Chapter 6 Problem Solutions

Printed by mathematica for students i np 2 m l

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Unformatted text preview: that Xy1 y\ 2 Xy2 y\ 2 1+Â 2 2 12 = =J 2 = N= 1 2 1 2 Therefore E1 + E2 2 = 1 4m I XE \ = 5 4m I ¬༼ En = p— 2 2 M I1 L p— 2 L 1 2m —2 p2 1 + 22 M M ü Part c For E= —2 p2 2 m L2 = 1 2m I p— 2 M L = E1 2 1 2 we have, as shown in (b) pHE1 L = 2px M L 0<x<L elsewhere ü Part b XE \ = sin I Xy1 y\ = so the probability that a measurement of energy on this state will yield 2 m L2 is 2 . Printed by Mathematica for Students I np— 2 M L Ch6ProblemSetKey.nb 7 ü Part d We know that ` `` Â ¶Ƞx = - — ZBH , xF^ + Z ¶Ƞt ^ ™ „ Xx\ „t =0 where `` `2 ` ``` ``` ` BH , xF µȡ B p , xF = p A p, xE + A p, xE p µȡ p ` µȡ 1ཽ ` µȡ 1ཽ Inside the well, the energy eigenstates are also momentum eigenstates (as V HxL = 0). These satisfy p2 E= 2m p= ¬༼ En = 2mE 1 2m I np— 2 M L np— L pn = Our state is in an equal superposition of such eigenstates, so the expectation value of momentum is given by ` X p\ = p— 2L 3p— 2L H1 + 2L = ¹Ȧ 0 Therefore „ Xx\ „t ¹Ȧ 0 so Xx\ is time-dependent. Problem 6.23 Given the square barrier potential 0 x<0 V0 0 § x § a 0 x...
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This note was uploaded on 02/01/2014 for the course PHYC 491/496 taught by Professor Akimasamiyake during the Fall '13 term at New Mexico.

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