 Solution

A real scalar field suffices to describe one neutral species of bosonic quasiparticle. Let $\pi (\mathbf{x})$ be its canonical momentum. In three spatial dimensions, choose the positive classical HamiltonianThe same construction works in any number of spatial dimensions by replacing $d^{3}x$ appropriately. Under a spatial Fourier transform, ${\nabla }^2\varphi$ becomes $-|\mathbf{p}{|}^2\tilde{\varphi }$, so the Hamiltonian is a collection of independent quantum harmonic oscillators with squared angular frequency ${\omega }_{\mathbf{p}}^2={\kappa }^2|\mathbf{p}{|}^4+m^2$. The usual reality condition pairs opposite momenta; it does not introduce a second species.

The canonical action is $S=\int dt[\int d^{3}x\pi \dot{\varphi }-H]$. Varying $\pi$ gives $\pi =\dot{\varphi }$; eliminating it by the Legendre transform givesThis Lagrangian density is local: only the field and finitely many derivatives at the same point occur. It contains second spatial derivatives but only first time derivatives, so no extra canonical time data are needed. A square root of a differential operator in the classical Hamiltonian is unnecessary. The fourth-order spatial term breaks Lorentz invariance while preserving spatial rotations and translations. This is a free scalar field with quartic spatial dispersion.