Question

Derive the Schrödinger's Time independent wave equation using kinetic energy and momentum.
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Answer 1

Consider,

`{:\implies \quad \displaystyle \sf \langle p\rangle =m\langle v(t)\rangle=m\int_(-\infty)^(\infty)x\bigg\{(\partial \Psi^(*)(x,t))/(\partial t)\Psi (x,t)+\Psi^(*)(x,t)(\partial \Psi (x,t))/(\partial t)\bigg\}dx}`

Multiply both sides by ih and simplification will yield

Some simplification, Then Integrate by parts and then knowing the fact that the wave function vanishes for
`{\bf x\to \pm \infty}` will yield:

`{:\implies \quad \displaystyle \sf \langle p\rangle =(ih)/(2)\int_(-\infty)^(\infty)\bigg\{(\partial \Psi^(*)(x,t))/(\partial x)\Psi (x,t)-\Psi^(*)(x,t)(\partial \Psi (x,t))/(\partial x)\bigg\}dx}`

Integrating by parts and knowing the same fact by some simplification will yield:

`{:\implies \quad \displaystyle \sf \langle p\rangle =-ih\int_(-\infty)^(\infty)\Psi^(*)(x,t)(\partial \Psi (x,t))/(\partial x)dx}`

The momentum is thus contained within the wave function, so we can then deduce that:

`{:\implies \quad \sf p\rightarrow -ih(\partial)/(\partial x)}`

`{:\implies \quad \sf p^(n)\rightarrow \bigg(-ih(\partial)/(\partial x)\bigg)^(n)}`

`{:\implies \therefore \quad \displaystyle \sf \langle p^(2)\rangle =-h^(2)\int_(-\infty)^(\infty)\Psi^(*)(x,t)(\partial^(2)\Psi (x,t))/(\partial x^2)dx}`

Now the kinetic energy

`{:\implies \quad \displaystyle \sf \langle K\rangle =(\langle p^(2)\rangle)/(2m)=(-h^2)/(2m)\int_(-\infty)^(\infty)\Psi^(*)(x,t)(\partial^(2)\Psi (x,t))/(\partial x^2)dx}`

The classical formula for the total energy

`{:\implies \quad \sf (p^2)/(2m)+V(x)=E}`

Multiplying this equation by
`{\sf \Psi (x,t)=\psi (x)exp\bigg((-iEt)/(h)\bigg)}` and use the above equations and simplify it we will be having

`{:\implies \quad \boxed{\bf{(-h^2)/(2m)(d^(2)\psi (x))/(dx^(2))+V(x)\psi (x)=E\psi (x)}}}`

This is the Famous Time-Independent Schrödinger wave equation

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