Question

Derive schrodingers equation

Answer 1

We know that wave equation

$\psi(x,t)=Ae^{i(kx-wt)}$

Hamilton equation

H=T+V

Now

$E=\frac{p^2}{2m}+V(x)$

Take first derivative of a function

$$\frac{d\psi(x,t)}{dt}=-iwAe^{(kx-wt)}=-iw\psi(x,t)$$

$$\frac{d^2\psi(x,t)}{dt^2}=-k^2Ae^{(kx-wt)}=-k^2\psi(x,t)$$

$\frac{d^2\psi(x,t)}{dt^2}=\frac{-p^2}{\hbar^2}\psi(x,t)$

$$E\psi(x,t)=\frac{p^2}{2m}\psi(x,t)+V(x)\psi(x, t)\rightarrow(1)$$

$P=\hbar k\\k=\frac{2\pi}{\lambda}$

$$E\psi(x,t)=\frac{-\hbar^2}{2m}\frac{d^2}{dt^2}\psi(x,t)+V(x)\psi(x,t)$$

$E=\hbar w$

$E\psi(x,t)=-\frac{\hbar w}{iw}\psi(x,t)$

$i\hbar\frac{d\psi}{dt}=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x,t)+V(x)\psi(x,t)$