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Massachusetts Institute of Technology Quantum Mechanics I (8.04) Spring 2006 Solutions to Problem Set 2 Kit Matan 1. Parseval’s theorem (10 points) Prove the following theorem: A function ψ(x) and its Fourier transform φ(k) have the same normalization, i.e. Z ∞ |ψ(x)|2 dx = −∞ Z ∞ |φ(k)|2 dk (1) −∞ § Z ∞ |ψ(x)|2 dx = −∞ Z ∞ ψ(x)ψ(x)dx −∞ Z ∞ ! ! 1 Z∞ 1 Z∞ 0 −ik0 x 0 ikx √ = dx √ φ(k )e dk φ(k)e dk −∞ 2π −∞ 2π −∞ Z ∞ Z ∞ Z ∞ 1 0 ix(k−k0 ) e dx dk 0 dk = φ(k )φ(k) 2π −∞ −∞ −∞ Z ∞ Z ∞ 1 = φ(k 0 )φ(k) (2πδ(k − k 0 )) dk 0 dk 2π −∞ −∞ Z ∞ = = φ(k)φ(k)dk −∞ Z ∞ |φ(k)|2 dk . (2) −∞ 1 4. The deBroglie wavelength of macroscopic objects. (10 points) What is the deBroglie wavelength of an automobile (2000 kg) traveling at 25 mile per hour? A dust particle of radius 1 µm and density 200 kg/m3 being jostled around by air molecules at room temperature (T = 300 K)? An 87 Rb atom that has been laser cooled to a temperature of T = 100 µK? Assume that the kinetic energy of the particle is given by 32 kB T . § The deBroglie wavelength, λdB , is equal to: λdB = h p (3) where h is Planck’s constant and p is a momentum. For the automobile, λdB = 6.626 × 10−34 J· s = 2.96 × 10−38 m . 25 miles/hour · 1609.344 meters/mile 2000 kg · 3600 s/hour (4) For the dust particle at room temperature. First calculate a momentum of the particle. =⇒ p = √ p2 2m = 32 kB T (5) 3 3mkB T = 3 · 200 kg/m3 43 π · (10−6 ) m3 · 1.38 × 10−23 J/K · 300 K 1/2 = 3.2 × 10−18 kg· m/s . (6) (7) Therefore, we have λdB = For the 87 6.626 × 10−34 J·s = 2.09 × 10−16 m . 3.2 × 10−18 kg·m/s (8) Rb atom. Its momentum at T = 100 µK is equal to, 86.909 × 10−3 kg/mole p = 3mkB T = 3 · · 1.38 × 10−23 J/K · 100 × 10−6 K 6.02 × 1023 atoms/mole = 2.44 × 10−26 kg·m/s . q !1/2 (9) (10) And, we have λdB = 6.626 × 10−34 J·s = 2.72 × 10−8 m . 2.44 × 10−26 kg·m/s (11) 5. Gaussian wavepacket in free space (30 points) A free particle of mass m has initial wave function ψ(x) = 1 − 1/2 (2π)1/4 w0 e x2 4w2 0 , where the initial width of the wavepacket w0 is a real, positive constant. 4 (12) (a) (15 points) Compute the Fourier transform of ψ(x, 0) and show that it is also a Guassian wavepacket: 2 − x2 1 4w φ(k) = e 0. (13) 1/2 (2π)1/4 k0 Determine the width k0 of the wavepacket in momentum space. § Calculate the Fourier transform of ψ(x, 0), ∞ 1 Z φ(k, 0) = √ dxψ(x, 0)e−ikx 2π −∞ ∞ 2 Z − x 2 −ikx 1 1 √ dxe 4w0 = 2 1/4 2π (2πw0 ) −∞ (14) Complete the square, x2 x + ikx = + ikw0 2 4w0 2w0 2 + k 2 w02 . (15) Substitute Eq. 15 into Eq. 14, and we obtain, 2 2 2 2 ∞ − e−k w0 Z 1 dxe φ(k, 0) = √ 2 1/4 2π (2πw0 ) −∞ x +ikw0 2w0 2 e−k w0 q 1 4πw02 = √ 2 1/4 (2πw ) 2π 0 = = where k0 = 4w02 2π !1/4 e−k 1 − 1/4 (2πk02 ) e 2 w2 0 k2 4k2 0 , (16) 1 . 2w0 (b) (15 points) § We have 1 ∆x∆k = w0 k0 = , (17) 2 which is the lower bound of the Heisenberg uncertainty relation ∆x∆k ≥ 1/2 (or equivalently ∆x∆p ≥ h̄/2). 5