Question

20 A homogeneous, isotropic, nonpermeable dielectric is characterized by an index of refraction n(), which is in general complex in order to describe absorptive processes. 348 Chapter 7 Plane Electromagnetic Waves and Wave Propagation-SI (a) Show that the general solution for plane waves in one dimension written where u(r, t) is a component of E or B. (b) If u(x, 1) is real, show that n(-ω-n*(e). (e) Show that, if u(0, t) and au(O, t)/ax are the boundary values of u and its deri ative at x - 0, the coefficients A(u) and B(a) are A( 1 1 a() ar 0, )

Answer 1

A free particle is not subjected to any forces its potential energy is constant. Set .t, ince the origin of the potential energy may bechosen arbitrarily. The Schroedinger equation then becomes (g(k) can be complex: g(k)-lg(k)lexp(ia(k)); ak) changes the phase of the plane Proper wave functions of free particles are wave packets. Let us first concentrate on one-dimensional systems. Plane waves are possible solutions, as long as o ik/(2m). is a superposition of plane waves traveling in thex direction. At some instance of time, say t-0, We have (2n) using the de Broglie relations. gk) is theFourier transform of vix,0) A plane wave represents a particle whose probability of presence is constant throughout space; (Fourier series: see appendix l, Cohen-Tannoudiü Any piece-wise regular periodic function (finite # of discontinuities, finite # of extreme values) can be written as a series of imaginary exponentials. Assume f(x) is a periodic function of xwith fundamental period L; ly(at) l 뉘 A 12-constant. But a plane wave is not square-integrable, it is not a proper solution. The Schroedinger equation is a linear equation, the principle of superposition applies. A linear combination of plane wave solutions is also a solution. The coefficients C are given by will be a solution as long as for each k we have ak/(2m). Since k is a continuous variable, the most general solution is not a sum, butan integral; In the limit L → this can be generalized to (27e) Such a wave function is called a three-dimensional wave packet and can represent any non-pathological sqare-intearable function. where

In the integral Ik-kel If we write expil(k-ka)(x-xz),-exp(i2찌k-ka)/t) with τ=2π/(x-Xc) we find that for It l <<ak or l x-xol >>ak.T the function exp (i(k-kalx-xd) oscillates very rapidly in the integration interval and the integral yields approximately zero. fx) and (Ware Fourier transforms of each other Note: A Fourier transformis a linear transtornm For xXs" Ο τ-> and the function exp (12π(k-k) /t) l in the integration interval. Then theintegral yields its maximum value. We therefore have xo Consider a wave packet ta) is the position of the center of the wave packet. As Ix-xol increases x,0 decreases. We expect the width of the wave packet Axin coordinate space to be、 Fak. Then Arak-21. This is the classical relation for and assume lgrk) lis centered at ko where it reaches a maximum and has a width Δk. Assume Ake<k Assumeg(k)-lg(k)lexpik)). Sucha wave packet representsa particle with fairly well defined momentump (Ap<p) two functions which are Fourier transforms of each other. The product AxAk has a lower bound. The exact value of this bound dependson the precise definition of Ax and ak. In the neighborhood of k-ko we may expand a(k)in a Taylor series expansion; Assume we want to build a wave packet representing a particle which has a high probability of being observed at position r-0in a small time interval At about to 10,4) We are only interested n (k) in a small range of k about k. Keeping only zeroth and first order terms we may write Here (a) and yo,t) are Fourier transforms of each other ot) is centered at to with small width At, then ()has width do such that Ath 2 1. The Heisenberg uncertainty relations: and setting X0 this becomes Here lgik)I is nonzero y in a small interval about ko. Therefore we need to integrate only over this small interval of width Ak.