Skip to main content

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.

Labels

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

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

Maxwell third equation

Also called the Faraday law of "electromagnetic induction". The Maxwell third equation has two statements. Statement I:  It states that whenever a magnetic flux link with a circuit changes, then induced electromotive force (emf)  is set up in the circuit. Statement II: The magnitude of induced emf is equal to the rate of magnetic flux linked with the circuit. Integral form: Therefore; $$Induced-emf = - \frac {d\psi_m}{dt}$$ where, $$\psi_m = \int B.ds$$   ......(5) The negative sign is because of Lentz law,  which states that the induced emf set up a current in such a direction that the magnetic effect produced by it opposes the cause producing it. Also, the definition of emf states that the emf is the closed line integral of the non conservative electric field generated by the battery. That is: $$emf = \int E.dL$$ ........(6) Comparing equ(5) and equ(6) we have: $$\int E.dL = -\int_{s} \frac{dB.ds}{dt}$$ ...(7) Differential form Applying Sto...
Post a Comment
Read more