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For $H=\int [a{\varphi }^2/2+\kappa |\nabla \varphi {|}^2/2+\gamma ({\nabla }^2\varphi {)}^2/2+B{\varphi }^2|\nabla \varphi {|}^2]$ and an even positive trial kernel $J(q)$, define $S_0={\sum }_q^{+}1/J(q)$ and $S_2={\sum }_q^{+}{q}^2/J(q)$ using a positive-wavevector sum for a real field. Isserlis theorem gives $⟨H_4{⟩}_0=4B{S}_0{S}_2/V$. The Feynman-Bogoliubov inequality therefore has stationary kernelsThus both the mass and gradient coefficient shift. In the thermodynamic limit, their self-consistency equations areA coarse-graining ultraviolet cutoff $\Lambda$ is necessary for the first integral in three dimensions: its large-$k$ radial integrand tends to a constant. Physical solutions require $J(q)>0$.

Let the stationary scattering potential fluctuation have autocorrelation function of a random fieldBecause $W$ is centered, this is also its covariance function. Multiplying the two real wave phase integrals and taking expectations givesThe two minus signs cancel. This is the wave phase variance, since the wave phase mean is zero. More generally the phase covariance in the first Rytov approximation replaces the first kernel by $b({\mathbf{r}}_1,{\mathbf{r}}^{\prime })$ and the second by $b({\mathbf{r}}_2,{\mathbf{r}}^{\prime \prime })$. Finite observation/scattering windows, or suitable weighted-integrability hypotheses, make these double integrals well-defined in a stationary infinite-medium model.

The correlation needed here is that of the scattering potential fluctuation. With the printed $V=n^2-1$, the covariance of a squared random field isIt involves a fourth moment of $n$, so its value is not generally determined by the ordinary two-point correlation $C_n=⟨n({\mathbf{r}}^{\prime })n({\mathbf{r}}^{\prime }+\mathbf{\rho })⟩$ alone. If one additionally assumes a zero-mean Gaussian random field, Isserlis theorem yields $C_W=2C_n^2$. That assumption is not printed and must not be inserted silently. Alternatively, for a physical weak fluctuation $n_{\mathrm{phys}}=1+\eta$, $W\simeq 2\eta$ gives $C_W\simeq 4C_{\eta }$. The general answer uses $C_W$; either reduction to a refractive index two-point correlation requires an extra assumption.

The printed strong white noise assumption does not imply a normal distribution or even a finite fourth moment. Thus it does not, by itself, determine the covariance of a quadratic transform. We give the intended Gaussian calculation, and then the general finite-fourth-moment answer.

If $\eta$ is Gaussian, $X$ is a centered Gaussian process. Put $v={\gamma }_X(0)=136/15$ and $U_t=X_t/\sqrt{v}$. Since $H_1(u)=u$ and $H_2(u)=u^2-1$, the centered transform isThe supplied Hermite polynomial identity makes the cross terms vanish and givesEquivalently,The constant $a$ has no effect on covariance.

For a general iid noise with $E{\eta }^4<\infty$, put $m_3=E{\eta }^3$ and ${\kappa }_4=E{\eta }^4-3$, its fourth cumulant. Independence and expansion of third and fourth moments give, for $k=|h|$,whereSumming the geometric series explicitly givesThe covariance of quadratic transforms of a linear process follows from a fourth-moment expansion consisting of the three Isserlis theorem pairings, plus the fourth-cumulant contribution when all four noise indices coincide. The third-moment contribution similarly requires three coincident indices. This proves the general formula without assuming a normal distribution. For Gaussian white noise $m_3={\kappa }_4=0$, recovering the simpler answer. If $c\ne 0$ and the noise has infinite fourth moment, $Z_t$ need not have finite variance, so an autocovariance function may not exist.

For a centered Gaussian random field, Wick theorem takes the classical form of the Isserlis theorem: every even correlator is a sum over pairings of two-point correlators, and every odd correlator vanishes. For example, with $G_i={\zeta }_G({\mathbf{x}}_i)$,A $2n$-point correlator has $(2n-1)!!$ pairings. Equivalently, all connected correlation functions beyond order two vanish. The covariance completely determines the centered Gaussian statistics; higher ordinary correlators can be nonzero, but contain no independent connected information. For a field with nonzero mean, apply these statements to the centered field and then restore its mean.

Choose an even positive $J(q)$. Each independent complex Fourier mode of the real field has density $(J(q)/\pi )e^{-J(q)|{\varphi }_q{|}^2}$, soThe quadratic part therefore has expectation ${\sum }_q^{+}G(q)/J(q)$. For the quartic part, apply Isserlis theorem to its four jointly centered Gaussian random variables, or to their real and imaginary components. Pairing the two differentiated fields together imposes $q_2=-q_1$, and its contribution isEach of the two other pairings instead gives $-(B/V)|{\sum }_{q}q/J(q){|}^2=0$, because the summand is odd under $q\mapsto -q$. Thus only the displayed contribution remains. Each full sum is twice its positive-wavevector sum for a real field; putting $S_0={\sum }_q^{+}1/J(q)$ and $S_2={\sum }_q^{+}{q}^2/J(q)$ gives $⟨H_4{⟩}_0=4B{S}_2{S}_0/V$.

Substitute these expectations, $F_0$ and $⟨H_0{⟩}_0$ into the Feynman-Bogoliubov inequality:The zero mode is absent by the composition constraint. The counting uses one representative of every nonzero $\{q,-q\}$ pair, so no extra factor of two is assigned to an independent complex amplitude.