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For bounded operators with trace-class commutator, the multiplicative commutator differs from identity by a trace-class operator. Along $F(t)=e^{tA}{e}^B{e}^{-tA}{e}^{-B}$, the trace of its logarithmic derivative is the constant $\mathrm{Tr}[A,B]$. This follows by writing $e^{B}A{e}^{-B}-A$ as the integral of conjugates of $[B,A]$ and using trace invariance under bounded similarities. Integrating the Fredholm determinant derivative from $F(0)=I$ proves the formula. Individual traces of $A$ and $B$ need not exist.

For a real standard Gaussian random vector $Z$, a real symmetric matrix $Q$, and $I-2Q$ strictly positive,Diagonalize $Q$ and complete the square in each coordinate. Failure of strict positivity makes the positive exponential integral divergent. The same formula extends to a self-adjoint trace-class quadratic form of an isonormal Gaussian process, with square-summable linear coefficients and a Fredholm determinant. A linear-plus-quadratic Gaussian random field functional is reduced to this form by its covariance square root.

Give $G(H)$ its natural trace-class topology: differentiability means $F(t+h)=F(t)+h\dot{F}(t)+o_1(h)$, with remainder small in trace norm. The Fredholm determinant is the continuous extension of finite-rank determinants; it is multiplicative by finite-rank approximation and satisfiesnear zero. For example its exterior-power series bounds the remainder by $e^{\Vert K{\Vert }_1}-1-\Vert K{\Vert }_1$, proving that estimate directly.

Factor $F(t+h)=F(t)[I+hF(t{)}^{-1}\dot{F}(t)+o_1(h)]$. Multiplicativity and the preceding expansion giveThe determinant is nonzero for invertible $F$, so division yields the required logarithmic derivative. Differentiability only in the bounded-operator norm would not justify this trace formula; the specified topology is essential.

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 $V=k_0^2(n^2-1)$.