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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.
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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?
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