# How to show that $\psi(x)=-A\exp(-\alpha^2x^2)$ satisfies TISE for $V(x)=\frac 1 2 m\omega_0^2x^2$? [closed]

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.

• 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. Aug 1, 2022 at 12:08

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.

• @ZeroTheHero I'd say it's a valid suggestion as the question indicates to "show that at large values of $x$..."
– nuwe
Aug 1, 2022 at 12:42
• @nuwe you’re right. Why didn’t I see this? I will delete my comment. Aug 1, 2022 at 12:43
• @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. Aug 1, 2022 at 12:49
• @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?
– user332426
Aug 1, 2022 at 13:43
• @NGodrich That's it, you are on the right track. You just forgot $A$ on the right side. Try to apply a limit $x\to\infty$. Aug 1, 2022 at 14:34