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?
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1$\begingroup$ It's not clear what you are asking $\endgroup$– RC_23Sep 12, 2022 at 22:59
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$\begingroup$ 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]? $\endgroup$– Adam El NairreeSep 12, 2022 at 23:11
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$\begingroup$ Energy splitting of what? What is $a$ physically? And what is $g$? $\endgroup$– GhosterSep 12, 2022 at 23:12
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1$\begingroup$ 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. $\endgroup$– GhosterSep 12, 2022 at 23:24
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1$\begingroup$ 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. $\endgroup$– GhosterSep 13, 2022 at 0:06
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1 Answer
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What you want to do with an equation like this is "de-dimensionalize" it, so that you are working with dimensionless parameters.
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.)
