Physics Asked on July 1, 2021
Consider the elastic scattering of a particle, which can be represneted by a plane wave with wave vector $mathbf k$, from a periodic potential. We know that $mathbf{k}$ can be scattered into $mathbf{k}’$ if
$$mathbf{k}’-mathbf{k}=mathbf{G}$$
$$|mathbf{k}|=|mathbf{k}’|$$
The first condition is the Laue condition and the second one is the conservation of energy.
Given this, Can I assert the following?
Particle with $mathbf{k}$ scatter into $mathbf{k}’$ where
$$mathbf{k}’=frac{|mathbf{k}|}{|mathbf{k}+mathbf{G}|}(mathbf{k}+mathbf{G})$$
The above condition follows both the conditions. But It looks like a particle will always going to scatter in some other direction. Is this correct?
Q: But It looks like a particle will always going to scatter in some other direction. Is this correct?
Generally, yes if the wave vector $vec k$ is not ended on the boundary of the first Brillouin zone (1st BZ).
For two points (end points of $vec k$ and $vec k'$) in a line and both are within the 1st BZ, their connection line is definitely shorter than any reciprocal lattice vector $vec G$. These two vectors cannot satisfied the Bragg's diffraction condition.
Since The 1st BZ is the Winger cell in the reciprocal lattice: the smallest possible region enclosed by the vertical middle planes of all reciprocal lattice vectors.
But, if the vector $vec k$ is ended at the boundary of 1st BZ. It is possible to make a back-on scatterring, as shown in the following figure:
$$ vec k' = -vec k$$
Answered by ytlu on July 1, 2021
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