Can QED be formulated in terms of independent vector potential and bivector fields?

Question

We can obtain the classical equations of motion for electromagnetism by considering the vector potential $A^\mu$ and the electric $\mathbf E$ and magnetic $\mathbf B$ fields as independent degrees of freedom.

$$\mathcal L = j^\mu A_\mu - A^0 (\nabla \cdot \mathbf E) - \mathbf A \cdot (\dot {\mathbf E} - \nabla \times \mathbf B) - \frac12 (\mathbf E^2 - \mathbf B^2).$$

Varying the action with respect to $A^\mu$ yields the inhomogeneous Maxwell equations in terms of $\mathbf E$ and $\mathbf B$ (up to possible sign errors):

$$-\frac{\delta S}{\delta A^0} = \nabla \cdot \mathbf E - j^0$$

$$-\frac{\delta S}{\delta \mathbf A} = \dot {\mathbf E} - \nabla \times \mathbf B + \mathbf j$$

while varying with respect to $\mathbf E$ and $\mathbf B$ yields the standard definitions of those fields in terms of derivatives of $A^\mu$:

$$-\frac{\delta S}{\delta \mathbf E} + \mathbf E = \dot {\mathbf A} - \nabla A^0$$

$$\frac{\delta S}{\delta \mathbf B} + \mathbf B = \nabla \times \mathbf A.$$

Under a gauge transformation $A_\mu \to A_\mu + \partial_\mu \lambda$, $\mathcal L$ is not invariant, but $\delta S/\delta \lambda \equiv 0$ when $\partial_{\mu}j^{\mu} = 0$.

In this picture, $A^\mu$ acts as an intermediary between the bivector field $(\mathbf E, \mathbf B)$ and the source of the current density $j^\mu$ (e.g. a spinor field). The relations between $(\mathbf E,\mathbf B)$ and the exterior derivative of the vector potential arise as classical equations of motion, rather than by definition.

1. Are there problems with this formulation that invalidate it at the classical level? (e.g. too many degrees of freedom, problems with a Hamiltonian formulation, etc.) It looks like the Hamiltonian wouldn't be bounded from below, but maybe there's a workaround.

2. Does this setup have an analogue in quantum field theory, where we normally consider the gauge field $A^\mu$ as the only fundamental degrees of freedom for electromagnetism?

Answer 1

I) OP's action in covariant notation$^1$

$$ S_1[A,F] ~:=~\int \! d^4x~{\cal L}_1, \qquad {\cal L}_1~:=~\frac{1}{4}F^{\mu\nu}F_{\mu\nu}- F^{\mu\nu}\partial_{\mu} A_{\nu} + j^{\mu}A_{\mu}, $$ $$ E_i~\equiv~F_{i0}, \qquad B_i~\equiv~\frac{1}{2}\epsilon_{ijk}F_{jk}, \tag{A} $$

is the first-order/Palatini formulation of E&M, cf. Ref. 1 & comment by Cosmas Zachos.

It is classically well-defined. The EL eqs. for the OP's action (A) read

$$ F_{\mu\nu}~\approx~\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}, \tag{B}$$ $$ d_{\mu} F^{\mu\nu}+j^{\nu}~\approx~0.\tag{C}$$

The quadratic potential term $${\cal V}~=~-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\ldots~=~\frac{1}{2}({\bf E}^2-{\bf B}^2)+\ldots\tag{D}$$ has a minus sign in front the independent ${\bf B}$-field, and is hence unbounded from below. Therefore OP's action (A) is quantum mechanically ill-defined.

II) However, if we integrate out the independent ${\bf B}$-field, we basically get the Hamiltonian formulation of E&M,

$$ S_H[A,{\bf E}] ~:=~\int \! d^4x~{\cal L}_H, \qquad {\cal L}_H~:=~ -{\bf E}\cdot \dot{\bf A}-{\cal H}, $$ $$ {\cal H}~:=~\frac{1}{2}({\bf E}^2+(\nabla \times {\bf A})^2)-{\bf J}\cdot {\bf A} +A_0{\cal G} \qquad {\cal G}~:=~\nabla \cdot {\bf E}-\rho, \tag{E}$$

cf. Ref. 2, which is quantum mechanically well-defined. (Minus) the independent ${\bf E}$-field plays the role of momentum for the magnetic gauge potential ${\bf A}$. Moreover, $A_0$ becomes the Lagrange multiplier for Gauss' law $${\cal G}~\approx~0.\tag{F}$$ To achieve QED, one should then proceed with quantization, gauge-fixing, etc.

References:

1. ADM, arXiv:gr-qc/0405109; eq. (3.5).

2. P.A.M. Dirac, Lectures on Quantum Mechanics, 1964; chapter 2.

--

$^1$ We use signature convention $(−,+,+,+)$ and $c=1$. Disclaimer: In this answer we have ignored some total space-time divergence terms in the action as they don't contribute to EL eqs.