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 22 3 12 2 xyz n nn E m LLL = x LL , == 2 yz L . Let π = h 0 2 8 E mL . Then ( 29 = 222 0123 4 EEnnn . 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 ( ( π - - -- × = = ×= ×× h 2 34 2 18 0 2 31 10 3 6.626 10 Js 33 4.52 10 J 28.2 eV 28 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 10 2 6 2 56.4 eV 8 h EE mL 8-3 = 2 11 n (a) ππ    hh 22 2 11 2 2 En mL mL (b) 1 n 2 n 3 n 1 1 3 1 3 1 3-fold degenerate 3 1 1 (c) π ψ  =   113 3 sin sin sin x yz A L LL π ψ πππ ψ  =    =   131 311 3 sin sin sin 3 sin sin sin x A LLL xyz A L 8-4 (a) ( 29 (29 ( 29 ψ ψ ψ = 12 , x y xy . In the two-dimensional case, ( 29 ( ψ= sin sin A kx ky where π = 1 1 n k L and π = 2 2 n k L . (b) ( 29 π+ = h 2 2 2 2 nn E mL If we let π = h 0 2 E mL , then the energy levels are: 1 n 2 n 0 E E 1 1 ψ 11 2 5 2 ψ 12 doubly degenerate 2 1 5 2 ψ 21 4 ψ 22 8-5 (a) === 123 1 nnn and ( ( ( - - -- × = = = ×≈ ×× 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
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MODERN PHYSICS 123 (b) States 211, 121, 112 have the same energy and ( 29 ++ = =≈ 2222 111 2 211 2 3.08 MeV 8 h EE mL and states 221, 122, 212 have the energy ( = 2 2 22 111 2 221 3 4.63 MeV 8 h mL . (c) Both states are threefold degenerate. 8-6 There is no force on a free particle, so that (29 Ur is a constant which, for simplicity, we take to be zero. Substituting ( 29 (29 ( 29 (29 (29 ψψψφ Ψ= 123 , t xyz t r into Schrödinger’s equation with (29 = 0 gives ( 29  ∂∂ - +  ∂∂∂  h h 2 222 ,, 2 t it mt rr . Upon dividing through by ( 29 (29 (29 t we obtain ψ ψ ψ φ ψ ψ ψφ ′′  - + +=   hh 2 2 3 1 2 y z x m x y zt . 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 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 mx x E x E t iE t This is subject to the condition that ++= EEEE . The equation for ψ 1 can be rearranged as ψ ψ =- h 2 11 1 22 2 d mE x dx , whereupon it is evident the solutions are sinusoidal ψ αβ =+ 1 1111 sin cos x kx with = h 2 1 1 2 2 mE k . However, the mixing coefficients α 1 and β 1 are indeterminate from this analysis. Similarly, we find ( 29 ( 29 ( 29 ψ ψ 2 3 33 sin cos sin cos y ky z kz
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124 CHAPTER 8 QUANTUM MECHANICS IN THREE DIMENSIONS with = h 2 2 2 2 2 mE k and = h 2 3 3 2 2 mE k . The equation for φ can be integrated once to get (29 ϖ φγ - = it te with ϖ = h E and γ another indeterminate coefficient. Since the energy operator is [] = h Ei t
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This note was uploaded on 11/05/2009 for the course PHSX 313 taught by Professor Staff during the Fall '09 term at Kansas.

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

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