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ps2ans_1976 - MIT OpenCourseWare http/ocw.mit.edu 5.80...

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MIT OpenCourseWare http://ocw.mit.edu 5.80 Small-Molecule Spectroscopy and Dynamics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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MASSACHUSETTS INSTITUTE OF TECHNOLOGY± Chemistry 5.76 ± Spring 1976 ± Problem Set #2 ANSWERS 1. (a) Construct the state | L = 2 , S = 1 , J = 1 , M J = 0 from the | LM L S M S basis using the ladder operator plus orthogonality technique. ANSWER: | L = 2 , S = 1 , J = 1 , M J = 0 We know that | L = 2 , S = 1 , J = 3 , M J = 3 = | L = 2 , S = 1 , M L = 2 , M S = 1 J | 2133 = ( L + S ) | 2121 6 | 2132 2 = 2 | 2111 + 1 2 | 2120 | 2132 = 6 | 2111 + 3 | 2120 (L S J M S ) (L S M L M S ) 2 1 2 2 = a 2 1 1 1 + b 2120 where a 2 + b 2 = 1 | | 2 | a b 1 and 2132 | 2122 = 0 = 6 2111 | 2111 + 3 2120 | 2120 = (2 a + b 2) 16 2122 = 1 3 2111 + 2 3 | 2120 J | 2122 = ( L + S ) 1 3 | 2121 + 3 2 | 2120 ⇒ | 2121 = 1 3 | 212 1 + 1 6 | 2110 � − 1 2 | 2101 Then J | L = 2 , S = 1 , J = 3 , M J = 2 = ( L + S ) 2 6 | 2111 + 1 3 | 2320 ⇒ | 2131 = 2 10 | 2101 + 4 30 | 2110 + 1 15 | 212 1 We know that | L = 2 , S = 1 , J = 1 , M S = 1 = a | 2101 + b | 2110 + c | 212 1 L = 2 , S = 1 , J = 1 , M J = 1 | L = 2 , S = 1 , J = 2 , M J = 1 = 0 and L = 2 , S = 1 , J = 1 , M J = 1 | L = 2 , S = 1 , J = 3 , M J = 1 = 0 2 a + 4 b + c = 0 1 10 30 15 a = 10 a 2 + b 2 + c 2 = 1 so b = 3 10 3 a 2 + b 6 + c 3 = 0 c = 5 1 3 3 so L = 2 , S = 1 , J = 1 , M J = 1 = 10 2101 + 10 | 2110 � − 5 | 212 1 then operating on both sides with J = S + L we get L = 2 , S = 1 , J = 1 , M J = 0 = 3 10 | | 21 11 � − 4 3 10 | 2100 + 10 | 211 1
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5.76 Problem Set #2 ANSWERS Spring, 1976 page 2 (b) Construct the states± L = 2 , S = 1 , J = 1 , M J = 0 3 D 1 ± | L = 2 , S = 2 , J = 1 , M J = 0 5 D 1 ± | L = 5 , S = 2 , J = 3 , M J = 1 5 H 3 ± | from the | L M L S M S basis using Clebsch-Gordan coe cients. The 3 D 1 function is the same as in Part (a) and is intended as a consistency check. ANSWER: 0 1 | | − � + = = 10 10 2 2 | | − � = 10 10 | L = 2 , S = 1 , J = 1 , M J = 0 Since S j 2 = 1 look in table 2 3 of Condon and Shortley, also J j = j 1 1 look in column 3 then | L = 2 , S = 1 , J = 1 , M J = 0 = (2 · 0)(2 · 0 + 1) | L = 2 , S = 1 , M L = 1 , M S = 1 2 2(2 2 + 1)± (2 0)(2 + 0) (2 + 0 + 1)(2 + 0)± + 2(2 · 2 + 1) | L = 2 , S = 1 , M L = 0 , M S = 0 + 2 · 2(2 · 2 + 1) | L = 2 , S = 1 , M L = 1 , M S = 1 10 + 2 L = 2 , S = 1 , M L = 0 , M S 3 L = 2 , S = 1 , M L = 1 , M S | L = 1 2 , S = 2 , J = 1 , M J = 0 = 10 | L = 2 , S = 2 , M L = 2 , M S = 2 + 10 | L = 2 , S = 2 , M L = 1 , M S = 1 + 0 | L = 2 , S = 2 , M L = 0 , M S = 0 3 L 2 S 1 J 1 M 0 L 2 S 1 M 1 M 1 | | J L S = = = = = = = = = , , , , , , 1 2 L 2 S 2 M 1 M 1 L 2 S 2 M 2 M − � + L S L S = = = = = = = , , , , , , | | 2 M 0 M L 5 S 2 M 1 M L 5 S 1 0 these are not correct −| � − | L S L S = = = = = = = = , , , , , , L 5 S 2 M 2 M 1 L 5 S 2 M 3 M 2 −| − � − | − � L S L S = = = = = = = = , , , , , , L = 5 , S = 2 , J = 3 , M S = 1 = L = 5 , S = 2 , M L = 1 , M S = 2
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5.76 Problem Set #2 ANSWERS
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