# When working with natural units, how do my other variables change? [closed]

I'm trying to plot energy splitting

as a function of $$a$$ (where $$g=1$$). When I set $$\hbar=1$$ such that $$a=[t^{-1}]$$, how does the value of this variable change to keep the equation consistent?

• It's not clear what you are asking Sep 12, 2022 at 22:59
• Say a=2. Since hbar=[Js] in SI units, when I set hbar=[1], does this impact the value of a? Is the unit of a=[1/s] in this context as hbar=[Js]? Sep 12, 2022 at 23:11
• Energy splitting of what? What is $a$ physically? And what is $g$? Sep 12, 2022 at 23:12
• You need to explain the non-natural dimensions of $a$ and $g$ before asking how to understand this equation in natural units. Give the symmetric double-well potential that you are using. Sep 12, 2022 at 23:24
• Your equation requires the dimensions of $a$ to be $ML^2T$ and the dimensions of $g$ to be $M^{-2}L^{-4}T^{-4}$. Both of these seem strange. Sep 13, 2022 at 0:06

Define $$\varepsilon \equiv E/(g^{1/6}\hbar^{4/3})$$ as a dimensionless energy parameter, and $$\alpha \equiv a/(g^{-1/3}\hbar^{1/3})$$ as a dimensionless form of your parameter $$a$$. Then your equation de-dimensionalizes to
$$\varepsilon_+ - \varepsilon_- = 8\sqrt\frac{2}{\pi}\alpha^{5/2}e^{-\tfrac43\alpha^3}$$
This really has nothing to do with "natural units" in the sense of setting $$\hbar = c = 1$$. Instead the "natural" energy scale here is $$g^{1/6}\hbar^{4/3}$$ and the "natural" scale for the parameter $$a$$ is $$g^{-1/3}\hbar^{1/3}$$.
(The peculiar dimensions of $$g$$ and $$a$$, however, make me wonder whether your equation is correct.)