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For a complex harmonic field use complex conjugation in its wave-field correlation: $m=⟨{\psi }_{\mathrm{sc}}(x_1,z){\psi }_{\mathrm{sc}}(x_2,z{)}^{\ast }⟩$. Put $A(\nu )={\hat{\psi }}_{\mathrm{sc}}(\nu )$ and let $C(\nu ,{\nu }^{\prime })=⟨A(\nu )A({\nu }^{\prime }{)}^{\ast }⟩$. Direct substitution of part (a), before imposing any homogeneity, givesThis is the general answer. It need not depend only on $\xi =x_1-x_2$: even a deterministic sum of two distinct outgoing plane waves has cross terms depending on the midpoint. The requested separation-only wave-field correlation therefore presumes translation-homogeneous two-point statistics.

Under that assumption define the uncentered spectrum by $C(\nu ,{\nu }^{\prime })=2\pi S(\nu )\delta (\nu -{\nu }^{\prime })$. Then the height propagation of an angular-spectrum correlation isEquivalently, for the spectral measure of a stationary random field $F(d\nu )=S(\nu )d\nu /(2\pi )$, write $m=\int e^{i\nu \xi -2\mathrm{Im}qz}F(d\nu )$. This version includes coherent spectral atoms and avoids squaring Dirac delta functions. The propagating spectrum $|\nu |\le k$ is independent of height, while an evanescent contribution is multiplied by $e^{-2\sqrt{{\nu }^2-k^2}z}$. At $\xi =0$ the formula gives the mean wave intensity.

If the brackets instead mean the spatial autocorrelation of a finite-energy deterministic field, define $m_L(\xi ,z)=\int {\psi }_{\mathrm{sc}}(x+\xi ,z){\psi }_{\mathrm{sc}}(x,z{)}^{\ast }dx$. Parseval's identity gives the same integral with $S(\nu )=|{\hat{\psi }}_{\mathrm{sc}}(\nu ){|}^2$. That finite-energy formula must not be applied unmodified to an infinite stationary field or an infinite plane wave, whose transforms require a spectral-measure convention.