# hw10 - 7 Boas Chapter 13 Section 7 Problem 18 1 8 In class...

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Homework 10 PHZ 3113 Due Friday, April 23, 2010 Chapters 12-13 1. Consider a single particle which is described in quantum mechanics by the Schrodinger equation - ~ 2 2 m 2 ψ ∂x 2 = i ~ ∂ψ ∂t Also, assume that ψ ( x,t ) has boundary conditions ψ ( x = 0 ,t ) = ψ ( x = l,t ) = 0. a) Using separation of variables, as described in class, show that the general solutions can be written ψ ( x,t ) = X n =1 a n sin nπx l e - iE n t/ ~ where E n = ~ 2 n 2 π 2 2 ml 2 . b) Now assume that we have an initial condition ψ ( x,t = 0) = δ ( x - x 0 ) where 0 < x 0 < l . Show that in this case, that for t > 0 the function ψ ( x,t ) is given by, ψ ( x,t ) = X n =1 ± 2 l ² sin nπx 0 l sin nπx l e - iE n t/ ~ Notice that the solution you found in part b) is actually the Green function G ( x,x 0 ; t, 0) for the particle in a box problem. 2. Boas, Chapter 13, Section 3, Problem 2 3. Boas, Chapter 13, Section 3, Problem 4 4. Boas, Chapter 13, Section 4, Problem 6 5. Boas, Chapter 13, Section 5, Problem 2 6. Boas, Chapter 13, Section 7, Problem 14

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Unformatted text preview: 7. Boas, Chapter 13, Section 7, Problem 18 1 8. In class, we found that in the case of diusion, if at t=0 we have u ( x,y,t = 0) = ( x-x ) ( y-y ), the subsequent distribution is given by the Green function G ( x,x ,y,y ; t ) = 1 4 2 t e-( x-x ) 2 +( y-y ) 2 4 2 t At t = 0, we place a point source in the rst quadrant at x = a and y = b (where a &amp;gt; 0 and b &amp;gt; 0). We also have boundary conditions u x = 0 for x = 0, and u = 0 for y = 0. Using the image method and the Green function given above, determine the distribution u ( x,y,t ) in the rst quadrant ( x &amp;gt; 0 and y &amp;gt; 0). Sketch the source and the images, using plus sign for sources and minus signs for any sinks. 2...
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hw10 - 7 Boas Chapter 13 Section 7 Problem 18 1 8 In class...

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