# Rewrite an equation by deriving the Schrondinger eigenvalue equation (linear momentum)

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.

• I still can not vote for your answer, but thank you, now I realized the reasoning. – Maria Barroso May 29 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.

• Why does the derivative of $\psi(x,t)$ give $ip\frac{2\pi}{h}$? – Maria Barroso May 29 at 9:05
• @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 $\hat{P} = - i \hslash \frac{\partial }{\partial x}$ – Sparsh Mishra May 30 at 4:37
• I don't think I wrote any equations – Aaron Stevens May 30 at 12:10
• @AaronStevens I don't quite understand. You wrote, "the equation goes ..." – Sparsh Mishra May 31 at 7:17
• You do realize that I didn't write this question, right? – Aaron Stevens May 31 at 11:23