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I know that Bragg reflection in solid states at the edge of e.g. the first Brillouin Zone causes standing waves at these edges, which creates a gap between the energy bands.
In this picture below you can see the probability density of a symmetric (+) and antisymmetric () standing wave. The energy gap is the difference of the energy of the symmetric wave and the energy of the antisymmetric wave.
And here you can see the energy gap:
It says that the first energy gap(which we see here between band 1 and band 2) occurs due to the bragg reflection at ##k = \pi/a## and ##k =  \pi/a##. The book from Kittel (Solid Sate Physics) also says:
"When the Bragg reflection condition ##k=\pm \pi/a## is satisfied by the wavevector ##\vec k## a wave traveling to the right is Braggreflected to travel to the left, and vica versa."
So the wave with the wave vector ##k=\pi/a## (exactly at a bragg plane, which is the edge of a BZ) will be reflected, which creates a standing wave. Or how can you imagine that?
In this picture below you can see the probability density of a symmetric (+) and antisymmetric () standing wave. The energy gap is the difference of the energy of the symmetric wave and the energy of the antisymmetric wave.
And here you can see the energy gap:
It says that the first energy gap(which we see here between band 1 and band 2) occurs due to the bragg reflection at ##k = \pi/a## and ##k =  \pi/a##. The book from Kittel (Solid Sate Physics) also says:
"When the Bragg reflection condition ##k=\pm \pi/a## is satisfied by the wavevector ##\vec k## a wave traveling to the right is Braggreflected to travel to the left, and vica versa."
So the wave with the wave vector ##k=\pi/a## (exactly at a bragg plane, which is the edge of a BZ) will be reflected, which creates a standing wave. Or how can you imagine that?
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