Can we use only the observables of Fermion fields?

There are legion ways to consider fermionic Dirac spinor fields, but is it possible to consider the asymptotic free field only in terms of observables, which in the case of the Dirac spinor field must be $U(1)$ invariant?

For a specific construction, to give some context, in a 4-dimensional formalism we might consider normal-ordered objects of the form $$\int:\hspace{-0.2em}\overline{\hat\psi_{\xi'}(x)}F_{\xi'\xi}(x)\hat\psi_\xi(x)\hspace{-0.2em}:\mathrm{d}^4x,$$ smeared by a Dirac matrix-valued test function $F_{\xi'\xi}(x)$. As we will see below, it's better if we instead first construct a 2-point formalism, for which we can then consider the limit as the two points approach each other. So, if we construct a 2-point Dirac operator as $$\hat\Phi_F=\int \overline{\hat\psi_{\xi'}(x_1)}F_{\xi'\xi}(x_1,x_2)\hat\psi_\xi(x_2)\mathrm{d}^4x_1\mathrm{d}^4x_2,$$ where the test function $F\in\mathcal{F}$ is Dirac algebra-valued for each pair of points $x_1,x_2$, we find that for the commutation relation $\left[\hat\Phi_F,\hat\Phi_G\right]$ we have $$\left[\hat\Phi_F,\hat\Phi_G\right]=\hat\Phi_{[F,G]},$$ where $[F,G]=F.\mathrm{i}\!\mathsf{S}.G-G.\mathrm{i}\!\mathsf{S}.F$, with $\mathrm{i}\!\mathsf{S}(x,x')_{\xi\xi'}=\left(\mathrm{i}\gamma^\mu\frac{\partial}{\partial x^\mu}+m\right)_{\xi\xi'}\mathrm{i}\!\Delta(x-x')$ as a Dirac spinor Green's function, and $$\left(F.\mathrm{i}\!\mathsf{S}.G\right)_{\xi'\xi}(x,y)=\int F_{\xi'a}(x,x_1)\mathrm{i}\!\mathsf{S}_{ab}(x_1,x_2)G_{b\xi}(x_2,y)\mathrm{d}^4x_1\mathrm{d}^4x_2,$$ so that the 2-point field $\hat\Phi_F$ generates a Lie algebra (and a justification of sorts for introducing a 2-point formalism). What we have constructed is effectively a matrix algebra that is a tensor product of both space-time indices and Dirac indices.

It's not too difficult to use Wick's theorem to show that the generating function for the time-ordered VEVs for these operators, in terms of the Feynman propagator $\mathrm{i}\mathsf{S_F}$ and using a Trace that contracts over both space-time points and Dirac indices, is $$\left<0\right|\mathrm{T}[\mathrm{e}^{\mathrm{i}\lambda\hat\Phi_F}]\left|0\right> =\exp{\left[\sum\limits_{n=1}^\infty \frac{(\mathrm{i}\lambda)^n}{n}\mathsf{Tr}{[(\mathrm{i}\mathsf{S_F}.F)^n]}\right]} =\exp\hspace{-0.4ex}{\Bigl(-\mathsf{Tr}[\ln{(1-\mathrm{i}\lambda\mathrm{i}\mathsf{S_F}.F)}]\Bigr)},$$ where the last equality is essentially formal, with the result being a quantum analogue of an infinite system of independent exponential distributions (without time-ordering, the result is straightforward but cannot be expressed so compactly).

We can also introduce a normal-ordered form, $\hat\Phi_F^{\hspace{-0.3ex}:\hspace{-0.1ex}:}$, for which $$\left[\hat\Phi^{\hspace{-0.3ex}:\hspace{-0.1ex}:}_F,\hat\Phi^{\hspace{-0.3ex}:\hspace{-0.1ex}:}_G\right]=\hat\Phi^{\hspace{-0.3ex}:\hspace{-0.1ex}:}_{[F,G]}+\mathsf{Tr}[F.\mathrm{i}\mathsf{S}_+.G.\mathrm{i}\mathsf{S}_- - G.\mathrm{i}\mathsf{S}_+.F.\mathrm{i}\mathsf{S}_-],$$ where $\mathrm{i}\mathsf{S}_\pm$ are the positive and negative frequency parts of $\mathrm{i}\mathsf{S}$. This construction changes the lowest degree VEV to $\left<0\right|\hat\Phi_F^{\hspace{-0.3ex}:\hspace{-0.1ex}:}\left|0\right>=0$ but all higher connected VEVs are unchanged; for this 2-point field operator, the limits of the VEVs as the two points approach each other exist. The commutator algebra of such 1-point objects is not closed, because $\hat\Phi_{[F,G]}^{\hspace{-0.3ex}:\hspace{-0.1ex}:}$ is a 2-point object, but the algebra of 2-point fields is closed.

There is a curiosity associated with this object, that $[F,G]$ is non-zero only when the test functions $F$ and $G$ have time-like separated supports, so that if we construct field operators using only merged 2-point field operators (operators such as $\hat\Phi^{\hspace{-0.3ex}:\hspace{-0.1ex}:}_F$, where the test function is $F(x,y)=\delta^4(x-y)F(x)\hspace{0.3ex}$), we can only obtain 2-point operators for which the two points are at time-like separation. By taking superpositions of terms of the form $\hat\Phi^{\hspace{-0.3ex}:\hspace{-0.1ex}:}_{[F,G]}$, we can construct terms of the form $\hat\Phi^{\hspace{-0.3ex}:\hspace{-0.1ex}:}_F$ for any $F$ for which $F(x,y)$ is zero whenever $x$ and $y$ are space-like separated. Consequently, gauge invariance amounts to requiring, in particle terms, that when constructing states, creation of an electron (or positron) must always be paired either with annihilation of an electron (or positron) at time-like separation or with the creation of a positron (or electron) at time-like separation. Equivalently, measurement operators cannot create a charged particle without introducing a compensating charged particle creation or annihilation at time-like separation.

This allows us more-or-less to reconcile the Lorentz and translation covariant 2-point formalism constructed above with the usual Lagrangian formalism, which is only barely covariant because it requires two space-like hyperplanes for its formulation, by taking all the test functions $F(x,y)$ to have support only on two time-like separated space-like hyperplanes.

This is enough different from anything I've ever seen on the Dirac spinor field that I can't believe that there's something out there already that's like it, but, if there is, I'd like to know about it! So this Question is partly a reference request.

An obvious Answer to the Question as asked would be no, or indeed of course not, because we need to be able to construct states using the action of an odd number of fermion field operators on the vacuum vector, however I take that objection to be nullified by the 4-space and two-time construction given above, which allows a state to be constructed in which an odd number of fermion field operators act on the 4-dimensional vacuum state at each of two times.

Feel free to comment that the formalism as given doesn't include a means to introduce interactions. I know, and I'm working on it, but comment anyway. If you think there's some other reason to save me wasting my time on this approach, feel free to be as substantively insulting as you like.

-