ps2ans_1976

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Unformatted text preview: 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. 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 |LML S MS � 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, ML = 2, MS = 1� ⇒ J− |2133� = (L− + S−√|2121� ) √ 6|2132� = 2|2111� + 2|2120� 2 1 |2132 = √ |2111� + √ |2120� 3 6 (L S J MS ) (L S ML MS ) |2 1 2 2� = a|2 1 1 1� + b|2120� where a2 + b2 = 1 √ 2 b and �2132|2122� = 0 = √a �2111|2111� + √ �2120|2120� = √1 (2a + b 2) 3 6 16 � �� � � � � √� ⇒ 2122 �= −1 � 2111 + 2 |2120� � 3 3 � � � √ J− |2122� = (L− + S− ) −1 |2121� + 2 |2120� 3 3 ⇒ |2121� = 1 √ |212 3 − 1� + 1 √ |2110� 6 − 1 √ |2101� 2 Then J− |L = 2, S = 1, J = 3, M J = 2� = (L− + S− ) � 2 √ |2111� 6 + � 1 √ |2320� 3 ⇒ |2131� = √2 |2101� + √4 |2110� + √1 |212 − 1� 10 30 15 We know that |L = 2, S = 1, J = 1, MS = 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 4 ⇒ √ a + √ b + √c = 0 ⎪ ⎪ ⎪ a = √1 ⎪ 10 30 15 ⎪ ⎪ 10 � ⎪ ⎪ ⎪ 3 a2 + b2 + c2 = 1 ⎬ b = 10 so ⎪ ⎪ ⎪ � ⎪ ⎪ ⎪ ⎪ ⎪ c =− 3 a b c 5 ⎭ −√ + √ + √ =0 ⎪ 2 6 3 � �� � � � � so L = 2, S = 1, J = 1, M J = 1 �= − √1 � 2101 + 130 |2110� − 3 |212 − 1� � � 5 10 � then operating on both sides with J− = S− + L− we get | = 2, S = 1, J = 1, M J = 0� = L � � 4 3 10 |2100� + 10 |211 − 1� � 3 10 |21 − 11� − 5.76 Problem Set #2 ANSWERS Spring, 1976 page 2 (b) Construct the states |L = 2, S = 1, J = 1, M J = 0� 3 D1 |L = 2, S = 2, J = 1, M J = 0� 5 D1 |L = 5, S = 2, J = 3, M J = 1� 5 H3 from the |L ML S MS � basis using Clebsch-Gordan coefficients. The 3 D1 function is the same as in Part (a) and is intended as a consistency check. ANSWER: |L = 2, S = 1, J = 1, M J = 0� Since S ≡ j2 = 1 look in table 23 of Condon and Shortley, also J ≡ j = j1 − 1 look in column 3 � −0)(2−0+ then |L = 2, S = 1, J = 1, M J = 0� = (22·2(2·2+1) 1) |L = 2, S = 1, ML = −1, MS = 1� � � + + + (2−0)(2+0) |L = 2, S = 1, ML = 0, MS = 0� + (22·0+1)(21)0) |L = 2, S = 1, ML = 1, MS = −1� 2(2·2+1) 2(2·2+ � |L = 2, S = 1, J = 1, M J = 0� = 130 |L = 2, S = 1, ML = −1, MS = 1� � � 2 3 + 10 |L = 2, S = 1, ML = 0, MS = 0� + 10 |L = 2, S = 1, ML = 1, MS = −1� |L = 2, S = 2, J = 1, M J = 0� = + − √1 | L 10 √1 | L 10 − √ 2 |L 10 = 2, S = 2, ML = −2, MS = 2� = 2, S = 2, ML = −1, MS = 1� + 0|L = 2, S = 2, ML = 0, MS = 0� = 2, S = 2, ML = 1, MS = −1� + √2 | L 10 = 2, S = 2, ML = 2, MS = −2� |L = 5, S = 2, J = 3, MS = 1� = |L = 5, S = 2, ML = −1, MS = 2� −|L = 5, S = 2, ML = 0, MS = 1� − |L = 5, S = 2, ML = 1, MS = 0� −|L = 5, S = 2, ML = 2, MS = −1� − |L = 5, S = 2, ML = 3, MS = −2� ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭ these are not correct 5.76 Problem Set #2 ANSWERS Spring, 1976 page 3 2. We know that the spin-orbit Hamiltonian, HSO = AL · S, is diagonal in the |L S J M J � basis but not in the |L ML S MS � basis. (a) Construct the full nine by nine HSO matrix in the |L = 1 ML S = 1 MS � basis. ANSWER: The nine basis functions are: |LS ML MS � |1� |11 − 1 − 1� |2� |1111� |3� |1100� |4� |111 − 1� |5� |11 − 11� |6� |110 − 1� |7� |1101� |8� |1110� |9� |11 − 10� � � The diagonal matrix elements: �i|AL · S|i� = i|A(Lz Sz + 1 [L+ S− + L− S+ ])|i = A �i|Lz Sz |i� 2 So H11 = H22 = A H33 = H66 = H77 = H88 = H99 = 0 H44 = H55 = −A The operators L± S� connect states with Δ ML = �1 and Δ MS = ±1. So H35 = H53 = A H34 = H43 = A H69 = H96 = A H78 + H87 = A All the rest are zero! (b) Construct the |L = 1, S = 1, J = 2, M J = 0� 3 P2 |L = 1, S = 1, J = 1, M J = 0� 3 P1 and |L = 1, S = 1, J = 0, M J = 0� 3 P0 functions in the |L ML S MS � basis. ANSWER: 1 |L = 1, S = 1, J = 2, M J = 0� = √ |L = 1, S = 1, ML = −1, MS = 1� 6 2 + √ |L = 1, S = 1, ML = 0, MS = 0� + 6 |L = 1, S = 1, J = 1, M J = 0� = − + 1 √ 2 − |L = 1, S = 1, ML = 1, MS = −1� |L = 1, S = 1, ML = −1, MS = 1� |L = 1, S = 1, ML = 1, MS = −1� |L = 1, S = 1, J = 0, M J = 0� = 1 √ 3 1 √ 2 1 √ 6 1 √ 3 |L = 1, S = 1, ML = −1, MS = 1� |L = 1, S = 1, ML = 0, MS = 0� + 1 √ 3 |L = 1, S = 1, ML = 1, MS = −1� 5.76 Problem Set #2 ANSWERS Spring, 1976 page 4 (c) Show that the matrix elements � � � � � � L = 1, S = 1, J = 2, M J = 0 �HSO � 1, 1, 2, 0 � � � � � � 1, 1, 2, 0 �HSO � 1, 1, 1, 0 � � � � � � 1, 1, 2, 0 �HSO � 1, 1, 0, 0 � � � � � � 1, 1, 1, 0 �HSO � 1, 1, 1, 0 � � � � � � 1, 1, 1, 0 �HSO � 1, 1, 0, 0 � � � � � � 1, 1, 0, 0 �HSO � 1, 1, 0, 0 � � expressed in terms of the |L ML S MS � basis in part (b) have the values expected from L·S = 1/2 J2 − L2 − S2 evaluated in the |L S J M J � basis. ANSWER: � � �1120|AL · S|1120� = 1 A 1120|J2 − L2 − S2 |1120 = 1 A[2(2 + 1) − 1(1 + 1) − 1(1 + 1)] = 1 A[2] = A 2 2 2 1 2 1 �1120| = √ �11 − 11| + √ �1100| + √ �111 − 1| 6 6 6 1 ⇒ �1120|AL · S|1120� = 1 �11 − 11|ALz Sz |11 − 11� + 2 �1100|ALz Sz |1100� + 6 �111 − 1|ALz Sz |111 − 1� + � �6 � � � � � � � �3 � � 2 2 2 �2 � �2 � �2 � 11 − 11 � 1 AL− S+ � 1100 + 1100 � 1 AL+ S− � 11 − 11 + 111 − 1 � 1 AL+ S− � 1100 + 6� 6 6 � � 1 AL S � 111 − 1� = A �− 1 − 1 + 2 + 2 + 2 + 2 � = A � 2 � − +� 6 1100 2 6 6 6 6 6 6 � � SO |1110 = 0 1120|H 1 �1120| = √ [�11 − 11| + 2�1100| + �111 − 1|] 6 −1 √ [�11 − 11| − �111 − 1|] 2 − so �1120|HSO |1110� = √ 1 �11 − 11|HSO |11 12 √2 �1100|HSO |111 − 1� = 0 � 12 � 1120|HSO |1100 = 0 �1110| = − 11� + √1 �111 12 − 1|HSO |111 − 1� − √2 �1100|HSO |11 12 − 11� + �1120| = 1 √ [�11 − 11| + 2�1100| + �111 − 1|] 6 1 �1100| = √ [�11 − 11| − �1100| + �111 − 1|] 3 so �1120|HSO |1100� = √1 �11 − 11|HSO |11 − 11� + √2 �1100|HSO |1100� − √1 �111 − 1|HSO |111 − 1� 18 18 18 √2 �1100|HSO |11 − 11� + √2 �1100|HSO |111 − 1� − √1 �11 − 11|HSO |1100� − √1 �111 − 1|HSO |1100� 18 18 18 18 A √ (−1 − 1 + 2 + 2 − 1 − 1) = 0 18 �1110|HSO |1110� = A [1(1 + 1) − 1(1 + 1) − 1(1 + 1)] = −A 2 √ �1110| = −1 [|11 − 11� − |111 − 1�] 2 So �1110|HSO |1110� = A [�11 − 11|HSO |11 − 11� + �111 − 1|HSO |111 − 1�] = A [−1 − 1] = −A 2 2 � � 1110|HSO |1100 = 0 |1110� = −1 √ [|11 − 11� 2 1 |1100� = √ [|11 − 11� 3 � � So 1110|HSO |1100 + = − |111 − 1�] − |1100� + |111 − 1�] � � � � � � √ = −A [ 11 − 11|HSO |11 − 11 − 111 − 1|HSO |111 − 1 − 11 − 11|HSO |1100 + 6 � � √ 111 − 1|HSO |1100 ] = −A (−1 + 1 − 1 + 1) = 0 6 � � SO |1100 = A (0(0 + 1) − 1(1 + 1) − 1(1 + 1)) = −2 A 1100|H 2 1 |1100� = √ [|11 − 11� − |1100� + |111 − 1�] 3 � � � � � � � � So 1100|HSO |1100 = 1 [ 11 − 11|HSO |11 − 11 + 111 − 1|HSO |111 − 1 − 1100|HSO |11 − 11 − 3 � �� �� � 1100|HSO |111 − 1 − 11 − 11|HSO |1100 − 111 − 1|HSO |1100 ] = A [−1 − 1 − 1 − 1 − 1 − 1] = −2A 3 5.76 Problem Set #2 ANSWERS Spring, 1976 page 5 3. Calculate the energies for the hydrogenic systems H and Li2+ in the following states: 2 2 P1/2 (means n = 2, s = 1/2, � = 1, j = 1/2) 2 2 P3/2 3 2 P1/2 3 2 P3/2 3 2 D3/2 3 2 D5/2 Please express “energies” in cm−1 : σ = E −1 hc cm and locate the zero of energy at n = ∞. ANSWER: ◦ En,�, s, j = En + Espin−orbit �� −�Z 2 µ ◦ En = (cm−1 ) me n2 � = Rydberg constant Z = nuclear charge µ me mN = me me (me + mN ) and � Espin−orbit = � 5.90 cm−1 Z 4 [ j( j + 1) − �(� + 1) − s( s + 1)] � � 2 n3 � + 1 (� + 1)� 2 µ For Hydrogen Z = 1, me = 0.999456, � = 109737.42 cm−1 µ For Li2+ Z = 3, me = 0.9999218. Term 22 P1/2 22 P3/2 32 P1/2 32 P3/2 32 D3/2 32 D5/2 ◦ En −27919.43 −27919.43 −12186.41 −12186.41 −12186.41 −12186.41 Hydrogen ES −0 E −.246 −27419.68 .123 −27419.31 −.073 −12186.48 .036 −12186.37 .022 −12186.39 .014 −12186.40 ◦ En −246889.89 −246889.89 −109728.89 −109728.89 −109728.89 −109728.89 Lithium ES 0 −19.91 9.96 −5.9 2.9 −1.77 1.18 E −109798.75 −109718.88 −109734.79 −109725.99 −109730.61 −109727.66 5.76 Problem Set #2 ANSWERS Spring, 1976 page 6 4. Consider the (nd)2 configuration. (a) There are 10 distinct spin-orbitals associated with nd; how many Pauli-allowed (nd)2 Slater determinants can you form using two of these spin-orbitals? ANSWER: (nd)2 available orbitals 2+ , 2− , 1+ , 1− , 0+ , 0− , −1+ , −1− , −2+ , −2− . M L \ MS 4 3 2 1 0 1 — (2+ , 1+ ) (2+ , 0+ ) (1+ , 0+ ) (2+ , −1+ ) (2+ , −2+ ) (1+ , −1+ ) 0 (2+ , 2− ) (2+ , 1− ) (2− , 1+ ) − , 0+ ) (2+ , 0− ) (1+ , 1− ) (2 + , 0− ) (1− , 0+ ) (2+ , −1− ) (2− , −1+ ) (1 (2+ , −2− ) (2− , −2+ ) (1+ , −1− ) (1− , −1+ ) (0+ , 0− ) −1 — − , 1− ) (2 (2− , 0− ) (1− , 0− ) (2− , 1− ) (2− , −2− ) (1− , −1− ) ⇒ there are 45 (nd)2 Slater Determinants. (b) What are the L − S states associated with the (nd)2 configuration? Does the sum of their degeneracies agree with the configurational degeneracy in part (a)? ANSWER: L − S states: 1 G, 3 F, 3 P, 1 D, 1 S. (c) What is the lowest energy triplet state (S = 1) predicted by Hund’s rules? Does Hund’s rule predict the lowest energy singlet state? ANSWER: Hund’s rules say 3 F will be lowest. Hund’s rules don’t really apply to other than the ground state so the lowest singlet state is not predicted. 5.76 Problem Set #2 ANSWERS Spring, 1976 page 7 (d) Calculate the energies of all states (neglecting spin-orbit splitting) which arise from (nd)2 in terms of the radial energy parameters F0 , F2 , and F4 . [This is a long and difficult problem. The similar (np)2 problem is worked out in detail in Condon and Shortley, pages 191-193, and in Tinkham, pages 177-178. The result for (nd)2 is also given, without explanation and in slightly different notation, Condon and Shortley, page 202.] What relationship between F2 and F4 is required by Hund’s rules? ANSWER: � � E (1G) = (2+ , 2− )|H|(2+ , 2− ) � � E (3 F ) = (2+ , 1+ )|H|(2+ , 1+ ) � �� � E (3 P) = (1+ , 0+ )|H|(1+ , 0+ ) + (2+ , −1+ )|H|(2+ , −1+ ) − E (3 F ) � �� �� � E (1 D) = (2− , 0+ )|H|(2− , 0+ ) + (2+ , 0− )|H|(2+ , 0− ) + (1+ , 1− )|H|(1+ , 1− ) − E (1G) − E (3 F ) � �� �� � E (1 S ) = (2+ , −2− )|H|(2+ , −2− ) + (2− , −2+ )|H|(2− , −2+ ) + (1+ , −1− )|H|(1+ , −1− ) � �� � + (1− , −1+ )|H|(1− , −1+ ) + (0+ , 0− )|H|(0+ , 0− ) − E (1G) − E (3 F ) − E (3 P) − E (1 D) � ⎛ p2 Z e2 ⎞ �1 ⎜i ⎟ ⎜ ⎟ ⎜ ⎟+ ⎜ ⎟ H= − ⎝ ⎠ 2m r r r> j i j �i������������������������������� ������ Equal for all terms since they come from same configuration we only need to evaluate these These 2-electron matrix elements are of the form �� �� �� �1� � � (a, b) = (a, b) � � (a, b|g|a, b� − �a, b|g|b, a�) � rab � a>b a>b since the number of electrons is 2 the sum is just one term: = �a, b|g|a, b� − �a, b|g|b, a� ≡ J (ab) − K (ab) 5.76 Problem Set #2 ANSWERS Spring, 1976 page 8 4D ANSWER, continued: where J (ab) = ∞ � ak (�a ma , �b mb )F k (na �a , nb �b ) � � k=0 ⎡∞ ⎤ ⎢� ⎥ ⎢ ⎢ ⎥ kaabb k aa bb⎥ ⎢ ⎥ K (ab) = ⎢ b (� m� , � m� )G (n � , n � )⎥ δma mb ⎢ ⎥ ss ⎣ ⎦ k=0 where bk = (ck )2 so E (1 G ) = F ◦ + E (3 F ) = F ◦ − E (3 P) = F ◦ + E (1 D) = F ◦ − E (1 S ) = F ◦ + 42 F+ 49 82 F− 49 72 F− 49 32 F+ 49 14 2 F+ 49 8 Since Hund’s rules say that E (3 F ) < E (3 P) ⇒ − 49 F 2 − 9 4 441 F 14 F 441 94 F 441 84 4 F 441 36 4 F 441 126 4 F 441 < 72 49 F − 84 4 441 F ∴ F2 F4 > 0.56 5. If an atom is in a (2 p)2 3 P0 state, to which of the following states is an electric dipole transition allowed? Explain in each case. (a) 2 p3d 3 D2 ANSWER: Forbidden since Δ J = 2. (b) 2 s2 p 3 P1 ANSWER: Allowed since Δ J = 1, Δ� = −1 & ΔL = 0, ΔS = 0. (c) 2 s3 s 3 S 1 ANSWER: Forbidden in the absence of configuration interaction since it is a 2–electron transition. (d) 2 s2 p 1 P1 ANSWER: Forbidden since ΔS � 0. ...
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