# Natural units and $\Delta E_n$ for an harmonic potential

For context, I am currently studying perturbation theory, as well as variational methods in quantum physics. My professor uses natural units when solving problems, and he states in every problem that $$\hbar / 2m = 1$$ will be assumed. This makes sense to me, as I can see that Schrödinger's time independent equation looks simpler this way, but I have trouble understanding how to derive certain things in this unit system.

For example, there's this perturbation theory problem where we're studying the potential $$V(x) = x^2$$, with the perturbation being $$V_1(x) = ax$$. Therefore:

$$H = H_0 + V_1(x) = -\frac{d^2}{dx^2} + x^2 + ax$$

($$H_0$$ being the original harmonic oscillator, and $$ax$$ the perturbation we're working with).

He states that $$\hbar / 2m = 1$$, and then he says that the energy associated to $$H_0$$ is:

$$E_n^{(0)} = \hbar \omega \left(n + \frac{1}{2}\right) = 2 \left(n+\frac{1}{2}\right)$$

I do not understand how you get $$\hbar\omega = 2$$ from $$\hbar / 2m = 1$$.

I guess this is usually covered in quantum physics textbooks, but I have checked the one I normally use (Messiah's) and I can't find an answer. If someone could help me understand this I'd be very thankful.

Note that the combination $$\sqrt{m\omega/\hbar}$$ has units of inverse length so use the dimensionless variable $$X=\sqrt{\frac{m\omega}{\hbar}}x$$ so that $$\frac{d}{dx}=\sqrt{\frac{m\omega}{\hbar}}\frac{d}{dX}\quad \Rightarrow \quad \frac{d^2}{dx^2}= \frac{m\omega}{\hbar}\frac{d^2}{dX^2}$$ and hence $$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m\omega^2 x^2\qquad \Rightarrow \qquad -\frac{\hbar\omega}{2}\frac{d^2}{dX^2}+\frac{\hbar \omega}{2}X^2$$ and thus, using the dimensionless energy $$\epsilon= E/\hbar\omega$$, we get the dimensionless Schrödinger equation $$\left(-\frac{1}{2}\frac{d^2}{dX^2}+\frac{1}{2}X^2\right)\psi(X)=\epsilon\psi(X)\, .$$ In other words, appropriate rescalings of length and energy, and multiplication by $$2$$, amount to taking $$\hbar^2/2m=1$$.
Thus, your perturbation $$V_1=ax$$ is really the dimensionfull perturbation $$a \sqrt{m\omega/\hbar} \hbar\omega x$$ where $$a$$ is dimensionless.