# How do derivative couplings affect canonical quantization?

Comments to the question (v2):

OP is considering the higher-derivative Lagrangian density $$ {\cal L}_1~=~ \frac{1}{2}(\partial\phi)^2 +\frac{g\phi}{2} (\partial^2\phi)^2,\tag{1} $$ where $g$ is a coupling constant. We use Minkowski sign convention $(+,-,-,-)$.

Quantum mechanically, the model is not unitary and therefore ill-defined, cf. the Ostrogradsky instability. However, classically (and as a formal perturbative expansion in Feynman diagrams), it makes sense.

One can in principle apply the Ostrogradsky procedure for higher-derivative theories. However, here we will just follow our nose: It seems natural to try to lower the number of derivatives by imposing a constraint $\Phi\approx \partial^2\phi$ via a Lagrange multiplier $B$. In other words, consider the Lagrangian density $$ {\cal L}_2~=~\frac{1}{2}(\partial\phi)^2 +B\Phi+\partial_\mu B~\partial^{\mu}\phi+\frac{g\phi}{2} \Phi^2 ~\sim~ \frac{1}{2}(\partial\phi)^2 +B(\Phi-\partial^2\phi)+\frac{g\phi}{2} \Phi^2 ,\tag{2}$$ where the $\sim$ symbol means equality modulo total spacetime divergence terms.

It is tempting to integrate out the $\Phi$ variable again. Then we arrive at the interesting Lagrangian density $$ {\cal L}_3~=~\frac{1}{2}(\partial\phi)^2 +\partial_\mu B~\partial^{\mu}\phi-\frac{B^2}{2g\phi} .\tag{3}$$

In other words, the Lagrangian densities (2) and (3) reduce to the original Lagrangian density (1) when we integrate out the new variables

$$ {\cal L}_2\quad\stackrel{\Phi}{\longrightarrow}\quad{\cal L}_3\quad\stackrel{B}{\longrightarrow}\quad{\cal L}_1 .\tag{4}$$ The Lagrangian densities (2) and (3) do not have higher-order derivatives. Therefore the usual techniques can be applied to find the Hamiltonian formulation and Feynman rules. (In the ${\cal L}_3$ case, the $\phi$ field should apparently be expanded around a non-zero classical solution. Perhaps a field redefinition $\phi=e^{\varphi}$ would be useful?) We leave that as an exercise to the reader.