Fermi's theory of beta decay - Does Fermi's Hamiltonian have the wrong transformation properties?

I'm studying the theory of beta decays as proposed by Fermi in the 30's, and I found an inconsistency between the transformation properties that he claims for his Hamiltonian and the transformation properties that Weisskopf and Blatt give about the very same Hamiltonian in the book "Theoretical Nuclear Physics". You'll find Fermi's original german article translated here (see in particular Section III).

Now, Fermi claims his Hamiltonian to be the time-component of a four-vector interaction, similar in form to the leptonic charged current that couples to the W boson in the standard model (it basically lacks, of course, the pseudovector interaction time component). He starts by enumerating four bilinear combinations of the components of the electron's ($\psi$) and antineutrino's ($\phi$) Dirac spinors, which altogether transform as a four-vector:

$$A_{0}=-\psi_{1}\phi_{2}+\psi_{2}\phi_{1}+\psi_{3}\phi_{4}-\psi_{4}\phi_{3}\\ A_{1}=\psi_{1}\phi_{3}-\psi_{2}\phi_{4}-\psi_{3}\phi_{1}+\psi_{4}\phi_{2}\\ A_{2}=i\psi_{1}\phi_{3}+i\psi_{2}\phi_{4}-i\psi_{3}\phi_{1}-i\psi_{4}\phi_{2}\\ A_{3}=-\psi_{1}\phi_{4}-\psi_{2}\phi_{3}+\psi_{3}\phi_{2}+\psi_{4}\phi_{1}\\$$

(see eq. 11 in Fermi's article; given the time of writing, the spinors should be taken to satisfy Dirac's equation in the Dirac's basis). Supposing that the decaying nucleon's velocity is much less than the speed of light, the Hamiltonian can be taken to contain only the time component of the vector (in analogy to the electromagnetic case, the spatial ones should couple weakly to the nucleon's wavefunction). So he explicitly writes:

$$H=g\,\left[\psi^{\dagger}\delta\,\phi^{*}\ Q^{\dagger}+\psi^{T}\delta\,\phi\ Q\right]$$

(see eq. 13) where the Hamiltonian is taken to act between the initial and final wavefunction of the nucleon, the operator $Q$ replaces a proton with a neutron (and $Q^{\dagger}$ does the opposite), $\delta$ is the matrix

$$\delta=\begin{pmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&1\\0&0&-1&0\end{pmatrix}$$

and $\psi$ and $\phi$ this time are the Dirac's bispinors multiplied by the creation/annihilation operators of the electron and neutrino, to be evaluated at the nucleon's position. Notice two things: first of all, as the proton's and neutron's creation/annihilation operators do not appear explicitly in the Hamiltonian, Fermi's formalism treats the heavy and the light particles on a different footing (but this is only a matter of writing) - moreover the familiar $\int d^{3}x$ integral must be taken only when calculating $H$'s matrix elements; second of all, Fermi's conventions about which neutrino (i.e. the neutrino or the antineutrino) should appear together with the electron/positron are opposite to the modern formalism: today we would call $\phi^{*}$ the neutrino's field instead of the antineutrino's creation operator (of course the two could be the same, but the distinction is useful if one wants to compare Fermi's Hamiltonian to the leptonic charged weak current).

On the other hand, in Blatt and Weisskopf's beautiful analysis of the possible interaction terms that can appear in the nucleus's Hamiltonian (see Ch. XIII, Sec. 5.C), the very same term, whose electron/neutrino portion for $\beta^{-}$ decays is expressed in Fermi's notation as

$$F_{S}=\psi^{\dagger}\delta\,\phi^{*}$$

is said to be a scalar ($S$) under the Poincaré group, while the time component of the four-vector interaction is given as

$$F_{V0}=\psi^{\dagger}\beta\delta\,\phi^{*}$$ where $\beta$ is Dirac's beta matrix, i.e. (in the Dirac's basis)

$$\beta=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$

The statement is accompanied with a footnote:

[In $F_{S}$] We might expect to find the familiar Dirac matrix $\beta$ instead of $\delta$, since $\psi^{\dagger}\beta\,\phi$ is known to be an invariant. However, there is a star upon $\phi$. The combination $\beta\delta\phi^{*}$ transforms like $\phi$.

Hence I redefined the various quantities to make them look more like the modern ones. As $\beta\delta\,\phi^{*}$ transform as a $\phi$ field, I put

$$\nu=\beta\delta\,\phi^{*}\,,\qquad\,e=\psi$$

Then

$$F_{S}=\bar{e}\nu\,,\qquad F_{V0}=\bar{e}\gamma^{0}\nu$$

as one would expect from a scalar and vector interaction, with $\gamma^{0}=\beta$. The latter appears explicitly as part of the time component of the leptonic charged weak current of the standard model. Fermi's time component, on the other hand, indeed looks like a scalar, and following Blatt and Weisskopf it $is$ a scalar.

So, in conclusion, did Fermi get the wrong transformation property for his weak interaction Hamiltonian?



P.S.: It is true that the time component of a four-vector field operator transforms like a scalar in the non-relativistic limit (i.e. when one is allowed to approximate the boosts to Galileian transformations), but here the derivation seems to be theoretically wrong from the very beginning. Thus I wouldn't regard such an approximate equivalence as a valid justification.