0
$\begingroup$

I am trying to include an analytical formula into a numerical program. In Planck units, the expression looks like this: $\frac{A}{B}$, where $A$ is in $m$ while $B$ is in $eV^{-1}$.

If we go to SI units, we have $\frac{Ae}{B \hbar c}$, where $e$ is the electric charge. In Hartree units, this give us $\frac{Ae\alpha}{B}$, where $\alpha$ is the fine structure constant, since in Hatree units $\hbar =1$ and $c=\frac{1}{\alpha}$. Am I right?

$\endgroup$
1
  • $\begingroup$ I don't think Planck units have anything to do with the unit of energy (or mass) called an "electron-volt". $\endgroup$ Commented Nov 20, 2023 at 18:25

1 Answer 1

1
$\begingroup$

"Converting from Planck units to Hartree units" is an ill-defined problem, because there are many SI quantities that correspond to a given Planck-unit quantity.

However, your problem is not quite what's in the title, because you have already identified which SI quantity you want to convert to Hartree atomic units, namely $q=Ae/\hbar c B$, where $A$ is a length and $B$ is an inverse energy. And, to do this, you (i) set $\hbar=1$, (ii) set $e=1$, and (iii) use the correct value of the speed of light, $c=1/\alpha$.

Thus, in Hartree atomic units, you would have $q=\alpha A/B$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.