# Converting a summation into an integration

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Amentia
Hello,

I want to convert a summation in reciprocal space and I am unsure about the integration volume. I have started with the formula:

$$\sum_{\vec{k}} \rightarrow \frac{V_{k}}{(2\pi)^{3}}\int\int\int \mathrm{d}V_{k}$$

where:

$$\mathrm{d}V_{k} = k^{2}\mathrm{d}k \sin{\theta_{k}}\mathrm{d}\theta_{k}\mathrm{d}\phi_{k}$$

Here k can go from 0 to infinity, so what should be the volume ##V_{k}##?

My second question is when I perform the integration with a delta function which restricts k to a finite set of values, should I redefine the volume a posteriori?

$$\frac{V_{k}}{(2\pi)^{3}}\int\int\int \mathrm{d}V_{k}f(\vec{k})\delta(k-k_{0})$$

Here should we have ##V_{k}= k_{0}4\pi## or even a "2D volume" ##V_{k}= 4\pi##? While the volume was supposed to be already defined before we make use of the delta function.

I hope my questions are clear!

Fred Wright
Q1: ##V_k## is the volume of the system. For a Fermi gas it would be the volume of the Fermi sphere.
Q2: Due to rotational symmetry, the preceding factor of your integral would become ##\frac{2\Omega}{(2\pi)^3}##. The 2 comes in if you have a Fermi gas(two energy states, spin up and spin down) the ##\Omega=4\pi## is the solid angle (scattering cross section) of the sphere of volume ##V_k##. Please see https://cpb-us-w2.wpmucdn.com/u.osu.edu/dist/3/67057/files/2018/09/density_of_states-vjqh7n.pdf for a detailed explanation.

Amentia
Thank you for the document. In equation (10), they get rid of ##V_{d}## by defining the density of states. But what if it was defined wihout dividing by this volume? And then assume I have a real system with a finite volume ##V = L_{x}L_{y}L_{z}##. Can I relate ##V_{d}## to V? E.g. ##V_{d}=(2\pi)^{3}/V## so that I will have in front of my integral:

$$\frac{V_{d}\Omega}{(2\pi)^{3}} = \frac{\Omega}{V} = \frac{4\pi}{L_{x}L_{y}L_{z}}$$

where I have assumed that the function inside the integral does not depend on the angles?