I want to rewrite the equation as follows:

$$\frac{\partial^2\psi(x,t)}{\partial x^2}=-\bigg(\frac{2\pi}{\lambda}\bigg)^2\psi(x,t)$$

The initial equation is as follows:

$$-i\frac{h}{2\pi}\frac{\partial \psi(x,t)}{\partial x}=p\,\psi(x,t)$$

How can I get to the equation: $\frac{\partial^2\psi(x,t)}{\partial x^2}=-\big(\frac{2\pi}{\lambda}\big)^2\psi(x,t)$, starting from the initial equation?

I don't know how I can derive the initial equation in order to x, because I don't know what the derivative of $\psi(x,t)$ is.


Sure you do - rearranging the "initial" equation you've provided gives $$\frac{\partial\psi(x,t)}{\partial x} = \frac{2\pi i p}{h}\psi(x,t)$$

Differentiating again gives you just what you're looking for - assuming that you are familiar with the concept of the de Broglie wavelength of a particle.

  • $\begingroup$ I still can not vote for your answer, but thank you, now I realized the reasoning. $\endgroup$ – Maria Barroso May 29 '19 at 9:47

You do! that's the initial equation that you have! $$-i \bigg(\frac{h}{2 \pi }\bigg)\frac{\partial \psi(x,t)}{\partial x} = p \psi(x,t)$$ So if you differentiate the equation again you get, $$-i \frac{h}{2 \pi }\frac{\partial^2 \psi(x,t)}{\partial x^2} = p \frac{\partial \psi(x,t)}{\partial x} = i p^2 \frac{2 \pi}{h} \psi(x,t) $$ The expression for the debroglie wavelength is, $\lambda = \frac{h}{p}$. So, we get, $$\frac{\partial^2 \psi(x,t)}{\partial x^2} = - \bigg(\frac{2\pi}{\lambda}\bigg)^2 \psi(x,t)$$ Which is the desired answer.

  • $\begingroup$ Why does the derivative of $\psi(x,t)$ give $ip\frac{2\pi}{h}$? $\endgroup$ – Maria Barroso May 29 '19 at 9:05
  • $\begingroup$ @MariaBarroso Oh, it was the initial equation that Aaron had written, But I think the assumption here is that $\psi(x,t)$ is a momentum eigenstate since the momentum operator $\hat{P} = - i \hslash \frac{\partial }{\partial x}$ $\endgroup$ – Sparsh Mishra May 30 '19 at 4:37
  • $\begingroup$ I don't think I wrote any equations $\endgroup$ – BioPhysicist May 30 '19 at 12:10
  • $\begingroup$ @AaronStevens I don't quite understand. You wrote, "the equation goes ..." $\endgroup$ – Sparsh Mishra May 31 '19 at 7:17
  • $\begingroup$ You do realize that I didn't write this question, right? $\endgroup$ – BioPhysicist May 31 '19 at 11:23

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