= Solution
Choose
$$
m(x_\parallel,\mathbf x_\perp)=\int\frac{dq_\parallel\,d^{d-1}q_\perp}{(2\pi)^d}\,
\widetilde m(q_\parallel,\mathbf q_\perp)e^{i(q_\parallel x_\parallel+\mathbf q_\perp\cdot\mathbf x_\perp)}.
$$
For a real scalar field $\widetilde m(-q)=\widetilde m(q)^*$. The parallel derivative becomes $iq_\parallel$, and the transverse <Laplacian> becomes $-q_\perp^2$. The latter is squared in the energy, producing $q_\perp^4$, not $q_\perp^2$. Thus
$$
\boxed{\beta H=\frac12\int_q[Kq_\parallel^2+Lq_\perp^4+t]|\widetilde m(q)|^2-h\widetilde m(0).}
$$
The 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> $[Kq_\parallel^2+Lq_\perp^4+t]^{-1}$.
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