Question

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

Answer 1

Consider,

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

Multiply both sides by ih and simplification will yield

${:\implies \quad \displaystyle \sf ih\langle p\rangle =m\int_{-\infty}^{\infty}x\bigg[\Psi (x,t)\bigg\{\dfrac{h^2}{2m}\dfrac{\partial^{2}\Psi^{*}(x,t)}{\partial x^2}-V(x)\Psi^{*}(x,t)\bigg\}+\Psi^{*}(x,t)\bigg\{V(x)\Psi (x,t)-\dfrac{h^2}{2m}\dfrac{\partial^{2}\Psi (x,t)}{\partial x^2}\bigg\}\bigg]dx}$

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

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

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

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

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

${:\implies \quad \sf p\rightarrow -ih\dfrac{\partial}{\partial x}}$

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

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

Now the kinetic energy

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

The classical formula for the total energy

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

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

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

This is the Famous Time-Independent Schrödinger wave equation

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