Condensed Matter – How to Calculate Integral of a Diverging Function in Condensed Matter Physics

Question

I'm reading the chapter 10 on excitons of "Quantum Theory of Optical and Electronic Properties of Semiconductors" by Hartmut Haug and and Stephan W. Koch. In the second section of this chapter (in my edition from equations (10.31) to (10.34)), at least as far as I understand, the authors say that

$$ \sum_{\textbf{q}\neq \textbf{k}} V_{\textbf{k}-\textbf{q}} P_{vc,\textbf{q}}(\omega) = \frac{1}{V} \int_V V(r) P_{vc}(\textbf{r},\omega) e^{-i \textbf{k} \cdot \textbf{r}} $$

where $ V(r) $ is the Coulomb potential between two point particles (and therefore diverges at $ r = 0 $), $ V $ is the volume of the crystal and $ P_{vc,\textbf{q}}(\omega) $ and $ P_{vc}(\textbf{r},\omega) $ are the Fourier transform of the polarization and its representation in terms of position.

Furthermore, the author says he gets from one result to the other by approximating the reciprocal space to a continuous space,

$$ \sum_{\textbf{q}} \rightarrow \frac{V}{(2\pi)^3} \int d^3q , $$

and by using the Fourier transforms in the form,

$$ f(\textbf{r}) = \frac{V}{(2\pi)^3} \int d^3q f_{\textbf{q}} e^{-i \textbf{q} \cdot \textbf{r}}, \ \ \ \ \ f_{\textbf{q}} = \frac{1}{V} \int_V f(\textbf{r}) e^{i \textbf{q} \cdot \textbf{r}}. $$

How did he do this?

EDIT:

I didn't explain very well what my problem was. Everything would be fine if the first expression was

$$ \sum_{\textbf{q}} V_{\textbf{k}-\textbf{q}} P_{vc,\textbf{q}}(\omega) = \frac{1}{V} \int_V V(r) P_{vc}(\textbf{r},\omega) e^{-i \textbf{k} \cdot \textbf{r}} $$

but if you look closely, in the first expression the sum doesn't include the term with $ \textbf{q} = \textbf{k} $. However, it seems like the author just does everything else like it did.

EDIT 2:

Sorry, I think you're right. Please see if you agree.

Notice that, before taking any limit, we can write the sum as

$$ \sum_{\textbf{q}\neq \textbf{k}} V_{\textbf{k}-\textbf{q}} P_{vc,\textbf{q}}(\omega) = \sum_{\textbf{q}} V_{\textbf{k}-\textbf{q}} P_{vc,\textbf{q}}(\omega)(1-\delta_{\textbf{k},\textbf{q}}). $$

Now, taking the thermodynamic limit, the first term yields the integral the author mentions. The second term yields

$$ V_{\textbf{0}} P_{vc,\textbf{k}}(\omega) $$

that should be null in the thermodynamic limit according to you and the author.

Evaluating this integral in spherical coordinates,

$$ V_{\textbf{0}} = \frac{1}{V} \int_V d^3r V(r) = \frac{e}{4 \pi \varepsilon_0 V} \int^{2\pi}_0 d\theta \int^\pi_0 \text{sin} \phi d\phi \int^L_0 r dr. $$

Since the volume is $ V \sim L^3 $ and the integral is of the order of $ L^2 $ this term goes to 0, as you said.

Answer 1

TLDR: The authors of the book imposed the thermodynamic limit, in which sums can be approximated by integrals, and some easy to prove identities of the Fourier transform.

1. The first part, $\sum_q \rightarrow \frac{V}{(2\pi)^3}\int \mathrm{d}^3 q$, can be motivated by taking the thermodynamic limit (assume $d=1$) $L\rightarrow \infty$, $\Delta q = \frac{2\pi}{L}\rightarrow 0$, in which a discrete sum can be approximated by an integral (and vice versa)

$$\int \mathrm{d}q \, f(q) \approx \sum_{n=-\infty}^\infty \Delta q f_n.$$

2. Now, any translationally invariant function $f(x+R)=f(x)$ can be written in terms of a discrete Fourier series $$f(x) = \sum_q f_q \mathrm{e}^{iqx}.$$ Using the orthogonality relation $\int \mathrm{d} r\, \mathrm{e}^{-i(q-q')r}=\delta_{qq'}$ is straight-forward to obtain $f_q$ in the form of the last equation of your question. This part is perfectly valid and one does not have to take the thermodynamic limit. In the thermodynamic limit, the discrete Fourier transform becomes a continuous one

$$f(x) \approx \frac{1}{\Delta q} \int \mathrm{d}q \, \mathrm{e}^{iqx}\tilde{f}(q).$$

Edit:

Taking

$$\begin{align} \sum_{q\neq k} V_{q-k}P_q &\rightarrow \frac{V}{(2\pi)^3}\int dq V(q-k)P(q)\\ &=\frac{V}{(2\pi)^3}\int dq \frac{1}{V}\int dr\, V(r)\mathrm{e}^{i(q-k)r}\frac{1}{V}\int dr'\, P(r')\mathrm{e}^{iqr'}\\ &=\frac{1}{(2\pi)^3 V}\int dq \int dr \int dr' V(r)P(r') \mathrm{e}^{-ikr}\mathrm{e}^{iq(r'+r)}\\ &=\frac{1}{(2\pi)^3 V} \int dr \int dr' V(r)P(r') \mathrm{e}^{-ikr}(2\pi)^3\delta(r'+r)\\ &= \frac{1}{V} \int dr V(r)P(-r) \mathrm{e}^{-ikr}, \end{align}$$

one obtains the given identity (however, note the differing sign in the argument of $P$, which probably can be fixed if $k-q\rightarrow \vert {\bf k}-{\bf q}\vert$ is taken more seriously than I did).