5.7 Bloch wave description 323
the following paragraphs, we will show that we do have enough information to
determine all the unknown parameters unambiguously.
Since the total energy of the incident electron is constant, all 2N Bloch waves
must have the same total energy; we say that the Bloch waves are degenerate. Since
the 2N wave vectors k
( j)
are, in general, different from each other, each Bloch
wave must correspond to a different kinetic energy, and therefore also a different
potential energy. This, in turn, means that different Bloch waves must travel at
different locations through the crystal. This is a very important observation, and
we will return to it in the next chapter.
The 2N wave vectors k
( j)
are ranked according to decreasing kinetic energy.
This means that the Bloch wave with the highest kinetic energy is number (1), and
that corresponding to the lowest kinetic energy is number (2N). The reader should
be warned that in some of the earlier literature on the Bloch wave theory the wave
vector with the lowest kinetic energy was taken to be k
(1)
.
We define the unit vector n as the surface normal to the sample, in the direction
opposite
†
to the incident beam, i.e. along the −z-direction. Since both the incident
wave and the wave inside the crystal are solutions to the Schr¨odinger equation,
which is a second-order equation in the spatial coordinates, both the functions and
their first-order derivatives must be continuous across the entrance plane of the
crystal. It can be shown quite generally (e.g. [Met75]) that this continuity condition
implies that the tangential component of the wave vector must be conserved across
the entrance plane. The only component of the wave vector that can change upon
entering the crystal is, therefore, the normal component. It is then convenient to
write the solution vectors k
( j)
as
k
( j)
= k
0
+ γ
( j)
n. (5.48)
This expression guarantees that all wave vectors k
( j)
have the same component
normal to the vector n. Let us use this expression to rewrite equation (5.45):
k
2
0
−
k
( j)
+ g
2
= k
2
0
−
k
0
+ γ
( j)
n + g
2
;
=−g · (2k
0
+ g) − 2n · (k
0
+ g)γ
( j)
−
γ
( j)
2
;
= 2k
0
s
g
− 2n · (k
0
+ g)γ
( j)
−
γ
( j)
2
. (5.49)
We have used the definition (2.89) of the excitation error s
g
in the last step. At this
point, we will make the assumption that the coefficients γ
( j)
are small compared
to k
0
, so that the quadratic term in γ
( j)
can be ignored. This is equivalent to the
high-energy approximation used in the previous sections, and as a result only N of
the 2N wave vectors k
( j)
remain.
‡
The N wave vectors which we chose to ignore
†
Note that some authors select the foil normal to point along the incident beam. The reader should be aware that
this leads to sign differences for the coefficients γ
( j )
.
‡
Equation (5.49) is now linear instead of quadratic in γ
( j )
.