Lattice wave dispersion relation

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SUMMARY

The discussion centers on the impossibility of wave propagation in a linear one-atomic chain when the wavelength equals the lattice constant. It is established that the dispersion relation indicates zero frequency under these conditions, leading to the conclusion that no energy eigenstates exist for traveling solutions. The Bragg equation for reflection is satisfied, resulting in a superposition of left and right traveling waves, which creates a band gap due to the energetic differences between the wave maxima at ionic cores and between them.

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Nikitin
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Hi. A very quick question. Why is it impossible for a wave to travel on a linear one-atomic chain if its wavelength equals the lattice constant? I.e. the lattice points vibrate with a wavelength equal to the distance between them? Here's what I mean:
http://www.lcst-cn.org/Solid%20State%20Physics/Ch42.files/image020.gif
http://www.lcst-cn.org/Solid%20State%20Physics/Ch42.html

The dispersion relation says that the "wave" will have zero frequency if the wavelength equals the lattice constant.

I can see why it must be so mathematically, but I can't understand intuitively why this must happen.
 
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The waves will fulfill the Bragg equation for reflection. Hence you get a (two to be precise) superposition of left and right traveling waves: sin(kx) and cos(kx). One of the two will have its maximum (of the squared function) at the ionic cores, the other one between the cores, so the first one will be energetically lower than the second one. That's the band gap. It also means that there are no energy eigenstates corresponding to traveling solutions ##(\cos(kx)\pm i \sin(kx))\exp(i\omega t)##.
 
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DrDu said:
The waves will fulfill the Bragg equation for reflection. Hence you get a (two to be precise) superposition of left and right traveling waves: sin(kx) and cos(kx). One of the two will have its maximum (of the squared function) at the ionic cores, the other one between the cores, so the first one will be energetically lower than the second one. That's the band gap. It also means that there are no energy eigenstates corresponding to traveling solutions ##(\cos(kx)\pm i \sin(kx))\exp(i\omega t)##.
as the cos and sin components are not degenerate.
 
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