I am reading (and trying to replicate the results of) a paper in the field of condensed matter physics. The system consists of a spatially separated electron and hole whose Coulomb interaction is the only source of potential energy. The authors write the Hamiltonian as

$$ H\Psi = \left[- \frac{\hbar^2}{2\mu}\nabla^2 - \frac{e^2}{\varepsilon\sqrt{r^2 + z^2}} \right]\Psi, $$

where all quantities have their standard definitions (and in any case are not relevant to my question). The authors do not explain their choice of units, nor do they provide a numerical value for any of the quantities listed, so there is no way for me to back-solve for their values.

This expression is deeply confusing to me because while it is certainly not given in SI, it does not appear to be given in natural units either. Because I am trying to replicate their numerical calculations, I need to understand the system of units employed here.

In natural (Planck) units, we define $\varepsilon_0=1$, $\mu_0 = 1$, $c=1$, $\hbar=1$, etc. I have tried looking online but I have not been able to find any information on a unit system in which $k_e=\frac{1}{4\pi\varepsilon_0}=1$, i.e. $\varepsilon_0 = \frac{1}{4\pi}$. But it must be the case that this is what is happening here, as they carry around factors of $\varepsilon$ to describe the permittivity of their substrates, and yet do not include factors of $k$ or $\varepsilon_0$.

The big sticking point for me is that it must be the case that $\hbar \neq 1$, but since $\hbar$ is itself a fundamental quantity, I have no way to backsolve for its numerical value given the starting point that $\varepsilon_0 = \frac{1}{4\pi}$.

I am also confused about what value they might be using for $e$ - disregarding the possibility that they are using $e = 1.6 * 10^{-19} C$, they still might be using the accepted value in natural units, $e = \sqrt{4\pi\alpha} = 0.303$ (where here $\alpha$ is the fine-structure constant), but I suspect the value for $e$ will not be 0.303, just as it is clearly the case that $\hbar\neq1$.

Edit: The paper I am reading is not publicly available, but the citation is: Lozovik, Y.E., Nishanov, V.N. Sov. Phys. Solid State, Vol. 18, No. 11. (1976). "Wannier-Mott excitons in layer structures and near an interface of two media"

  • $\begingroup$ For competleteness it would help if you reference the paper. Also, presumably epsilon is the medium's relative permittivity? $\endgroup$ Commented Sep 22, 2015 at 23:54
  • $\begingroup$ That aside, this simply looks like atomic units. $\endgroup$ Commented Sep 22, 2015 at 23:55
  • $\begingroup$ @EmilioPisanty is this what you mean by atomic units? I don't see how that could be the case, considering that $e$ and $\hbar$ appear throughout the paper, and considering that $k$ does not appear (ie it is implied to be set to unity), I do not think that they are using atomic units where $e,\hbar=1$ $\endgroup$ Commented Sep 23, 2015 at 2:52

2 Answers 2


You've been done a disservice if your earlier teachers didn't even mention the existence of Gaussian units (a cm-gram-sec system with "unrationalized" E&M). Not that I like them, but simply because they were very common in the mid twentieth century and they still have their adherents (some even on Physics SE).

The unit of charge goes by several names including "statCoulombs", and it folds the dimensionality of $k$ into the charge unit so that Columb's law takes on the form $$ \mathbf{F} = \frac{q_1 \,q_2}{r^2} \hat{\mathbf{r}}\,.$$ Several of Maxwell's equations take on the form a mathemtician would use. For instance Guass' Law becomes $$ \nabla \cdot \mathbf{E} = 4 \pi \rho \,,$$ and similar changes happen throughout. Adherents describe these forms as being nicer, but having grown up with SI units I still find them more natural.

  • $\begingroup$ "You've been done a disservice if your earlier teachers didn't even mention the existence of Gaussian units". Really? Outdated stuff that only creates confusion? I don't think so. $\endgroup$
    – Gert
    Commented Sep 23, 2015 at 0:16
  • 1
    $\begingroup$ @Gert I'm pretty sure we have here a case where $\textit{not}$ being familiar with Gaussian units was the cause of confusion. Furthermore, Gaussian units aren't outdated - theorists still use them, if they don't use Heaviside-Lorentz. $\endgroup$ Commented Sep 23, 2015 at 1:12
  • $\begingroup$ Some even on Physics SE Guilty! It's just too convenient! $\endgroup$
    – Kyle Kanos
    Commented Sep 23, 2015 at 1:27
  • $\begingroup$ @Kyle I couldn't recall any names, but I begin to have an inkling that you might be one of the here^H^H^H^H witc^H^H^H^H people who like Guassian units. Would you care to state the defense more clearly that my half-heated hand waving? $\endgroup$ Commented Sep 23, 2015 at 1:29
  • $\begingroup$ I gotta agree with Kyle here. Not having to keep the $1/4\pi\epsilon_0$ while writing infinite series for electrostatic potentials is a godsend. Even Feynman does this in his lectures, only he says "let's forget about the $1/4\pi\epsilon_0$ and stick it in at the end". $\endgroup$
    – Javier
    Commented Sep 23, 2015 at 1:52

I learned to keep track of the conversion from SI to Gaussian units for electromagnetism as \begin{align} \frac{e^2}{[4\pi\epsilon_0]} &= \alpha \hbar c \end{align} where the factor in [brackets] is unity in CGS units and isn't in SI. This is a nice way to remember things because it makes clear that Coulomb's law for two fundamental charges, $$ \vec F = \alpha\hbar c \frac{ \hat r }{r^2} , $$ is manifestly quantum-mechanical.

From $\hbar c = 3.162\times10^{-17} \rm\, erg\,cm$ and $\alpha^{-1} = 137.0$ I find \begin{align} e^2 &= 2.307\times10^{-19} \rm\,esu^2 \\ e &= 4.803\times10^{-10} \rm\,esu \end{align}

It's also cute that in Gaussian units $E^2$ and $B^2$ have units of energy density, but for real computations I use SI.

  • $\begingroup$ I very much like that conversion factor you use, can you elaborate on how to derive it? As I mentioned in a comment to a different answer, I get a little bit lost when we make the claim that $(4\pi\varepsilon_0)^{-1} = 1$ in CGS units because a straightforward unit conversion does not ensure that is the case, but it seems like we just throw our hands in the air and say "I hate writing Coulomb's constant so let's just claim it is equal to unity". The only thing I can think of is that after we convert to CGS-Gaussian units, we make $k=1$ by building those conversion factors into the statCoulomb. $\endgroup$ Commented Sep 23, 2015 at 17:50
  • $\begingroup$ I think this is the treatment in Wong's Nuclear Physics, the (second?) edition with the grey cover; there's a newer edition which may be all SI. My point in the numerical computation was to reassure you that yes, you've found the correct value for $e$ in CGS units, that $\hbar$ and $c$ transform like you'd expect, and that a statcoulomb has dimension $\rm (erg\,cm)^{1/2}$. $\endgroup$
    – rob
    Commented Sep 23, 2015 at 23:35

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