



\documentstyle[preprint,aps]{revtex}
\begin{document}
\draft
\title{How to generate families of spinors
}
\author{ N. Manko\v c Bor\v stnik}
\address{ Department of Physics, University of
Ljubljana, Jadranska 19,Ljubljana, 1111, \\
and Primorska Institute for Natural Sciences and Technology,\\
C. Mare\v zganskega upora 2, Koper 6000, Slovenia}
\author{ H. B. Nielsen}
\address{Department of Physics, Niels Bohr Institute,
Blegdamsvej 17,\\
Copenhagen, DK-2100
 }


\date{\today}

\maketitle
 
 
\begin{abstract} 


Using the technique \cite{holgernorma2002} to construct a basis for spinors and families of
spinors in terms of Clifford algebra objects, we define other Clifford algebra objects, 
which transform a state of one ''family'' of spinors into a state of another ''family'' of spinors,
changing nothing but the ''family'' number.  
The proposed transformation  works - as does the technique - for all dimensions and any signature,
opening a way for understanding families of quarks and leptons. 
\end{abstract}


\newpage

\section{Introduction}
\label{introduction}

\tightenlines
We presented in the paper\cite{holgernorma2002} the technique to construct a spinor 
basis as products of nilpotents and projections  formed from the objects $\gamma^a$ for which we only 
need to know that they obey the Clifford algebra. Nilpotents and projections are odd and even objects 
of $\gamma^a$'s, respectively and are chosen to be eigenstates of a Cartan subalgebra of the 
Lorentz group.  The technique can be used to construct a spinor basis for any dimension $d$
and any signature in an easy and transparent way. Equipped with the graphic presentation of basic states,  
the technique offers an elegant way to see all the quantum numbers of states with respect to the Lorentz 
group, as well as transformation properties of the states under the Clifford algebra objects. 

When multiplying products of nilpotents and projectors from the left   
with any of the Clifford algebra objects, we got a linear combination of these ``basic '' 
elements back: our basis spans a left ideal, and has $2^{d/2}$ elements for $d$ even and $2^{(d-1)/2}$
elements for $d$ odd. 

But there are $2^d$ products of nilpotents and projectors, all of them linearly independent. Mapping of 
ideals to spinor representations (treating
all  as the Hilbert space) leaded accordingly  to $2^{d/2}$ replicas of the usual spinor 
representation for $d$ even and $d^{(d+1)/2}$ for $d$ odd. We called these replicas ''families'' of  
representations.  


The proposed technique was initiated and developed by one of the authors of this paper, when  proposing an  
approach\cite{norma92,norma93,normaixtapa2001} in which all the internal degrees of freedom of either 
spinors or vectors can be described in the space of $d$-anticommuting (Grassmann) coordinates, if the 
dimension of ordinary space is also $d$. 


In the  approach of one of us, however, two kinds of $\gamma^a$ operators - two kinds of the Clifford algebra
objects - were defined, both fulfilling the same Clifford algebra relations, while one kind anticommutes with 
the other kind. When one
of the two kinds of $\gamma^a$'s is used to generate nilpotents and projectors  
(eigenstates of the Cartan subalgebra of the Lorentz group),  
products of which define when operating on a vacuum state  basic vectors for the spinor representation of the
Lorentz group, causes another kind transitions
among ''families'' of spinors, transforming one state of a ''family'' to a state with the same quantum numbers
with respect to the Lorentz group as the starting one but  belonging to anothers ''family''. 

We show in this short paper that there are  operators causing transitions among families\cite{pikanorma2002}
also within the simple technique, presented in the paper\cite{holgernorma2002}. While the left multiplication 
of nilpotents and projectors by the Clifford algebra objects generates all the states of one spinor
representation (section \ref{technique}), does {\em the right multiplication cause transitions among ''families''}
(section \ref{families}).
(The reader should see also
 Chapter 3 of the Chevalley's book \cite{chevalley}.) 

We demonstrate transitions among families for $d=4$ in subsection \ref{d4}.

In this paper, we  assume an arbitrary
signature of space time so that our metric tensor $\eta^{ab}$, with $a,b \in \{0,1,2,3,5,\cdots \d \}$ 
is diagonal with values $\eta^{aa} = \pm 1$, depending on the chosen signature ($+1$ for time like coordinates
and $-1$ for space like coordinates).





\section{Technique to generate spinor representations in terms of Clifford algebra objects}
\label{technique}


We shall briefly repeat the main points of the technique for generating spinor representations from the
Clifford algebra objects, following the
reference\cite{holgernorma2002}. We ask the reader to look for the details and the proofs in this reference.


We assume the objects $\gamma^a$, which fulfill the Clifford algebra
%
\begin{equation}
\{\gamma^a, \gamma^b \}_+ = I \;\;2\eta^{ab}, \quad {\rm for} \quad a,b \quad \in \{0,1,2,3,5,\cdots,d \},
\label{clif}
\end{equation}
%
for any $d$, even or odd.  $I$ is the unit element in the Cliffird algebra, while
 $\{\gamma^a, \gamma^b \}_{\pm} = \gamma^a \gamma^b \pm \gamma^b \gamma^a $.

We assume  the ``Hermiticity'' property for $\gamma^a$'s 
%
\begin{eqnarray}
\gamma^{a\dagger} = \eta^{aa} \gamma^a,
\label{cliffher}
\end{eqnarray}
%
in order that 
$\gamma^a$ are compatible with (\ref{clif}) and formally unitary, i.e. ${\gamma^a}^{\dagger}\gamma^a=I$.

We also define the Clifford algebra objects 
%
\begin{equation}
S^{ab} = \frac{i}{4} [\gamma^a, \gamma^b ] := \frac{i}{4} (\gamma^a \gamma^b - \gamma^b \gamma^a)
\label{sab}
\end{equation}
%
which close the Lie algebra of the Lorentz group  
$ \{S^{ab},S^{cd}\}_- = i (\eta^{ad} S^{bc} + \eta^{bc} S^{ad} - \eta^{ac} S^{bd} - \eta^{bd} S^{ac})$.
One finds from Eq.(\ref{cliffher}) that $(S^{ab})^{\dagger} = \eta^{aa} \eta^{bb}S^{ab}$ and that $\{S^{ab}, S^{ac} \}_+
= \frac{1}{2} \eta^{aa} \eta^{bc}$.

Recognizing from Eq.(\ref{sab}) and the Lorentz algebra relation that two Clifford algebra objects 
$S^{ab}, S^{cd}$ with all indices different 
commute, we  select (out of infinitely many possibilities) the Cartan subalgebra of the algebra of the 
Lorentz group as follows 
%
\begin{eqnarray}
S^{0d}, S^{12}, S^{35}, \cdots, S^{d-2\; d-1}, \quad {\rm if } \quad d &=& 2n,
\nonumber\\
S^{12}, S^{35}, \cdots, S^{d-1 \;d}, \quad {\rm if } \quad d &=& 2n +1.
\label{choicecartan}
\end{eqnarray}
%

It is  useful  to define one of the Casimirs of the Lorentz group -  
the  handedness $\Gamma$ ($\{\Gamma, S^{ab}\}_- =0$). 
%We shall speak about left handedness when $\Gamma= -1$ and about right
%handedness when $\Gamma =1$. 
(To see the definition of  $\Gamma$ for any spin in even-dimensional spaces see 
references\cite{norma92,norma93,normaixtapa2001,bojannorma,%
holgernormadk}.) 
%
\begin{eqnarray}
\Gamma :&=&(i)^{d/2}\; \;\;\;\;\;\prod_a \quad (\sqrt{\eta^{aa}} \gamma^a), \quad {\rm if } \quad d = 2n, 
\nonumber\\
\Gamma :&=& (i)^{(d-1)/2}\; \prod_a \quad (\sqrt{\eta^{aa}} \gamma^a), \quad {\rm if } \quad d = 2n +1,
\label{hand}
\end{eqnarray}
%
for any integer $n$. We understand the product of $\gamma^a$'s in the ascending order with respect to 
the index $a$: $\gamma^0 \gamma^1\cdots \gamma^d$. 
It follows from Eq.(\ref{cliffher})
for any choice of the signature $\eta^{aa}$ that
$\Gamma^{\dagger}= \Gamma,\;
\Gamma^2 = I.$

We also find that for $d$ even the handedness  anticommutes with the Clifford algebra objects 
$\gamma^a$ ($\{\gamma^a, \Gamma \}_+ = 0$) , while for $d$ odd it commutes with  
$\gamma^a$ ($\{\gamma^a, \Gamma \}_- = 0$). 

To make the technique simple we introduce the graphic presentation\cite{holgernorma2002}
as follows
%
\begin{eqnarray}
\stackrel{ab}{(k)}:&=& 
\frac{1}{2}(\gamma^a + \frac{\eta^{aa}}{ik} \gamma^b),\nonumber\\
\stackrel{ab}{[k]}:&=&
\frac{1}{2}(1+ \frac{i}{k} \gamma^a \gamma^b),\nonumber\\
\stackrel{+}{\circ}:&=& \frac{1}{2} (1+\Gamma),\nonumber\\
\stackrel{-}{\bullet}:&=& \frac{1}{2}(1-\Gamma),
\label{signature}
\end{eqnarray}
%
where $k^2 = \eta^{aa} \eta^{bb}$.
One can easily cheque by taking into account the Clifford algebra relation (Eq.\ref{clif}) and the
definition of $S^{ab}$ (Eq.\ref{sab})
that if one multiplies from the left hand side by $S^{ab}$ the Clifford algebra objects $\stackrel{ab}{(k)}$
and $\stackrel{ab}{[k]}$,
it follows that
%
\begin{eqnarray}
S^{ab}\stackrel{ab}{(k)}=\frac{1}{2}k \stackrel{ab}{(k)},\nonumber\\
S^{ab}\stackrel{ab}{[k]}=\frac{1}{2}k \stackrel{ab}{[k]},
\label{grapheigen}
\end{eqnarray}
%
which means that we get the same objects back multiplied by the constant $\frac{1}{2}k$. 
This also means that
when 
$\stackrel{ab}{(k)}$ and $\stackrel{ab}{[k]}$ acting from the left hand side on anything (on a
vacuum state $|\psi_0\rangle$, for example ) are eigenvectors of $S^{ab}$.

We further find 
%
\begin{eqnarray}
\gamma^a \stackrel{ab}{(k)}&=&\eta^{aa}\stackrel{ab}{[-k]},\nonumber\\
\gamma^b \stackrel{ab}{(k)}&=& -ik \stackrel{ab}{[-k]}, \nonumber\\
\gamma^a \stackrel{ab}{[k]}&=& \stackrel{ab}{(-k)},\nonumber\\
\gamma^b \stackrel{ab}{[k]}&=& -ik \eta^{aa} \stackrel{ab}{(-k)}
\label{graphgammaaction}
\end{eqnarray}
%
from where it follows
%
$
S^{ac}\stackrel{ab}{(k)}\stackrel{cd}{(k)} = -\frac{i}{2} \eta^{aa} \eta^{cc} 
\stackrel{ab}{[-k]}\stackrel{cd}{[-k]}, 
S^{ac}\stackrel{ab}{[k]}\stackrel{cd}{[k]} = \frac{i}{2}  
\stackrel{ab}{(-k)}\stackrel{cd}{(-k)}, 
S^{ac}\stackrel{ab}{(k)}\stackrel{cd}{[k]} = -\frac{i}{2} \eta^{aa}  
\stackrel{ab}{[-k]}\stackrel{cd}{(-k)}, 
S^{ac}\stackrel{ab}{[k]}\stackrel{cd}{(k)} = \frac{i}{2} \eta^{cc}  
\stackrel{ab}{(-k)}\stackrel{cd}{[-k]}.
$
%
It is useful to deduce the following relations
%
\begin{eqnarray}
\stackrel{ab}{(k)}^{\dagger}=\eta^{aa}\stackrel{ab}{(-k)},\quad
\stackrel{ab}{[k]}^{\dagger}= \stackrel{ab}{[k]},
\label{graphher}
\end{eqnarray}
%
and
%
\begin{eqnarray}
\stackrel{ab}{(k)}\stackrel{ab}{(k)}& =& 0, \quad \quad \stackrel{ab}{(k)}\stackrel{ab}{(-k)}
= \eta^{aa}  \stackrel{ab}{[k]}, \quad \stackrel{ab}{(-k)}\stackrel{ab}{(k)}=
\eta^{aa}   \stackrel{ab}{[-k]},\quad
\stackrel{ab}{(-k)} \stackrel{ab}{(-k)} = 0 \nonumber\\
\stackrel{ab}{[k]}\stackrel{ab}{[k]}& =& \stackrel{ab}{[k]}, \quad \quad
\stackrel{ab}{[k]}\stackrel{ab}{[-k]}= 0, \;\;\quad \quad  \quad \stackrel{ab}{[-k]}\stackrel{ab}{[k]}=0,
 \;\;\quad \quad \quad \quad \stackrel{ab}{[-k]}\stackrel{ab}{[-k]} = \stackrel{ab}{[-k]}
 \nonumber\\
\stackrel{ab}{(k)}\stackrel{ab}{[k]}& =& 0,\quad \quad \quad \stackrel{ab}{[k]}\stackrel{ab}{(k)}
=  \stackrel{ab}{(k)}, \quad \quad \quad \stackrel{ab}{(-k)}\stackrel{ab}{[k]}=
 \stackrel{ab}{(-k)},\quad \quad \quad 
\stackrel{ab}{(-k)}\stackrel{ab}{[-k]} = 0
\nonumber\\
\stackrel{ab}{(k)}\stackrel{ab}{[-k]}& =&  \stackrel{ab}{(k)},
\quad \quad \stackrel{ab}{[k]}\stackrel{ab}{(-k)} =0,  \quad \quad 
\quad \stackrel{ab}{[-k]}\stackrel{ab}{(k)}= 0, \quad \quad \quad \quad
\stackrel{ab}{[-k]}\stackrel{ab}{(-k)} = \stackrel{ab}{(-k)}.
\label{graphbinoms}
\end{eqnarray}
%
We recognize in  the first equation of the first row and the first equation of the second row
the demonstration of the nilpotent and the projector character of the Clifford algebra objects $\stackrel{ab}{(k)}$ and 
$\stackrel{ab}{[k]}$, respectively. 
%

{\em We pay attention to the reader that whenever the Clifford algebra objects apply from the left hand side
they always transform } $\stackrel{ab}{(k)}$ {\em to} $\stackrel{ab}{[-k]}$, {\em never to} $\stackrel{ab}{[k]}$,
{\em and similarly } $\stackrel{ab}{[k]}$ {\em to} $\stackrel{ab}{(-k)}$, {\em never to} $\stackrel{ab}{(k)}$.



According to ref.\cite{holgernorma2002},  we define a vacuum state $|\psi_0>$ so that one finds
%
\begin{eqnarray}
< \;\stackrel{ab}{(k)}^{\dagger}
 \stackrel{ab}{(k)}\; > = 1 \nonumber\\
< \;\stackrel{ab}{[k]}^{\dagger}
 \stackrel{ab}{[k]}\; > = 1.
\label{graphherscal}
\end{eqnarray}
%

Taking into account the above equations it is easy to find a Weyl spinor irreducible representation
for $d$-dimensional space, with $d$ even or odd. (We stimulate the reader to see the reference\cite{holgernorma2002}.) 

For $d$ even we simply make a starting state as a product of $d/2$, let us say, only nilpotents 
$\stackrel{ab}{(k)}$, one for each $S^{ab}$ of the Cartan subalgebra  elements (Eq.(\ref{choicecartan})),  applying it 
on an (unimportant) vacuum state\cite{holgernorma2002}. 
%For $d$ odd the basic states are products
%of $(d-1)/2$ nilpotents and a factor $(1\pm \Gamma)$.  
Then the generators $S^{ab}$, which do not belong 
to the Cartan subalgebra, being applied on the starting state from the left, 
 generate all the members of one
Weyl spinor.  
%
\begin{eqnarray}
\stackrel{0d}{(k_{0d})} \stackrel{12}{(k_{12})} \stackrel{35}{(k_{35})}\cdots \stackrel{d-1\;d-2}{(k_{d-1\;d-2})}
\psi_0 \nonumber\\
\stackrel{0d}{[-k_{0d}]} \stackrel{12}{[-k_{12}]} \stackrel{35}{(k_{35})}\cdots \stackrel{d-1\;d-2}{(k_{d-1\;d-2})}
\psi_0 \nonumber\\
\stackrel{0d}{[-k_{0d}]} \stackrel{12}{(k_{12})} \stackrel{35}{[-k_{35}]}\cdots \stackrel{d-1\;d-2}{(k_{d-1\;d-2})}
\psi_0 \nonumber\\
\vdots \nonumber\\
\stackrel{0d}{[-k_{0d}]} \stackrel{12}{(k_{12})} \stackrel{35}{(k_{35})}\cdots \stackrel{d-1\;d-2}{[-k_{d-1\;d-2}]}
\psi_0 \nonumber\\
\stackrel{od}{(k_{0d})} \stackrel{12}{[-k_{12}]} \stackrel{35}{[-k_{35}]}\cdots \stackrel{d-1\;d-2}{(k_{d-1\;d-2})}
\psi_0 \nonumber\\
\vdots 
\label{graphicd}
\end{eqnarray}
%
All the states have the handedness $\Gamma $, since $\{ \Gamma, S^{ab}\} = 0$, which one easily calculates 
by multiplying from the left hand side the starting
state by $\Gamma$ of Eq.(\ref{hand}). 
States, belonging to one multiplet  with respect to the group $SO(q,d-q)$, that is to one
irreducible representation of spinors (one Weyl spinor), can have any phase. We made a choice
of the simplest one, taking all  phases equal to one.

The above graphic representation demonstrated that for $d$ even 
all the states of one irreducible Weyl representation of a definite handedness follow from a starting state, 
which is, for example, a product of nilpotents $\stackrel{ab}{(k_{ab})}$, by transforming all possible pairs
of $\stackrel{ab}{(k_{ab})} \stackrel{mn}{(k_{mn})}$ into $\stackrel{ab}{[-k_{ab}]} \stackrel{mn}{[-k_{mn}]}$.
There are $S^{am}, S^{an}, S^{bm}, S^{bn}$, which do this.
The procedure gives $2^{(d/2-1)}$ states. A Clifford algebra object $\gamma^a$ being applied from the left hand side,
transforms  a 
Weyl spinor of one handedness into a Weyl spinor of the opposite handedness. Both Weyl spinors form a Dirac 
spinor. We call such a set of
states a ''family''. 

For $d$ odd a Weyl spinor has besides a product of $(d-1)/2$ nilpotents or projectors also either the
factor $\stackrel{+}{\circ}:= \frac{1}{2} (1+\Gamma)$ or the factor
$\stackrel{-}{\bullet}:= \frac{1}{2}(1-\Gamma)$. (See the ref.\cite{holgernorma2002}.) 
As in the case of $d$ even, all the states of one irreducible 
Weyl representation of a definite handedness follow from a starting state, 
which is, for example, a product of $(1 + \Gamma)$ and $(d-1)/2$ nilpotents $\stackrel{ab}{(k_{ab})}$, by 
transforming all possible pairs
of $\stackrel{ab}{(k_{ab})} \stackrel{mn}{(k_{mn})}$ into $\stackrel{ab}{[-k_{ab}]} \stackrel{mn}{[-k_{mn}]}$.
But $\gamma^a$'s, being applied from the left hand side, do not change the handedness of the Weyl spinor, 
since $\{ \Gamma,
\gamma^a \}_- =0$ for $d$ odd\cite{holgernorma2002}.
A Dirac and a Weyl spinor are for $d$ odd identical and a ''family'' 
has accordingly $2^{(d-1)/2}$ members of basic states of a definite handedness.

We shall speak about left handedness when $\Gamma= -1$ and about right
handedness when $\Gamma =1$ for either $d$ even or odd. 

{\em When the whole Clifford algebra is considered as states in a Hilbert space, then we get ''families''}.



\section{``Families'' }
\label{families}

When all $2^d$ states are considered as a Hilbert space, we recognize that there are for
$d$ even $2^{d/2}$ ''families'' and for $d$ odd $2^{(d+1)/2}$ ''families'' of spinors.

We prove  in this section (see also the ref.\cite{pikanorma2002}) that there exists an operation, 
which transforms a state of one ''family''
into a state of another ''family'', keeping all the  properties with respect to the Lorentz group unchanged.

We saw in the last section (\ref{technique}) that any Clifford algebra object when
multiplying from the left products of nilpotents and projectors - a state of a Dirac spinor -
transforms this state into a 
superposition of states of the same Dirac spinor. We  called a Dirac spinor a ''family''. Since there are $2^d$ 
linearly independent states, one finds for $d$ even  $2^{d/2}$ ''families''  and for $d$ odd $2^{(d+1)/2}$ 
''families'' of the Dirac spinors.
''Families'' form left ideals with respect to the multiplication with the Clifford algebra objects. 


{\em The question then arises: Which operation  transfroms a state of one ''family'' into a state of
another ''family''?}

{\em Theorem 1}: {\em It is the right multiplication with the Clifford algebra objects, which transforms a state of one
''family'' into a state of another ''family''}.

{\em Proof}: The Clifford algebra object $\stackrel{ab}{(k_{ab})}$ transforms, when being applied from the left
hand side 
by either $\gamma^a$ (or by $\gamma^b$), into $\stackrel{ab}{[-k_{ab}]}$ (Eq.(\ref{graphgammaaction})). 
One finds this by simply multiplying 
$\stackrel{ab}{(k_{ab})}$ from the left hand side by one of these two Clifford algebra objects and 
taking into accountEq.(\ref{clif})
%
\begin{eqnarray}
\gamma^a \stackrel{ab}{(k_{ab})} = \gamma^a \; \frac{1}{2}(\gamma^a + \frac{\eta^{aa}}{ik}\gamma^b)= \eta^{aa}
\frac{1}{2}(1  + \frac{i}{-k}\gamma^a \gamma^b) = \eta^{aa} \stackrel{ab}{[-k_{ab}]}.
\label{leftcheck}
\end{eqnarray}
%
(And similarly we get $\gamma^b \stackrel{ab}{(k)}= -ik \stackrel{ab}{[-k]}$. The product of $\gamma^a \gamma^b$ 
transforms $\stackrel{ab}{(k_{ab})}$ into itself, multiplying it by  $-ik$, since $\stackrel{ab}{(k_{ab})}$ is
the eigen vector of the Cartan subalgebra of the Lorentz algebra.)

Let us multiply now the same object $\stackrel{ab}{(k_{ab})}$ by  $\gamma^a$  {\em from
the right hand side}. It follows
%
\begin{eqnarray}
\stackrel{ab}{(k_{ab})}\gamma^a =  \frac{1}{2}(\gamma^a + \frac{\eta^{aa}}{ik}\gamma^b)\; \gamma^a = \eta^{aa}
\frac{1}{2}(1  + \frac{i}{k}\gamma^a \gamma^b) = \eta^{aa} \stackrel{ab}{[k_{ab}]}.
\label{rightcheck}
\end{eqnarray}
%
We saw in Eq.(\ref{leftcheck}) that the multiplication from the left hand side by any Clifford algebra object
transforms $\stackrel{ab}{(k_{ab})}$ into $\stackrel{ab}{[-k_{ab}]}$, never to $\stackrel{ab}{[k_{ab}]}$.
This means that $\stackrel{ab}{[k_{ab}]}$ is going to be a building block of a different ''family'' than
$\stackrel{ab}{[-k_{ab}]}$ is. This completes the proof.

{\em Theorem 2:} {\em The two operations: the left and the right multiplication by $\gamma^a$ anticommute}.

{\em Proof:} To see this we need to show that the two objects $\stackrel{ab}{(k_{ab})}$ 
and $\stackrel{ab}{[k_{ab}]}$, the second one obtained from the first one by the right multiplication,
have all the properties with respect to the Lorentz group (application of the Lorentz algebra 
objects concerns the left multiplicatio) equal and that they
differ accordingly only in the ''family'' name. The left multiplication by $\gamma^a $ of the object
$\stackrel{ab}{(k_{ab})}$ leads, as we know (Eq(\ref{grapheigen})), to $\stackrel{ab}{[-k_{ab}]}$, 
whose $S^{ab} \stackrel{ab}{(k_{ab})}= k_{ab}/2\stackrel{ab}{(k_{ab})}$.


To check the properties of the two Clifford algebra objects $\stackrel{ab}{(k_{ab})}$ and 
$\stackrel{ab}{[k_{ab}]}$ with respect to the Lorentz group we have to multiply each of the
two Clifford algebra objects  from 
the left hand side by $S^{ab}$, which is the Cartan subalgebra element. 
According to Eq.(\ref{grapheigen})
we find
$S^{ab}\stackrel{ab}{(k_{ab})} = \frac{i}{2} \frac{1}{2}(\gamma^a \gamma^b \gamma^a + \frac{\eta^{aa}}{ik}
\gamma^a \gamma^b \gamma^b) = \frac{1}{2} k \stackrel{ab}{(k_{ab})},
S^{ab}\stackrel{ab}{[k_{ab}]} = \frac{1}{2} k  \stackrel{ab}{[k_{ab}]}.
$

Both objects, $\stackrel{ab}{(k_{ab})}$ and $\stackrel{ab}{[k_{ab}]}$ have the same eigenvalue for the Cartan 
subalgebra element $S^{ab}$, namely $\frac{1}{2} k_{ab} $. Since the right multiplication of the object
$\stackrel{ab}{(k_{ab})}$ does not change the properties of the object with respect to the Lorentz group 
(the properties of which are determined by the left multiplication) it must be that the two operations - 
the left and the right multiplication
with $\gamma^a$'s, both fulfilling the Clifford algebra relation - anticommute and the proof is completed.

To simplify the technique, {\em we define}\cite{pikanorma2002} {\em the Clifford algebra objects}
$\tilde{\gamma}^a$'s {\em as operations, which operate formally from the left hand side} (as $\gamma^a$'s do)
{\em on objects } $\stackrel{ab}{(k_{ab})}$ {\em and} $\stackrel{ab}{[k_{ab}]}$, {\em transforming these states to}
$\stackrel{ab}{[k_{ab}]}$ {\em and} $\stackrel{ab}{(k_{ab})}$, respectively, {\em as} $\gamma^a$'s {\em would 
do if being applied from the right hand side}
%
\begin{eqnarray}
\tilde{\gamma^a} \stackrel{ab}{(k_{ab})}: = -\stackrel{ab}{(k_{ab})} \gamma^a = - \eta^{aa}
\stackrel{ab}{[k_{ab}]}.
\label{gammatilde}
\end{eqnarray}
%
One can easily check that it is also true
%
\begin{eqnarray}
\tilde{\gamma^a} \stackrel{ab}{[k_{ab}]}: = \stackrel{ab}{[k_{ab}]} \gamma^a = 
\stackrel{ab}{(k_{ab})}.
\label{gammatilde1}
\end{eqnarray}
%
 According to {\em Theorem 2} we may write
%
\begin{eqnarray}
\{\tilde{\gamma^a}, \gamma^b \}_+ = 0, \quad {\rm while} \nonumber\\
\{\tilde{\gamma}^a, \tilde{\gamma}^b \}_+ = 2 \eta^{ab}.
\label{gammatildegamma}
\end{eqnarray}
%
If we define
%
\begin{eqnarray}
\tilde{S}^{ab} = \frac{i}{4}\; [\tilde{\gamma}^a,\tilde{\gamma}^b] = \frac{1}{4} (\tilde{\gamma}^a
\tilde{\gamma}^b - \tilde{\gamma}^b\tilde{\gamma}^a),
\label{tildesab}
\end{eqnarray}
%
it follows then that $\tilde{S}^{ab}$ fulfil the Lorentz algebra relation as $S^{ab}$ do and that 
%
\begin{eqnarray}
\{\tilde{S}^{ab}, S^{ab}\}_- =0,\quad
\{\tilde{S}^{ab}, \gamma^c \}_-=0,\quad
\{S^{ab}, \tilde{\gamma}^c \}_-=0.
\label{sabtildesab}
\end{eqnarray}
%
One also finds that
%
\begin{eqnarray}
\{\tilde{S}^{ab}, \Gamma \}_- =0,\quad \{ \tilde{\gamma}^a, \Gamma \}_- = 0, \quad {\rm for \;\; d\;\; even,}\nonumber\\
\{\tilde{S}^{ab}, \Gamma \}_- =0,\quad \{ \tilde{\gamma}^a, \Gamma \}_+ = 0, \quad {\rm for \;\; d\;\; odd,}
\label{sabtildeGAMMA}
\end{eqnarray}
%
which means that when transforming in $d$ even one ''family'' into another with either $\tilde{S}^{ab}$ 
or $\tilde{\gamma}^a$ handedness $\Gamma$ stays unchanged, while the transformation to anotother 
''family'' in $d$ odd with $\tilde{\gamma}^a$ changes handedness of states, namely the factor
 $\frac{1}{2}(1\pm \Gamma)$ changes to  $\frac{1}{2}(1\mp \Gamma)$ in accordance with what we know from before:
 In spaces with odd $d$  changing the handedness means changing the ''family''.

We stimulate the reader to read also the refs.\cite{norma93,normaixtapa2001} where the two kinds of the
Clifford algebra objects follow as the two different superpositions of a Grassmann coordinate and its 
conjugate momentum.

We present  for $\tilde{\gamma}^a$'s some useful relations (as we did in Eqs.(\ref{graphgammaaction},
\ref{sab}) for $\gamma^a$'s) 
%
\begin{eqnarray}
\tilde{\gamma}^a \stackrel{ab}{(k)}&=&-\eta^{aa}\stackrel{ab}{[k]},\nonumber\\
\tilde{\gamma}^b \stackrel{ab}{(k)}&=& ik \stackrel{ab}{[k]}, \nonumber\\
\tilde{\gamma}^a \stackrel{ab}{[k]}&=& \stackrel{ab}{(k)},\nonumber\\
\tilde{\gamma}^b \stackrel{ab}{[k]}&=& ik \eta^{aa} \stackrel{ab}{(k)}
\label{tildegraphgammaaction}
\end{eqnarray}
%
and
%
\begin{eqnarray}
\tilde{S}^{ac}\stackrel{ab}{(k)}\stackrel{cd}{(k)}& = &-\frac{i}{2} \eta^{aa} \eta^{cc} 
\stackrel{ab}{[k]}\stackrel{cd}{[k]}, \nonumber\\
\tilde{S}^{ac}\stackrel{ab}{[k]}\stackrel{cd}{[k]}& = &\frac{i}{2}  
\stackrel{ab}{(k)}\stackrel{cd}{(k)}, \nonumber\\
\tilde{S}^{ac}\stackrel{ab}{(k)}\stackrel{cd}{[k]}& = &\frac{i}{2} \eta^{aa}  
\stackrel{ab}{[k]}\stackrel{cd}{(k)}, \nonumber\\
\tilde{S}^{ac}\stackrel{ab}{[k]}\stackrel{cd}{(k)}& = &-\frac{i}{2} \eta^{cc}  
\stackrel{ab}{(k)}\stackrel{cd}{[k]}.
\label{tildesac}
\end{eqnarray}
%

According to {\em Theorem 1} we transform a state of one ''family'' to a state of another ''family'' by 
the application of
$\tilde{\gamma}^a$ or $\tilde{S}^{ac}$ (formaly from the left hand side) on a state of the first
''family'' for a chosen $a$ or $a,c$. To transform all the states of one ''family'' into states
of another ''family'', we apply  $\tilde{\gamma}^a$ or $\tilde{S}^{ac}$ on each state of the starting ''family''.
It is, of course, enough to apply $\tilde{\gamma}^a$ or $\tilde{S}^{ac}$ 
on only one state of a ''family'' and then use generators of the Lorentz group ($S^{ab}$) and for $d$ even 
also $\gamma^a$'s 
to generate all the states of one Dirac spinor.

One must notice that nilpotents $\stackrel{ab}{(k)}$ and projectors $\stackrel{ab}{[k]}$ are eigenvectors
of not only the Cartan subalgebra $S^{ab}$ but also of $\tilde{S}^{ab}$. Accordingly only $\tilde{S}^{ac}$, which
do not carry the Cartan subalgebra indices, cause the transition from one ''family'' to another ''family''.

The starting state of Eq.(\ref{graphicd}) can change, for example, to
%
\begin{eqnarray}
\stackrel{0d}{[k_{0d}]} \stackrel{12}{[k_{12}]} \stackrel{35}{(k_{35})}\cdots \stackrel{d-1\;d-2}{(k_{d-1\;d-2})},
\label{tildegraphicd}
\end{eqnarray}
%
if $\tilde{S}^{01}$ was chosen to transform the Weyl spinor of Eq.(\ref{graphicd}) to the Weyl spinor of another
''family''.

In the next section \ref{d4} we demonstrate the appearance of ''families for $d=4$ with the Minkowski
signature.


\subsection{''Families'' for $d=4$}
\label{d4}



There are two  $(d/2=2)$ operators of the Cartan subalgebra of the Lorentz algebra (which is 
closed by the operators 
$S^{01}, S^{02}, S^{03}, S^{12}, S^{13}, S^{23}$), for which we made a choice, according to 
Eq.(\ref{choicecartan})
of $ S^{03}$ and $S^{12}$. Following Eq.(\ref{hand}) 
we find $\Gamma = 
i \gamma^0 \gamma^1 \gamma^2 \gamma^3. $ There are $2^4$ , that is sixteen basic states,
all of them  being  eigenstates of $S^{12}$ and $S^{03}$
%
\begin{eqnarray}
\stackrel{03}
{(\pm i)} \stackrel{12}{(\pm)}&=&(\frac{1}{2})^2 \;(\gamma^0 \mp \gamma^3) \;(\gamma^1 \pm i \gamma^2), \quad 
\stackrel{03}
{(\pm i)} \stackrel{12}{[\pm]}\;=\;(\frac{1}{2})^2 \;
(\gamma^0 \mp \gamma^3) (1 \pm i \gamma^1 \gamma^2),
\nonumber\\
\stackrel{03}
{[\pm i]} \stackrel{12}{(\pm)}&=&(\frac{1}{2})^2 (1 \pm \gamma^0 \gamma^3)\; (\gamma^1 \pm i \gamma^2), \quad 
\stackrel{03}
{[\pm i]} \stackrel{12}{[\pm]}\;=\;(\frac{1}{2})^2
(1 \pm \gamma^0 \gamma^3) (1 \pm i \gamma^1 \gamma^2),
\label{basisfour}
\end{eqnarray}
%
with the eigenvalues of the Cartan operator $S^{12}$ equal to
 $\pm 1/2$ for the $\pm$ sign in the second factor of the graphical presentation and the eigenvalues of the Cartan operator 
$S^{03}$ equal to
$\pm i/2$ for the $\pm i$ sign in the first factor of the graphical presentation. All sixteen basic states 
are orthonormal\cite{holgernorma2002}.


We arrange these sixteen states into four ``families'' so that we first make a choice of the starting state of
the first ''family'' as a product of two nilpotents $\stackrel{03}
{(+i)} \stackrel{12}{(+)} $.  Then we use $S^{01}$, for example, to find the second state 
$\stackrel{03}
{[-i]} \stackrel{12}{[-]} $ of one Weyl spinor. $\gamma^0$ generates from the first state of the first Weyl spinor
the first state of the second Weyl spinor, namely $\stackrel{03}{[-i]} \stackrel{12}{(+)} $ and $S^{01}$ then the
second state of the second spinor $\stackrel{03}{(+i)} \stackrel{12}{[-]} $. We choose all the phases of 
the states equal to one.

We transform this first ''family'' into the second ''family'' by the application of $\tilde{S}^{01}$ 
(or $\tilde{S}^{02}$, or $\tilde{S}^{31}$, or $\tilde{S}^{32}$) on each of the
state of the first ''family''. By the application of $\tilde{\gamma}^a$ on all the states of the first ''family''
we get the third ''family''. We also can generate the third ''family'' from the second one by the application of 
$\tilde{\gamma}^1$ (or $\tilde{\gamma}^{2}$ ). The fourth ''family'' can be reached from the first one by the application of
$\tilde{\gamma}^1$ (or $\tilde{\gamma}^2$), but it can be reached also from any other with the appropriate 
choice of operations. For all the ''families'' the simplest choice of the relative phases, namely the phase $1$ 
is made.


Each ``family'' includes two Weyl spinors, one left- and one right-handed. 
These four ``families'' are  presented in Table I.

\begin{center}
\begin{tabular}{|r|r||c|c|c|c||r|r|r||}
\hline
a&i&$(^a\psi_i)_1$&$(^a\psi_i)_2$&$(^a\psi_i)_3$&$(^a\psi_i)_4$&$ S^{12}$&$ S^{03}$&$
\Gamma$\\
\hline\hline
1&1&$\quad \stackrel{03}{(+i)} \stackrel{12}{(+)} $&$\quad \stackrel{03}{[+i]} \stackrel{12}{[+]} $&
$\quad \stackrel{03}{[+i]} \stackrel{12}{(+)}$&$\quad \stackrel{03}{(+i)} \stackrel{12}{[+]} $& 
$\frac{1}{2}$& $\frac{i}{2}$& -1\\
\hline 
1&2&$\quad \stackrel{03}{[-i]} \stackrel{12}{[-]} $&$\quad \stackrel{03}{(-i)} \stackrel{12}{(-)} $&
$\quad \stackrel{03}{(-i)} \stackrel{12}{[-]}$&$\quad \stackrel{03}{[-i]} \stackrel{12}{(-)} $&
 $-\frac{1}{2}$& $-\frac{i}{2}$& -1\\
\hline\hline
2&1&$\quad \stackrel{03}{[-i]} \stackrel{12}{(+)} $&$\quad \stackrel{03}{(-i)} \stackrel{12}{[+]} $&
$\quad \stackrel{03}{(-i)} \stackrel{12}{(+)}$&$\quad \stackrel{03}{[-i]} \stackrel{12}{[+]} $
& $\frac{1}{2}$& $-\frac{i}{2}$& 1\\
\hline 
2&2&$\quad \stackrel{03}{(+i)} \stackrel{12}{[-]} $&$\quad \stackrel{03}{[+i]} \stackrel{12}{(-)} $&
$\quad \stackrel{03}{[+i]} \stackrel{12}{[-]}$&$\quad \stackrel{03}{(+i)} \stackrel{12}{(-)} $
& $-\frac{1}{2}$& $\frac{i}{2}$& 1\\
\hline\hline
\end{tabular}
\end{center}
%SL(2,C)
Table I.-Four ``families'' of the two Weyl spinors of the Lorentz group $SO(1,3)$.
Basic vectors are eigenvectors of the two  operators of the Cartan subalgebra
$S^{12}$ and $S^{03}$. The eigenvalues of the operator of handedness $\Gamma $
are also presented. All the basic states are orthonormalized as discussed in ref.\cite{holgernorma2002}.
The simplest choice of  relative phases is used - 
all phases are assumed to be  equal to $+1$. 

We see in Table I that either $\tilde{\gamma}^a, \; a=0,1,2,3$, or $\tilde{S}^{01}, \tilde{S}^{02},
\tilde{S}^{31},\tilde{S}^{32},$ when being applied, change the ''family'' but do not change either 
the handedness or the
eigenvalues of the Cartan subalgebra of the Lorentz algebra.


Any of the four ``families'' can be used to present the solution of the Dirac  equation
for a massive spinor, while the massless spinors are either left- or right-handed, so that
only half of the space of the massive case is needed to find the solution. The chosen phases for basic states make
the matrix representation of $\gamma^a$'s and $S^{ab}$ equal to the usual ones.




\section{Conclusion}
\label{conclusion}



In the ref.\cite{holgernorma2002} we constructed the basis for a left ideal out of products of nilpotents
and projectors and identified the basis with 
the spinor space. But the spinor space (a Dirac spinor, which is a Weyl bi-spinor for $d$ even
and a Weyl spinor for $d$ odd) depends on a selection of a ``starting'' Clifford object. 
We have for $d$ even $2^{d/2}$ different starting states and for $d$ odd $2^{(d+1)/2}$ different 
starting states, which  with an appropriate 
choice of an (unimportant)
vacuum state can be made all orthogonal. We call different basis (that is different spinor spaces, which have 
all the same properties with respect to the Lorentz group) ''families'' of spinors.
%(
%To achieve the required orthogonality, we make   a choice of 
%phases for states belonging to different ``families'' in such a way that, when choosing the ``vacuum state'' 
% to be the sum of not only 
%all the states in one ``family'' but of all the states of all the ``families'', each state appearing with the same 
%coefficient, not only the expectation values of the operators $S^{ab}$, belonging
%to the Cartan subalgebra, but also the expectation values of $\gamma^a$ are for  such a ''vacuum state'' equal to zero, 
%for $d$ even and odd. )

We have shown in 
this paper that while the left multiplication with any Clifford algebra object makes from a ''starting'' state 
a superposition of basic states of one Dirac spinor, {\em the right multiplication} of a ''starting state''
with any Clifford algebra object makes a superposition of states belonging to different ''families'' but having the 
the same properties with respect 
to the Lorentz group (defined by the left multiplication of   Clifford algebra objects)  as the starting state.
One comes  accordingly from one ''family'' to another ''family'' (up to an over all factor) by multiplying 
the ''starting'' state from the right hand side by $S^{ab}$'s for $d$ odd and by $S^{ab}$'s and $\gamma^a$'s 
for $d$ even, if $S^{ab}$ does not belong to the Cartan subalgebra of the Lorentz group.

We have defined in this paper $\tilde{\gamma}^{a}$'s and $\tilde{S}^{ab}$ with the properties that both
transform nilpotents and projectors, when multiplying them from the left hand side as
$\gamma^{a}$'s and $S^{ab}$ would do if multiplying nilpotents and projectors from the right hand side. Since neither
$\tilde{\gamma}^{a}$'s nor $\tilde{S}^{ab}$  change properties of nilpotents and projectors (and
accordingly of a state) with respect to the Lorentz group and since both - $\gamma^a$'s and 
$\tilde{\gamma}^a$'s - are the Clifford algebra objects, 
 it follows that $\{ \tilde{\gamma}^{a},\gamma^b \}_+
=0$ and  $\{ \tilde{\gamma}^{a},\tilde{\gamma}^b \}_+ =
\eta^{ab}$. Consequently, $\tilde{S}^{ab}$ fulfill the Lorentz algebra relation as $S^{ab}$ do. This can 
be understood since one could construct the spinor basis for a right ideal instead for a left ideal and then use left
multiplication to generate families. (To understand this better we suggest the reader to see 
refs.\cite{norma93,normaixtapa2001,holgernormadk,pikanorma2002}.)

We have also demonstrated the application of $\tilde{\gamma}^a$'s and $\tilde{S}^{ab}$ on a basis with the help
of the graphic technique, introduced in ref.\cite{holgernorma2002}. We demonstrated the procedure 
for generating ''families'' on the case of $d =1 +3$, that is for $d=4$ and one time coordinate, where the number
of ''families'' is four.

CAt the end we have to ask ourselves whether or not the proposed generation of  ''families'' 
can be used for the description of the families of 
quarks and leptons? We believe that we have the right way to do this. According 
to refs.\cite{normaixtapa2001,pikanorma2002}, one can generate families of quarks and leptons dynamically if in the
covariant derrivative $\tilde{S}^{ab} $ appear as charges, accompanied by gauge fields like 
$\tilde{\omega}^{ab}{}_{c}$, so that 
%
\begin{eqnarray}
p^a{}_o = p^a - \frac{1}{2} \tilde{S}^{ab} \tilde{\omega}_{abc}- \tau^{Ai} A^{Ai a},
\label{covariantp}
\end{eqnarray}
%
with $\tau^{Ai}$ determining the known charges ($U(1), SU(2), SU(3) $) and $A^{Ai a}$ the corresponding gauge fileds.
This possibility has been discussed  in ref.\cite{pikanorma2002}.




 
 


\section{Acknowledgement } 

  
  
This work was supported by Ministry of Education, 
Science and Sport of Slovenia  as well as by funds CHRX -
CT - 94 - 0621, INTAS 93 - 3316, INTAS - RFBR 95
- 0567, SCI-0430-C (TSTS) of  Denmark. 


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%
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%\end{document}

\end{document}