused
mainly by the possible existence of the electric dipole moment of tau18.
If we ignore the CP violation in the production then w = w′; namely,
the polarization vectors of τ+ and τ− must be equal and parallel to
eachother in the center-of-masssystem4 of e±. When the energy is near
threshold such as in the case of the Tau-Charm Factory we have s wave
production. In the s wave production we have4
w +w
w =w′ =e 1 2 , (26)
z
1+w w
1 2
b
where w and w are longitudinal polarization of e− and e+ in the inci-
1 2
dent e− direction that is chosen as the z axis. For the B factory energy
w is given in Ref. (4).
5. From Eqs. (17)-(25), we see that there are much more polarization de-
pendenttermsthanthepolarizationindependentterms. Thusthepolar-
ization gives us more handles to find out whether CP is violated or how
it is violated. For example, in Eq. (17) the sine terms involve only A
12
given by Eq. (25) which vanishes when w = 0. The sign of τ± polar-
izationcanbe reversedby reversingthe polarizationofinitial (e−+e+).
This willhelp inisolatingdifferentpolarizationdependent terms suchas
the coefficients ofw q ,w q ,w (q q ), etc.
−→·−→1 −→·−→2 −→· −→1×−→3
3 Conclusions
This is a series of papers investigating new mechanisms of nonstandard CP
violation using τ decay and semileptonic decay of a hadron. If CP violation
is discovered in any one of these decays it will imply the existence of a spin
zero boson x transmitting a CP nonconserving force. After the discovery in
anyone channelwe canthensystematicallyinvestigatehowx is coupledto all
particles. ThiswillenableustoconstructacorrecttheoryofCPviolation. The
attractive feature ofthis mechanismis that it will allow leptons to participate
fully in CP violation. The standard theory is unnatural in the sense that it
will not allow leptons to participate in CP violation.
I am honored to be able to present this paper in memory of Professor
C.S.Wuwhofirstusedangulardistributionofadecayleptonfromapolarized
cobalt nucleus to discover the existence of σ p term and thus the
h−→icobalt ·−→e
existence of the parity violation. The phenomenon discussed in this paper is
a natural extension of her work in the sense that in the e± colliding machine
particle and antiparticle parents (τ±) are produced equally abundantly. By
10