\subsection{Heavy Quark Symmetry}
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\subsubsection{Heavy Flavour and Spin Symmetry}
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The main benefit of the heavy quark limit is the fact, that additional 
symmetries beyond the ones present in full QCD arise. In general, the infrared
or long distance behaviour is the same in the full and in the effective theory, 
and hence in its application to strong interactions the nonperturbative 
long distance contributions are the same in the full and in the effective theory.
The benefit of heavy quark effective theory is that it explicitly exhibits 
the additional symmetries which hold in the limit of infinit heavy quark masses. 

The first symmetry which arises is a heavy flavour symmetry. The interacton of the 
quarks with the gluons is determined by the color quantum numbers and the 
dependence on flavour enters in full QCD only throught the different quark masses. 
For the light quarks that fact that the light quark current masses are small 
compared to the QCD scale $\Lambda_{QCD}$ yields the well known flavour 
symmetry for the light quarks; in the heavy mass limit a flavour symmetry 
arises in a similar manner: once the heavy quark is replaced by a static 
source of color moving with a definite velocity the flavour does not matter 
any more. In other words, for two heavy flavours $b$ and $c$ an $SU(2)$ symmetry 
emerges which relates $b$ and $c$ quarks moving with the same velocity.

For the case of two heavy flavours $b$ and $c$ one has to leading order the 
Lagrangian   
\begin{equation}
{\cal L}_{heavy} = \bar{b}_v (v \cdot D) b_v + \bar{c}_v (v \cdot D) c_v ,
\end{equation}
where $b_v$ ($c_v$) is the field operator $h_v$ for the $b$ ($c$) quark 
moving with velocity $v$ and $D = \partial + i g A$ is the QCD covariant 
derivative.   
This Lagrangian is obviously invariant under the $SU(2)_{HF}$ rotations  
\begin{equation}
\left( \begin{array}{c} b_v \\ c_v \end{array} \right) \to 
U_v \left( \begin{array}{c} b_v \\ c_v \end{array} \right) \quad 
U_v \in SU(2)_{HF} .
\end{equation}
We have put a subscript $v$ for the transformation matrix $U$, since 
this symmetry only relates heavy quarks moving with the same velocity.  
To this end we shall be able to relate e.g. the matrix elements of 
mesonic transitions as 
 \begin{equation}
< D(v') | \bar{c}_{v'} \Gamma b_v | B (v) > \quad \mbox{ and } \quad
< B (v') | \bar{b}_{v'} \Gamma b_v | B (v) > \quad \mbox{ and } \quad
< D (v') | \bar{c}_{v'} \Gamma c_v | D (v) > 
\end{equation}
where $\Gamma$ is some arbitrary Dirac matrix. We shall discuss below that 
such matrix elements will be described in terms of a single universal 
form factor. Note that this symmetry is not a symmetry of the hadron spectrum 
(we have scaled out the masses) rather it is a symmetry relating form factors.  

The second symmetry emerging in the heavy mass limit is the so called spin 
symmetry. As is well known from electrodynamics, the spin coupling to 
the gauge field is proportional to the magnetic part of the gauge field and 
to the Bohr magneton of the corresponding particle. In the strong 
interaction case this coupling is proportional to the 
``strong Bohr magneton'' $g/2m_Q$ which vanishes as the mass goes to infinity. 
In the leading order Lagrangian both spin degrees of freedom couple in the same 
way to gauge field;  we may 
rewrite the leading-order Lagrangian as 
\begin{equation}
{\cal L} = \bar{h}_v^{+s} (iv D) h_v^{+s} + \bar{h}_v^{-s} (iv D) h_v^{-s},
\end{equation}
where $h_v^{\pm s}$ are the projections of the heavy quark field on a 
definite spin direction $s$
\begin{equation}
h_v^{\pm s} = \frac{1}{2} (1 \pm \gamma_5 \fmslash{s}) h_v, 
\quad s\cdot v = 0 .
\end{equation}
This Lagrangian has a symmetry under the rotations of the heavy quark 
spin. In Hilbert space 
this symmetry is generated by operators $S_v (\epsilon)$ as
\begin{equation} \label{cr1}
[ h_v , S_v (\epsilon) ] = i \fmslash{\epsilon} \fmslash{v} \gamma_5 h_v
\end{equation}
where $\epsilon$ with $\epsilon^2 = -1$ is the rotation axis.  
Note that the Dirac matrix $ \fmslash{\epsilon} \fmslash{v} \gamma_5 $
is in fact the generator of a spin rotation around the axis $\epsilon$
in the space of Dirac spinors. 

Now consider a heavy meson built of the heavy quark and a light antiquark. The
mass $M$ of the meson in the spin symmetry limit tends to the quark mass
$M=m_Q(1 +
{\cal O}({\Lambda_{QCD}\over m_Q}))$. Such a meson is moving with an
``infinite'' momentum $P^\mu=M v^\mu$ and obviously the only heavy quark
states present in the meson wave-function come from the $v^\mu$ sector of the
theory. In this sector we have the symmetry generated by the operators
$S (v,\epsilon)$
and all hadron states in the sector must occur in multiplets of
SU(2$)_{spin}$. On the other hand, the operators
$S (v,\epsilon)$ do not commute with
total angular momentum, which means that they in general
connect states of different
spins.

As an example, consider a heavy pseudoscalar meson in its rest frame.
The configuration of the spins may be written as
\begin{equation}
| P (0) > = | \Uparrow \downarrow - \Downarrow \uparrow >
\end{equation}
where the double lined arrow indicates the heavy quark spin, while
the other arrow indicates the spin of the light degrees of freedom.
A spin rotation of the heavy quark spin is archived by the matrix
$\Sigma_3$ defined by
\begin{equation}
\Sigma_3 | \Uparrow > = | \Uparrow > \quad
\Sigma_3 | \Downarrow > = - | \Downarrow >
\end{equation}
Applying $\Sigma_3$ to the pseudoscalar meson state
\begin{equation}
\Sigma_3 | P (0) > =
\Sigma_3 | \Uparrow \downarrow - \Downarrow \uparrow >
= | \Uparrow \downarrow + \Downarrow \uparrow > = | V(0, \epsilon = e_z)>
\end{equation}
one obtains a symmetric combination of the spins which corresponds to
a longitudinally polarized vector meson (i.e. with the polarization vector pointing into 
the $z$ direction) at rest.
In a general frame one has the relation
\begin{equation} \label{states}
S (v,\epsilon) |P(v)> = \eta |V(v,\epsilon)> ,
\end{equation}
where $\eta$ is a phase factor which may be chosen to be $\eta = 1$.

Thus the spin symmetry rotates a pseudoscalar meson into the
corresponding vector meson. Consequently spin symmetry predicts equal masses
for the $0^-$ states and the corresponding $1^-$ state. In the $D$ system
$m_D \approx 1870$ MeV and $m_D^* \approx 2010$ MeV, while in
the $B$ system $m_B \approx 5271$ MeV and $m_B^* \approx 5330$ MeV.
The coupling of the spin to the gluons is of the order $1/m$;
consequently one would expect the relative splittings in the spin
symmetry doublets to be of the order of $\Lambda_{QCD} / m$. Using
$\Lambda_{QCD} = 200$ MeV we have for the above example
$\Lambda_{QCD}/m_c \approx 0.1$ and $\Lambda_{QCD}/m_b \approx 0.04$
respectively. Comparing this to the measured splitting between the
$D$ and the $D^*$ or the $B$ and the $B^*$ one gets some confidence
that for the $b$ and the $c$ quark the
heavy quark symmetries may work reasonably well.

Furthermore, since the spin symmetry violating terms are the ones proportional 
to the ``strong Bohr magneton'' $g/(2m_Q)$ one would expect that the splitting 
between the spin symmetry doublets scale as $1/m_Q$. In fact we may consider the 
quantity 
\begin{equation}
\Delta = M^2 (1^-) - M^2 (0^-) = (M (1^-) + M (0^-))(M (1^-) - M (0^-))
       \approx 2m_Q (M (1^-) - M (0^-)) 
\end{equation}
which we should expect to be a constant in the heavy quark systems. As can be 
seen in table \ref{masstab1} $\Delta$ indeed turns out to be constant in the 
$B$ and $D$ meson systems, but to some surprise one obtains also the 
same constant looking into light quark systems. The interpretation of this is 
not clear; while in the heavy quark systems one is led to say that one 
understands the constant $\Delta$ one could as well interpret this in a negative 
way and say that the $1/m_Q$ dependence of 
splitting between  the $1^-$  and the $0^-$ mesons 
is not an indication for the fact that the heavy mass limit is useful. 
\begin{table}
\begin{center}
\begin{tabular}{|c|c|}
\hline
System & $\Delta$ in GeV${}^2$ \\
\hline
$(B^*, B)$    & 0.53 \\
$(D^*, D)$    & 0.54 \\
$(K^*, K)$    & 0.55 \\
$(\rho, \pi)$ & 0.57 \\
\hline
\end{tabular}
\end{center}
\caption{Value of the splitting $\Delta$ or the different systems}
\label{masstab1}
\end{table} 
\subsubsection{Mesons} 
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The two additional symmetries have interesting consequences for matrix elements. 
We shall study the case of heavy meson transitions first. 
Starting from full QCD we may consider the bottom current $\bar{b} \gamma_\mu b$, 
which is conserved in full QCD. Hence the matrix element of this current 
is given in terms of a single normalized form factor
\begin{equation}
< B^- (p') | \bar{b} \gamma_\mu b | B^- (p) > = (p_\mu^\prime + p_\mu) F(q^2), 
\quad \mbox{with } F(q^2 = 0) = 1
\end{equation}
where $q = p' - p$. We may now switch to the heavy mass limit and obtain 
the relation
\begin{equation}
< B^- (v_1) | \bar{b}_{v_1} \gamma^\mu b_{v_2} | B^- (v_2) > = 
 m_B (v_1^\mu + v_2^\mu) \xi(v_1 \cdot v_2) 
\quad \mbox{with } \xi(v_1 \cdot v_2 = 1) = 1
\end{equation}
In this limit we may employ the heavy flavour symmetry for heavy quark 
moving with velocity $v_1$ to rotate the $B^-$ meson into a $D^0$ 
and obtain
\begin{equation}
< D^0 (v_1) | \bar{c}_{v_1} \gamma^\mu b_{v_2} | B^- (v_2) >
= \sqrt{m_B m_D} \xi (v_1 \cdot v_2)  (v^\mu_1 + v^\mu_2 )
\end{equation}
Due to the symemtry the same form factor appears which is called the Isgur Wise 
function, and which is normalized due to heavy quark symmetry for equal velocity 
$\xi(v_1 \cdot v_2 = 1) = 1$. 

Similarly we may employ heavy quark spin symmetry for the heavy quark moving 
with velocity $v_1$ to rotate the pseudoscalar meson into a vector meson. 
In this way we may relate the above matrix element of a vector
current to a matrix element of left handed current between a heavy  
pseudoscalar and a heavy vector meson  
\begin{eqnarray} \label{eight}
&& < D^{*0} (v_1, \epsilon_1) |
                \bar{c}_{v_1} \gamma^\mu (1-\gamma_5) b_{v_2}  | B^- (v_2) > \\
&& = - < D^0 (v_1) | \bar{c}_{v_1} \gamma_5
                            \fmslash{v}_1 \fmslash{\epsilon}^{*}_1 \gamma^\mu
(1-\gamma_5) b_{v_2} | B^- (v_2) >
     \nonumber \\
&& =
\sqrt{m_B m_D} \xi (v_1 \cdot v_2)
({\epsilon^{*}_1}^\mu (1 + v_1\cdot v_2) - v^\mu_1 v_2\cdot
\epsilon^{*}_1 -
i {\varepsilon^\mu}_{abc} v^a_1 {\epsilon^{*}_1}^b v_2^c )
\nonumber
\end{eqnarray}
where we have used eqn.(\ref{states}) and eq.(\ref{cr1})
and the gamma matrix relations
\begin{eqnarray} \label{nine}
\gamma_5 \fmslash{v} \fmslash{\epsilon}^* \gamma^\mu (1-\gamma_5)
&=& (v^\mu \epsilon^{* \nu} - v^\nu \epsilon^{* \mu} -
 i \varepsilon^{\mu \nu a b} v_a \epsilon_b^* ) \gamma_\nu (1-\gamma_5) \\
\nonumber 
\gamma^\mu (1-\gamma_5) \gamma_5 \fmslash{v} \fmslash{\epsilon}
&=& (v^\mu \epsilon^\nu - v^\nu \epsilon^\mu + \label{ninea}
 i \varepsilon^{\mu \nu a b} v_a \epsilon_b ) \gamma_\nu (1-\gamma_5) 
\end{eqnarray}

These arguments may be further generalized by using the group theory
of spin symmetry. In order to implement both spin symmetry and
Lorentz covariance it is convenient to represent the mesons by
appropriate matrices carrying a heavy quark spinor index $A$ and a
light quark index $\alpha$. In fact, the matrix
\begin{equation} \label{mesonrep0}
H(v) = H_{A \alpha} (v) = \frac{1}{2} \sqrt{m_H} \gamma_5
       (\fmslash{v}-1)
\end{equation}
has the correct Lorentz transformation properties for a $0^-$ particle
composed of two spinors: For a connected Lorentz transformation
$\Lambda$ we have
\begin{equation}
H(v) \to S (\Lambda) H(v) S^{-1} (\Lambda) = H(\Lambda v)
\end{equation}
while a parity transformation $P$ gives
\begin{equation}
H(v) \to P H(v) P^{-1} = - H(\hat{v})
\end{equation}
with $\hat{v} = (v_0 , - \vec{v} )$.
Furthermore, the heavy quark is on shell in the limit $m_H \to \infty$
thus we must require
\begin{equation}
(\fmslash{v}-1) H(v) = 0
\end{equation}
Finally, spin rotations of the heavy quark spin rotate the matrix
$H(v)$ into the corresponding matrix for the vector meson
\begin{equation} \label{mesonrep1}
\Sigma (v,\epsilon) H(v) = \gamma_5 \fmslash{v} \fmslash{\epsilon} H(v)
= \frac{1}{2} \sqrt{m_H} \fmslash{\epsilon}
  (\fmslash{v}-1) =  H^*(v,\epsilon)
\end{equation}
where the matrix $\Sigma$ has been defined in (\ref{cr1}).
It is easy to check that $H^* (v,\epsilon)$ has the correct Lorentz
transformation properties for the $1^-$ state and fulfills the Dirac
equation for the heavy quark. 

Thus these two matrices have the correct properties under the
symmetries involved and may be used to formulate the analogue of the Wigner
Eckhard theorem for spin symmetry. If ${\cal H} (v)$ denotes either
$H(v)$ or $H^*(v,\epsilon)$ and if $| {\cal  H} (v) >$ denotes the
corresponding state , we have
\begin{equation} \label{wet}
< {\cal H} (v^\prime) | \bar{h}_{v^\prime} \Gamma h_v
                      | {\cal  H} (v) > =
\xi (v \cdot v^\prime ) \mbox{Tr }
\left\{ \overline{{\cal H}} (v^\prime) \Gamma {\cal H} (v) \right\}
\end{equation}
where $\Gamma$ is some arbitrary combination of Dirac matrices.
Eq.(\ref{wet}) is the main result of heavy quark symmetry in the
mesonic sector, since it relates {\it every} matrix element of
bilinear heavy to heavy currents between two heavy mesons
to a single form factor, the Isgur Wise function $\xi (v \cdot v')$ .
Furthermore, since the current
\begin{equation}
j_\mu =  \bar{h}_{v} \gamma_\mu h_v
\end{equation}
generates the heavy flavor symmetry, we have a normalization statement
for the Isgur Wise function
\begin{equation} \label{xinorm}
\xi (v \cdot v^\prime = 1 ) = 1
\end{equation}
Note, finally, that the Isgur Wise function in a group theoretical
language is just the reduced matrix element which is universal for
the whole spin flavor symmetry multiplet. Furthermore, the trace in
(\ref{wet}) is the Clebsch Gordan coefficient which is entirely
determined by the current operator and the states of the multiplet.

One may formulate the Wigner Eckart Theorem also for the spin symmetry 
doublets for excited heavy mesons; the corresponding relations may be
found in \cite{Falkexc}.

\subsubsection{Baryons}
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In a similar way one may derive the consequences of spin symmetry for 
heavy baryons \cite{IWbary,MRRbary}. The suggested naming scheme for these states in 
\begin{eqnarray} \nonumber
\Lambda_h &=& [(qq^\prime)_0 h]_{1/2} \\  \nonumber
\Xi_h^\prime &=& [(qs)_0 h]_{1/2} \\  \nonumber
\Sigma_h &=& [(qq^\prime)_1 h]_{1/2} \\  \nonumber
\Xi_h &=& [(qs)_1 h]_{1/2} \\  \nonumber
\Omega_h &=& [(ss)_1 h]_{1/2} \\  \nonumber
\Sigma_h^* &=& [(qq^\prime)_1 h]_{3/2} \\  \nonumber
\Xi_h^* &=& [(qs)_1 h]_{3/2} \\  \nonumber
\Omega_h^* &=& [(ss)_1 h]_{3/2}.   \nonumber
\end{eqnarray}
Here, $q, q^\prime$ refer to $u$ and $d$ quarks, $q \neq q^\prime$ for the
$\Lambda_h$, but $q$ may be the same as $q^\prime$ for the $\Sigma_h$ and
$\Sigma_h^*$. The first subscript (0, 1) is the total spin of the light
degrees of freedom,
while the second subscript (1/2, 3/2) is the total spin of the baryon. For
later convenience, $\xi_h$ may refer to any of the two baryons $\Lambda_h$ or
$\Xi_h^\prime$. Similarly, we denote any of the three baryons $\Sigma_h,
\Xi_h$ and $\Omega_h$ as $\omega_h$, and $\Sigma_h^*, \Xi_h^*$ and
$\Omega_h^*$ as $\omega_h^*$.


Let us now exhibit
the group theoretical framework of the spin symmetry
$(SU(2)_{spin})$ of heavy baryons which arises in the heavy quark limit
and identify the representations that correspond
to the light baryons. The simplest baryons are the $\xi_h$'s. The
nontrivial Lorentz structure of $\xi_h$ is due solely to the two
polarization states of the heavy quark. This means that the $SU(2)_{spin}$
rotations constitute a subset of the Lorentz transformations, i.e., that
$SU(2)_{spin}$ connects polarization states of the same particle. It follows
that the representation which contains $\xi_h$ can be described by an
ordinary Dirac spinor $u_{\xi_h}(v)$. The Lorentz transformations are
\begin{equation}
u_{\xi_h}(v) \rightarrow  D (\Lambda) u_{\xi_h}(\Lambda v),
\end{equation}
where $\Lambda$ denotes the fundamental representation matrix of the given
Lorentz transformation and $D(\Lambda)$ is its spinorial representation.
Note that the spinor indices have been suppressed. The spin transformations
are
\begin{equation}
u_{\xi_h}(v) \rightarrow - \gamma_5 \fmslash{v} \fmslash{\epsilon}
u_{\xi_h}(v),
\end{equation}
which transform one polarization state of $\xi_h$ into another.

On the other hand, consider the baryon states in which the light
degrees of freedom are in a spin one state.
These are the $\omega_h$ and $\omega_h^*$ states which
states have nontrivial Lorentz structure coming from both the light system
polarizations and the heavy quark polarizations. In this representation 
there are spin 1/2 states ($\omega_h$), and spin 3/2 states ($\omega_h^*$)
\footnote{%
There is a slight ambiguity as to which representation describes
the $\omega_h$ and $\omega_h^*$ states. This is because under the Lorentz
group the light quarks' indices can transform either as a pseudovector or as
an antisymmetric tensor. The second possibility will not be considered
here, details on it may be found in \cite{IWbary,MRRbary,Falkexc}.}.
To represent these states we introduce a
pseudovector-spinor object $R^\mu$, where $v_\mu R^\mu=0$ and the spinor
indices have been suppressed. Note that in general $\gamma_\mu R^\mu \neq 0$
because $R^\mu$ describes a spin 1/2 particle together with a spin 3/2
particle. In other words, $R^\mu$ contains a Rarita-Schwinger field as well as
a Dirac field. Under the Lorentz transformation
\begin{equation}
R^\mu(v) \rightarrow {\Lambda^\mu}_\nu  D(\Lambda) R^\nu(\Lambda v),
\end{equation}
and under the spin transformation
\begin{equation}
R^\mu(v) \rightarrow - \gamma_5 \fmslash{v} \fmslash{\epsilon} R^\mu(v).
\end{equation}
The parity transformation of $R^\mu$ is
\begin{equation}
R^\mu(v^\alpha) \rightarrow - \gamma^0 R_\mu(v_\alpha).
\end{equation}

Before we
continue it is important to identify the parts of $R$ that correspond
to spin 1/2 and those that correspond to spin 3/2 states. Spin 3/2 states can
be easily projected by contraction with $\gamma_\mu$
\begin{equation}
\gamma_\mu R^\mu _{\omega_h^*} = 0,
\end{equation}
The rest of the independent components of $R$ correspond to $\omega_h$:
\begin{equation}
R^\mu_{\omega_h} = {1 \over \sqrt{3}} (\gamma^\mu + v^\mu) \gamma_5
u_{\omega_h},
\end{equation}
where $u_{\omega_h}$ is the Dirac spinor of the $\omega_h$ state.

The power of the spin symmetry formalism lies in the observation that a
generic operator $\bar{h}_{{v_1}} \Gamma h_{v_2}$ transforms as a spinor under
$SU(2)_{v_1}$ (and of course independently as a spinor under $SU(2)_{v_2}$).
Suppose we look at the matrix element of $\bar{h}_{{v_1}} \Gamma h_{v_2}$
sandwiched between heavy baryon  states. One of the baryon states must
correspond to the $v_1$ velocity sector (and the other one to the $v_2$
sector). The irreducible representation of which describes the heavy
baryon state of velocity $v_1$ decomposes into a direct sum of the spinors of
$SU(2)_{v_1}$. In such a representation the  different $SU(2)_{v_1}$ spinors
correspond to all the independent polarization states of the light quark
system (and for this reason the indices which label the spinors are termed
light quark indices). The Wigner-Eckart theorem tells us that for each
independent $SU(2)_{v_1}$ spinor in the given representation  there
is only one $SU(2)_{v_1}$ invariant amplitude that describes the matrix
element of $\bar{h}_{{v_1}} \Gamma h_{v_2}$%
\footnote{Likewise, for each spinor in
          the representation that describes the hadron moving with velocity
          $v_2$ there is only one $SU(2)_{v_2}$ invariant amplitude.}.

On the other hand, Lorentz transformations mix the light quark indices,
which means that the set of all the $SU(2)_{spin}$ invariant amplitudes
transforms as some tensor representation (not necessarily irreducible) of the
Lorentz group. Such a Lorentz tensor can only be built from the velocity
vectors $v^\mu_1$, $v^\mu_2$ and invariant tensors $g^{\mu \nu}$,
$\epsilon^{\mu \nu \alpha \beta}$. In other words, the requirement that the
hadron matrix element transforms properly under the Lorentz group
means that the light quark indices must be contracted
with a Lorentz tensor built out of  $v^\mu_1$, $v^\mu_2$, $g^{\mu \nu}$
and $\epsilon^{\mu \nu \alpha \beta}$. The number of irreducible
components in such a tensor corresponds to the number of independent
form factors describing the heavy hadron matrix element of $\bar{h}_{{v_1}}
\Gamma h_{v_2}$.

As we have discussed above, $\xi_h$ baryons are described by the simplest
possible representation, a single spinor. This means that the matrix
elements which describe
processes involving $\xi_h$ particles will be most constrained by
the requirement of the spin symmetry. For example let us look at the matrix
element
\begin{equation}\label{lambda}
< \xi_{h}(v) | \bar{h}\Gamma h' | \xi_{h'}(v') > = A(v\cdot v')
\bar{u}_{\xi_h}(v) \Gamma u_{\xi_{h'}}(v'),
\end{equation}
where we have allowed for the possibility of two heavy quark species $h$
and $h'$. Heavy quark symmetry tells us that there is only one independent amplitude
$A(v\cdot v')$.
 Moreover, if we  put $\Gamma=\gamma_\mu$ then the matrix element
(\ref{lambda})
corresponds to the matrix element of the generator of flavor symmetry
and in the forward limit it is uniquely specified
\begin{equation}
A(1)= \sqrt{m_{\xi_h} m_{\xi_{h'}}} \cdot 1,
\end{equation}
where the factor $\sqrt{m_{\xi_h} m_{\xi_{h'}}}$ means that the hadron states
in (\ref{lambda}) are normalized relativistically. The case of $\xi_h$ is
special because it is the only hadron for which spin symmetry connects
different polarization states of the same particle. This means that the number
of independent Lorentz amplitudes will be reduced not only for $heavy \to
heavy$ transitions. As we show later in this section, in the case of the
semileptonic decays $\xi_h \to light\; baryon$ the {\it a priori} six
independent amplitudes describing such a transition get reduced to two.


Now let us discuss a matrix element related to a $\xi_{h'}\to \omega_h
(\omega_h^*)$ transition. Again, the constraint of heavy quark symmetry tells us
that only one
amplitude consistent with Lorentz invariance {\it and} spin symmetry can
exist:
\begin{equation}\label{sl1}
<\omega^{(*)}_h(v) | \bar{h}_v \Gamma h_{v'} |\xi_h > =
G(v\cdot v')\, \bar{R}^{\mu}_{\omega_h^{(*)}} (v) \Gamma u_{\xi_{h'}}
(v')\, v'_\mu ,
\end{equation}
In fact equation (\ref{sl1}) is consistent only with
the subset of the proper Lorentz transformations. If we use the parity
properties of $R$ we see that parity forces $G(v\cdot v')
= 0$. In ref. \cite{Po90} spin symmetry arguments in combination with
parity and angular momentum conservation explain that (\ref{sl1})
in fact had to be equal to zero.

Now consider the matrix element between $\omega^{(*)}_h$ and
$\omega^{(*)}_{h'}$
\begin{eqnarray} \label{ss1}
&& <\omega^{(*)}_h(v) | \bar{h}_v \Gamma h_{v'} |\omega^{(*)}_h > \\
&& = \bar{R}^{\mu}_{\omega_h^{(*)}} (v) \Gamma {R}^{\nu}_{\omega_h^{(*)}}(v')
\, \lbrace B(v\cdot v')g_{\mu \nu} + C(v\cdot v')v'_\mu v_\nu \rbrace .
\nonumber
\end{eqnarray}
Heavy quark symmetry allows two independent form factors (without parity there could
be a third one).

Finally we derive also a relation for heavy to light transitions \cite{MRRbary}. 
Note that,
in general, spin symmetry multiplets contain particles which decay
electromagnetically or strongly. This means that even if spin symmetry
relates some weak heavy to light amplitudes for the particles in a
given representation, many of these relations have little chance of
being tested experimentally. One important exception corresponds to
the $\xi_h$ particles. Here spin symmetry relates different
polarizations of the same particle and imposes interesting
constraints. Consider for example the matrix element of an operator
$\bar{\ell} \Gamma h_v$ between a heavy $\xi_h$ and a light
spin-1/2 baryon $B_\ell$. It is described by only two form
factors,
\begin{equation}
< B_\ell (p) | \bar{\ell} \Gamma h_v | \xi_h (v) > =
\bar{u}_\ell (p) \{ F_1 (v \cdot p ) + \fmslash{v} F_2 (v \cdot p) \} \Gamma
u_{\xi_h} (v) .
\end{equation}
Thus in this particular case spin symmetry greatly reduces the
number of independent Lorentz-invariant amplitudes which may
describe the heavy to light transitions.
