{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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 π π ψ -∞ - -  = = = +   4 4 4 4 2 2 2 2 2 4 1 cos 1 cos 2 L L L L x x A dx A dx dx L L so = 2 A L . (b) ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 π π ψ π π π = = = + = + = + = 8 8 8 8 2 2 2 0 0 0 0 2 4 4 1 cos 1 cos 2 4 2 2 1 1 sin 0.409 8 4 4 2 L L L L x x P dx A dx dx L L L x L L L L L 6-3 (a) ( 29 π λ = × 10 2 sin sin 5 10 x A A x so ( 29 π λ - = × 10 1 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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
82 CHAPTER 6 QUANTUM MECHANICS IN ONE DIMENSION (c) - = = × 2 31 9.11 10 kg 2 p K m m ( 29 ( 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 ( 29 { } ( 29 ( 29 { } ϖ α ϖ α π - - - -∞ Ψ = = 2 2 , ikx i k t k i kx k t C x t a k e dk e dk , with ( 29 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 ( 29 ϖ = h 2 2 t k t k m , we group this term together with α 2 2 k by introducing β α = + h 2 2 2 i t m to get ( 29 ϖ α β β β - - = - - - 2 2 2 2 2 2 4 ix x ikx k t k k . Then, changing variables to β β = - 2 ix z k gives ( 29 β β α α β β π - - - -∞ Ψ = = 2 2 2 2 2 4 4 , x x z C C x t e e e . To interpret this result, we must recognize that β is complex and separate real and imaginary parts. Thus, β α α = + = + h h 2 2 2 2 2 4 2 2 i t t m m and the exponent for Ψ is ( 29 ( 29 ( 29 α α β β α - = = + + h h 2 2 2 2 2 2 2 2 2 2 2 imaginary terms 4 4 4 i t m t m x x x then ( 29 ( 29 ( 29 ( 29 { } { } α α α α - + Ψ = + h h 2 2 2 4 2 1 4 2 4 2 , x t m t m C x t e .
Background image of page 2
MODERN PHYSICS 83 We see that apart from a phase factor, ( 29 Ψ , x t is still a gaussian but with amplitude diminished by ( 29 ( 29 α α + h 1 4 2 4 2 t m and a width ( 29 α α = + h 1 2 2 2 2 t x t m where ( 29 α = ∆ 0 x is the initial width. 6-5 (a) Solving the Schrödinger equation for U with = 0 E gives ( 29 ψ ψ = h 2 2 2 2 d dx U m . If ψ - = 2 2 x L Ae then ( 29 ( 29 ψ - = - 2 2 2 3 2 2 4 1 4 6 x L d Ax AxL e dx L ,  = -   h 2 2 2 2 4 6 2 x U mL L . (b) ( 29 U x is a parabola centered at = 0 x with ( 29 - = < h 2 2 3 0 0 U mL : U x 3 2 2 h mL 6-6 ( 29 ( 29 ( 29 ( 29 ( 29 ψ ψ ψ ψ = + = - + = - - - - - = + h h 2 2 2 2 2 2 cos sin sin cos cos sin 2 2 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 ( 29 ( 29 ψ ψ - = - h 2 2 2 2 m E U x or ( 29 ( 29 ( 29 2 - - + = + h 2 2 cos sin cos sin mE k A kx B kx A kx B kx .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}