PC Chapter 06

PC Chapter 06 - 6 Quantum Mechanics in One Dimension 6-1...

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81 6 Quantum Mechanics in One Dimension 6-1 (a) Not acceptable – diverges as →∞ x . (b) Acceptable. (c) Acceptable. (d) Not acceptable – not a single-valued function. (e) Not acceptable – the wave is discontinuous (as is the slope). 6-2 (a) Normalization requires ππ ψ - --   = = =+     ∫∫ 44 2 22 2 24 1 cos 1 cos 2 LL xx A dx A dx dx so = 2 A L . (b) ( 29 ψ π = = = + = += 8 88 8 2 0 00 0 41 cos 1 cos 2 4 2 2 11 sin 0.409 8 4 42 L L P dx A dx dx L x L 6-3 (a) ( 29 π λ 10 2 sin sin 5 10 x A Ax so π λ - 1 01 2 5 10 m , π λ - = × 10 10 2 1.26 10 m 5 10 . (b) λ - - - × = = × 34 24 10 6.626 10 Js 5.26 10 kg m s 1.26 10 m h p
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82 CHAPTER 6 QUANTUM MECHANICS IN ONE DIMENSION (c) - = 2 31 9.11 10 kg 2 p Km m ( 29 ( - - - - - × = ×× × == × 2 24 17 31 17 19 5.26 10 kg m s 1.52 10 J 2 9.11 10 kg 1.52 10 J 95 eV 1.6 10 J eV K K 6-4 The time development of Ψ is given by Equation 6.8 or (29 { } ( 29 { } ϖα ϖ α π -- - -∞ Ψ ∫∫ 22 , ikx i k t k ikx kt C x t a k e dk e dk , with ϖ= h 2 2 k k m for a free particle of mass m . As in Example 6.3, the integral may be reduced to a recognizable form by completing the square in the exponent. Since ϖ  =   h 2 2 t k tk m , we group this term together with α k by introducing βα =+ h 2 it m to get ϖ αβ β β = - 2 2 2 2 4 i xx i k x k t kk . Then, changing variables to β β =- 2 ix zk gives ββ αα β βπ - -∞ Ψ 2 44 , z CC x te ee . To interpret this result, we must recognize that β is complex and separate real and imaginary parts. Thus, β = + hh 2 2 24 i tt mm and the exponent for Ψ is ( 29 α α β β α - =  +   h h 2 2 2 2 2 imaginary terms 4 4 4 m t m x then ( 29 ( { } { } α α -+ Ψ= + h h 2 42 14 2 4 2 , x tm t m C x .
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MODERN PHYSICS 83 We see that apart from a phase factor, (29 Ψ , xt is still a gaussian but with amplitude diminished by ( 29 ( 29 α α+ h 14 2 4 2 t m and a width α α  ∆=+   h 12 2 2 2 t m where α=∆ 0 x is the initial width. 6-5 (a) Solving the Schrödinger equation for U with = 0 E gives ψ ψ = h 2 2 2 2 d dx U m . If ψ - = 22 xL Ae then ( 29 ψ - =- 2 32 24 1 46 d Ax AxL e d , h 4 6 2 x U m LL . (b) Ux is a parabola centered at = 0 x with - =< h 2 2 3 00 U mL : U x 3 2 2 h mL 6-6 ( 29 ( 29 ( 29 ψ ψ ψ ψ =+ = -+ = -- - hh 2 2 cos sin sin cos cos sin cos sin x A kx B kx kA kx kB kx x k A kx k B kx x m mE E U A kx B kx The Schrödinger equation is satisfied if ψ ψ ∂- h 2 2 m EU x or ( 29 ( 29 2 - - + h 2 2 cos sin cos sin mE k A kx B kx A kx B kx . Therefore = h 2 k E m .
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84 CHAPTER 6 QUANTUM MECHANICS IN ONE DIMENSION 6-7 Since the particle is confined to the box, x can be no larger than L , the box length. With = 0 n , the particle energy = 22 2 8 n nh E mL is also zero. Since the energy is all kinetic, this implies = 2 0 x p . But = 0 x p is expected for a particle that spends equal time moving left as right, giving = -= 2 2 0 x xx p pp . Thus, for this case ∆ ∆= 0 x px , in violation of the uncertainty principle.
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PC Chapter 06 - 6 Quantum Mechanics in One Dimension 6-1...

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