Skip to main content

Time independent 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:
$$-\frac{\hbar^{2}}{2m} \nabla^2 \psi (r) = E \psi (r) - V(r) \psi (r)$$
$$-\frac{\hbar^{2}}{2m} \nabla^{2} \psi(r) + V(r) \psi(r) = E \psi(r)$$ ..............(6)
Equ (6) is the time independent Schrodinger equation.

Labels:

Comments

Post a Comment

Popular posts from this blog

Schrodinger equation as a law in physics

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...
Post a Comment
Read more

Auxiliary time dependent Schrodinger equation

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...
Post a Comment
Read more

Maxwell first equation

The Maxwell first equation in electrostatics is called the Gauss law in electrostatics. Statement:  It states that the total electric flux \(\psi_E\) passing through a closed hypothetical surface is equal to \(\frac{1}{\epsilon_0}\) enclosed by the surface. Integral Form: $$\phi_E = \int E.ds = \frac{q}{\epsilon_0}$$ $$\int D.ds = q$$ where, $$D = \epsilon_0 E = displacement-vector$$ Let the change be distributed over a volume v and \(\rho\) be the volume charge density. Hence, $$q = \int \rho dv$$ Therefore; $$\int D.ds = \int_{v} \rho dv$$ .........(1) Equ(1) is the integral form of Maxwell first law Differential form: Apply Gauss divergence theorem to the L.H.S of equ(1) from surface integral to volume integral. $$\int D.ds = \int (\nabla.D)dv$$ Substituting this equation to equ(1) $$\int(\nabla.D)dv = \int_{v} \rho dv$$ As two volume integrals are equal only if their integrands are equal. Thus; $$\nabla.D = \rho v$$ ............(2) Equ(2) is the dif...
Post a Comment
Read more