 Solution

Define $E_{\mathbf{p}}=\sqrt{{\kappa }^2|\mathbf{p}{|}^4+m^2}$ and ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$. The Hermitian quantum field has the mode expansion of a free fieldwhere the annihilation operators and creation operators obey the bosonic canonical commutation relationsBecause $\pi =\dot{\varphi }$, its mode expansion of a free field isThe two nonzero cross commutators then giveand the equal-time field-field and momentum-momentum commutators vanish. This verifies the factor $1/\sqrt{2E_{\mathbf{p}}}$ without requiring Lorentz invariance.

Substitution into the Hamiltonian, followed by normal ordering, yieldsConsequently $[:H:,a_{\mathbf{p}}^{†}]=E_{\mathbf{p}}{a}_{\mathbf{p}}^{†}$: one application of a creation operator adds precisely the required quasiparticle energy. Before normal ordering there is the usual constant zero-point vacuum energy. If $m=0$, the exactly zero-momentum mode has $E_0=0$ and is a free canonical coordinate, not an oscillator with a normalizable oscillator vacuum state; in a finite box handle that mode separately or introduce a small positive mass. The formula applies directly to all modes with $E_{\mathbf{p}}>0$.