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For a trace-class operator $K$ on a Hilbert space, the Fredholm determinant is the convergent product over its eigenvalues, counted with algebraic multiplicity. For self-adjoint $K$, spectral theorem for compact self-adjoint operators reduces the product to a diagonal calculation. For $\Vert K\Vert <1$, $\log \det (I+K)={\sum }_{n\ge 1}(-1{)}^{n+1}\mathrm{tr}(K^n)/n$. This identity exposes the successive cumulants in a Gaussian quadratic exponential moment.

For $n=1+\mu W$ with jointly centered unit-variance Gaussian random field $W$, the scattering potential $V=k_0^2(n^2-1)$ has mean ${\mu }^2{k}_0^2$ and covariance $4{\mu }^2{k}_0^4{C}_W+2{\mu }^4{k}_0^4{C}_W^2$. It is not Gaussian. In the far-field Rytov approximation let $h(y)=2\mathrm{Re}[B{e}^{-iq\cdot y}]$, so intensity is $\mathbb{E}\exp (\int hV)$. Expansion through order ${\mu }^2$ includes both the mean term and the covariance term shown. The remainder starts at order ${\mu }^4$ under suitable exponential integrability. Keeping a covariance correction while omitting the same-order mean term is inconsistent. The exact quadratic model uses the Gaussian quadratic exponential moment.

The zero-mean fluctuation is $W$, not the refractive index $n$: $⟨n⟩=1$. Also the full scattering potential has a nonzero mean and is not Gaussian:The usual Gaussian closure requires $W$ to be a jointly Gaussian random field, not merely to have a Gaussian distribution at each point. We first give the intended leading weak-fluctuation result with the linearized potential $V_1=2\mu k_0^{2}W$, then account for the quadratic term.

Keep the definitions $q,B$ from the preceding solution and let ${\chi }_1=B{\int }_D{V}_1(y)e^{-iq\cdot y}dy$. Because the incident field has unit modulus,The exponent is a real Gaussian linear functional $Z={\int }_{D}h{V}_1$. If $C_W(y-z)=⟨W(y)W(z)⟩$, the linearized potential has autocorrelation function of a random field $C_1(s)=4{\mu }^2{k}_0^4{C}_W(s)$. Complete the square in the one-dimensional Gaussian density of $Z$: $⟨e^Z⟩=\exp (⟨Z⟩+\mathrm{Var}Z/2)$. Here its mean is zero, soThis is Gaussian intensity in the first Rytov approximation. In a more explicit complex notation, putThenThe second term is the pseudo-covariance of a complex random variable contribution. It does not vanish for an arbitrary finite real medium. The real-kernel expression shows that the combined exponent is nonnegative, even though its two complex-form contributions need not be separately nonnegative.

For the full potential, Gaussian moment pairing theorem gives its centered covariance functionIts uncentered autocorrelation is $R_V(s)=C_V(s)+{\mu }^4{k}_0^4$. The first Rytov approximation is linear in $V$, but its intensity is an exponential average; applying a Gaussian formula exactly to this quadratic potential would be incorrect. A consistent expansion through order ${\mu }^2$ isThe first term is the coherent mean-potential correction. Odd powers vanish by the symmetry of the joint Gaussian field. The variance term can equally use $C_V=R_V-{\overline{V}}^2$ through this order. Thus the mean shift must not be discarded while retaining the order-${\mu }^2$ fluctuation intensity.

An exact expression within the far-field, first-Rytov approximation model is also available when the full $n^2$ is retained. Assume the restriction of $W$ to $D$ is a Gaussian element of the real Hilbert space $L^2(D)$, with covariance operator $C$ of kernel $C_W(y-z)$. Let $H$ denote multiplication by $h$, ${\ell }_h=2\mu k_0^{2}h$, $Q={\mu }^2{k}_0^2{C}^{1/2}H{C}^{1/2}$ and $b_h=C^{1/2}{\ell }_h$. The Gaussian quadratic exponential moment isHere the determinant is the Fredholm determinant; require $I-2Q$ strictly positive for finiteness. Diagonalize the self-adjoint trace-class operator $Q$ and complete the square in each Gaussian coordinate to derive the formula. On a bounded $D$, unit pointwise variance makes $\mathrm{tr}C=|D|$, so the covariance is trace class under the stated measurable $L^2$ hypothesis. Small enough $\mu$ satisfies the positivity condition. Expanding the determinant and inverse recovers the preceding mean and covariance terms. For small $\mu$ this is also determined by the potential autocorrelation: since $C_W\in [-1,1]$,using the branch near zero. This makes explicit the extra joint-Gaussian assumption behind an exact correlation-based answer.

Finally, Gaussian marginals alone are insufficient. Take $X\sim N(0,1)$ and independent random signs $S_j$. The stationary sequence $W_j=S_{j}X$ has Gaussian marginals and covariance $⟨W_i{W}_j⟩={\delta }_{ij}$, just like independent standard Gaussian variables. However $⟨e^{t(W_1+W_2)}⟩=(1+e^{2t^2})/2$, whereas for the independent Gaussian pair it is $e^{t^2}$. The different intensities show why the joint hypothesis is needed. The mean-zero hint is interpreted for $W$, and the linearized Gaussian formula is not claimed to be exact for $V=k_0^2(n^2-1)$.