SchrodingerEquation

SchrodingerEquation - Schr odinger Equation Sudarshan...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Schr odinger Equation Sudarshan Balakrishnan September 1, 2011 Analytic Solution The time-dependent Schr odinger Equation in one dimension given by : h 2 2 m 2 ( ( x,t )) x 2 + U ( x,t ) ( x,t ) = ih ( x,t ) t (1) Here, we consider h , m both be equal to one and U ( x,t ) = 0. Now, for x 2 we have: i ( t,x ) t = 2 ( t,x ) x 2 , (0 ,x ) = f ( x ) (2) The initial conditions for ( x,t ) are the same as those observed for the linearly dispersive wave equation in [3]. ( t, 0) = ( t, 2 ) (3) x ( t, 0) = x ( t, 2 ) (4) We take as initial data the unit step function: f ( x ) = ( x ) = 1 , 0 < x < ; 0 , < x < 2 . (5) Note that the boundary conditions allow us to extend the initial data and solution to be 2 periodic functions in x . In order to construct the solution, we have to first have to express it as a time-dependent(complex) Fourier series: ( t,x ) X k =- a k ( t ) e ikx (6) 1 where, we use rather than = to indicate that the Fourier series is for- mal and, without additional assumptions or analysis, its convergence is not guaranteed. As we substitute (6) in (2), we get: e ikx da k dt = ik 2 a k ( t ) e ikx (7) As we equate the coefficients of the individual exponentials on both sides, we notice that a k ( t ) satisfies an elementary linear ordinary differential equation: da k dt = ik 2 a k ( t ) (8) Now, solving (8) using the separation of variables we get: a k ( t ) = a k (0) e ik 2 t (9) Re-substituting (9) in (6): ( t,x ) = X k =- a k (0) e ik 2 t + ikx (10) Observe that the solution (10) is 2 periodic in both t and x . Moreover, since the k-th summand is a function of , periodic waves of frequency k move with wave speed- k 2 ie., ( k ) =- k 2 , thereby justifying the dispersive nature of the system. As we solve for the initial value problem, we note the initial condition for the Fourier series given by: f ( x ) = X k =- b k e ikx (11) In order to compute the coefficients of this fourier series, we note that in { e ikx } is an orthonormal basis for the initial data.We will implement this property in our computation[1], for k,l N we have: Z 2 e ikx e- ilx dx = 2 , k = l ; , k 6 = l . (12) Multiplying (11) by e- ilx on both sides, we notice: f ( x ) e- ilx = { X k =- b k e ikx } e- ilx (13) 2 Integrating both sides from 0 to 2 we get: Z 2 f ( x ) e- ilx dx = Z 2 { X k =- b k e ikx } e- ilx dx. (14) Z 2 f ( x ) e- ilx dx = Z 2 b l e ikx- ilx dx. (15) b k = 1 2 Z 2 f ( x ) e- ikx dx. (16) Equating (0 ,x ) = f ( x ), we notice that a k (0) = b k are the Fourier coeffi- cients of the given initial data. Evaluating this integral:cients of the given initial data....
View Full Document

This note was uploaded on 11/19/2011 for the course MATH 101 taught by Professor Wormer during the Spring '08 term at UCSD.

Page1 / 10

SchrodingerEquation - Schr odinger Equation Sudarshan...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online