1159_Physics ProblemsTechnical Physics

# 1159_Physics ProblemsTechnical Physics - 500 P41.23 Quantum...

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500 Quantum Mechanics P41.23 Problem 43 in Chapter 16 helps students to understand how to draw conclusions from an identity (a) ψ xA x L af =− F H G I K J 1 2 2 d dx Ax L 2 2 d dx A L 2 2 2 Schrödinger’s equation d dx m EU 2 22 2 = a f becomes −= F H G I K J + −− 1 2 1 2 2 2 22 2 A L m EA x L m xA xL mL L x == = ej e j + 1 2 2 4 L mE mEx L x L . This will be true for all x if both 1 L mE = = and mE LL = 4 1 0 both these conditions are satisfied for a particle of energy E Lm = = 2 2 . (b) For normalization, 11 1 2 2 2 2 2 2 2 2 4 4 F H G I K J + F H G I K J zz A x L dx A x L x L dx L L L L 1 2 35 2 2 16 15 15 16 2 3 2 5 4 + L N M O Q P + + + L N M O Q P = F H G I K J = Ax x L x L AL L L L A L A L L L . (c) Pd x L x L x L dx L x x L x L L L L L L L L L + F H G I K J + L N M O Q P + L N M O Q P 2 3 3 2 2 4 4 3 3 3 2 5 5 3 3 15 16 1 21 5 16 2 30 16 3 2 81 1 215 P 47 81 0580 . P41.24 (a) Setting the total energy E equal to zero and rearranging the Schrödinger equation to isolate the potential energy function gives
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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