In the textbook Understanding Molecular Simulation by Frenkel and Smit (Second Edition), the authors represent a function $f(\textbf{r})$ (which depends on the coordinates of a periodic system) as a Fourier series. I quote from page 295 of the text:

Let us consider a periodic system with a cubic box of length $L$ and volume $V$. Any function $f(\textbf{r})$ that depends on the coordinates of our system can be represented by a Fourier series:

$$f(\textbf{r}) = \frac{1}{V} \sum_{\boldsymbol{\ell} = -\infty}^{\infty} \tilde{f}(\textbf{k}) e^{i \textbf{k} \cdot \textbf{r}} \; \; \; \; \textbf{(12.1.6)}$$

where $\textbf{k} = \frac{2\pi}{L}\boldsymbol{\ell}$ with $\boldsymbol{\ell} = (\ell_x, \ell_y, \ell_z)$ are the lattice vectors in Fourier space. The Fourier coefficients $\tilde{f}(\textbf{k})$ are calculated using

$$\tilde{f}(\textbf{k}) = \int_V d\textbf{r} \; f(\textbf{r}) e^{-i\textbf{k} \cdot \textbf{r}} \; \; \; \; \textbf{(12.1.7)}$$

Now, the authors use equation (12.1.6) to write the electric potential $\phi(\textbf{r})$ in Fourier space:

$$\phi(\textbf{r}) = \frac{1}{V} \sum_{\textbf{k}} \tilde{\phi}(\textbf{k}) e^{i\textbf{k} \cdot \textbf{r}}$$

The authors write:

In Fourier space, Poisson's equation has a much simpler form. We can write for the Poisson equation:

$$-\nabla^2 \phi(\textbf{r}) = -\nabla^2 \left( \frac{1}{V} \sum_{\textbf{k}} \tilde{\phi}(\textbf{k}) e^{i\textbf{k} \cdot \textbf{r}} \right) = \frac{1}{V} \sum_{\textbf{k}} k^2 \tilde{\phi}(\textbf{k}) e^{i\textbf{k} \cdot \textbf{r}} \; \; \; \; \textbf{(12.1.8)}$$

My question is, why is the $\frac{1}{V}$ factor present in equations (12.1.6) and (12.1.8)? What is the significance of the $\frac{1}{V}$ factor in $\phi(\textbf{r}) = \frac{1}{V} \sum_{\textbf{k}} \tilde{\phi}(\textbf{k}) e^{i\textbf{k} \cdot \textbf{r}}$?

In contrast, the article on Wikipedia does not include this prefactor. I realize that that article is dealing with the general case, whereas here we are considering a system with a cubic box of volume $V$. But shouldn't the units of $\phi(\textbf{r})$ be the same as those of $\tilde{\phi}(\textbf{k})$? The $\frac{1}{V}$ seems to preclude $\phi(\textbf{r})$ and $\tilde{\phi}(\textbf{k})$ having the same units.

Do you have any advice? Thanks.


2 Answers 2


I) Let us just consider $1$ dimension for simplicity. (The generalization to higher dimensions is straightforward). Then the volume factor $V$ is just a length factor $L$.

II) The standard Fourier series formulas can be derived from $(12.1.7)$ and $(12.1.6)$ by taken the length $L$ to be $L=2\pi$. Then $(12.1.7)$ and $(12.1.6)$ become the standard Fourier series formulas

$$\tag{12.1.7'} c_{n} ~=~ \frac{1}{2\pi}\int_{-\pi}^{\pi} \! dx~ f(x) e^{-in x}, $$ $$\tag{12.1.6'} f(x)~=~\sum_{n\in\mathbb{Z}} c_n~e^{in x} ~=~f(x+2\pi), $$

via the identifications

$$\ell~=~ n~\in~\mathbb{Z}, \qquad \tilde{f}(\ell) ~=~2\pi c_{n} . $$

III) Going back to $3$ dimensions, the $1/V$ normalization in $(12.1.6)$ is important. Of course, in another convention, it could be put in $(12.1.7)$ instead, or alternatively, symmetrically as $1/\sqrt{V}$ in both formulas $(12.1.6)$ and $(12.1.7)$.

  • $\begingroup$ Thanks. But, given the $1/V$ normalization, what are the units of $\tilde{\phi}(\textbf{k})$ with respect to those of $\phi(\textbf{r})$ in the equation $$\phi(\textbf{r}) = \frac{1}{V} \sum_{\textbf{k}} \tilde{\phi}(\textbf{k}) e^{i\textbf{k}\cdot\textbf{r}}$$? $\endgroup$
    – Andrew
    Aug 9, 2012 at 20:42
  • 1
    $\begingroup$ $[\tilde{f}]= [V] [f]= [L]^3 [f]$, cf. eq. $(12.1.7)$, or eq. $(12.1.6)$. $\endgroup$
    – Qmechanic
    Aug 9, 2012 at 20:52

The usual Fourier series formula states that a function $g(x)$ with period $2\pi$ can be expressed as

$$ g(x) = \sum_{k\in {\mathbb Z}} c_k \, e^{ikx}$$

where $c_n$ (the coefficients of the Fourier series) are given by

$$c_k = \frac{1}{2 \pi} \int_{0}^{2 \pi} g(x) \, e^{-i x \, k}dx$$

If we have $f(x)$ with period $L$, we can relate with the above by $f(x) = g(2 \pi x/L)$, so we get the equivalent formulas:

$$ f(x) = \sum_{k\in {\mathbb Z}} c_k \, e^{i k x}$$

$$c_k = \frac{1}{L} \int_{0}^{L} f(x) \, e^{-i x \, k}dx$$

One can, alternatively move the normalization factor $1/L$ to the first equation. It's just a matter a convention.

The extension to 3 dimensions is straightforward.


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