500 Quantum MechanicsP41.23Problem 43 in Chapter 16 helps students to understand how to draw conclusions from an identity(a)ψxAxLaf=−FHGIKJ122ddxAxL22ddxAL222Schrödinger’s equationddxmEU2222−=afbecomes−=−−FHGIKJ+−−−12122 222 2ALmEAxLmxA xLmL Lx===ejej−+−1224LmEmExLxL.This will be true for all xif both1LmE==andmELL=410both these conditions are satisfied for a particle of energyELm==22.(b)For normalization,1112222222244FHGIKJ+FHGIKJzzAxLdxAxLxLdxLLLL1235221615151623254+LNMOQP++−+LNMOQP=FHGIKJ=−AxxLxLAL LLLALALLL.(c)PdxLxLxLdxLxxLxLLLLLLLLL−+FHGIKJ+LNMOQP+LNMOQP−23322443332553315161215162301632811 215P47810580.P41.24(a)Setting the total energy Eequal to zero and rearranging the Schrödinger equation to isolatethe 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 .