Cap. 1
01-10-2018
Prof. Dr. Elias Oliveira Serqueira 31
[email protected]
β2 π 2 Ξ¨ π₯, π‘ πΞ¨ π₯, π‘ β + π π₯ Ξ¨ π₯, π‘ = πβ 2π ππ₯ 2 ππ‘
Ξ¨ π₯, π‘ = Ο π₯ π π‘
β2 π 2 Ο π₯ π π‘ β 2π ππ₯ 2
π Ο π₯ π π‘ + π π₯ Ο π₯ π π‘ = πβ ππ‘
β2 π2 Ο π₯ ππ π‘ β π π‘ + π π₯ Ο π₯ π π‘ = πβΟ π₯ 2π ππ₯ 2 ππ‘ 32
β2 π2 Ο π₯ ππ π‘ β π π‘ + π π₯ Ο π₯ π π‘ = πβΟ π₯ 2π ππ₯ 2 ππ‘ Dividindo toda equação por Ξ¨ π₯, π‘ = Ο π₯ π π‘ π2 Ο π₯ β2 ππ π‘ π π‘ πβΟ π₯ π π₯ Ο π₯ π π‘ 2 2π ππ₯ ππ‘ β + = Ο π₯ π π‘ Ο π₯ π π‘ Ο π₯ π π‘ β2 1 π 2 Ο π₯ 1 ππ π‘ β + π π₯ = πβ 2π Ο π₯ ππ₯ 2 π π‘ ππ‘
β2 1 π 2 Ο π₯ β +π π₯ =π΄ 2 2π Ο π₯ ππ₯ 1 ππ π‘ πβ =π΄ π π‘ ππ‘ 33
1 ππ π‘ πβ =π΄ π π‘ ππ‘
ππ π‘ π΄ = π π‘ ππ‘ πβ
ππ π‘ ππ΄ = 2 π π‘ ππ‘ π β
ππ π‘ π΄ = βπ π π‘ ππ‘ β O mΓ©todo de solução:
ππ π‘β² π΄ = βπ ππ‘β² π π‘β² β
ln π π‘
π π0
π΄ = βπ π‘β² β
π π‘ π΄ ln = βπ π‘ π0 β
π‘ π‘0 =0
π
ππ π‘β² π΄ = βπ π π‘β² β π0
π‘
ππ‘β² π‘0 =0
π΄ ln π π‘ β ln π0 = βπ π‘ β π‘0 β π π‘ π΄ = exp βπ π‘ π0 β
π΄ π π‘ = π0 exp βπ π‘ β
π π‘ =
π΄ βπ π‘ π0 π β
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π π‘ =
π΄ βπ π‘ π0 π β
Note que o expoente deve ser adimensional
π΄ = π β1 β
Assim, essa relação leva a entender que a grandeza A seja energia da forma
π΄ = πΈ = βπ Onde a parte temporal da função de onda
π π‘ = π0 π βπππ‘ 35
Retornando as equaçáes: β2 1 π 2 Ο π₯ β +π π₯ =π΄ 2π Ο π₯ ππ₯ 2 πβ
Γ
Ο π₯
1 ππ π‘ =π΄ π π‘ ππ‘
Rescrevendo a 1ΒΊ ao realizar a devida operação, obtΓ©m-se β2 π 2 Ο π₯ β + π π₯ Ο π₯ = πΈΟ π₯ 2π ππ₯ 2 Equação de Schrodinger independente do Tempo
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No limite, vΓ£o para zero
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1 0
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