Chapter 3 Problem Solutions

5 sy 2 js s n s 0 2 b z basis 0 0 2 0 0 0 2 0 0 0 2 0

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Unformatted text preview: ee that ` 1, 1n \ = RHq j L 1, 1z \ ` ` 1 We derive the spin-1 representation of RHq j L in the Sz basis in a similar manner with the spin- 2 case (problem 3.5) ` ` Â` ` Sy = - 2 JS+ - S- N ØS 0 Â - 2 B— z basis 0 0 2 0 0 0 -— 2 0 0 0 2 0 0 F= 0 0 2 0 — 2 0 -Â 0 Â 0 -Â 0Â0 This gives `0 ` ` Sy = 1ཽ ØS ` ` Sy ØS z basis — z basis 2 `2 ` Sy ØS z basis `3 ` Sy ØS z basis 2 —2 2 —3 2 1 0 0 0 Â 0 1 0 -1 00 10 01 -Â 0 0 -Â Â0 0 -1 20 01 0 -2 Â 0 2 Â 0 -2 Â 0 2Â 0 ¬༼ be careful of this term ` = —2 S y ` `n So Sy does not square to the identity, but Sy as a function of n does have a period of 2 (up to normalization and excepting n = 0, which is the identity). Thus Âq ` ` 1 -S RHq j L = ‰ — y = ⁄¶= 0 n! H-Â qLn n ` 1 = 1ཽ + ⁄¶ = 1 H2 mL! H-Â qL2 m m `n Sy ¬༼ Split into identity term and the rest into n-even/odd — ` 2m Sy — 1 + ⁄¶ = 0 H2 m + 1L! H-Â qL2 m + 1 m = cos q - 1 ` 100 `2 Sy — — ` JS y ê—N ` 2 = JS y ê—N ` H-1Lm = 1ཽ + B⁄¶ = 1 H2 mL! q2 m F m ` 2m+1 Sy H-1Lm ` Sy Printed M - Â B⁄¶ = 0 by m athematica + 1 F tudents q2 m for S — m H2 + 1L! = sin q 1 0 -1 0 -Â 0 ` `n So Sy does not square to the identity, but Sy as a function of n does have a period of 2 (up to normalization and excepting n = 0, which is the identity). Thus Ch3ProblemSetKey.nb `n Sy Âq ` ` 1 -S RHq j L = ‰ — y = ⁄¶= 0 n! H-Â qLn n ¬༼ Split into identity term and the rest into n-even/odd — ` 2m Sy ` 1 = 1ཽ + ⁄¶ = 1 H2 mL! H-Â qL2 m m 1 + ⁄¶ = 0 H2 m + 1L! H-Â qL2 m + 1 m — ` 2m+1 Sy — ` JS y ê—N ` 2 = JS y ê—N ` H-1Lm = 1ཽ + B⁄¶ = 1 H2 mL! q2 m F m `2 Sy H-1Lm - Â B⁄¶ = 0 H2 m + 1L! q2 m + 1 F m — = cos q - 1 — = sin q 100 010 001 ` RHq j L Ø8 j, mz \< basis ` Sy + 1 2 1 0 -1 Hcos q - 1L 0 2 0 -1 0 1 Â - 2 0 -Â 0 sin q Â 0 -Â 0Â0 Therefore 1 2 ` RHq j L Ø8 j, mz \< basis 1 2 1 2 1 H1 + cos qL sin q 1 2 sin q 2 cos q 1 H1 - cos qL 2 H1 - cos qL - 1 2 sin q 1 2 sin q H1 + cos qL and 1 2 ` 1, 1n \ ØS H1 + cos qL 1 z basis 2 1 2 sin q H1 - cos qL 1 2 sin q 1 2 cos q 1 2 H1 - cos qL - sin q 1 2 1 2 sin q H1 + cos qL 1 0 0 1 2 1 = Printed by Mathematica for Students H1 + cos qL 2 1 2 sin q H1 - cos qL 7...
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