One can "derive" the Klein-Gordon (KG) equation as follows (in natural units and $1+1$ dimensions, for simplicity). Starting from the relativistic relation $E=\sqrt{p^2+m^2}$, we use the quantum mechanical prescription $E\to i\partial_t$, $p\to-i\partial_x$, giving $i\partial_t=\sqrt{-\partial_x^2 + m^2}$. Squaring the result, acting on $\psi$, and rearranging gives the KG equation.
Many QFT books (e.g., Srednicki) start out by doing this as a first stab at relativistic quantum mechanics, but then show that there are negative energy solutions (and other problems), and hence the approach must be abandoned in favor of QFT.
My question is, what if we don't square both sides of $i\partial_t=\sqrt{-\partial_x^2 + m^2}$ ? Instead we define $$\sqrt{-\partial_x^2 + m^2}\,\psi(x)=\int dk\, \hat\psi(k)\sqrt{k^2 + m^2}e^{ikx}.$$
I think that $\sqrt{-\partial_x^2 + m^2}$ is then a genuine square root of $-\partial_x^2 + m^2$ as operators on the Hilbert space $L^2$. Defining $H = \sqrt{-\partial_x^2 + m^2}$, we see that the problematic "negative energy states" arise because $(i\partial_t)^2 = H^2$ has some solutions corresponding to $i\partial_t = H$ and some corresponding to $i\partial_t = -H$. From this viewpoint, the latter can hardly be said to be "negative energy" at all! They are not eigenstates of the Hamiltonian, but of the negative Hamiltonian, which we inadvertently let allowed into the picture by squaring everything.
Most references mention that squaring is what leads to the negative states, but they don't mention that $H = \sqrt{-\partial_x^2 + m^2}$ can also be defined as above. Does a good theory result from $\pmb{i\partial_t\psi = \sqrt{-\partial_x^2 + m^2}\,\psi}$ ? In particular, does this equation allow for a relativistic theory of a spin-0 particle without resorting to QFT?
I should add that a similar question was asked here, but that was more concerned with the idea of expanding the square root. The top answer there mentions something similar to the Fourier-transform idea I gave here, but just says "it's a pain to work with ... because all our lovely tools from linear algebra don't really work too well."
