there is anon
We have seen that the wavefunction of a free particle of mass (m) satisfies [\label{e2.78} \psi(x,t)=\int_{-\infty}^{\infty}\bar{\psi}(k)\,{\rm e}^{\,{\rm i}\,(k\,x-\omega\,t)}\,dk,] where (\bar{\psi}(k)) is determined by (\psi(x,0)), and [\label{e2.79} \omega(k) = \frac{\hbar\,k^{\,2}}{2\,m}.] It follows from Equation ([e2.78]) that [\label{e2.80} \frac{\partial\psi}{\partial x} = \int_{-\infty}^{\infty}({\rm i}\,k)\,\bar{\psi}(k)\,{\rm e}^{\,{\rm i}\,(k\,x-\omega\,t)}\,dk,] and [\frac{\partial^{\,2}\psi}{\partial x^{\,2}} = \int_{-\infty}^{\infty}(-k^{\,2})\,\bar{\psi}(k)\,{\rm e}^{\,{\rm i}\,(k\,x-\omega\,t)}\,dk,] whereas [\frac{\partial \psi}{\partial t} = \int_{-\infty}^{\infty}(-{\rm i}\,\omega)\,\bar{\psi}(k)\,{\rm e}^{\,{\rm i}\,(k\,x-\omega\,t)}\,dk.] Thus, [{\rm i}\,\frac{\partial\psi}{\partial t} +\frac{\hbar}{2\,m}\,\frac{\partial^{\,2}\psi}{\partial x^{\,2}} = \int_{-\infty}^{\infty}\left(\omega-\frac{\hbar\,k^{\,2}}{2\,m}\right)\bar{\psi}(k)\,{\rm e}^{\,{\rm i}\,(k\,x-\omega\,t)}\,dk= 0,] where use has been made of the dispersion relation ([e2.79]). Multiplying through by (\hbar), we obtain [\label{e2.84} {\rm i}\,\hbar\,\frac{\partial\psi}{\partial t} = -\frac{\hbar^{\,2}}{2\,m}\frac{\partial^{\,2}\psi}{\partial x^{\,2}}.] This expression is known as Schrödinger’s equation,