Midterm 7 Solutions

# Nb statistics mean 574 median 6 stddev 185 problem

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Unformatted text preview: Ø2x Bp I m w 3 1ê4 MF — YHx, tL = 1 - YHx, tL = 1 4 x‰ - m w x2 = Xx 1\ 2— and we have 2 2 B‰ Âwt 2 m w 1ê4 I p— M m w 1ê 4 I p— M ‰ - Âwt 2 ‰ - m w x2 2— + Â‰ - B1 + Â ‰- Â w t I 3Âwt 2 4 Bp I 2 m w 1ê2 M — mw 3 MF — xF ‰ - 1ê4 x‰ - m w x2 2— F m w x2 2— Printed by Mathematica for Students mw 2— ` Ix - Â mw ` px M mw — MidtermExam7Solutions.nb ü Statistics mean = 5.74 median = 6 stddev = 1.85 Problem 2 ü Part a Let a\ = ⁄¶= 0 cn n\ n ` We desire that a\ be an eigenstate of a, i.e. that it satisfies ` a a] = a a] This gives ` ⁄¶= 0 cn a n] = a ⁄¶= 0 cn n] n n ⁄¶= 1 cn n n n - 1] = ⁄¶= 0 cn a n] n ⁄¶= 0 cn + 1 n n + 1 n^ = ⁄¶= 0 cn a n^ n Setting corresponding terms equal give the recurrence relation cn + 1 = cn = a n+1 an n! cn c0 where we determine c0 through normalization as Xa a\ = B⁄¶ = 0 m Ha* Lm m! *m c* Xm FB⁄¶= 0 n 0 n an n! Printed by Mathematica for Students c0 n\F 5 ⁄¶= 0 cn + 1 n n + 1 n^ = ⁄¶= 0 cn a n^ n Setting corresponding terms equal 6 MidtermExam7Solutions.nb a cn + 1 = cn n...
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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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