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Unformatted text preview: Schr odinger Equation Sudarshan Balakrishnan September 1, 2011 Analytic Solution The timedependent 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 timedependent(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) Resubstituting (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 kth 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....
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This note was uploaded on 11/19/2011 for the course MATH 101 taught by Professor Wormer during the Spring '08 term at UCSD.
 Spring '08
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