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I'm struggling to approach this 'show that' question:

Write down the time-independent Schrödinger differential equation for $\psi(x)$ in a one-dimensional and time-independent potential $V(x)$. In the case that $V(x)=\frac 1 2 m\omega_0^2x^2$, show that at large values of $x$, $\psi(x)=A\exp(-\alpha^2x^2)$ is a solution to this differential equation, where $\alpha$ and $A$ are constants.

I've tried substituting $\psi$ into the TISE, but (to my untrained eye) this didn't offer up any evidence that $\psi$ is a solution.

I would be grateful if someone might suggest an approach, not a complete solution.

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    $\begingroup$ It will be great if you can show your work, and it will be more evident to see in which step you had gone wrong. $\endgroup$
    – Andy Chen
    Aug 1, 2022 at 12:08

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Write the TISE as $$\frac{d^2\psi}{dx^2}=\frac{m^2\omega_0^2x^2-2mE}{\hbar^2}\psi$$ and observe that for large $x$, $2mE$ is negligible compared to $m^2\omega_0^2 x^2$.

Therefore, you can write it as $$\frac{d^2\psi}{dx^2}=\frac{m^2\omega_0^2x^2}{\hbar^2}\psi$$ Now prove that the function $\psi(x)=Ae^{-\alpha^2x^2}$ satisfies this equation. First, you should differentiate $\psi(x)$ twice to get to the left side of the equation, and discard some terms that are negligible when $x$ is very large. Then, insert the $\psi(x)$ to the right side and some common terms should cancel out. You should also be able to find the value of $\alpha$.

If you still can't solve the problem using these suggestions, tell me in the comments and I'll post further advices.

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  • $\begingroup$ @ZeroTheHero I'd say it's a valid suggestion as the question indicates to "show that at large values of $x$..." $\endgroup$
    – nuwe
    Aug 1, 2022 at 12:42
  • $\begingroup$ @nuwe you’re right. Why didn’t I see this? I will delete my comment. $\endgroup$ Aug 1, 2022 at 12:43
  • $\begingroup$ @User123. This college HW question brings no value to this website (thought of as a knowledge database), so we generally discourage providing answers. The HW policy has exceptions, but this Q is none of them. $\endgroup$
    – DanielC
    Aug 1, 2022 at 12:49
  • $\begingroup$ @User123 Thank you so much for your help - it's really appreciated. So I have (and I don't know if maths text will work in the comments section but here it is) $4A\alpha^4x^2-2A\alpha^2=\frac{m^2\omega_0^2x^2}{\hbar^2}$. I could see how, in limit where x is large, this equivalence sort of holds - am I on the right track? $\endgroup$
    – user332426
    Aug 1, 2022 at 13:43
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    $\begingroup$ @NGodrich That's it, you are on the right track. You just forgot $A$ on the right side. Try to apply a limit $x\to\infty$. $\endgroup$
    – User123
    Aug 1, 2022 at 14:34