Chapter 6 Problem Solutions

# Therefore yhx t l i yhx t l 1 et m 1 2 i y1

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Unformatted text preview: = 2 Bn 1 2 L 2 L 2 Ÿ 0 „ x sin I Bn 2 Bn npx M L L 2npx ME L 2 Ÿ 0 „ x A1 - cosI ¬༼ sinH2 n pL = sinH0L = 0, so cosine Ÿ is 0 Setting equal to 1 and choosing Bn to be real gives Bn = 2êL thus 2 ê L sinI yn HxL = npx M L 0 0&lt;x&lt;L elsewhere and we have yHxL = I 1+Â M y1 HxL 2 + 1 2 y2 HxL Time evolution imparts a phase onto each of the energy eigenstates. Therefore Â yHx, t L = I yHx, t L = 1+Â - Et M‰ — 1 2 I y1 HxL + 1+Â -Â M expB 2 m — 2 0 I 1 p— 2 M L 2 tF Â ‰ - E2 t — 2 L y2 HxL sin I px M+ L ¬༼ En = 1 2 1 2m -Â I np— 2 M, L expB 2 m — I and substitute in y81, 2&lt; from above 2p— 2 M L Printed by Mathematica for Students tF 2 L sin I 2px M L 0&lt;x&lt;L elsewhere Time evolution imparts a phase onto each of the energy eigenstates. Therefore 6 Ch6ProblemSetKey.nb Â 1+Â - Et yHx, t L = I 2 M ‰ — 1 I yHx, t L = y1 HxL + 1+Â -Â M expB 2 m — 2 I 1 p— 2 M L 2 tF Â ‰ - E2 t — 2 L y2 HxL sin I px M+ L ¬༼ En = 1 2 1 2m I -Â np— 2 M, L expB 2 m — I and substitute in y81, 2&lt; from above 2p— 2 M L tF 2 L 0 We have XE\ = ⁄¶= 1 En pHEn L = ⁄¶= 1 En Xyn y\ n n 2 From our expression for yHxL in part (a), we see...
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