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Use $\hat{\psi }(\nu )=\int \psi (x,0)e^{-i\nu x}dx$ and $q(\nu )=\sqrt{k^2-{\nu }^2}$ on the upward branch. If the uncentered field correlation is translationally homogeneous and $⟨\hat{\psi }(\nu )\hat{\psi }({\nu }^{\prime }{)}^{\ast }⟩=2\pi S(\nu )\delta (\nu -{\nu }^{\prime })$, its correlation at height $z$ is the displayed integral. The propagating spectrum is unchanged; evanescent contributions decrease as $e^{-2\sqrt{{\nu }^2-k^2}z}$. A spectral measure of a stationary random field permits discrete coherent components. Without correlation homogeneity a double spectral covariance integral depends on both observation coordinates rather than only their separation.

For the Fourier transform ${\hat{E}}_L(\nu )=(2\pi {)}^{-1}{\int }_{-L/2}^{L/2}E(z)e^{-i\nu z}dz$ of a stationary random field, the periodogram measure $(2\pi /L)\mathbb{E}|{\hat{E}}_L(\nu ){|}^{2}d\nu$ converges weakly to the spectral measure of a stationary random field. The expected squared modulus is $(4{\pi }^2{)}^{-1}{\int }_{-L}^L(L-|\zeta |)C(\zeta )e^{-i\nu \zeta }d\zeta$. The factor $2\pi /L$ follows from Parseval identity, and preserves the total power per unit length. It also avoids meaningless squares of Dirac delta distributions for coherent plane-wave components.

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.

Inverse Fourier transform of the expected value givesindependent of both coordinates. Its squared modulus is the coherent scalar wave intensity $e^{-{\sigma }^2}$, which must not be confused with the mean wave intensity.

For the latter use the stationary two-point statistics. Let $F_0(d\nu )$ be the uncentered spectral measure of a stationary random field of $E(0,z)$, normalized bySince $|E(0,z)|=1$, its total mass is $\int F_0(d\nu )=1$. Under free paraxial propagation each spectral component is multiplied by $e^{-i{\nu }^{2}x/(2k)}$; in its wave-field correlation this wave phase multiplies its complex conjugate and cancels. The stationary intensity under free paraxial propagation is consequentlyStationarity is essential to this pointwise statement: it makes the spectral covariance diagonal. Conservation of integrated power alone would not show that a particular spatial point keeps the same scalar wave intensity.

The same result follows locally from $I_x+J_z=0$, where $J=\mathrm{Im}(E^{\ast }{E}_z)/k$. Taking expectations makes $⟨J⟩$ independent of $z$, so ${\partial }_x⟨I⟩=0$. Thus coherent scalar wave intensity and mean wave intensity are both constant in propagation, with different values. The incoherent contribution has mean $1-e^{-{\sigma }^2}$. Individual realizations can show focusing and changing scalar wave intensity even though the expected value remains constant. All these propagation statements concern the free parabolic wave equation from part (a).

Put $\beta =k\xi \mu$. The random phase screen producesInterpret the normality assumption as a jointly Gaussian process, not merely Gaussian one-point marginal distributions. Write its normalized autocorrelation as $\rho (\zeta )=⟨W(z+\zeta )W(z)⟩$. The difference $W(z+\zeta )-W(z)$ is a zero-mean Gaussian random variable with variance $2(1-\rho (\zeta ))$. Its characteristic function therefore gives the Gaussian phase-screen correlationWithout joint Gaussianity the autocorrelation alone does not determine this expectation.

For an acoustic pressure amplitude $\psi$, with background mass density ${\rho }_0$, the time-averaged acoustic energy flux isAn acoustic potential convention changes the dimensional prefactor. The physical field includes the carrier $e^{ikx}$: a transverse Fourier transform mode has axial wavenumber ${\kappa }_{\nu }=k-{\nu }^2/(2k)$ in the parabolic wave equation. Thus its spectral acoustic flux is proportional to ${\kappa }_{\nu }|\hat{E}(x,\nu ){|}^2$, or to $k|\hat{E}(x,\nu ){|}^2$ at leading paraxial order. Differentiating the reduced field alone would omit the carrier contribution.

The infinite stationary illumination has a spectral measure of a stationary random field, rather than a finite ordinary value of $⟨|\hat{E}{|}^2⟩$ at each $\nu$. To specify the normalization, truncate the screen to an aperture of length $L$, propagate this truncated initial field, and setThe propagation multiplier has modulus one. Using the Gaussian phase-screen correlation and the overlap of the two aperture intervals givesMultiplication by ${\kappa }_{\nu }/(2{\rho }_0\omega )$ gives the ensemble-averaged pressure-normalized modal acoustic energy flux for this finite-aperture convention.

The infinite-aperture spectral normalization consistent with the specified Fourier transform isThe integral and limit are understood as spectral measures of a stationary random field when necessary. By Parseval's identity, this is power per unit transverse length: ${\int }_{\mathbb{R}}{S}_E(d\nu )=C_E(0)=1$. The ensemble-averaged axial spectral acoustic flux is independent of $x$:Here the notation includes Dirac delta distributions in the spectral measure of a stationary random field; the leading-order total flux is $I_{\mathrm{inc}}$. The ${\kappa }_{\nu }$ correction is meaningful within $|\nu |\ll k$, the range of the paraxial approximation.

If $\rho (\zeta )\to 0$ and $e^{{\beta }^2\rho (\zeta )}-1$ is integrable, the stationary phase-screen power spectrum separates into coherent and diffuse wave fields:The coherent fraction is $e^{-{\beta }^2}$ and the diffuse fraction is $1-e^{-{\beta }^2}$. If additionally $|\beta |\ll 1$, the stationary phase-screen power spectrum is $(1-{\beta }^2)\delta +{\beta }^2\hat{\rho }+O({\beta }^4)$ with the same Fourier transform convention. Small $\mu$ alone does not imply small accumulated phase $k\xi \mu$.

At each point the Gaussian process has a standard normal marginal distribution. Its characteristic function gives the coherent screen fieldIt is constant in $z$, so its Fourier transform is $m\delta (\nu )$. Applying the Fresnel propagator to this zero-transverse-wavenumber component gives the ensemble-averaged reduced spectrumThus the reduced mean is independent of $x$. The physical mean $⟨\psi (x,z)⟩=e^{ikx}{e}^{-{\beta }^2/2}$ still has its carrier phase. The coherent attenuation by a Gaussian phase screen takes place at the screen, with no further coherent attenuation in homogeneous space.

The full spectral acoustic flux depends on the second moment, not the square of this mean. For a stationary process, the generalized cross-spectral correlation isIndeed its propagation factors cancel on $\nu ={\nu }^{\prime }$. This explains why the ensemble-averaged spectral acoustic flux in (b) is also independent of $x$, even for nonzero transverse wavenumbers whose individual complex amplitudes change phase. The coherent atom has weight $|m{|}^2=e^{-{\beta }^2}$; the full stationary phase-screen power spectrum also contains fluctuations. Under the decay assumptions in (b), their integrated weight is $1-e^{-{\beta }^2}$. One must use these weights in a spectral measure of a stationary random field, rather than square a Dirac delta distribution. A phase-only screen preserves the leading-order total acoustic energy flux, although it reduces its coherent part.

For the random phase screen $E(z)=e^{i\beta W(z)}$, the Gaussian phase-screen correlation and spectral measure of a stationary random field giveIf $\rho (\zeta )\to 0$ and $e^{{\beta }^2\rho }-1$ is integrable, this is $e^{-{\beta }^2}\delta (\nu )$ plus the continuous density $(2\pi {)}^{-1}{e}^{-{\beta }^2}\int (e^{{\beta }^2\rho (\zeta )}-1)e^{-i\nu \zeta }d\zeta$. The total mass is one, split into coherent mass $e^{-{\beta }^2}$ and diffuse mass $1-e^{-{\beta }^2}$. Without the decay and integrability assumptions there may be additional spectral atoms or singular components.

For a complex random wave amplitude, this two-point expectation retains the coherent field and uses a complex conjugate on the second amplitude. It is not a normalized wave-field correlation coefficient. Subtracting $\mathbb{E}[E(\mathbf{r})]\mathbb{E}[E({\mathbf{r}}^{\prime }){]}^{\ast }$ gives the centered covariance function. At coincident points the uncentered wave-field correlation equals the mean wave intensity. Under second-order stationarity it depends only on the separation and has a spectral measure of a stationary random field. A deterministic field is included by taking a degenerate ensemble.