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As in the title, how to convert $\rm cm^{-1}$ to $\rm eV/Å^2$? Å stands for angstrom.

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    $\begingroup$ If you explain the context of the question, why you want to do that, or tell us who said that and for what reason, we may be able to figure out what is really going on. $\endgroup$
    – garyp
    Commented Nov 8, 2021 at 11:52
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    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Nov 8, 2021 at 12:11
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    $\begingroup$ Note that, in atomic spectroscopy, it’s common to report energies/wavenumbers in “inverse centimeters,” with an implicit factor of $\hbar c$. For a single photon, it’s totally reasonable to convert separately between inverse centimeters, electron-volts, and angstroms. If you’re using the “/“ to mean “or,” like I did in the first sentence of this comment, that’s confusing because of the mathematical meaning of the slash. If you really mean energy per unit area, please clarify. $\endgroup$
    – rob
    Commented Nov 8, 2021 at 12:25

2 Answers 2

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This should not be possible. Notice that $cm^{-1}$ has dimensions of 1/length, as does $A^{-1}$.

$eV/A^2$ has dimensions of energy/length$^2$. This makes no sense, given the available information.

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    $\begingroup$ There is an important difference between “this makes no sense” versus “I cannot immediately think of a context where this would make sense.” $\endgroup$
    – rob
    Commented Nov 8, 2021 at 12:36
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    $\begingroup$ With the information available in the question this kind of conversion is not sensible. However I understand your point and, in some contexts where dimensional factors are implicit, this might make some sense. $\endgroup$ Commented Nov 8, 2021 at 14:01
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    $\begingroup$ @HenriqueCalazansPrates it is, unfortunately, an ill-posed question, however, it should not come as a surprise that those two untis are somehow related. This is because, for instance, cm-1 and eV shouldn't be related, using your reasoning, but it is possible to switch and convert one to the other. See physics.stackexchange.com/questions/423969 $\endgroup$
    – TheVal
    Commented Nov 8, 2021 at 16:49
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The only use I can think of for $\left[eV/Å^2\right]$ is in the quadratic potential constant $\alpha$ such that

$$ V(x)=\alpha x^2 $$

as it appears in a generic hamiltionian. $V$ should be in $eV$ (energy) and $x$ should be in $Å$ (length). It's my assumption that you have something like this

$$ V'(x)=\beta \xi^2(x) $$

where $V'$ is in $cm^{-1}$, and $\xi$ is dimensionless. Thus, I infer you want to convert from $\beta$ to $\alpha$, and I also infer that the dimensionless distance is

$$ \xi(x)=\frac{x}{A_0} $$

where $A_0=a_0/(1\cdot 10^{-10})$ is the Bohr radius in $Å$ (length), and $a_0$ is the Bohr radius in SI units. Proceeding à-la engineer, the conversion is:

$$ V(x)\left[eV\right]=\left[\frac{eV}{cm^{-1}}\right]\cdot V'(x)\left[cm^{-1}\right]\tag{1}\label{one} $$

where the conversion factor $\mathit{CF}$ is obtained from the definition of eV (see this for info):

$$ \left[\frac{eV}{cm^{-1}}\right]=\frac{|e|}{100|h||c|}\cdot hc=\mathit{CF}\cdot hc\approx 8066\cdot hc $$

with $|e|$, $|h|$, $|c|$ are the absolute values of the electron charge, Planck constant and speed of light in vacuum in SI units. Also, $hc$ is in SI units.

The conversion (eq. \ref{one}) becomes, once substituting each expression, as

$$ \alpha x^2=\mathit{CF}\cdot hc\cdot \beta\frac{x^2}{A_0^2} $$

and after simplifying $x$ and expliciting $\alpha$, the following conversion relation is obtained:

$$ \alpha\left[eV/Å^2\right]=\mathit{CF}'\cdot\frac{hc}{a_0^2}\cdot\beta\left[cm^{-1}\right]\qquad \mathit{CF}'=\frac{|e|}{100|h||c|}\cdot10^{-20} $$

where all physical constants are in SI units. Thus, after calculating the dimensionless unit conversion factor $\mathit{CF}'$ and plugging in all the physical constants in their SI units, you should be able to get the desired unit of measure from the initial one.

DISCLAIMER: I am assuming that you're in the middle of converting a hamiltonian from a wavenumber formulation to a atomic-unit formulation, but I could be wrong. The unit $\left[cm^{-1}\right]$ usually refers to an energy, and there are countless usages that could correspond to different physical interpretations of your target unit of measure. I have chosen the simplest one it came to mind, and I've shown how it is possible to connect the two.

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