We're looking at the Kronig-Penney model in class and one of the conundrums is related to the Kronig-Penney potential for a chain of N atoms. I'm supposed to squeeze out some expression for the Fourier components, UG, but I don't end up with the right expression, although I think I did the right stuff.
The following is given for the delta potential function, U(x)=A∞∑n=−∞δ(x−na), of which I have to show that the Fourier components are given by UG=Aa,G∈Z, using the hint that ∑Nn=0cos(Gna)=N.
I'll start with the generalized form of UG (I am not sure whether this expression is correct), UG=1a∫aU(x)e−iGxdx, in which I can substitute my potential function U(x) (U(x)=A∞∑n=−∞δ(x−na)), to find UG=1a∫aA∞∑n=−∞δ(x−na)e−iGxdx. A is just a constant and the sum and integral signs can be swapped, so UG=Aa∞∑n=−∞∫aδ(x−na)e−iGxdx. Now by the definition of the delta function, ∫aδ(x−na)e−iGxdx=∫aδ(x−na)f(x)dx=f(na)=e−iGna Back to our original equation, UG=AaN∑n=−Ne−iGna=2AaN∑n=0e−iGna+eiGna2=2AaN∑n=0cos(Gna)=2AaN. ...which doesn't add up?
The wikipedia article on this derivation simply skips a few steps but I'm having some trouble filling those in: UG=1a∫a/2−a/2dxU(x)e−iGx=1a∫a/2−a/2dx∞∑n=−∞A⋅δ(x−na)e−iGx=Aa
See this link for more info.
Answer
From this step,
UG=1a∫a/2−a/2dxU(x)e−iGx=1a∫a/2−a/2dx∞∑n=−∞A⋅δ(x−na)e−iGx
note that the summation is an impulse train with spacing a. Since the integral is from −a2 to a2, just the impulse at x=0 is integrated over so only the n=0 term in the summation contributes to the integral:
UG=1a∫a/2−a/2dx∞∑n=−∞A⋅δ(x−na)e−iGx=1a∫a/2−a/2dxA⋅δ(x)e−iGx=Ae−iG0a=Aa
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