Energy of multiple particles in quantum field theory

Question

Consider the free scalar theory with the zeroed Hamiltonian (i.e. such that the vacuum energy is zero). What is the energy of a multi-particle state $\phi^n|\Omega\rangle$ or (perhaps the more physically meaningful normal-ordered) $:\phi^n:|\Omega\rangle$? (where $\phi$ is appropriated smeared such that it is localized over some small compact region.)

I reckon that the Hamiltonian is simply proportional to the number operator, so the energy would just scale linearly with $n$.

More generally, for any relativistic QFT, is it universally true that the energy, excited by some compactly supported operator $O$ applied $n$ times, of $O^n|\Omega\rangle$ grows linearly with $n$ at the leading order?

Answer 1

This is only necessarily true for non-interacting quantum field theories. A really simple example is the second-quantized description of atoms in an optical lattice. If the atoms don't interact, then we just have $H=\sum_i \delta_i n_i$, where $n_i$ is the number operator for site $i$. But if they do, then you can get extra terms like $c\sum_i n_i^2$, which describes attractive it repulsive interactions depending on the sign of $c$. The eigenstates are the same, but the spectrum is no longer linear in $n$.