Simply Lecture

Recall the relationship: $$k = \frac{E}{\hbar}$$ ........(1) And also the relationship: $$(\nabla^{2} + k^{2}) \psi (r)=0$$ ........(2) Substitute equ (1) into equ (2) $$(\nabla^{2} + \frac{E^{2}}{\hbar^{2}}) \psi (r) = 0$$ Opening the brackets: $$\nabla^{2} \psi(r) + \frac{E^{2}}{\hbar^{2}} \psi(r) =0$$ Therefore: $$\nabla^{2} \psi(r) = -(\frac{E}{\hbar})^{2} \psi(r)$$ ............(3) Multiply both sides of equ (3) by \(-\frac{\hbar^{2}}{2m}\), we have: $$-\frac{\hbar^{2}}{2m} \nabla^{2} \psi(r) = \frac{\hbar^{2}}{2m} (\frac{E^{2}}{\hbar^{2}}) \psi(r)$$ $$-\frac{\hbar^{2}}{2m} \nabla^{2} \psi(r) = (\frac{E^{2}}{2m}) \psi(r)$$ ..........(4) Recall, the relationship from the energy form of the Schrodinger equation, we have: $$\frac{E^{2}}{2m} = E - V(r)$$ .............(5) Substitute equ (5) into equ (4): $$-\frac{\hbar^{2}}{2m} \nabla^{2} \psi(r) = (E - V(r)) \psi(r)$$ Opening bracket,  we have: $$-\frac{\hbar^{2}}{2m} \nabla^2 \psi (r) = E \psi (r) - V(r) \psi (r)...