# New Covariant QED representation of the E.M. field

Gold Member
90 years have gone by since P.A.M. Dirac published his equation in 1928. Some of its most basic consequences however are only discovered just now. (At least I have never encountered this before). We present the Covariant QED representation of the Electromagnetic field.

1 - Definition of the fundamental operator fields

This representation is arguably the most fundamental one because it is constructed using the generators that act on the Dirac spinors (in the chiral form).
$$\begin{array}{lrcl} \mbox{mass dimension 1:}~~~~ & \mathbf{A} &=& \gamma^\mu A^\mu \\ \mbox{mass dimension 2:}~~~~ & \mathbf{F} &=& \vec{K}\cdot\vec{E}-\vec{J}\cdot\vec{B} \\ \mbox{mass dimension 3:}~~~~ & \mathbf{J}\, &=& \gamma^\mu\,j^\mu \\ \end{array}$$
Using the standard chiral gamma matrices and the boost and rotation generators.
$$\gamma^o ~=~ \left( \begin{array}{rr} ~0&I~ \\ ~I&0~ \end{array} \right) ~~~~~~~~ \gamma^i\, ~=~ \left(\, \begin{array}{cc} \!0&\!\sigma_i\! \\ \!\!\!\!-\sigma_i&\!\!0\! \end{array} \right) ~~~~~~~~ \gamma^5 ~=~ \left(\! \begin{array}{rr} \!-I&0~ \\ ~0&I~ \end{array}\! \right)$$
$$~J^i ~=~ \left(\, \begin{array}{rr} \!\!\!\!-i\sigma_i\!\!&0 \\ ~0&\!\!\!\!-i\sigma_i\!\! \end{array}~ \right) ~~~~~~~~ K^i ~=~ \left(\, \begin{array}{cc} \!\!\!-\sigma_i\!\!&0 \\ ~0&\sigma_i\! \end{array}~\right)$$
The Lorentz force operator field ##\mathbf{F}## that acts on the bi-spinor field is defined combining the boost generator ##\vec{K}## with the electric field ##\vec{E}## and the rotation generator ##\vec{J}## with the magnetic field ##\vec{B}##, with a (historical) minus sign.

2 - Covariant representation of the E.M. field laws

We can now write the laws of the Electromagnetic field, going from one mass dimension to another, by applying the differential operator matrix ##/\!\!\!{\partial} =\gamma^\mu\partial_\mu = \sqrt{\Box}## on the operator field matrices defined above.
$$/\!\!\!{\partial}\mathbf{A}^{\!\dagger} = \mathbf{F}~~~~~~ ~~~/\!\!\!{\partial}\mathbf{F} = \mathbf{J}^\dagger$$
The complex conjugate transpose ##\mathbf{A}^{\!\dagger}=\gamma^\mu A_\mu ## is the covariant form of ##\mathbf{A}##.

In the first step we have applied the conservation law ##\partial_\mu A^\mu\!=\!0## on the diagonal and in the second step we find all four of Maxwell's laws, the inhomogeneous ##\partial_\mu F^{\mu\nu}\!=\!j^\nu## as well as the homogeneous ##~\partial_\mu\! *\!\!F^{\mu\nu}\!=j^\nu_{^A}=\!0##.

The equations above do not only provide a complete description of the electromagnetic field. They also symmetrize the transitions between mass dimensions. One can say that the expressions takes the square root of,
$$\Box\,\mathbf{A} ~=~ \mathbf{J}$$
and halfway we find the Lorentz Force operator field. Going in the opposite direction from higher to lower mass dimensions we see that it splits the propagator ##\Box^{-1}## into two more convenient ##(\gamma^\mu\partial_\mu)^{-1}## propagators.

We can write for the Lagrangian of the Electromagnetic field in vacuum.
$$\mathcal{L} ~=~ \tfrac12\mathbf{F}\mathbf{F}$$
Where ##\mathcal{L}## is a matrix operator field invariant under Lorentz transform.

To find the equations of motions: The four Maxwell equations. We write:
$$\mathcal{L} ~=~ \tfrac12/\!\!\!{\partial}\mathbf{A}^{\!\dagger}/\!\!\!{\partial}\mathbf{A}^{\!\dagger}~~~~ \underrightarrow{\mbox{ Euler Lagrange }}~~~~ /\!\!\!{\partial}/\!\!\!{\partial}\mathbf{A}^{\!\dagger} ~=~ /\!\!\!{\partial}\mathbf{F} ~=~0$$
Thus the equations of motion, the four Maxwell equations, are given by ##/\!\!\!{\partial}\mathbf{F}=0##.

If we work out the Lorentz invariant Lagrangian operator field we get.
$$\mathcal{L} ~~=~~ \tfrac12\mathbf{F}\mathbf{F} ~~=~~ \tfrac12\left(E^2-B^2\right)\!I ~+~ \left(\vec{E}\cdot\vec{B}\right)i\gamma^5$$
The Lorentz scalar ##\tfrac12(E^2-B^2)## of the electromagnetic field is associated with the diagonal matrix ##I## while the pseudo scalar ##\vec{E}\cdot\vec{B}## of the electromagnetic field is associated with the pseudo scalar generator ##i\gamma^5##. If we include sources to the Lagrangian we can write.
$$\mathcal{L} ~=~ \tfrac12( \mathbf{F}\mathbf{F} + (\mathbf{J}\mathbf{A} + \mathbf{A}\mathbf{J})^\dagger) ~=~ \left(\tfrac12(E^2-B^2) + j_\mu A_\mu\right)\! I + \left(\vec{E}\cdot\vec{B}\right)i\gamma^5$$
The Lorentz scalars on the diagonal ##I## lead to the inhomogeneous Maxwell equations while the pseudo-scalar associated with ##i\gamma^5## provides the homogeneous Maxwell equations.
$$\begin{array}{lll} \tfrac12(E^2-B^2) + j_\mu A_\mu & \rightarrow & \mbox{inhomogenious Maxwell equations}\\ ~~~(\,\vec{E}~\,\cdot~\vec{B}~) & \rightarrow & \mbox{homogenious Maxwell equations} \end{array}$$

3 - Lorentz transform of the operator fields

The Lorentz transform operator used for chiral bi-spinor fields is.
$$\mbox{\Lambda} ~=~ \exp\left(~\tfrac12 \vec{\phi}\cdot\vec{J} ~+~\tfrac12 \vec{\varpi}\cdot\vec{K}~\right)$$
This operator is also used to transform all the electromagnetic operator fields in the same way, including the axial current field ##\mathbf{J}_{^{\!A}}\!=i\gamma^\mu\gamma^5\,j^\mu_{^{\!A}}## associated with the two homogenious Maxwell equations.
$$\begin{array}{lcccll} \mbox{Bi-spinor fermion field:} && \psi^{'} & = & ~~\mbox{\Lambda\,\mathbf{\psi}} & \\ \\ \mbox{Potential operator field:} && \mathbf{A}^{\!'} & = & \mbox{\Lambda^{^{\!-\dagger}} \,\mathbf{A} \,~\Lambda^{^{\!\dagger}}} & \\ \mbox{Force operator field:} && \mathbf{F}^{ '} & = & \mbox{\Lambda^{^{\!-\dagger}} \,\mathbf{F} \,~\Lambda^{^{\!\dagger}}} & \\ \mbox{Current operator field:} && \mathbf{J}^{ '} & = & \mbox{\Lambda^{^{\!-\dagger}} \,\mathbf{J} ~~\Lambda^{^{\!\dagger}}} & \\ \mbox{Axial operator field:} && \,\mathbf{J}^{ '}_{^{\!A}} & = & \mbox{\Lambda^{^{\!-\dagger}} \,\mathbf{J}_{^{\!A}}\,\Lambda^{^{\!\dagger}}} & \\ \mbox{Matrix Lagrangian:} && \mathcal{L}^{ '} & = & \mbox{\Lambda^{^{\!-\dagger}} \,\mathcal{L}\,~\Lambda^{^{\!\dagger}}} & =~\mathcal{L}\\ \end{array}$$
Where the ##\dagger## and -##\dagger## denote the conjugate transpose of ##\Lambda## and the conjugate transpose of ##\Lambda^{\!-1}##. The generators used to construct the operator field determine how the field behaves under a Lorentz transform.

There is a PDF file here, and a Mathematica file here.

Last edited:

## Answers and Replies

Related High Energy, Nuclear, Particle Physics News on Phys.org
DrDu
In how far does this differ from the better known representation in terms of differential forms?

Gold Member
In how far does this differ from the better known representation in terms of differential forms?
There is one crucial difference, if you compare:
• exterior product and the exterior derivative
• (anti-) commutation rules of the gamma and Pauli matrices
A gamma or Pauli matrix does commute with itself instead of anti-commute with itself.

For a totally anti-symmetric (exterior) product they would have to anti-commute. So for instance ##/\!\!\!\partial \mathbf{A}^{\!\dagger}## does have the conservation term ##\partial_\mu A^\mu## (on the diagonal) which the differential form ##dA## does not have.

This difference becomes essential in the case of the Lagrangian which produces the Lorentz invariants that become the four Maxwell equations when using Euler-Lagrange to obtain the equations of motion:
$$\mathcal{L} ~~=~~ \tfrac12\mathbf{F}\mathbf{F} ~~=~~ \tfrac12\left(E^2-B^2\right)\!I ~+~ \left(\vec{E}\cdot\vec{B}\right)i\gamma^5$$
The right hand terms are all the results of anticommutators of equal Pauli matrices like ##\{\sigma^x,\sigma^x\}## or ##\{\sigma^y,\sigma^y\}##. These terms would be zero if we would use a totally anti-symmetric exterior product.

The unique thing is here that the Lagrangian of the electromagnetic field produces all four Maxwell equations of motion instead of just a subset.

DrDu