 Outgoing branch of the angular spectrum

The outgoing branch of the angular spectrum selects the longitudinal wavenumber that represents forward radiation and bounded evanescent decay in a specified source-free half-space. For real $k>0$, time dependence $e^{-i\omega t}$ and propagation into $z>0$,The first root is positive real and the second positive imaginary. They accompany the factor $e^{i{k}_{z}z}$.^[ref]

The scalar Helmholtz equation permits both signs. For $q<k$, $e^{i{k}_{z}z}$ and $e^{-i{k}_{z}z}$ describe opposite longitudinal phase propagation under the stated time convention. For $q>k$, the corresponding factors are decaying and growing exponentials. A source-free upper half-space with no incident field from infinity uses the forward, bounded choice.

For an original transverse spatial frequency $q=1.25k$, the decaying root is $k_z=0.75ik$. At $z=\lambda$, amplitude changes by $e^{-1.5\pi }\simeq 0.00898$. Choosing the other sign would instead multiply it by approximately 111.3. A sign error can therefore become an enormous near-field propagation error without changing the input data.

This choice is a statement about the physical region and boundary conditions, not an instruction to delete reflected waves from every problem. An interface or cavity can require both propagation directions in a finite region. Likewise, propagation towards negative $z$ must use a convention consistent with that region. At $q=k$, the root is zero and the component is grazing; formulas that divide by $k_z$ need separate attention.

Backward reconstruction of a decayed evanescent angular-spectrum component multiplies measurement error by an increasing exponential. That is an inverse problem's conditioning issue, rather than evidence that the forward outgoing branch should grow. Stating the time convention, physical half-space and radiation condition prevents the square-root symbol from hiding these choices.