To determine the absolute hand, one has to perform two dynamical simulations for the original model and the inverted model. This doubles the computational cost.
This assumes that the absorption effects can be ignored and the incident beam is a plane wave.
From the latter, the incident wavefunction can be assumed to be $(1, 0, 0, \cdots, 0)$ without loss of generality, because the global phase factor and the order of the beams are arbitrary.
Set up the structure matrix $A$ and diagonalize it as usual. Calculate the dynamical intensities by using a negative thickness $-t$. This corresponds to the dynamical intensities of the inverted structure at thickness $t$! In this way, one can avoid the computationally-heavy matrix diagonalization step.
In the Bloch wave method, the exit wavefunction at thickness $z$, $\psi(z)$, can be written as
$$ \psi(z) = \exp(2 \pi i z A) \psi(0),$$
where $\psi(0)$ is the incident wavefunction.
The matrix $A$ consists of real diagonal elements related to the excitation error and complex off-diagonal elements related to the structure factors $F$.
Without absorption effects, the structure factor of an inverted structure $F_{inv}$ is simply the complex conjugate of the structure factor of the original structure $F$ as $$F_{inv} = F^*,$$ where the asterisk denotes complex conjugate. Thus, the matrix of the inverse structure is also the complex conjugate of the original matrix: $$A_{inv} = A^*.$$
Therefore, $$\psi_{inv}(z) = \exp(2 \pi i z A_{inv})\psi(0) = \exp (2 \pi i z A^*)\psi(0) = \exp\left(\left(- 2 \pi i z A\right)^*\right)\psi(0) = \exp(- 2 \pi i z A)^* \psi(0) = \psi(-z)^*. $$
Noting that the observed intensities are the squared magnitude of this wavefunction, $$ I_{inv}(z) = |\psi_{inv}(z)|^2 = |\psi(-z)^*|^2 = I(-z),$$ this completes the proof.
Written by Takanori Nakane at the Institute for Protein Research, The University of Osaka.