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\noindent\leftline{Preprint UGVA-DPT 1994/04-847}
\title{New aspects of field theory\footnote{$^\ast$}{{\rm Talk presented at
the Conference on Frontiers in Theoretical Physics at Erdine.}}}
\author{C. Piron and D. J. Moore}
\address{D\'epartement de Physique Th\'eorique, Universit\'e de Gen\`eve
CH-1211 Gen\`eve 4, Switzerland}
\abstract{
After a brief reminder of the Cartan 1-form and its application to the
mechanics of point particles we review the application of Cartan 4-forms
to field theory.}
%
\section{Introduction}
%
The beginning of this century saw two revolutions in physics -- relativity and
quantum theory. In spite of this fact, which should have changed our
fundamental conception of nature, consciously or unconsciously most physicists
still hold to nineteenth Century ideas on the nature of the physical world. In
this philosophy the world is nothing more than a collection of point particles
moving in the void and only interacting by contact. Relativity is seen as just
modifying the Newtonian laws of motion and quantum theory is regarded as
forbidding precise measurement and so introducing probabilities and statistics.
A clear image of physical reality disappears in a cloud of many hypothetical
and probabilistic worlds, all in fact built in the same spirit as the schemes
proposed by Leibnitz or Descartes.

Such a view, where one attempts to graft quantum theory and relativity onto
a preconceived mechanistic worldview, must be forsaken once and for all -- we
must take quantum theory and relativity seriously (1). The empty space is not
the void, it possesses a reality in of itself. Matter is not a system of point
particles but is a collection of entities acting on space in a more or less
local way (2). Such entities are created or annihilated individually but as a
totality. One can describe entities
first of all by integers such as leptonic, hadronic or charge quantum numbers
and also by their non-local actions on the physical space. These actions are described
by fields.

The new generation of experiments involving single neutrons, atoms,
electrons or photons demonstrate the reality of non-local
phenomena. Such phenomena are completely outside the current philosophy
inherited from the nineteenth Century. They require completely new
concepts since the notions of position and velocity of small particles
of some substance are now definitively inadequate. Calling forth
statistical effects or measurement imprecision will not save the situation.

It was the great merit of D. Aerts (3) to introduce the concepts of property
and state in physics in a new way well adapted to this new aspect of
reality. Given a system, a part of reality, we associate to each
possible experimental project a particular property of the system. By
experimental project we mean an experiment that can be performed on the
system and where one has precisely defined in advance what is to be
considered as a positive result. The corresponding property is said to
be actual if, were one to realise the experiment, the result would be
certainly positive. As we can see, an actual property is nothing more
than what Einstein called an element of reality of the system.

We can give a very simple example of an experimental project which,
while outside the usual physics, is very precise and adequate for our
purposes. The regulations of golf state that the size of a golf ball
should not be less that 1.680 inches in diameter. They go on to state that
``This requirement will be satisfied if, under its own weight, a ball
falls through a 1.680 inches diameter ring gauge in fewer than 25 out of
100 randomly selected positions, the test being carried out at a
temperature of $23\pm1$ ${}^0$C.'' This is a practical example where the
positive result is not just one event but a collection of events with a
statistical criterion, illustrating very well the generality of the concept of
experimental project.

The experimental project just described is non-destructive, one has the
same ball after performing the experiment. The opposite case is
illustrated by the following experimental project. Consider a chalk. The
chalk is said to be breakable if I can snap it into two pieces. However
the property of being breakable is a property of the chalk before we
decide to perform the experiment. If we verify the breakability of a
given chalk then after the experiment we only have pieces of it left.

As a third example let us consider a field in the vacuum. To say that
there is an electric field in the vacuum means the following. If a
charged particle were introduced at rest into the vacuum then it would be
accelerated according to the direction and the intensity of the field.
The element of reality, here the electric field, is a property of the
vacuum in the absence of the charged particle and has nothing to do with
the new complicated situation which results from the introduction of
the charged particle.

%
\section{The electromagnetic field}
%
The electromagnetic field can be simply described in four dimensions in
terms of differential forms (4). The quadripotential
$A_\mu(x)=\big(-V(x),\vec{A}(x)\big)$, with $A_0(x)=-V(x)$ the potential and
$\vec{A}(x)$ the potential vector, defines a 1-form
$$A&=A_\mu(x)dx^\mu=-V(x)dt+A_i(x)dx^i.\cr$$
Note that since $V(x)$ is like an energy and $\vec{A}(x)$ like a
momentum, the negative sign in $A_0(x)$ comes from the thermodynamic
convention that the energy increases when work is provided to the
system.

The exterior derivative of $A$ gives a 2-form
$$B&=dA=\partial_\mu A_\nu(x)dx^\mu\wedge dx^\nu\cr
&=\frac12\left(\partial_\mu A_\nu(x)-\partial_\nu A_\mu(x)\right)dx^\mu\wedge
dx^\nu.\cr$$
Writing
$$B&=E_i(x)dx^i\wedge dt+B^i(x)dx^j\wedge dx^k+B^j(x)dx^k\wedge dx^i
+B^k(x)dx^i\wedge dx^j\cr$$
we recognise the electric field $E_i=-\partial_iV(x)-\partial_tA_i(x)$
and the magnetic flux $B^k(x)=\partial_iA_j(x)-\partial_jA_i(x)$. From
$dB=ddA=0$ we then derive the first two Maxwell equations
$$\partial_\rho B_{\mu\nu}(x)+\partial_\mu B_{\nu\rho}(x)+\partial_\nu
B_{\rho\mu}(x)&=0,\cr$$
or equivalently
$$\partial_t\curlvec{B}(x)+{\rm curl}\,\vec{E}(x)&=0\cr
{\rm div}\,\curlvec{B}(x)&=0.\cr$$

In Maxwell's theory we also have another differential form, a 2-form of
the second kind (an odd form), which changes sign with the sign
of the Jacobian of the transformation:
$$H&=\frac14H^{\mu\nu}(x)\epsilon_{\mu\nu\rho\lambda}dx^\rho\wedge
dx^\lambda,\cr$$
where $H^{ij}(x)=-H^{ji}(x)=H_k(x)$ is the magnetic field and
$H^{0i}(x)=-H^{i0}(x)=D^i(x)$ is the electric displacement. Here
$\epsilon_{\mu\nu\rho\lambda}$ is the completely antisymmetric tensor
such that $\epsilon_{0123}=1$ in any coordinate system. Upon changing
coordinates it is fourfold covariant and transforms with an extra factor,
the inverse of the Jacobian of the transformation. The second pair
of Maxwell equations can then be written
$$dH&=J,\cr$$
where $J=\frac16J^\mu(x)\epsilon_{\mu\nu\rho\lambda}dx^\nu\wedge dx^\rho\wedge
dx^\lambda$ and $J^\mu(x)=\big(\rho(x),\vec{J}(x)\big)$ is the
quadricurrent (or more simply the current). Here
$\rho(x)$ is the charge density and $\vec{J}(x)$ the usual current
density. The forms $H$ and $J$ must both be of the second kind in such a
way that the total charge $\int\rho(x)\,\d v$ does not depend on some
orientation which is not even defined, but is positive or negative depending
on the nature of the charge.

The second pair of Maxwell equations can also be written
$$\partial_\nu H^{\mu\nu}(x)&=J^\mu(x),\cr$$
or equivalently
$${\rm div}\,\vec{D}(x)&=\rho(x)\cr
{\rm curl}\,\curlvec{H}(x)-\partial_t\vec{D}(x)&=\vec{J}(x).\cr$$

>From $dJ=ddH=0$ we derive the conservation of the charged current:
$$\partial_\mu J^\mu(x)&=0,\cr$$
which is equivalent to
$$\partial_t\rho(x)+{\rm div}\,\vec{J}(x)&=0.\cr$$
Finally, in the linear Maxwell theory the field $H^{\rho\lambda}(x)$ creates,
via the vacuum, the field $B_{\mu\nu}(x)$ according to the phenomenological
relations
$$B_{\mu\nu}(x)&=\frac12\mu_{\mu\nu\rho\lambda}(x)H^{\rho\lambda}(x)\cr$$
or more explicitly
$$\curlvec{B}(x)&=\mu_0(x)\curlvec{H}(x),\cr
\vec{E}(x)&=\epsilon_0^{-1}(x)\vec{D}(x).\cr$$
This means that $\mu_{\mu\nu\rho\lambda}(x)$ is in fact a $6\times6$
diagonal matrix in the two pairs of indices $\mu\nu$ and $\rho\lambda$.
It can be shown (see appendix I) that the phenomenological relations can
also be written in an essentially unique way as
$$B_{\mu\nu}(x)&=\frac12\left(\hat{g}_{\mu\rho}(x)\hat{g}_{\nu\lambda}(x)
-\hat{g}_{\mu\lambda}(x)\hat{g}_{\nu\rho}(x)\right)H^{\rho\lambda}(x)\cr
&=\hat{g}_{\mu\rho}(x)\hat{g}_{\nu\lambda}(x)H^{\rho\lambda}(x).\cr$$

It is important to remark that while any diffeomorphism of
${\Bbb R}^4$ leaves the Maxwell equations in the same form, only Lorentz
conformal diffeomorphisms also conserve the phenomenological relations.

%
\section{Mechanics in the Cartan formalism}
%
The state space for a point particle in classical mechanics is given by
$\Sigma=(\vec{p},\vec{q},t)$. A motion of the system is given by
functions $t\mapsto\vec{p}\,(t)$ and $t\mapsto\vec{q}\,(t)$ satisfying the
canonical equations of motion
$$
\dot{p}_i(t)&=-{\partial H\over\partial q^i}(\vec{p}\,(t),\vec{q}\,(t),t),\cr
\dot{q}^i(t)&=+{\partial H\over\partial p_i}(\vec{p}\,(t),\vec{q}\,(t),t),\cr$$
where $H$ is the Hamiltonian. In this section we will show that the
canonical equations can be written in a compact way in terms of
differential forms.

Let us define the 1-forms
$$
\alpha_i&=dp_i+{\partial H\over\partial q^i}(\vec{p},\vec{q},t)dt,\cr
\beta^i&=dq^i-{\partial H\over\partial p_i}(\vec{p},\vec{q},t)dt\cr$$
and choose $X=(\dot{p}_i(t),\dot{q}^i(t),1)$ as tangent vector to the motion.
Then since $dt(X)=1\not=0$ the canonical equations are equivalent to
$$i_X\alpha_i&\equiv\alpha_i(X)=dp_i(X)+{\partial H\over\partial
q^i}(\vec{p}\,(t),\vec{q}\,(t),t)dt(X)\cr
&=\dot{p}_i(t)+{\partial H\over\partial q^i}(\vec{p}\,(t),\vec{q}\,(t),t)=0\cr
i_X\beta^i&\equiv\beta^i(X)=dq^i(X)-{\partial H\over\partial
p_i}(\vec{p}\,(t),\vec{q}\,(t),t)dt(X)\cr
&=\dot{q}^i(t)-{\partial H\over\partial p_i}(\vec{p}\,(t),\vec{q}\,(t),t)
=0.\cr$$

We can further simplify the expression for the canonical equations by defining
the 2-form
$$\Omega&=\sum_i\alpha_i\wedge\beta^i.\cr$$
Then for $X$ tangent to the motion we have that
$$i_X\Omega(Y)&=\Omega(X,Y)=
\sum_i\left(\alpha_i(X)\beta^i(Y)-\alpha_i(Y)\beta^i(X)\right)=0\cr$$
for all tangent vectors $Y$.
In fact this condition is equivalent to the canonical
equations. Let us write $\partial_{p_1}$ for the tangent vector
$(1,0,0,0,0,0,0)$, $\partial_{q^1}$ for the tangent vector
$(0,0,0,1,0,0,0)$ and so on. Then the
condition $i_X\Omega(Y)=0$ gives
$$i_X\Omega(\partial_{q^i})&=\sum_j\left(
\alpha_j(X)\beta^j(\partial_{q^i})-\alpha_j(\partial_{q^i})\beta^j(X)\right)\cr
&=\alpha_i(X)=0,\cr
i_X\Omega(\partial_{p_i})&=\sum_j\left(\alpha_j(X)\beta^j(\partial_{p_i})
-\alpha_j(\partial_{p_i})\beta^j(X)\right)\cr
&=-\beta^i(X)=0.\cr$$
We have that
$$
\Omega&=\sum_i(dp_i+{\partial H\over\partial q^i}dt)\wedge(dq^i
-{\partial H\over\partial p_i}dt)\cr
&=\sum_idp_i\wedge dq^i-{\partial H\over\partial q^i}dq^i\wedge dt-
{\partial H\over\partial p_i}dp_i\wedge dt\cr
&=\sum_idp_i\wedge dq^i-dH\wedge dt\cr
&=d\omega,\cr$$
where the 1-form $\omega$, called the Cartan form, is given by
$$\omega&=\sum_ip_idq^i-H(\vec{p},\vec{q},t)dt.\cr$$
To summarise, the canonical equations are equivalent to the requirement that
we set the exterior derivative $d\omega$ of the Cartan form to zero on the
tangent vector $X=(\dot{p}_i(t),\dot{q}^i(t),1)$ to the motion.

This compact formalism has many advantages. For example, let us give a
simple proof of Noether's theorem. First we need a preliminary
definition. Let $\mu:\Sigma\rightarrow\Sigma$ be a diffeomorphism. Then
we can define the pullback (reciprocal image) $\mu^\ast\alpha$ of a
differential form $\alpha$: for a function $f$ we have that $\mu^\ast
f=f\circ\mu$ and $\mu^\ast(df)=d(\mu^\ast f)$. If $\mu_\lambda$ is a
one-parameter group of such diffeomorphisms then $\mu_\lambda$ will leave the
motion invariant if, for example, $\mu_\lambda^\ast\omega=\omega$. It is easy
to show that if $Y$ is the generator of the one-parameter group then
$\omega$ is invariant if and only if ${\cal L}_Y\omega=0$ where
${\cal L}_Y=(i_Yd+di_Y)$ is the Lie derivative. Let
us consider a 1-parameter group of transformations generated by $Y$ such
that ${\cal L}_Y\omega=0$. Then Noether's theorem states that
$di_Y\omega=0$ on the motion. This is very easy to prove. We have that
$di_Y\omega={\cal L}_Y\omega-i_Yd\omega$. However by hypothesis ${\cal
L}_Y\omega=0$ and on the motion $i_Yd\omega=0$. Finally, as discussed in
appendix II, the Cartan formalism provides a direct and rigorous
proof of variational principles.

%
\section{Field theory in the Cartan formalism}
%
The same formalism can be adapted to the treatment of fields (5). In the
case of mechanics we lifted one degree of freedom, the time, into the
state space to provide the motion of the system. This was done with the
use of a 1-form. In the case of field theory we must lift four degrees
of freedom. We will therefore search a 4-form to provide the equations
of motion.
\vfill\eject
%
\subsection{The scalar field}
%
We start with the simple case of a free scalar field $\phi$ whose motion
is given by the equation $(\square-m)\phi=0$, where $\square=g^{-1/2}
\partial_\mu g^{1/2}g^{\mu\nu}\partial_\nu$ is the generalised
Laplace-Beltrami operator acting on scalars. Here $g^{\mu\nu}$ is the
inverse matrix of the symmetric metric tensor $g_{\mu\nu}$ 
and $g$ the absolute value of the determinant of $g_{\mu\nu}$.
If we introduce the
auxiliary variables $\pi^\mu$ we can rewrite the equations of motion in
the equivalent form
$$
\partial_\mu\phi(x)-g^{-1/2}g_{\mu\nu}\pi^\nu(x)&=0,\cr
-\partial_\mu\pi^\mu(x)+mg^{1/2}\phi(x)&=0.\cr$$
Note that as $\phi$ is a scalar $\pi^\mu$ is a contravariant density.

To find the corresponding Cartan form we must write these equations in
terms of differential forms. For simplicity we introduce the notation
$$
\eta&={1\over4!}\epsilon_{\mu\nu\rho\lambda}dx^\mu\wedge dx^\nu\wedge
dx^\rho\wedge dx^\lambda,\cr
\eta_\mu&={1\over3!}\epsilon_{\mu\nu\rho\lambda}dx^\nu\wedge
dx^\rho\wedge dx^\lambda\cr$$
and so on. With these definitions we can cast the equations of motion into
the form
$$
s^\ast(d\phi-g^{-1/2}g_{\mu\nu}\pi^\nu dx^\mu)&=0,\cr
s^\ast(-d\pi^\mu\wedge\eta_\mu+mg^{1/2}\phi\eta)&=0.\cr$$
Here $s:{\Bbb R}^4\rightarrow\Sigma$ is an immersion and $s^\ast$ is the
corresponding pullback. As $s$ is an immersion
we have that $s^\ast\eta\not=0$ so that $s$ provides a parameterisation
of the motion.

As in the case of mechanics the equations of motion are then given by
the requirement that $i_X\Omega=0$, where here $X$ is the geometrical object
defined by the antisymmetric tensor product of four linearly independent vectors
tangent to the motion and
$$
\Omega&=(d\phi-g^{-1/2}g_{\mu\nu}\pi^\nu dx^\mu)\wedge(-d\pi^\mu\wedge\eta_\mu
+mg^{1/2}\phi\eta)\cr
&=d\pi^\mu\wedge d\phi+(mg^{1/2}\phi d\phi-g^{-1/2}g_{\mu\nu}\pi^\nu
d\pi^\mu)\wedge\eta\cr
&=d\left(\pi^\mu d\phi\wedge\eta_\mu+\hbox{$1\over2$}(mg^{1/2}\phi^2-g^{-1/2}
g_{\mu\nu}\pi^\mu\pi^\nu)\eta\right)\cr
&\equiv d\omega.\cr$$
This condition is equivalent to the requirement that
$s^\ast(i_Y\Omega)=0$ for all tangent vectors $Y$ since $s$ is an immersion.

We generalise this Cartan 4-form to the complex scalar field by
introducing new ``conjugate'' variables $\phi^\ast$ and
$\pi^{\ast\mu}$. We define
$$\omega&=\pi^{\ast\mu}d\phi\wedge\eta_\mu+\pi^\mu d\phi^\ast\wedge\eta_\mu
+(mg^{1/2}\phi^\ast\phi-g^{-1/2}g_{\mu\nu}\pi^{\ast\mu}\pi^\nu)\eta.\cr$$

We now discuss the nature of the 4-form $\omega$. First, using the
transformation properties of $\phi$, $\pi^\ast$, $\eta$ and $\eta_\mu$ it is
easy to see that $\omega$ changes sign with the Jacobian of the transformation.
It is thus a form of the second kind (an odd form). This is very
important as the conserved currents obtained from Noether's theorem are then
3-forms of the second kind which can be integrated over a hypersurface. (As we
will see later, the fact that $\omega$ must be of second kind can be
exploited, for example, to find possible interaction terms between
different fields).
Secondly we have chosen an expression for $\omega$ which makes explicit
the invariance of the Cartan 4-form under the changes of phase
$$\phi&\mapsto{\rm e}^{{\rm i}\lambda}\phi,\ \ \phi^\ast\mapsto{\rm
e}^{-{\rm i}\lambda}\phi^\ast,\ \ \pi^\mu\mapsto{\rm e}^{{\rm i}\lambda}
\pi^\mu\ \ {\rm and}\ \ \pi^{\ast\mu}\mapsto{\rm e}^{-{\rm
i}\lambda}\pi^{\ast\mu}.\cr$$
This is also very important as allows us to interpret such a field as a
charged field with a conservation law via Noether's theorem. In fact,
since the one parameter group of changes of phase is generated by the
tangent vector
$$Y&=\i\phi\partial_\phi-\i\phi^\ast\partial_{\phi^\ast}+\i\pi^\mu
\partial_{\pi^\mu}-\i\pi^{\ast\mu}\partial_{\pi^{\ast\mu}},\cr$$
the corresponding conserved current $J^\mu(x)\eta_\mu\equiv s^\ast(i_Y\omega)$
is given by
$$J^\mu(x)\eta_\mu&=\i s^\ast\big((\pi^{\ast\mu}\phi-\pi^\mu\phi^\ast)\eta_\mu
\big)\cr
&=\big(\pi^{\ast\mu}(x)\phi(x)-\pi^\mu(x)\phi^\ast(x)\big)\eta_\mu.\cr$$

If $g_{\mu\nu}(x)$ is constant then the Cartan 4-form $\omega$ is also invariant
under translations, transformations generated by the four vectors
$\partial_{x^\mu}$. The corresponding four conserved currents
$\tau_\mu{}^\nu(x)\eta_\nu\equiv s^\ast(i_{\partial_{x^\mu}}\omega)$ are given
by
$$\tau_\mu{}^\nu(x)\eta_\nu&=s^\ast(-\pi^{\ast\nu}d\phi\wedge\eta_{\mu\nu}
-\pi^\nu d\phi^\ast\wedge\eta_{\mu\nu}
-(g^{-1/2}g_{\nu\rho}\pi^{\ast\nu}\pi^\rho-mg^{1/2}\phi^\ast\phi)\eta_\mu)\cr
&=-\pi^{\ast\nu}(x)\partial_\mu\phi(x)\eta_\nu+\pi^{\ast\rho}(x)
\partial_\rho\phi(x)\eta_\mu-\pi^\nu(x)\partial_\mu\phi^\ast(x)\eta_\nu
+\pi^\rho(x)\partial_\rho\phi^\ast(x)\eta_\mu\cr
&\phantom{\qquad}-\big(g^{-1/2}(x)g_{\nu\rho}(x)
\pi^{\ast\nu}(x)\pi^\rho(x)-mg^{1/2}(x)\phi^\ast(x)\phi(x)\big)\eta_\mu).\cr$$
This is the energy-momentum tensor for the scalar field. In particular
$\tau_0~{}^0(x)$ is the energy density, $\tau_i~{}^0(x)$ the momentum density,
$\tau_0~{}^i(x)$ the energy flux and $\tau_i~{}^j(x)$ the momentum flux.

%
\subsection{The fermion field}
%
Let us now consider the free fermion field. This is like the complex
scalar field, but with the variables being a four component spinor
$\psi$ and its adjoint $\psi^{\dagger}$ By analogy the Cartan 4-form is
given by
$$\omega&=-\i\hbar\psi^{\dagger}\alpha^\mu d\psi\wedge\eta_\mu+\psi^{\dagger}
u\psi\omega,\cr$$
where the constant four-by-four matrices $\alpha^\mu$ and $u$ depend on the
dynamical symmetry group that we consider. Since
$$\Omega&=d\omega=-\i\hbar d\psi^{\dagger}\wedge\alpha^\mu d\psi\wedge\eta_\mu
+d(\psi^{\dagger}u\psi)\wedge\eta\cr$$
the equations of motion are then the following:
$$s^\ast(i_{\partial_{\psi^{\dagger}}}d\omega)&=s^\ast(-\i\hbar\alpha^\mu
d\psi\wedge\eta_\mu+u\psi\eta)=0,\cr
s^\ast(i_{\partial_\psi}d\omega)&=s^\ast(\i\hbar d\psi^{\dagger}\wedge
\alpha^\mu\eta_\mu+\psi^{\dagger}u\eta)=0,\cr$$
or
$$-\i\hbar\alpha^\mu\partial_\mu\psi(x)+u\psi(x)&=0,\cr
\i\hbar\partial_\mu\psi^{\dagger}(x)\alpha^\mu+\psi^{\dagger}(x)u&=0.\cr$$

As for the complex scalar field the Cartan 4-form $\omega$ is invariant
under the one-parameter change of phase
$\psi\mapsto\e^{\i\hbar^{-1}\lambda}\psi$ and $\psi^{\dagger}\mapsto
\e^{-\i\hbar^{-1}\lambda}\psi^{\dagger}$. Such an action is generated by
the tangent vector $Y=\i\hbar^{-1}\psi\partial_\psi-\i\hbar^{-1}\psi^{\dagger}
\partial_{\psi^{\dagger}}$, the corresponding conserved current being
given by
$$J^\mu(x)\eta_\mu&\equiv s^\ast(i_Y\omega)\cr
&=s^\ast(\psi^{\dagger}\alpha^\mu\psi\eta_\mu)\cr
&=\psi^{\dagger}(x)\alpha^\mu\psi(x)\eta_\mu.\cr$$

Since $\alpha^\mu$ and $u$ do not depend on $x$ the Cartan 4-form is also
invariant under the translations generated by the four vectors
$\partial_{x^\mu}$. The four corresponding conserved quantities are given by
$$\tau_\mu{}^\nu(x)\eta_\nu&\equiv s^\ast(i_{\partial_{x^\mu}}\omega)\cr
&=s^\ast(\i\hbar\psi^{\dagger}\alpha^\nu d\psi\wedge\eta_{\nu\mu}
+\psi^{\dagger}u\psi\eta_\mu)\cr
&=\i\hbar\psi^{\dagger}(x)\alpha^\nu\partial_\mu\psi(x)\eta_\nu-\i\hbar
\psi^{\dagger}(x)\alpha^\rho\partial_\rho\psi(x)\eta_\mu+\psi^{\dagger}(x)u
\psi(x)\eta_\mu\cr
&=\left(\i\hbar\psi^{\dagger}(x)\alpha^\nu\partial_\mu\psi(x)
+\delta_\mu{}^\nu\big(-\i\hbar\psi^{\dagger}(x)\alpha^\rho\partial_\rho\psi(x)
+\psi^{\dagger}(x)u\psi(x)\big)\right)\eta_\nu.\cr$$

It remains to give specific forms for the matrices $\alpha^\mu$ and $u$. This
choice is motivated by the required dynamical covariance. We obtain the
usual Dirac equation with its dynamical Lorentz covariance by choosing
$$\alpha^0&=\mat{I&0\cr0&I},\qquad\alpha^i=c
\mat{0&\sigma^i\cr\sigma^i&0\cr},\qquad u=mc^2\mat{I&0\cr0&-I\cr}.\cr$$

On the other hand, for the Schr\"odinger case with its dynamical Galilei
covariance (6) we must consider two sets of matrices to be able to interpret
the non-relativistic limit of the Dirac equation. We therefore choose
$$\alpha^0_I&=\mat{I&0\cr0&0},\qquad\alpha^i_I=
\mat{0&\sigma^i\cr\sigma^i&0\cr},\qquad u_I=2m\mat{0&0\cr0&-I},\cr$$
leading to the usual Schr\"odinger equation, or
$$\alpha^0_{II}&=\mat{0&0\cr0&I},\qquad\alpha^i_{II}=
\mat{0&\sigma^i\cr\sigma^i&0\cr},\qquad u_{II}=2m\mat{I&0\cr0&0},\cr$$
which leads to the ``negative energy'' equivalent of the usual
Schr\"odinger equation.

%
\subsection{The Maxwell field}
%
The Maxwell theory of the electromagnetic field can also be derived from
a Cartan 4-form. As discussed above, the electromagnetic field is
described by the even 1-form $A=A_\mu(x)dx^\mu$ and the odd 2-form
$H={1\over4}H^{\mu\nu}(x)\epsilon_{\mu\nu\rho\lambda}dx^\rho\wedge
dx^\lambda$. The four $A_\mu$ and the six $H^{\mu\nu}=-H^{\nu\mu}$ with the
four $x^\mu$ are then the field variables. The derived even 2-form $B$ is
related to $H$ via the phenomenological relations $B_{\mu\nu}
={\partial\over\partial H^{\mu\nu}}L$, where $L$ is the function
$L(x,H)=\frac18\mu_{\mu\nu\rho\lambda}(x)H^{\mu\nu}
H^{\rho\lambda}$. Note that we could take a more general function $L$ in
order to formulate, for example, non-linear electrodynamics.
Let $J=J^\mu(x)\eta_\mu$ be a given conserved current. Then Maxwell's
equations can be derived from the Cartan 4-form
$$\omega&=dA\wedge H-L\eta-eA\wedge J\cr$$
since
$$d\omega&=dA\wedge dH-B\wedge dH-edA\wedge J=(dA-B)\wedge(dH-eJ),\cr$$
which leads directly to the Maxwell equations following the same
reasoning as for particle mechanics.

If the function $L$ does not depend explicitly on $x$ and if $J^\mu(x)=0$
(the free field) then the Cartan 4-form
is invariant under translations generated by the four vectors
$\partial_{x^\mu}$. As for the scalar field the four corresponding conserved
currents yield the energy momentum tensor. We have
$$\tau_\mu{}^\nu(x)\eta_\nu&\equiv s^\ast(i_{\partial_{x^\mu}}\omega)\cr
&=B_{\rho\mu}(x)H^{\nu\rho}(x)\eta_\nu+\frac14B_{\rho\lambda}(x)
H^{\rho\lambda}(x)\eta_\mu.\cr$$
In particular, it is easy to show that the
free-field energy density is given by
$$\tau_0{}^0(x)&=\frac12(E_i(x)D^i(x)+B(x)^iH_i(x))\cr$$
with corresponding flux (the Poynting vector)
$\tau_0{}^i(x)=\epsilon^{ijk}E_j(x)H_k(x)=(\vec{E}(x)\times\vec{H}(x))^i$.
We stress that one must not confuse the Poynting vector $\tau_0{}^i$ with the
momentum density
$$\tau_i{}^0(x)&=\epsilon_{ijk}B^j(x)D^k(x)=(\vec{B}(x)\times\vec{D}(x))_i\cr$$
with corresponding flux $\tau_i{}^j(x)$.

%
\subsection{Electrodynamics}
%
The current $J^\mu(x)=\psi^{\dagger}(x)\alpha^\mu\psi(x)$ is conserved.
Hence we take $eA\wedge J=eA_\mu\psi^{\dagger}\alpha^\mu\psi\eta$ as the
interaction giving the coupled Cartan 4-form
$$\omega&=dA\wedge H-L\eta-eA\wedge J-\i\hbar\psi^{\dagger}\alpha^\mu d\psi
\wedge\eta_\mu +\psi^{\dagger} u\psi\eta.\cr$$
This leads not only to the Maxwell equations but also to the equations
of motion of the coupled fermion fields
$$s^\ast(i_{\partial_{\psi^{\dagger}}}d\omega)&=\alpha^\mu(-\i\hbar
\partial_\mu-eA_\mu(x))\psi(x)+u\psi(x)=0,\cr
s^\ast(i_{\partial_\psi}d\omega)&=(\i\hbar\partial_\mu-eA_\mu(x))
\psi^{\dagger}(x)\alpha^\mu+\psi^{\dagger}(x)u=0.\cr$$

The invariance of the Cartan form under the one-parameter group of
changes of phase leads to the conservation of the current $J$ on the
motion. Further, if $L$ does not depend explicitly on $x$ the Cartan form
$\omega$ will also be invariant under translations. The corresponding energy
momentum tensor is the sum
of three terms. The first two are the energy-momentum tensors for the
free fermion and free Maxwell fields respectively, whereas the third
$$s^\ast(i_{\partial_{x^\mu}}eA\wedge J)&=e\psi^{\dagger}(x)\alpha^\nu\psi(x)
A_\nu(x)\eta_\mu\cr$$
arises from the interaction. We stress that it is only the total energy
momentum tensor that is conserved.

%
\section{Conclusion: a new aspect of the Dirac equation}
%

If we write $\psi_I(x)=\mat{\phi_I(x)\cr\chi_I(x)\cr}$ for the solutions of
the first Schr\"odinger equation we have, supressing the dependence on $x$ for
simplicity,
$$\i\hbar\partial_0\phi_I&=\sigma^i(-\i\hbar\partial_i-eA_i)\chi_I-eA_0
\phi_I\cr
0&=\sigma^i(-\i\hbar\partial_i-eA_i)\phi_I-2m\chi_I.\cr$$
Substituting for $\chi_I$ we have
$$\i\hbar\partial_0\phi_I&=(\frac1{2m}g^{ij}(-\i\hbar\partial_i-eA_i)(-\i\hbar
\partial_j-eA_j)+\frac{\hbar e}{2m}\sigma_i B^i-eA_0)\phi_I,\cr$$
that is the Schr\"odinger-Pauli equation for a particle with magnetic
moment $g=2$.

On the other hand, writing the solutions of the second Schr\"odinger equation
as $\psi_{II}=\mat{\phi_{II}\cr\chi_{II}\cr}$ we have
$$0&=\sigma^i(-\i\hbar\partial_i-eA_i)\chi_{II}+2m\phi_{II}\cr
\i\hbar\partial_0\chi_{II}&=\sigma^i(-\i\hbar\partial_i-eA_i)\phi_{II}
-eA_0\chi_{II}.\cr$$
Substituting for $\phi_{II}$ gives
$$\i\hbar\partial_0\chi_{II}&=(-\frac1{2m}g^{ij}(-\i\hbar\partial_i-eA_i)
(-\i\hbar\partial_j-eA_j)-\frac{\hbar e}{2m}\sigma_i B^i-eA_0)\chi_{II
}.\cr$$
This is the ``negative-energy equivalent'' of the usual Schr\"odinger-Pauli
equation.

One can simply see how that non-relativistic limit arises in this formalism.
If we write $\psi=\mat{\phi\cr c^{-1}\chi\cr}$ and perform the transformation
$\psi\mapsto\e^{-\i mc^2t/\hbar}\psi$ so that we label the energy in
agreement with the Galilei convention, the Dirac equation takes the form
$$
\i\hbar\partial_0\phi&=\sigma^i(-\i\hbar\partial_i-eA_i)\chi-eA_0\phi,\cr
\i\hbar c^{-2}\partial_0\chi&=\sigma^i(-\i\hbar\partial_i-eA_i)\phi
+(-ec^{-2}A_0-2m)\chi.\cr$$
If the solution has positive energy the component $\chi$ is small with respect
to $\phi$ and we can ignore the ``mass'' due to its ``frequency and potential
energies'' with respect to $2m$. We then recover the first
Schr\"odinger equation.  Note that in the non-relativistic limit the small
components do not approach zero, rather they approach a definite relationship
with the large components:
$\chi\rightarrow\frac1{2m}\sigma^i(-\i\hbar\partial_i-eA_i)\phi$ which is the
same as in Schr\"odinger.
If we have a solution of negative energy
then the same argument allows us to recover the second Schr\"odinger equation.

In the first Schr\"odinger case the Hilbert space which describes the
state of the particle must be constructed with the first two components
of $\psi$ since it is the integral of $J^0_I=\phi^{\dagger}_I
\phi^{\phantom{\dagger}}_I$ over space which is conserved according to
Noether's theorem. In this case the two other components only play a role
in the dynamics. For the second Schr\"odinger equation with the
opposite energy sign we must do the contrary. The Hilbert space
must be built on the second two components and the first two only play a
dynamical role.

The positive energy solutions for the Dirac case must be compared with
solutions of the first Schr\"odinger equation and so the corresponding
Hilbert space must be constructed with the first two components, the
other two only playing a role in the dynamics. For negative energy
solutions it is just the contrary. There is then a superselection
variable between these two spaces since the primitive observables
position, momentum and spin preserve this separation. In this way many
paradoxes such as the Klein paradox and Zitterbewegung disappear. Further the
spectroscopic convention of labelling eigenvectors by the quantum numbers of
the first two components is then fully justified.

\vfill\eject
\leftline{\sectionfont I: The phenomenological relations}
\bigskip
In this appendix we show that there is an essentially unique symmetric
matrix $\hat{g}_{\mu\nu}$ such that $B_{\mu\nu}=\frac12(\hat{g}_{\mu\rho}
\hat{g}_{\nu\lambda}-\hat{g}_{\mu\lambda}\hat{g}_{\nu\rho})H^{\rho\lambda}$.
The equations $B_{ij}=\frac12(\hat{g}_{i\rho}\hat{g}_{j\lambda}
-\hat{g}_{i\lambda}\hat{g}_{j\rho})H^{\rho\lambda}=\mu_0H^{ij}$ and
$B_{0i}=\frac12(\hat{g}_{0\rho}\hat{g}_{i\lambda}
-\hat{g}_{0\lambda}\hat{g}_{i\rho})H^{\rho\lambda}=\varepsilon_0^{-1}H^{i0}$
give first the following two sets of relations

\item{$(i)$} $\hat{g}_{ii}\hat{g}_{jj}-\hat{g}_{ij}^2=\mu_0$ for $i\not=j$
\item{$(ii)$} $\hat{g}_{00}\hat{g}_{ii}-\hat{g}_{0i}^2=-\epsilon_0^{-1}$

and for the off-diagonal part we find

\item{$(iii)$} $\hat{g}_{0j}\hat{g}_{ik}-\hat{g}_{0k}\hat{g}_{ij}=0$
\item{$(iv)$} $\hat{g}_{00}\hat{g}_{ij}-\hat{g}_{0i}\hat{g}_{0j}=0$ for
$i\not=j$.

>From $(iii)$ we have that
$\hat{g}_{0j}^2\hat{g}_{ik}^2=\hat{g}_{0k}^2\hat{g}_{ij}^2$. If we
eliminate the off-diagonal $\hat{g}_{\mu\nu}$ using $(i)$ and
$(ii)$ we find that
$$(\hat{g}_{00}\hat{g}_{ii}+\varepsilon_0^{-1})(\hat{g}_{ii}\hat{g}_{kk}
-\mu_0)&=(\hat{g}_{00}\hat{g}_{kk}+\varepsilon_0^{-1})(\hat{g}_{ii}
\hat{g}_{jj}-\mu_0).\cr$$
Expanding and regrouping leads to
$$(\hat{g}_{ii}-\hat{g}_{kk})(\varepsilon_0^{-1}\hat{g}_{jj}+
\mu_0\hat{g}_{00})&=0.\cr$$
Hence either $\hat{g}_{ii}=\hat{g}_{kk}$ or
$\hat{g}_{jj}=-\varepsilon_0\mu_0\hat{g}_{00}$. Thus at least three of the
values $\alpha=\hat{g}_{11}$, $\beta=\hat{g}_{22}$, $\gamma=\hat{g}_{33}$ and
$\delta=-\varepsilon_0\mu_0\hat{g}_{00}$ must be equal.

Suppose that $\alpha=\beta=\gamma$ (the other cases being treated by
permutation). Then from $(i)$ the three $\hat{g}_{ij}^2$ with $i\not=j$ must be
equal and from $(iii)$ so must the three $\hat{g}_{0i}^2$. Now, setting $i=j$ in
$(iii)$ we have
$$\hat{g}_{0i}^2\hat{g}_{ik}^2&=\hat{g}_{0k}^2\hat{g}_{ii}^2.\cr$$
However from $(i)$ we see that $g_{ik}^2\not=g_{ii}^2$ since $\mu_0$ does
not vanish and so $\hat{g}_{0i}=0$. Since $\epsilon_0^{-1}$ does not
vanish either, from $(ii)$ we find that $\hat{g}_{00}\not=0$. Using $(iv)$ we
then have that $\hat{g}_{ij}=0$ for $i\not=j$. Finally $(i)$ and $(ii)$ give
$\hat{g}_{ii}=\pm\mu_0^{1/2}$ and $\hat{g}_{00}=\mp\varepsilon_0^{-1}
\mu_0^{-1/2}$. It is trivial to verify that this does indeed give a
solution. Hence there is a diagonal matrix $\hat{g}_{\mu\nu}$, unique up to a
sign, such that $\mu_{\mu\nu\rho\lambda}=(\hat{g}_{\mu\rho}\hat{g}_{\nu\lambda}
-\hat{g}_{\mu\lambda}\hat{g}_{\nu\rho})$.

\vfill\eject
\leftline{\sectionfont II: Variational principles}
\bigskip
It is well known that the canonical equations for particle mechanics can
be derived from the variational principle: among all curves
$\Gamma$ between two points $P_1$ and $P_2$ of the state space the one
which realises the motion (if such a path exists) renders extremal
(stationary) the integral
$$s(\Gamma)&=\int_\Gamma\omega=\int_\Gamma\left(p_idq^i-H(p,q,t)dt
\right).\cr$$
The proof is very simple in the Cartan formalism. If $\Gamma'$ is
another curve between $P_1$ and $P_2$ obtained by moving each point of
$\Gamma$ along an infinitesimal tangent vector $Y$ then the stationarity of
$s(\Gamma)$ follows directly from
$$\int_{\Gamma'}\omega-\int_\Gamma\omega&=\int_{\Xi}d\omega=0\cr$$
since for $X$ tangent to the motion $d\omega(X,Y)=0$ for all tangent
vectors $Y$. Note that $\Xi$ is the infinitesimal surface bounded by
$\Gamma'-\Gamma$.

It is clear that such a variational theorem can also be formulated in
field theory, where the motions are characterised by
$s^\ast(i_Yd\omega)=0$ for the Cartan 4-form $\omega$. We do not utilise such
principles since the Cartan condition leads directly to the equations of motion
and the Noether theorems.


%
\references
%
\item{(1) }C. Piron in ``New Frontiers in Quantum Electrodynamics and Quantum
Optics'' ed. A. O. Barut (Plenum, New York, 1990) p. 495.

\item{(2) }C. Piron in ``Note di matematica e fisica'' vol.5 (Cerfim, CH-6601
Locarno, 1991) p. 139.

\item{(3) }D. Aerts {\it Found. Phys.} {\bf12} (1982) 1131.

\item{(4) }A. O. Barut, D. J. Moore and C. Piron ``Spacetime models and field
theory'' in preparation.

\item{(5) }A. O. Barut, D. J. Moore and C. Piron {\it Helv. Phys. Acta\/}
{\bf66} (1993) 471.

\item{(6) }J.-M. L\'evy-Leblond {\it Commun. Math. Phys.\/} {\bf6} (1967) 286.

\bye