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5.80 Small-Molecule Spectroscopy and Dynamics
Fall 2008
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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 |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�
|2132 = √2 |2111� + √1 |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
√
2a
and �2132|2122� = 0 = √
�2111|2111� + √b �2120|2120� = √1 (2a + b 2)
3
6
16
�
� �
��
��
⇒ 2122 ��= −√1 �� 2111 + 23 |2120�
3
�
�
�
J− |2122� = (L− + S− ) −√1 |2121� + 23 |2120�
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
⎫
⇒ √2a + √4b + √c = 0 ⎪
⎪
⎪
a = √1
⎪
10
30
15
⎪
⎪
�10
⎪
⎪
⎪
3
a2 + b2 + c2 = 1 ⎬
b = 10
so
⎪
⎪
⎪
�
⎪
⎪
⎪
⎪
⎪
c = − 35
a
b
c
⎭
−√ + √ + √ =0 ⎪
2
6
3
�
� �
��
��
3
so L = 2, S = 1, J = 1, M J = 1 ��= − √1 �� 2101 + 10
|2110� − 35 |212 − 1�
10
�
then operating on both sides with J− = S− + L− we get |L = 2, S = 1, J = 1, M J = 0� =
�
�
4
3
|2100�
+
10
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+1)
then |L = 2, S = 1, J = 1, M J = 0� = (2−
2·2(2·2+1)
|L = 2, S = 1, ML = −1, MS = 1�
�
�
(2+0+1)(2+0)
+ (2−0)(2+0)
|L
=
2,
S
=
1,
M
=
0,
M
=
0�
+
L
S
2(2·2+1)
2·2(2·2+1) |L = 2, S = 1, ML = 1, MS = −1�
�
3
|L = 2, S = 1, J = 1, M J = 0� = 10
|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 + 12 [L+ S− + L− S+ ])|i = A �i|Lz Sz |i�
So
H11 = H22 = A
H33 = H66 = H77 = H88 = H99 = 0
H44 = H55 = −A
The operators L± S� connect states with ΔML = �1 and ΔMS = ±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:
|L = 1, S = 1, J = 2, M J = 0� = √1 |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� = 12 A 1120|J2 − L2 − S2 |1120 = 12 A[2(2 + 1) − 1(1 + 1) − 1(1 + 1)] = 12 A[2] = A
�1120| = √1 �11 − 11| + √2 �1100| + √1 �111 − 1|
6
6
6
⇒ �1120|AL · S|1120� = 16 �11 − 11|ALz Sz |11 − 11� + 23 �1100|ALz Sz |1100� + 61 �111 − 1|ALz Sz |111 − 1� +
�
�
�
�
�
�
��
��
��
��
��
��
2
1
2
1
2
1
�
�
�
�
�
�
1100
11
1100
11
−
11
AL
S
+
1100
AL
S
−
11
+
111
−
1
AL
S
+
−
+
+
−
+
−
6�
6
2
6
2
�
�
�
�� 2
��
2
1
1
1
2
2
2
2
1100 � 2 AL− S+ � 111 − 1 = A − 6 − 6 + 6 + 6 + 6 + 6 = A
�6
�
1120|HSO |1110 = 0
�1120| = √1 [�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
�1100| = √1 [�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� = A2 [1(1 + 1) − 1(1 + 1) − 1(1 + 1)] = −A
�1110| = −√1 [|11 − 11� − |111 − 1�]
2
SO |1110� = A [�11 − 11|HSO |11 − 11� + �111 − 1|HSO |111 − 1�] = A [−1 − 1] = −A
So
�1110
H
|
2
2
�
�
1110|HSO |1100 = 0
|1110� =
−
√1 [|11 − 11�
2
|1100� = √1 [|11 − 11�
3
�
�
So 1110|HSO |1100
+
=
− |111 − 1�]
− |1100� + |111 − 1�]
�
�
�
�
�
�
√ [ 11 − 11|HSO |11 − 11 − 111 − 1|HSO |111 − 1 − 11 − 11|HSO |1100 +
= −A
6
�
�
√ (−1 + 1 − 1 + 1) = 0
111 − 1|HSO |1100 ] = −A
6
�
�
A
SO
1100|H |1100 = 2 (0(0 + 1) − 1(1 + 1) − 1(1 + 1)) = −2A
|1100� = √1 [|11 − 11� − |1100� + |111 − 1�]
3
�
�
�
�
�
�
�
�
So 1100|HSO |1100
= 13 [ 11 − 11|HSO |11 − 11 + 111 − 1|HSO |111 − 1 − 1100|HSO |11 − 11 −
�
� �
� �
�
1100|HSO |111 − 1 − 11 − 11|HSO |1100 − 111 − 1|HSO |1100 ] = A3 [−1 − 1 − 1 − 1 − 1 − 1] = −2A
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
For Hydrogen Z = 1, mµe = 0.999456, � = 109737.42 cm−1
For Li2+ Z = 3, mµe = 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− .
ML \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+ )
−
(2 , 0+ ) (2+ , 0− ) (1+ , 1− )
(1+ , 0− ) (1− , 0+ ) (2+ , −1− ) (2− , −1+ )
+
(2 , −2− ) (2− , −2+ ) (1+ , −1− ) (1− , −1+ ) (0+ , 0− )
−1
—
−
(2 , 1− )
(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 Ze2 ⎞⎟⎟
� 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
⎡∞
⎤
⎢⎢⎢�
⎥⎥⎥
k
a
a
b
b
k
a
a
b
b
K(ab) = ⎢⎢⎢⎣ b (� m� , � m� )G (n � , n � )⎥⎥⎥⎦ δmas mbs
k=0
where bk = (ck )2
so
E(1G) = F ◦ +
E(3 F) = F ◦ −
E(3 P) = F ◦ +
E(1 D) = F ◦ −
E(1 S ) = F ◦ +
4 2
F +
49
8 2
F −
49
7 2
F −
49
3 2
F +
49
14 2
F +
49
8 2
Since Hund’s rules say that E(3 F) < E(3 P) ⇒ − 49
F −
9
4
441 F
1 4
F
441
9 4
F
441
84 4
F
441
36 4
F
441
126 4
F
441
<
7 2
49 F
−
84 4
441 F
∴
F2
F4
> 0.56
5. If an atom is in a (2p)2 3 P0 state, to which of the following states is an electric dipole transition allowed?
Explain in each case.
(a) 2p3d 3 D2
ANSWER: Forbidden since ΔJ = 2.
(b) 2s2p 3 P1
ANSWER: Allowed since ΔJ = 1, Δ� = −1 & ΔL = 0, ΔS = 0.
(c) 2s3s 3 S 1
ANSWER: Forbidden in the absence of configuration interaction since it is a 2–electron transition.
(d) 2s2p 1 P1
ANSWER: Forbidden since ΔS � 0.