\boldmath
  \subsubsection{Baryon Decays: $\Lambda_b \to \Lambda_c \ell \nu$}  
\unboldmath
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Exclusive decays of the heavy-to-heavy type of baryons may also be considered 
using the heavy mass limit. The easiest example of this type is the decay 
$\Lambda_b \to \Lambda_c \ell \nu$, since in both baryons the light degrees of 
freedom are spinless. 

In general the vector current between two spin-1/2/ baryons is bgiven in terms 
of three form factors
\begin{eqnarray} \label{gene}
<B^\prime, p^\prime, s^\prime |\bar q \gamma_\mu q|B,p,s>
&=&\bar u(p^\prime, s^\prime) [f_1(q^2) \gamma_\mu - i
f_2(q^2)\sigma_{\mu\nu} q^\nu \\
&+& f_3(q^2) q_\mu] u(p,s), \nonumber
\end{eqnarray}
where $f_1$, $f_2$ and $f_3$ are real. 

Employing heavy quark symmetry we find for the $\Lambda$'s, (c.f. eq. \ref{lambda})
\begin{equation}
f_1=A, \,\,\, f_2=f_3=0.
\end{equation}

Finally, for the form factors for the transition between the
$\omega_h$'s, we find (c.f. eq. \ref{ss1})
\begin{eqnarray}
f_1 &=& -\frac{1}{3} \left[B(v \cdot v^\prime - 4)
        + C (v \cdot v^\prime - 1)(v \cdot v^\prime -3)\right], \\
f_2 &=& -\frac{4}{3(M+M^\prime)} \left[B + C(v \cdot v^\prime - 1)
       \right], \nonumber \\
f_3&=& 0. \nonumber
\end{eqnarray}
where $M$ and $M^\prime$ are the masses of the initial and
final baryon.

Let us now discuss the semileptonic decays of a heavy $\Lambda$, say
$\Lambda_b$. Its semileptonic weak decay into $\Lambda _c$ involves the matrix
element of the charged current which, due to spin symmetry, may be
parametrized in terms of a single form factor:
\begin{equation}
< \Lambda_c (v^\prime) | \bar{c} \gamma_\mu (1-\gamma_5) b | \Lambda_b (v) >
= A(v \cdot v^\prime ) \bar{u}_{\Lambda_c} (v^\prime) \gamma_\mu (1-\gamma_5)
                        u_{\Lambda_b} (v).
\end{equation}
As has been pointed out previously,  the form factor $A$ is
normalized at $v \cdot v^\prime=1$. Consequently, as in the case of the
semileptonic decays of heavy mesons, this fact may be used to extract $V_{cb}$
of the Cabibbo-Kobayashi-Maskawa (CKM) matrix using these semileptonic decays.

However, data on this type of decay is still sparse compared to the 
corresponding meson decays. Semileptonic decays of the $\Lambda_b$ have
up to now only been observed at LEP from $b$ quark hadronization into a 
$\Lambda_b$ and only one out of about ten $b$ quarks fragment into a
$\Lambda_b$ \cite{LEPlam}

From the theoretical side the $\Lambda_Q$ type heavy baryons are the 
simplest objects, since here the light degrees of freedom are in 
a spinless state. Due to this one has also a strong restriction on 
the possible $1/m_Q$ corrections for semileptonic 
$\Lambda_b \to \Lambda_c$ decays. 

Let us study a transition mediated
by a heavy quark bilinear current between $\Lambda_b$ and
$\Lambda_c$. Our representation follows the paper by Georgi,
Grinstein and Wise \cite{GG90}; following this reference
we shall discuss the matrix element
\begin{equation}
< \Lambda_c | \bar{c} \Gamma b | \Lambda_b >.
\end{equation}
To leading order in heavy quark effective theory
this matrix element is given in terms of a single form
factor $A$
\begin{equation} \label{lllo}
< \Lambda_c (v') |
\bar{h}_{v'}^{(c)} \Gamma h_v^{(b)}
| \Lambda_b (v) > = A (v \cdot v' )
\bar{u}_c  (v') \Gamma u_b (v)
\end{equation}
To order $1/m_c$ we shall start with the non-local
corrections originating from the Lagrangian. The operators are
given in (\ref{HarvLag}) or (\ref{FWlag}). To be specific, we 
use the Lagrangian (\ref{HarvLag}) and write the corrections
as 
\begin{eqnarray}
{\cal L}_1 &=& \frac{1}{2m_c} \bar{h}_v (iD)^2 h_v  \label{l1} \\ 
{\cal L}_2 &=& \frac{1}{2m_c} \bar{h}_v (ivD)^2 h_v \label{l2} \\
{\cal L}_3 &=& \frac{1}{2m_c} \bar{h}_v (-i\sigma_{\mu \nu}) (iD^\mu)(iD^\nu) h_v
\label{l3}
\end{eqnarray}
The first two pieces
${\cal L}_1$ and ${\cal L}_2$ transform trivially under the spin
symmetry and thus their contributions look exactly like the one
from the lowest order (\ref{lllo}), i.e.\ they just renormalize the
form factor $A$
\begin{eqnarray} \label{l1l2}
&& (-i) \int d^4 x \,  \sum_{k=1}^2 C_k (\mu)
< \Lambda_c (v') |
T \left[ {\cal L}_k (x)
\bar{h}_{v'}^{(c)} \Gamma h_v^{(b)} (0) \right]
| \Lambda_b (v) >  \\ \nonumber
&& = \Delta A (v \cdot v' )
\bar{u}_c  (v') \Gamma u_b (v)
\end{eqnarray}
Finally, we shall study the contribution of
the chromomagnetic moment operator ${\cal L}_3$. From spin symmetry
this matrix element is given in terms of an antisymmetric tensor
$\xi_{\mu \nu}$
\begin{eqnarray}
&& (-i) \int d^4 x \, C_3 (\mu)
< \Lambda_c (v') |
T \left[ {\cal L}_3 (x)
\bar{h}_{v'}^{(c)} \Gamma h_v^{(b)} (0) \right]
| \Lambda_b (v) > \\ \nonumber
&& = \xi^{\mu \nu} (v' \cdot v' )
\bar{u}_c  (v') \sigma_{\mu \nu}
\left( \frac{\fmslash{v}'+1}{2} \right) \Gamma u_b (v)
\end{eqnarray}
There are only two vectors available and thus $\xi^{\mu \nu}$ has to
be of the form
\begin{equation}
\xi_{\mu \nu} (v' \cdot v' ) = \left( v_\mu v_\nu^\prime
- v_\mu^\prime  v_\nu \right)
\eta (v \cdot v')
\end{equation}
This shows that the chromomagnetic operator does not contribute, since
one has
\begin{equation}
\bar{u}_c  (v') \sigma_{\mu \nu}
\left( \frac{\fmslash{v}'+1}{2} \right) v^{\mu \prime} = 0.
\end{equation}
and thus from spin symmetry
\begin{equation}
(-i) \int d^4 x \, C_3 (\mu)
< \Lambda_c (v') |
T \left[ {\cal L}_3 (x)
\bar{h}_{v'}^{(c)} \Gamma h_v^{(b)} (0) \right]
| \Lambda_b (v) > = 0
\end{equation}

The second class of contributions are the matching contributions.
In the following we shall neglect QCD scaling and discuss only the
operators appearing at the matching scale $m_c$. From spin symmetry
one obtains for the following matrix element
\begin{equation}
< \Lambda_c (v') |
\bar{h}_{v'}^{(c)} i \stackrel{\leftarrow}{{D}_\mu} \Gamma h_v^{(b)}
| \Lambda_b (v) > =
\bar{u}_c  (v') \Gamma u_b (v)
\left[ E v_\mu + F v_\mu^\prime \right]
\end{equation}
where the arrow over the covariant derivative indicates that it acts
to the left on the $c$ quark.
Here two form factors $E$ and $F$ appear. These are not independent,
since one has from the equations of motion for the $c$ quark
\begin{eqnarray}
0 &=& < \Lambda_c (v') |
\bar{h}_{v'}^{(c)} i (v' \cdot \stackrel{\leftarrow}{D}) \Gamma h_v^{(b)}
| \Lambda_b (v) > \\
&=& \bar{u}_c  (v') \Gamma u_b (v)
\left[ E (v \cdot v') + F \right]
\end{eqnarray}
and thus $E (v \cdot v') = - F$. Furthermore, momentum conservation
relates the the form factor $E$ (or $F$) to the lowest order form
factor $A$, since we have
\begin{eqnarray} \label{momcon}
&& i \partial_\mu < \Lambda_c (v') |
\bar{h}_{v'}^{(c)} \Gamma h_v^{(b)}
| \Lambda_b (v) > \\ \nonumber
&& =  \left[(m_{\Lambda_b} - m_b) v_\mu
      - (m_{\Lambda_c} - m_c) v_\mu^\prime \right]
A(v \cdot v') \bar{u}_c  (v') \Gamma u_b (v) \\
&& = \bar{\Lambda} ( v_\mu - v_\mu^\prime )
A(v \cdot v') \bar{u}_c  (v') \Gamma u_b (v) \nonumber
\end{eqnarray}
where we may equate $m_{\Lambda_b} - m_b$ with
$m_{\Lambda_c} - m_c$ to the order in the $1/m$ expansion we are
working. We denote the mass difference between the baryon and the
quark by $\bar{\Lambda}$ which is typical scale from the binding
of the light degrees of freedom and thus is of the order of
$\Lambda_{QCD}$.

Pulling the derivative inside one obtains from
(\ref{momcon})
\begin{eqnarray} \label{mc2}
&& i \partial_\mu < \Lambda_c (v') |
\bar{h}_{v'}^{(c)} \Gamma h_v^{(b)}
| \Lambda_b (v) > = \\
&& < \Lambda_c (v') |
\bar{h}_{v'}^{(c)} i \stackrel{\leftarrow}{{D}_\mu} \Gamma h_v^{(b)}
| \Lambda_b (v) > +
< \Lambda_c (v') |
\bar{h}_{v'}^{(c)} \mu \Gamma i D_\mu h_v^{(b)}
| \Lambda_b (v) > \nonumber
\end{eqnarray}
Combining this with (\ref{momcon}), contracting with $v_\mu$ and
using the equation of motion for the $b$ quark the second term on the
right side of (\ref{mc2}) does not contribute and one finds a relation
between $E$ and $A$
\begin{equation}
E (1+ v \cdot v') = \bar{\Lambda} A
\end{equation}
Combining the two contributions from the Lagrangian and from the
matching one finds that at the $\Lambda_{QCD}/m_c$ level there is still only
one form factor. This is entirely due to the fact that the $\Lambda$
baryons are in fact the simplest objects from the point of view of
heavy quark effective theory. For mesons and for more complicated
baryons there are additional independent form factors at order
$\Lambda_{QCD}/m_c$ \cite{Lu90,BB91}.

Finally may check Luke's theorem for the present case. in wich it 
states that the shift $\Delta A$
of the form factor $A$ due to the two operators ${\cal L}_1$
and ${\cal L}_2$ vanishes at the normalization point; i.e.\ the single 
form factor to order $\Lambda_{QCD} / m_c$ is still normalized.
This may be seen very similar to the discussion in section 2.3.2 
by splitting the two operators  ${\cal L}_1$ and ${\cal L}_2$ into 
their singlet and triplet piece
with respect to heavy flavor symmetry:
\begin{eqnarray}
&& {\cal L}_1 + {\cal L}_2
= \frac{1}{2 m_c} \bar{h}_v^{(c)} [(i D)^2 - (i v \cdot D)^2] h_v^{(c)}
\\
&& = \frac{1}{4 m_c} \left\{
\bar{h}_v^{(c)} [(i D)^2 - (i v \cdot D)^2] h_v^{(c)}
+ \bar{h}_v^{(b)} [(i D)^2 - (i v \cdot D)^2] h_v^{(b)} \right\}
\nonumber \\
&& + \frac{1}{4 m_c} \left\{
 \bar{h}_v^{(c)} [(i D)^2 - (i v \cdot D)^2] h_v^{(c)}
- \bar{h}_v^{(b)} [(i D)^2 - (i v \cdot D)^2] h_v^{(b)} \right\}
\end{eqnarray}
where the first combination is singlet under heavy flavor symmetry,
while the second one behaves like a triplet.

Inserting the triplet piece into the the matrix element (\ref{l1l2})
one easily sees that its matix element  vanishes due to heavy flavor 
symmtry and thus
only the singlet piece contributes to the matrix element under
consideration. However, since only the singlet piece gives a
contribution the vector current for equal velocities of the heavy quark
$\bar{h}_v^{(c)} \gamma_\mu h_v^{(b)}$ still generates a symmetry, even
including the $1/m_c$ terms. Thus the form factor -- including the
correction $\Delta A$ -- is still normalized:
\begin{equation}
( A + \Delta A )|_{v \cdot v'=1}  = 1
\end{equation}
which is Luke's theorem for the present case.

