problem40_13 - 40.13 a Eq(40.1 2 d 2 U E 2m dx2 2 d 2...

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40.13: a) dx ψ d m = + - 2 2 2 2 : Eq.(40.1) ) Left-hand side: kx A U kx A dx d m sin ) sin ( 2 0 2 2 2 + - kx A U kx A m k sin sin 2 0 2 2 + = . 2 0 2 2 ψ U m k + = But E U U m k + 0 0 2 2 2 ) for constant . k But 0 2 2 2 U m k + should equal E no solution. b) If , 0 U E then
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This note was uploaded on 04/17/2008 for the course PHYS 2306 taught by Professor Ykim during the Spring '06 term at Virginia Tech.

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