 Solution

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.