Chapter 6 Problem Solutions

# J 2 x2 n 2 2p a2 2 ch6problemsetkeynb 3 f nf 1 2 m

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Unformatted text preview: m a2 —t 2 2 x2 2 F NF ma p Ba2 + J yHx, tL 2 m a2 —t 2 a2 + J = - N B1 + J The uncertainty in x is given by ‰ —t m a2 2 a2 2 B1 + —t m a2 2 F 2 NF Printed by Mathematica for Students 2 2 ‰- x ë 2 a è X p y\ = yH pL p Ba2 + J a2 + J 2 yHx, tL NF ma = —t 2 ma p Ba2 + J 2 x2 - N ‰ 2 2p a2 2 Ch6ProblemSetKey.nb 3 F NF 1 = 2 m a2 —t 2 2 ma x2 - yHx, tL —t 2 a2 B 1 + ‰ B1 + J 2 a2 2 —t B1 + 2 m a2 F 2 —t NF m a2 The uncertainty in x is given by ` `2 `2 ` HD xL2 = ZIx - Xx\M ^ = Xx \ - Xx\2 so x2 - ` Xx\ = ¶ Ÿ -¶ „ x 1 2p a2 2 B1 + J 2 —t ma x‰ 2 a2 2 B1 + 1 2p a2 2 B1 + J 2 F =0 ¬༼ integrand is odd NF 2 x2 - `2 Xx \ = —t m a2 2 —t m a2 ¶ 2 Ÿ -¶ „ x x ‰ 2 a2 2 B1 + —t m a2 2 F NF Using the relation ¶ 1 Ÿ-¶ „ x x2 ‰ 2ps - x2 = s2 2 s2 2 gives `2 Xx \ = a2 B1 2 +J —t m a2 2 NF Therefore Dx = a 1+J 2 2 ht 2 ma N ü Part b We have ` `2 `2 ` HD px L2 = ZI px - X px \M ^ = X px \ - X px \2 Thus ` ¶ X px \ = Ÿ -¶ p y H p, tL `2 ¶ X px \ = Ÿ -¶ p2 yH p, tL 2 2 = = ¶ a — p ¶ a — Ÿ-¶ „ p p ‰ p - 2 Ÿ-¶ „ p p ‰ a2 p2 =0 —2 - a2 ¬༼ integrand...
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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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