Converting Between K-Space Sum and Integral for Macroscopic Solids"

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To convert between a k-space sum and an integral for macroscopic solids, one must consider the periodic boundary conditions that define k-space states. Each state occupies a volume of (2π/L)³, leading to a density of V/(2π)³. To count states with wavevector k < k0, the volume of a sphere of radius k0 can be used, which translates into an integral form. The relationship can be expressed as ∑_k = (V/(2π)³) ∫ d³k, where the sum can also be represented as an integral of delta functions. For slowly varying test functions, these delta functions can be replaced by their density, facilitating the conversion process.
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How is it exactly i convert between a k-space sum an integral?
Assume that we have some macroscopic solid. Periodic boundary conditions leads to kx,ky,kz = 2π/L, so each k-space state fills a volume (2π/L)3 or has a density of V/(2π)3. To then count for instance the number of state with wavevector k<k0, what do you then do?
Intuitively I would multiply the volume of a cube of radius k0, but how does this translate into an integral exactly?
 
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##∑_k=\frac{V}{(2\pi)^3}∫d^3k##
 
You can also write the sum as an integral over a sum of delta functions.
For slowly varying test functions, the delta functions may then be replaced by their density ##V/(2\pi)^3##.
 
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