Past exam of the quantum field theory course of the University of Oxford / 2019 / 1

A real scalar field suffices to describe one neutral species of bosonic quasiparticle. Let $\pi (\mathbf{x})$ be its canonical momentum. In three spatial dimensions, choose the positive classical HamiltonianThe same construction works in any number of spatial dimensions by replacing $d^{3}x$ appropriately. Under a spatial Fourier transform, ${\nabla }^2\varphi$ becomes $-|\mathbf{p}{|}^2\tilde{\varphi }$, so the Hamiltonian is a collection of independent quantum harmonic oscillators with squared angular frequency ${\omega }_{\mathbf{p}}^2={\kappa }^2|\mathbf{p}{|}^4+m^2$. The usual reality condition pairs opposite momenta; it does not introduce a second species.

The canonical action is $S=\int dt[\int d^{3}x\pi \dot{\varphi }-H]$. Varying $\pi$ gives $\pi =\dot{\varphi }$; eliminating it by the Legendre transform givesThis Lagrangian density is local: only the field and finitely many derivatives at the same point occur. It contains second spatial derivatives but only first time derivatives, so no extra canonical time data are needed. A square root of a differential operator in the classical Hamiltonian is unnecessary. The fourth-order spatial term breaks Lorentz invariance while preserving spatial rotations and translations. This is a free scalar field with quartic spatial dispersion.

For a Lagrangian density depending on $\varphi$, ${\partial }_t\varphi$, ${\partial }_i\varphi$ and ${\partial }_i{\partial }_j\varphi$, direct variation givesTake $\delta \varphi$ to have compact support, so every boundary term vanishes. One integration by parts for the first-derivative terms and two for the second-derivative term yield the Euler-Lagrange field equation with second derivativesThe plus sign on the last term is essential. For the action in a, it is particularly transparent to vary the Laplacian directly:Thus the classical equation of motion isHere ${\nabla }^4=({\nabla }^2{)}^2$ is the biharmonic operator. Substitution of the plane wave $\varphi =A{e}^{-i\omega t+i\mathbf{p}\cdot \mathbf{x}}$ givesA real classical solution combines complex-conjugate plane waves. Quantization assigns the positive branch to the excitation energy, giving the required dispersion relation. The negative-frequency branch supplies the creation operator part of the field, rather than negative excitation energy. Independently, the Hamiltonian in a gives $\dot{\varphi }=\pi$, $\dot{\pi }=-{\kappa }^2{\nabla }^4\varphi -m^2\varphi$, confirming the same sign.

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 $E_{\mathbf{p}}>0$.