5.5 Slice methods 313
If a general incident amplitude is described by the column vector S(0), then the
general solution at the exit plane of the crystal is given by
S(z
0
) = SS(0).
Hence, the scattering matrix S can be computed by explicitly solving the DHW
equations for N independent initial conditions. However, the exponential expression
for the matrix S in terms of the structure matrix A shows that S depends in a non-
linear way on the crystal thickness z
0
. This means that the previous equation should
actually read as
S(z
0
) = S(z
0
)S(0).
The scattering matrix must, therefore, be recalculated for every crystal thickness.
In Section 5.7 on Bloch waves we will describe how this matrix can be written as
the product of three matrices, with only one of them being dependent on the crystal
thickness.
In the absence of absorption, the sum of the intensities of the beams leaving the
crystal must equal the incident intensity, and the matrix relation above expresses a
unitary transformation, i.e. a transformation that leaves the norm of the vector S
unchanged. Consequently, the dynamical scattering process can be regarded as a
rotation in an N -dimensional space, and S is a hermitian matrix. When absorption
occurs, the transformation is no longer unitary, and the total intensity decreases
with increasing crystal thickness.
5.5 Slice methods
The exponential of a matrix, as defined in equation (5.27), converges after a small
number of terms if the crystal thickness z
0
is sufficiently small. We can make use
of the properties of exponential functions and rewrite relation (5.25) as follows:
S(z
0
) = S(z
0
)S(0);
= S(n)S(0);
=
[
S()
]
n
S(0).
We have made use of the fact that (x)
ab
= (x
a
)
b
. The total crystal thickness equals
z
0
= n, and, if we take sufficiently small, then the series expansion for S()
converges rapidly; the scattering matrices for thicknesses that are a multiple of
are obtained by repeated multiplication of S() with itself, and matrix multiplication
can be carried out efficiently on fast processors.
We have subdivided the crystal into a stack of slices of equal thickness ,as
illustrated in Fig. 5.1(a). Now it becomes apparent how we can use the matrix