Simply Lecture

The unified theory of wave-particle duality has been used to derive the Schrödinger equations. The Schrodinger equations are generally accepted, by postulate rather than derivation, to be laws of physics. The Schrodinger equations provide a basis for analyzing many kinds of systems (molecular, atomic, and nuclear) in a particular inertial reference frame. The success of the Schrödinger equations constitutes a basis for accepting them, their derivations, and the unified theory of wave-particle duality which makes such derivations possible. This acceptance is completely justified in the favored inertial reference frame. In accord with the principle of relativity, all physical laws must be the same in all inertial reference frames, i.e., all physical laws must be Lorentz invariant. Recall, the relationship: $$\nabla^{2} \psi = \frac{\partial^{2} \psi}{\partial t^{2}}$$ ...........(1) Equ (1) is Lorentz invariant and reduces, by means of the procedure presented in the previou...

Recall the relationship: $$\phi(t) = C_+ exp[-i(\frac{E}{\hbar})t] + C_- exp[i(\frac{E}{\hbar})t]$$ .............(1) Recall, from the present special case: $$C_+ = 0$$ ........(2) Substitute equ (2) into equ (1), we have that: $$\phi(t) = C_- exp[i(\frac{E}{\hbar})t]$$ .......(3) Also,  recall the relationship: $$\Psi(r,t) = \psi(r) \phi(t)$$ ..........(4) Substitute equ (3) into equ (4): $$\Psi(r,t) = C_- exp \psi (r)  [i(\frac{E}{\hbar})t]$$ ..............(5) Equ (5) can be differentiated with respect to t, and by rearranging we have: $$\Psi(r,t) = -i(\frac{\hbar}{E}) \frac{d \Psi (r, t)}{dt}$$ ..............(6) Recall the relationship: $$-\frac{\hbar^{2}}{2m} \nabla^{2} \Psi(r,t) + V(r) \Psi(r,t) = E \Psi(r,t)$$ ...........(7) Substituting equ (6) into the right hand side of equ (7) we have: $$-\frac{\hbar^{2}}{2m} \nabla^{2} \Psi(r,t) + V(r) \Psi(r,t) = i\hbar \frac{d \Psi(r,t)}{dt}$$ .........(8) Equ (8) is the auxiliary time dependent Schrodinger equation...

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)...

Recall from the time independent Schrodinger equation: $$-\frac{\hbar^{2}}{2m} \nabla^{2} \psi(r) + V(r) \psi(r) = E \psi(r)$$ ........(1) Multiply both sides of equ (1) by \(\phi(t)\), we have: $$-\frac{\hbar^{2}}{2m} \nabla^{2} \psi(r) \phi(t) + V(r) \psi(r) \phi(t) = E \psi(r) \phi(t)$$ ..........(2) Recall: $$\psi(r) \phi(t) = \Psi(r,t)$$ .............(3) Substitute equ (3) into equ (2): $$-\frac{\hbar^{2}}{2m} \nabla^{2} \Psi(r,t) + V(r) \Psi(r,t) = E \Psi(r,t)$$ .........(4) Recall that the well known Planck-Einstein relation is given by: $$E = hv$$ ...........(5) And also, the reduced Planck's constant is given by: $$\hbar = \frac{h}{2\pi}$$ ................(6) Also recall that time dependence is given by: $$\phi(t) = C_+ exp(-i2\pi vt) + C_- exp(i2\pi vt)$$ ..........(7) Substituting equ (6) into (5), we have: $$V = \frac{E}{\hbar 2 \pi}$$ ..........(8) Substitute equ (8) into equ (7) $$\phi(t) = C_+ exp(-i2 \pi \frac{E}{\hbar 2 \pi} t)  + C_- exp(i2 \pi \f...