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Energy form of the Schrodinger equation

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:
$$\frac{E^{2}}{2m} = E_c - E_c + \frac{m}{2} - \frac{m}{2} - V(r) + E$$
$$\frac{E^{2}}{2m} = E - V(r)$$ .........(5)
Equ (5) is the energy form of the Schrodinger equation.
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