Hw208a - Physics 512 Homework Set#8 Solutions Winter 2003 1 This is based on Sakurai Chapter 7 Problem 9 Consider scattering by a repulsive-shell

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Physics 512 Winter 2003 Homework Set #8 – Solutions 1. This is based on Sakurai, Chapter 7, Problem 9. Consider scattering by a repulsive δ -shell potential 2 mV ( r ) ¯ h 2 = γδ ( r a ) > 0 a ) Set up an equation that determines the s -wave phase shift, δ 0 , as a function of k (where E h 2 k 2 / 2 m ). Because of the δ -function, we expect the Frst derivative of the wavefunction to jump at r = a . ±or this reason, we cannot use the standard logarithmic derivative matching conditions. However this problem is easy to solve for the ` =0 phase shift. ±or r 6 = a , all we have are free spherical waves. So we may write the exact solution for the radial wavefunction as R ( r )= ± j 0 ( kr ) ,r < a B [ h (2) 0 ( )+ e 2 0 h (1) 0 ( )] a (1) where B is a normalization constant, and δ 0 is the s -wave phase shift. Actually, for the δ -function jump, it is better to work with the function u ( r rR ( r ) , which satisFes the simple one-dimensional equation ² d 2 dr 2 ` ( ` +1) r 2 2 mV ¯ h 2 + k 2 ³ u ( r )=0 ±or ` and the δ -shell potential, this becomes simply u 0 + k 2 u = ( r a ) u (2) In terms of u ( r ) , the radial wavefunction (1) becomes u ( r ± k 1 sin( ) < a ik 1 B [ e ikr e 2 0 e ikr ] a ±or convenience, we may choose a di²erent normalization in order to eliminate the k 1 factor. Thus we write u ( r ± sin( ) < a B sin( + δ 0 ) a (where we have additionally absorbed a phase e 0 into B ). This clearly solves (2) for r 6 = a , and moreover retains the correct deFnition of the phase shift δ 0 . ±rom (2) , we now match the wavefunction and Frst derivative at r = a . The two conditions give B sin( ka + δ 0 ) = sin( ) kB cos( + δ 0 k cos( γ sin( ) 1
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Dividing the two eliminates B , and yields the result cot( ka + δ 0 ) = cot( )+ γ k Using the relation cot( α + β ) = (cot α cot β 1) / (cot α + cot β ) , and solving for cot δ 0 , we Fnd the equivalent condition cot δ 0 = cot( ) k γ csc 2 ( ) (3) b ) Assume that γ is very large γ ± 1 a ± k Show that if tan is not close to zero, the s -wave phase shift resembles the hard-sphere result, δ 0 = . Since tan is not close to zero, sin is not close to zero, and csc 2 ( ) is bounded. Thus for γ ± k , we may drop the last term in (3) and Fnd cot δ 0 cot( ) . This demonstrates that δ 0 ≈− , which is the hard sphere result. Note that this result does not depend on being small, (although it does depend on the condition ² γa being satisFed). A comparison of the exact result with the hard sphere approximation is shown by 5 10 15 20 -3 -2.5 -2 -1.5 -1 -0.5 δ 0 δ 0 ka = - ka (mod π ) exact Note, of course, that the phase shift is only deFned up to factors of . This shows that, for the most part, the spherical shell behaves just like a hard sphere. However part c ) below shows us that it is not just a hard sphere ... c ) Also show that for tan close to (but not exactly equal to) zero, resonance behavior is possible; that is, cot δ 0 goes through zero from the positive side as k increases (so that δ 0 is increasing as k is increasing). Determine approximately the positions of the resonances keeping terms of order 1 .
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This note was uploaded on 02/15/2011 for the course PHYS 512 taught by Professor Unknow during the Winter '03 term at Cornell College.

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Hw208a - Physics 512 Homework Set#8 Solutions Winter 2003 1 This is based on Sakurai Chapter 7 Problem 9 Consider scattering by a repulsive-shell

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