Obtain one dimensional time dependent time independent Schrodinger equation.

#### Solution

For one dimensional case, the classical wave is described by the wave equation

`(dy^2)/(dx^2)=1/ (v^2) × (d^2 y)/( dt^2)`

where y is the displacement and v is the velocity of the wave travelling in a direction. The displacement of the particle at any instant ‘t’ , at any point ‘x’ in space

`y(x, t ) = Ae^{j(kx-ωt)}

where ω = 2πϑ and k = 2π/λ

in analogy with this the wave function which describes the behaviour of the matter particle at any instant ‘t’, at any point ‘x’ in space can be written as

Ψ(x,t) = Ae^{j(kx-ωt) }

Where , ω = 2πϑ = 2π`E/h = E/h`

And `k = (2π)/λ = (2π)/h ×p = p/h`

The total energy of the particle is given by

E = kinetic energy + potential energy

`= 1/2(mv^2) + V = ((mv)^2)/(2m) + V`

`E = (p^2)/(2m )+ V`

Operating this on the wave function Ψ(x, t) it is found that

`EΨ(x ,t) = (p^2)/(2m) Ψ(x, t) + VΨ(x, t)`

Differentiating equation with respect to ‘x’ and ‘t’ it is obtained that

`(Ə^2 Ψ(x, t))/ (Əx^2) = -p^2Ae^(j(kx-ωt)) = - k ^2Ψ(x, t)`

Hence

`(p^2)/(Ət) Ψ(x, t) = - jAωe^(j(kx-ωt)) = - jωΨ(x, t)`

Hence the final equation is as follows :-

`jħ(ƏΨ(x, t))/(Ət ) = - ħ/(2m) ×(Ə ^2Ψ(x, t) ) /(Əx^2) + VΨ(x, t)`

Or `- ħ /( 2m) ×(Ə^2 Ψ(x, t) )/(Əx^2) + VΨ(x, t) = jħ (ƏΨ(x, t))/(Ət)`

The first and the second term on the left hand side represents the kinetic and potential energies respectively of the particle and the right hand side represents the total energy.

This is called as the one dimensional time dependent Schrodinger equation.

ONE DIMENSIONAL TIME INDEPENDENT SCHRODINGER EQUATION.

`- (ħ^2)/(2m) × (d^2 Ψ(x))/(dx^2) + V(x)Ψ(x) = EΨ(x)`