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set5 - Hydrogenatom:(proton)(electron 2 e V(r = 4 0 r p r...

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Hydrogen atom: Interac0on of a posi0vely charged par0cle (proton) with a nega0vely charged par0cle (electron) p r V r ( ) = e 2 4 πε 0 r
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If we assume that the proton (much heavier by about a factor of 2000) is at rest while the electron is mobile, we can write approximate hamiltonian for the energy levels of the electron H = 2 2 m e d 2 dx 2 + d 2 dy 2 + d 2 dx 2 e 2 4 πε 0 r = 2 2 m e 2 e 2 4 πε 0 r 2 = 1 r 2 r r 2 r + 1 r 2 1 sin θ ( ) θ sin θ ( ) θ + 1 sin 2 θ ( ) 2 φ 2 multiplying H ψ r , θ , φ ( ) = E ψ r , θ , φ ( ) by 2 m e r 2 we have 2 r r 2 r ψ r , θ , φ ( ) 2 1 sin θ ( ) θ sin θ ( ) θ + 1 sin 2 θ ( ) 2 φ 2 ψ r , θ , φ ( ) 2 m e r 2 e 2 4 πε 0 r + E ψ r , θ , φ ( ) = 0 Note that only the middle term depends on θ and ϕ sugges0ng separa0on of variables ψ r , θ , φ ( ) = R r ( ) ? Y θ , φ ( ) = R r ( ) Θ θ ( ) Φ φ ( )
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Mathema0cal detour: Separa0on of variables I Two non‐interac0ng par0cles in a box H = 2 2 m 1 d 2 dx 1 2 2 2 m 2 d 2 dx 2 2 H 1 + H 2 If the hamiltonian is wriGen as a sum of independent terms then the wavefunc0on is a product of the independent components. i.e. ψ x 1 , x 2 ( ) = φ x 1 ( ) χ x 2 ( ) H ψ = 2 2 m 1 d 2 dx 1 2 + 2 2 m 2 d 2 dx 2 2 φ x 1 ( ) χ x 2 ( ) = E ψ 2 2 m 1 φ '' x 1 ( ) χ x 2 ( ) 2 2 m 2 φ x 1 ( ) χ '' x 2 ( ) = E φ x 1 ( ) χ x 2 ( ) dividing by ψ = φχ 2 2 m 1 φ '' x 1 ( ) φ x 1 ( ) 2 2 m 2 χ '' x 2 ( ) χ x 2 ( ) = E
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Mathema0cal detour II The final equa0on derived is a sum of two terms one depends only on x 1 and only on x 2 and their sum is equal a constant. This can be right only if each term is a constant 2 2 m 1 φ '' x 1 ( ) φ x 1 ( ) 2 2 m 2 χ '' x 2 ( ) χ x 2 ( ) = E ⇒ − 2 2 m 1 φ '' x 1 ( ) φ x 1 ( ) = E 1 2 2 m 2 χ '' x 2 ( ) χ x 2 ( ) = E 2 We therefore have two independent one dimensional par0cle in a box problem to solve 2 2 m 1 φ '' x 1 ( ) = E 1 φ x 1 ( ) 1 2 2 m 2 χ '' x 2 ( ) = E 2 χ x 2 ( ) & E 1 + E 2 = E
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Back to hydrogen atom 2 r r 2 r ψ r , θ , φ ( ) 2 1 sin θ ( ) θ sin θ ( ) θ
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