Fractional exponent in a scalar quantum field: Is energy and momentum conserved in this case? Assuming that I would have the following term in the Lagrangian for a scalar boson field $$L=\int d^4x g (\phi^{2-p} \phi^{\dagger 2+p}+\phi^{\dagger 2-p} \phi^{2+p}))$$
with a fractional number $p$. Now I am inserting the Fourier Expansion for the scalar boson:
$$\phi=(2 \pi)^{-4}\int d^4kA_\vec{k}e^{i \vec{k} \vec{x}}$$ Because of the fractional exponents I cannot use the relation $$\int f(x) dx)^q=\int \prod_{i=1}^q f(x_i)dx_i$$
How can I express above term in the Lagrangian in terms of multiple integrals over $k$?
Does it happen that I have not the coefficients $e^{i(\vec{k_1}+...+\vec{k_n}) \vec{x}}$ (which lead to energy/momentum-conservation Delta distributions $\delta(\vec{k_1}+...+\vec{k_n}) $ after integration over the spacetime)? Will there occur a energy/momentum-nonconserving term?
 A: The $e^{\vec{x}\cdot(\sum \vec{k})}$ which leads to momentum conservation at each vertex when we go to momentum space feynman diagrams is a perturbative result at heart. Your model as it stands is not cast in a way conducive of perturbation theory..
Let's try this instead, if we want to have something amenable to perturbation. 
(if $p<1$), then let's do a field redefinition $\phi\equiv \rho e^{i\phi}$, then we will have
$L = |\partial (\rho e^{i\phi})|^2 + 2\rho^4 \text{cos }(2p\phi)$
this can only be seen as a perturbative model if we are to expand in $p$ maybe? which would lead to an infinite number of vertices between $\rho^4$ and $n$ $\phi$s after expanding the cosine.. only then you can say that momentum is conserved at each vertex
Otherwise you'd have to solve the model nonperturbatively. With the guaranteed result that momentum will be conserved in the final answer, but without insight to what happens at each vertex.. because the picture of vertices and such is not valid to begin with!
