Quantum Mechanics: Substitution and Coulomb Potential in Schrodinger Equation

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SUMMARY

The discussion focuses on the derivation of the Schrödinger equation through the substitution of operators corresponding to dynamical variables, leading to the equation \left( \frac{-ħ^{2}}{2m} ∇^{2} + V \right)ψ = Eψ. It also addresses the Coulomb potential for an electron in a hydrogen atom, defined as V(r) = \frac{-e^{2}}{4πε₀r}. The participants confirm the correctness of the derivation and the potential equation, emphasizing the application of quantum mechanics principles in these calculations.

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Homework Statement



(i) Briefly indicate how substitution of operators corresponding to dynamical variables in an eigenvalue equation leads to the Schrödinger equation \left( \frac{-ħ^{2}}{2m} ∇^{2} + V \right)ψ = Eψ.

(ii) What is the Coulomb potential, V(r), of an electron, charge e, in a hydrogen atom at distance r from the nucleus?

(iii), (iv), (v) left out for the moment

Homework Equations



The Attempt at a Solution



(i) (T + V) = E : law of conservation of energy

Multiply by ψ to obtain an eigenvalue equation: (T + V)ψ = Eψ

Substitute operators \widehat{T} and \widehat{V} corresponding to the dynamical variables T and V in the eigenvalue equation: ( \widehat{T} + \widehat{V} ) ψ = Eψ

\widehat{T} = \frac{\widehat{p}^{2}}{2m} = \frac{(-iħ∇)^{2}}{2m} = \frac{-ħ^{2}}{2m} ∇^{2}

\widehat{V} = V

So, the eigenvalue equation becomes the Schrödinger equation \left( \frac{-ħ^{2}}{2m} ∇^{2} + V \right)ψ = Eψ.(ii) V(r) = \frac{-e^{2}}{4πε₀r}

Any comments would be greatly appreciated.
 
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