The calculation of electronic structure of infinite periodic material using Plane-waves is the object of study here. The cutoff energy $E_{cutoff}$ provides information about the accuracy of the simulation in the inequality $$|k+G|^2<E_{cutoff}$$ where $G$ is a reciprocal lattice vector and $k$ a vector in the reference Brillouin zone (if I understand correctly).

Isn't there a problem with the dimensions of the term in the inequality? Since $k$ and $G$ are of the inverse of length and $E_{cutoff}$ an energy?

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    $\begingroup$ I think you are correct in what you are pointing at, there is incosistency in this inequality. I think this could be modified to |𝑘+𝐺|² < 2𝐸_cut𝑜ff/ℏ². I think By dividing cut off energy by Planck's constant ℏ², the resulting expression has units of length squared, making the inequality dimensionally consistent. $\endgroup$ Commented Nov 20, 2023 at 1:30

1 Answer 1


You’re quite right: as written, the equation is dimensionally inconsistent. There are actually a bunch of constants that make the units work out, but it is customary in matter modeling to define units such that the most common constants end up being equal to 1, and then we simply neglect to write them down.* In particular, the so-called atomic units are such that the reduced Planck’s constant ($h/2\pi)$, the charge $e$ and mass $m$ of the electron, and also the dielectric permittivity of the vacuum, are all numerically equal to 1 of their respective unit:

$$ \hbar = e = m = \frac{1}{4\pi \varepsilon_0} = 1. $$

Then any conversion between them doesn’t change the number and can be “safely” skipped over in the actual paper. This also shows up in high-energy physics: the reason units like $\text{MeV}$ are used as energy scales and length scales interchangeably is that the units make their conversions have the same value (or at most a reciprocal).

As for the specific case in your question: Consider the free electron, which has the Hamiltonian

$$ H = - \frac{\hbar^2}{2m} \nabla^2, $$ where $\nabla^2$ is the Laplacian operator $\nabla^2 = \partial_{xx} + \partial_{yy} + \partial_{zz}$.

It’s a classic result, and pretty easy to demonstrate, that the resulting wavefunctions are proportional to plane waves

$$ \psi(\mathbf{r}) \propto e^{i\mathbf{k}\cdot\mathbf{r}}, \quad \mathbf{k} \in \mathbb{R}^3. $$

Here $\mathbf{k}$ has units of length$^{-1}$, as you mentioned, but is actually the particle’s momentum! Thus the energy of a free electron with momentum $\mathbf{k}$ is

$$ E(\mathbf{k}) = \frac{\hbar^2}{2m} p(\mathbf{k})^2 = \frac{\hbar^2 \mathbf{k}^2}{2m} \implies \mathbf{k}^2 = \frac{2m}{\hbar^2} E(\mathbf{k}). $$

But in atomic units, $\hbar = m = 1$, and we can just write $E = \mathbf{k}^2/2$. This is how you should interpret the energy cutoff—$\lvert\mathbf{k}\rvert^2 < E_{\text{cut}}$ means to use all the plane-waves with momentum up to $\mathbf{k}(/2)$ as basis functions.**

* This makes equations faster to read, but only after you know about the practice already.

** We have to add in $\mathbf{G}$, which has the same units as $\mathbf{k}$, because we force the plane waves to live in a box with a finite size and impose periodic boundary conditions. This also discretizes $\mathbf{k}$, as it turns out; the derivation of this is worth a separate question, though.


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