Lattice wave dispersion relation

  • Thread starter Nikitin
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  • #1
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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 [Broken]
http://www.lcst-cn.org/Solid%20State%20Physics/Ch42.html [Broken]

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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DrDu
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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 travelling 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 travelling solutions ##(\cos(kx)\pm i \sin(kx))\exp(i\omega t)##.
 
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  • #3
DrDu
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805
The waves will fulfill the Bragg equation for reflection. Hence you get a (two to be precise) superposition of left and right travelling 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 travelling solutions ##(\cos(kx)\pm i \sin(kx))\exp(i\omega t)##.
as the cos and sin components are not degenerate.
 
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