Periodic Boundary Conditions proof

Wminus
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Hi! When we model bloch-waves in a solid we assume that there exist some kind of periodic boundary conditions such that the wave function is periodic. In 1D, ##\psi(x)## repeats itself for every ##L##, ##\psi(x) = \psi(x+L)##, such as here:
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OK, fine, we get pretty wave solutions if we assume the existence of the PBC. But what ##L##? As far as I know the only repeating unit in a crystal is the Wieger Seitz cell, which is sized on the atomic scale.

Is ##L## just the wavelength?
 

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No, L is the size of your crystal, so e.g. 1cm. In the end, this size doesn't matter to much and you often take ##L \to \infty##.
 
DrDu said:
No, L is the size of your crystal, so e.g. 1cm. In the end, this size doesn't matter to much and you often take ##L \to \infty##.

How can that be? if that is true, it means ##\psi## is a standing wave... But bloch waves are traveling!
 
That's why you use periodic boundary conditions, the waves then move on a circle and you can have left and right moving waves.
 
this makes no sense physically. if you have a crystal cube of 1cm, you are assuming that its left side is connected to its right?
 
No, it doesn't, but there is a theorem by Wigner that the influence of the boundary conditions on the states vanishes like 1/N, where N is the number of particles (or elementary cells). So if you are interested in the bulk states only, you can pick the boundary conditions as seem convenient.
 
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DrDu said:
No, it doesn't, but there is a theorem by Wigner that the influence of the boundary conditions on the states vanishes like 1/N, where N is the number of particles (or elementary cells). So if you are interested in the bulk states only, you can pick the boundary conditions as seem convenient.
Ah, OK, I see. So you can pick whatever PBC you like? What about if L = length of unit cell?
 
No, I said that the boundary conditions become unimportant when L is much larger than the elementary cell.
 
DrDu said:
No, I said that the boundary conditions become unimportant when L is much larger than the elementary cell.

Yes you did, my apologies. And thanks for the help!
 
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