- #1

Amentia

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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!