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

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?

• This is the first order formulation of electromagnetism; what's the advantage of not writing it covariantly, $\frac{1}{4}F^{\mu\nu}F_{\mu\nu}- F^{\mu\nu}\partial_\mu A_\nu$, for independent $A_\mu$ and $F_{\mu\nu}$, which is more conventional? – Cosmas Zachos Jan 6 '18 at 1:29
• There's no advantage; I just think of $F_{\mu\nu} \equiv \partial_\mu A_\nu - \partial_\nu A_\mu$ as being the stronger association, so it might have been clearer what I was asking if I used $\mathbf E$ and $\mathbf B$. Does it matter whether the derivative acts on $A_\nu$ or $F_{\mu\nu}$? It only determines which of $\mathbf A$ or $\mathbf E$ have conjugate momenta, right? – rossng Jan 6 '18 at 5:59

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.

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$^1$ 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.

• So we can vary $A_\mu$ and $E_i$ independently (with their respective equations of motion holding only in the classical limit), but must assert $\delta S/\delta B_i \equiv 0$ in order to get a sensible quantum theory? With gauge fixing that leaves 6 nominal degrees of freedom, but the conjugate momenta for $E_i$ vanish, leaving only 3 dynamical DoFs. Is that correct? Thanks! – rossng Jan 6 '18 at 19:01
• In the traditional counting, (3+1)D E&M has 3 off-shell DOF and 2 on-shell DOF, cf. e.g. this Phys.SE post. – Qmechanic Jan 7 '18 at 20:58