Aula 09 - Teoria De Shrodinger.pdf

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Cap. 1

01-10-2018

Prof. Dr. Elias Oliveira Serqueira 31 [email protected]

ℏ2 𝑑 2 Ξ¨ π‘₯, 𝑑 πœ•Ξ¨ π‘₯, 𝑑 βˆ’ + 𝑉 π‘₯ Ξ¨ π‘₯, 𝑑 = 𝑖ℏ 2π‘š 𝑑π‘₯ 2 πœ•π‘‘

Ξ¨ π‘₯, 𝑑 = ψ π‘₯ πœ‘ 𝑑

ℏ2 𝑑 2 ψ π‘₯ πœ‘ 𝑑 βˆ’ 2π‘š 𝑑π‘₯ 2

πœ• ψ π‘₯ πœ‘ 𝑑 + 𝑉 π‘₯ ψ π‘₯ πœ‘ 𝑑 = 𝑖ℏ πœ•π‘‘

ℏ2 𝑑2 ψ π‘₯ πœ•πœ‘ 𝑑 βˆ’ πœ‘ 𝑑 + 𝑉 π‘₯ ψ π‘₯ πœ‘ 𝑑 = π‘–β„Οˆ π‘₯ 2π‘š 𝑑π‘₯ 2 πœ•π‘‘ 32

ℏ2 𝑑2 ψ π‘₯ πœ•πœ‘ 𝑑 βˆ’ πœ‘ 𝑑 + 𝑉 π‘₯ ψ π‘₯ πœ‘ 𝑑 = π‘–β„Οˆ π‘₯ 2π‘š 𝑑π‘₯ 2 πœ•π‘‘ Dividindo toda equação por Ξ¨ π‘₯, 𝑑 = ψ π‘₯ πœ‘ 𝑑 𝑑2 ψ π‘₯ ℏ2 πœ•πœ‘ 𝑑 πœ‘ 𝑑 π‘–β„Οˆ π‘₯ 𝑉 π‘₯ ψ π‘₯ πœ‘ 𝑑 2 2π‘š 𝑑π‘₯ πœ•π‘‘ βˆ’ + = ψ π‘₯ πœ‘ 𝑑 ψ π‘₯ πœ‘ 𝑑 ψ π‘₯ πœ‘ 𝑑 ℏ2 1 𝑑 2 ψ π‘₯ 1 πœ•πœ‘ 𝑑 βˆ’ + 𝑉 π‘₯ = 𝑖ℏ 2π‘š ψ π‘₯ 𝑑π‘₯ 2 πœ‘ 𝑑 πœ•π‘‘

ℏ2 1 𝑑 2 ψ π‘₯ βˆ’ +𝑉 π‘₯ =𝐴 2 2π‘š ψ π‘₯ 𝑑π‘₯ 1 πœ•πœ‘ 𝑑 𝑖ℏ =𝐴 πœ‘ 𝑑 πœ•π‘‘ 33

1 πœ•πœ‘ 𝑑 𝑖ℏ =𝐴 πœ‘ 𝑑 πœ•π‘‘

πœ•πœ‘ 𝑑 𝐴 = πœ‘ 𝑑 πœ•π‘‘ 𝑖ℏ

πœ•πœ‘ 𝑑 𝑖𝐴 = 2 πœ‘ 𝑑 πœ•π‘‘ 𝑖 ℏ

πœ•πœ‘ 𝑑 𝐴 = βˆ’π‘– πœ‘ 𝑑 πœ•π‘‘ ℏ O mΓ©todo de solução:

π‘‘πœ‘ 𝑑′ 𝐴 = βˆ’π‘– 𝑑𝑑′ πœ‘ 𝑑′ ℏ

ln πœ‘ 𝑑

πœ‘ πœ‘0

𝐴 = βˆ’π‘– 𝑑′ ℏ

πœ‘ 𝑑 𝐴 ln = βˆ’π‘– 𝑑 πœ‘0 ℏ

𝑑 𝑑0 =0

πœ‘

π‘‘πœ‘ 𝑑′ 𝐴 = βˆ’π‘– πœ‘ 𝑑′ ℏ πœ‘0

𝑑

𝑑𝑑′ 𝑑0 =0

𝐴 ln πœ‘ 𝑑 βˆ’ ln πœ‘0 = βˆ’π‘– 𝑑 βˆ’ 𝑑0 ℏ πœ‘ 𝑑 𝐴 = exp βˆ’π‘– 𝑑 πœ‘0 ℏ

𝐴 πœ‘ 𝑑 = πœ‘0 exp βˆ’π‘– 𝑑 ℏ

πœ‘ 𝑑 =

𝐴 βˆ’π‘– 𝑑 πœ‘0 𝑒 ℏ

34

πœ‘ 𝑑 =

𝐴 βˆ’π‘– 𝑑 πœ‘0 𝑒 ℏ

Note que o expoente deve ser adimensional

𝐴 = 𝑇 βˆ’1 ℏ

Assim, essa relação leva a entender que a grandeza A seja energia da forma

𝐴 = 𝐸 = β„πœ” Onde a parte temporal da função de onda

πœ‘ 𝑑 = πœ‘0 𝑒 βˆ’π‘–πœ”π‘‘ 35

Retornando as equaçáes: ℏ2 1 𝑑 2 ψ π‘₯ βˆ’ +𝑉 π‘₯ =𝐴 2π‘š ψ π‘₯ 𝑑π‘₯ 2 𝑖ℏ

Γ—

ψ π‘₯

1 πœ•πœ‘ 𝑑 =𝐴 πœ‘ 𝑑 πœ•π‘‘

Rescrevendo a 1ΒΊ ao realizar a devida operação, obtΓ©m-se ℏ2 𝑑 2 ψ π‘₯ βˆ’ + 𝑉 π‘₯ ψ π‘₯ = 𝐸ψ π‘₯ 2π‘š 𝑑π‘₯ 2 Equação de Schrodinger independente do Tempo

36

37

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39

40

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No limite, vΓ£o para zero

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1 0

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