# Converting from Planck units to Hartree units

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?

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

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 $$q=\alpha A/B$$.