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 Apr 25 awarded Popular Question Apr 8 awarded Notable Question Mar 11 awarded Popular Question Feb 22 awarded Notable Question Feb 18 accepted About unit definition in numerical computation Feb 17 comment About unit definition in numerical computation Thanks Emilio. It is a good explanation and I follow most of the math here. But I still have one question. Here we define hbar in terms of meV and ps. So everywhere in the computation I should make time in unit of ps, right? If I want the length in unit of nm, I should keep everywhere length has the same unit. Now get back to $\hbar$, SI unit gives J s, but we can break J in terms of kg m^2/s^2, so there is length as well. I am so confusing if we have to take care of the m^2 in $\hbar$ so to have something nm^2 as well? I read your example but I didn't see that conversion. Feb 15 comment About unit definition in numerical computation Thanks garyp. I corrected my question. I add more explanation on how to convert meter to nanometer. But if I do that way, which will make kg m^2/s^2 for J converted to kg (nm)^2/s^2, so is it still good to multiply $(1.6\times10^{-19})^{-1}$ to get eV? Feb 15 revised About unit definition in numerical computation deleted 19 characters in body Feb 15 asked About unit definition in numerical computation Jan 22 awarded Popular Question Jan 5 awarded Popular Question Jan 4 awarded Popular Question Dec 31 awarded Popular Question Nov 28 awarded Notable Question Nov 26 awarded Popular Question Nov 22 awarded Yearling Nov 16 awarded Popular Question Nov 3 awarded Popular Question Aug 29 awarded Notable Question Jun 9 awarded Notable Question