# How do derivative couplings affect canonical quantization?

Consider a Lagrangian for a scalar field $\phi$ with an interaction term $$\mathcal{L}_{int} = (\partial^2 \phi)^2 \phi.$$ Here I'm suppressing all indices for brevity. Now, this is just a three-point interaction with some extra momentum factors, so there seems to be no problem writing down the Feynman rule.

However, actually trying to carry out the procedure with canonical quantization looks tricky. The interaction term can't be written as a function of $\phi$ and $\partial \phi$, even if you use integration by parts, so we must let the Lagrangian have $\partial^2 \phi$ dependence. But then I don't know how to produce the Hamiltonian, because that's based on a Legendre transformation from $(\phi, \partial \phi)$ to the fields and momenta $(\phi, \pi)$. The new dependence on $\partial^2 \phi$ seems to totally mess this up; I don't know how to handle it in classical mechanics, let alone QFT!

How do you actually obtain the Feynman rule for this interaction, in canonical quantization?

• This term has dimension 7 and is non-renormalizable. But in EFT, you should write down any terms that obey the symmetries of your theory, and this term should show up. Feynman rule for it is just $p^4$. – Nahc Apr 3 '16 at 18:39
• @HChan I know what the Feynman rule is, but it looks like actually deriving it is sort of tricky. Do you know how to? – knzhou Apr 3 '16 at 19:48
• You just need to use the Feynman rule for free scalar field theory. Say the coupling constant in your interaction term is $\lambda$, then the first order correction is $\lambda \times (#)$, while the term (#) should come from the zero order result. – Nahc Apr 4 '16 at 1:25
• I...don't think one can canonically quantize terms with higher derivatives because those do not have an associated Hamiltonian picture. – ACuriousMind Apr 15 '16 at 21:30
• @ACuriousMind: It is possible to obtain a Hamilton for higher-derivative theories; you can use the techniques that Ostrogradski used to prove that such theories are unstable. There's a nice explanation of how the derivation proceeds in Section 2 of this paper by R.P. Woodard. – Michael Seifert Apr 15 '16 at 21:47

1. 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 $(+,-,-,-)$.

2. 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.

3. 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.

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

5. 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.

Given a Lagrangian the states of the theory are given by the solutions to the linearised equations of motion, i.e. the quadratic part of the Lagrangian. Assuming that you have a standard kinetic term the solutions are plane waves $e^{i k x}$. Then an efficient way to find the Feynman rule for a vertex is to replace each field by its plane wave solutions. In your case: $$\lambda (\partial^2 \phi)^2 \phi \longrightarrow \lambda (\partial^2 e^{i k_1 x}) (\partial^2 e^{i k_2 x}) e^{i k_3 x} \sim \underbrace{\lambda k_1^2 k_2^2}_{vertex} \times \text{plane waves}$$ The same result can be found using path integral, Wick theorem, etc., but this approach is particularly efficient. A simple and well-known example where agreement can be checked is scalar electrodynamics (see for example these notes).

Note that in general canonical quantization in QFT is performed by finding the physical states as solutions to the free equations of motion in order to get the free field creation and annihilation operators, and then constructing amplitudes perturbatively using Wick theorem, Green's functions, etc. In this case the fact that your vertex contains derivatives has no impact (again as an example see the canonical quantization of scalar QED). Indeed trying to quantize the full Hamiltonian directly is not possible in general.

Nonetheless if you insist for trying to quantize the full Lagrangian, or at least to write an Hamiltonian, then you should look at the Ostrogradski formalism that was set up to deal with higher-order Lagrangians, as pointed out in comments. A nice recent review is arxiv:1506.02210.