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.


1 Answer 1


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.

  • 1
    $\begingroup$ What a great answer! Thank you very much, I wouldn't have found out by myself and this is just crystal clear. Thanks! $\endgroup$
    – Manuel
    Jul 1, 2022 at 20:02

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