6
$\begingroup$

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?

$\endgroup$
4
  • 1
    $\begingroup$ The question is actually about math. $\endgroup$ Jan 24 at 13:36
  • $\begingroup$ Can you give the source of this? $\endgroup$ Jan 24 at 13:38
  • 2
    $\begingroup$ @RogerVadim I think this question still fits best at physics stack exchange. Many more physicists than mathematicians have worked with the Schrödinger equation and just because a physics concept contains math doesn't mean that it is a math question. $\endgroup$ Jan 24 at 14:04
  • 2
    $\begingroup$ @AccidentalTaylorExpansion I think that question is about general mathematical technique, which is why proposed moving it to the Math community. But I do agree that here it arises in a specific context, which is mostly of interest to physicists. $\endgroup$ Jan 24 at 14:09

2 Answers 2

6
$\begingroup$

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 $$\psi(x)=\psi_{new}(\frac{x}{a})$$

$\endgroup$
6
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.