Simply Lecture

Recall that classical energy is given by the equation: $$E_c = \frac{p^{2}}{2m} + V(r) + V_0$$ ........(1) Where, V( r ) = spatially dependent potential              \(V_0\) = constant potential energy Recall from the square of the magnitude of the momenergy 4-vector associated with quantum object is given by: $$E^{2} - p^{2} = m^{2}$$ ..........(2) Divide equ (2) by 2m $$\frac{E^{2}}{2m} - \frac{p^{2}}{2m} = \frac{m^{2}}{2m}$$ $$\frac{E^{2}}{2m} - \frac{p^{2}}{2m} = \frac{m}{2}$$ Making \(\frac{p^{2}}{2m}\) subject of the formula, we have: $$\frac{p^{2}}{2m} = \frac{E^{2}}{2m} - \frac{m}{2}$$ .........(3) Also let: $$V_0 = \frac{m}{2} + E_c - E$$ ............(4) Substitute equ (4) and equ (3) into equ (1), we have: $$E_c = \frac{E^{2}}{2m} - \frac{m}{2} + V(r) + \frac{m}{2} + E_c - E$$ Making \(\frac{E^{2}}{2m}\) subject of the formula, we have: $$\frac{E^{2}}{2m} = E_c - E_c + \frac{m}{2} - \frac{m}{2} - V(r) + E$$ $$\frac{E^{2}}{2m}...