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{\bf\Huge Laboratoire d'Annecy-le-Vieux de }\\
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{\bf\Huge Physique Th\'eorique}
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{\large {\bf Improvement of the Calculation of Scattering Amplitudes
with External Fermions}}
\vspace*{0.5cm}
E.~Chopin$^{1}$\\
{\it Laboratoire d'Annecy-le-Vieux de Physique Th\'eorique
LAPTH.}\footnote{ URA 14-36 du CNRS, associ\'ee \`a l'Universit\'e de
Savoie.}\\
{\it B.P.110, 74941 Annecy-Le-Vieux Cedex, France} \\
{\tenrm 1. e-mail:chopin@lapp.in2p3.fr} \\
{\tenrm PACS numbers: 11.80-Cr, 11.30-Cp, 11.55-m}
\end{center}
\vspace*{\fill}
\centerline{ {\bf Abstract} }
\baselineskip=14pt
\noindent
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{\small
In this paper, we present an improvement of a method for
computing scattering amplitudes that include external
(polarized) fermions with the following features: the formulas are
quite general and work for different kinematic configurations and
different external masses, they are explicitly covariant, they do
not depend on a specific representation of the Dirac matrices and
they have a meaningful limit when the masses tend to $0$.
The results presented make use of some well known formulas
describing the density matrices in terms of projection operators
within a more general formalism. Since our formulas intend to
be as general as possible, we also take into account the
possibility of a transverse polarization for massless fermions.
}
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\rightline{LAPTH-682/98}
%\rightline{hep-ph/9805203}
%\rightline{May. 1998}
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%*********************** INTRODUCTION *******************************
The notion of spin is somewhat difficult to express in a general,
covariant formalism which would have also a meaningful limit for
$m\ra 0$, since the Poincar\'e little groups are different for
$p^2>0$ and $p^2=0$. Therefore, for the computation of Feynman
amplitudes, we cannot obtain general formulas that are exactly
continuous in the limit $m\ra 0$. Furthermore, when fermions are
involved, one has to deal with spinors and Dirac matrices,
and no particular representation of these matrices has some physical
meaning. But for practical calculations involving spinors, a lot of
people still make reference to a specific representation
of these matrices, which leads to analytical results that are
not explicitly covariant. Therefore, for the explicit
calculation of some elements of the ${\cal S}$ matrix
which involve fermions,
there could be an improvement if one can express them within a very
general, covariant, representation independent formula, where
the spin degrees of freedom clearly appear, and where the
limit $m\ra 0$ is meaningful and rather straightforward.
There already exist in the literature some very useful
formulas (see~\cite{maina,bouchiat,wudka}),
and the purpose of this paper is to generalize them, keeping in
mind that we want to respect the criteria described above.
For instance, the definition of the ``conjugate'' spinor $\bar \psi$ by
$\psi^\dagger\gamma^0$ is purely conventional~\cite{jauch},
assuming this way that we take a unitary representation of the
Dirac matrices. This actually is not imposed by a physical principle
but is rather a way to normalize the lagrangian.
These matrices are even not supposed to be hermitian or
anti-hermitian, and most of the calculations in this paper
are completely representation independent, except
for the sake of illustration.
It is not claimed however that the formulas presented in this
paper lead to faster algorithms for the calculation of huge
Feynman amplitudes, and we must recall the reader that for this
purpose, there are some quite fast calculation
techniques~\cite{xzc,vincent} using spinor inner products. However,
these algorithms are very fast only for massless fermions and
introducing the case of massive fermions requires much more
complex calculations. It is possible that in some cases, the formulas
presented in this paper could be competitive for writing Monte-Carlo
programs. Our purpose here is mostly to give simple tools for
calculating analytically some amplitudes that are not very large,
in such a way that one can possibly see the main physical
features of the amplitude just by looking at its expression.
For spinor inner products, one makes
use of polar coordinates which need to set several geometrical
conventions, yielding calculations that are not explicitly Lorentz
covariant. This is why, although these tools exist and have proved
to be efficient, we have found also useful to present how to
calculate some amplitudes in a way which respect the symmetries
(physical or not) of the problem and which includes as few
conventions as possible.
The paper is organized as follows. In the first section, we review
some generalities about the Dirac equation, focusing especially
on the subtleties that are very scarcely found in elementary
textbooks, and which we found useful to gather here. Then
we show on a simple example of calculation involving neutrinos, how
one can in general express an amplitude in a representation
independent and explicitly covariant way. The next part
is devoted to the mathematical derivation of our general
formulas. For this purpose, we will
redemonstrate the well-known formula giving the density
matrix of a pure spinor state $u\bar u = (p\sla+m)(1+
s\sla\gamma^5)/2$, using some representation independent
calculations. We then show how to compute Feynman amplitudes
or its square using these projection operator, focusing especially
on the possible singularities appearing in the phase space which
must be taken under consideration for an implementation in a
Monte-Carlo program. The advantage of the demonstration we used is
that it shows the uniqueness of the decomposition of the
density matrix, it exhibits clearly the spin degrees of freedom,
and especially the transverse degrees of freedom naturally
emerge for massless fermions.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% GENERALITIES %%%%%%%%%%%%%%%%%%%%%%%
\section{Generalities about the Dirac Equation}
\label{generalities}
\subsection{From Klein-Gordon to Dirac}
In this section, we will review some basic things about the
Dirac equation, and we will also focus on some points that
are present in the literature, but unfortunately in very few papers.
First, one must recall that the Dirac equation is obtained by
looking for a factorization of the Klein Gordon equation:
\be
[\partial_\mu g^{\mu\nu} \partial_\nu +m^2 ]\Psi =0
\ee
provided we choose the ($+$\,$-$\,$-$\,$-$) metric convention, and
which can be factored into:
\be
(i\partial\sla +m) (i\partial\sla -m)\Psi
\ee
where $\partial\sla = \partial_\mu \gamma^\mu$, the $\gamma^\mu$ are
the Dirac matrices ($4\times4$ complex matrices which must obey
the anticommutation rules $\{\gamma^\mu,\gamma^\nu\} = 2g^{\mu\nu}$).
Our choice of a specific sign convention
for the metric is of importance here. If we choose the
($-$\,$+$\,$+$\,$+$) signature, we must then replace $m$
by $im$ or, to keep the mass term real, $\gamma^\mu$
by $i\gamma^\mu$, which is not algebraically equivalent to
$\gamma^\mu$. This could lead to some confusions when one reads
the large literature about the differences in the structure of the
two Clifford algebras corresponding to the two possible sign
convention for the metric, $(1,3)$ or $(3,1)$
(see~\cite{pezzaglia} and also some more mathematical
references~\cite{tucker,okubo,dauns,sternberg}).
However, the resulting physics fortunately does
not depend on this metric convention, and to show this, we shall
include the two possible sign conventions by rewriting the
Klein-Gordon equation in the following way:
\be
[\eta\partial_\mu g^{\mu\nu} \partial_\nu +m^2 ]\Psi =0
\ee
where $\eta = +1$ if the signature is $(1,3)$ and
$\eta = -1$ if the signature is $(3,1)$. Thus the factorization now
reads:
\be
(i\omega\partial\sla +m) (i\omega\partial\sla -m)\Psi
\ee
where $\omega$ is one square root of $\eta$, i.e. up to a sign.
Then, we can also choose between the two operators
$(i\omega\partial\sla +m)$ and $(i\omega\partial\sla -m)$
to define the Dirac equation which gives another sign ambiguity.
Therefore, to reabsorb these ambiguities, we define
the matrices $\tilde \gamma^\mu = \omega\gamma^\mu$. We will
consequently denote $\tilde v = v_\mu \tilde \gamma^\mu$ where
$v_\mu$ is any 4-vector. The new anticommutation rules are therefore
$\{\tilde\gamma^\mu,\tilde\gamma^\nu\} = 2\eta g^{\mu\nu}$.
That is to say, the $\tilde \gamma^\mu$ matrices are in a
representation of the $C(1,3)$ Clifford algebra whatever the
sign convention for the metric is. The study of $C(3,1)$,
with a completely different
complex structure becomes therefore irrelevant on the physical
point of view. We can now set the Dirac equation to be
$(i\tilde \partial -m)\Psi=0$, and since we have just shown
that the only relevant signature is $(1,3)$, we will assume
throughout this paper that this metric is chosen (or
equivalently that $\gamma^\mu = \tilde\gamma^\mu$).
Since the $\gamma^\mu$ matrices (or $\tilde\gamma^\mu$) do not
transform like a four-vector, there must be a specific
transformation for $\Psi$ to ensure the Lorentz covariance
of this equation\footnote{We recall the reader that by Lorentz
group is understood the subgroup of $O(1,3)$ which is connected
to the identity. It is also called the orthochronous
Lorentz group and denoted $L_+^{\uparrow}$. Spinors are
in a representation of its covering group $SL(2,C)$
(see~\cite{pct,bogoliubov}), but are not in general in a representation
of the full Lorentz group, covered by what one call the
$Pin(1,3)$ group~\cite{cdewitt} (and $Pin(3,1)$ for $O(3,1)$).
The fact that one consider only $L_+^{\uparrow}$ comes from
the possibility of parity or time reversal violations
(see~\cite{pct,sachs,tomnio,www,wolfenstein,aharonov.susskind}
and related experiments~\cite{rauch.al,werner.al}), but
some authors argue that, for systems that are symmetric under
the full Lorentz group, one can in principle see experimentally
some differences between $Pin(1,3)$ and $Pin(3,1)$~\cite{cdewitt}.}:
\be
\Psi(x) \stackrel{\Lambda}{\longrightarrow}
S_\Lambda\Psi(\Lambda^{(-1)}x)
=\exp\left(i\frac{\omega_{\mu\nu}}{2}
\sigma^{\mu\nu}\right)\Psi((\Lambda^{(-1)})^\mu_{~\nu}x^\nu)
\ee
with $\sigma^{\mu\nu} = \frac{i}{2}[\gamma^\mu,\gamma^\nu]$ (or
equivalently $\frac{i}{2}[\tilde\gamma^\mu,\tilde\gamma^\nu]$), and
$\omega_{\mu,\nu}$ is an antisymmetric tensor, depending on the
Lorentz transformation one performs\footnote{$S_\Lambda$ obeys
$S_\Lambda^{(-1)}\gamma^\mu S_\Lambda = \Lambda^\mu_{~\nu}\gamma^\nu$.
Using the expansion $e^A B e^{-A} = B + [A,B] +\frac{1}{2!}[A,[A,B]]
+ \ldots$ and $[\sigma^{\mu,\nu},\gamma^\rho] =
2i(g^{\nu\rho}\gamma^\mu - g^{\mu\rho}\gamma^\nu)$,
one gets this relation:
$\Lambda^\mu_{~\nu} = \exp[-2(\omega)]^\mu_{~\nu}$. }. Thanks to
this transformation, the Dirac equation
($(i\tilde\partial -m)\Psi =0$) is Lorentz invariant.
Now one can question how arbitrary are the Dirac matrices.
First, it is easy to check that if we have a set of 4 Dirac
matrices $\gamma^\mu$, and $S$ is an invertible $4\times 4$ matrix,
then the new set ${\gamma'}^\mu = S\gamma^\mu S^{-1}$ still
obey the anticommutation rules, and can be used as well as
the first set of matrices, provided we operate the
transformation $\Psi' = S\Psi$ on the wavefunction. The very
interesting property is that the converse is true, and
we have summarized in the appendix the old proof given
in~\cite{jauch}. This property will be used in the following section,
where we will define the notion of charge conjugation. For this
purpose, we will also need the general solution of the
Dirac equation, expressed in momentum space:
\be
\psi = \int \frac{d^3 \vec p}{(2\pi)^3 2p^0}
\left( a(p)u(p)e^{-ip\cdot x} + b^\dagger(p)v(p)e^{ip\cdot x} \right)
\ee
where we have included the creation and annihilation operators
($a(p)$ annihilates a fermion with momentum $p$ and $b^\dagger$
creates an antifermion). $u$ and $v$ are the associated spinors that
are respectively solutions of $(p\sla -m)u(p)=0$ and
$(p\sla +m)v(p)=0$\footnote{We can remark that we could have chosen
to write the Dirac equation with a $+m$ instead of $-m$,
which is equivalent to make the transformation
$\psi \rightarrow \gamma^5\psi$ on the spinor, and it is in fact
just a change in the representation of the gamma
matrices (because it is equivalent to make the transformation
$\gamma^\mu \rightarrow -\gamma^\mu = \gamma^5 \gamma^\mu \gamma^5
= \gamma^5 \gamma^\mu (\gamma^5)^{(-1)}$.}.
\subsection{Spinors and Charge Conjugation}
\label{conjugation}
The Dirac equation being given, we must now construct the
corresponding lagrangian. For this purpose, we have to face
the definition of adjoint spinors. Indeed, when one calculate a
specific Feynman amplitude, one commonly use the definition
$\bar \psi = \psi^\dagger \gamma^0$ and the property
${\gamma^\mu}^\dagger = \gamma^0\gamma^\mu\gamma^0$. One must
be careful to the fact that the use of $\gamma^0$ here
is only conventional, and has very little to do with physics.
The main point is to see that whenever
$\{\gamma^\mu, \mu \in [\![0..3]\!]\}$ obey the anticommutation
rules, then $\{{\gamma^\mu}^\dagger, \mu \in [\![0..3]\!]\}$
verify also the same anticommutation relations. We have then
demonstrated (see in appendix) that these two basis are related
by an interior automorphism, i.e. there exists an invertible
matrix $H_+$ such that:
\be
\forall \mu \in [\![0..3]\!]~~{\gamma^\mu}^\dagger = H_+\gamma^\mu
H_+^{-1}
\ee
and also it implies (using the Schur lemma) that $H_+$ verify the
property $H_+^\dagger = {\bf h_+} H_+$ with ${\bf h_+}$ a
complex number of modulus one. Then the adjoint spinor is defined as
$\bar \psi = \psi^\dagger H_+$. The Dirac lagrangian can
then be constructed as follows:
\be
{\cal L} = \bar \psi (i\partial\sla -m) \psi
\ee
For the sake of unitarity of the scattering matrix, we need the
classical lagrangian to be real, which implies ${\bf h_+}=1$,
that is to say $H_+$ is hermitian\footnote{When one changes
the representation of the Dirac matrices through $\gamma^\mu =
S{\gamma'}^\mu S^{-1}$, $H_+$ changes into
${H'}_+ = \lambda S^\dagger H_+ S$ like an
hermitian form (its signature is $(2,2)$).}.
For the class of unitary
representations, we can take $H_+ = \gamma^0$. However, unitary
representations have no more physical relevance than
any other representation, and therefore we shall not suppose in the
following that $H_+$ is equal to $\gamma^0$. Yet, since
$H_+$ is defined up to a complex constant, imposing the unitarity
of $H_+$ is a good way to normalize the lagrangian. We will
rather impose a normalization on the density matrices at the
end of section~\ref{solutions}. Although it leads to the
same constraint at the end, it allows us to keep $H_+$ undetermined.
In other words, we will normalize the density matrices
instead of $H_+$.
Now, the notion of charge conjugation needs the introduction of the
gauge field and we get $(i\partial\sla -eA\sla -m)\psi = 0$ for
the equation of motion of the interacting fermion field.
Taking now the complex conjugate we obtain the charge conjugated
equation $(i\partial\sla +eA\sla -m)\psi^c=0$, provided
we take the following definition for $\psi^c$:
\bea
\psi^c &=& C_-^{-1}\psi^* \\
\forall \mu \in [\![0..3]\!]~~{\gamma_\mu^*} &=&
-C_-\gamma_\mu C_-^{-1}
\label{defC}
\ena
(Note that when one conjugates $\psi$, one shall not forget to
take the hermitian conjugate of the creation and annihilation
operators). This definition can be seen at the level of free
spinors in the momentum representation: if we have $(p\sla -m)u(p)=0$,
then after complex conjugation
one gets $(p\sla + m)C_-^{-1}u^*(p)=0$. In other words, the charge
conjugated spinor of a so-called {\bf u} spinor becomes a {\bf v}
spinor, which is the intuitive definition of charge conjugation
(by computing $\psi^c$ one can make the following
identifications: $u^c = C_-^{-1}v^*$ and $v^c = C_-^{-1}u^*$).
The existence of $C_-$ is actually given
by the theorem we have proved in appendix relating sets of matrices
that obey the anticommutation relations, which is the case of
$-\gamma_\mu^*$\footnote{In the so-called Majorana representation of
the Dirac matrices, $C_-$ is just a multiple of the identity matrix,
and if we look at another representation
$\gamma'_\mu = S\gamma_\mu S^{-1}$ then the matrix $C_-$
transforms with the formula $C_-' = \alpha S^* C_- S^{-1}$,
where $\alpha$ can be any complex number
different from $0$.}.
$C_-$ must satisfy $C_-^* C_- = {\bf c_-}I$ where ${\bf c_-}$ is a
real constant\footnote{In fact, one can scale the matrix $C_-$ such
that ${\bf c_-} =\pm 1$. The sign of ${\bf c_-}$ depends on
the convention we choose
for the metric, but not on the representation of the Dirac matrices.
Actually, one has ${\bf c_-} =+1$ for the convention
($+$\,$-$\,$-$\,$-$) (see~\cite{tucker}) and ${\bf c_-} =-1$ if
one uses the ($-$\,$+$\,$+$\,$+$) convention,
like in the book of Jauch and Rohrlich~\cite{jauch}.
In this latter case, charge conjugation becomes an anti-involution,
and we could wrongly state that Majorana spinors (spinors which
are identical to their charge-conjugated version) do not exist
in this case, and the physics would depend on the convention for the
metric. The point is in this case that the rule given to define
charge conjugation would change, and we should have with
this convention $\psi^c = C_+^{-1}\psi$ where
$\gamma_\mu^* = C_+\gamma_\mu C_+^{-1}$ and $C_+^* C_+ =
{\bf c_+}I$ (${\bf c_+}>0$). One can obviously relate $C_+$ and
$C_-$ by $C_+ = C_-\gamma^5$.}.
Having defined charge conjugation, we can take the opportunity
to define a Majorana spinor. A Majorana spinor
is its own charge conjugated particle, which necessarily
constrains this kind of particle to be neutral, and
it must therefore obey the relation $\psi^c = \alpha \psi$
($\alpha$ being a possible phase). It implies the following
relations between creation-annihilation operators and {\bf u}
and {\bf v} spinors:
\bea
b&=&a \label{majorana0} \\
C_-^{-1}{\bf v}^* &=& \alpha {\bf u} \label{majorana1} \\
C_-^{-1}{\bf u}^* &=& \alpha {\bf v} \label{majorana2}
\ena
And for the sake of consistency, one should have
$C_-^*C_-=\frac{1}{|\alpha|^2}I$. This happens to be possible because
the sign of ${\bf c_-}$ is independent of the representation
of the Dirac matrices, and in the Majorana
representation of the Dirac matrices\footnote{Note that the Majorana
representation is only a kind of ``eigenvector'' for complex
conjugation (up to a real inner automorphism), which is
different from the notion of Majorana spinors. The existence of a
Majorana representation (which depends only on the metric signature),
only implies the possible existence of Majorana spinors.}, $C_-$
is proportional to the identity (and thus $C_-$ can always be
written in the form $C_-= \sqrt{{\bf c_-}} S^*S^{-1}$ where $S$ is any
$4\times 4$ invertible matrix).
In fact, eq.~\ref{majorana1} and~\ref{majorana2}
don't tell us much more about the structure of the wave function
of a Majorana fermion, and Majorana spinors are nothing but
standard spinors. The main difference comes from the quantization
of the field, and especially from eq.~\ref{majorana0} which
allows to construct specific lagrangians that may distinguish
between Majorana and Dirac spinors. We won't give more details
about Majorana fermions here and refer the reader to the
literature on this topic.
In a similar way, we could also have defined an invertible matrix
$T_+$ or $T_-$ for the transposition of the Dirac matrices
(which also obey the anticommutation rules), but the three operations
are related and since $T_-$ can be obtained by combining
$H_+$ and $C_-$, we will not have to use it. However, it can be useful
to get a relation between $C_-$ and $H_+$ using the fact that
hermiticity and complex conjugation are two commuting operations.
For this purpose we write:
\bea
{^t}\gamma_\mu = (\gamma_\mu^\dagger)^* &=& -H_+^* C_-\gamma_\mu
C_-^{-1}(H_+^*)^{-1}~~~~(= -T_-\gamma_\mu T_-^{-1}) \\
{^t}\gamma_\mu = (\gamma_\mu^*)^\dagger &=& -(C_-^{-1})^\dagger H_+
\gamma_\mu H_+^{-1}C_-^\dagger
\ena
And using the Schur lemma, we know that there exists a non-vanishing
constant $\kappa$ such that:
\bea
H_+^*C_- &=& \kappa (C_-^{-1})^\dagger H_+ \label{rel1} \\
H_+^{-1} C_-^\dagger &=& \kappa C_-^{-1} (H_+^*)^{-1} \label{rel2}
\ena
Then, from a density matrix like $\rho = \psi\bar \psi$ (or more
generally a rank~1 matrix like $\psi\bar\psi'$),
we can compute its charge conjugated version $\psi^c \bar \psi^c$,
and using the relations written in this section we may obtain easily:
\bea
\rho^c = \psi^c \bar \psi^c &=& C_-^{-1} \rho^* (H_+^*)^{-1}
(C_-^{-1})^\dagger H_+ \\
&=& (\kappa)^{-1} C_-^{-1} \rho^* C_- \label{rho-conj}
\ena
The last equation being obtained thanks to eq.~\ref{rel2}, and
if we require the lagrangian to be invariant under charge
conjugation we must set $\kappa = 1$. We will see beyond
(section~\ref{rhoc}) that charge conjugation, given by
eq.~\ref{rho-conj} can be expressed as a condition on some
Lorentz tensors which define completely the density matrix $\rho$.
%%%%%%%%%%%%%%%%%%%%% GENERIC AMPLITUDE AND EXAMPLE %%%%%%%%%%%%%
\section{General fermionic amplitudes}
In this section, we sketch the way we intend to compute
any scattering amplitude with at least one fermionic current.
We shall first note that there are four different possible
configurations corresponding to {\it a priori} four different
expressions for the amplitudes (see fig.~\ref{ffx}, where
the overlined letters correspond to the antiparticles).
%------------------------- BEGINNING OF THE FIGURE -----------------
\begin{figure}
\begin{center}
\caption{\label{ffx}{\em Amplitudes involving some fermion pairs}}
\mbox{
\begin{picture}(20000,12000)
\drawline\fermion[3 0](9000,10000)[5000]
\drawarrow[3\ATTIP](\pmidx,\pmidy)
\global\Xone=\pfrontx \global\Yone=\pfronty
\global\advance\Xone by -2500 \put(\Xone,\Yone){$f$, $p$}
\put(0,6000){${\cal M}_{in,~\overline{in}}(p,p') =$}
\global\Xone=\pbackx \global\Yone=\pbacky
\put(\Xone,\Yone){\circle*{3000}}
\drawline\fermion[5 0](\pbackx,\pbacky)[5000]
\global\Xtwo=\pbackx \global\Ytwo=\pbacky
\global\advance\Xtwo by -2500 \put(\Xtwo,\Ytwo){$\bar f$, $p'$}
\drawarrow[1\ATTIP](\pmidx,\pmidy)
\put(\Xone,\Yone){\line(1,1){4000}}
\put(\Xone,\Yone){\line(2,1){5000}}
\put(\Xone,\Yone){\line(2,0){5500}}
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\put(\Xone,\Yone){\line(1,-1){4000}}
\end{picture}
\begin{picture}(20000,12000)
\drawline\fermion[3 0](9000,10000)[5000]
\drawarrow[3\ATTIP](\pmidx,\pmidy)
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\global\advance\Xone by -2500 \put(\Xone,\Yone){$f$, $p$}
\put(0,6000){${\cal M}_{in,~out}(p,p') =$}
\global\Xone=\pbackx \global\Yone=\pbacky
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\drawline\fermion[5 0](\Xone,\Yone)[5000]
\drawline\fermion[1 0](\Xone,\Yone)[5000]
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\global\advance\Xtwo by 1000 \global\advance\Ytwo by 500
\put(\Xtwo,\Ytwo){$f$, $p'$}
\drawarrow[1\ATBASE](\pmidx,\pmidy)
\put(\Xone,\Yone){\line(2,1){5000}}
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\put(\Xone,\Yone){\line(2,-1){5000}}
\put(\Xone,\Yone){\line(1,-1){4000}}
\end{picture}
} \mbox{
\begin{picture}(20000,12000)
\drawline\fermion[3 0](9000,10000)[5000]
\drawarrow[3\ATTIP](\pmidx,\pmidy)
\global\Xone=\pfrontx \global\Yone=\pfronty
\global\advance\Xone by -2500 \put(\Xone,\Yone){$\bar f$, $p'$}
\put(0,6000){${\cal M}_{\overline{in},~\overline{out}}(p,p') =$}
\global\Xone=\pbackx \global\Yone=\pbacky
\put(\Xone,\Yone){\circle*{3000}}
\drawline\fermion[5 0](\Xone,\Yone)[5000]
\drawline\fermion[1 0](\Xone,\Yone)[5000]
\global\Xtwo=\pbackx \global\Ytwo=\pbacky
\global\advance\Xtwo by 1000 \global\advance\Ytwo by 500
\put(\Xtwo,\Ytwo){$\bar f$, $p$}
\drawarrow[1\ATBASE](\pmidx,\pmidy)
\put(\Xone,\Yone){\line(2,1){5000}}
\put(\Xone,\Yone){\line(2,0){5500}}
\put(\Xone,\Yone){\line(2,-1){5000}}
\put(\Xone,\Yone){\line(1,-1){4000}}
\end{picture}
\begin{picture}(20000,12000)
\drawline\fermion[3 0](9000,10000)[5000]
\put(0,6000){${\cal M}_{out,~\overline{out}}(p,p') =$}
\global\Xone=\pbackx \global\Yone=\pbacky
\put(\Xone,\Yone){\circle*{3000}}
\drawline\fermion[5 0](\pbackx,\pbacky)[5000]
\drawline\fermion[1 0](\Xone,\Yone)[5000]
\global\Xtwo=\pbackx \global\Ytwo=\pbacky
\global\advance\Xtwo by 1000 \global\advance\Ytwo by 500
\put(\Xtwo,\Ytwo){$f$, $p'$}
\drawarrow[1\ATBASE](\pmidx,\pmidy)
\drawline\fermion[3 0](\Xone,\Yone)[5000]
\global\Xtwo=\pbackx \global\Ytwo=\pbacky
\global\advance\Xtwo by 1000 \global\advance\Ytwo by -500
\put(\Xtwo,\Ytwo){$\bar f$, $p$}
\drawarrow[3\ATBASE](\pmidx,\pmidy)
\put(\Xone,\Yone){\line(1,1){4000}}
\put(\Xone,\Yone){\line(2,1){5000}}
\put(\Xone,\Yone){\line(2,0){5500}}
\put(\Xone,\Yone){\line(2,-1){5000}}
\end{picture}
}
\end{center}
\end{figure}
%----------------------------- END OF FIGURE ----------------------
In a condensed notation, one may note a general
amplitude as:
\be
{\cal M} = \bar \psi' O \psi = tr[{\cal O}\psi \bar \psi']
= tr[{\cal O}\Sigma]
\ee
With $\Sigma = u(-cp)\bar u(-c'p')$ where $u(p)$ is a
solution of $(p\sla-m)u=0$, and $c$, $c'$ are the
fermionic charge (\ie $-1$ for a fermion, $+1$ for
an antifermion), provided that $p$ is the
momentum corresponding to an {\bf incoming} fermionic current,
and $p'$ is the momentum corresponding to an {\bf outgoing}
fermionic current (see fig~\ref{ffx}).
Of course, we could have chosen to write an amplitude by reference
to an antifermionic current, or an annihilation or a
pair production, but the four type of amplitudes are related
and it is just a matter of convention. Also, we will
parameterize the generic amplitudes using $u(-cp)$ for the
two possible choices (fermions or antifermions) instead of
using charge conjugation because we will see beyond that
charge conjugation can be easily expressed in terms of
changing the sign of the constant $c$ or $c'$ and also
the sign of the spin degree of freedom which will be introduced
in the next section. The general expression of the amplitude
can be specialized to the four kinematic configurations
of figure~\ref{ffx} in this way:
\bea
{\cal M}_{in,~\overline{in}} &=& {\bar v}(p_+^i){\cal O}
u(p_-^i) =
tr[{\cal O}u(p_-^i){\bar u}(-p_+^i)]\\
{\cal M}_{in,~out} &=& {\bar u}(p_-^o){\cal O}u(p_-^i) =
tr[{\cal O}u(p_-^i){\bar u}(p_-^o)]\\
{\cal M}_{\overline{in},~\overline{out}} &=&
{\bar v}(p_+^o){\cal O}v(p_+^i)
= tr[{\cal O}u(-p_+^i){\bar u}(-p_+^o)] \\
{\cal M}_{out,~\overline{out}} &=& {\bar u}(p_-^o)
{\cal O}v(p_+^o)
= tr[{\cal O}u(-p_+^o){\bar u}(p_-^o)]
\ena
where $p_-$ denotes the momentum of the fermion,
$p_+$ is for the antifermion, and
${\cal O}$ is the interaction operator.
Instead of decomposing spinors in their components, we
will express the amplitudes without having to
specify a specific representation for the Dirac
matrices\footnote{Some variants of this formulation exist
already in the literature, see~\cite{maina,bouchiat,wudka}. Another
calculation technique was developed by Hagiwara and
Zeppenfeld~\cite{algo.hagiwara}, decomposing spinors
in the chiral representation into their Weyl spinors. This
method yields a quick algorithm for computation but here, we are
looking for a formulation that is explicitly covariant
and representation independent.}.
We shall therefore try to express the rank~1 matrix
$\Sigma = u(-cp)\bar u(-c'p')$ in the basis of Dirac matrices, and
we can do the same for the operator ${\cal O}$. Then the
trace can be easily computed using simple algorithms\footnote{Symbolic
calculation programs can do this very easily nowadays. The fastest
being probably FORM, and a rather convenient one is
``M''\cite{M}.}. If one prefer to compute directly the square
of the amplitude, one has:
\bea
|{\cal M}|^2 &=& \left[\bar u(-c'p') {\cal O} u(-cp)\right]
\left[\bar u(-c'p') {\cal O} u(-cp)\right]^\dagger \nonumber \\
&=& \left[\bar u(-c'p') {\cal O} u(-cp)\right]
\left[\bar u(-cp)H_+^{-1}{\cal O}^\dagger H_+ u(-c'p')\right]
\nonumber \\
&=& tr\left[ \rho'{\cal O}\rho \bar{\cal O} \right] \label{ampl-sq}
\ena
with $\rho = u(-cp)\bar u(-cp)$, $\rho' = u(-c'p')\bar u(-c'p')$
and $\bar {\cal O} = H_+^{-1}{\cal O}^\dagger H_+$. If we decompose
${\cal O}$ on the basis of the Dirac matrices we have:
\be
\label{o-decomp}
{\cal O} = s_o+\bar s_o\gamma^5+{v\sla}_o
+{a\sla}_o\gamma^5 + S^o_{\mu\nu}\sigma^{\mu\nu}
\ee
and\footnote{By ${v\sla}^*$ we denote $v^*_\mu \gamma^\mu$. The Dirac
matrices are not conjugated in this notation.}:
\be
\label{obar-decomp}
\bar{\cal O} = s_o^*-\bar s_o^*\gamma^5+{v\sla}^*_o
+{a\sla}^*_o\gamma^5 + {S^o_{\mu\nu}}^*\sigma^{\mu\nu}
\ee
Then, in the following sections we will give the decomposition
on the same basis for the density matrix $\rho$,
which will give the final result for $\Sigma$ and ${\cal M}$.
Before, we shall discuss a simple example to show the practical
procedure we will use in the general case.
\section{A simple example}
\label{example}
We will thus consider in this section the scattering of a neutrino,
for instance in an electroweak process. The neutrino spinors can
be easily expressed in terms of Weyl spinors if we take the chiral
representation for the Dirac matrices. We get:
\be
\Phi_L=\left(\begin{array}{c} -e^{-i\varphi/2}\sin(\theta/2)\\
e^{i\varphi/2}\cos(\theta/2) \end{array}\right),~\Phi_R=
\left(\begin{array}{c} e^{-i\varphi/2}\cos(\theta/2)\\
e^{i\varphi/2}\sin(\theta/2) \end{array}\right)
\ee
and using the notations $p_{(in,out)} =
E_{(in,out)}(1, \vec n_{(in,out)})$, the $\Sigma$ matrix can
be computed with a little bit of algebra:
\bea
u(p_{in})\bar u(p_{out})&=&\sqrt{E_{in}E_{out}}\left(\begin{array}{cc}
0 &0 \\ \Phi_L(p_{in})\Phi^\dagger_L(p_{out})
=\Omega & 0 \end{array}\right) \\
&=& \sqrt{E_{in}E_{out}}\gamma^0\left(\begin{array}{cc}
\Omega &0 \\ 0 & 0 \end{array}\right)
= \ldots = \frac{1-\gamma^5}{2}v\sla
\ena
One may then define the vector $v^\mu$ such
that $v^0=(\Omega_{11}+\Omega_{22})/2$,
$v^1=-(\Omega_{12}+\Omega_{21})/2$,
$v^2=-i(\Omega_{12}-\Omega_{21})/2$,
$v^3=-(\Omega_{11}-\Omega_{22})/2$. When the two fermions are
not back-to-back, $v^0$ doesn't vanish and one can impose
the condition that $v^0$ is real, multiplying by the
suitable phase. One finally obtains the following form
for $v^\mu$:
\be
\label{v-exp}
v^\mu =\sqrt{E_{in}E_{out}}\left(\begin{array}{c}
\sqrt{2}\sqrt{1+\vec n_{in}\cdot\vec n_{out}}\\
\ds -\frac{\sqrt{2}}{\sqrt{1+\vec n_{in}\cdot\vec n_{out}}}
(\vec n_{in}+\vec n_{out}
+i\vec n_{in}\wedge\vec n_{out}) \end{array}\right)
\ee
One can see that $v^\mu$ is orthogonal to both momenta
of the fermions, and moreover, that the space-like part
of this vector represents an elliptic polarization
associated to the 3-vector $\vec n_{in}-\vec n_{out}$.
We have therefore combined the two spin~$(1/2)$ particles to obtain
a spin-1 current.
But we will show that the expression of $v^\mu$ can be
expressed in a covariant way through the formula:
\be
v^\mu =\frac{(p_{out}\cdot k)p_{in}^\mu + (p_{in}\cdot k)p_{out}^\mu
-(p_{out}\cdot p_{in})k^\mu -i\eps^{\mu\nu\rho\sigma}
k_\nu p_{in\,\rho} p_{out\,\sigma}}{\sqrt{2}\sqrt{2(p_{out}\cdot k)
(p_{in}\cdot k)-(p_{out}\cdot p_{in})k^2}}
\ee
Where the four-vector $k^\mu$ serves as a reference.
Expression~\ref{v-exp} is actually obtained by setting
$k^\mu = (1,\vec 0)$, but one can note that $v^\mu$ is
invariant under the transformation
$k~\ra~k+\lambda\,p_{in}~+\mu\,p_{out}$, which also implies
that $k$ must not lie in the plane generated by $p_{in}$ and
$p_{out}$. This will be rather obvious with the following. Note
also that a rescaling of $k$ does not affect $v$. Therefore, there
is only one degree of freedom of $k$ that may influence $v$,
through a phase multiplication.
To obtain this expression, let's consider the density
matrix for both the incoming and outgoing momenta (we will
derive its expression in the following section):
\be
u {\bar u} = {p\sla}\frac{1+\gamma^5}{2}
\ee
then one has for a given vector $k^\mu$:
\be
\label{density-prod}
u_{in} {\bar u}_{in} k\sla u_{out} {\bar u}_{out}
= ({\bar u}_{in} k\sla u_{out}) \Sigma =
{p\sla}_{in} k\sla{p\sla}_{out}\frac{1+\gamma^5}{2} =
v\sla\frac{1+\gamma^5}{2}
\ee
with:
\be
v^\mu = (p_{out}\cdot k)p_{in}^\mu + (p_{in}\cdot k)p_{out}^\mu
-(p_{out}\cdot p_{in})k^\mu -i\eps^{\mu\nu\rho\sigma}
k_\nu p_{in\,\rho} p_{out\,\sigma}
\ee
If $k$ lies in the plane generated by $p_{in}$ and $p_{out}$,
$v$ vanishes and the latter relations are useless. So let's
take $k$ outside this plane. Then we obtain the normalization
factor in eq.~\ref{density-prod} by multiplying once again
by ${k\sla}^*$\footnote{${k\sla}^* \stackrel{def}{=}
k^{*\mu}\gamma_\mu$ (see previous section).} and taking the trace,
we get:
\bea
|{\bar u}_{in} k\sla u_{out}|^2
&=& ({\bar u}_{in} k\sla u_{out}) ({\bar u}_{out} {k\sla}^* u_{in})
\nonumber \\
&=& tr\left[v\sla\frac{1+\gamma^5}{2}{k\sla}^*\right] = 2(v\cdot k^*)
\nonumber \\
&=& 2\left(2\Re((p_{in}\cdot k)(p_{out}\cdot k^*))
-(p_{out}\cdot p_{in})(k\cdot k^*)\right. \nonumber \\
&& \left. +i\eps^{\mu\nu\rho\sigma}
k_\mu k^{*}_\nu p_{in\,\rho} p_{out\,\sigma}\right)
\ena
\bea
\lefteqn{\Rightarrow {\bar u}_{in} k\sla u_{out} =} \nonumber \\
&& e^{i\varphi} \sqrt{2}\sqrt{2\Re((p_{in}\cdot k)(p_{out}\cdot k^*))
-(p_{out}\cdot p_{in})(k\cdot k^*) +i\eps^{\mu\nu\rho\sigma}
k_\mu k^{*}_\nu p_{in\,\rho} p_{out\,\sigma}}
\ena
In this example, we have been obliged to introduce another
4-vector which has {\it a priori} no physical meaning (its role is
only to set a reference for the phase of the amplitude), and we
can be led into trouble if this 4-vector is not suitably chosen.
This problem would not occur if we had computed directly the square
of the amplitude, for which the introduction of this arbitrary
4-vector wouldn't be necessary. This also enlighten the impossibility
to find a formalism for the amplitude which would be continuous in the
massless limit, since this kind of singularity does not exist
if one of the fermions is massive and $k$ is chosen on the light
cone. We will discuss this point further
at the end of the paper, and we now turn to the general case
of an amplitude that includes a fermionic current, represented
by what we have called the $\Sigma$ matrix.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%% SOLUTIONS OF THE CONSTRAINTS %%%%%%
\section{The general case}
\subsection{General properties of the $\Sigma$ matrix}
In this section, we will specify the properties of
the $\Sigma$ matrix. Instead of expressing the spinors
in a specific representation of the the Dirac matrices,
we will express the general equations that determine
$\Sigma$ without having to take any specific representation.
These equations are therefore more physically meaningful.
This kind of formalism have already been studied a very long time
ago by Pauli and others. It was called the bispinor algebra,
and their intrinsic relations were studied in the context
of Fierz identities (see~\cite{crawford}).
We just give here another demonstration, which does not use
Fierz identities which contain far too much information
for our purpose. The main point of our proof being in fact a
representation independent characterization
of a rank-1 $4\times 4$ matrix. To proceed in the same way
as in the previous example, we shall deal first with
the density matrices $\rho$, for which we will use three
constraints. The first is the equation of motion
$(p\sla-m)\rho = 0 = \rho(p\sla-m)$. Then we will
use a conjugation constraint $\bar \rho\, \stackrel{def}{=} \,
H_+^{-1}\rho^\dagger H_+ = \rho$, and finally
a criterion to say that $\rho$ is a rank~1 matrix (using
a characterization described in appendix). In principle,
after these three steps, there should only remain a normalization
condition to determine. So, let us first write the consequences
of the equations of motion on $\rho$, provided we have
decomposed $\rho$ on a typical basis of Dirac matrices, that is:
\be
\label{deux}
\rho = s+\bar s\gamma^5+{v\sla}+{a\sla}\gamma^5 +
S_{\mu\nu}\sigma^{\mu\nu}
\ee
where $S_{\mu\nu}$ is an antisymmetric tensor,
$\sigma^{\mu\nu}=\frac{i}{2}[\gamma^\mu,\gamma^\nu]$,
and one has the relations:
$\begin{array}{cccccc}
\\
s &=&\frac{1}{4}tr[\rho] & \bar s &= &
\frac{1}{4}tr[\rho\gamma^5] \\
\\
v^\mu & = & \frac{1}{4}tr[\rho\gamma^\mu] & a^\mu & = &
\frac{1}{4}tr[\rho\gamma^5\gamma^\mu] \\
\\
S_{\mu\nu} & = & \frac{1}{8}tr[\rho\sigma^{\mu\nu}] & & & \\
\end{array}$
Then the equations of motion can be also projected on the same basis
and it gives \footnote{$[p,v]^{\mu\nu} \stackrel{def}{=} p^\mu v^\nu
- p^\nu v^\mu$}:
\bea
(p\sla-m)\rho &=& 0 \Leftrightarrow \\
p\cdot v &=& m s \\
p\cdot a &=& m\bar s \\
sp^\mu -2iS^{\mu\nu}p_\nu &=& mv^\mu \\
\bar s p^\mu -\eps^{\alpha\beta\mu\nu}S_{\alpha\beta}p_\nu &=&
ma^\mu \\
\frac{1}{2}\eps^{\mu\nu\rho\sigma}a_\rho p_\sigma
-\frac{i}{2}[p,v]^{\mu\nu} &=& mS^{\mu\nu}
\ena
and:
\bea
\rho(p\sla-m) &=& 0 \Leftrightarrow \\
p\cdot v &=& m s \\
p\cdot a &=& -m\bar s \\
sp^\mu +2iS^{\mu\nu}p_\nu &=& mv^\mu \\
-\bar s p^\mu -\eps^{\alpha\beta\mu\nu}S_{\alpha\beta}p_\nu &=&
ma^\mu \\
\frac{1}{2}\eps^{\mu\nu\rho\sigma}a_\rho p_\sigma
+\frac{i}{2}[p,v]^{\mu\nu} &=& mS^{\mu\nu}
\ena
These two sets of equations can be rearranged into a simpler one:
\bea
\bar s &=& 0 \label{simple1} \\
\ [p,v]^{\mu\nu} &=& 0 \Leftrightarrow v^\mu = \lambda p^\mu
\label{simple2} \\
sp^\mu &=& mv^\mu \Rightarrow s = \lambda m \label{simple3} \\
p\cdot a &=& 0 \label{simple4} \\
p_\mu S^{\mu\nu} &=& 0 \label{simple5} \\
\eps^{\mu\nu\alpha\beta}S_{\alpha\beta}p_\nu &=& -ma^\mu
\label{simple6} \\
\eps^{\mu\nu\alpha\beta} a_\alpha p_\beta &=& 2m S^{\mu\nu}
\label{simple7}
\ena
In the massive case, the two last equations are indeed equivalent.
We can already see that the structure of the density matrix is
quite constrained. Since $\bar s =0$, the conjugation constraint
simply says that all the remaining coefficients ($s$, $v^\mu$,
$a^\mu$ and $S_{\mu\nu}$) must be real (see eq.~\ref{obar-decomp}).
We shall now use the
characterization of $\rho$ as a rank~1 matrix, which is the last
constraint on $\rho$. In order to find the necessary and sufficient
conditions for $\rho$ to be a rank~1 matrix, we will apply the
corollary of the second theorem demonstrated in appendix~B, and
in the first place we shall not make reference to the constraints
we have already obtained with the equations of motion. We will add
them at the end. The corollary tells us that we must have
$tr(\rho Q\rho Q') = tr(\rho Q)tr(\rho Q')$
for every matrix $Q$ and $Q'$. We will therefore choose for
$Q$ and $Q'$ the different kind of Dirac matrices, which leads
to\footnote{We use here the convention $\eps_{0123}=1$.}:
$\star$ For $Q=I$ we obtain respectively for
$Q'=I,~\gamma^5,~\gamma^\mu, ~\gamma^\mu\gamma^5,~\sigma^{\mu\nu}$:
\bea
\label{un-1}
2S_{\alpha\beta}S^{\alpha\beta} &=& 3s^2-\bar s^2-(v^2-a^2) \\
\label{un-2}
2s\bar s &=& -i\eps^{\mu\nu\rho\sigma}S_{\mu\nu}S_{\rho\sigma} \\
\label{un-3}
sv^\mu &=& \eps^{\lambda\alpha\beta\mu}a_\lambda S_{\alpha\beta} \\
\label{un-4}
sa^\mu &=& \eps^{\lambda\alpha\beta\mu}v_\lambda S_{\alpha\beta} \\
\label{un-5}
2sS^{\mu\nu} &=& -\eps^{\alpha\beta\mu\nu}(v_\alpha a_\beta +
i{\bar s}S_{\alpha\beta})
\ena
Note that if $s\ne 0$, the lemma demonstrated in the appendix tells
us that these 5 conditions~\ref{un-1} to~\ref{un-5} are sufficient
to ensure that $\Sigma$ is a rank 1 matrix. However, to treat also
the case where $s=0$, we must set $Q$ to the other kind of ``Dirac
matrices''.
$\star$ For $Q=\gamma^5$ we obtain ($Q'=\gamma^5,~\gamma^\mu,
~\gamma^\mu\gamma^5,~\sigma^{\mu\nu}$):
\bea
\label{deux-2}
2S_{\alpha\beta}S^{\alpha\beta} &=& 3{\bar s}^2-s^2+(v^2-a^2) \\
\label{deux-3}
{\bar s}v^\mu &=& 2i a_\alpha S^{\alpha\mu} \\
\label{deux-4}
{\bar s}a^\mu &=& 2i v_\alpha S^{\alpha\mu} \\
\label{deux-5}
2{\bar s}S^{\mu\nu} &=&-is\eps^{\mu\nu\rho\sigma}S_{\rho\sigma}
+i(v^\mu a^\nu-v^\nu a^\mu) \nonumber \\
&& +\frac{1}{4}(S^\mu_{~\alpha}S_{\rho\sigma}\eps^{\nu\alpha\rho\sigma}
-S^\nu_{~\alpha}S_{\rho\sigma}\eps^{\mu\alpha\rho\sigma})
\ena
$\star$ For $Q=\gamma^\alpha$ and
$Q'=\gamma^\beta,~\gamma^\beta\gamma^5,~\sigma^{\mu\nu}$ we get:
\bea
\label{trois-3}
0&=&g^{\alpha\beta}(-a^2-v^2+s^2-{\bar s}^2
+2S_{\rho\sigma}S^{\rho\sigma})
-2v^\alpha v^\beta+2a^\alpha a^\beta
+8S^\alpha_{~\lambda}S^{\lambda\beta} \\
\label{trois-4}
0&=& g^{\alpha\beta}(a\cdot v)+(v^\alpha a^\beta -v^\beta a^\alpha)
-s\eps^{\alpha\beta\rho\sigma}S_{\rho\sigma}
+2i{\bar s}S^{\alpha\beta} \\
\label{trois-5}
0&=& i{\bar s}(a^\mu g^{\alpha\nu}-a^\nu g^{\alpha\mu})
+sa_\lambda\eps^{\lambda\alpha\mu\nu} \nonumber \\
&&+2v_\lambda(S^{\lambda\mu}g^{\alpha\nu}-S^{\lambda\nu}g^{\alpha\mu})
+2(-v^\alpha S^{\mu\nu}+v^\mu S^{\alpha\nu}-v^\nu S^{\alpha\mu})
\ena
$\star$ For $Q=\gamma^\alpha\gamma^5$ and $Q'=\gamma^\beta\gamma^5,
~\sigma^{\mu\nu}$, we obtain:
\bea
\label{quatre-4}
0&=&g^{\alpha\beta}(-a^2-v^2-s^2+{\bar s}^2-2
S_{\rho\sigma}S^{\rho\sigma}) +2v^\alpha v^\beta
-2a^\alpha a^\beta -8S^\alpha_{~\lambda}S^{\lambda\beta} \\
\label{quatre-5}
0&=& i{\bar s}(g^{\alpha\mu}v^\nu-g^{\alpha\nu}v^\mu)
-sv_\lambda\eps^{\lambda\alpha\mu\nu} \nonumber \\
&&+2a_\lambda(g^{\alpha\mu}S^{\lambda\nu}-g^{\alpha\nu}S^{\lambda\mu})
+2(a^\alpha S^{\mu\nu} -a^\mu S^{\alpha\nu}+a^\nu S^{\alpha\mu})
\ena
$\star$ For $Q=\sigma^{\mu\nu}$ and $Q'=\sigma^{\rho\sigma}$,
we obtain:
\bea
\label{cinq-5}
0&=& (g^{\mu\rho}g^{\nu\sigma}-g^{\mu\sigma}g^{\nu\rho})
(s^2+{\bar s}^2+v^2-a^2+2S_{\alpha\beta}S^{\alpha\beta}) \nonumber \\
&&+2\left( g^{\mu\rho}(a^\nu a^\sigma-v^\nu v^\sigma)
-g^{\mu\sigma}(a^\nu a^\rho-v^\nu v^\rho)
-g^{\nu\rho}(a^\mu a^\sigma-v^\mu v^\sigma)
+g^{\nu\sigma}(a^\mu a^\rho-v^\mu v^\rho) \right)
\nonumber \\
&& +8( g^{\nu\sigma}S^{\mu}_{~\lambda}S^{\lambda\rho}
-g^{\nu\rho}S^{\mu}_{~\lambda}S^{\lambda\sigma}
-g^{\mu\sigma}S^{\nu}_{~\lambda}S^{\lambda\rho}
+g^{\mu\rho}S^{\nu}_{~\lambda}S^{\lambda\sigma} )
\nonumber \\
&& -8(S^{\mu\nu}S^{\rho\sigma}+S^{\mu\sigma}S^{\nu\rho}
-S^{\mu\rho}S^{\nu\sigma} ) -2is{\bar s}\eps^{\mu\nu\rho\sigma}
\ena
In eq.~\ref{trois-5} we can insert eq.~\ref{deux-4} and it
simplifies into:
\be
v^\alpha S^{\mu\nu}-v^\mu S^{\alpha\nu}+v^\nu S^{\alpha\mu}
= \frac{s}{2}a_\lambda\eps^{\lambda\alpha\mu\nu}
\ee
And similarly, inserting eq.~\ref{deux-3} into eq.~\ref{quatre-5}
we get:
\be
a^\alpha S^{\mu\nu}-a^\mu S^{\alpha\nu}+a^\nu S^{\alpha\mu}
= \frac{s}{2}v_\lambda\eps^{\lambda\alpha\mu\nu}
\ee
Realizing the sum of eq.~\ref{trois-3} and
eq.~\ref{quatre-4} we obtain $a^2=-v^2$, which then leads to
$2S_{\alpha\beta}S^{\alpha\beta} = 2(s^2+{\bar s}^2)$ and
$v^2 = s^2-{\bar s}^2$ using eq.~\ref{un-1} and eq.~\ref{deux-2}.
The latter relations can be inserted in eq.~\ref{trois-3} and
eq.~\ref{quatre-4}, the difference of which leads now to the
simpler relation:
\be
8S^\alpha_{~\lambda}S^{\lambda\beta} =
2(v^\alpha v^\beta -a^\alpha a^\beta -s^2 g^{\alpha\beta})
\label{eps-normalization}
\ee
Inserting this equation and the previous ones
in eq.~\ref{cinq-5} we obtain:
\be
S^{\mu\nu}S^{\rho\sigma}+S^{\mu\sigma}S^{\nu\rho}
-S^{\mu\rho}S^{\nu\sigma} = -i\frac{s{\bar s}}{4}
\eps^{\mu\nu\rho\sigma}
\ee
\subsection{Solutions of the rank~1 equations}
If we have the conditions $s\ne 0$ and $s^2-{\bar s}^2\ne 0$, the
solutions of equations~\ref{un-1} to~\ref{un-5} are given by:
\bea
\label{f1}
a\cdot v &=&0 \\
\label{f2}
v^2 &=& s^2-{\bar s}^2 = -a^2 \\
\label{f3}
S^{\mu\nu} &=& \frac{-i}{2(s^2-{\bar s}^2)}\left[
{\bar s}(v^\mu a^\nu -v^\nu a^\mu) -is\eps^{\mu\nu\alpha\beta}
v_\alpha a_\beta \right]
\ena
The dual of $S^{\mu\nu}$ then reads:
\be
S^{*\mu\nu} = \frac{i}{2}\eps^{\mu\nu\rho\sigma}S_{\rho\sigma}
= \frac{-i}{2(s^2-{\bar s}^2)}\left[
-s(v^\mu a^\nu -v^\nu a^\mu) +i{\bar s}\eps^{\mu\nu\alpha\beta}
v_\alpha a_\beta \right]
\ee
However, the relation~\ref{f1} must always be true thanks to
equation~\ref{trois-4}. The sum of eq.~\ref{trois-3} and
eq.~\ref{quatre-4} yields $a^2=-v^2$ in any case, and the
difference between eq.~\ref{deux-2} and eq.~\ref{un-1} shows
that the fundamental relation~\ref{f2} must also be true in any case.
Note also that in eq.~\ref{deux-5}, the last term could be
surprising, but using~\ref{f3}, one finds that
its contribution is $0$. The remaining part of this equation
is actually the ``dual'' equation of equation~\ref{un-5}.
Now what happens if ${\bar s} = \pm s$. We can rewrite
equation~\ref{un-5} using the new notations $S_\pm = S\pm S^*$
and $\omega_\pm = s\pm {\bar s}$:
\be
\omega_\pm S_\pm^{\mu\nu} =
\pm \frac{1}{2}\left[ i(v^\mu a^\nu-v^\nu a^\mu)
\mp \eps^{\mu\nu\rho\sigma}v_\rho a_\sigma\right]
\ee
Then if $\omega_+$ or $\omega_-$ vanish, the corresponding
self-dual tensor is left undetermined by this equation and a
constraint appear between $v$ and $a$.
\subsection{Solution with the whole set of constraints}
\label{solutions}
At this stage we can summarize all the constraints we have obtained.
The equations of motion for the density matrix led us in the massive
case to:
\be
\rho = \lambda(m+p\sla)+a\sla\gamma^5
+ \frac{1}{2m}\eps^{\mu\nu\alpha\beta}a_\alpha p_\beta \sigma_{\mu\nu}
\ee
with $a\cdot p=0$ and the rank~1 characterization only adds
the condition $a^2 = -\lambda^2 m^2$ because of course,
a lot of the previous equations are redundant. The last two
equations show that the 4-vectors $a/\lambda \pm p$
lie on the light cone. That is to say
we can write $a = \lambda(p-k_1) = \lambda (k_2-p)$ with
$k_1^2=k_2^2 =0$, $k_1\cdot p = k_2\cdot p = m^2$. Actually, only
one of these two vectors is useful since, if $k_1$ is given,
one can take $2p-k_1$ for $k_2$. Therefore,
let us consider any non-vanishing 4-vector $k$ lying on the
light cone. We can therefore set:
\be
a^\mu = h\lambda\left( p^\mu -\frac{m^2}{p\cdot k}k^\mu\right)
\label{a-mu}
\ee
Where $h=\pm 1$.
The scale of $k$ is of no importance in this expression, whereas
its direction becomes a good candidate for defining the direction
of reference for the polarization of the fermion. We can also look
at the light-cone limit of this expression. If the space-like
projection of $p^\mu$ does not tend to the one of $k^\mu$, the limit
is obviously the limit of $h\lambda p^\mu$, and the possible
singularity occurs when $p$ becomes almost collinear to $k$. In order
to study this limit, we can set $p^\mu = k^\mu + \epsilon \delta^\mu$
with $\epsilon \ra 0$ (we must have chosen $k$ suitably here, and
especially $k^0 >0$). The expression of $a^\mu$ then reads:
\be
\frac{a^\mu}{h\lambda} = -k^\mu\left(1+\epsilon \frac{\delta^2}
{\delta\cdot k}\right) +\epsilon\delta^\mu
\ee
And the limit for $a^\mu$ exists ($a^\mu \simeq -h\lambda p$).
Rewriting the whole density matrix we get:
\bea
\rho &=& \lambda\left(m+p\sla+h\left(p\sla-\frac{m^2}{p\cdot k}
k\sla\right)\gamma^5 + \frac{mh}{2(p\cdot k)}\eps^{\mu\nu\alpha\beta}
p_\alpha k_\beta \sigma_{\mu\nu}\right) \label{rho} \\
&=& \lambda\left(m+p\sla+h\left(p\sla-\frac{m^2}{p\cdot k}k\sla\right)
\gamma^5 + mh\left(1-\frac{p\sla k\sla}{p\cdot k}\right)
\gamma^5\right) \\
&=& \lambda(p\sla +m)\left(1+ h\left(1-\frac{mk\sla}{p\cdot k}\right)
\gamma^5\right)
\ena
We recover for the massive case the well known
result ($\rho = (p\sla+m)\frac{1+a\sla\gamma^5}{2}$ with
$a\cdot p =0$ and $a^2 = -1$), in a slightly
different form. This last formula is already very
often used in the literature, and one
may question about the interest in redemonstrating it.
The first interest is to use this formula (especially in the form
of eq.~\ref{rho}) to compute scattering amplitudes or the modulus
square of these amplitudes with external fermions,
(as we will see in the following) in a way that respects the
criteria we have defined in introduction. And also, it shows
naturally how the spin degrees of freedom emerge and the possibility of
some transverse degrees of freedom in the massless limit. In fact,
when one looks at the limit $m \ra 0$ in eq.~\ref{rho}, the
last term vanish\footnote{This limit is obvious if $p\cdot k$ does
not tend to $0$. Otherwise, if $p\cdot k\ra 0$ when $m\ra 0$, one
has $p^\mu \ra C k^\mu$. Then one can see that the limit is $0$
by taking for instance $p^\mu=(E=\sqrt{m^2+p^2},0,0,p)$ and
$k = (1,0,0,1)$. Then $\eps^{1203}(p_0k_3-p_3k_0) = E-p = p\cdot k$
and the limit is clearly $0$ also in this case.}, and if
$p\cdot k$ doesn't tend to $0$,
the term containing $\gamma^5$ becomes equivalent
to $p\sla\gamma^5$, whereas when
$p\cdot k$ also tends to $0$ when $m$ vanish (i.e. $p^\mu\ra
C k^\mu$), then the $\gamma^5$ term is equivalent
to $-p\sla\gamma^5$. The $m\ra 0$ limit allows us to recover
the two possible density matrices used in the literature for
massless fermions ($\rho = p\sla(1\pm\gamma^5)/2$). However, in the
massive case we had more freedom for this density matrix because the
direction of $k$ is free. In some sense, this freedom seems
``frozen'' in the limit $m\ra 0$, but it is not really the case,
and we shall study in details the solutions of the constraint
equations for $m=0$ exactly. We can actually deduce from
eq.~\ref{simple6} that the generic solution for $S^{\mu\nu}$
when $m=0$ is $S^{\mu\nu} = \lambda[p,\eps]^{\mu\nu}/2$ with
$p\cdot \eps =0$ (eq.~\ref{simple5}).
Therefore, $S^{\mu\nu}$ is not necessarily
$0$, contrary to what we have deduced from the limit $m\ra 0$.
The rank~1 conditions then add another constraint, but relax also one.
Since $a^2=0$ in our case, the normalization of $a^\mu$ is no
more imposed (the coefficient $\lambda$ in factor of eq.~\ref{a-mu}
is no more valid) and from eq.~\ref{simple7} one has only
$a^\mu = \lambda'p^\mu$ with {\it a priori}
$\lambda'\ne \lambda$. From eq.~\ref{eps-normalization} we
get a constraint on the normalization of $\eps^\mu$:
\be
{\lambda}^2(1+\eps^2) = {\lambda'}^2 \label{eps-norm}
\ee
We can therefore write the density matrix in the massless limit
in the form:
\bea
\rho &=& p\sla\left(\lambda +\lambda'\gamma^5+i\eps\sla \right)
\label{massless0} \\
&=& \lambda p\sla\left( 1+h\sqrt{1+\eps^2}\gamma^5 +i\eps\sla\right)
\label{massless1}
\ena
The emergence of the ``transverse'' degrees of freedom in the
massless case is not really a surprise. Their origin comes
from the fact that the little group of a massless momentum
($ISO(2)$) is of the same dimension ($3$) as the one of a massive
momentum ($SO(3)$), and we should therefore obtain the same
number of degrees of freedom. However, the transverse degrees
of freedom of massless particles are not observed in experiments, and
we are led to discard them for neutrinos (see \cite{weinberg-book}
eq.~2.5.38). In the context of the study of some new physics where
some massless fermions not yet observed could exist, there is
{\it a priori} no reason to discard them and it is also a
reason why we present the most general situation in this paper.
Finally, for the global normalization of these density matrices,
we choose $\lambda = 1/2$ to conform to a common normalization
used in the literature.
\subsection{Charge conjugation in this formulation}
\label{rhoc}
Since we have obtained the decomposition of the density matrices
on the basis of Dirac matrices, we shall now study how charge
conjugation operates on these decompositions. From what we have
seen in section~\ref{conjugation} we can write:
\bea
\rho^c &=& C_-^{-1}\rho^*C_- \\
&=& s^* I - {\bar s}^*\gamma^5 - {v\sla}^* + {a\sla}^*\gamma^5
-S_{\mu\nu}^*\sigma^{\mu\nu} \\
&=& s I - {v\sla} + {a\sla}\gamma^5
-S_{\mu\nu}\sigma^{\mu\nu}
\ena
The last equation comes from the hermiticity of $\rho$ and the
fact that we have $\bar s =0$ in any case. Applying this on the
explicit expression of $\rho$ (eq.~\ref{rho}), the charge
conjugation acts rather simply on $\rho$:
\be
\rho(p,k,h) \ra \rho^c(p,k,h) = \rho(p'=-p,k'=k,h'=-h) \label{conj}
\ee
or also this way:
\be
\rho(p,k,h) \ra \rho^c(p,k,h) =
\rho\left(p'=-p,k'=2p-\frac{m^2}{p\cdot k}k,h'=h\right)
\ee
We can see that it is equivalent to change the sign of
$h$, keeping the vector of reference constant, or
to keep $h$ the same and to enforce a symmetry operation
on $k$. Also eq.~\ref{conj} has the advantage to apply in the
massless case where we can also write from eq.~\ref{massless1}:
\be
\rho(p,\eps,h) \ra \rho^c(p,\eps,h) = \rho(-p,\eps,-h)
\ee
\subsection{Computation of a generic amplitude}
We have now all the elements necessary to compute
a generic amplitude of the form ${\cal M} = tr[{\cal O}\Sigma]$.
We therefore wish to proceed in a similar way as we did in the
simple example of section~\ref{example}. Thus we can write
the equivalence:
\be
u\bar u' \equiv \frac{u \bar u k\sla u' \bar u'}
{\sqrt{tr[u \bar u k\sla u' \bar u' k\sla]}}
\ee
where the 4-vector $k$ serves as a reference for the phase of
the amplitude and the equivalence means ``up to a phase''. It
is important to note at this stage that we can be led into trouble
if an amplitude contains several identical particles. In this
case we should have the same reference for the phase of two
diagrams where two identical fermions are permuted, because of the
possible interferences. In this case, this method shouldn't work
because we have arbitrarily changed the relative phase between the
diagrams. In this case, a covariant expression can be obtained
through the computation of the square of the amplitude, which
often means huge analytical expressions. We shall not discuss
further this problem. Coming back to our single fermionic current,
we can generically write the $\Sigma$ matrix in the form:
\be
\Sigma = \frac{\rho_u k\sla \rho_{u'}}{\sqrt{tr[\rho_u k\sla \rho_{u'}
k\sla]}} \label{generic}
\ee
where one can replace $\rho_u$ and $\rho_{u'}$ by their expression
given in formula~\ref{rho} for massive fermions or eq.~\ref{massless1}
for massless fermions. Then, using eq~\ref{generic}, the amplitude
can be written:
\be
{\cal M} = \frac{tr[{\cal O}\rho_u k\sla \rho_{u'}]}
{\sqrt{tr[\rho_u k\sla \rho_{u'} k\sla]}} \label{amp}
\ee
and one has to compute two simpler traces, instead of one possibly
huge trace if one wants to compute the square of the
amplitude using eq.~\ref{ampl-sq}. As for the choice of $k^\mu$, we
have seen in the simple example presented in section~\ref{example}
that we may have some problems if this vector is not suitably
chosen, especially in the massless case, because of a
kinematic singularity appearing in the plane generated by the
two momenta. And if one intends to implement this formula inside
a Monte-Carlo program, where the events are generated randomly,
it is better to avoid this kind of singularity to be sure that
none of the momentum configurations will be close to the singularity.
Therefore in the massless case, one may choose $k^\mu$
such that $k^2 = 1$, and in the massive case, the explicit
computation of the normalization factor in eq.~\ref{amp} shows
that one may preferably choose $k^\mu$ on the light cone. Since
the explicit calculation of these traces can be done easily
by some symbolic calculation programs, we shall let the
reader do them if he is interested in such calculations,
because it is often important to take advantage of the
particular situations to choose $k^\mu$, or the spin axes properly.
We may end this section by mentioning that we can also
choose a simpler operator than $k\sla$ to be inserted
into the normalization factor when at least one of the fermions
is massive. We can simply use the identity matrix instead
of $k\sla$ and we get:
\be
{\cal M} = \frac{tr[{\cal O}\rho_u \rho_{u'}]}
{\sqrt{tr[\rho_u \rho_{u'}]}}
\ee
Which is simpler to compute, and shall not lead to kinematical
singularities in most cases. Since the denominator can be
of order $mm'$, it will be numerically better to use it for
heavy fermions.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% CONCLUSION %%%%%%%%%%%%%%%%%%%%%
\section{Conclusion}
In this paper, we have shown a quite general method for
the calculation of Feynman amplitudes or its square with external
fermions. The mass of the fermions can be of any value, even
in the same fermionic line. The formulas given are also independent
of the representation of the Dirac matrices, explicitly covariant
and have a meaningful massless limit. The formalism
is not exactly continuous in $m=0$ in the strict
sense since there are some degrees of freedom that appear in the
case $m=0$ that compensate the transverse degrees that are
``frozen'' into the two helicity modes when $m\ra 0$. However,
these transverse degrees of freedom are not observed for neutrinos
and they are therefore not taken under consideration for these
particles. For the study of some New Physics,
it can be interesting to keep them. The other trouble with massless
fermions comes when one computes a scattering amplitude
with a current composed of two of these massless fermions. There
can be some singularity in phase space for the normalization
condition if one takes a fixed momentum for the reference phase,
which can lay in the plane of the two external momenta
for some kinematic configurations. This singularity does
not exist for the computation of the square of the amplitude,
which fortunately in the massless case may lead to
expressions of reasonable size, contrary to the massive case.
To obtain the general expression of an amplitude, we started
from well known formulas giving the density matrices
that we have redemonstrated in the most general case.
Using this kind of demonstration, the possibility of a transverse
polarization for massless fermions naturally emerges, and we
have also characterized within this formalism the notion of charge
conjugation and Majorana spinors. We expect that these results
can be useful to get simpler analytic expressions for
short Feynman amplitudes, or not-so-short amplitudes computed
using symbolic calculation programs. We also expect a
simpler implementation of fermions in Monte-Carlo programs, thanks
to the fact that the formalism presented in this paper requires
very few conventions, which was also one of the goals of this work.
\subsubsection*{Acknowledgments}
I must thank Pr.~W.~Pezzaglia and Pr.~I.~Benn
for their interesting answers to my questions, and P.~Overmann
for his advice in the practical use of his symbolic
calculation program ``M''.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% APPENDIX %%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Appendix A: some properties of the Dirac matrices}
Suppose that we have one set of $\gamma^\mu$ matrices that obey
the anticommutation rules. From this set of $4$ matrices, one can
explicit a basis of ${\cal M}_4(C)$\footnote{${\cal M}_4(C)$ denotes
the set of $4\times 4$ complex matrices throughout the paper.},
$\gamma^r$ ($r~\in~[\![1..16]\!]$) in this way:
\be
\gamma_r = {\gamma_0}^{\alpha_0(r)}{\gamma_1}^{\alpha_1(r)}
{\gamma_2}^{\alpha_2(r)}{\gamma_3}^{\alpha_3(r)}
\ee
with $\alpha_\mu(r) = 0,1$. We may also sometimes denote
$n_\gamma(r) = \alpha_0(r)+\alpha_1(r) +\alpha_2(r) + \alpha_3(r)$.
When $n_\gamma(r)$ is even, the matrix is said to be in the
even subalgebra of the Clifford algebra.
This basis is chosen essentially for the proof of the theorem
relating two different sets of Dirac matrices (see beyond),
but most of the time one uses the more convenient basis
($I=\gamma_{(r=1)}$, $\gamma^5 = i\gamma^0\gamma^1\gamma^2\gamma^3=
i\gamma_{(r=16)}$, $\gamma^\mu$, $\gamma^\mu\gamma^5$,
$\sigma^{\mu\nu} = \frac{i}{2}[\gamma^\mu,
\gamma^\nu]$)\footnote{Sometimes we will have to
consider $\gamma_{(r=1)}$ which is just the identity, different from
the usual $\gamma_{(\mu=1)}$. We will therefore write the ``$r=...$''
when necessary, in order to avoid some confusions about the
meaning of the indices.}.
With these definitions one can prove that $tr[\gamma_r] = 0$
if $\gamma_r$ is not the identity matrix ($r\ne 1$) and
that $\gamma_p \gamma_q = \eta_{p,q}
\gamma_{I(p,q)}$, where $\eta_{p,q} =\pm 1$, and more
importantly, for a fixed $q$, $p\mapsto I(p,q)$ is a permutation
of $[\![1..16]\!]$ (and similarly for a fixed $p$,
$q\mapsto I(p,q)$ is also a permutation). Also $\eta_{p,q}$ depends
only on the anticommutation relations and the metric convention,
and not on the specific set (representation) of Dirac matrices
chosen. Since we have also shown in the beginning of the paper
that the only relevant signature is $(1,3)$, we also assume
this metric to be chosen. We have also
$tr[\gamma_i\gamma_j] = C_{ij}$, where the coefficients $C_{i,j}$
vanish if and only if $i\ne j$.
\subsubsection*{Demonstration of the relation between different
representations}
Suppose now that we have two sets of Dirac matrices $\gamma^\mu$
and ${\gamma'}^\mu$ that obey the anticommutation rules. As before,
we can define a basis for the $4\times 4$ complex matrices
${\gamma}_r$, and another similar basis ${\gamma'}_r$ defined in
the same way but using ${\gamma'}^\mu$ instead of $\gamma^\mu$.
We shall also need another $4\times4$ complex matrix, $F$ upon
which we do not impose any constraint at this time. Then we
define the matrix $S$ as:
\be
S = \sum_{r=1..16} {\gamma'}_r F {\gamma_r}^{(-1)}
\ee
We will show that $S$ must be $0$ or invertible,
and then that in this latter case one has
${\gamma'}^r = S\gamma^r S^{-1}$ (it is sufficient to check
this property for $r\equiv \mu~\in~[\![0..3]\!]$).
The proof comes from the following relation:
\bea
S\gamma_t &=& \sum_{r=1..16} {\gamma'}_r F {\gamma_r}^{(-1)}\gamma_t
\nonumber \\
&=& \sum_{s=1..16}{\gamma'}_{I(t,s)} F {\gamma_{I(t,s)}}^{(-1)}
\gamma_t~~~~~~~~~~~~(\Leftarrow~~~ r \ra I(t,s) ) \nonumber \\
&=& \sum_{s=1..16}{\gamma'}_{I(t,s)} F {\gamma_s}^{(-1)}
{\gamma_t}^{(-1)}{\gamma_t} \eta_{t,s}~~~~~~~~~~~~~(\Leftarrow~~
def.~~of~\gamma_{I(t,s)}) \nonumber \\
&=& \sum_{s=1..16} \eta_{t,s}{\gamma'}_{I(t,s)} F {\gamma_s}^{(-1)}
\nonumber \\
&=& \sum_{s=1..16} {\gamma'}_t {\gamma'}_s F {\gamma_s}^{(-1)}
~~~~~~~~~~~~~(\Leftarrow~~ def.~~of~\gamma_{I(t,s)})\nonumber \\
&=& {\gamma'}_t S
\label{fondam-th}
\ena
This is the very fundamental property we need to prove our latest
assertion. Now if $x$ is a four vector such that $Sx=0$, then
every $\gamma_r x$ is in the kernel of $S$. It means that
$Ker\, S$ is stable through the action of the whole algebra, that
is to say every $4\times 4$ complex matrix (${\cal M}_4(C)$). If
we consider an irreducible representation of the Dirac matrices,
it implies that $Ker\, S$ must be an improper subspace, i.e. $0$ or
the whole spinor space $C^4$. Therefore, $S$ is invertible or
$0$.
Now, why is the four dimensional representation irreducible?
Suppose that $K = Ker\, S$ is not $0$, and let $L$ be a subspace
of $C^4$ such that $C^4 = K\oplus L$. Let ${\cal M}$ be an
endomorphism of $C^4$ such that the image of one (non-zero) vector
in $K$ lies in $L$ (it only needs $L$ to be different from $0$).
Now, the key point is that ${\cal M}$ can be decomposed on
the $\gamma_r$ basis. Since $K$ is stable by all the $\gamma_r$,
we get a contradiction and $L$ must be $0$,
$\Rightarrow K =C^4 \Rightarrow S=0.~\Box$
We have now demonstrated that two sets of Dirac matrices are
related by an inner automorphism if we can find $S \ne 0$. And it is
possible to find $F$ such that $S$ is not $0$. Otherwise, for
all $x,y\,\in\,C^4$, posing $F=yx^\dagger$, one has
$\sum_{r} (y^\dagger {\gamma'}_r y) x^\dagger {\gamma_r}^{(-1)} =0$,
thus $\sum_{r} (y^\dagger {\gamma'}_r y) {\gamma_r}^{(-1)} =0$ and
taking the trace, one obtains $y^\dagger{\gamma'}_{(r=1)} y =
4y^\dagger y$ which cannot be $0$ if $F\ne 0$. So one can
find $F$ such that $S$ is non-zero, then $S$ is invertible
and the two representations are related through an inner
automorphism.~$\Box$
\subsubsection*{Unitary representations}
In this paragraph, we will show that one can construct a unitary
representation of the Dirac matrices from any irreducible
representation.
So, if we denote $H = \sum_{r} \gamma_r^\dagger\gamma_{r}$,
one can use some arguments similar to the ones used
in eq.~\ref{fondam-th}
(with also the fact that $\eta_{p,q}^2=1$) to show that
$\gamma_\mu^\dagger H\gamma_\mu = H$. Now, it is clear that
$H$ is hermitian definite and positive, which allows us to
write it as a ``square'', i.e. $H=h^\dagger h$ and $h$ is
invertible\footnote{$h$ is also unique if it is taken hermitian
positive definite.}. Thus we have
$\gamma_\mu^\dagger h^\dagger h\gamma_\mu = h^\dagger h$ and therefore
the representation given by $h\gamma_\mu h^{-1}$ is unitary.~$\Box$
\section*{Appendix B: mathematical proof of some elementary theorems}
In this appendix we will demonstrate two interesting properties.
The second one is used in the body of this paper. The first one
is rather trivial if one uses a specific representation
for the Dirac matrices. The aim here is just to illustrate that most
of the important properties within the Dirac formalism can be shown
without the need of any specific representation for the $\gamma$
matrices.
\noindent{\bf Theorem 1:} {\it Let $p$ be a non vanishing four
momenta.
We denote $m=\sqrt{p^2}$ if $p^2\geq 0$ and $m=i\sqrt{-p^2}$
if $p^2<0$. Let $c$ be a constant which can be set
to $\pm 1$ (if the convention for the metric is ($+$\,$-$\,$-$\,$-$)),
or $\pm i$ (if the convention for the metric is ($-$\,$+$\,$+$\,$+$),
case which can be avoided if we replace $p\sla$ by $\tilde p$,
as shown in the beginning of the paper).
Then $Ker(p\sla-cmI) = Im(p\sla+cmI)$.}
{\bf Proof:}
Let $M_4$ be the four dimensional Minkowski space, and
$C^4$ a four dimensional complex vector space, in which
we define spinors. \\
Since $p\sla^2 = c^2m^2I$, it is clear that
$Im(p\sla+cmI) \subset Ker(p\sla-cmI)$.
In the case $m \neq 0$, if $u \in Ker(p\sla-cmI)$, then
$u = (p\sla+cmI)(u/(2cm))$, and the theorem is proved. In fact,
the eigenvalues of $p\sla$ are of order two, because $p\sla$
has at most 2 non-vanishing eigenvalues, and its trace must be
zero. Thus $p\sla-cmI$ are two rank two matrices,
and we recover the elementary result that the space of states for
a spin~1/2 particle is of dimension 2. This is also true when
$m=0$ as we shall see.\\
Now suppose that $m=0$. We know that $dim~Ker(p\sla) +
dim~Im(p\sla) =4$, and we have seen that
$Im(p\sla) \subset Ker(p\sla)$, and since
$p^\mu$ is a non vanishing vector, $p\sla \neq 0$ and one has:
$1 \leq dim~Im(p\sla) \leq dim~Ker(p\sla)$. Thus we have two
possibilities:
\begin{itemize}
\item{$dim~Im(p\sla) = 2 = dim~Ker(p\sla)$. Then the theorem is
proved.}
\item{$dim~Im(p\sla) = 1$ and $dim~Ker(p\sla)=3$.}
\end{itemize}
We will therefore show that the second case is not possible.
Let $u_i$, $i \in \{0,1,2,3\}$ be a basis of $C_4$, such that
$\{u_1,u_2,u_3\}$ is a basis of $Ker(p\sla)$ and $u_1 = p\sla u_0$.
The vector $p^\mu$ is of the form ${\bf p}(1,\vec n)$
(with ${\vec n}^2=1$). Let $\vec n'$ and
$\vec n''$ two vectors such that $\vec n'$, $\vec n''$ and $\vec n$
form an orthonormal basis of $R^3$.
Then if $q' = (0,\vec n')$ and $q'' = (0,\vec n'')$,
one has $(q'\cdot p) = (q''\cdot p) = (q'\cdot q'')=0$,
and ${q\sla'}^2 =
{q'}^2 = \pm 1 = {q\sla''}^2$. Therefore, the eigenvalues of $q\sla'$
and $q\sla''$ are in the set $\{i,-i\}$ or $\{1,-1\}$ depending on the
sign convention for the metric. \\
One has $u_1 = p\sla(u_0)$, then $q\sla' u_1 = p\sla (-q\sla' u_0)$,
but if $dim~Im(p\sla) = 1$, we must have $q\sla' u_1 = \lambda' u_1$
with $\lambda' = \pm i$ (or $\pm 1$), and similarly for $q\sla'' u_1$.
Now $q\sla' (q\sla'' u_1) = \lambda'\lambda'' u_1 =
-{q\sla'' ({q\sla'} u_1)} = -\lambda'\lambda'' u_1$. This would
imply $\lambda'\lambda'' u_1=0$, which is impossible.
Thus the only possibility is $dim~Im(p\sla) = 2 = dim~Ker(p\sla)$,
which implies $Im(p\sla) = Ker(p\sla)$, and the theorem is proved
also for $m=0$.~$\Box$ \\
\noindent{\bf Theorem 2:} {\it Let $M$ be a $n\times n$ complex matrix.
then one has:}
\be
rank(M)=1 \Leftrightarrow M\neq 0~and~\forall Q\in {\cal M}_n(C)~~
MQM = tr(MQ)M
\ee
{\bf Lemma:} {\it Let $M$ be a $n\times n$ complex matrix, such
that $tr(M)\neq 0$. Then :}
\be
rank(M)=1 \Leftrightarrow M^2 = tr(M)M
\ee
\indent{\bf Proof:}
If $M$ is a rank~1 matrix, then one can write $M=xy^\dagger$ where
$x$ and $y$ are two complex vectors. Since we have $tr(M) =
y^\dagger\cdot x$, one has $M^2=tr(M)M$.\\
Conversely, since $tr(M)\neq 0$ then $X\wedge (X-tr(M)) =1$, and
the condition $M(M-tr(M)) =0$ implies that
$C^n = Ker(M)\oplus Ker(M-tr(M)I)$ ($n=4$), thanks to the
kernel decomposition theorem. Thus, using a basis compatible
with this decomposition, $M$ can be written in the form:
\be M=P^{-1}\left( \begin{array}{cccc} 0&&0& \\
&tr(M)&& \\ 0&&\ddots& \\ &&&tr(M) \end{array} \right)P
\ee
Where $P$ is an invertible matrix. It leads
to $tr(M) = dim~Ker(M-tr(M)I)\times tr(M)$, thus
$dim~Ker(M-tr(M)I) =1\Rightarrow rank(M)=1$.~$\Box$ \\
\indent{\bf Proof of the theorem:}
If $rank(M)=1$, then $\forall Q,~rank(MQ)=0~or~1$, and
applying the same reasoning as in the lemma, we easily conclude that
$MQM = tr(MQ)M$.\\
Conversely, if $\forall Q,~tr(MQ)=0$ then $M=0$. Therefore, in
our case, we can find $Q$ such that $tr(MQ)\neq 0$. And
since $Q\mapsto tr(MQ)$ is a continuous function, and
the group of invertible matrices is dense in ${\cal M}_n(C)$,
we can choose $Q$ invertible. Then, multiplying on the right side
by the matrix $Q$ we get $(MQ)^2=tr(MQ)MQ$ and the lemma tells us
then that $MQ$ is a rank~1 matrix, and thus $M=(MQ)Q^{-1}$ is also
a rank~1 matrix.~$\Box$
\noindent{\bf Corollary:} {\it Let $M$ be a $n\times n$ complex
matrix ($M\ne 0$). We have the following equivalence:}
\be
rank(M)=1 \Leftrightarrow \forall (Q,Q')~~ tr(MQMQ')=tr(MQ)tr(MQ')
\ee
\indent{\bf Proof:}
If one knows that $M$ is a rank 1 matrix, the conclusion is a direct
consequence of theorem~2. Conversely, if we know that
$\forall (Q,Q')~~tr(MQMQ')=tr(MQ)tr(MQ')$, then the linear form
$Q'\mapsto tr(MQMQ')-tr(MQ)tr(MQ')$ vanishes, which implies
that $MQM =tr(MQ)M$ for every matrix $Q$. Then, the theorem 2
tells us that $M$ is a rank~1 matrix.~$\Box$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% BIBLIOGRAPHY %%%%%%%%%%%%%%%%%%%%%%%
\begin{thebibliography}{10}
\bibitem{maina}
{\bf A.~Ballestrero, E.~Maina}, ``A new method for helicity
calculations'', \pl {\bf B350} (1995) 225-233.
\bibitem{bouchiat}
{\bf C.~Bouchiat, L.~Michel}, ``Mesure de la polarisation des
\'electrons relativistes'', \np {\bf 5} (1958) 416-434.
\bibitem{wudka}
{\bf R.~Vega, J.~Wudka}, ``A covariant method for calculating helicity
amplitudes'', {\em hep-ph/9511318},
\pr {\bf D53} (1996) 5286-5292
(Erratum-ibid. \pr {\bf D56} (1997) 6037).
\bibitem{jauch}
{\bf J.~M.~Jauch, F.~Rohrlich}, ``The theory of photons and
electrons'', ed. Addison-Wesley, New-York (1959).
\bibitem{xzc}
{\bf Z.~Xu, D.~Zhang, L.~Chang}, ``Helicity amplitudes for multiple
bremsstrahlung in massless non-abelian gauge theories'',
\np {\bf B291} (1987) 392-428.
\bibitem{vincent}
{\bf V.~Lafage}, ``Etats \`a plusieurs particules dans les futurs
collisionneurs $e^+e^-$ et $\gamma\gamma$: techniques de calcul
et effet d'une Nouvelle Physique'', Th\`ese de doctorat,
22 oct. 1996 (Paris XI - ENSLAPP).
\bibitem{pezzaglia}
{\bf W.~M.~Pezzaglia~Jr., J.~J.~Adams}, ``Should Metric Signature
Matter in Clifford Algebra Formulations of Physical Theories'',
({\em gr-qc/9704048}).
\bibitem{tucker}
{\bf I.~M.~Benn, R.~W.~Tucker}, ``An introduction to spinors
and geometry with applications in physics'', Adam~Hilger~ed.,
Bristol \& Philadelphia (1987).
\bibitem{okubo}
{\bf S.~Okubo}, ``Representations of Clifford Algebras And
Its Applications'', ({\em hep-th/9408165}).
\bibitem{dauns}
{\bf J.~Dauns}, ``Metrics Are Clifford Algebra Involutions'',
{\em Int. J. Th. Phys.} {\bf 27} (1988) 183-192.
\bibitem{sternberg}
{\bf S.~Sternberg}, ``On Charge Conjugation'', {\em Comm. Math. Phys.}
{\bf 109} (1987) 649-679.
\bibitem{pct}
{\bf R.~F.~Streater, A.~S.~Wightman}, ``PCT, Spin and Statistics,
and All that'', {\em Addison-Wesley} (1964-1978-1980), in the
collection {\em Advanced Book Classics}.
\bibitem{bogoliubov}
{\bf N.~N.~Bogolubov, A.~A.~Logunov, A.~I.~Oksak, I.~T.~Todorov},
``General Principles of Quantum Field Theory'', {\em Kluwer
Academic Publisher} (1990) in the collection
{\em Mathematical Physics and Applied Mathematics}.
\bibitem{cdewitt}
{\bf C.~DeWitt-Morette, S.~Gwo Jr., E.~Kramer}, \\ ``Spin or Pin?'',
preprint submitted to {\em Rev. Mod. Phys.}, available at
http://wwwrel.ph.utexas.edu/Members/cdewitt/SpinOrPin1.ps (see also
the references therein).
\bibitem{sachs}
{\bf R.~G.~Sachs}, ``The Physics of Time Reversal'', {\em Univ. of
Chicago Press} (1987).
\bibitem{tomnio}
{\bf C.~N.~Yang, J.~Tomnio}, ``Reflection Properties of Spin-1/2
Fields and a Universal Fermi-Type Interaction'',
\pr {\bf 79} (1950) 495-498.
\bibitem{www}
{\bf G.~C.~Wick, A.~S.~Wightman, E.~P.~Wigner}, ``The Intrinsic
Parity of Elementary Particles'', \pr {\bf 88} (1952) 101-105.
\bibitem{wolfenstein}
{\bf L.~Wolfenstein, D.~G.~Ravenhall}, ``Some consequences of
Invariance under Charge Conjugation'', \pr {\bf 88} (1952) 279-282.
\bibitem{aharonov.susskind}
{\bf Y.~Aharonov, L.~Susskind}, ``Charge Superselection Rule'',
\pr {\bf 155} (1967) 1428-1431.
\bibitem{rauch.al}
{\bf H.~Rauch et al}, ``Verification of Coherent Spinor Rotation
of Fermions'', \pl {\bf A54} (1975) 425-427.
\bibitem{werner.al}
{\bf S.~A.~Werner et al}, ``Observation of the Phase Shift of
a Neutron Due to Precession in a Magnetic Field'',
\prl {\bf 35} (1975) 1053-1055.
\bibitem{algo.hagiwara}
{\bf K.~Hagiwara, D.~Zeppenfeld}, ``Helicity amplitudes
for heavy lepton production in $e^+e^-$ annihilations'',
\np {\bf B274} (1986) 1-32.
\bibitem{M}
{\bf P.~Overmann}, ``M symbolic calculation program'', available
at this home-page:
{\tt http://www.thphys.uni-heidelberg.de/$\sim$overmann/M.html}\\
(some compiled versions are available for different
kind of computers).
\bibitem{crawford}
{\bf J.~P.~Crawford}, ``On the algebra of Dirac bispinor densities:
Factorization and inversion theorems'', {\em J. Math. Phys.}
{\bf 26} (1985) 1439-1441.
\bibitem{weinberg-book}
{\bf S.~Weinberg}, ``The Quantum Theory of Fields, vol1:
foundations'', (1995) {\it Cambridge Univ. Press}.
\end{thebibliography}
\end{document}