Converting a summation into an integration

[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?