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PC Chapter 08

# PC Chapter 08 - 8 Quantum Mechanics in Three Dimensions 8-1...

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121 8 Quantum Mechanics in Three Dimensions 8-1 π = + + h 2 2 2 2 2 3 1 2 2 x y z n n n E m L L L = x L L , = = 2 y z L L L . Let π = h 2 2 0 2 8 E mL . Then ( 29 = + + 2 2 2 0 1 2 3 4 E E n n n . Choose the quantum numbers as follows: 1 n 2 n 3 n 0 E E 1 1 1 6 ground state 1 2 1 9 * first two excited states 1 1 2 9 * 2 1 1 18 1 2 2 12 * next excited state 2 1 2 21 2 2 1 21 2 2 2 24 1 1 3 14 * next two excited states 1 3 1 14 * Therefore the first 6 states are ψ 111 , ψ 121 , ψ 112 , ψ 122 , ψ 113 , and ψ 131 with relative energies = 0 6 E E , 9, 9, 12, 14, 14. First and third excited states are doubly degenerate. 8-2 (a) = 1 1 n , = 2 1 n , = 3 1 n ( 29 ( 29( 29 π - - - - × = = = = × = × × h 2 34 2 2 2 18 0 2 2 2 31 10 3 6.626 10 Js 3 3 4.52 10 J 28.2 eV 2 8 8 9.11 10 kg 2 10 m h E mL mL (b) = 1 2 n , = 2 1 n , = 3 1 n or = 1 1 n , = 2 2 n , = 3 1 n or

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122 CHAPTER 8 QUANTUM MECHANICS IN THREE DIMENSIONS = 1 1 n , = 2 1 n , = 3 2 n = = = 2 1 0 2 6 2 56.4 eV 8 h E E mL 8-3 = 2 11 n (a) π π = = h h 2 2 2 2 2 2 2 11 2 2 E n mL mL (b) 1 n 2 n 3 n 1 1 3 1 3 1 3-fold degenerate 3 1 1 (c) π π π ψ = 113 3 sin sin sin x y z A L L L π π π ψ π π π ψ = = 131 311 3 sin sin sin 3 sin sin sin x y z A L L L x y z A L L L 8-4 (a) ( 29 ( 29 ( 29 ψ ψ ψ = 1 2 , x y x y . In the two-dimensional case, ( 29 ( 29 ψ = 1 2 sin sin A k x k y where π = 1 1 n k L and π = 2 2 n k L . (b) ( 29 π + = h 2 2 2 2 1 2 2 2 n n E mL If we let π = h 2 2 0 2 E mL , then the energy levels are: 1 n 2 n 0 E E 1 1 1 ψ 11 1 2 5 2 ψ 12 doubly degenerate 2 1 5 2 ψ 21 2 2 4 ψ 22 8-5 (a) = = = 1 2 3 1 n n n and ( 29 ( 29( 29 - - - - × = = = × × × 2 34 2 13 111 2 27 28 3 6.63 10 3 2.47 10 J 1.54 MeV 8 8 1.67 10 4 10 h E mL
MODERN PHYSICS 123 (b) States 211, 121, 112 have the same energy and ( 29 + + = = 2 2 2 2 111 2 2 1 1 2 3.08 MeV 8 h E E mL and states 221, 122, 212 have the energy ( 29 + + = = 2 2 2 2 111 2 2 2 1 3 4.63 MeV 8 h E E mL . (c) Both states are threefold degenerate. 8-6 There is no force on a free particle, so that ( 29 U r is a constant which, for simplicity, we take to be zero. Substituting ( 29 ( 29 ( 29 ( 29 ( 29 ψ ψ ψ φ Ψ = 1 2 3 , t x y z t r into Schrödinger’s equation with ( 29 = 0 U r gives ( 29 ( 29 - + + Ψ = Ψ h h 2 2 2 2 2 2 2 , , 2 t i t m t x y z r r . Upon dividing through by ( 29 ( 29 ( 29 ( 29 ψ ψ ψ φ 1 2 3 x y z t we obtain ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ψ ψ ψ φ ψ ψ ψ φ ′′ ′′ ′′ - + + = h h 2 2 3 1 1 2 3 2 y z x i t m x y z t . Each term in this equation is a function of one variable only. Since the variables x, y, z, t are all independent, each term, by itself, must be constant, an observation leads to the four separate equations ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ψ ψ ψ ψ ψ ψ φ φ ′′ - = ′′ - = ′′ - = = h h h h 2 1 1 1 2 2 2 2 2 3 3 3 2 2 2 x E m x x E m x x E m x t i E t

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PC Chapter 08 - 8 Quantum Mechanics in Three Dimensions 8-1...

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