Given a potential $$U(x)=\frac{\epsilon}{2}\left (\left(\frac{x}{a}\right)^2 +\left(\frac{a}{x}\right)^2\right )$$
where $x$ is the particle's position along the $x$-axis, what kind of units can I claim that $a$ and $\epsilon$ have?
I know that this is an equation for potential energy, so the units must work out to $$Joules=\frac{kg \cdot m^2}{s^2}=\frac{M \cdot L^2}{T^2}$$ Therefore:
Let $[d]$ symbolize the unknown units of $d$ and $M=$mass, $L=$length, $T=$time. Then,
$$\frac{M \cdot L^2}{T^2}=[\epsilon]\left(\left(\frac{L^2}{[a]^2}\right)+\left(\frac{[a]^2}{L^2}\right)\right)$$
$$=[\epsilon]\left(\left(\frac{L^2}{[a]^2}\right)+\left(\frac{[a]^2}{L^2}\right)\right)$$
$$=[\epsilon]\left(\frac{L^4+[a]^4}{L^2[a]^2}\right)$$
At this point, I reason that if the numerator can add together, then $[a]$ must have units of length. Therefore, the bottom must also be of units of length in its entirety. But then the numerator will cancel with the denominator. Therefore, $\epsilon$ has units of $\frac{M \cdot L^2}{T^2}$. Is there a flaw in my reasoning?
We can let $[\epsilon ] : = [U] $ were $[U]$ is in Joules of electron-Volts and then we will have that $a$ has dimensions of length.