Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 69 / 2 / a / Solution

ChooseFor a real scalar field $\tilde{m}(-q)=\tilde{m}(q{)}^{\ast }$. The parallel derivative becomes $i{q}_{\parallel }$, and the transverse Laplacian becomes $-q_{\perp }^2$. The latter is squared in the energy, producing $q_{\perp }^4$, not $q_{\perp }^2$. ThusThe zero-momentum source is understood in a finite box or as a uniform-field distribution before taking the thermodynamic limit. With an orthonormal box transform the source is $-h\sqrt{V}{m}_0$. The positive coefficients and $t>0$ give a normalizable Gaussian field theory, with mean uniform field $\overline{m}=h/t$ and connected Fourier covariance $[K{q}_{\parallel }^2+L{q}_{\perp }^4+t{]}^{-1}$.