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http://ocw.mit.edu 5.80 SmallMolecule 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�
)
√
62132� = 22111� + 22120�
2
1
2132 = √ 2111� + √ 2120�
3
6
(L S J MS ) (L S ML MS )
2 1 2 2� = a2 1 1 1� + b2120� where a2 + b2 = 1
√
2
b
and �21322122� = 0 = √a �21112111� + √ �21202120� = √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� = a2101� + b2110� + c212 − 1�
�L = 2, S = 1, J = 1, M J = 1L = 2, S = 1, J = 2, M J = 1� = 0
and �L = 2, S = 1, J = 1, M J = 1L = 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 ClebschGordan 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� + 0L = 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 spinorbit 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: �iAL · Si� = iA(Lz Sz + 1 [L+ S− + L− S+ ])i = A �iLz 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:
�
�
�1120AL · S1120� = 1 A 1120J2 − 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
⇒ �1120AL · S1120� = 1 �11 − 11ALz Sz 11 − 11� + 2 �1100ALz Sz 1100� + 6 �111 − 1ALz 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
1120H
1
�1120 = √ [�11 − 11 + 2�1100 + �111 − 1] 6
−1
√ [�11 − 11 − �111 − 1]
2
−
so �1120HSO 1110� = √ 1 �11 − 11HSO 11
12
√2 �1100HSO 111 − 1� = 0
� 12
�
1120HSO 1100 = 0 �1110 = − 11� + √1 �111
12 − 1HSO 111 − 1� − √2 �1100HSO 11
12 − 11� + �1120 = 1
√ [�11 − 11 + 2�1100 + �111 − 1]
6
1
�1100 = √ [�11 − 11 − �1100 + �111 − 1]
3
so �1120HSO 1100� = √1 �11 − 11HSO 11 − 11� + √2 �1100HSO 1100� − √1 �111 − 1HSO 111 − 1�
18
18
18
√2 �1100HSO 11 − 11� + √2 �1100HSO 111 − 1� − √1 �11 − 11HSO 1100� − √1 �111 − 1HSO 1100�
18
18
18
18
A
√ (−1 − 1 + 2 + 2 − 1 − 1) = 0
18
�1110HSO 1110� = A [1(1 + 1) − 1(1 + 1) − 1(1 + 1)] = −A
2
√
�1110 = −1 [11 − 11� − 111 − 1�]
2
So �1110HSO 1110� = A [�11 − 11HSO 11 − 11� + �111 − 1HSO 111 − 1�] = A [−1 − 1] = −A
2
2
�
�
1110HSO 1100 = 0 1110� = −1
√ [11 − 11�
2
1
1100� = √ [11 − 11�
3
�
�
So 1110HSO 1100 +
= − 111 − 1�]
− 1100� + 111 − 1�]
�
�
�
�
�
�
√
= −A [ 11 − 11HSO 11 − 11 − 111 − 1HSO 111 − 1 − 11 − 11HSO 1100 + 6
�
�
√
111 − 1HSO 1100 ] = −A (−1 + 1 − 1 + 1) = 0
6
�
�
SO 1100 = A (0(0 + 1) − 1(1 + 1) − 1(1 + 1)) = −2 A
1100H
2
1
1100� = √ [11 − 11� − 1100� + 111 − 1�]
3
�
�
�
�
�
�
�
�
So 1100HSO 1100
= 1 [ 11 − 11HSO 11 − 11 + 111 − 1HSO 111 − 1 − 1100HSO 11 − 11 −
3
�
��
��
�
1100HSO 111 − 1 − 11 − 11HSO 1100 − 111 − 1HSO 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 spinorbitals associated with nd; how many Pauliallowed (nd)2 Slater determinants
can you form using two of these spinorbitals?
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 spinorbit 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 191193, and in Tinkham, pages 177178. 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 2electron matrix elements are of the form
��
��
��
�1�
� � (a, b) =
(a, b) � �
(a, bga, b� − �a, bgb, a�)
� rab �
a>b
a>b
since the number of electrons is 2 the sum is just one term:
= �a, bga, b� − �a, bgb, 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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