The problem at hand has been discussed in loads of previous questions (1, 2, 3)), and my version can be stated as follows.

Consider the sum $$\sum_{\mathbf k} \ln(1+e^{-(\alpha+\beta \varepsilon_k)}) \ .$$ We are summing over, say, two dimonsional k-space lattice $\mathbf k = \frac{2\pi}{L}(n_x,n_y),$ where the $n_x,n_y$ run through the positive integers. I want to understand how we can write this as an integral. Our energy is given by $\varepsilon_k = \frac{\hbar^2 k^2}{2m}$.

  • $\begingroup$ -1. Not clear what you are asking. What is your question? How does it differ from those you have cited? "Is this a situation where hand-waving arguments are used in physics?" is a question teaching styles rather than the content of physics. Why this procedure works is a question about mathematics rather than physics. $\endgroup$ – sammy gerbil Feb 18 '17 at 15:25
  • $\begingroup$ @sammygerbil My question is, indeed, why and how this procedure works. Since it is used extensively in statistical mechanics, I thought it would be on-topic at physics.stackexchange. $\endgroup$ – Mussé Redi Feb 18 '17 at 15:35
  • $\begingroup$ @sammygerbil I removed the phrase about hand-waving, to avoid distraction from my main problem. $\endgroup$ – Mussé Redi Feb 18 '17 at 15:54
  • $\begingroup$ Is your problem in the steps you show, or in the remaining steps until you get the energy integral? $\endgroup$ – anonymous Feb 18 '17 at 16:44
  • $\begingroup$ @anonymous My problem is in the step where we subsitute a function $f(\mathbf k)$ in the sum $\sum_{\mathbf k} f(\mathbf k)$ and the assumptions made from here on to get the integral. I would be most satisfied if I could follow each (implicit) step made in this procedure. I've had (have been avoiding) this confusion for some time. $\endgroup$ – Mussé Redi Feb 18 '17 at 17:58

Why does it work? By assuming the separation distance between the points in the discrete space is negligibly small compared to the total volume, we can make use of the definition of the Riemann integral.

Say we are summing over discrete points $\mathbf k$ in k-space $$\sum_\mathbf{k}\ f(\mathbf k)\ .$$

If we want to rewrite this as an integral we have to assume that the separation between the points in k-space $\Delta k$ is negligible in comparison to the volume of the k-space $V$.

In the following we can then use the definition of the Riemann integral \begin{align} \sum_{\mathbf k} f(\mathbf k) &= \frac{1}{\Delta k} \sum_{\mathbf k} f(\mathbf k) \Delta k \\ &\equiv \frac{1}{\Delta k} \int_{\mbox{all space}} f(\mathbf k) d\mathbf k \ , \end{align} where, in the last step we used our assumption that our seperation distance is negligibly small compared to the volume of the whole space.

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