Why do we get standing waves at the Brillouin zone boundary?
Isn't a standing wave a result of superposition of two waves travelling in opposite directions? What waves travel in opposite direction and meets at the Brillouin zone boundary?
Not really into solid state physics, but i'll try to think from an NVH perspective.I imagine you can consider the bond between atoms as springs, thus, you would get a linear mass-spring system, with multiple DOF. This results in an eigenvalue problem of the system and thus, a standing wave analysis with resulting eigenfunctions and its corresponding eigenvalues.
Since we are dealing with a bounded medium, it is expected to have standing wave patterns. Probably the Brillouin zone boundary is defining that up to its limit, you have an approximate elastic medium, so following the mass-spring-mass representation, a standing wave pattern might arise.
I hope it might give some intuition, however i still comment that i'm not related to this topic.
Good Luck!
I believe the Brillouin zone boundary is not a physical boundary, but is in "k" space - it is the wavenumber that coincides with one half-wavelength across the unit cell. This means that waves going in opposite directions look the same (because the plot is periodic the waves going in opposite directions are actually at the same place on the k axis. This means that you can't have a wave of that wavenumber in one direction without the wave in the other direction, so waves of the same wavenumber but opposite direction exist simultaneously, which is a standing wave.
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