Quantum harmonic oscillator substitution

Question

The Schrödinger equation for the harmonic oscillator is $$\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+\frac{m\omega^2}{2}x^2\right)\psi(x)=E\psi(x).$$ Then often $x$ gets substituted with $x=\xi a$, where $a=\sqrt{\hbar/m\omega}$, which leads to $$\frac{\hbar\omega}{2}\left(-\frac{\partial^2}{\partial \xi^2}+\xi^2\right)\psi(\xi)=E\psi(\xi). $$ So my question is if now it should not be $\psi(a\xi)$, or do they define $\psi^\prime(\xi)= \psi(a\xi) $ and then rename it?

Answer 1

Yes, you are right, one just redefines the $\psi(x)=\psi_{new}(\xi)=\psi(a\xi)$. The Schrödinger equation is thus $$\frac{\hbar\omega}{2}\left(-\frac{\partial^2}{\partial \xi^2}+\xi^2\right)\psi_{new}(\xi)=E\psi_{new}(\xi) $$ The basic motivation is to non dimensionalize the independent variable. In order to recover the original $\psi(x)$, we have $$\psi(x)=\psi_{new}(\frac{x}{a})$$

Answer 2

Indeed, to perform such an operation properly, one has to define a new function $$\phi(\xi)=\phi(ax)=\psi(x).$$ More generally, when defining new variable(s) $\zeta(x)$ one has $$\phi(\xi)=\phi(\xi(x))=\psi(x).$$ One can then calculate the derivatives as, e.g., $$\frac{d\psi(x)}{dx}=\frac{d\phi(\xi)}{dx}=\frac{d\phi(\xi)}{d\xi}\frac{d\xi}{dx},\\ \frac{d^2\psi(x)}{dx^2}=\frac{d^2\phi(\xi)}{dx^2}=\frac{d^2\phi(\xi)}{d\xi^2}\left(\frac{d\xi}{dx}\right)^2+\frac{d\phi(\xi)}{d\xi}\frac{d^2\xi}{dx^2},$$ and so on.