Midterm 7 Solutions

Nb where h t y h0l 1 y h0l 1 2 h

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Unformatted text preview: Âwt 2 0\ +  ‰ So now we calculate Xx 0\ and Xx 1\, using - ¬༼ ‰ - ` ÂH t — 3Âwt 2 1\F Printed by Mathematica for Students 1 n\ = ‰ - w t Jn + N 2 n\ 3 YHx, tL = Xx yHtL\ = Xx ‰ 4 ` ÂH t - y H0L\ — MidtermExam7Solutions.nb where ‰ - ‰ - ` ÂH t — y H0L\ = 1 y H0L\ = 1 2 ` ÂH t — ‰ - B‰ 2 ` ÂH t B 0\ +  1\F — Âwt - 0\ +  ‰ 2 - ¬༼ ‰ - ` ÂH t — 1 n\ = ‰ - w t Jn + N 2 n\ 3Âwt 1\F 2 So now we calculate Xx 0\ and Xx 1\, using ` a 0] = 0 ` Yx a 0] = 0 ` Xx Ix + mw 2— ` ¬༼ a = ` px M 0\ = 0  mw „ „x mw — Xx 0\ = A ‰ - ` Ix +  mw ` px M ` Yx x 0] = x Xx 0\ ` „ Yx px 0] = - — „ x Xx 0\ ¬༼ Xx 0\ = - mw 2— x Xx 0\ m w x2 2— where A is imposed by normalization ¶ 2 Ÿ-¶ Xx 0\ A ¶ 2 Ÿ-¶ ‰ - ¶ p — 1ê 2 ImwM 2 ¬༼ Ÿ-¶ ‰- b x = „x = 1 — 2 A „x = 1 m w x2 pê b ; b = ¬༼ Choose A œ Rཻ, therefore =1 m w 1ê4 A = I p— M Thus m w 1ê4 X x 0\ = I p — M ‰ - m w x2 2— We calculate Xx 1\ by applying the raising operator `† a 0^ = ` Xx Ix - mw 2— mw 2— m w 1ê4 I p— M Ix -  mw —„ M‰ m w „x `† ¬༼ a = 1^ ` px M 0\ = Xx 1\ m w x2 = Xx 1\ 2— therefore...
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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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