How does Clifford algebra show the way to the second quantized fermions with
unified spins, charges and families, and with vector and scalar gauge
fields beyond the standard model

N.S. Mankoč Borštnik,^1 H.B. Nielsen^2

^1Department of Physics, University of Ljubljana

SI-1000 Ljubljana, Slovenia

^2Niels Bohr Institute, University of Copenhagen

Blegdamsvej 17, CopenhagenØ, Denmark

Fifty years ago the standard model offered an elegant new step
towards understanding elementary fermion and boson fields, making several
assumptions, suggested by experiments. The assumptions are still waiting
for an explanation.
There are many proposals in the literature for the next step.
The spin-charge-family theory of one of us (N.S.M.B.) is offering the
explanation for not only all by the standard model
assumed properties of quarks and leptons and antiquarks and antileptons, with the
families included, of the vectors gauge fields, of the Higgs's scalar and Yukawa
couplings, but also for the second quantization postulates of Dirac and
for cosmological observations, like there are the appearance of the dark matter,
of matter-antimatter asymmetry, making several predictions. This theory
proposes a simple starting action in $ d\ge (13+1)$-dimensional space with fermions
interacting with the gravity only (the vielbeins and the two kinds of the spin
connection fields), what manifests in $d=(3+1)$ as the vector and scalar gauge
fields, and uses the odd Clifford algebra to describe the internal space of
fermions, what enables that the creation and annihilation operators for fermions
fulfill the anticommutation relations for the second quantized fields without Dirac's
postulates: Fermions single particle states already anticommute.
We present in this review article a short overview of the spin-charge-family
theory, illustrating shortly on the toy model the breaks of the starting symmetries
in $d=(13+1)$-dimensional space, which are triggered either by scalar fields —
the vielbeins with the space index belonging to $d>(3+1)$ — or by the
condensate of the two right handed neutrinos, with the family quantum
number not belonging to the observed families.
We compare properties and predictions of this theory with the properties and
predictions of $SO(10)$ unifying theories.

§ INTRODUCTION

Physicists are gaining knowledge about the laws, that apply to the smallest
constituents of nature and thus to the whole universe, by observations,
experiments, we make mathematical models, predict with models new events,
check predictions by experiments, observations,  all these lead
to new ideas, new models.

Repeating this circle again and again we ask ourselves questions like:
Is the law of
nature simple and elegant,  manifesting complexity on many fermion and
boson states in particular at low energies, or ”nature uses mathematics,
when just appears to be needed”?
Is the space-time $(3+1)$-dimensional? Why? Or has it
many more dimensions? How many? There are namely only two simple and
elegant choices, zero and infinity.
What are elementary constituents and interactions among constituents in our
universe, in any universe?
How has the space-time of our universe started?
v. And many others.

Fifty years ago the standard model offered an elegant new step towards
understanding elementary fermion and boson fields by postulating:

a. The existence of massless fermion family members with the  spins
and charges in the fundamental representation of the groups, a.i. the
quarks as colour triplets and colourless leptons, a.ii the left
handed  members as the weak doublets, the right handed
weak chargeless members, a.iii. the left handed quarks differing from
the left handed leptons in the hyper charge, a.iv. all the right
handed members differing among themselves in hyper charges,
a.v. antifermions carrying the corresponding anticharges of fermions and
opposite handedness, a.vi. the  families of massless fermions,
suggested by experiments  and required by the gauge invariance of the boson
fields (there is no right handed neutrino postulated,
since it would carry none of the so far observed charges, and correspondingly
there is also no left handed antineutrino allowed in the standard model).

b. The existence of massless vector gauge fields to the observed
charges of quarks and leptons, carrying charges in the adjoint representations
of the corresponding charged groups and manifesting the gauge invariance.

c.  The existence of the massive weak doublet scalar higgs, c.i.
carrying the weak charge $\pm \frac{1}{2}$  and the hyper charge
$\mp \frac{1}{2}$ (as it would be in the fundamental representation of the
two groups), c.ii. gaining at some step of the expanding universe the
nonzero vacuum expectation value, c.iii. breaking the weak and the
hyper charge and correspondingly breaking the mass protection, c.iv.
taking care of the properties of fermions and of the weak bosons masses,
c.v. as well as the existence of the Yukawa couplings.

d. The presentation of fermions and bosons as second quantized fields.

e. The gravitational field in $d=(3+1)$ as independent gauge field.
(The standard model is defined  without gravity in
order that it be renormalizable, but yet the standard model particles
are ”allowed” to couple to gravity in the ”minimal” way.)

The standard model assumptions have been experimentally confirmed
without raising any doubts so far, except for some few and possibly statistical
fluctuations anomalies [
We think here on the improved standard model,
in which neutrinos have non-zero masses, and has no ambition to explain
severe cosmological problems], but also by offering no explanations for
the assumptions.
The last among the fields postulated by the standard model,  the
scalar higgs, was detected in June 2012,  the gravitational waves
were detected in February 2016.

The standard model  has in the literature several  explanations, mostly
with many new not explained assumptions. The most  popular seem to be
the grand unifying theories <cit.>.
At least $SO(10)$-unifying theories offer the explanation for the postulates from
a.i. to a.iv, partly
to b. by assuming that to all the fermion charges there exist the
corresponding vector gauge fields — but does not explain the  assumptions
a.v. up to a.vi., c. and d., and does not connect gravity
with gauge vector and scalar fields.

What questions should we ask ourselves to be able to find a trustworthy next step
beyond the standard models of elementary particle physics and cosmology,
which would offer understanding of not yet understood phenomena?

i. Where do fermions, quarks and leptons, originate and why do they differ
from the boson fields in spins, charges and statistics?

ii. Why are charges of quarks and leptons so different, why have the
left handed family members so different charges from the right handed ones and
why does the handedness relate charges to anticharges?

iii. Isn't the first quantization an approximation to the second quantization?
How should one describe the internal degrees of fermions to explain the
Dirac's postulates of the second quantization?

iv. Where do families of quarks and leptons originate and how many
families do exist?

v. Why do family members – quarks and leptons — manifest so different
masses if they all start as massless?

vi. How is the origin of the scalar field (the Higgs's scalar) and the
Yukawa couplings connected with the origin of families and how many scalar
fields determine properties of the so far (and others possibly be) observed
fermions and masses of weak bosons? (The Yukawa couplings certainly speak
for the existence of  several scalar fields with the properties of Higgs's
Why is  the Higgs's scalar, or are all scalar fields of similar properties
as the higgs,  if there are several, doublets with respect to the weak and the
hyper charge?

vii. Do possibly exist also scalar fields with the  colour charges in the
fundamental representation (like the Higgs's scalars are doublets with respect
to the weak charge) and where, if they are, do they manifest?

viii. Where do the so far observed  (and others possibly non observed)
vector gauge fields originate? Do they have anything in common with the scalar
fields and the gravitational fields?

ix. Where does the dark matter originate?

x.  Where does the "ordinary"  matter-antimatter asymmetry originate?

xi. Where does the  dark energy originate and why is it so small?

xii. What is the dimension of space? $(3+1)?$, $((d-1)+1)?$, $\infty?$

xiii. And many others.

The assumptions of the standard model  are still waiting for an explanation.

Motivation for using Clifford algebra objects to  describe the internal
space of fermions:

In a long series of works we, mainly one of us N.S.M.B. (<cit.> and the references therein), have found
phenomenological success with the model named
the spin-charge-family  theory,  with fermions, the internal space of which
is described with the Clifford algebra of all linear
combinations of odd products of $\gamma^a$'s in $d=(13+1)$,
Sect. <ref>,

interacting with only gravity ([14] and references
therein), Sect. <ref>. The spins of fermions from
higher dimensions, $d>(3+1)$, manifest
in $d=(3+1)$ as charges of the standard model, Sect. <ref>,
gravity in higher dimensions manifest as the standard model gauge vector
fields as well as the Higgs's scalar and Yukawa
couplings, Sect. <ref> [21, 17].

Let be added that one irreducible representation of $SO(13,1)$ contains,
if looked from the point of view of $d=(3+1)$,

all the quarks and leptons and antiquarks and antileptons just with the
properties, required by the standard model, including the relation between
quarks and leptons and handedness  and antiquarks and antileptons and the
opposite handedness, Table <ref>.

All what in the standard model have to be assumed (extremely efficiently
"read" from experiments), in the
spin-charge-family theory appear as a possibility
from the starting simple action, Eq. (<ref>).
This simple starting action offers the explanation for not only the properties
of quarks and leptons and antiquarks and antileptons, but also for vector
gauge fields, scalar gauge fields, which represent higgs and explain the
Yukawa couplings and offers also the scalars, which cause matter/antimatter
asymmetry, proton decay, the appearance of the dark matter,
Sect. <ref>.

It has happened so many times in the history of science that the simpler,
more elegant, model has shown up as a more ”powerful” one.
Examples  can be found in nonrelativistic classical physics, after embedded it
into relativistic physics, in Newton's laws after embedded them in general theory
of gravity, in symmetries of fermion and gauge fields when embedded
symmetries into larger groups, etc..

The working hypotheses of at least one of us (N.S.M.B.) is that the laws of
nature are simple and correspondingly elegant and that the many body
systems around the phase transitions look to us complicated at least from
the point of view of the elementary constituents of fermion and boson fields.

To this working hypotheses belongs also the description of the internal space
of fermions with the Clifford algebra.

There appear in the literature two anticommuting kinds of algebras, the
Grassmann algebra [9]  and the Clifford algebra <cit.>.
We recognized <cit.> that the Grassmann algebra includes
two  Clifford algebras, anticommuting with each other, each of them expressible
with  the Grassmann algebra elements and opposite.

The Grassmann algebra, with elements $\theta^a$,
and their Hermitian conjugated partners
$\frac{\partial}{\partial \theta^a}$ [22],
can  be used to describe the internal space of fermions with the integer spins
and charges in the adjoint representations, each of the two Clifford algebras,
we denote their elements with $\gamma^a$ and
$\tilde{\gamma}^a$ [8, 9, 10, 11, 66, 67],
can be used to describe the half integer spins and charges in fundamental
The Grassmann algebra  is expressible with the two Clifford algebras and
opposite, Sects. <ref>, <ref>.

In both algebras, the Grassmann algebra and the two Clifford algebras, the odd
products of objects, arranged to be eigenstates of the Cartan subalgebras of the
corresponding Lorentz algebras, form creation and annihilation operators, which
fulfill the anticommutation relations postulated by Dirac for the second quantized
fermions. These creation operators can be used to describe the internal space of
fermions, Sects <ref>, <ref>.

Steps to the second quantized states with the Clifford or Grassmann algebra:

i.  The internal space of a fermion,
Sects. <ref>,$\,$ <ref>, is described by either Clifford or
algebra of an odd Clifford character (superposition of an odd number of Clifford
"coordinates" — operators — $\gamma^{a}$'s or of an odd number of Clifford
"coordinates" — operators — $\tilde{\gamma}^{a}$'s)
or of an odd Grassmann character (superposition of an odd number of Grassmann
"coordinates"  — operators — $\theta^a$'s).

ii. The eigenvectors of all the (chosen) Cartan subalgebra members of the
corresponding Lorentz algebra

are used to define the "basis vectors" of an odd character.
(The Cartan subalgebra is in all three cases chosen in the way to be in agreement
with the ordinary choice.)
The algebraic application of this "basis vectors" on the corresponding vacuum state (either

Clifford $|\psi_{oc}>$, defined in Eq. (<ref>), or Grassmann $|\phi_{og}>$,
Eq. (<ref>), which  is in the Grassmann case just the identity) generates
the "basis states", usable for describing the internal space of fermions. The members
of the "basis vectors" manifest together with their Hermitian conjugated partners
the properties of creation and annihilation operators,  which anticommute,
Eq. (<ref>) in the Grassmann case and Eq. (<ref>) in the
Clifford case, when
applying on the corresponding vacuum state, due to the algebraic properties of the
odd products of the algebra elements.

iii. The plane wave solutions of the corresponding  Weyl equations (either Clifford,
Eq. (<ref>) or Grassmann, Eq. (<ref>)) for free massless fermions are
the tensor products, $*_{T}$, of the superposition of the members of the "basis vectors"
and of the momentum basis. The coefficients of the superposition correspondingly
depend on a
chosen momentum $\vec{p}$, with $p^0 =|\vec{p}|$, for any of the continuous
values of moments $\vec{p}$.

iv. The creation operators defined on the tensor products, $*_{T}$, of
superposition of the finite number of  ”basis vectors” defining the internal space and
of the infinite (continuous) momentum space, Eq. (<ref>) in the Clifford case
and  Eq. (<ref>) in the Grassmann case, define the infinite basis.

v. Applied on the vacuum state these creation operators form anticommuting
single fermion states of an odd Clifford/Grassmann character.

vi. The second quantized Hilbert space ${\cal H}$ consists of all possible
tensor products, $*_{T_{H}}$,  of any number of single fermion states, starting
with no single single state,

"Slater determinants" with no single particle state occupied (with no creation operators
applied on the vacuum state), with one single particle state occupied (with one creation
operator applied on the vacuum state), with two single particle states occupied  (with
two creation operator applied on the vacuum state), and so on.

vii. The creation operators together with their Hermitian conjugated partners
annihilation operators fulfill, due to the oddness of the ”basis vectors”, while the momentum
part commutes, the anticommutation relations,
postulated by Dirac for second quantized fermion fields, not only when they apply on the
vacuum state, but also when they apply on the Hilbert space ${\cal H}$,

Eq. (<ref>) in the Clifford case and Eq. (<ref>) in the Grassmann

In the Clifford case this happens only after ”freezing out” half of the Clifford  space,

what brings besides the correct anticommutation
relations also  the ”family” quantum number to each irreducible representation of
the Lorentz group of the remaining internal space.
The oddness of the "basis vectors" is transfered  to the creation operators
forming the single fermion states and further  to the creation operators forming
the whole Hilbert space of the second quantized fermions in the Clifford case,
Sect. <ref>, Eq. (<ref>), and in the Grassmann case,
Sect. <ref>,  Eq. (<ref>).

viii. Correspondingly the creation and annihilation operators with the internal
space described by either odd Clifford or odd Grassmann algebra, since fulfilling
the anticommutation relations required for the second quantized fermions without
postulates, explain the Dirac's postulates for the second quantized fermions.

In Sect. <ref> we clarify the notation used in the second quantization of
fermions in literature and in our description of  the internal space of fermions when
using either the Grassmann or the Clifford algebras. The comparison of the
Dirac and our way of the second quantization of fermions is presented in
Sect. <ref>.

In Sect. <ref> we present in steps our use of the odd Grassmann
algebra and the odd Clifford algebras for description of the internal
space of fermions — their spins and charges in both algebras, and families in the
Clifford algebra,  in a tensor product with
the external space of coordinates or momenta,  relating our way of describing the
internal space of fermons in any $d$, Sect. <ref>, to the usual
way   using the group theory, Sect. <ref>.

Sect. <ref> presents the Grassmann odd algebra, that is
the algebra of odd products of coordinates $\theta^a$'s, determining with their
Hermitian conjugated partners $\frac{\partial}{\partial \theta_a}$'s
$(\theta^{a \dagger} = \eta^{a a} \frac{\partial}{\partial \theta_{a}}$,
$\eta^{a b}=diag\{1,-1,-1,\cdots,-1\}$) the internal space of  second
quantized "fermions" with the integer spin.  Corresponding creation operators
determine in $d=(3+1)$ spins and charges of "fermions" in adjoint representations.

In Sect. <ref> the two anticommuting Clifford odd
algebras, $\gamma^a$'s and $\tilde{\gamma}^a$'s, are discussed, each of them
are the superposition of $\theta^a$'s and $\frac{\partial}{\partial \theta_a}$'s
( $\gamma^{a}= (\theta^{a} + \frac{\partial}{\partial \theta_{a}})$, $\tilde{\gamma}^{a}= i \,(\theta^{a} -
\frac{\partial}{\partial \theta_{a}})$ [9, 66, 67]). They are, up to
$\eta^{aa}$, Hermitian operators.

Each of the two Clifford algebras describes half integer spins, manifesting in $d=(3+1)$
spins, charges and families in  the fundamental representations of the corresponding

In Sect. <ref> we explain the
reduction of the two Clifford algebras to only one, we make a choice of
$\gamma^a$'s, while $\tilde{\gamma}^a$'s determine the ”family charges”
to each irreducible representation of  the generators of the Lorentz algebra in the
remaining Clifford space  of $\gamma^a$'s and insure that the corresponding
creation and annihilation operators
have all the properties  of the second quantized fermion fields.

The reduction of the Clifford two algebras to only one reduces also the
Grassmann algebra space.

In Sect. <ref> we discuss the tensor products of the internal space of
fermions and the space of coordinates or momenta in the ordinary second quantization
procedure, as well as in the case when the internal space is described by the
Clifford algebra.

In Sect. <ref>   the ordinary second quantization procedure
is presented.

In Sect. <ref>  the second quantization with the
Clifford algebra is presented.

Sect. <ref> demonsrates the solutions of equations of motion
for free massless chargeless fermions in $d=(3+1)$ for the case
that the internal space is described by the Clifford algebra with $\gamma^a$'s.

Sect. <ref>  relates  our way of
second quantization and the usual, Dirac's, way of second quantization,
Refs. [24, 26].

In Sect. <ref> the discrete symmetry operators for fermions and  bosons
(vielbeins and spin connections) in $d=((d-1)+1)$-dimensional space and
as observed in $d=(3+1)$  as fermion fields, vector gauge fields and
scalars gauge fields in the spin-charge-family, Ref. [136], are presented.
App. <ref> adds more details about discrete symmetries of vector and
scalar gauge fields.

In Sect. <ref> the action and the corresponding equations of motion
for free massless  fermions in the standard model, Sect. <ref> and
in the spin-charge-family theory, Sect. <ref>, is discussed and
solutions expressed with the creation operators in $d=(5+1)$,
Sect. <ref>,  in $d=(9+1)$ and $d=(13+1)$ and in
Sect. <ref>, with families included.
The proposal for the action for free massless "Grassmann fermions" can be found in
App. <ref>, where also the solutions are presented, again expressed with
the creation operators.

In Sect. <ref> the Hilbert space determined by all possible numbers of
tensor products of all possible odd Clifford creation operators applying on the
vacuum state is presented and the anticommutation properties of the
application of creation and annihilation operators on the Hilbert space is demonstrated,
Sect. <ref>.

In Sect. <ref> the simple starting action for massless
fermions, the internal space of which is described by the odd Clifford algebra, and
which interact with the vielbeins and the two kinds of the spin connection fields only,
and for the gravitational fields is presented and discussed.

In Sect. <ref> the action for massless fermions as seen in
$d=(3+1)$ when fermions interact with the dynamical vectors gauge fields,
Sect. <ref>
and the scalar  gauge fields, which for particular scalar space indexes manifest properties
of Higgs's scalar, explaining the appearance of the scalar higgs and the Yukawa
couplings of the standard model, Sect. <ref>, while for the rest of the
scalar indexes the scalar triplets and antitriplet fields cause the transition of
antileptons into quarks and antiquarks into quarks, offering explanation for the
matter/antimatter asymmetry in the expanding universe and predict the proton decay.

In Sect. <ref> we compare the assumptions of the standard
model, Sect. <ref>, with in the literature proposed unifying theories,
Sect. <ref>, mostly $SO(10)$-unifying theories, Sect. <ref>,
and the spin-charge-family theory, Sect. <ref>, discussing besides the
assumptions also the predictions of these theories,
Sects. <ref>,
<ref>,  searching for the next step beyond
the standard model.

In Sect. <ref> the  toy model is discussed in $d=(5+1)$ of massless odd
Clifford fermions, interacting with the zweibein  and spin connection fields in $d=(5,6)$,
of particular properties, which cause the break of the $5+1$-dimensional space into
$3+1$ times an almost $S^2$ sphere as an introduction into the study of the
spin-charge-family theory case in $d=(13+1)$, which is under
consideration [
The mass of fermions is in this toy model determined either
with the dynamics in the $6^{th}$ and the $7^{th}$ dimension or, at low energies,
by the constant values of the spin connection fields, which break the mass protection.].

In Sect. <ref>  we shortly present other
possibilities for which we have our own understanding of the  topics.

In Sect. <ref> we mention possibilities that the realization and
predictions of the ıspin-charge-family theory could influence hadron physics, if at all.

In Sect. <ref> we shortly overview what one can learn from this
review article.

In App. <ref> the trial to find the action for the massless integer
spin "Grassmann fermions" is presented.

In  App. <ref> properties of the Clifford even commuting  "basis vectors" are
discussed, App. evenclifford, and the matrices for $\gamma^a$'s,
$\tilde{\gamma}^a$'s, $S^{ab}$, $\tilde{S}^{ab}$ are presented and discussed,
App. <ref>
In App. <ref> the triangle anomaly cancellation of the standard
model is explained in the ıspin-charge-family theory and in the $SO(10)$-unifying

Most of the rest of appendixes offer detail explanations or proofs needed in the
main text.

§ SECOND QUANTIZATION IN TEXT BOOKS AND RELATION TO OUR WAY OF SECOND QUANTIZATION

This section is to present in a short way how does the main assumption,

the decision to describe the internal space of fermions with
the ”basis vectors” expressed with the superposition of odd products of
the anticommuting members of the algebra, as discussed in details in
Sect. <ref>,  either the Clifford one or
the Grassmann one, acting algebraically, $*_{A}$, on the internal
vacuum state  $|\psi_{o}>$ in each of these algebras, relate to the creation
and annihilation anticommuting operators of the second quantized  fermion fields,
discussed in details in Sect. <ref>, Sect. <ref>,
Sect. <ref>.

Let us first  tell that the algebraic product $*_A$ is
usually not present in other works, and thus has no
well known physical meaning. It is at first a product by which one can
multiply two internal "basis vectors", describing internal space of
fields, $\hat{b}^{\dagger}_{i}$ and
$\hat{b}^{\dagger}_{j}$, with each other,

\begin{eqnarray}
\label{HNA}
\hat{c}^{\dagger}_k&=& \hat{b}^{\dagger}_i *_{A} \hat{b}^{\dagger}_j\,,\nonumber\\
\hat{b}^{\dagger}_i  *_{A} \hat{b}^{\dagger}_j&=& \mp \hat{b}^{\dagger}_j  *_{A}
\hat{b}^{\dagger}_i \,, %\nonumber
\end{eqnarray}

the sign $\mp$ depends on whether $\hat{b}^{\dagger}_i$ and $\hat{b}^{\dagger}_j$
are products of odd or even number of algebra elements: The sign is $-$ if both
are (superposition of) odd products of algebra elements, in all other cases
the sign is $+$.

Let ${\bf R}^{d-1}$ define the external spatial or momentum basis $|\vec{p}>$.
Then the tensor product $*_{T}$ extends the internal "basis vectors"
into the creation operator $ \hat{\bf b}^{\dagger}_{\vec{p} ,\,i}$ defined in both spaces,
Sect. <ref>, App. <ref>

\begin{eqnarray}
\label{HNT}
\hat{\bf b}^{\dagger}_{\vec{p}, \, i} =|\vec{p}>*_{T}\, \hat{b}^{\dagger}_{i}\,,%\nonumber
\end{eqnarray}

where again $\hat{b}^{\dagger}_i $ represent the superposition of products of elements
of the  anticommuting algebras, in our case either
$\theta^a$ or $\gamma^a$ or $\tilde{\gamma}^a$,
used in this paper.

We can make a choice of the orthogonal and normalized basis so that
$<\hat{\bf b}_{\vec{p}, i} | \hat{\bf b}^{\dagger}_{\vec{p'}, j} > =
\delta(\vec{p} - \vec{p'}) \,\delta_{ij}$. Let us point out that either
$\hat{b}^{\dagger}_{i}$ or $\hat{\bf b}^{\dagger}_{\vec{p}, \, i}$ apply algebraically
on the corresponding vacuum state, $\hat{b}^{\dagger}_i *_{A} |\psi_o>$
and $\hat{\bf b}_{\vec{p}, \, i} *_{A}

To give to  the algebraic product, $*_A$,  and to the tensor product, $*_{T}$,
defining the space of single particle wave functions depending on the internal space
and external space, the understandable meaning,
we postulate the connection  between the anticommuting/commuting properties
of the ”basis  vectors”, expressed with  the odd/even products of the anticommuting
algebra elements and the corresponding creation operators, creating second
quantized single fermion/boson states.

\begin{eqnarray}
\label{HNb}
\hat{\bf b}^{\dagger}_{\vec{p}, i} *_{A}|\psi_o> &=& |\psi_{\vec{p}, i}>\,,\nonumber\\
\hat{\bf b}^{\dagger}_{\vec{p}, i}  *_{T}\,|\psi_{\vec{p'}, j}> &=& 0\,,\nonumber\\
{\rm if \,} \vec{p} &=& \vec{p'}\, {\rm and} \, i=j\,,\nonumber\\
{\rm in\; all\;other\;cases\;} & &{\rm it\;follows\,}\nonumber\\
\hat{\bf b}^{\dagger}_{\vec{p}, i}  *_{T}\,
\hat{\bf b}^{\dagger}_{\vec{p'}, j} *_{A}|\psi_o>
&=& \mp \,\hat{\bf b}^{\dagger}_{\vec{p'},j}  *_{T} \,
\hat{\bf b}^{\dagger}_{\vec{p}, i}  *_{A}|\psi_o>\,,%\nonumber
% C_{\vec{p}, i} *_T \Psi\,.
% {\hat b}^{\dagger}(C_{\vec{p}}, i) \Psi  &=& C_{\vec{p}, i} *_T \Psi\,.
\end{eqnarray}

with the sign $\pm $ depending on whether  $\hat{\bf b}^{\dagger}_{\vec{p}, i}$
have both an odd character, the sign is $-$, if this is not the case the sign is $+$.

Not necessarily both, odd and even products of anticommuting algebra elements
have any physical meaning (any realization in the observed phenomena), if
any at all. It will be shown that in Grassmann algebra both, odd and even products
of anticommuting algebra elements exist, behaving as either anticommuting or
commuting  creation and annihilation operators, respectively,  Sect. <ref>,

while in the Clifford algebra
anticommuting odd products of anticommuting elements of the Clifford algebra
exist with the Hermitian conjugated partners annihilation operators belonging to
the independent group, Sect. <ref> [26].
The Clifford even products of elements commute and have their Hermitian conjugated
partners within the same group, the paper is in preparation.

To each creation operator $\hat{\bf b}^{\dagger}_{\vec{p}, i}$ its Hermitian
conjugated partner  represents the annihilation operator $\hat{\bf b}_{\vec{p}, i}$

\begin{eqnarray}
\label{HNC}
\hat{\bf b}_{\vec{p}, i} &=&
(\hat{\bf b}^{\dagger}_{\vec{p}, i})^{\dagger}\,,
\nonumber\\
{\rm with \; the}&&{\rm property}\nonumber\\
\hat{\bf b}_{\vec{p}, i}\, *_{A}\,|\psi_o> &=&0\,,\nonumber\\
{\rm defining\; the\; } && {\rm vacuum \; state\; as\;}\nonumber\\
|\psi_o> :&=&\sum_i \,{\hat b}_{i}\, *_{A} \,{\hat b}^{\dagger}_{i}
\,|\;I>|0_{\vec{p}}>\,,%\nonumber
\end{eqnarray}

where summation $i$ runs over all different products of annihilation operator
$\times$ its Hermitian conjugated  creation operator, no matter what
$\vec{p}$ is chosen, and $|\;I>\,|0_{\vec{p}}>$ represents the identity with the
starting momentum state $|0_{\vec{p}}>$, from where all the momentum
states $|\vec{p}>$ follow by translation in momentum space, ${\hat b}_{i}$ represents
the Hermitian conjugated annihilation operator to  $({\hat b}^{\dagger}_{i})^{\dagger}$,
Sect. <ref>.

Let the tensor multiplication $*_{T_{H}}$ denotes   the multiplication of any number of
any possible single particle states, and correspondingly of any number of any possible
creation operators, Sect. <ref>.

What further means that to each single particle wave function we define
the creation operator $\hat{\bf b}^{\dagger}_{\vec{p}, i} $, applying
in a tensor product from the left hand side on the second quantized
Hilbert space — consisting of all possible products
of any number of  single particle wave functions — adding to the Hilbert
space the single particle wave function created by this particular creation
operator. In the case of the second quantized fermions, if this particular
wave function with the quantum numbers ${}_{i}$ and momentum $\vec{p}$  of
$\hat{\bf b}^{\dagger}_{\vec{p}, i} $ is already
among the single fermion wave functions of a particular product of the
fermion wave functions, the action of the creation operator gives zero,
otherwise the number of the fermion wave functions increases for one,
Sect. <ref>.
(In the boson case the number of boson second quantized wave functions
increases always  for one.)

If we apply with the annihilation operator $\hat{\bf b}_{\vec{p}, i}$
on the second quantized Hilbert space, then the application gives a
nonzero contribution only if  the particular products of the single particle
wave functions do include the wave function with the quantum number
$i$ and $\vec{p}$, Sect. <ref>.

In a Slater determinant formalism the single particle wave functions define
the empty or occupied places of any of infinite numbers of Slater determinants.
The creation operator $\hat{\bf b}^{\dagger}_{\vec{p}, i} $ applies
on a particular Slater determinant from the left hand side. Jumping over
occupied states to the place with its $i$ and $\vec{p}$. If this state is
occupied, the application gives in the fermion case zero. (In the boson case
the number of particles increase for one.)
The particular Slater determinant changes sign in the
fermion case if
$\hat{\bf b}^{\dagger}_{\vec{p}, i}$ jumps over odd numbers of occupied
states. (In the boson case the sign of the Slater determinant does not change.)

When annihilation operator $\hat{\bf b}_{\vec{p}, i} $ applies on
particular Slater determinant, is jumping over occupied states to its own place,
giving zero, if this place is empty and decreasing the number of occupied states
if this place is occupied. The Slater determinant changes sign in the fermion
case, if the number of occupied states before its own place is odd. (In the boson
case the sign does not change.)

Let us stress again that choosing antisymmetry or symmetry of products of
the algebra elements is possible only in the case of the Grassmann algebra —
odd algebra describes second quantized integer spin "fermions",
Sect. <ref>, while the commuting Grassmann even algebra can be
used to describe second quantized integer spin bosons.

In the Clifford case Clifford odd choice, describing the
half integer fermions, is the only choice, since the Clifford even algebra does not
offer the description of the second quantized boson states, as explained in
App. <ref>.

To describe the second quantized fermion states we use the "basis vectors", which are
the superposition of the odd numbers of algebra elements in both algebras, in the two
Clifford algebras and in the Grassmann algebra.

The creation operators and their Hermitian conjugated partners annihilation
operators therefore in the fermion case anticommute. The single fermion
states, which are the application of the creation operators on the vacuum
state $|\psi_o>$, manifest correspondingly as well the oddness.
The vacuum state, defined as the sum over all different products of
annihilation $\times$ the corresponding creation operators, have an even

One usually means antisymmetry when talking about Slater-determinants
because otherwise one would not get determinants.

In the present paper [8, 9, 13, 66] the choice of
the symmetrizing versus antisymmetrizing  relates indeed the commutation versus
anticommutation with respect to the a priori completely different product $*_A$,
of anticommuting members of the Clifford or Grassmann algebra. The oddness or
evenness  of these products transfer to quantities to which these algebras

Let us add up with the recognition: Oddness and eveness of the "basis vectors" to
describe "fermions" and bosons is meaningful
only in the case of Grassmann algebra, while the  Clifford algebra offers only the
description of fermions, offering at the same time family quantum numbers for
different irreducible representations of the corresponding Lorentz algebra. This is
uniqum of the Clifford algebra that it offers the unification of spins, charges and
families. The approaches with the group theory description of the
internal space of fermions do not offer this kind of unification.

§ INTERNAL AND EXTERNAL SPACE OF FERMIONS

Single fermion states are functions of  external coordinates
and of internal space of fermions.

The basis for single fermion states are usually chosen to have some symmetries in both spaces,  external and internal.

If $M^{ab}$  denote infinitesimal generators of the Lorentz algebra in both spaces,
$M^{ab}= L^{ab} + S^{ab}$, with $L^{ab}= x^a p^b - x^b p^a$, $p^a=
i \frac{\partial}{\partial x_a}$, determining operators in ordinary space, while
$S^{ab}$ are equivalent operators in internal space of fermions [
We here mainly think of the algebra and do not for the moment go into
the problem that if we take these  operators to be represented on
the space of states for a particle then  $x^0$ is not the operator,
it is instead a moment of time, what we do have in mind.]

\begin{eqnarray}
\label{Lorentzcom}
\{M^{ab}, M^{cd}\}_{-} &=&
i \{M^{ad} \eta^{bc} + M^{bc} \eta^{ad} -  M^{ac} \eta^{bd}- \bf M^{bd} \eta^{ac}\}\,,\nonumber\\
\{M^{ab}, p^{c}\}_{-} &=&  -i  \eta^{ac}p^b + i \eta^{cb}p^a\,, \nonumber\\
\{M^{ab}, S^{cd}\}_{-} &=&i \{S^{ad} \eta^{bc} + S^{ad} -  S^{ac} \eta^{bd}- \bf S^{bd} \eta^{ac}\}\,, ,,
% M^{ab \dagger} &=& \eta^{aa} \eta^{bb} {\cal {\bf M}}^{ab}\,.
\end{eqnarray}

with $L^{ab\,\dagger}=L^{ab}$, provided that $p^{a \dagger} =p^a$, since they
apply on states which are zero on the boundary [7], while $S^{ab \dagger}
=\eta^{aa}\eta^{bb} S^{ab}$.
Commutation relation  of Eq. (<ref>) are valid for either $L^{ab}$ or
$S^{ab}$ in any dimension of space-time, $\{L^{ab}, S^{cd}\}_{-}=0 $.

We use the metric tensor $\eta^{ab}=diag(1,-1,-1,\dots,-1,-1) $ for

The  usual choice of the Cartan subalgebra of the commuting operators of the
Lorentz algebra is

\begin{eqnarray}
M^{03}, M^{12}, M^{56}, \dots, M^{d-1\, d}\,,
\label{cartanM}
\end{eqnarray}

defining correspondingly the members of the irreducible representations of the
Lorentz group to be the eigenvectors of all  the members of the Cartan subalgebra
presented in Eqs. (<ref>, <ref>).

In the literature the single fermion states usually do not anticommute.
Dirac postulated anticommuting properties for fermion fields only in the second
quantization, by postulating anticomutation rules for creation and annihilation operators,
which operating on the vacuum state generate the second quantized states.
Although all the elementary fermion constituents, quarks and leptons,
do carry besides spins also charges and appear in families,  and although
charges and families are treated as internal degrees of freedom of fermions, yet
internal space of fermions is not built from spins, charges and families in equivalent
way, unifying spins, charges and families.

One finds in the literature [27, 28] several trials to unify all the charges,
but not  the spin with the charges. We comment in App. <ref> that the
appearance of spins and charges in one irreducible representation of the Lorentz
group explains the triangle anomalies cancellation in the standard model better
than the unification of only charges, since in this first case the handedness and
charges are already related, this is happening in the spin-charge-family theory,
while in the unification of only  charges the relation between spins and charges must
be postulated like in the standard model.

We show in what follows that anticommuting properties of spinor states, brought into
the theory by using the anticommuting "basis vectors", bring simplicity, transparency
and elegance in the theoretical description of fermion states, if spins and  charges
appear in the same irreducible representation of the group.

Let us point out that in the literature the anticommuting properties of the second
quantized fermion fields are just postulated. In the ”Grassmann fermions” and the two
”Clifford fermions” the ”basis vectors” anticommute due to the oddness of the ”basis
vectors”. The Hermitian conjugated partners of the ”basis vectors” belong to independent
representations, while in the usual second quantized fields they are just postulated to

We shall use in what follows two kinds of anticommuting algebras, the Grassmann
algebra and  the two anticommuting kinds of the Clifford algebras, denoting them by
$\theta^{a}$'s and $p^{\theta a}$'s $= \frac{\partial}{\partial \theta_a}$'s, for the
Grassmann algebra objects, and $\gamma^a$'s  and $\tilde{\gamma}^{a}$'s
for the two kinds of the  Clifford algebra objects.

The existence of the two kinds of the Clifford algebras is our recognition [8, 9, 11, 66, 67].

We denote the corresponding infinitesimal Lorentz generators as ${\cal {\bf S}}^{ab}$
for the Grassmann algebra, and as $S^{ab}$ and $\tilde{S}^{ab}$ for the two kinds
of the Clifford algebras.

\begin{eqnarray}
{\cal {\bf S}}^{ab}= i \, (\theta^{a} \frac{\partial}{\partial \theta_{b}} - \theta^{b}
\frac{\partial}{\partial \theta_{a}})\,, \quad
&& S^{ab} = \frac{i}{4}(\gamma^a \gamma^b - \gamma^b \gamma^a)\,,\quad
\tilde{S}^{ab} =\frac{i}{4}(\tilde{\gamma}^a \tilde{\gamma}^b - \tilde{\gamma}^b
\tilde{\gamma}^a)\, , \nonumber\\
{\cal {\bf S}}^{ab} =S^{ab} + \tilde{S}^{ab}\,, \quad&&
\{S^{ab}, \tilde{S}^{ab}\}_{-}=0\,,\nonumber\\
\{{\cal {\bf S}}^{ab}, \, \theta^e\}_{-} = - i \,(\eta^{ae} \,\theta^b - \eta^{be}\,\theta^a)\,, \quad
&&\{{\cal {\bf S}}^{ab}, \, p^{\theta e}\}_{-} = - i \,(\eta^{ae} \,p^{\theta b} -
\eta^{be}\,p^{\theta a})\,,
\nonumber\\
\{ S^{ab}, \gamma^c\}_{-}&=& i (\eta^{bc} \gamma^a - \eta^{ac} \gamma^b)\,, \nonumber\\%\quad
\{ \tilde{S}^{ab}, \tilde{\gamma}^c \}_{-}&=& i (\eta^{bc} \tilde{\gamma}^a -
\eta^{ac }\tilde{\gamma}^b)\,,\nonumber\\
\{ S^{ab}, \tilde{\gamma}^c\}_{-}&=&0\,,\quad \{\tilde{S}^{ab}, \gamma^c\}_{-}=0\,,
\label{sabtildesab}
\end{eqnarray}

the proof for the relation ${\cal {\bf S}}^{ab} =S^{ab} + \tilde{S}^{ab}$ is
presented in App. <ref>, Statement 2..

It then follows for the  Cartan subalgebra of the commuting operators of
the Lorentz algebra for each of the two kinds of the operators of the Clifford algebra,
$S^{ab}$ and $\tilde{S}^{ab}$ and for the Grassmann algebra ${\cal {\bf S}}^{ab}$

\begin{eqnarray}
{\cal {\bf S}}^{03}, {\cal {\bf S}}^{12}, {\cal {\bf S}}^{56}, \cdots,
{\cal {\bf S}}^{d-1 \;d}\,, \nonumber\\
S^{03}, S^{12}, S^{56}, \cdots, S^{d-1 \;d}\,, \nonumber\\
\tilde{S}^{03}, \tilde{S}^{12}, \tilde{S}^{56}, \cdots,  \tilde{S}^{d-1\; d}\,.
\label{cartangrasscliff}
\end{eqnarray}

We shall see that the Grassmann algebra  is uniquely expressible with the two Clifford
algebras  and opposite.

§.§ Internal space of fermions  in text books

Fermion states can be represented as a superposition of tensor products of basis in
ordinary (coordinate or momentum) space and in internal space. Fermions carry in
general as a part of the internal degrees of freedom besides spins also charges.

The spinor part of the internal space of massless fermions (without charges) can be
represented by the spinor representation of the Lorentz group $S^{mn}$, $(m,n)
=(0,1,2,3)$. Making the usual choice of the Cartan subalgebra of the Lorentz group,
Eq. (<ref>),

\begin{eqnarray}
\label{cartaninternal}
S^{03}, S^{12}\,,
\end{eqnarray}

one usually  choses  internal part of states to be the eigenstates of
the spin, $S^{12}$, and the handedness,
$\Gamma^{(3+1)}=-\frac{i}{3!} \varepsilon_{mnop} S^{mn}S^{op}$, which is
in the case of the Cartan subalgebra of Eq. (<ref>) equal to
$\Gamma^{(3+1)}=i (-2i)^2 S^{03 } S^{12 }$.

There are two possibilities for spin, $S^{12}=\pm \frac{1}{2}$, and two for
$\Gamma^{(3+1)}=\pm 1$, for right and left handedness, respectively.
Due to the relation $\Gamma^{(3+1)}=i (-2i)^2 S^{03 } S^{12 }$ the spinor
part can be represented as well as ”eigenstates” of the Cartan subalgebra members.

These four spinor states, which are the eigenstates of the Cartan subalgebra,
can be  written as presented in Table (<ref>).

[t]16.5 cm
Four spinor states in $d=(3+1)$,
two left handed (chiral) and two right handed (chiral), each with spin
$\pm \frac{1}{2}$,  are the eigenstates of the Cartan subalgebra of
the Lorentz algebra in the internal space of fermions.

$ h =\Gamma^{(3+1)}
$   $\psi_{S^{03}, S^{12}}$
$S^{03}$   $ S^{1 2}$   $\Gamma^{3+1}$

$R$   $\psi_{+\frac{1}{2}, +\frac{1}{2}}$    $+\frac{1}{2}$   $+\frac{1}{2}$   $+1$

$R$   $\psi_{-\frac{1}{2},- \frac{1}{2}}$    $-\frac{1}{2}$   $-\frac{1}{2}$   $+1$

$L$   $\psi_{-\frac{1}{2}, +\frac{1}{2}}$    $-\frac{1}{2}$   $\frac{1}{2}$   $-1$

$L$   $\psi_{+\frac{1}{2},- \frac{1}{2}}$    $\frac{1}{2}$   $-\frac{1}{2}$   $-1$

These two representations, the right and  the left one, can be written each
in terms of Pauli spin matrices $\vec{\sigma}$ and the unit matrix $\sigma^0$.

\begin{equation}\label{pauli}
\sigma^1=
\left(  \begin{array}{cc}
0 & 1 \\
1 & 0
\end{array} \right)\, , \qquad
\sigma^2=
\left(  \begin{array}{cc}
0 & -i\\
i & 0
\end{array} \right)\, , \qquad
\sigma^3=
\left(  \begin{array}{cc}
1 & 0\\
0 & -1
\end{array} \right)\, ,  \qquad
\sigma^0=
\left(  \begin{array}{cc}
1 & 0\\
0 & 1
\end{array} \right)\,.
\end{equation}

Let us write down the $4 \times 4$ matrix of handdedness/chirality,
$\Gamma^{(3+1)}$, and the corresponding  matrices  of the infinitesimal
generators   of the Lorentz transformations in   internal space of fermions, $\vec{S}$
($\vec{S}=\frac{1}{2}  \varepsilon_{ijk} S^{jk}$) and   $\vec{K}$ ($K^{i}= S^{0i}$),
defined in the space of four spinor states,

\begin{equation}\label{sigma4x4}
\mathbf{\Gamma^{(3+1)}}_=
\left(  \begin{array}{cc}
\mathbf{1} & 0 \\
0 &-  \mathbf{1}
\end{array} \right)\, ,\qquad
\vec{S}=
\frac{1}{2} \left(  \begin{array}{cc}
\vec{\sigma} & 0 \\
0 &   \vec{\sigma}
\end{array} \right)\, ,\qquad
\vec{K}=
\frac{i}{2} \left(  \begin{array}{cc}
\vec{\sigma} & 0 \\
0 &  - \vec{\sigma}
\end{array} \right)\, .%\qquad
\end{equation}

Each of the two  ”basis states” represents two decoupled irreducible representations
of the Lorentz group. They only can be connected when introducing Dirac matrices
$(\gamma^0, \gamma^1,\gamma^2,\gamma^3)$ with properties

\begin{equation}\label{gamma4x4}
\mathbf{\gamma^{0}}_=
\left(  \begin{array}{cc}
0 & \sigma^0 \\
\sigma^0& 0
\end{array} \right)\, ,\qquad
\mathbf{\gamma^{1}}_=
\left(  \begin{array}{cc}
0 & \sigma^1 \\
- \sigma^1& 0
\end{array} \right)\, ,\qquad
\mathbf{\gamma^{2}}_=
\left(  \begin{array}{cc}
0 & -\sigma^2 \\
\sigma^2& 0
\end{array} \right)\, ,\qquad
\mathbf{\gamma^{3}}_=
\left(  \begin{array}{cc}
0 & \sigma^3 \\
-\sigma^3& 0
\end{array} \right)\, .
\end{equation}

The eigenstates of the Cartan subalgebra, Table <ref>, are in the literature
not assumed to anticommute, they are rather assumed to be commuting objects.  In Ref. ([44], page 17)  P. Ramond mentions
the possibility to give to spinors the Grassmann  anticommuting character.

We point out in this paper that the internal space of fermions do have anticommuting
character, either with the integer spins, if represented with the odd Grassmann algebra,
or with the half integer spins, if represented with any of the two kinds of the odd
Clifford algebras. All the ”basis vectors” in internal space, either described by
the Grassmann algebra or in any of the two Clifford algebras, are taken to be
superposition of products of an odd number of algebra elements and to be at the same
time eigenstates of all the Cartan subalgebra members, to be comparable with the
usual choice of the spin representation.

We show in App. <ref> that using any of the two odd Clifford
algebras to describe the internal space of fermions,  offers  several irreducible
representations. However,  the left and the right handed representation in $d=(3+1)$
appear in the same irreducible representation (what means that  both
have the odd handedness)  provided that $d>(3+1)$.
In the case of higher dimensions than $d=(3+1)$ not only the spin and
handedness but also all the charges contribute to the internal space of fermions,
determining the anticommuting ”basis vectors”.

In $d=(13 +1)$, for example, in one irreducible representation of the Lorentz group
there are all the quarks and all the leptons and the antiquarks and the antileptons
with all the properties postulated by the standard model for quarks and
antiquarks and for leptons and antileptons, all carrying the required charges,
handedness and spins.

To the oddness of each member of the irreducible representation the spin part
and the charge part contribute.
All  the irreducible representations are equivalent with respect to the Lorentz group.

§.§ Internal space of fermions described by Grassmann algebra and two kinds
of Clifford algebras

We shall pay attention only to even dimensional spaces: $d=2(2n+1)$ and $d=4n$,
$n$ is a non negative integer.

In Grassmann $d$-dimensional space there are $d$ anticommuting operators
$\theta^{a}$, $\{\theta^{a}, \theta^{b}\}_{+}=0$, $a=(0,1,2,3,5,..,d)$, and
$d$ anticommuting derivatives with respect to $\theta^{a}$,
$\frac{\partial}{\partial \theta_{a}}$, $\{\frac{\partial}{\partial \theta_{a}},
\frac{\partial}{\partial \theta_{b}}\}_{+} =0$, offering
together $2\cdot2^d$ operators, the half of which are superposition of products of
$\theta^{a}$ and another half corresponding superposition of
$\frac{\partial}{\partial \theta_{a}}$.

\begin{eqnarray}
\label{thetaderanti0}
\{\theta^{a}, \theta^{b}\}_{+}=0\,, \, && \,
\{\frac{\partial}{\partial \theta_{a}}, \frac{\partial}{\partial \theta_{b}}\}_{+} =0\,,
\nonumber\\
\{\theta_{a},\frac{\partial}{\partial \theta_{b}}\}_{+} &=&\delta_{ab}\,, (a,b)=(0,1,2,3,5,\cdots,d)\,.
\end{eqnarray}

We define [22]

\begin{eqnarray}
(\theta^{a})^{\dagger} &=& \eta^{a a} \frac{\partial}{\partial \theta_{a}}\,,\nonumber\\
{\rm leading  \, to}\nonumber\\
(\frac{\partial}{\partial \theta_{a}})^{\dagger}&=& \eta^{a a} \theta^{a}\,,
\label{thetaderher0}
\end{eqnarray}

the identity is the self adjoint member. We make a choice for the complex properties
of $\theta^a$, and correspondingly
of $\frac{\partial}{\partial \theta_{a}}$, as follows

\begin{eqnarray}
\label{complextheta}
\{\theta^a\}^* &=&  (\theta^0, \theta^1, - \theta^2, \theta^3, - \theta^5,
\theta^6,...,- \theta^{d-1}, \theta^d)\,, \nonumber\\
\{\frac{\partial}{\partial \theta_{a}}\}^* &=& (\frac{\partial}{\partial \theta_{0}},
\frac{\partial}{\partial \theta_{1}}, - \frac{\partial}{\partial \theta_{2}},
\frac{\partial}{\partial \theta_{3}}, - \frac{\partial}{\partial \theta_{5}},
\frac{\partial}{\partial \theta_{6}},..., - \frac{\partial}{\partial \theta_{d-1}},
\frac{\partial}{\partial \theta_{d}})\,.
\end{eqnarray}

In $d$-dimensional space of anticommuting Grassmann
coordinates and of their Hermitian conjugated partners derivatives,
Eqs. (<ref>, <ref>),
there exist two kinds of the Clifford coordinates (operators) —
$\gamma^{a}$ and $\tilde{\gamma}^{a}$ — both are expressible in terms of
$\theta^{a}$ and their conjugate momenta $p^{\theta a}= i \,
\frac{\partial}{\partial \theta_{a}}$  [9].

\begin{eqnarray}
\label{clifftheta}
\gamma^{a} &=& (\theta^{a} + \frac{\partial}{\partial \theta_{a}})\,, \quad
\tilde{\gamma}^{a} =i \,(\theta^{a} - \frac{\partial}{\partial \theta_{a}})\,,\nonumber\\
\theta^{a} &=&\frac{1}{2} \,(\gamma^{a} - i \tilde{\gamma}^{a})\,, \quad
\frac{\partial}{\partial \theta_{a}}= \frac{1}{2} \,(\gamma^{a} + i \tilde{\gamma}^{a})\,,
%(\gamma^{a})^{\dagger} &=&\eta^{aa}\, \gamma^{a}\,, \quad
%(\tilde{\gamma}^{a})^{\dagger} = \eta^{aa} \,\tilde{\gamma}^{a}\,,
\end{eqnarray}

offering together  $2\cdot 2^d$  operators: $2^d$ of those which are products of
$\gamma^{a}$  and  $2^d$ of those which are products of $\tilde{\gamma}^{a}$.

Taking into account Eqs. (<ref>, <ref>) ($\{\theta^{a}, \theta^{b}\}_{+}=0$,
$\{\frac{\partial}{\partial \theta_{a}}, \frac{\partial}{\partial \theta_{b}}\}_{+} =0$,
$\{\theta_{a},\frac{\partial}{\partial \theta_{b}}\}_{+} = \delta_{ab}$,
$\theta^{a \dagger}
=\eta^{aa}\, \frac{\partial}{\partial \theta_{a}} $ and
$(\frac{\partial}{\partial \theta_{a}})^{\dagger}=\eta^{aa} \theta^{a}$) one finds

\begin{eqnarray}
\label{gammatildeantiher0}
\{\gamma^{a}, \gamma^{b}\}_{+}&=&2 \eta^{a b}= \{\tilde{\gamma}^{a},
\tilde{\gamma}^{b}\}_{+}\,, \nonumber\\
\{\gamma^{a}, \tilde{\gamma}^{b}\}_{+}&=&0\,,\quad
(a,b)=(0,1,2,3,5,\cdots,d)\,, \nonumber\\
(\gamma^{a})^{\dagger} &=& \eta^{aa}\, \gamma^{a}\, , \quad
(\tilde{\gamma}^{a})^{\dagger} =  \eta^{a a}\, \tilde{\gamma}^{a}\,,\nonumber\\
% S^{ab} &=& \frac{i}{2}\, (\gamma^a \gamma^b - \gamma^b \gamma^a)\,, \quad
% \tilde{S}^{ab} = \frac{i}{2}\, (\tilde{\gamma}^a \tilde{\gamma}^b - \tilde{\gamma}^b
% \tilde{\gamma}^a)\,,
\gamma^a \gamma^a &=& \eta^{aa}\,, \quad
\gamma^a (\gamma^a)^{\dagger} =I\,,\quad
\tilde{\gamma}^a  \tilde{\gamma}^a = \eta^{aa} \,,\quad
\tilde{\gamma}^a  (\tilde{\gamma}^a)^{\dagger} =I\,,
\end{eqnarray}

where $I$ represents the unit operator.

Making a choice for the $\theta^a$ properties as presented in
Eq. (<ref>), it follows for the Clifford objects

\begin{eqnarray}
\label{complexgamatilde}
\{\gamma^a\}^* &=&  (\gamma^0, \gamma^1, - \gamma^2, \gamma^3, - \gamma^5,
\gamma^6,...,- \gamma^{d-1}, \gamma^d)\,, \nonumber\\
\{\tilde{\gamma}^a\}^* &=&  (- \tilde{\gamma}^0, - \tilde{\gamma}^1, \tilde{\gamma}^2,
- \tilde{\gamma}^3,  \tilde{\gamma}^5,- \tilde{\gamma}^6,..., \tilde{\gamma}^{d-1},
- \tilde{\gamma}^d)\,,
\end{eqnarray}

The Cartan subalgebra of the Lorentz algebra is presented in
Eqs. (<ref>, <ref>).

§.§.§ Grassmann algebra and ”basis vectors” describing internal space

It appears useful to arrange $2^d$ products of $\theta^{a}$ into  irreducible
representations with
respect to the Lorentz group with the generators [9],
Eq. (<ref>),

\begin{eqnarray}
{\cal {\bf S}}^{a b} &=& i \, (\theta^{a} \frac{\partial}{\partial \theta_{b}} - \theta^{b}
\frac{\partial}{\partial \theta_{a}})\,, \quad  ({\cal {\bf S}}^{a b})^{\dagger} = \eta^{a a}
\eta^{b b} {\cal {\bf S}}^{a b}\, .
\label{thetasab}
\end{eqnarray}

$2^{d-1}$ members of the representations  have an odd Grassmann character (those which are
superposition of  odd products of $\theta^{a}$'s).  All the members of any particular odd
irreducible representation follow from any starting member by the application of
${\cal {\bf S}}^{a b}$'s.

If we exclude the self adjoint identity there is $(2^{d-1}-1)$ members of an even Grassmann
character, they are even products of $\theta^{a}$'s. All the members of any particular even
representation follow from any starting member by the application of ${\cal {\bf S}}^{a b}$'s.

The Hermitian conjugated $2^{d-1}$ odd partners of  odd representations of $\theta^a$'s
and $(2^{d-1}-1)$  even partners of even representations of $\theta^a$'s are reachable from
odd and even representations, respectively, by the application of Eq. (<ref>).

It  appears useful to make the choice of the Cartan subalgebra of the commuting operators of the Lorentz algebra,

${\cal {\bf S}}^{03}, {\cal {\bf S}}^{12}, {\cal {\bf S}}^{56}, \cdots,
{\cal {\bf S}}^{d-1 \;d}$ , Eq. (<ref>),

and choose the members of the irreducible representations of the Lorentz group to be the
eigenvectors of all  the members of the Cartan subalgebra of Eq. (<ref>),
${\cal {\bf S}}^{a b} = i \, (\theta^{a} \frac{\partial}{\partial \theta_{b}} - \theta^{b}
\frac{\partial}{\partial \theta_{a}})$

\begin{eqnarray}
{\cal {\bf S}}^{ab} \,\frac{1}{\sqrt{2}}\, (\theta^a + \frac{\eta^{aa}}{i k} \theta^b) &=&
k\,\frac{1}{\sqrt{2}} (\theta^a + \frac{\eta^{aa}}{ik} \theta^b) \,, \nonumber\\
{\cal {\bf S}}^{ab} \,\frac{1}{\sqrt{2}}\, (1+ \frac{i}{k}  \theta^a \theta^b) &=&0\,,
\nonumber\\ &{\rm or} &\;  \nonumber\\
{\cal {\bf S}}^{ab} \,\frac{1}{\sqrt{2}}\,  \frac{i}{k}  \theta^a \theta^b &=&0\,,
\label{eigengrasscartan}
\end{eqnarray}

with $k^2=\eta^{aa}\eta^{bb}$.  The  eigenvector $\frac{1}{\sqrt{2}}\,
(\theta^0 \mp  \theta^3)$ of   ${\cal {\bf S}}^{03}$ has the eigenvalue $k=\pm i$,
the eigenvalues of all the other eigenvectors of the rest of the Cartan subalgebra
members, Eq. (<ref>), are  $k=\pm 1$.
In App. <ref>, Statement 2a. the proof for Eq. (<ref>) is presented.

We choose the "basis vectors" to be products of an odd number of nilpotents
$\frac{1}{\sqrt{2}}\, (\theta^a + \frac{\eta^{aa}}{i k} \theta^b)$ and the
rest of the even objects $  \frac{i}{k}  \theta^a \theta^b $, with eigenvalues
$k=\pm i$  and $0$, respectively, so that all the Cartan subalgebra eigenvectors
appear in the "basis vestor".

Let us check how does ${\cal {\bf S}}^{ac}= i (\theta^a \frac{\partial}{\partial \theta_c} -
\theta^c \frac{\partial}{\partial \theta_a})$ transform the product of two "nilpotents"
$\frac{1}{\sqrt{2}}\, (\theta^a + \frac{\eta^{aa}}{i k} \theta^b)$ and
$\frac{1}{\sqrt{2}}\, (\theta^c + \frac{\eta^{cc}}{i k'} \theta^d)$.

One finds that ${\cal {\bf S}}^{ac}$
$\frac{1}{\sqrt{2}}\, (\theta^a + \frac{\eta^{aa}}{i k} \theta^b)$
$\frac{1}{\sqrt{2}}\, (\theta^c + \frac{\eta^{cc}}{i k'} \theta^d)$ $ =
- \frac{\eta^{aa} \eta^{cc}}{2k}$ $(\theta^a \theta^b  +
\frac{k}{k'} \theta^c \theta^d)$.
${\cal {\bf S}}^{ac}$ transforms the product of two Grassmann odd
eigenvectors of the Cartan
subalgebra into the superposition of two Grassmann even eigenvectors.

"Basis vectors" have an odd or an even Grassmann character, if their products contain
an odd  or an even number of "nilpotents", $\frac{1}{\sqrt{2}}\, (\theta^a +
\frac{\eta^{aa}}{i k} \theta^b)$, respectively.
"Basis vectors" are normalized, up to a phase, in accordance with Eq. (<ref>) of App. <ref>.

The Hermitian conjugated representations of (either an odd or an even) products of $\theta^a$'s
can be obtained by taking into account  Eq. (<ref>) for each "nilpotent"

\begin{eqnarray}
\frac{1}{\sqrt{2}} (\theta^a + \frac{\eta^{aa}}{i k} \theta^b)^{\dagger} &=&
\eta^{aa}\,\frac{1}{\sqrt{2}} (\frac{\partial}{\partial \theta_{a}} + \frac{\eta^{aa}}{- i k}
\frac{\partial}{\partial \theta_{b}}) \,, \nonumber\\
(\frac{i}{k}\, \theta^a \theta^b)^{\dagger} &=&
\frac{i}{k} \,\frac{\partial}{\partial \theta_{a}}\,\frac{\partial}{\partial \theta_{b}}\,.
\label{grasscartanher}
\end{eqnarray}

Making a choice of the identity for the vacuum state,

\begin{eqnarray}
\label{vactheta}
|\phi_{og}> &=& |\,1>\,,
\end{eqnarray}

we see that algebraic products — we shall use  a dot, ${\,}\cdot{\,}$, or without  a dot
for an algebraic product of eigenstates of the Cartan subalgebra forming ”basis vectors”
and $*_{A}$ for the algebraic product of  ”basis vectors”  — of different $\theta^a$'s,
if applied on such a vacuum state, give always nonzero contributions,

$$ (\theta^0 \mp \theta^3) \cdot (\theta^1 \pm i \theta^2)
\cdots  (\theta^{d-1} \pm i \theta^d) |\,1>\ne {\rm zero},$$

(this is true also, if we substitute any of  nilpotents
$ \frac{1}{\sqrt{2}} (\theta^a + \frac{\eta^{aa}}{i k} \theta^b)$ or all of them
with the corresponding  even operators $(\frac{i}{k}\, \theta^a \theta^b)$; in the case
of odd Grassmann irreducible representations at least one nilpotent must remain in the product).
The Hermitian conjugated partners, Eq. (<ref>), applied on $|\,1>$, give
always zero

$$ (\frac{\partial}{\partial \theta_{0}} \pm
\frac{\partial}{\partial \theta_{3}}) \cdot (\frac{\partial}{\partial \theta_{1}} \mp i
\frac{\partial}{\partial \theta_{2}})\cdots ( \frac{\partial}{\partial \theta_{d-1}} \mp i
\frac{\partial}{\partial \theta_{d}}) |\,1> = 0.$$

Let us notice the properties of the odd products of $\theta^a$'s and of their Hermitian
conjugated partners:

i.  Superposition of
products of different  $\theta^a$'s,  applied on the vacuum state
$|\,1>$, give nonzero contribution.  To create on the vacuum state the  ”fermion” states
we make a choice of the ”basis vectors” of the odd number of $\theta^a$'s, arranging them
to be the eigenvectors of all the Cartan subalgebra elements, Eq. (<ref>).

ii.  The Hermitian conjugated partners of the “basis vectors”, they are
products of derivatives $\frac{\partial}{\partial \theta_{a}}$'s,  give, when applied
on the vacuum state  $|\,1>$, Eq. (<ref>),  zero. Each annihilation operator
annihilates the corresponding creation operator, due to :
$\frac{\partial}{\partial \theta_{a}}\theta^b = \eta^{ab}$.

iii. The algebraic product, $*_{A}$, of  a “basis vector” by itself gives zero,

the algebraic anticommutator of any two ”basis vectors” of an odd Grassmann character
(superposition of an odd products of $\theta^a$'s) gives zero, due to Eq. (<ref>).

iv.  The algebraic  application of  any annihilation operator on the corresponding Hermitian conjugated ”basis vector” gives identity, on all the rest of  ”basis vectors” gives
zero. Correspondingly the algebraic  anticommutators of the creation operators and their Hermitian conjugated partners, applied on the vacuum state, give identity,  all the rest anticommutators of creation and annihilation operators applied on the vacuum state,
give zero.

v. Correspondingly the “basis vectors” and their Hermitian conjugated
partners, applied on the vacuum state $|\,1>$, Eq. (<ref>), fulfill the properties of  creation  and annihilation operator, respectively, for the second quantized ”fermions”
on the level of one ”fermion” state.

From the eigenvectors of the members of the Cartan subalgebra,
Eq. (<ref>), we construct  $ 2^{d-1}$ Grassmann odd "basis vectors"
and $ 2^{d-1}-1$ (we skip self adjoint identity, which we use to describe the vacuum
state $|\,1>$)  Grassmann even "basis vectors" as superposition of odd and even
products of $\theta^a$'s, respectively. Their Hermitian conjugated $ 2^{d-1}$ odd
and $ 2^{d-1}-1$ even partners are, according to Eqs. (<ref>, <ref>), determined by the corresponding superposition of  odd and even  products of  $\frac{\partial}{\partial \theta_{a}}$'s, respectively [
Relations among operators and their Hermitian conjugated partners in both kinds of
the Clifford algebra objects are more complicated than in the Grassmann case, where the
Hermitian conjugated operators follow by taking into account Eq. (<ref>).

In the Clifford case
$\frac{1}{2} (\gamma^a + \frac{\eta^{aa}}{i\,k} \gamma^b)^{\dagger}$  is
proportional to $\frac{1}{2} (\gamma^a + \frac{\eta^{aa}}{i \,(- k)} \gamma^b) $,
while $ \frac{1}{\sqrt{2}} (1 +  \frac{i}{k}  \gamma^a \gamma^b)$ are self adjoint.
This is the case also for representations in the sector of $\tilde{\gamma}^a$'s.].

We present in this Sect. <ref> the Grassmann anticommuting
odd "basis vectors".
In App. <ref> the reader can find illustration of the Grassmann even and
correspondigly commuting ”basic vectors”.

Let us  in $d=2(2n+1)$-dimensional space-time, $n\ge 0$,  make a choice of the starting
Grassmann odd "basis vector" $\hat{b}^{\theta 1\dagger}_{1}$, which is the
eigenvector of the Cartan subalgebra of Eqs. (<ref>, <ref>)
with the egenvalues
$(+i, +1, +1, \cdots, +1 )$, respectively, and has the Hermitian conjugated partner
equal to $(\hat{b}^{\theta 1\dagger}_{1})^{\dagger}=$ $\hat{b}^{\theta 1}_{1}$,

\begin{eqnarray}
% |\phi^{1}_{1}> &=&(\frac{1}{\sqrt{2}})^{\frac{d}{2}} \,
% (\theta^0 - \theta^3) (\theta^1 + i \theta^2) (\theta^5 + i \theta^6) \cdots (\theta^{d-1} +
%  i \theta^{d}) \hat{b}^{\theta 1}_{1} \, |1\,>\,,\nonumber\\
% \hat{b}^{\theta 1\dagger}_{1} \,, \quad {\rm with} \nonumber\\
&&\hat{b}^{\theta 1\dagger}_{1}{\bf :}= (\frac{1}{\sqrt{2}})^{\frac{d}{2}} \,
(\theta^0 - \theta^3) (\theta^1 + i \theta^2) (\theta^5 + i\theta^6) \nonumber\\
&& {} \cdots (\theta^{d-1} + i \theta^{d}) \,, \nonumber\\
%&&(\hat{b}^{\theta 1 \dagger}_{1})^{\dagger}=   \nonumber\\
&& \hat{b}^{\theta 1}_{1}  {\bf :} =
(\frac{\partial}{\partial \theta^{d-1}} - i \frac{\partial}{\partial \theta^{d}})%\nonumber\\
{}\cdots (\frac{\partial}{\partial \theta^{0}}
+ \frac{\partial}{\partial \theta^3})\,,\nonumber\\
d &=& 2(2n+1) \,.
\label{start(2n+1)2theta}
\end{eqnarray}

In the case of $d=4n$, $n\ge 0$, the corresponding starting Grassmann odd "basis vector"

can be chosen as

\begin{eqnarray}
%|\phi^{ 1}_{1}> &=& b^{\theta 1 \dagger}_{1}  |1\,>\,,\nonumber\\
{\hat b}^{\theta 1 \dagger}_{1}{\bf :} &=&(\frac{1}{\sqrt{2}})^{\frac{d}{2}-1} \,
(\theta^0 - \theta^3) (\theta^1 + i \theta^2) (\theta^5 + i \theta^6)\cdots
\nonumber\\
&&{}\cdots (\theta^{d-3} +
i \theta^{d-2})  \theta^{d-1} \theta^d\,,\nonumber\\
d &=& 4n \,.
\label{start4ntheta}
\end{eqnarray}

All the rest of "basis vectors", belonging to the same irreducible representation of the
Lorentz group, follow by the application of ${\cal {\bf S}}^{ab}$'s.

Let us denote the members $m$ of this starting irreducible representation $f$ by
$\hat{b}^{\theta m \dagger}_{f}$, $m=1, f=1$,
and their Hermitian conjugated partners by $\hat{b}^{\theta m}_{f}$, with
$m=1, f=1$.

"Basis vectors", belonging to different irreducible representations $f'$, will be
denoted by $\hat{b}^{\theta m' \dagger}_{f'}$, $f'$ and their Hermitian
conjugated partners by
$\hat{b}^{\theta m' }_{f'}=(\hat{b}^{\theta m' \dagger}_{f'})^{\dagger} $.

${\cal {\bf S}}^{ac}$'s, which do not belong to the Cartan subalgebra, transform
step by step the two by two "nilpotents", no matter how many "nilpotents" are
between the chosen two, up to a constant,

as follows:

${\cal {\bf S}}^{ac}
\frac{1}{\sqrt{2}}\, (\theta^a + \frac{\eta^{aa}}{i k} \theta^b) \cdots $
$\frac{1}{\sqrt{2}}\, (\theta^c + \frac{\eta^{cc}}{i k'} \theta^d)$
$\propto - \frac{\eta^{aa} \eta^{cc}}{2k} (\theta^a \theta^b  +
\frac{k}{k'} \theta^c \theta^d) \cdots$,

leaving at each step at least one "nilpotent" unchanged, so that the whole irreducible
representation remains odd.

The superposition of ${\cal {\bf S}}^{bd}$ and $ i {\cal {\bf S}}^{bc}$
transforms $- \frac{\eta^{aa} \eta^{cc}}{2k}$ $(\theta^a \theta^b  +
\frac{k}{k'} \theta^c \theta^d)$ into $\frac{1}{\sqrt{2}} $
$(\theta^a - \frac{\eta^{aa}}{i k} \theta^b)
\frac{1}{\sqrt{2}}\, (\theta^c - \frac{\eta^{cc}}{i k'} \theta^d)$, and not into
$\frac{1}{\sqrt{2}}\, (\theta^a + \frac{\eta^{aa}}{i k} \theta^b)$
$\frac{1}{\sqrt{2}}\, (\theta^c - \frac{\eta^{cc}}{i k'} \theta^d)$ or not into
$\frac{1}{\sqrt{2}}\, (\theta^a - \frac{\eta^{aa}}{i k} \theta^b)$
$\frac{1}{\sqrt{2}}\, (\theta^c + \frac{\eta^{cc}}{i k'} \theta^d)$.

Therefore we can start another odd representation with the "basis vector"
$\hat{b}^{\theta m \dagger }_{2}$  as follows

\begin{eqnarray}
% |\phi^{1}_{1}> &=&(\frac{1}{\sqrt{2}})^{\frac{d}{2}} \,
% (\theta^0 - \theta^3) (\theta^1 + i \theta^2) (\theta^5 + i \theta^6) \cdots (\theta^{d-1} +
%  i \theta^{d}) \hat{b}^{\theta 1}_{1} \, |1\,>\,,\nonumber\\
% \hat{b}^{\theta 1\dagger}_{1} \,, \quad {\rm with} \nonumber\\
\hat{b}^{\theta m=1\dagger}_{2}{\bf :} &=& (\frac{1}{\sqrt{2}})^{\frac{d}{2}} \,
(\theta^0 + \theta^3) (\theta^1 + i \theta^2) (\theta^5 + i\theta^6)
%\nonumber\\ &&
{} \cdots (\theta^{d-1} + i \theta^{d}) \,, \nonumber\\
(\hat{b}^{\theta m=1 \dagger}_{2})^{\dagger} &=& \hat{b}^{\theta 1}_{2} {\bf :} =
(\frac{\partial}{\partial \theta^{d-1}} - i \frac{\partial}{\partial \theta^{d}})
% \nonumber\\&&
{}\cdots (\frac{\partial}{\partial \theta^{0}}
-\frac{\partial}{\partial \theta^3})\,.
\label{start(2n+1)2thetasecond}
\end{eqnarray}

The application of ${\cal {\bf S}}^{ac}$'s determines the whole second irreducible representation
$\hat{b}^{\theta 2 \dagger }_{j}$.

One finds that each of these two irreducible representations has
${\bf \frac{1}{2}\frac{d!}{\frac{d}{2}! \frac{d}{2}!}}$ members, Ref. [22].

Taking into account Eq. (<ref>), it follows that odd products of $\theta^a$'s
anticommute and so do the odd products of $\frac{\partial}{\partial \theta_{a}}$'s.
It then follows:

Statement 1.
The oddness of the products of  $\theta^a$'s guarantees the
anticommuting properties of all objects which include odd number of $\theta^a$'s.

One further sees that $\frac{\partial}{\partial \theta^{a}} \theta^b = \eta^{ab}$,
while $\frac{\partial}{\partial \theta_{a}}|\,1>=0$, and $\theta^a |\,1>=\theta^a  |\,1>$.

We can therefore conclude

\begin{eqnarray}
\{ \hat{b}^{\theta m}_f, \hat{b}^{\theta m' \dagger}_{f'} \}_{*_{A}+}
|\,\,1> &=& \delta_{f f'}\; \delta^{m m'}\;|\,\,1>\,,\nonumber\\
\{ \hat{b}^{\theta m}_f, \hat{b}^{\theta m'}_{f'} \}_{*_{A}+}  |\,\,1>
&=& 0\;\cdot\, |\,\,1> \,,\nonumber\\
\{\hat{b}^{\theta m \dagger}_f,\hat{b}^{\theta m' \dagger}_{f'}\}_{*_{A}+} \;|\,\,1>
&=&0\;\cdot\, |\,\,1> \,,\nonumber\\
% \hat{b}^{\theta k \dagger}_{i} \;|\,\,1>& =& \, |\phi^{k}_{i}>\,, \nonumber\\
\hat{b}^{\theta m}_{f} \,*_{A}\,|\,\,1>& =&0\;\cdot\, |\,\,1> \,, \nonumber\\
|\phi^{m}_{o\, f }> &=& \hat{b}^{\theta m\dagger}_{f} \,|\,1>\,,
%(k,l) &=&(1,2) \,.
\label{ijthetaprod}
\end{eqnarray}

where $\{ \hat{b}^{\theta m}_f, \hat{b}^{\theta m' \dagger}_{f'} \}_{*_{A}+}=$
$ \hat{b}^{\theta m}_f *_{A} \hat{b}^{\theta m' \dagger}_{f'} +
\hat{b}^{\theta m'}_{f'} *_{A} \hat{b}^{\theta m \dagger}_{f} $.

These anticommutation relations of the ”basis vectors” of the odd Grassmann character,
manifest on the level of the Grassmann algebra the  anticommutation relations

required by Dirac [3] for second quantized fermions.

The creation operators, determining the ”Grassmann fermions” carrying the integer
spins (in adjoint representations, Eq. (<ref>)), are tensor products,
$*_{T}$, of ”basis vectors” in internal space and of the (momentum or coordinate)
”basis” in ordinary space. Since the oddness of the ”basis vectors”  transfers to
the creation operators, then the creation operators and their Hermitian conjugated
partners annihilattion operators, fulfill the anticommutation relations postulated
by Dirac for the second quantized ”Grassmann fermions” with the integer spin.

The integer spin fermions have not been observed, all the observed  fermions carry the
half integer spin.

$\;\;$ Grassmann integer spin fermions in $d=(5+1)$:

Let us illustrate  anticommuting ”basis vectors” with spins and charges in adjoint
representation in $d=(5+1)$-dimensional space.

Table <ref> represents two decuplets, which are the
"egenvectors" of the Cartan subalgbra (${\cal {\bf S}}^{03}$,
${\cal {\bf S}}^{12}$, ${\cal {\bf S}}^{56}$), Eq. (<ref>),
of the Lorentz algebra ${\cal {\bf S}}^{ab}$.
The two decuplets represent two Grassmann odd  irreducible representations
of $SO(5,1)$.

One can read on the same table, from the first to the third and from the
fourth to the sixth line in each of the two decuplets, two Grassmann even
triplet representations of $SO(3,1)$, if paying attention on the
eigenvectors of ${\cal {\bf S}}^{03}$ and  ${\cal {\bf S}}^{12}$ alone,
while the  eigenvector of ${\cal {\bf S}}^{56}$ has, as a "spectator", the
eigenvalue  either $+1$ (the first triplet in both decuplets) or $-1$ (the
second triplet in both decuplets).  Each of the two decuplets contains
also one "fourplet"  with the "charge" ${\cal {\bf S}}^{56}$ equal to zero
($(7^{th}, 8^{th}, 9^{th}, 10^{th})$ lines in each of the two decuplets
(Table II in Ref. [9])).

Paying attention on the eigenvectors of ${\cal {\bf S}}^{03}$ alone one
recognizes as well even and odd representations of $SO(1,1)$:
$\theta^0 \theta^3$

and $\theta^0 \pm \theta^3$, respectively.

The Hermitian conjugated "basis vectors" follow  by using
Eq. (<ref>) and is for the first "basis vector" of
Table <ref> equal to $(-)^2
(\frac{1}{\sqrt{2}})^3 (\frac{\partial}{\partial \theta_5} -i
\frac{\partial}{\partial \theta_6})\, (\frac{\partial}{\partial \theta_1} - i
\frac{\partial}{\partial \theta_2})\, (\frac{\partial}{\partial \theta_0} +
\frac{\partial}{\partial \theta_3})$. One correspondingly finds that when $(\frac{1}{\sqrt{2}})^3
(\frac{\partial}{\partial \theta_5} -i \frac{\partial}{\partial \theta_6})\, (\frac{\partial}{\partial \theta_1} - i
\frac{\partial}{\partial \theta_2})\, (\frac{\partial}{\partial \theta_0} +
\frac{\partial}{\partial \theta_3}$) applies on $(\frac{1}{\sqrt{2}})^3
(\theta^{0} - \theta^{3}) (\theta^{1} + i \theta^{2}) (\theta^{5} + i \theta^{6}) $ the result is
identity. Application of $ (\frac{1}{\sqrt{2}})^3 (\frac{\partial}{\partial \theta_5} -i
\frac{\partial}{\partial \theta_6})\, (\frac{\partial}{\partial \theta_1} - i
\frac{\partial}{\partial \theta_2})\, (\frac{\partial}{\partial \theta_0} +
\frac{\partial}{\partial \theta_3})$ on all the rest of "basis vectors" of the decuplet $I$ as well as
on all the "basis vectors" of the decuplet $II$ gives zero. "Basis vectors" are orthonormalized with respect to
Eq. (<ref>) of App. <ref>.

[t]16.5 cm
The two decuplets, the odd eigenvectors of the
Cartan subalgebra, Eq. (<ref>),
(${\cal {\bf S}}^{0 3}, {\cal {\bf S}}^{1 2}$, ${\cal {\bf S}}^{5 6}$)
of the Lorentz algebra $SO(5,1)$ in Grassmann
$(5+1)$-dimensional space, forming two irreducible representations, are presented.
Table is taken from Refs. [22, 26].
The "basis vectors" within each decuplet are reachable from any member by
${\cal {\bf S}}^{ab}$'s and are decoupled from another decuplet.

The two operators of handedness,  $\Gamma^{((d-1)+1)}$ for $d=(6,4)$, are invariants
of the Lorentz algebra, Eq. (<ref>) of App. <ref>, $\Gamma^{(5+1)}$ for the whole decuplet,
$\Gamma^{(3+1)}$ for the "triplets" and "fourplets".

$I$   $i$    $\rm{decuplet\; of\; eigenvectors} $   ${\cal {\bf S}}^{03}$   ${\cal {\bf S}}^{1 2}$
${\cal {\bf S}}^{5 6}$   $\Gamma^{(5+1)}$   $\Gamma^{(3+1)}$

$1$      ($\frac{1}{\sqrt{2}})^3 (\theta^{0} - \theta^{3}) (\theta^{1} + i \theta^{2})
(\theta^{5} + i \theta^{6})$    $ i$   $ 1$   $1$   $1$   $1$

$2$      ($\frac{1}{\sqrt{2}})^2 (\theta^{0} \theta^{3} + i \theta^{1} \theta^{2})
(\theta^{5} + i \theta^{6})$     $ 0$     $0 $    $1$   $1$   $1$

$3$      ($\frac{1}{\sqrt{2}})^3 (\theta^{0} +  \theta^{3}) (\theta^{1} - i \theta^{2})
(\theta^{5} + i \theta^{6})$    $-i $   $-1$   $1$   $1$   $1$

$4$      ($\frac{1}{\sqrt{2}})^3 (\theta^{0} -  \theta^{3}) (\theta^{1} - i \theta^{2})
(\theta^{5} - i \theta^{6})$    $ i $   $-1$   $-1$   $1$   $-1$

$5$      ($\frac{1}{\sqrt{2}})^2 (\theta^{0} \theta^{3} - i \theta^{1} \theta^{2})
(\theta^{5} - i \theta^{6})$     $ 0 $    $0 $   $-1$   $1$   $-1$

$6$      ($\frac{1}{\sqrt{2}})^3 (\theta^{0} +  \theta^{3}) (\theta^{1} + i \theta^{2})
(\theta^{5} - i \theta^{6})$    $-i $   $ 1$   $-1$   $1$   $-1$

$7$      ($\frac{1}{\sqrt{2}})^2 (\theta^{0} - \theta^{3}) (\theta^{1} \theta^{2} +
\theta^{5} \theta^{6})$    $ i$   $ 0$   $0$   $1$   $0$

$8$      ($\frac{1}{\sqrt{2}})^2 (\theta^{0} + \theta^{3}) (\theta^{1} \theta^{2} -
\theta^{5} \theta^{6})$    $- i$   $ 0$   $0$   $1$   $0$

$9$      ($\frac{1}{\sqrt{2}})^2 (\theta^{0}  \theta^{3} +i \theta^{5} \theta^{6})
(\theta^{1}+i \theta^{2})
$    $ 0$   $ 1$   $0$   $1$   $0$

$10$      ($\frac{1}{\sqrt{2}})^2 (\theta^{0}  \theta^{3} - i \theta^{5} \theta^{6})
(\theta^{1}-i \theta^{2})
$    $ 0$   $- 1$   $0$   $1$   $0$

$II$   $i$    $\rm{decuplet\; of\; eigenvectors}$   ${\cal {\bf S}}^{03}$   ${\cal {\bf S}}^{1 2}$
${\cal {\bf S}}^{5 6}$   $\Gamma^{(5+1)}$   $\Gamma^{(3+1)}$

$1$      ($\frac{1}{\sqrt{2}})^3 (\theta^{0} + \theta^{3}) (\theta^{1} + i \theta^{2})
(\theta^{5} + i \theta^{6})$    $- i$   $ 1$   $1$   $-1$   $-1$

$2$      ($\frac{1}{\sqrt{2}})^2 (\theta^{0} \theta^{3} - i \theta^{1} \theta^{2})
(\theta^{5} + i \theta^{6})$     $ 0$   $ 0 $   $1$   $-1$   $-1$

$3$      ($\frac{1}{\sqrt{2}})^3 (\theta^{0} -  \theta^{3}) (\theta^{1} - i \theta^{2})
(\theta^{5} + i \theta^{6})$    $ i $   $-1$   $1$   $-1$   $-1$

$4$      ($\frac{1}{\sqrt{2}})^3 (\theta^{0} +  \theta^{3}) (\theta^{1} - i \theta^{2})
(\theta^{5} - i \theta^{6})$    $- i $   $-1$   $-1$   $-1$   $1$

$5$      ($\frac{1}{\sqrt{2}})^2 (\theta^{0} \theta^{3} + i \theta^{1} \theta^{2})
(\theta^{5} - i \theta^{6})$     $ 0$    $0 $   $-1$   $-1$   $1$

$6$     ($\frac{1}{\sqrt{2}})^3 (\theta^{0} -  \theta^{3}) (\theta^{1} + i \theta^{2})
(\theta^{5} - i \theta^{6})$    $ i $   $ 1$   $-1$   $-1$   $1$

$7$      ($\frac{1}{\sqrt{2}})^2 (\theta^{0} + \theta^{3}) (\theta^{1} \theta^{2} +
\theta^{5} \theta^{6})$    $- i$   $ 0$   $0$   $-1$   $0$

$8$      ($\frac{1}{\sqrt{2}})^2 (\theta^{0} - \theta^{3}) (\theta^{1} \theta^{2} -
\theta^{5} \theta^{6})$    $ i$   $ 0$   $0$   $-1$   $0$

$9$      ($\frac{1}{\sqrt{2}})^2 (\theta^{0} \theta^{3} - i \theta^{5} \theta^{6})
(\theta^{1}+i \theta^{2})
$    $ 0$   $ 1$   $0$   $-1$   $0$

$10$      ($\frac{1}{\sqrt{2}})^2 (\theta^{0}  \theta^{3} + i \theta^{5} \theta^{6})
(\theta^{1}-i \theta^{2})
$    $ 0$   $- 1$   $0$   $-1$   $0$

Let us notice that $\frac{\partial}{\partial \theta_{a}}$ on a "state" which is just  an identity, $|\,1>$,
gives zero, $\frac{\partial}{\partial \theta_{a}}\,|\,1>=0$, while $\theta^a \,|\,1>$, or
any superposition of products of different $\theta^a$'s, applied on $|\,1>$, gives the
"vector" back.

One easily sees that application of
products of superposition of $\theta^a$'s on $|\,1>$  gives nonzero contribution, while
application of products of superposition of $\frac{\partial}{\partial \theta^a}$'s on $|\,1>$ gives zero.

The two by ${\cal {\bf S}}^{ab}$ decoupled Grassmann decuplets of
Table <ref> are the largest two
irreducible representations of  odd products of $\theta^a$'s. There are $12$  additional Grassmann
odd "vectors", arranged into irreducible representations of six singlets and six sixplets

\begin{eqnarray} %\[
&& (\frac{1}{2}\,(\theta^0\mp \theta^3), \frac{1}{2}\,(\theta^1\pm i \theta^2),
\frac{1}{2}\,(\theta^5 \pm i \theta^6)\,,\nonumber\\
&&\frac{1}{2}\,(\theta^0 \mp \theta^3)\, \theta^1 \theta^2 \theta^5 \theta^6,%\]
\frac{1}{2}\,(\theta^1\pm i \theta^2)\, \theta^0 \theta^3 \theta^5 \theta^6\,,
\frac{1}{2}\,(\theta^5 \pm i \theta^6)\,\theta^0 \theta^3 \theta^1 \theta^2)\,.
%\] %Proove sixplet
\label{nondecuplet}
\end{eqnarray}

The algebraic application of products of superposition of
$\frac{\partial}{\partial \theta^a}$'s on the corresponding Hermitian conjugated partners, which are products of superposition of $\theta^a$'s,
leads to the identity for either even or odd Grassmann character [
We shall see that the vacuum states are in the Clifford case, similarly as in the
Grassmann case, for both kinds of the Clifford algebra objects, $\gamma^a$'s
and  $\tilde{\gamma}^a$'s, normalized sums of products of the annihilation
$\times$, $*_{A}$, its Hermitian conjugated creation operators, and correspondingly self adjoint operators,
but they are not the identity.].

Besides $32$ Grassmann odd eigenvectors of the Grassmann Cartan subalgebra,
Eq. (<ref>), there are $(32-1)$ Grassmann "basis vectors",
which we  arrange into irreducible representations, which are superposition of
even products of $\theta^a$'s. The even self adjoint operator identity
(which is indeed the normalized product of all the annihilation times, $*_{A}$,
creation operators)  is used to represent the vacuum state.

It is not difficult to see that  Grassmann "basis vectors" of an odd Grassmann character

anticommute among themselves and so do odd products of superposition of
$\frac{\partial}{\partial \theta^a}$'s, while equivalent even
products commute.

We shall demonstrate the matrix representations of the operators of the
Lorentz transformations and of the elements of the algebras for the Clifford
algebra elements $\gamma^a$'s and $\tilde{\gamma}^a$'s when applying
on ”basis vectors” of the Clifford algebra of $\gamma^a$'s in
App. <ref>.

The Grassmann odd algebra (as well as the two odd Clifford algebras) offers, due to the
oddness of the internal space giving oddness as well to the elements of the tensor products
of the internal space and of the momentum space, the
description of the anticommuting second quantized fermion fields, as postulated by Dirac.
But  the Grassmann "fermions" carry the integer spins, while the observed fermions —
quarks and leptons — carry half integer spin.

§.§.§ Clifford algebras and ”basis vectors” describing internal space

We learn in Sect. <ref>, Eq. (<ref>), that in
$d$-dimensional space of anticommuting Grassmann coordinates (and of their
Hermitian conjugated partners — derivatives)

there exist two kinds of the Clifford coordinates (operators) —
$\gamma^{a}$ and $\tilde{\gamma}^{a}$ — both are expressible in terms of
$\theta^{a}$ and their conjugate momenta $p^{\theta a}= i \,
\frac{\partial}{\partial \theta_{a}}$  [9].

\begin{eqnarray}
\label{clifftheta1}
\gamma^{a} &=& (\theta^{a} + \frac{\partial}{\partial \theta_{a}})\,, \quad
\tilde{\gamma}^{a} =i \,(\theta^{a} - \frac{\partial}{\partial \theta_{a}})\,,\nonumber\\
\theta^{a} &=&\frac{1}{2} \,(\gamma^{a} - i \tilde{\gamma}^{a})\,, \quad
\frac{\partial}{\partial \theta_{a}}= \frac{1}{2} \,(\gamma^{a} + i \tilde{\gamma}^{a})\,,
\nonumber\\
%(\gamma^{a})^{\dagger} &=&\eta^{aa}\, \gamma^{a}\,, \quad
%(\tilde{\gamma}^{a})^{\dagger} = \eta^{aa} \,\tilde{\gamma}^{a}\,,
\end{eqnarray}

offering together  $2\cdot 2^d$  operators: $2^d$ of those which are products of
$\gamma^{a}$  and  $2^d$ of those which are products of $\tilde{\gamma}^{a}$.

Let us repeat the anticommuting properties of both Clifford algebras, presented
already  in Eq. (<ref>).

\begin{eqnarray}
\label{gammatildeantiher}
\{\gamma^{a}, \gamma^{b}\}_{+}&=&2 \eta^{a b}= \{\tilde{\gamma}^{a},
\tilde{\gamma}^{b}\}_{+}\,, \nonumber\\
\{\gamma^{a}, \tilde{\gamma}^{b}\}_{+}&=&0\,,\quad
(a,b)=(0,1,2,3,5,\cdots,d)\,, \nonumber\\
(\gamma^{a})^{\dagger} &=& \eta^{aa}\, \gamma^{a}\, , \quad
(\tilde{\gamma}^{a})^{\dagger} =  \eta^{a a}\, \tilde{\gamma}^{a}\,,\nonumber
% S^{ab} &=& \frac{i}{2}\, (\gamma^a \gamma^b - \gamma^b \gamma^a)\,, \quad
% \tilde{S}^{ab} = \frac{i}{2}\, (\tilde{\gamma}^a \tilde{\gamma}^b - \tilde{\gamma}^b
% \tilde{\gamma}^a)\,,
\end{eqnarray}

with $\eta^{a b}=diag\{1,-1,-1,\cdots,-1\}$, leading to the statement that the
two Clifford algebras form the independent algebras.

Statement 2. $\gamma^a$'s  and $\tilde{\gamma}^{a}$'s define two independent Clifford algebras.

The proof of this statement can be found in App. <ref>, Statement 1.

As in the Grassmann case we chose the ”basis vectors” in each of the two
spaces to beproducts of eigenstates of the Cartan subalgebra members,
Eq. (<ref>), of the Lorentz algebras,
($S^{ab} =\frac{i}{4}(\gamma^a \gamma^b - \gamma^b \gamma^a),
\tilde{S}^{ab} =\frac{i}{4}(\tilde{\gamma}^a \tilde{\gamma}^b -
\tilde{\gamma}^b  \tilde{\gamma}^a)$).

First we look for ”eigenstates” of each of the Cartan subalgebra members,
Eq. (<ref>), for each of the two kinds of the Clifford algebras

\begin{eqnarray}
S^{ab} \frac{1}{2} (\gamma^a + \frac{\eta^{aa}}{ik} \gamma^b) &=& \frac{k}{2}  \,
\frac{1}{2} (\gamma^a + \frac{\eta^{aa}}{ik} \gamma^b)\,,\quad
S^{ab} \frac{1}{2} (1 +  \frac{i}{k}  \gamma^a \gamma^b) = \frac{k}{2}  \,
\frac{1}{2} (1 +  \frac{i}{k}  \gamma^a \gamma^b)\,,\nonumber\\
\tilde{S}^{ab} \frac{1}{2} (\tilde{\gamma}^a + \frac{\eta^{aa}}{ik} \tilde{\gamma}^b) &=&
\frac{k}{2}  \,\frac{1}{2} (\tilde{\gamma}^a + \frac{\eta^{aa}}{ik} \tilde{\gamma}^b)\,,
\quad
\tilde{S}^{ab} \frac{1}{2} (1 +  \frac{i}{k}  \tilde{\gamma}^a \tilde{\gamma}^b) =
\frac{k}{2}  \, \frac{1}{2} (1 +  \frac{i}{k} \tilde{\gamma}^a \tilde{\gamma}^b)\,,
\label{eigencliffcartan}
\end{eqnarray}

$k^2=\eta^{aa} \eta^{bb}$.
The proof of Eq. (<ref>) is presented in App. (<ref>),
Statement 2a.
The Clifford "basis vectors" — nilpotents $\frac{1}{2} (\gamma^a + \frac{\eta^{aa}}{ik} \gamma^b), (\frac{1}{2} (\gamma^a + \frac{\eta^{aa}}{ik} \gamma^b))^2=0$ and projectors $ \frac{1}{2} (1 +  \frac{i}{k}  \tilde{\gamma}^a \tilde{\gamma}^b),
( \frac{1}{2} (1 +  \frac{i}{k}  \tilde{\gamma}^a \tilde{\gamma}^b))^2 =
\frac{1}{2} (1 +  \frac{i}{k}  \tilde{\gamma}^a \tilde{\gamma}^b)$ —
of both algebras are normalized,  up to a phase, with respect to
Eq. (<ref>) of App. <ref>.

Both, nilpotents and projectors, have half integer spins. The "eigenvalues" of the operator $S^{03}$, for example,
for the "vector" $\frac{1}{2} (\gamma^0 \mp \gamma^3)$ are equal to
$\pm\, \frac{i}{2}$, respectively, for the "vector" $\frac{1}{2} (1\pm \gamma^0  \gamma^3)$
are $\pm\, \frac{i}{2}$, respectively, while all the rest "vectors" have "eigenvalues"
$\pm\, \frac{1}{2}$. One finds equivalently for the "eigenvectors" of the operator
$\tilde{S}^{03}$: for $\frac{1}{2} \,( \tilde{\gamma^0} \mp \tilde{\gamma}^3)$
the "eigenvalues"
$\pm\, \frac{i}{2}$, respectively, and for the "eigenvectors" $\frac{1}{2}
(1\pm \tilde{\gamma}^0  \tilde{\gamma}^3)$ the "eigenvalues" $k=\pm\,  \frac{i}{2}$,
respectively, while all the rest "vectors" have $k=\pm\,  \frac{1}{2}$.

To make discussions easier let us introduce the notation for the "eigenvectors" of the
two Cartan subalgebras, Eq. (<ref>), Ref. [66, 9].

\begin{eqnarray}
\stackrel{ab}{(k)}:&=&
\frac{1}{2}(\gamma^a + \frac{\eta^{aa}}{ik} \gamma^b)\,,\quad
\stackrel{ab}{(k)}^{\dagger} = \eta^{aa}\stackrel{ab}{(-k)}\,,\quad
(\stackrel{ab}{(k)})^2 =0\,, \quad \stackrel{ab}{(k)}\stackrel{ab}{(-k)}
\stackrel{ab}{[k]}:&=&
\frac{1}{2}(1+ \frac{i}{k} \gamma^a \gamma^b)\,,\quad \;\,
\stackrel{ab}{[k]}^{\dagger} = \,\stackrel{ab}{[k]}\,, \quad \quad \quad \quad
(\stackrel{ab}{[k]})^2 = \stackrel{ab}{[k]}\,,
\quad \stackrel{ab}{[k]}\stackrel{ab}{[-k]}=0\,,
\nonumber\\
\stackrel{ab}{(k)}\stackrel{ab}{[k]}& =& 0\,,\qquad \qquad \qquad
\stackrel{ab}{[k]}\stackrel{ab}{(k)}=  \stackrel{ab}{(k)}\,, \quad \quad \quad
\stackrel{ab}{(k)}\stackrel{ab}{[-k]} =  \stackrel{ab}{(k)}\,,
\quad \, \stackrel{ab}{[k]}\stackrel{ab}{(-k)} =0\,,
\nonumber\\
\stackrel{ab}{\tilde{(k)}}:&=&
\frac{1}{2}(\tilde{\gamma}^a + \frac{\eta^{aa}}{ik} \tilde{\gamma}^b)\,,\quad
\stackrel{ab}{\tilde{(k)}}^{\dagger} = \eta^{aa}\stackrel{ab}{\tilde{(-k)}}\,,\quad
\stackrel{ab}{\tilde{[k]}}:&=&
\frac{1}{2}(1+ \frac{i}{k} \tilde{\gamma}^a \tilde{\gamma}^b)\,,\quad \;\,
\stackrel{ab}{\tilde{[k]}}^{\dagger} = \,\stackrel{ab}{\tilde{[k]}}\,,
\quad \quad \quad \quad
\stackrel{ab}{\tilde{(k)}}\stackrel{ab}{\tilde{[k]}}& =& 0\,,\qquad \qquad \qquad
\stackrel{ab}{\tilde{[k]}}\stackrel{ab}{\tilde{(k)}}=  \stackrel{ab}{\tilde{(k)}}\,,
\quad \quad \quad
\stackrel{ab}{\tilde{(k)}}\stackrel{ab}{\tilde{[-k]}} =  \stackrel{ab}{\tilde{(k)}}\,,
\quad \, \stackrel{ab}{\tilde{[k]}}\stackrel{ab}{\tilde{(-k)}} =0\,,
\nonumber\\
\label{graficcliff}
\end{eqnarray}

with $k^2 = \eta^{aa} \eta^{bb}$ for both algebras.
Let us notice that the “eigenvectors” of  the Cartan subalgebras and the eigenvalues
are equivalent  in both algebras. Both algebras have projectors and nilpotents:
($ (\stackrel{ab}{[k]})^2= \stackrel{ab}{[k]}\,,(\stackrel{ab}{(k)})^{2}=0$),
($ (\stackrel{ab}{\tilde{[k]}})^2= \stackrel{ab}{\tilde{[k]}}\,,
We pay attention on even dimensional spaces, $d=2(2n+1)$ or $d=4n$, $n\ge0$ only.

The "basis vectors", which are products of $\frac{d}{2}$ either of nilpotents or
of projectors or of both, are “eigenstate” of all the members of the Cartan
subalgebra, Eq. (<ref>), of the corresponding Lorentz algebra, forming
$2^{\frac{d}{2}-1}$ irreducible representations with $2^{\frac{d}{2}-1}$ members
in each of the two  Clifford algebras cases.

The "basis vectors", which appear  below in  Eq. (<ref>),
are "eigenvectors"  of all the Cartan subalgebra members, Eq. (<ref>), in
$d=2(2n+1)$-dimensional space of $\gamma^a$'s.

The first one is the product of  nilpotents only and correspondingly a superposition
of an odd products of $\gamma^a$'s. The second one belongs to the same
irreducible representation as the first one, since it follows from the first one by the
application of $S^{01}$, for example, the third one follows from the second one by
the application of $S^{d-3\,d-1}$.

\begin{eqnarray}
\label{allcartaneigenvec}
&&\stackrel{03}{(+i)}\stackrel{12}{(+)}\cdots \stackrel{d-1 \, d}{(+)}\,,\;\;\quad
\stackrel{03}{[-i]}\stackrel{12}{[-i]} \stackrel{56}{(+)} \cdots
\stackrel{d-1 \, d}{(+)}\,,  \nonumber\\
&&\stackrel{03}{[-i]} \stackrel{12}{[-]} \stackrel{56}{(+)} \cdots
\stackrel{d-3\,d-2}{[-]}\;\stackrel{d-1\,d}{[-]}\,.
\end{eqnarray}

The number of nilpotents in any of the members remains all the time odd, due to
the eveness of the generators of the Lorentz transformations.
One finds for their Hermitian conjugated partners, up to a sign,

\begin{eqnarray}
\label{allcartaneigenvecher}
&&\stackrel{03}{(-i)}\stackrel{12}{(-)}\cdots \stackrel{d-1 \, d}{(-)}\,,\;\;\quad
\ \stackrel{03}{[-i]}\stackrel{12}{[-i]} \stackrel{56}{(-)}\cdots
\stackrel{d-1 \, d}{(-)}\,,\nonumber\\
&& \stackrel{03}{[-i]} \stackrel{12}{[-]} \stackrel{56}{(-)}\cdots
\stackrel{d-3\,d-2}{[-]}\;\stackrel{d-1\,d}{[-]}\,.
\nonumber
\end{eqnarray}

The members of the above irreducible representation for the $\tilde{\gamma}^a$'s
follow from Eq. (<ref>) by replacing $\gamma^a$'s with

The "basis vectors" form an orthonormal basis within each of the irreducible

\begin{eqnarray}
&& \stackrel{d-1 \, d}{(-)}\cdots \stackrel{12}{(-)} \stackrel{03}{(-i)} *_{A}
\stackrel{03}{(+i)}\stackrel{12}{(+)}\cdots \stackrel{d-1 \, d}{(+)} =1\,.
\label{Cliffnormalization}
\end{eqnarray}

This is,  due to Eq. (<ref>), equal to
$\stackrel{03}{[-i]}\stackrel{12}{[-]}\cdots \stackrel{d-1 \, d}{[-]}$, what we
normalize to one, as we show in Eq. (<ref>).

Usually  in the literature, the operators $\gamma^a$'s are represented as matrices.
We use here $\gamma^a$'s as Clifford objects, which form the ”basis vectors”.
One can calculate,  as seen in Ref. [25] and in App. <ref>,
the matrix representations of $\gamma^a$'s when apply on "basis vectors" and
the generators of the Lorentz transformations $S^{ab}$ when apply on
"basis vectors"

defined in $d$-dimensional space. In App. <ref> the
matrix representations  of $\gamma^a$'s and $S^{ab}$'s in
$d=(3+1)$-dimensional space-time are presented.

In the Grassmann case the $ 2^{d-1}$ odd and $ 2^{d-1}-1$ (we skip the selfadjoint
identity defining the vacuum state, Eq. (<ref>)) even Grassmann
”basis vectors” (operators), which are superposition of either odd or even products
of $\theta^a$'s, respectively, are well distinguishable from their $ 2^{d-1}$ odd and
$ 2^{d-1}-1$ even Hermitian conjugated partners, which are superposition of
odd and even  products of $\frac{\partial}{\partial \theta_{a}}$'s.

In the Clifford case the relation

between "basis vectors" and their Hermitian conjugated partners  (made
of products of nilpotents ($\stackrel{ab}{(k)}$ or  $\stackrel{ab}{\tilde{(k)}}$) and
projectors  ($\stackrel{ab}{[k]}$ or  $\stackrel{ab}{\tilde{[k]}}$), Eq. (<ref>),
are less transparent (although still easy to be evaluated).

This can be noticed in Eq. (<ref>), since for nilpotents we notice that
$\frac{1}{\sqrt{2}} (\gamma^a + \frac{\eta^{aa}}{i\,k} \gamma^b)^{\dagger}$
$=\eta^{aa}\,\frac{1}{\sqrt{2}} (\gamma^a + \frac{\eta^{aa}}{i \,(- k)}
\gamma^b) $, while for projectors we find that they are self adjoint objects,
$ (\frac{1}{\sqrt{2}} (1 +  \frac{i}{k}  \gamma^a \gamma^b))^{\dagger}=$
$ \frac{1}{\sqrt{2}} (1 +  \frac{i}{k}  \gamma^a \gamma^b)$.
This is the case also for representations in the sector of $\tilde{\gamma}^a$'s.

One easily sees that in even dimensional spaces, either in $d=2(2n+1)$ or in $d=4n$,
the Clifford odd "basis vectors" (they are products of an odd number of nilpotents and
an even number of projectors) have their Hermitian conjugated partners in another
irreducible representation, since Hermitian conjugation changes an odd number of
nilpotents (changing at the same time the handedness of the "basis vectors"), while the generators of the Lorentz transformations change two nilpotents at
the same time (keeping the handedness unchanged).

The Clifford even "basis vectors" have an even number of nilpotents and can have
an odd or an even number of projectors. Correspondingly an irreducible representation
of even "basis vectors" have among "basis vectors" the product of projectors only,
which  is  therefore  selfadjoint.

Taking into account Eq. (<ref>) one finds that the product of
Hermitian conjugated partner of a ”basis vector”  from Eq. (<ref>)
and the corresponding ”basis vector” is nonzero and normalized to identity, as
presented in Eq. (<ref>)

($\stackrel{03}{(-i)}\stackrel{12}{(-)}\cdots \stackrel{d-1 \, d}{(-)} *_{A}
\stackrel{03}{(+i)}\stackrel{12}{(+)}\cdots \stackrel{d-1 \, d}{(+)}$).
This is the case for any irreducible
representation in the case of the Clifford algebra of $\gamma^a$'s as well in
the Clifford algebra of $\tilde{\gamma}^a$'s.

Statement 3.
The product of any Hermitian conjugated ”basis vector” with the corresponding
”basic vector” is the same for all the members of an irreducible representation.

The proof is presented in App. <ref>, Statement 4.

Statement 4.
Changing a pair of any two nilpotents $\stackrel{ab}{(k)}\stackrel{cd}{(k')}$ to a pair
of projectors with the same ($k,k'$) $\stackrel{ab}{[k]}\stackrel{cd}{[k']} $, or of
any two projectors $\stackrel{ab}{[k']}\stackrel{cd}{[k'']}$ two a pair of two
nilpotents with the
same ($k',k''$), or  a pair of any nilpotent (projector) and projector (nilpotent)
$\stackrel{ab}{(k)}\stackrel{cd}{[k']}$ to a pair of the projector (nilpotent) and the
nilpotent (projektor) with the same ($k,k'$) $\stackrel{ab}{[k]}\stackrel{cd}{(k')}$,
the obtained ”basis vector” belongs to different irreducible representations.,
since the Lorentz transformations  $S^{ce}$  transform
$\stackrel{ab}{(k)}\stackrel{cd}{(k')}$ into $\stackrel{ab}{[-k]}\stackrel{cd}{[-k']}$,
never to $\stackrel{ab}{[k]}\stackrel{cd}{[k']} $, and equivalently for other  pairs.

This statement is proven in App. <ref>, Statement 2b. Due to
Eq. (<ref>) the irreducible representations are equivalent with respect
to the eigenvalues of the Cartan subalgebra members and also with respect to the
generators of the Lorentz transformations $S^{ab}$.

Statement is valid also if we  replace $\gamma^a$ with $\tilde{\gamma}^a$.

It can be checked, however, that there are Hermitian conjugated partners of the
”basis vectors” of different irreducible representations, which applying on
$\stackrel{03}{(+i)}\stackrel{12}{(+)}\cdots \stackrel{d-1 \, d}{(+)}$ from
the left hand side, give also nonzero contributions, not the
identity. Like it is
$\stackrel{03}{[+i]}\stackrel{12}{[+]}\cdots \stackrel{d-1 \, d}{(-)}*_{A}$

$\stackrel{03}{(+i)}\stackrel{12}{(+)}\cdots \stackrel{d-1 \, d}{(+)}$, giving (due to
Eq. (<ref>))
$\stackrel{03}{(+i)}\stackrel{12}{(+)}\cdots \stackrel{d-1 \, d}{[-]}$.
The ”basis vector” of $\stackrel{03}{[+i]}\stackrel{12}{[+]}\cdots \stackrel{d-1 \, d}{(-)}$
is $\stackrel{03}{[+i]}\stackrel{12}{[+]}\cdots \stackrel{d-1 \, d}{(+)}$, belonging
to another irreducible representation.

There are several other choices, like

\begin{eqnarray}
\label{notunique0}
\stackrel{03}{[+i]}\stackrel{12}{(-)}\stackrel{56}{[+]}\cdots \stackrel{d-1 \, d}{(-)} *_{A} %\cdot
\stackrel{03}{(+i)}\stackrel{12}{(+)}\cdots \stackrel{d-1 \, d}{(+)}\,,\nonumber\\
\stackrel{03}{[+i]}\stackrel{12}{[+]} \stackrel{56}{[+]}\cdots
\stackrel{d-1 \, d}{(-)} *_{A} %\cdot
\stackrel{03}{(+i)}\stackrel{12}{(+)}\cdots \stackrel{d-1 \, d}{(+)}\,,
\end{eqnarray}

which also give nonzero contributions duo to Eq. (<ref>).

Let us recognize:

i. The two Clifford spaces, the one spanned by $\gamma^{a}$'s and the second one
spanned by $\tilde{\gamma}^{a}$'s, are independent vector spaces, each with
$2^{d}$ "basis vectors" (App. <ref>, Statement 1.).

ii. The Clifford odd "basis vectors" (the superposition of products of
odd numbers of $\gamma^a$'s or of $\tilde{\gamma}^{a}$'s, respectively)
can be arranged for each kind of the Clifford algebras into two groups, Hermitian
conjugated to each other, of
$2^{\frac{d}{2} - 1}$ members of $2^{\frac{d}{2} - 1} $ irreducible
representations of the corresponding Lorentz group.

iii. Different irreducible representations are indistinguishable with respect to the
"eigenvalues" of the corresponding  Cartan subalgebra members, Statement 4.

iv. The Clifford even part (made of superposition of products of even numbers of
$\gamma^a$'s or  $\tilde{\gamma}^{a}$'s, respectively) splits as well into
twice $2^{\frac{d}{2} - 1} \cdot 2^{\frac{d}{2} - 1} $ irreducible representations
of the Lorentz group. One member of each Clifford  even representation, the one
which is the product of projectors only, is self adjoint. Members of one irreducible
representation are with respect to the Cartan subalgebra indistinguishable from
all the other irreducible representations for each of the two algebras.

v. The  odd $2^{\frac{d}{2}-1}$ members of each of the $ 2^{\frac{d}{2}-1}$
irreducible representations are among themselves orthogonal and so are
orthogonal $2^{\frac{d}{2}-1} \cdot$ $2^{\frac{d}{2}-1}$
members of their Hermitian conjugated part. For illustration of the
orthogonality one can look at Table <ref>,  and recognizes
that any ”basis vector” of the first four multiplets (odd I, II, III, IV),  the
third column, gives zero, if it is multiplied from the left hand side or from the
right hand side with any other ”basis vector” from the third column,
according to Eq. (<ref>).
The same is true for any ”basis vector” from the fourth column, Hermitian
conjugated to the third column.

Generalization to any even dimension $d$ is straightforward.

vi.  Let us denote the Clifford odd "basis vectors" of the Clifford algebra $\gamma^a$
by $\hat{b}^{m \dagger}_{f}$, where $f$ defines different irreducible
representations and $m$ a member in the representation $f$. Then
their Hermitian conjugate partners is denoted by  $\hat{b}^{m}_{f}$
$=(\hat{b}^{m \dagger}_{f})^{\dagger}$, let us start for $d=2(2n+1)$.

Let us repeat Statement 2.:
The algebraic product of $\hat{b}^{m}_{f}{}_{*_A}
\hat{b}^{m \dagger}_{f}$ is the same for all the members of one
irreducible representation.

vii. Let us look for the vacuum state $|\psi_{oc}>$ as the sum of the
products $\hat{b}^{m}_{f}{}_{*_A}
\hat{b}^{m \dagger}_{f}$ over  all the irreducible representations

\begin{eqnarray}
\label{vaccliff}
|\psi_{oc}>= \sum_{f=1}^{2^{\frac{d}{2}-1}}\,\hat{b}^{m}_{f}{}_{*_A}
\hat{b}^{m \dagger}_{f} \,|\,1\,>\,,
\end{eqnarray}

for one of the members $m$, anyone, of the odd irreducible representation $f$,
with $|\,1\,>$, which is the vacuum without any structure
($\hat{b}^{m}_{f}{} |\,1\,>=0$).

If we make a choice of the starting member of the starting irreducible representation

\begin{eqnarray}
\label{bmfdagerbmf}
\hat{b}^{m=1 \dagger}_{f=1} \, {\bf :} &=&\stackrel{03}{(+i)}\stackrel{12}{(+)}
\cdots \stackrel{d-1 \, d}{(+)}\,,\nonumber\\
(\hat{b}^{m=1 \dagger}_{f=1})^{\dagger}=\hat{b}^{m=1 }_{f=1} \, {\bf :} &=& \,
\stackrel{d-1 \, d}{(-)} \cdots \stackrel{12}{(-)}\stackrel{03}{(-i)}\,,
\end{eqnarray}

we recognize for the "basis vectors" of an
odd Clifford character for each of the two Clifford algebras the properties

\begin{eqnarray}
\label{almostDirac}
\hat{b}^{m}_{f} {}_{*_{A}}|\psi_{oc}>&=& 0\, |\psi_{oc}>\,,\nonumber\\
\hat{b}^{m \dagger}_{f}{}_{*_{A}}|\psi_{oc}>&=&  |\psi^m_{f}>\,,\nonumber\\
\{\hat{b}^{m}_{f}, \hat{b}^{m'}_{f'}\}_{*_{A}+}|\psi_{oc}>&=&
0\,|\psi_{oc}>\,, \nonumber\\
\{\hat{b}^{m \dagger}_{f}, \hat{b}^{m \dagger}_{f}\}_{*_{A}+}|\psi_{oc}>
%%<\psi_{oc}| \hat{b}^{m' \dagger}_{f'}\, *_{A}\,\hat{b}^{m \dagger}_{f}
%*_{A}|\psi_{oc}> &=& \delta^{m m'} \delta_{f f'}\,.
\end{eqnarray}

where  ${*_{A}}$ represents the algebraic multiplication of
$\hat{b}^{m \dagger}_{f}$  and $ \hat{b}^{m'}_{f'} $  among themselves and
with the vacuum state  $|\psi_{oc}>$ of Eq.(<ref>), which takes into account
Eq. (<ref>).

All the products of Clifford algebra elements are up to now the algebraic ones and so
are also the products in Eq. (<ref>). Since we use  here
anticommutation relations, we want to point out with ${}_{*_{A}}$ this algebraic
of the products, to be later distinguished from the tensor product ${}_{*_{T}}$,
when the creation and annihilattion operators  are defined on an extended basis,
which is the tensor product of the superposition of the ”basis vectors” of the Clifford
space and of the momentum basis, applying on the Hilbert space of
”Slater determinants”.
The tensor product $*_{T_{H}}$ will be  used  as the product
mapping a pair of the fermion wave functions into two fermion wave functions
and further to many fermion wave functions — that is to the extended algebra
of many fermion system.

Obviously, $\hat{b}^{m \dagger}_{f}$ and $\hat{b}^{m}_{f}$ have on the
level of the algebraic products, when applying on the vacuum state $|\psi_{oc}>$,
almost the properties of creation and annihilation operators of the second
quantized fermions in the postulates of Dirac,  as it is discussed in the next items.

We illustrate properties of "basis vectors"  and their Hermitian conjugated
partners on the example of $d=(5+1)$-dimensional space in
Sect. <ref>.

${\quad}$ viii.  There is, namely, the  property, which the  second quantized
fermions should
fulfill in addition to the relations of Eq. (<ref>). The anticommutation
relations of creation and annihilation operators should include:

\begin{eqnarray}
\label{should}
\{\hat{b}^{m}_{f}, \hat{b}^{m'\dagger}_{f'}\}_{*_{A}+}|\psi_{oc}>&=&
\delta^{m m'} \delta_{f f'} |\psi_{oc}>\,.
\end{eqnarray}

For  any $\hat{b}^{m}_{f}$ and any $\hat{b}^{m'\dagger}_{f'}$ this is not
the case, as we demonstrated in Eq. (<ref>)
(besides $\hat{b}^{m=1}_{f=1} = \,
\stackrel{d-1 \, d}{(-)} \cdots  \stackrel{56}{(-)} \stackrel{12}{(-)}\stackrel{03}{(-i)}$, for example, also

$\hat{b}^{m'}_{f'} = \, \stackrel{d-1 \, d}{(-)} \cdots  \stackrel{56}{(-)}
\stackrel{12}{[+]}\stackrel{03}{[+i]} $

and several others give, when applied on $\hat{b}^{m=1\dagger}_{f=1}$,
nonzero contributions).
There are namely $2^{\frac{d}{2}-1}-1$ too many annihilation
operators for each creation operator, which give, applied on the creation operator,
nonzero contribution.

${\quad}$ viii. a. To use the Clifford algebra objects to describe second quantized fermions,
the observed quarks and leptons as well as the antiquarks and antileptons
<cit.>,  the families should exist.

${\quad}$ viii. b.
The operators should exist, which connect one irreducible
representation of fermions with all the other irreducible representations.

${\quad}$ viii. c.
Two independent choices for describing the internal degrees
of freedom of the observed quarks and leptons are not in agreement with the observed
properties of fermions.

We solve these problems, cited in viii. a., viii. b., viii. c.,
by reducing  the degrees
of freedom offered by  the two kinds of the Clifford algebras, $\gamma^a$'s
and  $\tilde{\gamma}^a$'s, making a choice of one — $\gamma^a$'s — to describe
the internal space of fermions, and using the other one —
$\tilde{\gamma}^a$'s — to describe the "family" quantum number of each irreducible
representation of $S^{ab}$'s in  space  defined by $\gamma^a$'s.

§.§.§ Reduction of Clifford and Grassmann space and
appearance of family quantum number

The creation and annihilation operators of an odd Clifford  algebra of both kinds, of
either $\gamma^a$'s or  $\tilde{\gamma}^{a}$'s,   would obviously
obey the anticommutation relations for the second quantized fermions,
postulated by Dirac, at least on the vacuum state, which is a sum of all the products
of annihilation times, $*_{A}$, the corresponding creation operators, provided that
each of the irreducible representations would carry a different quantum number.

But we know that a particular member $m$ has for all the irreducible representations
the same quantum numbers, that is the same "eigenvalues" of the
Cartan subalgebra (for the vector space of either $\gamma^a$'s or
$\tilde{\gamma}^{a}$'s), Eq. (<ref>).

The only possibility to "dress" each irreducible representation of one kind of the
two independent vector spaces with a new, let us say   "family"  quantum number,
is that we "sacrifice" one of the two vector spaces, let us make a choice of
$\tilde{\gamma}^{a}$'s,  and use $\tilde{\gamma}^{a}$'s
to define the "family" quantum number for each irreducible representation of the
vector space of $\gamma^a$'s, while  keeping the relations of
Eq. (<ref>) unchanged:  $\{\gamma^{a},
\gamma^{b}\}_{+}=2 \eta^{a b}= \{\tilde{\gamma}^{a},
\tilde{\gamma}^{b}\}_{+}$, $\{\gamma^{a}, \tilde{\gamma}^{b}\}_{+}=0$,
$ (\gamma^{a})^{\dagger} = \eta^{aa}\, \gamma^{a}$,
$(\tilde{\gamma}^{a})^{\dagger} =  \eta^{a a}\, \tilde{\gamma}^{a}$,

We therefore postulate:

Let  $\tilde{\gamma}^{a}$'s operate on $\gamma^a$'s as follows
[67, 9, 19, 21, 22]

\begin{eqnarray}
\{\tilde{\gamma}^a B &=&(-)^B\, i \, B \gamma^a\}\, |\psi_{oc}>\,,
\label{tildegammareduced}
\end{eqnarray}

with $(-)^B = -1$, if $B$ is (a function of) an odd product of $\gamma^a$'s,
otherwise $(-)^B = 1$ [67], $|\psi_{oc}>$ is defined in
Eq. (<ref>).

After this postulate the vector  space of $\tilde{\gamma}^{a}$'s is
"frozen out". No vector space of $\tilde{\gamma}^{a}$'s
needs to be taken into account  any longer, in agreement with the observed properties
of fermions. This solves the problems viii.a -  viii. c. of Sect. <ref>.

Taking into account Eq. (<ref>)  we can check that
(App. <ref>, Statement 3, 3a, 3b):

a. Relations of Eq. (<ref>) remain unchanged [
Let us show that the relation $ \{\tilde{\gamma}^{a},
\tilde{\gamma}^{b}\}_{+}= 2 \eta^{ab}$ remains valid when applied on $B$, if $B$
is either an odd or an even product of $\gamma^a$'s:
$ \{\tilde{\gamma}^{a}, \tilde{\gamma}^{b}\}_{+}$ $ \gamma^c=$
$-i \,(\tilde{\gamma}^{a} \gamma^c \gamma^b + \tilde{\gamma}^{b} \gamma^c \gamma^a)=$ $-i\,i\,\gamma^c ( \gamma^b \gamma^a +  \gamma^a \gamma^b)
= 2 \eta^{ab} \gamma^c$, while $ \{\tilde{\gamma}^{a}, \tilde{\gamma}^{b}\}_{+}$
$ \gamma^c \gamma^d=$ $ i \,(\tilde{\gamma}^{a} \gamma^c \gamma^d  \gamma^b + \tilde{\gamma}^{b} \gamma^c \gamma^d \gamma^a)=$ $i(-i)\,\gamma^c \gamma^d (\gamma^b \gamma^a +  \gamma^a \gamma^b)
= 2 \eta^{ab} \gamma^c \gamma^d$. The relation is valid for any $\gamma^c$ and
$\gamma^d$, even if  $c=d$.].

b. Relations of Eq. (<ref>)
remain unchanged [One easily checks
that $\tilde{\gamma}^{a \dagger} \gamma^c= -i \gamma^c \gamma^{a \dagger}=
-i \eta^{aa}\gamma^c \gamma^a$ $= \eta^{aa} \tilde{\gamma}^a \gamma^c =$
$-i \eta^{aa}\gamma^c \gamma^a$.].

c. The eigenvalues of the operators $S^{ab}$  and $\tilde{S}^{ab}$
on nilpotents and projectors of $\gamma^a$'s are after the reduction of Clifford space
equal to

\begin{eqnarray}
\label{signature0}
S^{ab} \,\stackrel{ab}{(k)} = \frac{k}{2}  \,\stackrel{ab}{(k)}\,,\quad && \quad
\tilde{S}^{ab}\,\stackrel{ab}{(k)} = \frac{k}{2}  \,\stackrel{ab}{(k)}\,,\nonumber\\
S^{ab}\,\stackrel{ab}{[k]} =  \frac{k}{2}  \,\stackrel{ab}{[k]}\,,\quad && \quad
\tilde{S}^{ab} \,\stackrel{ab}{[k]} = - \frac{k}{2}  \,\,\stackrel{ab}{[k]}\,,
\end{eqnarray}

demonstrating that the eigenvalues of $S^{ab}$ on nilpotents and projectors of
$\gamma^a$'s differ from the eigenvalues of $\tilde{S}^{ab}$, so that
$\tilde{S}^{ab}$ can be used to denote irreducible representations of $S^{ab}$
with the ”family” quantum number, what solves the problems viii. a. and viii. b.
of Subsubsect. <ref>.

d. We further recognize
that $\gamma^a$ transform  $\stackrel{ab}{(k)}$ into  $\stackrel{ab}{[-k]}$, never
to $\stackrel{ab}{[k]}$, while $\tilde{\gamma}^a$ transform  $\stackrel{ab}{(k)}$
into $\stackrel{ab}{[k]}$, never to $\stackrel{ab}{[-k]}$

\begin{eqnarray}
&&\gamma^a \stackrel{ab}{(k)}= \eta^{aa}\stackrel{ab}{[-k]},\; \quad
\gamma^b \stackrel{ab}{(k)}= -ik \stackrel{ab}{[-k]}, \; \quad % \nonumber\\
\gamma^a \stackrel{ab}{[k]}= \stackrel{ab}{(-k)},\;\quad \;\;
\gamma^b \stackrel{ab}{[k]}= -ik \eta^{aa} \stackrel{ab}{(-k)}\,,\nonumber\\
&&\tilde{\gamma^a} \stackrel{ab}{(k)} = - i\eta^{aa}\stackrel{ab}{[k]},\quad
\tilde{\gamma^b} \stackrel{ab}{(k)} =  - k \stackrel{ab}{[k]}, \;\qquad  \,%\nonumber\\
\tilde{\gamma^a} \stackrel{ab}{[k]} =  \;\;i\stackrel{ab}{(k)},\; \quad
\tilde{\gamma^b} \stackrel{ab}{[k]} =  -k \eta^{aa} \stackrel{ab}{(k)}\,.
\label{snmb:gammatildegamma}
\end{eqnarray}

One finds, using  Eq. (<ref>),

\begin{eqnarray}
\stackrel{ab}{\tilde{(k)}} \, \stackrel{ab}{(k)}& =& 0\,,
\qquad %\;
\stackrel{ab}{\tilde{(-k)}} \, \stackrel{ab}{(k)} = -i \,\eta^{aa}\,
\stackrel{ab}{[k]}\,,\quad  %\nonumber\\
\stackrel{ab}{\tilde{(k)}} \, \stackrel{ab}{[k]} = i\, \stackrel{ab}{(k)}\,,\quad
\stackrel{ab}{\tilde{(k)}}\, \stackrel{ab}{[-k]} = 0\,, \nonumber\\
% \stackrel{ab}{(k)} \, \stackrel{ab}{\tilde{(k)}} &=&0\,,
% \quad\;
% \stackrel{ab}{(k)}\,\stackrel{ab}{\tilde{(-k)}} = i \eta^{aa}\,  \stackrel{ab}{[k]}\,,
% \quad \;
% \stackrel{ab}{[k]} \,\stackrel{ab}{\tilde{(k)}} = i \,\stackrel{ab}{(k)}\,,
% \quad\;
% \stackrel{ab}{[-k]} \, \stackrel{ab}{\tilde{(k)}}= 0\,,\nonumber\\
\stackrel{ab}{\tilde{[k]}} \, \stackrel{ab}{(k)}& =& \, \stackrel{ab}{(k)}\,,
\quad
\stackrel{ab}{\tilde{[-k]}} \, \stackrel{ab}{(k)} = \, 0 \,,  \qquad  \quad %\nonumber\\
\quad \;\;\;\,
\stackrel{ab}{\tilde{[k]}} \, \stackrel{ab}{[k]} =  0\,,\qquad \;\;\;
\stackrel{ab}{\tilde{[- k]}} \, \stackrel{ab}{[k]} =  \, \stackrel{ab}{[k]}\,.
\label{graphbinomsfamilies}
\end{eqnarray}

From Eq. (<ref>) it follows

\begin{eqnarray}
\label{stildestrans}
S^{ac}\stackrel{ab}{(k)}\stackrel{cd}{(k)} &=& -\frac{i}{2} \eta^{aa} \eta^{cc}
\stackrel{ab}{[-k]}\stackrel{cd}{[-k]}\,,\qquad %\,\nonumber\\%quad\quad
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} \eta^{cc}
\stackrel{ab}{[k]}\stackrel{cd}{[k]}\,,\qquad \qquad \;\;%\,\nonumber\\%\quad\quad
\tilde{S}^{ac}\stackrel{ab}{[k]}\stackrel{cd}{[k]} = -\frac{i}{2}
\stackrel{ab}{(k)}\stackrel{cd}{(k)}\,,\,\nonumber\\
S^{ac}\stackrel{ab}{(k)}\stackrel{cd}{[k]}  &=& -\frac{i}{2} \eta^{aa}
\stackrel{ab}{[-k]}\stackrel{cd}{(-k)}\,,\,\qquad \quad%\nonumber\\%\quad\quad
S^{ac}\stackrel{ab}{[k]}\stackrel{cd}{(k)} = \frac{i}{2} \eta^{cc}
\stackrel{ab}{(-k)}\stackrel{cd}{[-k]}\,,\,\nonumber\\%\quad\quad
\tilde{S}^{ac}\stackrel{ab}{(k)}\stackrel{cd}{[k]} &=& -\frac{i}{2} \eta^{aa}
\stackrel{ab}{[k]}\stackrel{cd}{(k)}\,,\qquad \qquad\;\;\,%\,\nonumber\\
\tilde{S}^{ac}\stackrel{ab}{[k]}\stackrel{cd}{(k)} = \frac{i}{2} \eta^{cc}
\stackrel{ab}{(k)}\stackrel{cd}{[k]}\,.
\end{eqnarray}

Each irreducible representation
has now the "family" quantum
number, determined by $\tilde{S}^{ab}$ of the Cartan subalgebra of
Eq. (<ref>).
Correspondingly the creation and annihilation operators fulfill algebraically the
anticommutation relations, postulated by Dirac for the second quantized fermions,
since: g.a Different irreducible representations carry different
"family" quantum numbers and to each "family" quantum member only one
Hermitian conjugated partner with the same "family" quantum number belongs.
g.b Each summand of the vacuum state,  Eq. (<ref>), belongs to
a particular  "family".
This solves the problem viii. a. -  viii. c. of Sect. <ref>.

The anticommutation relations of Dirac, postulated for fermions, are
fulfilled on the vacuum state,  Eq. (<ref>), on the algebraic level, without
postulating them.
They follow by themselves due to  the fact that the creation and
annihilation operators are superposition of odd products of $\gamma^{a}$'s.

Statement 5.
The oddness of superpositions of products of  $\gamma^a$'s, representing the
"basis vectors", guarantees the anticommuting properties of objects which are
tensor, $*_{T}$, products of these "basis vectors" and the (commuting) basis of
the ordinary space.

We show in Sect. <ref> that solutions of equations of motion
must be  the tensor, $*_{T}$, product of the finite number of the
"basis vectors", $2^{\frac{d}{2}-1}\times 2^{\frac{d}{2}-1}$
(determining properties of the internal space of fermions), and
the (continuously)  infinite number of basis vectors in ordinary space.
The corresponding creation and annihilation operators fulfill the
Dirac's postulates for the second quantized fermions. We shall see
that for positive energy $p^0=|\vec{p}|$ the solutions of equations of motion
for massless fermions of particular handedness and charges
the solutions for their antifermions of opposite charges and handedness

with $-\vec{p}$ correspond, reducing the (continously infinite) number of
creation operators for the factor of $2$.

In Sect. <ref> we show that the anticommuting property of
creation and annihilation operators  manifest also on
the Hilbert space, formed as the tensor products, $*_{T_{H}}$, of all
possible numbers of all possible fermion and antifermion states solving the
equations of motion, the creation operators of which are

the tensor products, $*_{T}$, of the
finite number, $2^{\frac{d}{2}-1}\times 2^{\frac{d}{2}-1}$, of Clifford odd
”basis vectors” and of the continuous infinite basis of the ordinary (momentum or
coordinate) space,

applying on the vacuum state.

Let us write down the anticommutation relations of Clifford odd "basic vectors",
representing the creation operators in internal space of fermions with half integer
spin represented by $\gamma^a$'s and of the corresponding Hermitian
conjugated partners annihilation operators again. After the reduction of the
Clifford algebra any irreducible representation carry the family quantum number,
distinguishing families among themselves.

\begin{eqnarray}
\{ \hat{b}^{m}_{f}, \hat{b}^{m' \dagger}_{f'} \}_{*_{A}+}\, |\psi_{oc}>
&=& \delta^{m m'} \, \delta_{ff'} \,  |\psi_{oc}>\,,\nonumber\\
\{ \hat{b}^{m}_{f}, \hat{b}^{m'}_{f'} \}_{*_{A}+}  \,  |\psi_{oc}>
&=& 0 \,\cdot\,  |\psi_{oc}>\,,\nonumber\\
\{\hat{b}^{m  \dagger}_{f},\hat{b}^{m' \dagger}_{f'}\}_{*_{A}+} \, |\psi_{oc}>
&=& 0 \, \cdot\, |\psi_{oc}>\,,\nonumber\\
\hat{b}^{m \dagger}_{f} \,{}_{*_{A}} |\psi_{oc}>&=& |\psi^{m}_{f}>\,, \nonumber\\
\hat{b}^{m}_{f}   \,{*_{A}}  |\psi_{oc}>&=& 0 \,\cdot\,  |\psi_{oc}>\,,
\label{alphagammatildeprod}
\end{eqnarray}

with ($m,m'$) denoting the "family" members and ($f,f'$) denoting "families",
${*_{A}}$ represents the algebraic multiplication of $ \hat{b}^{\dagger m}_{f} $
and  $ \hat{b}^{m}_{f} $ with the vacuum state $|\psi_{oc}>$ of
Eq. (<ref>) and among themselves, taking into account
Eq. (<ref>).

Let us make a choice of  the starting creation operator $\hat{b}^{m=1 \dagger}_{f=1}$

of an odd Clifford character and correspondingly of its Hermitian conjugated partner in $d=2(2n+1)$ and $d=4n$, respectively, as we start in Eq. (<ref>),
but now, after reduction of the two kinds of the Clifford algebras into only one, each irreducible representation carries the family quantum number.

\begin{eqnarray}
%|\psi^{1}_{1}>&=&{\hat b}^{1\dagger}_{1} |\psi_{oc}>\,,\nonumber\\
{\hat b}^{1 \dagger}_{1}\, {\bf :} &=& \stackrel{03}{(+i)} \stackrel{12}{(+)}
\stackrel{56}{(+)}\cdots
\stackrel{d-3\;d-2}{(+)}\;\;\stackrel{d-1\;d}{(+)}\,,\nonumber\\
({\hat b}^{1 \dagger}_{1})^{\dagger} &=& {\hat b}^{1}_{1}\, {\bf :}=
\stackrel{d-1\;d}{(-)} \;\;
\stackrel{d-3\;d-2}{(-)}\cdots  \stackrel{56}{(-)}  \stackrel{12}{(-)}  \stackrel{01}{(-i)}\,,
\nonumber\\
\nonumber\\
{\hat b}^{1 \dagger}_{1} \, {\bf :} &=& \stackrel{03}{(+i)} \stackrel{12}{(+)}
\stackrel{56}{(+)}\cdots
\stackrel{d-3\;d-2}{(+)}\;\;\stackrel{d-1\;d}{[+]}\,,\nonumber\\
({\hat b}^{1 \dagger}_{1})^{\dagger}&=& {\hat b}^{1}_{1}\, {\bf :} =
\stackrel{d-1\;d}{[+]} \;\;
\stackrel{d-3\;d-2}{(-)}\cdots  \stackrel{56}{(-)}  \stackrel{12}{(-)}  \stackrel{01}{(-i)}\,,
\nonumber\\
\label{start(2n+1)2cliffgammatilde4n}
\end{eqnarray}

All the rest  creation operators in the internal space  described by $\gamma^a$'s
(the "basis vectors"), belonging to the same Lorentz representation, follow by
the application of the Lorentz generators $S^{ab}$'s.

i. The representations with different "family" quantum numbers are reachable
by $\tilde{S}^{ab}$, since, according to Eq. (<ref>), we recognize
that $\tilde{S}^{ac}$ transforms two nilpotents
$ \stackrel{ab}{(k)} \stackrel{cd}{(k)}$ into two projectors
$ \stackrel{ab}{[k]} \stackrel{cd}{[k]}$, without changing $k$
($\tilde{S}^{ac}$ transforms $ \stackrel{ab}{[k]} \stackrel{cd}{[k]}$ into
$ \stackrel{ab}{(k)} \stackrel{cd}{(k)}$, as well as
$ \stackrel{ab}{[k]} \stackrel{cd}{(k)}$ into $ \stackrel{ab}{(k)} \stackrel{cd}{[k]}$).
All the "family" members are reachable from one member of a new family by the
application of $S^{ab}$'s.

In this way, by starting with the creation operator ${\hat b}^{1 \dagger}_{1}$,
Eq. (<ref>), $2^{\frac{d}{2}-1}$ "families", each with
$2^{\frac{d}{2}-1}$ "family" members follow.

Let us find the starting member of the next "family" to the "family" of
Eq. (<ref>) by the application of $\tilde{S}^{01}$

\begin{eqnarray}
{\hat b}^{1 \dagger}_{2}\, {\bf :} &=& \stackrel{03}{[+i]} \stackrel{12}{[+]}
\stackrel{56}{(+)}\cdots \stackrel{d-3\;d-2}{(+)}\;\;\stackrel{d-1\;d}{(+)}\,,\nonumber\\
{\hat b}^{1}_{2}\,{\bf :} &=&
\stackrel{d-1\;d}{(-)} \;\;
\stackrel{d-3\;d-2}{(-)}\cdots  \stackrel{56}{(-)}  \stackrel{12}{[+]}  \stackrel{01}{[+i]}\,.
\label{d=2(2n+1)}
\end{eqnarray}

The corresponding annihilation operators, that is the Hermitian conjugated partners of
$2^{\frac{d}{2}-1}$ "families", each with $2^{\frac{d}{2}-1}$ "family" members,
following from the starting creation operator ${\hat b}^{1 \dagger}_{1}$ by the application of $S^{ab}$'s — the family members — and the application of
$\tilde{S}^{ab}$ — the same family member of another family —
can be obtained by Hermitian conjugation.

The creation and annihilation operators of an odd Clifford character, expressed by
nilpotents and projectors of  $\gamma^a$'s, obey  anticommutation relations of$\,$
Eq. (<ref>),
without postulating the second quantized anticommutation relations.

j. The vacuum state  $|\psi_{oc}>$ for the vector space determined by
which is the sum over family quantum numbers of
products of an  annihilation operator with its Hermitian conjugated partner
creation operator of any family member $m$,
\hat{b}^{m \dagger}_{f}$.

\begin{eqnarray}
|\psi_{oc}>&=& \frac{1}{\sqrt{2^{\frac{d}{2}-1}}} (\stackrel{03}{[-i]} \stackrel{12}{[-]} \stackrel{56}{[-]}\cdots
\stackrel{d-1\;d}{[-]} + \stackrel{03}{[+i]} \stackrel{12}{[+]} \stackrel{56}{[-]} \cdots
\stackrel{d-1\;d}{[-]} \nonumber\\
+ &&\stackrel{03}{[+i]} \stackrel{12}{[-]} \stackrel{56}{[+]}\cdots
\stackrel{d-1\;d}{[-]} + \cdots) |1>\,, \quad \nonumber\\
% |\psi_{o\tilde{\gamma}}>&=& \stackrel{03}{\tilde{[-i]}} \stackrel{12}{\tilde{[-]}}
% \stackrel{56}{\tilde{[-]}}\cdots
% \stackrel{d-1\;d}{\tilde{[-]}} + \stackrel{03}{\tilde{[+i]}} \stackrel{12}{\tilde{[+]}}
% \stackrel{56}{\tilde{[-]}} \cdots
% \stackrel{d-1\;d}{\tilde{[-]}} + \stackrel{03}{\tilde{[+i]}} \stackrel{12}{\tilde{[-]}}
% \stackrel{56}{\tilde{[+]}}\cdots
%\stackrel{d-1\;d}{\tilde{[-]}} + \cdots |1>\,, \quad \nonumber\\
&&{\rm for}\; d=2(2n+1)\,,
\nonumber\\
|\psi_{oc}>&=&\frac{1}{\sqrt{2^{\frac{d}{2}-1}}} (\stackrel{03}{[-i]} \stackrel{12}{[-]} \stackrel{35}{[-]}\cdots
\stackrel{d-3\;d-2}{[-]}\stackrel{d-1\;d}{[+]} \nonumber\\
+ &&\stackrel{03}{[+i]} \stackrel{12}{[+]} \stackrel{56}{[-]}
\cdots \stackrel{d-3\;d-2}{[-]} \;\,\stackrel{d-1\;d}{[+]} + \cdots )|1>\,,
\quad  \nonumber\\
% |\psi_{o\tilde{\gamma}}>&=& \stackrel{03}{\tilde{[-i]}} \stackrel{12}{\tilde{[-]}}
% \stackrel{35}{\tilde{[-]}}\cdots
% \stackrel{d-3\;d-2}{\tilde{[-]}}\stackrel{d-1\;d}{\tilde{[+]}} + \stackrel{03}{\tilde{[+i]}}
% \stackrel{12}{\tilde{[+]}} \stackrel{56}{\tilde{[-]}} \cdots \stackrel{d-3\;d-2}{\tilde{[-]}} \;\,
% \stackrel{d-1\;d}{\tilde{[+]}} + \cdots |1>\,, \quad  \nonumber\\
&&{\rm for}\; d=4n\,, \quad n \,{\rm is \,a\, positive\, integer}\,.
\label{vac1}
\end{eqnarray}

The Hermitian conjugated part of the space in the Grassmann case is "freezed out"
together with the "basis vector" space of $\tilde{\gamma}^a$'s.

The even partners of the Clifford odd creation and annihilation operators follow by either the
application of $\gamma^a$ on the creation operators, leading to  $2^{\frac{d}{2}-1}$
"families", each with $2^{\frac{d}{2}-1}$ members,  or with the application of
$\tilde{\gamma}^a$ on the creation  operators,  leading to another group of the
Clifford even operators, again with the  $2^{\frac{d}{2}-1}$ "families", each with
$2^{\frac{d}{2}-1}$ members.

It is not difficult to recognize, that each of the Clifford even "families", obtained by
the application of $\gamma^a$ or by $\tilde{\gamma}^a$ on the creation operators,
contains one selfadjoint operator, which is the product  of projectors only, contributing
as a summand to the vacuum state, Eq. (<ref>).

$\;\;$ Clifford half integer spin fermions in $d=(5+1)$:

We illustrate properties of the Clifford odd, and correspondingly anticommuting,
creation and their Hermitian conjugated partners annihilation operators, belonging
to $2^{\frac{6}{2}-1}=4$ "families", each with  $2^{\frac{6}{2}-1}=4$
members in $d=(5+1)$-dimensional space. The spin in  the fifth and the sixth
dimension manifests as the charge in $d=(3+1)$.

Half of the Clifford odd "basis vectors" are
creation operators
${\hat b}^{m \dagger}_{f}$, appearing in Table <ref>  in the
fourth column. Denoted by odd I, odd II, odd III and odd IV
they represent four "families",
each family having four members. Their Hermitian conjugated partners
${\hat b}^{m}_f$ are presented in the fifth column.

All the families have the same eigenvalues  of the Cartan subalgebra
members ($S^{03}, S^{12}, S^{56}$), the handedness $\Gamma^{(3+1)}
=i (2i)^2 S^{03}S^{12}$, written in the sixth, seventh, eighth and  ninth

The rest half of the Clifford odd "basis vectors" are their Hermitian conjugated
partners  ${\hat b}^{m}_f$, presented in the fifth column of
Table <ref>. The last three  columns  of this table determine
the family quantum numbers, the eigenvalues of
($\tilde{S}^{03}, \tilde{S}^{12}, \tilde{S}^{56}$), the same for all family

The normalized vacuum state is the product of ${\hat b}^{m}_f \cdot$
${\hat b}^{m \dagger}_{f}$,

summed over four families.

\begin{eqnarray}
\label{vac5+1}
|\psi_{oc}> &=&\frac{1}{\sqrt{2^{\frac{6}{2}-1}}}\,
+ \stackrel{03}{[+i]}\stackrel{12}{[+1]}\stackrel{56}{[-1]}
\nonumber\\
+ \stackrel{03}{[-i]}\stackrel{12}{[+1]}\stackrel{56}{[+1]}).
\end{eqnarray}

One easily checks, by taking into account Eq. (<ref>), that
the creation operators ${\hat b}^{m \dagger}_{f}$
and the annihilation operators ${\hat b}^{m}_{f}$ fulfill the
anticommutation relations of Eq (<ref>)
[If we would choose for creation operators those from the fifth column,
then the annihilation operators would be the present creation operators.
The vacuum state would be  the sum of  products of  the present creation
operators of the fourth column  times the present annihilation operators of the
fifth column.].

All the Clifford even "families" with "family" members of  Table <ref>
can be obtained as algebraic products, $*_{A}$, of the Clifford odd "basis vectors"
of Table <ref> ([26], Table I).

[t]16.5 cm

The basic creation operators —
($ch$ (charge), the eigenvalue of $S^{56}$, and $s$ (spin), the eigenvalues of
$S^{03}$ and $S^{12}$, explain the index $m$) — and
their annihilation partners — $\hat{b}^{m=(ch,s)}_{f}$ — are presented for
$d= (5+1)$-dimensional case. Each basic creation operator is a product of
projectors and an odd product of nilpotents, and is the "eigenstate" of the
Cartan subalgebra members, ($S^{03}$, $S^{12}$, $S^{56}$). The
eigenvalues of  ($\tilde{S}^{03}$, $\tilde{S}^{12}$, $\tilde{S}^{56}$),
Eq. (<ref>), presented in the last three columns of the table,
determine the family quantum numbers.
Operators $\hat{b}^{m=(ch,s) \dagger}_{f}$
and $\hat{b}^{m=(ch, s)}_{f}$
fulfill the anticommutaion relations of Eq. (<ref>).

${\rm family}\, f $   $m $   $=(ch,s)$   $\hat{b}^{ m=(ch,s) \dagger}_f$
$\hat{b}^{m=(ch,s) }_{f}$
$S^{03}$   $ S^{1 2}$   $S^{5 6}$   $\Gamma^{3+1}$
$\tilde{S}^{03}$   $\tilde{S}^{1 2}$    $\tilde{S}^{5 6}$

$I$   $1$   $(\frac{1}{2},\frac{1}{2})$   $%\hat{b}^{ \dagger}_{\frac{1}{2},\frac{1}{2}} =
\stackrel{03}{(+i)}\,\stackrel{12}{(+)}| \stackrel{56}{(+)}$
$%\hat{b}_{\frac{1}{2},\frac{1}{2}} =
{\scriptstyle (-)} \stackrel{56}{(-)}|{\scriptstyle (-)}
\stackrel{12}{(-)} \stackrel{03}{(-i)}$   $\frac{i}{2}$   $\frac{1}{2}$   $\frac{1}{2}$   $1$
$\frac{i}{2}$   $\frac{1}{2}$   $\frac{1}{2}$

$I$    $2$   $(\frac{1}{2},-\frac{1}{2})$   $
%\hat{b}^{ \dagger}_{\frac{1}{2}, - \frac{1}{2}} =
\stackrel{03}{[-i]}\,\stackrel{12}{[-]}|\stackrel{56}{(+)}$   $
%\hat{b}_{\frac{1}{2},-\frac{1}{2}} =
{\scriptstyle (-)} \stackrel{56}{(-)}| \stackrel{12}{[-]} \stackrel{03}{[-i]}$
$-\frac{i}{2}$   $-\frac{1}{2}$   $\frac{1}{2}$   $1$
$\frac{i}{2}$   $\frac{1}{2}$   $\frac{1}{2}$

$I$    $3$   $(-\frac{1}{2},\frac{1}{2})$   $
%\hat{b}^{ \dagger}_{- \frac{1}{2},\frac{1}{2}} =
\stackrel{03}{[-i]}\,\stackrel{12}{(+)}|\stackrel{56}{[-]}$   $
%\hat{b}_{-\frac{1}{2},\frac{1}{2}} =
\stackrel{56}{[-]}|{\scriptstyle (-)} \stackrel{12}{(-)} \stackrel{03}{[-i]}$
$-\frac{i}{2}$   $ \frac{1}{2}$   $-\frac{1}{2}$   $-1$
$\frac{i}{2}$   $\frac{1}{2}$   $\frac{1}{2}$

$I$    $1$   $(-\frac{1}{2},-\frac{1}{2})$   $
%\hat{b}^{ \dagger}_{-\frac{1}{2},- \frac{1}{2}} =
\stackrel{03}{(+i)}\,\stackrel{12}{[-]}|\stackrel{56}{[-]}$   $
%\hat{b}_{-\frac{1}{2},-\frac{1}{2}} =
\stackrel{56}{[-]}| \stackrel{12}{[-]} \stackrel{03}{(-i)}$
$\frac{i}{2}$   $- \frac{1}{2}$   $-\frac{1}{2}$   $-1$
$\frac{i}{2}$   $\frac{1}{2}$   $\frac{1}{2}$

$II$   $1$   $(\frac{1}{2},\frac{1}{2})$   $
%\hat{b}^{ \dagger}_{\frac{1}{2},\frac{1}{2}} =
\stackrel{03}{[+i]}\,\stackrel{12}{[+]}| \stackrel{56}{(+)}$
$%\hat{b}_{\frac{1}{2},\frac{1}{2}} =
{\scriptstyle (-)} \stackrel{56}{(-)}|\stackrel{12}{[+]} \stackrel{03}{[+i]}$   $\frac{i}{2}$
$\frac{1}{2}$   $\frac{1}{2}$   $1$   $-\frac{i}{2}$   $-\frac{1}{2}$   $\frac{1}{2}$

$II$    $2$   $(\frac{1}{2},-\frac{1}{2})$   $
%\hat{b}^{ \dagger}_{\frac{1}{2}, - \frac{1}{2}} =
\stackrel{03}{(-i)}\,\stackrel{12}{(-)}|\stackrel{56}{(+)}$   $
%\hat{b}_{\frac{1}{2},-\frac{1}{2}} =
{\scriptstyle (-)} \stackrel{56}{(-)}| {\scriptstyle (-)} \stackrel{12}{(+)}
\stackrel{03}{(+i)}$
$-\frac{i}{2}$   $-\frac{1}{2}$   $\frac{1}{2}$   $1$
$-\frac{i}{2}$   $-\frac{1}{2}$   $\frac{1}{2}$

$II$    $3$   $(-\frac{1}{2},\frac{1}{2})$   $
%\hat{b}^{ \dagger}_{- \frac{1}{2},\frac{1}{2}} =
\stackrel{03}{(-i)}\,\stackrel{12}{[+]}|\stackrel{56}{[-]}$   $
%\hat{b}_{-\frac{1}{2},\frac{1}{2}} =
\stackrel{56}{[-]}| \stackrel{12}{[+]} \stackrel{03}{(+i)}$
$-\frac{i}{2}$   $ \frac{1}{2}$   $-\frac{1}{2}$   $-1$
$-\frac{i}{2}$   $-\frac{1}{2}$   $\frac{1}{2}$

$II$    $4$   $(-\frac{1}{2},-\frac{1}{2})$   $
%\hat{b}^{ \dagger}_{-\frac{1}{2},- \frac{1}{2}} =
\stackrel{03}{[+i]}\, \stackrel{12}{(-)}|\stackrel{56}{[-]}$   $%\hat{b}_{-\frac{1}{2},-\frac{1}{2}} =
\stackrel{56}{[-]}|{\scriptstyle (-)}  \stackrel{12}{(+)} \stackrel{03}{[+i]}$
$\frac{i}{2}$   $- \frac{1}{2}$   $-\frac{1}{2}$   $-1$
$-\frac{i}{2}$   $-\frac{1}{2}$   $\frac{1}{2}$

$III$   $1$   $(\frac{1}{2},\frac{1}{2})$   $
%\hat{b}^{ \dagger}_{\frac{1}{2},\frac{1}{2}} =
\stackrel{03}{[+i]}\,\stackrel{12}{(+)}| \stackrel{56}{[+]}$
$%\hat{b}_{\frac{1}{2},\frac{1}{2}} =
\stackrel{56}{[+]}|{\scriptstyle (-)}\stackrel{12}{(-)} \stackrel{03}{[+i]}$   $\frac{i}{2}$
$\frac{1}{2}$   $\frac{1}{2}$   $1$   $-\frac{i}{2}$   $\frac{1}{2}$   $-\frac{1}{2}$

$III$    $2$   $(\frac{1}{2},-\frac{1}{2})$   $
%\hat{b}^{ \dagger}_{\frac{1}{2}, - \frac{1}{2}} =
\stackrel{03}{(-i)}\,\stackrel{12}{[-]}|\stackrel{56}{[+]}$   $
%\hat{b}_{\frac{1}{2},-\frac{1}{2}} =
\stackrel{56}{[+]}|  \stackrel{12}{[-]} \stackrel{03}{(+i)}$
$-\frac{i}{2}$   $-\frac{1}{2}$   $\frac{1}{2}$   $1$
$-\frac{i}{2}$   $\frac{1}{2}$   $-\frac{1}{2}$

$III$    $3$   $(-\frac{1}{2},\frac{1}{2})$   $
%\hat{b}^{ \dagger}_{- \frac{1}{2},\frac{1}{2}} =
\stackrel{03}{(-i)}\,\stackrel{12}{(+)}|\stackrel{56}{(-)}$   $
%\hat{b}_{-\frac{1}{2},\frac{1}{2}} =
{\scriptstyle (-)}\stackrel{56}{(+)}| {\scriptstyle (-)} \stackrel{12}{(-)} \stackrel{03}{(+i)}$
$-\frac{i}{2}$   $ \frac{1}{2}$   $-\frac{1}{2}$   $-1$
$-\frac{i}{2}$   $\frac{1}{2}$   $-\frac{1}{2}$

$III$    $4$   $(-\frac{1}{2},-\frac{1}{2})$   $
%\hat{b}^{ \dagger}_{-\frac{1}{2},- \frac{1}{2}} =
\stackrel{03}{[+i]} \stackrel{12}{[-]}|\stackrel{56}{(-)}$   $%\hat{b}_{-\frac{1}{2},-\frac{1}{2}} =
{\scriptstyle (-)} \stackrel{56}{(+)}|  \stackrel{12}{[-]} \stackrel{03}{[+i]}$
$\frac{i}{2}$   $- \frac{1}{2}$   $-\frac{1}{2}$   $-1$
$-\frac{i}{2}$   $\frac{1}{2}$   $-\frac{1}{2}$

$IV$   $1$   $(\frac{1}{2},\frac{1}{2})$   $
%\hat{b}^{ \dagger}_{\frac{1}{2},\frac{1}{2}} =
\stackrel{03}{(+i)}\,\stackrel{12}{[+]}| \stackrel{56}{[+]}$
$%\hat{b}_{\frac{1}{2},\frac{1}{2}} =
\stackrel{56}{[+]}|\stackrel{12}{[+]} \stackrel{03}{(-i)}$   $\frac{i}{2}$
$\frac{1}{2}$   $\frac{1}{2}$   $1$   $\frac{i}{2}$   $-\frac{1}{2}$   $-\frac{1}{2}$

$IV$    $2$   $(\frac{1}{2},-\frac{1}{2})$   $
%\hat{b}^{ \dagger}_{\frac{1}{2}, - \frac{1}{2}} =
\stackrel{03}{[-i]}\,\stackrel{12}{(-)}|\stackrel{56}{[+]}$   $
%\hat{b}_{\frac{1}{2},-\frac{1}{2}} =
\stackrel{56}{[+]}| {\scriptstyle (-)} \stackrel{12}{(+)}
\stackrel{03}{[-]}$
$-\frac{i}{2}$   $-\frac{1}{2}$   $\frac{1}{2}$   $1$
$\frac{i}{2}$   $-\frac{1}{2}$   $-\frac{1}{2}$

$IV$    $3$   $(-\frac{1}{2},\frac{1}{2})$   $
%\hat{b}^{ \dagger}_{- \frac{1}{2},\frac{1}{2}} =
\stackrel{03}{[-i]}\,\stackrel{12}{[+]}|\stackrel{56}{(-)}$   $
%\hat{b}_{-\frac{1}{2},\frac{1}{2}} =
{\scriptstyle (-)} \stackrel{56}{(+)}| \stackrel{12}{[+]} \stackrel{03}{[-i]}$
$-\frac{i}{2}$   $ \frac{1}{2}$   $-\frac{1}{2}$   $-1$
$\frac{i}{2}$   $-\frac{1}{2}$   $-\frac{1}{2}$

$IV$    $4$   $(-\frac{1}{2},-\frac{1}{2})$   $
%\hat{b}^{ \dagger}_{-\frac{1}{2},- \frac{1}{2}} =
\stackrel{03}{(+i)}\, \stackrel{12}{(-)}|\stackrel{56}{(-)}$   $
%\hat{b}_{-\frac{1}{2},-\frac{1}{2}} =
{\scriptstyle (-)} \stackrel{56}{(+)}|\stackrel{12}{(+)} \stackrel{03}{(-i)}$
$\frac{i}{2}$   $- \frac{1}{2}$   $-\frac{1}{2}$   $-1$