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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 anti-symmetric (-) standing wave. The energy gap is the difference of the energy of the symmetric wave and the energy of the anti-symmetric 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 Bragg-reflected 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 anti-symmetric (-) standing wave. The energy gap is the difference of the energy of the symmetric wave and the energy of the anti-symmetric 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 Bragg-reflected 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?