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\def\bea{\begin{eqnarray}}
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\def\nn{\nonumber}
\def\sla#1{\raise.15ex\hbox{$/$}\kern-.57em #1}% Feynman slash
\def\ss{\scriptstyle}
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\def\bd{B_d^0}
\def\bdbar{{\overline{B_d^0}}}
\def\bs{B_s^0}
\def\bsbar{{\overline{B_s^0}}}
\def\barp{{\raise.35ex\hbox
{${\sss (}$}}---{\raise.35ex\hbox{${\sss )}$}}}
\def\bdbarp{\hbox{$B_d$\kern-1.4em\raise1.4ex\hbox{\barp}}}
\def\bsbarp{\hbox{$B_s$\kern-1.4em\raise1.4ex\hbox{\barp}}}
\def\ks{K_{\sss S}}
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\def\roughly#1{\mathrel{\raise.3ex\hbox
{$#1$\kern-.75em\lower1ex\hbox{$\sim$}}}}
\def\lsim{\roughly<}
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\def\mK{m_{\sss K}}
\def\tildeP{{\tilde P}}
\def\ZI{Z_{\sss I}}
\def\ZR{Z_{\sss R}}
\def\tildeZI{{\tilde Z}_{\sss I}}
\def\tildeZR{{\tilde Z}_{\sss R}}
\def\calP{{\cal P}}
%
% Journal and other miscellaneous abbreviations for references
\def\anp#1#2#3{{\it Ann.\ Phys. (NY)} {\bf #1} (19#2) #3}
\def\arnps#1#2#3{{\it Ann.\ Rev.\ Nucl.\ Part.\ Sci.} {\bf #1}, (19#2) #3}
\def\cmp#1#2#3{{\it Comm.\ Math.\ Phys.} {\bf #1} (19#2) #3}
\def\epjc#1#2#3{{\it Eur.\ Phys.\ J.}\ {\bf C#1} (19#2) #3} 
\def\ijmp#1#2#3{{\it Int.\ J.\ Mod.\ Phys.} {\bf A#1} (19#2) #3}
\def\jetp#1#2#3{{\it JETP Lett.} {\bf #1} (19#2) #3}
\def\jetpl#1#2#3#4#5#6{{\it Pis'ma Zh.\ Eksp.\ Teor.\ Fiz.} {\bf #1} (19#2) #3
[{\it JETP Lett.} {\bf #4} (19#5) #6]}
\def\jpb#1#2#3{{\it J.\ Phys.} {\bf B#1} (19#2) #3}
\def\mpla#1#2#3{{\it Mod.\ Phys.\ Lett.} {\bf A#1}, (19#2) #3}
\def\nci#1#2#3{{\it Nuovo Cimento} {\bf #1} (19#2) #3}
\def\npb#1#2#3{{\it Nucl.\ Phys.} {\bf B#1} (#2) #3}
\def\plb#1#2#3{{\it Phys.\ Lett.} {\bf #1B} (#2) #3}
\def\pla#1#2#3{{\it Phys.\ Lett.} {\bf #1A} (19#2) #3}
\def\prb#1#2#3{{\it Phys.\ Rev.} {\bf B#1} (19#2) #3}
\def\prc#1#2#3{{\it Phys.\ Rev.} {\bf C#1} (19#2) #3}
\def\prd#1#2#3{{\it Phys.\ Rev.} {\bf D#1} (19#2) #3}
\def\newprd#1#2#3{{\it Phys.\ Rev.} {\bf D#1}: #3 (19#2)}
\def\newprdtwo#1#2#3{{\it Phys.\ Rev.} {\bf D#1}: #3 (20#2)}
\def\pr#1#2#3{{\it Phys.\ Rev.} {\bf #1} (19#2) #3}
\def\prep#1#2#3{{\it Phys.\ Rep.} {\bf C#1} (19#2) #3}
\def\prl#1#2#3{{\it Phys.\ Rev.\ Lett.} {\bf #1} (19#2) #3}
\def\prltwo#1#2#3{{\it Phys.\ Rev.\ Lett.} {\bf #1} (20#2) #3}
\def\rmp#1#2#3{{\it Rev.\ Mod.\ Phys.} {\bf #1} (19#2) #3}
\def\sjnp#1#2#3#4#5#6{{\it Yad.\ Fiz.} {\bf #1} (19#2) #3
[{\it Sov.\ J.\ Nucl.\ Phys.} {\bf #4} (19#5) #6]}
\def\zpc#1#2#3{{\it Zeit.\ Phys.} {\bf C#1} (#2) #3}
\def \stone{{\it B Decays}, edited by S. Stone (World Scientific, Singapore,
1994)}

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%=======================================================================
% Start of document
% -----------------
\pagestyle{plain}

\begin{document}

\begin{flushright}  
UdeM-GPP-TH-02-96\\
\end{flushright}
\vskip0.5truecm

\begin{center} 

{\large \bf
\centerline{T-violating Triple-Product Correlations in Charmless $\Lambda_b$ 
 Decays}}
\vspace*{1.0cm}
%\vskip1cm
{\large Wafia Bensalem\footnote{email: wafia@lps.umontreal.ca},
  Alakabha Datta\footnote{email: datta@lps.umontreal.ca} 
and David
  London\footnote{email: london@lps.umontreal.ca}} \vskip0.3cm
{\it  Laboratoire Ren\'e J.-A. L\'evesque, Universit\'e de
  Montr\'eal,} \\
{\it C.P. 6128, succ.\ centre-ville, Montr\'eal, QC, Canada H3C 3J7} \\
\vskip0.5cm
%
\bigskip
(\today)
\vskip0.5cm
{\Large Abstract\\}
\vskip3truemm
%
\parbox[t]{\textwidth} {Using factorization, we compute, within the
standard model, the T-violating triple-product correlations in the
charmless decays $\Lambda_b \to F_1 F_2$, where $F_1$ is a light
spin-$1\over 2$ baryon and $F_2$ is a pseudoscalar ($P$) or vector
($V$) meson. We find a large triple-product asymmetry of 18\% for the
decay $\Lambda_b \to p K^-$. However, for other classes of $\Lambda_b
\to F_1 P$ decays, the asymmetry is found to be at most at the percent
level. For $\Lambda_b \to F_1 V$ decays, we find that all
triple-product asymmetries are small (at most $O(1\%)$) for a
transversely-polarized $V$, and are even smaller for longitudinal
polarization. Our estimates of the nonfactorizable contributions to
these decays show them to be negligible, and we describe ways of
testing this.}
%%
%\vskip2cm
\end{center}
\thispagestyle{empty}
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% Decrease texheight (for preprint numbers) again
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\baselineskip=14pt

Over the past two decades, there has been a great deal of theoretical
work examining CP violation in the $B$ system. Most of this work has
focussed on the decays of $B$ mesons. The main reason is that the
indirect CP-violating asymmetries in $B$-meson decays can be used to
extract the interior angles of the unitarity triangle ($\alpha$,
$\beta$ and $\gamma$) with no hadronic uncertainty \cite{CPreview}.
The knowledge of these angles will allow us to test the standard model
(SM) explanation of CP violation. In order to make such measurements,
the $B$-factories BaBar and Belle have been built. These machines
produce copious numbers of $B^0$--${\bar B}^0$ pairs, and have now
provided the first definitive evidence for CP violation outside the
kaon system: $\sin 2\beta = 0.78 \pm 0.08$ \cite{betameas}.

On the other hand, in the coming years machines will be built which
are capable of producing large numbers of $\Lambda_b$ baryons. These
include hadron machines, such as the Tevatron, LHC, etc., as well as
possibly a high-luminosity $e^+ e^-$ machine running at the $Z$ pole.
People have therefore started to examine the SM predictions for a
variety of $\Lambda_b$ decays. This is a worthwhile effort, since it
is conceivable that certain types of new physics will be more easily
detectable in $\Lambda_b$ decays than in $B$ decays. For example,
$\Lambda_b$'s can be used to probe observables which depend on the
spin of the $b$-quark, whereas such observables will be unmeasurable
in $B$-meson decays.

One class of observables which may involve, among other things, the
$b$-quark spin is triple-product correlations. These take the form
$\vec v_1 \cdot (\vec v_2 \times \vec v_3)$, where each $v_i$ is a
spin or momentum. These triple products are odd under time reversal
(T) and hence, by the CPT theorem, also constitute potential signals
of CP violation. By measuring a nonzero value of the asymmetry
%
\beq
A_{\sss T} \equiv 
{{\Gamma (\vec v_1 \cdot (\vec v_2 \times \vec v_3)>0) - 
\Gamma (\vec v_1 \cdot (\vec v_2 \times \vec v_3)<0)} \over 
{\Gamma (\vec v_1 \cdot (\vec v_2 \times \vec v_3)>0) + 
\Gamma (\vec v_1 \cdot (\vec v_2 \times \vec v_3)<0)}} ~,
\label{Toddasym}
\eeq
%
where $\Gamma$ is the decay rate for the process in question, one can
establish the presence of a nonzero triple-product correlation. Note
that there is a well-known technical complication: strong phases can
produce a nonzero value of $A_{\sss T}$, even if there is no CP
violation (i.e.\ if the weak phases are zero). Thus, strictly
speaking, the asymmetry $A_{\sss T}$ is not in fact a T-violating
effect. Nevertheless, one can still obtain a true T-violating signal
by measuring a nonzero value of
%
\beq
{\cal A}_{\sss T} \equiv {1\over 2}(A_T-{\bar A}_T) ~,
\label{Tviolasym}
\eeq
%
where ${\bar A}_{\sss T}$ is the T-odd asymmetry measured in the
CP-conjugate decay process.

Recently, T-violating triple-product correlations were calculated for
the inclusive quark-level decay $b \to s \bar{u} u$ \cite{Wafia}. In
that calculation, all final-state masses were neglected. Ignoring
triple products which involve three spins, only two non-negligible
triple-product asymmetries were found. They are: (i) ${\vec p}_u \cdot
( {\vec s}_u \times {\vec s}_{\bar u} )$ or ${\vec p}_{\bar u} \cdot (
{\vec s}_u \times {\vec s}_{\bar u} )$, and (ii) $\vec s_b \cdot (\vec
p_u\times \vec p_s)$. While the former triple product can be probed in
$B \to V_1 V_2$ decays, where $V_1$ and $V_2$ are vector mesons, the
latter can only be measured in $\Lambda_b$ decays, since the spin of
the $b$ quark is involved.

In this paper we study, within the SM, the triple products in
charmless two-body $\Lambda_b$ decays which are generated by the
quark-level transitions $b \to s \bar{u} u$ or $b \to d \bar{u} u$.
These decays are of the type $\Lambda_b \to F_1 F_2$, where $F_1$ is a
light spin-$1\over 2$ baryon, such as $p$, $\Lambda$, etc., and $F_2$
is a pseudoscalar ($P$) or vector ($V$) meson. Such T-violating
triple-product correlations, along with other P-violating asymmetries,
have been studied for hyperon decays \cite{Pak}, but relatively little
work has been done to study CP violation in $\Lambda_b$ decays.

The decays $\Lambda_b \to F_1 P$ are similar to hyperon decays. As we
will see, there is a triple-product correlation in such decays of the
form ${\vec s}_{\Lambda_b} \cdot ({\vec s}_{F_1} \times {\vec p})$,
where ${\vec s}_{\Lambda_b}$ and ${\vec s}_{F_1}$ are the
polarizations of the $\Lambda_b$ and $F_1$, respectively, and ${\vec
p}$ is the momentum of one of the final-state particles in the rest
frame of the $\Lambda_b$. On the other hand, $\Lambda_b$ decays can
also include a vector meson in the final state, which is not
kinematically accesible for hyperon decays. The decay $\Lambda_b \to
F_1 V$ can give rise to a variety of triple-product correlations
involving the spin of the $\Lambda_b$ and/or $V$.

Many of these triple products involve the spin of the $\Lambda_b$.
Perhaps the easiest way to obtain this quantity is to produce the
$\Lambda_b$ baryons in the decay of an on-shell $Z$ boson. This is
because, in the decay $Z\to b{\bar b}$, the $b$-quarks have a large
average longitudinal polarization of about $-$94\%.  According to
heavy-quark effective theory, this polarization is retained when a
$b$-quark hadronizes into a $\Lambda_b$, and recent measurements of
the average longitudinal polarization of $b$-flavored baryons produced
in $Z^0$ decays (measured through their decay to $ \Lambda_c \ell
\nu_{\ell} X$) is consistent with this conclusion \cite{Opal}. Thus,
the so-called GigaZ option ($2 \times 10^9$ $Z$ bosons per year
\cite{tesla,usaLC}) of a high-luminosity $e^+ e^-$ collider running at
the $Z$ peak would be a particularly good environment for measuring
triple-product correlations in $\Lambda_b$ decays. However, even if
the spin of the $\Lambda_b$ cannot be measured at a given machine,
some of the triple-product correlations in $\Lambda_b \to F_1 V$ do
not involve the polarization of the initial state. Thus, triple
products can be measured at a variety of facilities in which a large
number of $\Lambda_b$ baryons is produced.

We begin our analysis by studying the nonleptonic decay
$\Lambda_b\to F_1 P$. The general form for this amplitude can
be written as
%
\beq
{\cal M}_P= A(\Lambda_b\to F_1 P) = i {\bar u}_{F_1} (a +
b\gamma_{5}) u_{\Lambda_b} ~.
\label{pscalar}
\eeq
%
In order to make contact with the conventional notation for hyperon
decay, we note that, in the rest frame of the parent baryon, the decay
amplitude reduces to 
%
\beq 
A(\Lambda_b\to F_1 P) = i \chi_{F_1} (S+P {\vec\sigma} \cdot
{\hat p}) \chi_{\Lambda_b} ~,
\eeq
%
where ${\hat p}$ is the unit vector along the direction of the
daughter baryon momentum, and
$S=\sqrt{2m_{\Lambda_b}(E_{F_1}+m_{F_1})}a$ and
$P=-\sqrt{2m_{\Lambda_b}(E_{F_1}-m_{F_1})}b$, where $E_{F_1}$ and
$m_{F_1}$ are, respectively, the energy and mass of the final-state
baryon $F_1$. The decay rate and the various asymmetries are given by
%
\beq
\Gamma=\frac{{\vec p}}{8\pi{m_{\Lambda_b}}^{2}}(|S|^2 +|P|^2) ~,~~
\alpha=\frac{2 \, {\rm Re}(S^{*}P)}{|S|^2 +|P|^2} ~,~~
\beta=\frac{2 \, {\rm Im}(S^{*}P)}{|S|^2 +|P|^2} ~,~~
\gamma=\frac{|S|^2 -|P|^2}{|S|^2 +|P|^2} ~.
\label{asymp}
\eeq
%
(Note: above, the quantities $\alpha$, $\beta$ and $\gamma$ should not
be confused with the CP phases of the unitarity triangle, which have
the same symbols.)

The calculation of $|{\cal M}_P|^2$ in Eq.~(\ref{pscalar}) yields
%
\bea
|{\cal M}_P|^2 & = & (|a|^2 - |b|^2) \left(m_{F_1} m_{\Lambda_b} +
 p_{F_1} \cdot s_{\Lambda_b} \, p_{\Lambda_b} \cdot s_{F_1} -
 p_{F_1} \cdot p_{\Lambda_b} \, s_{F_1} \cdot s_{\Lambda_b}
 \right) \nn\\
& &
+ (|a|^2 + |b|^2) \left( p_{F_1} \cdot p_{\Lambda_b} -
m_{F_1} m_{\Lambda_b} s_{F_1} \cdot s_{\Lambda_b} \right) \nn\\
& & 
+ 2 \, {\rm Re}(a b^*) \left( m_{\Lambda_b} p_{F_1} \cdot
s_{\Lambda_b} - m_{F_1} p_{\Lambda_b} \cdot s_{F_1}
\right) \nn \\ 
& & + 2 \, {\rm Im}(a b^*) \epsilon_{\mu\nu\rho\sigma}
p^\mu_{F_1} s^\nu_{F_1} p^\rho_{\Lambda_b}
s^\sigma_{\Lambda_b} ~.
\label{m2scalar}
\eea
%
It is the last term above which gives a triple-product correlation.
(It corresponds to $\beta$ in Eq.~(\ref{asymp}).) In the rest frame of
the $\Lambda_b$, it takes the form ${\vec p}_{F_1} \cdot ({\vec
s}_{F_1} \times {\vec s}_{\Lambda_b})$.

In order to estimate the size of this triple product, we will use
factorization to calculate ${\rm Im}(a b^*)$ at the hadron level. The
starting point is the SM effective hamiltonian for charmless hadronic
$B$ decays \cite{BuraseffH}:
%
\beq 
H_{eff}^q = {G_F \over \protect \sqrt{2}} [V_{ub}V^*_{uq}(c_1O_{1}^q
+ c_2 O_{2}^q) - \sum_{i=3}^{10} V_{tb}V^*_{tq} c_i^t O_i^q] + h.c.,
\label{H_eff}
\eeq
%
where
%
\bea
O_1^q = {\bar q}_\alpha \gamma_\mu L u_\beta \, {\bar u}_\beta
\gamma^\mu L b_\alpha &,& 
O_2^q = {\bar q} \gamma_\mu L u \, {\bar u} \gamma^\mu L b ~, \nn\\
O_{3(5)}^q = {\bar q} \gamma_\mu L b \, \sum_{q'} {\bar q}' \gamma^\mu L (R)
q' &,&
O_{4(6)}^q = {\bar q}_\alpha \gamma_\mu L b_\beta \, \sum_{q'} {\bar
q}'_\beta \gamma^\mu L (R) q'_\alpha ~, \\
O_{7(9)}^q = {3\over 2} {\bar q} \gamma_\mu L b \, \sum_{q'}
e_{q'} {\bar q}' \gamma^\mu R (L) q' &,&
O_{8(10)}^q = {3\over 2} {\bar q}_\alpha \gamma_\mu L b_\beta \,
\sum_{q'} e_{q'} {\bar q}'_\beta \gamma^\mu R (L) q'_\alpha ~. \nn
\label{H_effops}
\eea
%
In the above, $q$ can be either a $d$ or an $s$ quark, depending on
whether the decay is a $\Delta S = 0$ or a $\Delta S = -1$ process,
$q' = d$, $u$ or $s$, with $e_{q'}$ the corresponding electric charge,
and $R(L)= 1 \pm \gamma_5$. The values of the Wilson coefficients
$c_i$ evaluated at the scale $\mu = m_b = 5$ GeV, for $m_t = 176$ GeV
and $\alpha_s(m_Z) = 0.117$, are \cite{FSHe}:
%
\bea
c_1 = -0.324 &,& c_2 = 1.151 ~, \nonumber\\
c^t_3 = 0.017 ~,~~ c^t_4 = -0.037 &,& c^t_5 = 0.010 ~,~~
c^t_6 =-0.045 ~, \nonumber\\
c^t_7 = -1.24\times 10^{-5} ~,~~ c_8^t = 3.77\times 10^{-4} &,&
c_9^t = -0.010 ~,~~ c_{10}^t = 2.06\times 10^{-3} ~.
\eea

In our analysis we will also consider the gluonic dipole operator in
which the gluon splits into two quarks, giving the effective operator
%
\beq
H_{11} = i \, \frac{G_F}{\sqrt{2}} \, \frac{\alpha_s(\mu)}{2\pi k^2}
\, m_b(\mu) \, c_{11} \, V_{tb} V_{ts}^* \, \bar{s}(p_s)
\sigma_{\mu\nu} R T^a b(p_b) \bar{q}(p_2) \gamma^{\mu} T^a
q(p_1)k^{\nu} ~,
\label{H11d}
\eeq
%
where $k=p_b-ps$ and $c_{11} = 0.2$ \cite{Dipole}. It is often useful
to write this in the Fierz-transformed form
%
\beq
H_{11} = -\frac{G_F}{\sqrt{2}} \, \frac{\alpha_s(\mu)}{16\pi} \,
\frac{m_b^2(\mu)}{k^2} \, c_{11} \, \frac{N_c^2-1}{N_c^2} \, V_{tb}
V_{ts}^* \, \left[ \delta_{\alpha \beta}
\delta_{\alpha^{\prime}\beta^{\prime}} -\frac{2N_c}{N_c^2-1}
T^a_{\alpha \beta} T^a_{\alpha^{\prime} \beta^{\prime}}\right]\sum_i
T_i ~,
\eeq
%
where
%
\bea
T_1 & = & 2\bar{s}_{\alpha} \gamma_{\mu}Lq_{\beta}
\bar{q}_{\alpha^{\prime}} \gamma^{\mu}Lb_{\beta^{\prime}} -
4\bar{s}_{\alpha} Rq_{\beta}
\bar{q}_{\alpha^{\prime}}Lb_{\beta^{\prime}} ~, \nonumber\\
T_2 &= & 2\frac{m_s}{m_b}\bar{s}_{\alpha} \gamma_{\mu}Rq_{\beta}
\bar{q}_{\alpha^{\prime}} \gamma^{\mu}Rb_{\beta^{\prime}} -
4\frac{m_s}{m_b}\bar{s}_{\alpha} Lq_{\beta}
\bar{q}_{\alpha^{\prime}}Rb_{\beta^{\prime}} ~, \nonumber\\
T_3 &= & \frac{(p_b+p_s)_{\mu}}{m_b}[ \bar{s}_{\alpha}
\gamma^{\mu}Lq_{\beta} \bar{q}_{\alpha^{\prime}} Rb_{\beta^{\prime}} +
\bar{s}_{\alpha} Rq_{\beta}
\bar{q}_{\alpha^{\prime}}\gamma^{\mu}Rb_{\beta^{\prime}}] ~, \nonumber\\
T_4 &= & \frac{(p_b+p_s)_{\mu}}{m_b}[ i\bar{s}_{\alpha} \sigma^{\mu
\nu}Rq_{\beta} \bar{q}_{\alpha^{\prime}}
\gamma_{\nu}Rb_{\beta^{\prime}} - i\bar{s}_{\alpha}\gamma_{\nu}
Lq_{\beta} \bar{q}_{\alpha^{\prime}}\sigma^{\mu
\nu}Rb_{\beta^{\prime}}] ~,
\eea
%
in which we have defined $\sigma_{\mu \nu} =
\frac{i}{2}[\gamma_{\mu}\gamma_{\nu} -\gamma_{\nu}\gamma_{\mu}]$.

We now apply the effective hamiltonian to specific exclusive
$\Lambda_b$ decays. We will focus on those processes for which
factorization is expected to be a good approximation, namely
colour-allowed decays. We begin with $\Lambda_b \to p K^-$, which is a
$b \to s \bar{u} u$ transition. Factorization allows us to write
%
\beq
A(\Lambda_b\to pK^-)= \sum_{O,O'} \bra{K^-} O \ket{0} \, \bra{p} O'
\ket{\Lambda_b} ~.
\eeq
%
It is straightforward to show that the operators in $H_{eff}^s$ and
$H_{11}$ lead to two classes of terms in the decay amplitude: (a)
$\bra{K^-} {\bar s} \gamma^\mu (1 \pm \gamma_5) u \ket{0} \, \bra{p}
{\bar u} \gamma_\mu (1 \pm \gamma_5) b \ket{\Lambda_b}$, and (b)
$\bra{K^-} {\bar s} (1 \pm \gamma_5) u \ket{0} \, \bra{p} {\bar u} (1
\pm \gamma_5) b \ket{\Lambda_b}$. For the first of these, we define
the pseudoscalar decay constant $f_K$ as
%
\beq
i f_K q^{\mu}= \bra{K} \bar{s} \gamma^{\mu}(1 - \gamma_5)u \ket{0} ~,
\eeq
%
where $q^\mu \equiv p^\mu_{\Lambda_b} - p^\mu_p = p^\mu_K$ is the
four-momentum transfer. For the second, one can show that
%
\beq
\bra{K^-} {\bar s} (1 \pm \gamma_5) u \ket{0} = \mp {f_K m_K^2 \over
m_s + m_u} ~~,~~
\bra{p} {\bar u} (1 \pm \gamma_5) b \ket{\Lambda_b} = {q^\mu\over m_b} 
\bra{p} {\bar u} \gamma_\mu (1 \mp \gamma_5) b \ket{\Lambda_b} ~.
\label{LRtrans}
\eeq
%
(In the second matrix element, we have neglected $m_u$ compared to
$m_b$.) Thus, factorization leads to the following form for the
$\Lambda_b \to p K^-$ amplitude:
%
\beq
A(\Lambda_b\to pK^-)= if_K q^{\mu} \bra{p} \bar{u}
\gamma_\mu(1-\gamma_5) b\ket{\Lambda_b} X_K + i f_K q^{\mu}
\bra{p}\bar{u} \gamma_\mu(1+\gamma_5) b\ket{\Lambda_b} Y_K ~.
\label{ampdef}
\eeq 

Like any CP-violating observable, a nonzero triple product can arise
only if there are two interfering amplitudes. This will occur only if
both $X_K$ and $Y_K$ are nonzero. Since all the operators
$O_1$--$O_{10}$ involve a left-handed $b$-quark, it is clear that $X_K
\ne 0$ in the SM. Furthermore, though it is less obvious, one can also
have $Y_K \ne 0$. Consider, for example, the operator $O_6$ of
Eq.~(\ref{H_effops}). After performing Fierz transformations, this can
be written as
%
\beq
O_6 \sim {\bar s} (1 + \gamma_5) u \, {\bar u} (1 - \gamma_5) b ~.
\label{O6}
\eeq
%
However, according to Eq.~(\ref{LRtrans}), $\bra{p} {\bar u} (1 -
\gamma_5) b \ket{\Lambda_b}$ can be related to $\bra{p} {\bar u}
\gamma_\mu (1 + \gamma_5) b \ket{\Lambda_b}$. Thus, $Y_K$ receives
contributions from operators such as $O_6$. We find
%
\bea
X_K & = & \frac{G_F}{\sqrt{2}} \left[ V_{ub}V_{us}^* a_2-
\sum_{q=u,c,t}V_{qb}V_{qs}^* (a_4^q +a_{10}^q) - V_{tb}V_{ts}^* a_d
\left( 1+\frac{2E_K}{m_b} \right) \right] ~, \nonumber\\
Y_K & = & -\frac{G_F}{\sqrt{2}} \left[ \sum_{q=u,c,t}V_{qb}V_{qs}^*
(a_6^q +a_8^q)+\frac{5}{4} V_{tb} V_{ts}^* a_d \right] \chi_K ~,
\label{pk}
\eea
%
with
%
\beq
\chi_K = \frac{2 m_K^2}{(m_s+m_u)m_b} ~~,~~~~
a_d = \frac{\alpha_s(\mu)}{16\pi} \langle \frac{m_b^2(\mu)}{k^2}
\rangle c_{11} \frac{N_c^2-1}{N_c^2} ~.
\label{pk2}
\eeq
%
In the above, we have defined $a_i^q=c_i^q+ \frac{c_{i+1}^q}{N_c}$ for
$i$ odd and $a_i^q=c_i^q+ \frac{c_{i-1}^q}{N_c}$ for $i$ even. We
estimate the average gluon momentum in the dipole operator to be
$\langle m_b^2/k^2 \rangle = \int \phi_K(x)m_b^2/k^2 dx$, where the
gluon momentum in the heavy-quark limit is $k^2=m_b^2(1-x)$ and
$\phi_K$ is the kaon light-cone distribution. Choosing the asymptotic
form $\phi_K=6x(1-x)$, we find $\langle m_b^2/k^2 \rangle = 3$, which
leads to $a_d = 0.0021$.

Now, the vector and axial-vector matrix elements between the
$\Lambda_b$ and $p$ can be written in the general form
%
\begin{eqnarray}
\bra{p} \bar{u} \gamma^\mu b \ket{\Lambda_b} & =& \bar{u}_{p} \left[
f_1 \gamma^\mu + i \frac{f_2}{m_{\Lambda_b}}\sigma^{\mu\nu} q_\nu +
\frac{f_3}{m_{\Lambda_b}} q^\mu \right] u_{\Lambda_b} \nonumber \\
\bra{p} \bar{u} \gamma^\mu \gamma_5 b \ket{\Lambda_b} & =& \bar{u}_{p}
\left[ g_1 \gamma^\mu + i \frac{g_2}{m_{\Lambda_b}}\sigma^{\mu\nu}
q_\nu + \frac{g_3}{m_{\Lambda_b}} q^\mu \right] \gamma_5 u_{\Lambda_b}
~,
\label{genamps}
\end{eqnarray}
%
where the $f_i$ and $g_i$ are Lorentz-invariant form
factors. Heavy-quark symmetry imposes constraints on these form
factors. A systematic expansion of these form factors, including
$1/m_b$ corrections, has been calculated \cite{Dattaff}: in the
$m_b\to \infty$ limit, one obtains the relations
%
\beq
f_1 = g_1 ~~,~~~~ f_2 = g_2 = f_3 = g_3 ~.
\label{hqet}
\eeq
%
Using the above expressions, we find that the parameters $a$ and $b$
of Eq.~(\ref{pscalar}) can be written as
%
\bea
a_K & = & f_K (X_K+Y_K)\left[(m_{\Lambda_b}-m_p)f_1
+f_3\frac{m_K^2}{m_{\Lambda_b}} \right] ~, \nonumber\\
b_K & = & f_K (X_K-Y_K)\left[(m_{\Lambda_b}+m_p)g_1
-g_3\frac{m_K^2}{m_{\Lambda_b}} \right] ~.
\label{abTP}
\eea

According to Eq.~\ref{m2scalar}, the triple product in $\Lambda_b\to
pK^-$ is proportional to ${\rm Im}(a_K b_K^*)$, which is in turn
proportional to ${\rm Im}(X_K Y_K^*)$. Since $X_K$ and $Y_K$ are both
nonzero, and have different weak phases [Eq.~(\ref{pk})], we expect a
nonzero triple-product asymmetry in $\Lambda_b\to pK^-$ of the form
${\vec p}_{p} \cdot ({\vec s}_{p} \times {\vec s}_{\Lambda_b})$. At
first sight, this appears to contradict the results of
Ref.~\cite{Wafia}, since no triple products involving two spins were
found in the quark-level decay $b \to s \bar{u} u$. However, note that
$Y_K$ is proportional to $\chi_K$, which is formally suppressed by
$1/m_b$.  Thus, in the limit $m_b \to \infty$, one has $Y_K=0$, so
that the triple-product correlation will vanish. This agrees with the
conclusions of Ref.~\cite{Wafia}, which neglects the masses of the
final-state quarks (i.e.\ the limit $m_b \to \infty$ is implicitly
assumed).

However, the key point is that, for finite $m_b$, $\chi_K$ is not
small because of the presence of the chiral enhancement term
$m_K^2/(m_s+m_u)$. In fact, for $m_s=100$ MeV and $m_b= 5$ GeV,
$\chi_K \sim 1$, and hence is clearly non-negligible. The
triple-product asymmetry of ${\vec p}_{p} \cdot ({\vec s}_{p} \times
{\vec s}_{\Lambda_b})$ may therefore be sizeable. Note that this
triple product requires the measurement of both the $\Lambda_b$ and
the $p$ polarizations. If the measurement of the proton polarization
is not possible, one can instead consider a final state with an
excited nucleon, such as $\Lambda_b \to N(1440) K^-$. In this case the
polarization of the $N(1440)$ can be determined from its decay
products. (Alternatively, one can consider the decay $\Xi_b \to
\Sigma^+ K^-$, where $\Xi_b$ has quark content $bus$.)

Note also that in Eq.~(\ref{pk}) we have included the up- and
charm-quark penguin pieces, proportional to $V_{ub}V_{us}^*$ and
$V_{cb}V_{cs}^*$ respectively. These are generated by rescattering of
the tree-level operators in the effective Hamiltonian in
Eq.~(\ref{H_eff}). As we will see, the contributions from these
rescattering terms are very important. The coefficients associated
with these terms are given by
%
\bea
c_{3,5}^i = -c_{4,6}^i/N_c = P^i_s/N_c &,&
c_{7,9}^i = P^i_e ~,~~
c_{8,10}^i = 0 ~,~~
i=u,c ~,
\label{coeffs}
\eea
%
where $N_c$ is the number of colours. The leading contributions to
$P^i_{s,e}$ are given by $P^i_s = ({\frac{\alpha_s}{8\pi}}) c_2
({\frac{10}{9}} +G(m_i,\mu,q^2))$ and $P^i_e =
({\frac{\alpha_{em}}{9\pi}}) (N_c c_1+ c_2) ({\frac{10}{9}} +
G(m_i,\mu,q^2))$, in which the function $G(m,\mu,q^2)$ takes the form
%
\begin{eqnarray}
G(m,\mu,q^2) = 4\int^1_0 x(1-x) \mbox{ln}{m^2-x(1-x)q^2\over
\mu^2} ~\mbox{d}x ~,
\label{rescatt}
\end{eqnarray}
%
where $q$ is the momentum carried by the virtual gluon in the penguin
diagram. Of course, we are really interested in the matrix elements
of the various operators for the decay $\Lambda_b \to p K^-$, and so
the coefficients in Eq.~(\ref{coeffs}) should be understood to be
%
\beq
{\bar{c}}_i^{u,c} = \frac{\bra{p K^-}c_i^{u,c}(q^2)O_i\ket{\Lambda_b}}
{\bra{p K^-}O_i\ket{\Lambda_b}} ~.
\eeq
%
We will henceforth drop the distinction between ${\bar{c}}_i^{u,c}$
and $c_i^{u,c}$, with the understanding that it is the
${\bar{c}}_i^{u,c}$ which appear in the amplitude.

The analysis of other colour-allowed $\Lambda_b$ decays follows
straightforwardly from that for $\Lambda_b \to p K^-$. For example,
consider $\Lambda_b \to p \pi^-$, which is generated by the
quark-level decay $b \to d \bar{u} u$. The amplitude for $\Lambda_b
\to p \pi^-$ is given by Eq.~(\ref{pscalar}), with
%
\bea
a_\pi & = & f_{\pi} (X_\pi+Y_\pi)\left[(m_{\Lambda_b}-m_p)f_1
+f_3\frac{m_\pi^2}{m_{\Lambda_b}} \right] ~, \nonumber\\
b_\pi & = & f_{\pi} (X_\pi-Y_\pi)\left[(m_{\Lambda_b}+m_p)g_1
-g_3\frac{m_\pi^2}{m_{\Lambda_b}} \right] ~,
\eea
%
where
%
\bea
X_\pi & = & \frac{G_F}{\sqrt{2}} \left[ V_{ub}V_{ud}^* a_2-
\sum_{q=u,c,t}V_{qb}V_{qd}^* (a_4^q +a_{10}^q) - V_{tb}V_{td}^*
a_d(1+\frac{2E_{\pi}}{m_b}) \right] ~, \nonumber\\
Y_\pi & = & -\frac{G_F}{\sqrt{2}} \left[ \sum_{q=u,c,t}V_{qb}V_{qd}^*
(a_6^q +a_8^q)+\frac{5}{4} V_{tb} V_{td}^* a_d \right] \chi_{\pi} ~,
\label{ppi}
\eea
%
with
%
\beq
\chi_{\pi} = \frac{2 m_{\pi}^2}{(m_d+m_u)m_b} ~.
\eeq

Finally, we consider the decay $\Lambda_b \to \Lambda \eta
(\eta^{\prime})$ \cite{Dattaeta}, which is dominated by a
colour-allowed $b\to s$ penguin transition (there is also a small
colour-suppressed tree contribution). For the decay $\Lambda_b \to
\Lambda \eta$ we get
%
\bea
a_\eta & = & f_\pi (X_\eta+Y_\eta)\left[(m_{\Lambda_b}-m_{\Lambda})f_1
+f_3\frac{m_{\eta}^2}{m_{\Lambda_b}} \right] ~, \nonumber\\
b_\eta & = & f_\pi (X_\eta-Y_\eta)\left[(m_{\Lambda_b}+m_{\Lambda})g_1 
-g_3\frac{m_{\eta}^2}{m_{\Lambda_b}}
\right] ~,
\eea
%
where
%
\bea
X_\eta & = & \frac{G_F}{\sqrt{2}} \left[ V_{ub}V_{us}^* a_1r_1-
\sum_{q=u,c,t}V_{qb}V_{qs}^* (r_1A_q +r_2B_q) - V_{tb}V_{ts}^* r_2 a_d
\left( 1+\frac{2E_{\eta}}{m_b} \right) \right] ~, \nonumber\\
Y_\eta & = & -\frac{G_F}{\sqrt{2}} \left[ \sum_{q=u,c,t}V_{qb}V_{qs}^*
(a_6^q -\frac{1}{2}a_8^q)+\frac{5}{4} V_{tb} V_{ts}^* a_d \right]
r_2 \chi_{\eta} ~,
\label{leta}
\eea
%
with
%
\bea
A_q & = & 2a_3^q-2a_5^q-\frac{1}{2}a_7^q +\frac{1}{2}a_9^q ~,
\nonumber\\
B_q & = & a_3^q+a_4^q-a_5^q+\frac{1}{2}a_7^q -\frac{1}{2}a_9^q
-\frac{1}{2}a_{10}^q ~, \nonumber\\
\chi_{\eta} & = & \frac{m_{\eta}^2}{m_s m_b} ~.
\eea 
%
In the above, we have defined $r_1=f_{\eta}^u/f_{\pi}$ and
$r_2=f_{\eta}^s/f_{\pi}$, with
%
\bea
if_{\eta}^u p_{\eta}^{\mu} & = &
\bra{\eta}\bar{u}\gamma^{\mu}(1-\gamma_5)u\ket{0}
=\bra{\eta}\bar{d}\gamma^{\mu}(1-\gamma_5)d\ket{0} ~, \nonumber\\
if_{\eta}^sp_{\eta}^{\mu} & = &
\bra{\eta}\bar{s}\gamma^{\mu}(1-\gamma_5)s\ket{0} ~.
\eea
%
The amplitude for $\Lambda_b \to \Lambda \eta^{\prime}$ has the same
form as Eq.~(\ref{leta}) with the replacement $\eta \to
\eta^{\prime}$. Note that the polarization of the final-state
$\Lambda$ can be measured via its decay $\Lambda \to p \pi^{-}$.

The above analysis has been performed within the framework of
factorization. Before turning to estimates of the size of the
triple-product asymmetries, it is useful at this point to address the
issue of nonfactorizable corrections. Nonfactorizable effects are
known to be important for hyperon and charmed-baryon nonleptonic
decays, but are expected to be negligible for non-leptonic $\Lambda_b$
decays. An unambiguous signal for the presence of nonfactorizable
effects would be the observation of the decay $\Lambda_b \to \Delta^+
K^- (\pi^-)$, $\Lambda_b \to \Sigma \eta (\eta^\prime)$, or $\Lambda_b
\to \Sigma \phi$. This is because, for the factorizable contribution,
the light diquark in the $\Lambda_b$ baryon remains inert during the
weak decay. Thus, since the light diquark is an isosinglet, and since
strong interactions conserve isospin to a very good approximation, the
above $\Lambda_b$ decays are forbidden within factorization
\cite{Dattalip}.

One way to estimate the size of nonfactorizable corrections is by
using the pole model. In this model, one assumes that the
nonfactorizable decay amplitude receives contributions primarily from
one-particle intermediate states, and that these contributions then
show up as simple poles in the decay amplitude. An example of
intermediate single-particle states is the ground-state
positive-parity baryons. Consider the decay $\Lambda_b \to p K^-$. One
nonfactorizable contribution is described by the diagram in which
there is a $\Lambda_b \to \Sigma^0$ weak transition through a $W$
exchange, followed by the strong decay $\Sigma^0 \to p K^-$. The pole
contribution to the parity-violating amplitude, $a$, in
Eq.~(\ref{pscalar}) is known to be small for charmed-baryon decays
\cite{Cheng1}, and we assume this to be the case here as well. For the
parity-conserving amplitude, $b$, in Eq.~(\ref{pscalar}), we can then
write
%
\beq
b_{nonfac} \sim V_{ub}V_{us}^* \frac{\bra{\Sigma^0}H_w\ket{\Lambda_b}}
{m_{\Lambda_b}-m_{\Sigma_0}} g_{\Sigma^0 pK^{-}} ~,
\eeq
%
where $g_{\Sigma^0 pK^{-}}$ is the strong-coupling vertex which will
depend on the energy of the emitted kaon. We can use heavy-quark and
flavour $SU(3)$ symmetry to set $\bra{\Sigma^0}H_w\ket{\Lambda_b} \sim
\bra{\Sigma^+}H_w\ket{\Lambda_c}$. Writing the weak matrix element
$\bra{\Sigma^+}H_w\ket{\Lambda_c}= \frac{G_F}{\sqrt{2}}m^3$, we obtain
%
\beq
{b_{nonfac} \over b_{fac}} \sim \frac{m}{f_K}\frac{m^2}
{(m_{\Lambda_b}-m_{\Sigma^0})(m_{\Lambda_b}+m_p)}g_{\Sigma^0 pK^{-}}
~,
\label{nf}
\eeq
%
where we have chosen the tree-level term for $A_{fac}$. Since the
emitted kaon is hard and since the quarks inside it are energetic, the
strong coupling $g_{\Sigma^0 p K^-}\sim \alpha_s(\mu \sim E_K \sim
m_b)$. In other words, the offshell $\Sigma^0$ has to emit a hard
gluon to create a $\bar{u} u$ pair to form the $p K^-$ final
state. The matrix element $m$ can either be estimated using a model
\cite{Cheng1}, or obtained from a fit to the charmed baryon decay
$\Lambda_c \to \Sigma^0 \pi^+$ \cite{Dattanl}. In both cases one
obtains $m \sim 0.1-0.2$ GeV, so that, from Eq.~(\ref{nf}), the
nonfactorizable corrections are found to be tiny. Arguments for small
nonfactorizable effects in $\Lambda_b$ decays can also be made based
on the total width calculations \cite{Cheng2}.

To summarize: for colour-allowed $\Lambda_b\to F_1 P$ decays, we find
that the triple-product correlation ${\rm Im}(a b^*) {\vec p}_{F_1}
\cdot ({\vec s}_{F_1} \times {\vec s}_{\Lambda_b})$ can be nonzero.
The next step is to calculate the size of the asymmetry ${\cal
A}_{\sss T}$ in Eq.~(\ref{Tviolasym}) for the various decays.

We begin with $\Lambda_b \to p K^-$. For this decay, we use the
expressions for $a_K$ and $b_K$ found in Eq.~(\ref{abTP}). We note
that the $f_3$ ($g_3$) term is suppressed relative to the $f_1$
($g_1$) term by a factor $m_K^2/m_{\Lambda_b}^2 \sim 0.01$, and so can
be neglected (and similarly for the $g_3$ piece). Furthermore, we take
$f_1 = g_1$ [Eq.~(\ref{hqet})], in which case all dependence on this
form factor cancels in $A_{\sss T}$ [Eq.~(\ref{Toddasym})]. The
quantities $a_K$ and $b_K$ depend on the parameters of the
Cabibbo-Kobayashi-Maskawa (CKM) matrix, whose values are taken to be
%
\beq
\rho = 0.17 ~~,~~~~ \eta = 0.39 ~.
\eeq
%
For the chiral enhancement term $\chi_K$ [Eq.~(\ref{pk2})], we
take $\chi_K = 1$.

In order to estimate the value of rescattering terms, one has to
choose a value of $q^2$. We consider two possibilities:
%
\beq
{\rm Model~1:}~~q^2 = {m_b^2 \over 4} ~~,~~~~ 
{\rm Model~2:}~~q^2 = {m_b^2 \over 2} ~.
\eeq
%
Taking $m_c = 1.4$ GeV, $m_u = 6$ MeV and $m_b = 5$ GeV, and writing
$c_4^c = |c_4^c| e^{i \delta^c}$, we find
%
\bea
& {\rm Model~1:} & |c_4^u| = 0.02 ~~,~~ |c_4^c| = 0.02 ~~,~~ \delta^c
= 51^\circ ~, \nn\\
& {\rm Model~2:} & |c_4^u| = 0.021 ~~,~~ |c_4^c| = 0.015 ~~,~~ \delta^c
= 0 ~.
\eea
%
(In accordance with CPT, we set the phase of $c_4^u$ to zero
\cite{Hou}.)

Before presenting the numerical analysis, it is useful to anticipate
the results. Referring to Eqs.~(\ref{m2scalar}), (\ref{abTP}) and
(\ref{pk}), we expect the triple-product asymmetry to be of order
%
\beq
{2 {\rm Im} (a_Kb_K^*) \over |a_K|^2 + |b_K|^2 + 2 {\rm Re}(a_Kb_K^*)}
\simeq {{\rm Im}(X_K Y_K^*) \over |X_K|^2} \simeq {a_2 a_6 \eta
\lambda^2 \over a_2^2 \lambda^4 + a_4^2} = 24\% ~.
\label{Pestimate}
\eeq
%
Of course, this is a back-of-the-envelope estimate, but it does
indicate that we can expect a reasonably large asymmetry, even when
the rescattering effects are included. The fundamental reason for this
is the following: the triple product is due mainly to the interference
of the $V_{ub}V_{us}^*$ piece of $X_K$ (we refer to this as $T$, the
``tree'') and the $V_{tb}V_{ts}^*$ piece of $Y_K$ ($P$, the
``penguin''). Like any CP-violating quantity, the asymmetry will
therefore be maximized when the two interfering amplitudes are of
comparable size. A quick calculation of these two quantities in Model
2 above shows that $|T/P| = 0.35$. The two amplitudes are therefore
similar in size, leading to the sizeable asymmetry estimate above.

We have performed the phase-space integration for $\Lambda_b \to p
K^-$ using the computer program RAMBO. For Model 1, we find that
$A_{\sss T} = -20.8\%$ and ${\bar A}_{\sss T} = +15.0\%$, leading to a
T-violating asymmetry of ${\cal A}_{\sss T}^{pK} = -17.9\%$. In Model
2, since the strong phase vanishes, one necessarily has ${\bar
A}_{\sss T} = -A_{\sss T}$, and we find ${\cal A}_{\sss T}^{pK} =
-19.1\%$. (We note in passing that the rescattering effects are quite
important. Without them, the asymmetry would be ${\cal A}_{\sss
T}^{pK} = -26.1\%$. Thus, their inclusion leads to a correction in the
asymmetry of about 25\%.) These numbers are all consistent with the
estimate in Eq.~(\ref{Pestimate}). We therefore conclude that the SM
predicts a sizeable triple-product asymmetry in the decay $\Lambda_b
\to p K^-$. (Note that, since the estimate in Eq.~(\ref{Pestimate})
uses only the values of the Wilson coefficients and the CKM matrix
elements, we expect a large asymmetry even if nonfactorizable
contributions are present.)

There is one digressionary remark which is worth making here. From the
measurement of $\epsilon_K$, the CP-violating parameter in the kaon
system, we know that the product $B_K \eta$ is positive, where $B_K$
is the kaon bag parameter and $\eta$ is the CP-violating CKM
parameter. It is usually assumed that $B_K > 0$, so that $\eta$ is
also positive, and the unitarity triangle points up. However, there is
no experimental evidence yet that $B_K > 0$. The T-violating
triple-product asymmetry in $\Lambda_b \to p K^-$ is proportional to
$\eta \cos(\delta)$, where $\delta$ is a strong phase. If one assumes
that $|\delta| < 90^\circ$, which is strongly favoured theoretically,
then the triple product asymmetry measures the sign of $\eta$. This
provides a cross check to the information obtained from the kaon
system.

Turning now to the decay $\Lambda_b \to p \pi^-$, we have applied this
same analysis as above. Taking $m_d = m_u = 6$ MeV, we have $\chi_\pi
= 0.65$. In this case, the tree amplitude $T$ is larger than the
penguin amplitude $P$, with $|P/T| = 0.08$. Because these two
interfering amplitudes are less comparable in size than was the case
for $\Lambda_b \to p K^-$, we expect a correspondingly smaller
asymmetry. This is indeed what is found. In Model 1, we have $A_{\sss
T} = 6.3\%$ and ${\bar A}_{\sss T} = -4.5\%$, so that ${\cal A}_{\sss
T}^{p\pi} = 5.4\%$. Model 2 gives a similar asymmetry: ${\cal A}_{\sss
T}^{p\pi} = 5.6\%$.

For the decays $\Lambda_b \to \Lambda \eta$ and $\Lambda_b \to \Lambda
\eta'$, we have to define the quark content and mixing of the physical
$\eta$ and $\eta'$ mesons. We use the Isgur mixing \cite{Isgur}:
%
\beq
\bra{\eta} = \frac{1}{\sqrt{2}}[N-S] ~~,~~~~
\bra{\eta^{\prime}}= \frac{1}{\sqrt{2}}[N+S] ~,
\eeq
%
where $N= [\bra{u\bar{u}} + \bra{d\bar{d}}]/\sqrt{2}$ and $S=
\bra{s\bar{s}}$. $SU(3)$ symmetry then gives
%
\beq
f_{\eta}^u=f_{\pi}/2 ~~,~~~~
f_{\eta}^s=-f_{\pi}/\sqrt{2} ~~,~~~~
f_{\eta^{\prime}}^u=f_{\pi}/2 ~~,~~~~
f_{\eta^{\prime}}^s=f_{\pi}/\sqrt{2} ~,
\eeq
%
where $f_{\pi}=131$ MeV. We also take $\chi_\eta = 0.6$ and
$\chi_{\eta'} = 1.8$. For both decays the interfering amplitudes are
very different in size: $|T/P| = 0.03$ and 0.01 for the $\Lambda\eta$
and $\Lambda\eta'$ final states, respectively. We can therefore expect
to obtain tiny triple-product asymmetries, and this should hold even
if nonfactorizable effects are present. For $\Lambda_b \to \Lambda
\eta$ we have ${\cal A}_{\sss T}^{\Lambda\eta} = 0.6\%$ (Model 1) or
0.9\% (Model 2), while for $\Lambda_b \to \Lambda \eta'$, ${\cal
A}_{\sss T}^{\Lambda\eta'} = -0.6\%$ (Model 1) or $-0.5\%$ (Model
2). It is unlikely that such tiny asymmetries can be
measured. However, this also suggests that these processes might be
good areas to search for new physics \cite{newpaper}.

We now turn to the decays $\Lambda_b\to F_1 V$. The general decay
amplitude can be written as \cite{SPT}
%
\beq
{\cal M}_V= Amp(\Lambda_{F_1}\to B V) = {\bar u}_{F_1}
\varepsilon^*_\mu \left[ (p_{\Lambda_b}^\mu + p_{F_1}^\mu)
(a+b\gamma_{5}) + \gamma^{\mu} (x+y\gamma_{5}) \right] u_{\Lambda_b}
~,
\label{vector}
\eeq
%
where $\varepsilon^*_\mu$ is the polarization of the vector meson. In
the rest frame of the $\Lambda_b$, we can write $p_V =
(E_V,0,0,|\vec{p}|)$ and $p_{F_1} = (E_{F_1},0,0,-|\vec{p}|)$. Thus,
it is clear that $\varepsilon^*_V \cdot (p_{\Lambda_b} + p_{F_1})$
will be nonzero only for a longitudinally-polarized $V$. This will be
important in what follows.

The calculation of $|{\cal M}_V|^2$ gives the following triple-product
terms:
%
\bea
|{\cal M}_V|^2_{t.p.} & = & 2 \, {\rm Im}(a b^*) \left\vert
\varepsilon_V \cdot (p_{\Lambda_b} + p_{F_1}) \right\vert^2 \,
\epsilon_{\mu\nu\rho\sigma} p^\mu_{F_1} s^\nu_{F_1} p^\rho_{\Lambda_b}
s^\sigma_{\Lambda_b} \nn\\
&& + 2 \, {\rm Im} \left(x y^*\right) \epsilon_{\alpha\beta\mu\nu}
\left[ \varepsilon_V \cdot s_{F_1} p_{F_1}^\alpha p_{\Lambda_b}^\beta
s_{\Lambda_b}^\mu \varepsilon_V^\nu - \varepsilon_V \cdot p_{F_1}
s_{F_1}^\alpha p_{\Lambda_b}^\beta s_{\Lambda_b}^\mu \varepsilon_V^\nu
\right. \nn\\
&& \hskip1.2truein \left. + \varepsilon_V \cdot s_{\Lambda_b}
p_{F_1}^\alpha s_{F_1}^\beta \varepsilon_V^\mu p_{\Lambda_b}^\nu -
\varepsilon_V \cdot p_{\Lambda_b} p_{F_1}^\alpha s_{F_1}^\beta
\varepsilon_V^\mu s_{\Lambda_b}^\nu \right] \nn\\
&& + 2 \, \varepsilon_V \cdot \left( p_{\Lambda_b} + p_{F_1} \right)
\epsilon_{\alpha\beta\mu\nu} \left[ {\rm Im} \left( a x^* + b y^*
\right) p_{F_1}^\alpha s_{F_1}^\beta p_{\Lambda_b}^\mu
\varepsilon_V^\nu \right. \nn\\
&& \hskip1.8truein + m_{\Lambda_b} {\rm Im} \left( b x^* + a y^*
\right) p_{F_1}^\alpha s_{F_1}^\beta s_{\Lambda_b}^\mu
\varepsilon_V^\nu \nn\\
&& \hskip1.8truein - {\rm Im} \left( a x^* - b y^* \right)
p_{F_1}^\alpha p_{\Lambda_b}^\beta s_{\Lambda_b}^\mu \varepsilon_V^\nu
\nn\\ && \hskip1.8truein \left. - m_{F_1} {\rm Im} \left( a y^* - b
x^* \right) s_{F_1}^\alpha p_{\Lambda_b}^\beta s_{\Lambda_b}^\mu
\varepsilon_V^\nu \right] ~.
\label{vecTPs}
\eea
%
Note that if we sum over the polarization of the vector meson, we
essentially reproduce the results found for $\Lambda_b\to F_1 P$. That
is, there is only one triple product, which takes the form
$\epsilon_{\mu\nu\rho\sigma} p^\mu_{F_1} s^\nu_{F_1}
p^\rho_{\Lambda_b} s^\sigma_{\Lambda_b}$.

As usual, we use factorization to calculate the coefficients $a$, $b$,
$x$ and $y$. Consider first the decay $\Lambda_b \to p K^{*-}$. We
define the decay constant ${g_{K^*}}$ as
%
\beq
m_{K^*}g_{K^*}\varepsilon_{\mu}^{*} = \bra{K^*}
\bar{s}\gamma_{\mu}u\ket{0} ~.
\label{K*vector}
\eeq
%
In general, factorization allows us to write
%
\bea
A(\Lambda_b\to pK^{*-}) &=& m_{K^*} g_{K^*} \left\{
\varepsilon_{\mu}^{*} \bra{p} \bar{u} \gamma^\mu(1-\gamma_5)
b\ket{\Lambda_b} X_{K^*} \right. \nn\\
&& \qquad\qquad + \, \varepsilon_{\mu}^{*} \bra{p}\bar{u}
\gamma^\mu(1+\gamma_5) b\ket{\Lambda_b} Y_{K^*} \\
&& \qquad\qquad + \, \varepsilon \cdot (p_{\Lambda_b} + p_p) \, q_\mu \,
\bra{p}\bar{u} \gamma^\mu(1-\gamma_5) b\ket{\Lambda_b} A_{K^*} \nn\\
&& \qquad\qquad \left.  + \, \varepsilon \cdot (p_{\Lambda_b} + p_p) \,
q_\mu \, \bra{p}\bar{u} \gamma^\mu(1+\gamma_5) b\ket{\Lambda_b}
B_{K^*} \right\}~. \nn
\eea
%
The coefficients $X_{K^*}$, $Y_{K^*}$, $A_{K^*}$ and $B_{K^*}$ can be
calculated using the effective hamiltonian. As noted earlier, $A_{K^*}$
and $B_{K^*}$ are nonzero only for a longitudinally-polarized
$K^{*-}$.

Consider first the operators $O_1$--$O_{10}$. Since all of these lead
to $K^{*-}$ matrix elements of the form in Eq.~(\ref{K*vector}), none
of them can contribute to $A_{K^*}$ and $B_{K^*}$. Furthermore, one
can show that none of these give $Y_{K^*} \ne 0$ either. For example,
consider again the operator $O_6$, which led to $Y_K \ne 0$. Because
$\bra{K^{*-}} {\bar s} (1 + \gamma_5) u \ket{0} = 0$, $O_6$ will not
contribute to $Y_{K^*}$. Thus, within factorization, if we restrict
ourselves only to the operators $O_1$--$O_{10}$, the only nonzero
coefficient is $X_{K^*}$, which means that all triple products vanish,
since there is only a single decay amplitude.

In order to generate triple products in $\Lambda_b \to p K^{*-}$, it
is necessary to consider the dipole operator $O_{11}$, whose effective
coefficient is rather small (Eq.~(\ref{pk2}): $a_d = 0.0021$).
However, there is an important observation one can make. The
contributions of $O_{11}$ to $Y_{K^*}$, $A_{K^*}$ and $B_{K^*}$ all
involve the tensor matrix element for $K^{*-}$, which we define as
%
\beq
-ig_{K^*}^T\left[\varepsilon_{\mu}^{*}p_{\nu}^{K^*}-
  \varepsilon_{\nu}^{*}p_{\mu}^{K^*}\right] = \bra{K^*}
\bar{s}\sigma_{\mu \nu}u\ket{0} ~.
\label{K*tensor}
\eeq
%
Now, in the rest frame of the $\Lambda_b$, we can write
$p_{K^{*}}=(E_{K^*},0,0,|\vec{p}_{K^*}|)$. In the heavy-quark limit,
in which $E_{K^*} \gg m_{K^*}$, the longitudinal polarization vector
can be written approximately as
%
\beq
\varepsilon_\mu^{\lambda=0} \simeq {1 \over m_{K^*}} \left(
p_{K^*}^\mu + {m_{K^*}^2 \over 2 E_{K^*}} n^\mu \right),
\eeq
%
with $n^\mu = (-1,0,0,1)$. From Eq.~(\ref{K*tensor}), we see that the
piece of $\varepsilon_\mu^{\lambda=0}$ which is proportional to
$p_{K^*}^\mu$ will not contribute to the matrix element. Thus, in the
heavy-quark limit, we have $A_{K^*} \simeq B_{K^*} \approx 0$.
Furthermore, we expect that the value of $Y_{K^*}$ for a
longitudinally-polarized $K^{*-}$ meson will be suppressed relative to
that for a transversely-polarized $K^{*-}$ by about $m_{K^*} / 2
E_{K^*} = 16\%$. As we will see, $Y_{K^*}$ is already small for a
transversely-polarized $K^{*-}$, so that $Y_{K^*} \approx 0$ for
longitudinal polarization. Therefore, any triple products in
$\Lambda_b \to p K^{*-}$ should be largest for a
transversely-polarized $K^{*-}$, although we expect even these to be
small.

Considering separately the longitudinal ($\lambda = 0$) and transverse
($\lambda = \perp$) polarizations of the final-state vector meson, we
find
%
\bea
X_{K^*}^{\lambda=\perp} & = & \frac{G_F}{\sqrt{2}} \left[ V_{ub}V_{us}^*
a_2- \sum_{q=u,c,t}V_{qb}V_{qs}^* (a_4^q +a_{10}^q) - V_{tb}V_{ts}^*
\frac{a_d}{2} \right] ~, \nonumber\\
X_{K^*}^{\lambda=0} & = & \frac{G_F}{\sqrt{2}} \left[ V_{ub}V_{us}^* a_2-
\sum_{q=u,c,t}V_{qb}V_{qs}^* (a_4^q +a_{10}^q) - V_{tb}V_{ts}^* a_d
\left( 1+\frac{2E_{K^*}}{m_b} \right) \right] ~, \nonumber\\
Y_{K^*}^{\lambda=\perp} & = & \frac{G_F}{\sqrt{2}}V_{tb}V_{ts}^* a_d ~, \nn\\
Y_{K^*}^{\lambda=0} & \approx & 0 ~,
\label{pkstar}
\eea
%
where
%
\beq
z \equiv \frac{E_{K^*}}{m_{K^*}}\frac{g_{K^*}^T}{g_{K^*}} ~.
\eeq
%
For $g_{K^*} = 226$ MeV and $g_{K^*}^T = 160$ MeV \cite{Ball}, $z =
2.23$. To a good approximation, the quantities $a$, $b$, $x$ and $y$
of Eq.~(\ref{vector}) can then be expressed as
%
\bea
a_{K^*}^\lambda &=& m_{K^{*}} g_{K^{*}}\frac{f_{2}}{m_{\Lambda_b}}
[X_{K^*}^{\lambda}+ z Y_{K^*}^{\lambda}] ~, \nn\\
b_{K^*}^\lambda &=& -m_{K^{*}} g_{K^{*}}\frac{g_{2}}{m_{\Lambda_b}}
[X_{K^*}^{\lambda}- z Y_{K^*}^{\lambda}] ~, \nn\\
x_{K^*}^\lambda &=& m_{K^{*}} g_{K^{*}} [f_{1}-
\frac{m_p+m_{\Lambda_b}}{m_{\Lambda_b}}f_{2}][X_{K^*}^{\lambda}+ z
Y_{K^*}^{\lambda}] ~, \nonumber\\
y_{K^*}^\lambda &=& -m_{K^{*}} g_{K^{*}} [g_{1}+
\frac{m_{\Lambda_b}-m_p}{m_{\Lambda_b}}g_{2}][X_{K^*}^{\lambda}- z
Y_{K^*}^{\lambda}] ~.
\label{abxyvec}
\end{eqnarray}

There are several points to be deduced from the above results. First,
since $Y_{K^*}^{\lambda=0} \approx 0$, the coefficients $a$, $b$, $x$
and $y$ all have the same phase for a longitudinally-polarized
$K^*$. Thus all triple products involving a longitudinal $K^*$ are
expected to vanish. Furthermore, since $\varepsilon_V \cdot
p_{\Lambda_b} = 0$ for a transversely-polarized $K^*$, most of the
triple products in Eq.~(\ref{vecTPs}) are expected to vanish in the
SM. The only potential nonzero triple-product correlations are
%
\beq
2 \, {\rm Im} \left(x y^*\right) \epsilon_{\alpha\beta\mu\nu} \left[
\varepsilon_{K^*} \cdot s_p \, p_p^\alpha p_{\Lambda_b}^\beta
s_{\Lambda_b}^\mu \varepsilon_{K^*}^\nu + \varepsilon_{K^*} \cdot
s_{\Lambda_b} \, p_p^\alpha s_p^\beta \varepsilon_{K^*}^\mu
p_{\Lambda_b}^\nu \right] ~.
\eeq
%
Since these both require the measurement of all three spins, this
result is consistent with the results of Ref.~\cite{Wafia}. 

Second, and more importantly, both of these asymmetries only arise due
to the interference between the (small) dipole term
$Y_{K^*}^{\lambda=\perp}$ and the $V_{ub} V_{us}^*$ piece of
$X_{K^*}^{\lambda=\perp}$. Thus, by analogy with
Eq.~(\ref{Pestimate}), we estimate the size of the asymmetries to be
roughly
%
\beq
{2 {\rm Im} (xy^*) \over |x|^2 + |y|^2 + 2 {\rm Re}(xy^*)} \simeq {z
{\rm Im}(X_{K^*}^{\lambda=\perp} Y_{K^*}^{{\lambda=\perp}^*}) \over
|X_{K^*}^{\lambda=\perp}|^2} \simeq {z a_2 a_d \eta \lambda^2 \over
a_2^2 \lambda^4 + a_4^2} \sim 2\% ~,
\label{Vestimate}
\eeq
%
which would be very difficult to measure. (Essentially, the asymmetry
is reduced compared to that in $\Lambda_b \to p K^-$ by the factor $z
|a_d/a_6| = 0.11$.) Furthermore, the decay $\Lambda_b \to p K^{*-}$ is
dominated by the longitudinally-polarized $K^{*-}$; the rate for the
production of a transversely-polarized $K^{*-}$ is suppressed by the
factor $(m_{K^*}/E_{K^*})^2 = 0.1$.  Thus, even if the asymmetry were
larger, it would still be difficult to detect, given the small rate.

We therefore conclude that {\it any} measurement of a sizeable
triple-product asymmetry in the decay $\Lambda_b \to p K^{*-}$ is an
unequivocal signal of new physics \cite{newpaper}. (As noted in the
case of $\Lambda_b \to p K^{-}$ decay, if the measurement of the
proton polarization is difficult, one can consider a final state with
an excited nucleon such as $\Lambda_b \to N(1440) K^{*-}$. In this
case, the polarization of the $N(1440)$ can be determined from its
decay products. Alternatively, one can consider $\Xi_b \to \Sigma^+
K^{*-}$, for which the above conclusions should also hold.)

The decay $\Lambda_b \to p\rho^{-}$ is similar to $\Lambda_b \to p
K^{*-}$, and its amplitude can be obtained from Eqs.~(\ref{vector}),
(\ref{abxyvec}) and (\ref{pkstar}) with the replacements $V_{is} \to
V_{id}$ and $K^* \to \rho$. However, here too the asymmetry is
expected to be smaller than that in $\Lambda_b \to p \pi^-$ by the
factor $z |a_d/a_6| = 0.11$, yielding an asymmetry of less than 1\%.
Finally, the pure penguin process $\Lambda_b \to \Lambda \phi$, is
dominated by a single weak amplitude, so that all its triple-product
asymmetries vanish. (As mentioned earlier, the observation of the
decay $\Lambda_b \to \Sigma \phi$ would indicate the existence of
nonfactorizable contributions and the possible presence of a
significant $V_{ub}V_{us}^*$ piece in the amplitude.)

To summarize, we have examined the predictions of the standard model
for T-violating triple-product asymmetries in $\Lambda_b \to F_1 F_2$
decays, where $F_1$ is a light spin-$1\over 2$ baryon, and $F_2$ is a
pseudoscalar ($P$) or vector ($V$) meson. In $\Lambda_b \to F_1 P$
decays, there is only a triple product possible. In the rest frame of
the $\Lambda_b$, it takes the form ${\vec p}_{F_1} \cdot ({\vec
s}_{F_1} \times {\vec s}_{\Lambda_b})$, where ${\vec p}_{F_1}$ is the
3-momentum of the $F_1$, and ${\vec s}_{F_1}$ and ${\vec
s}_{\Lambda_b})$ are the spins of the $F_1$ and $\Lambda_b$,
respectively. On the other hand, in $\Lambda_b \to F_1 V$ decays,
since all three particles have a non-zero spin, there are several
possible triple products.

Using factorization, we find the following results. First, for
$\Lambda_b \to F_1 P$ decays, the SM predicts a large asymmetry ($\sim
18\%$) only for $\Lambda_b \to p K^-$. This is due to the presence of
the chiral enhancement term $m_K^2/(m_s+m_u)$ in the amplitude, which
compensates the $1/m_b$ suppression. The asymmetry in $\Lambda_b \to p
\pi^-$ is smaller ($\sim 5\%$), and for the decays $\Lambda_b \to
\Lambda \eta, \Lambda \eta'$, it is less than 1\%. Second, for
$\Lambda_b \to F_1 V$ decays with a transversely-polarized $V$, the
asymmetries are quite small: for $\Lambda_b \to p K^{*-}$ and
$\Lambda_b \to p \rho^-$ they are $O(1\%)$ and $<1\%$,
respectively. The asymmetries involving a longitudinally-polarized $V$
are expected to be roughly 15\% smaller than those for a
transversely-polarized $V$, so that they are effectively unmeasurable.
There are no asymmetries in $\Lambda_b \to \Lambda \phi$ since this
decay is dominated by a single weak amplitude. The fact that, within
the SM, the triple-product asymmetries in many decays are tiny
suggests that this is a good area to search for new physics.

\bigskip
\noindent
{\bf Acknowledgements}:
%\bigskip
We thank S. Pakvasa and R. MacKenzie for helpful discussions. This
work was financially supported by NSERC of Canada.

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\end{thebibliography}
\end{document}


