Aula 13 - Potenciais Unidimensionais..pdf

Cap. 1

23-10-2018

Prof. Dr. Elias Oliveira Serqueira 54 [email protected]

𝑎 𝑘 tan 𝑘 =𝑞 2

𝑘=

2𝑚𝐸 ℏ2

𝑞=

2𝑚 𝑉0 − 𝐸 2 ℏ

𝑎 𝑎 𝑎 𝑘 tan 𝑘 =𝑞 2 2 2 2𝑚𝐸 𝑎 𝑎 tan 𝑘 = ℏ2 2 2

𝑚𝐸𝑎2 tan 2ℏ2

2 𝑚𝐸𝑎 𝜀2 = 2ℏ2

𝑚𝐸𝑎2 2ℏ2

2 𝑚𝑎 𝑉0 𝑅2 = 2ℏ2

2𝑚 𝑎 𝑉 −𝐸 ℏ2 0 2

=

𝑚𝑎2 𝑉0 𝑚𝑎2 𝐸 − 2ℏ2 2ℏ2

55

𝜀 tan 𝜀 =

𝑦 𝜀 =

𝑅2

−

𝑅2 − 𝜀 2

𝑦 2 = 𝑅2 − 𝜀 2

𝜀2

𝑓 𝜀 = 𝜀 tan 𝜀

𝑦

𝑦 2 + 𝜀 2 = 𝑅2

𝜀

56

𝑦 𝜀 =

𝑅2 − 𝜀 2

𝑓 𝜀 = 𝜀 tan 𝜀

57

𝑎 𝑘 cot 𝑘 = −𝑞 2 𝑘 cot 𝑘

𝑘=

𝑞=

2𝑚𝐸 ℏ2 2𝑚 𝑉0 − 𝐸 2 ℏ

𝑎 = −𝑞 2

2𝑚𝐸 𝑎 𝑎 − cot 𝑘 = ℏ2 2 2

𝑚𝐸𝑎2 − cot 2ℏ2

2𝑚 𝑎 𝑉 −𝐸 ℏ2 0 2

𝑚𝐸𝑎2 = 2ℏ2

−𝜀 cot 𝜀 =

𝑚𝑎2 𝑉0 𝑚𝑎2 𝐸 − 2ℏ2 2ℏ2

𝑅2 − 𝜀 2 58

−𝜀 cot 𝜀 =

𝑅2

−

𝜀2

𝑔 𝜀 = −𝜀 cot 𝜀 𝑦 𝜀 = 𝑅2 − 𝜀 2

59

60

61

2ℏ2 𝑅2 =𝜎 𝑚

A relação da geometria do poço e os níveis de energia

𝑅2

𝑚𝑎2 𝑉0 = 2ℏ2

𝑎2 𝑉0

2ℏ2 𝑅2 = 𝑚

𝑎2 =

𝜎 𝑉0

𝑎2 𝑉0 = 𝜎 𝑉0 𝑎 − 2

𝑎 2

𝑥

𝑉1

−

𝑏 2

𝑏 2

𝑥

62

𝑅2

𝑚𝑎2 𝑉0 = 2ℏ2

63

64

65

Voltando ao poço

… 𝑐𝑜𝑠 𝑘𝑥

… sin 𝑘𝑥

… 𝑐𝑜𝑠 𝑘𝑥

66

67

68

𝑉 𝑥

𝐸

𝑉0

𝑎 − 2 ℏ 2 𝑑 2 ψ𝐼 𝑥 − + 𝑉0 ψ𝐼 𝑥 = 𝐸ψ𝐼 𝑥 2𝑚 𝑑𝑥 2 𝑑 2 ψ𝐼 𝑥 2𝑚 = − 𝐸 − 𝑉0 ψ𝐼 𝑥 𝑑𝑥 2 ℏ2 𝑑 2 ψ𝐼 𝑥 = −𝑞 2 ψ𝐼 𝑥 𝑑𝑥 2 ψ𝐼 𝑥 = 𝐶𝑒 𝑖𝑞𝑥 + 𝐷𝑒 −𝑖𝑞𝑥

𝑎 2 ℏ2 𝑑 2 ψ𝐼𝐼 𝑥 − = 𝐸ψ𝐼𝐼 𝑥 2𝑚 𝑑𝑥 2 𝑑 2 ψ𝐼𝐼 𝑥 2𝑚𝐸 = − ψ 𝑥 𝑑𝑥 2 ℏ2 𝐼𝐼

𝑑 2 ψ𝐼𝐼 𝑥 = −𝑘 2 ψ𝐼𝐼 𝑥 𝑑𝑥 2 ψ𝐼𝐼 𝑥 = 𝐴′𝑒 𝑖𝑘𝑥 + 𝐵′𝑒 −𝑖𝑘𝑥

𝑥

ℏ2 𝑑 2 ψ𝐼𝐼𝐼 𝑥 − + 𝑉0 ψ𝐼𝐼𝐼 𝑥 = 𝐸ψ𝐼𝐼 𝑥 2𝑚 𝑑𝑥 2 𝑑 2 ψ𝐼𝐼𝐼 𝑥 2𝑚 = − 𝐸 − 𝑉0 ψ𝐼𝐼𝐼 𝑥 𝑑𝑥 2 ℏ2 𝑑 2 ψ𝐼𝐼𝐼 𝑥 = −𝑞 2 ψ𝐼𝐼𝐼 𝑥 𝑑𝑥 2 ψ𝐼𝐼𝐼 𝑥 = 𝐹𝑒 𝑖𝑞𝑥 + 𝐺𝑒 −𝑖𝑞𝑥

ψ𝐼𝐼 𝑥 = 𝐴 cos 𝑘𝑥 + 𝐵 sin 𝑘𝑥

𝑞=

2𝑚 𝐸 − 𝑉0 ℏ2

𝑘=

2𝑚𝐸 ℏ2

𝑞=

2𝑚 𝐸 − 𝑉0 ℏ2

69

ψ𝐼 𝑥 = 𝐵 cos 𝑞𝑥 ψ𝐼𝐼 𝑥 = 𝐴 cos 𝑘𝑥 ψ𝐼𝐼𝐼 𝑥 = 𝐶 cos 𝑞𝑥 70

ψ𝐼

𝑎 𝑎 𝑥=− = ψ𝐼𝐼 𝑥 = − 2 2

𝑎 𝑎 = 𝐴 cos −𝑘 2 2 𝑎 𝑎 𝐵 cos 𝑞 = 𝐴 cos 𝑘 2 2 𝑎 cos 𝑘 2 𝐵=𝐴 𝑎 cos 𝑞 2

𝐵 cos −𝑞

ψ𝐼𝐼 𝑥 =

𝑎 𝑎 = ψ𝐼𝐼𝐼 𝑥 = 2 2

𝑎 𝑎 𝐴 cos 𝑘 = 𝐶 cos 𝑞 2 2 𝑎 cos 𝑘 2 =𝐶 𝐴 𝑎 cos 𝑞 2

𝜕ψ𝐼 𝑥 𝜕𝑥

𝑎 𝑥=− 2

−𝑞𝐵 sin −𝑞

𝜕ψ𝐼𝐼 𝑥 𝜕𝑥

=

𝜕ψ𝐼𝐼 𝑥 𝜕𝑥

𝑎 𝑥=− 2

𝑎 𝑎 = −𝑘𝐴 𝑠𝑖n −𝑘 2 2

𝑎 𝑥=2

=

𝜕ψ𝐼𝐼𝐼 𝑥 𝜕𝑥

𝑎 𝑥=2

𝑎 𝑎 −𝑘 𝐴 𝑠𝑖𝑛 𝑘 = −𝑞 𝐶 𝑠𝑖𝑛 𝑞 2 2 71

𝑎 𝑎 𝑎 2 −𝑞𝐴 𝑎 sin −𝑞 2 = −𝑘𝐴 𝑠𝑖n −𝑘 2 cos 𝑞 2 𝑎 𝑎 sin 𝑞 𝑠𝑖n 𝑘 2 =𝑘 2 𝑞 𝑎 𝑎 cos 𝑞 cos 𝑘 2 2 𝑎 𝑎 𝑞 tan 𝑞 = 𝑘 tan 𝑘 2 2 𝑎 𝑎 𝑎 𝑎 𝑞 tan 𝑞 = 𝑘 tan 𝑘 2 2 2 2 cos 𝑘

2𝑚 𝑎 𝐸 − 𝑉0 tan 2 ℏ 2 𝑚𝐸𝑎2 𝑚𝑉0 𝑎2 − tan 2ℏ2 2ℏ2

2𝑚 𝑎 𝐸 − 𝑉0 = 2 ℏ 2 𝑚𝐸𝑎2 𝑚𝑉0 𝑎2 − 2ℏ2 2ℏ2

𝜀 2 − 𝑅2 tan

2𝑚𝐸 𝑎 tan 2 ℏ 2 =

𝑚𝐸𝑎2 tan 2ℏ2

2𝑚𝐸 𝑎 ℏ2 2 𝑚𝐸𝑎2 2ℏ2

𝜀 2 − 𝑅2 = 𝜀 tan 𝜀

𝑦 tan 𝑦 = 𝜀 tan 𝜀

72

𝑔=

𝜀 2 − 𝑅2 tan

𝜀 2 − 𝑅2

𝑓 = 𝜀 tan 𝜀

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