Quantum Mechanics I - Solutions for Homework 2 | PHY 851, Assignments of Quantum Mechanics

Download Quantum Mechanics I - Solutions for Homework 2 | PHY 851 and more Assignments Quantum Mechanics in PDF only on Docsity! SOLUTIONS for Homework #2 1. The given wave function can be normalized to the total probability equal to 1, ψ(x) = Ne−λ|x|. (1) To get ∫ ∞ −∞ dx |ψ(x)|2 = 2|N |2 ∫ ∞ 0 dx e−2λx = 1, (2) we choose N = √ λ. (3) This state corresponds to the average position 〈x〉 = 0 and the coordinate uncertainty 〈x2〉 = ∫ ∞ −∞ dx |ψ(x)|2x2 = 1 2λ2 . (4) The wave function in the momentum representation is given by the Fourier expansion, φ(k) = ∫ ∞ −∞ dxψ(x)e−ikx, (5) or, after simple calculation, φ(k) = √ λ ( 1 λ+ ik + 1 λ− ik ) = 2λ3/2 λ2 + k2 , (6) the Lorentzian in momentum space is the Fourier image of the exponential packet in coordinate space. The centroid of φ(k) is, obviously, at 〈k〉 = 0, while its variance can be defined as 〈k2〉 = ∫ ∞ −∞ dk 2π |φ(k)|2k2 = λ2. (7) Thus, we have ∆x = 1√ 2λ , ∆p = h̄∆k = h̄λ, (8) and therefore (∆x)(∆p) = h̄√ 2 > h̄ 2 , (9) in accordance with the uncertainty relation. 2. To find the lower bound we assume that the quantum spreading during given time τ is much greater than the original position uncertainty. Then ∆x ' √ h̄τ m . (10) 1 For τ ≈ 1010 years ≈ 3×1017 sec this gives ∆x ' 6×108 cm for an electron, ' 1.5× 107 cm for a proton, ' 3× 10−6 cm for an object of m = 1 g. For the Universe we can make an estimate starting with an equivalent number of protons 10−5 cm−3 which would give a critical density Ω = 1 of the Universe. This would give, in the volume of c× t = 1010 light years, total mass M ≈ 1082me, or the spreading distance by 41 order of magnitude smaller than for an electron... 3. Let us consider the components φ1,2(k) of the initial wave function in the momentum representation, ψ1,2(x, 0) = ∫ dk 2π φ1,2(k)eikx. (11) /We always define the Fourier transformation with the factor (2π)−d, where d is the dimension of space, in the momentum integral ∫ ddk./ If the energy of a plane wave with wave vector k is (k), for example h̄2k2/2m, the time evolution of these components is given by ψ1,2(x, t) = ∫ dk 2π φ1,2(k)eikx−(i/h̄)(k)t. (12) Therefore we can find the time dependence of the overlap, γ(t) = ∫ dx [∫ dk1 2π φ1(k1)eik1x−(i/h̄)(k1)t ]∗ [∫ dk2 2π φ2(k2)eik2x−(i/h̄)(k2)t ] . (13) First we notice that the coordinate integral cancels all interference terms except for those with coherent spatial phases, k1 = k2,∫ dx e−ik1x+ik2x = 2πδ(k1 − k2). (14) Now we can take into account this δ-function to integrate over one of the wave vectors and see that the energy dependent phases cancel, γ(t) = ∫ dk 2π φ∗1(k)φ2(k) = γ(0). (15) The overlap is constant in time. The reason for this, as can be seen from the derivation, is that only the mutually coherent parts of the packet in- terfere in infinite space, and they have the same energy dependent phases. 4. In the frame K ′ the wave function has a standard form being expressed through the coordinate x′ with respect to this frame, Ψ′(x′, t) = Aeik ′x′−iω′t, (16) 2