Research: Science & Education
Reducible Representations for Linear Molecules Ian J. McNaught School of Chemistry, University of Sydney, Sydney, NSW 2006, Australia A number of methods have been published in the literature for reducing a representation in an infinite point group. Most methods are based on the formula χ (R) = Σ ai χi (R)
(1)
where χ (R) is the character in the reducible representation corresponding to the operation R, χi (R) the corresponding character in the irreducible representation i, and ai the required coefficient of the irreducible representation. Schäfer and Cyvin (1) give a method for determining the coefficients in eq 1; however they claim that it is necessary to guess which of the coefficients are nonzero. Strommen and Lippincott (2) use a finite subgroup of the required group and then correlation tables to find the coefficients without using eq 1. However it is necessary to be careful in the choice of subgroup: not all subgroups of the given infinite group give the correct result. Alvariño (3) used eq 1 to derive a method based on exhaustive elimination of possible solutions. For the example treated below it would be necessary to generate the characters of hundreds of combinations before finally finding the correct set of coefficients. Lie (4) introduced fictitious point group character tables for Cnv and Dnh (n → ∞) to derive the coefficients by taking limits as n → ∞ in the standard reduction formula for finite groups. The required mathematical demands are more than most of our students could handle. Flurry (5) uses eq 1 but works in the three-dimensional rotation group. Although the technique is elegant it uses techniques not usually treated in a first course in group theory. Kettle (6) uses eq 1 to generate sufficient equations to solve for all the coefficients. When applied to the example treated below it yields three sets of two equations in two unknowns (for the degenerate irreducible representations) plus one set of four equations in four unknowns (for the nondegenerate irreducible representations). In the procedure introduced here the algebraic equations are put into a more tractable form before any attempt is made to solve them. In the first example, this produces one set of four equations in one unknown (for the degenerate irreducible representations) plus two sets of two equations in two unknowns (for the nondegenerate irreducible representations). These equations all contain coefficients of ±1 so they can be solved very easily. Consider the point group D ∞h . When eq 1 is applied to the operations C∞φ, S∞φ, σv , and C2 the following equations are generated: χ (C∞φ) = a (Σg+ ) + a(Σg {) + a(Σu+ ) + a(Σu {) + 2 Σ [a(E ng) + a (Enu)] cos (nφ)
n≥1
χ (S∞φ) = a (Σg+ ) + a(Σg {) – a(Σu +) – a (Σu {) + 2Σ ({1) n [a (Eng) – a(Enu)] cos (nφ)
n≥1
χ(σv) = a(Σg
+)
–
a(Σg {) + a(Σu +)
–
a(Σu{ )
tions that are simple to solve. The first is to take the following linear combinations of the above equations: χ (C∞φ) + χ (S ∞φ ) = 2[a(Σg +) + a(Σg {)] + 4Σa (E2n - 1u) cos[(2n – 1) φ] + 4Σa(E 2ng) cos(2nφ)
(2)
χ (C∞φ) – χ (S ∞φ ) = 2[a(Σu +) + a(Σu {)] + 4Σa (E2n - 1g) cos[(2n – 1) φ] + 4Σa(E2nu) cos(2nφ)
(3)
χ (σv ) + χ (C2) = 2[a (Σg+ ) – a (Σg{ )]
(4)
χ (σv ) – χ (C 2) = 2[a (Σu+) – a (Σu{)] Σg±
(5) Σu±
from and It can be seen that this has separated Eng from Enu. The second step is to realize that term-by-term identification of the coefficients of the trigonometric functions immediately gives the number of each doubly degenerate irreducible representation in the reducible representation. The constant term in χ (C∞φ) + χ (S∞φ ) taken with χ (σv ) + χ (C2) gives a (Σg +) and a (Σg{), while the constant term in χ (C∞φ) – χ (S∞φ) taken with χ (σv) – χ (C2) gives a (Σu +) and a(Σu{ ). An example shows the ease with which the most complicated reducible representation can be reduced: D∞h E Γ
2C ∞φ
∞σv i
3 + 6 cos φ + 2 cos 2φ 13 {1 + 2 cos 3φ
2S ∞φ
-1 - 2 cos φ + 2 cos 2φ 5 - 2 cos 3φ
∞C 2 3
From the characters of the operations in the reducible representation χ (C∞φ) + χ (S∞φ) = 2 + 4 cosφ + 4cos 2φ
(6)
Application of eq 2 produces χ (C∞φ) + χ (S∞φ) = 2[a (Σg+) + a (Σg {)] + 4a (E1u) cosφ + 4a (E2g ) cos2φ
(7)
immediately giving a(E 1u ) ≡ a (Πu) = 1, a(E2g ) ≡ a (∆g) = 1. From eq 4, χ (σv) + χ (C2) = 2 = 2[ a(Σg+ ) – a (Σg {)], while equating the constant terms in eqs 6 and 7 gives 2 = 2[a (Σg+ ) + a(Σg {)], so a (Σg +) = 1, a(Σg {) = 0. χ (C∞φ) – χ (S ∞φ ) = 4 + 8cos φ + 4 cos 3 φ
(8)
Application of eq 3 produces χ (C∞φ ) – χ (S∞φ) = 2 [a(Σu +) + a(Σu {)] + 4 a (E 1g ) cos φ + 4a (E3g) cos 3φ
(9)
so that a (E1g ) ≡ a(Πg ) = 2, a(E 3g ) ≡ a (Φg) = 1. From eq 5, χ (σv) – χ(C2) = {4 = 2[a(Σu+) – a(Σu{)], while equating the constant terms in eqs 8 and 9 gives 4 = 2 [a (Σu+ ) + a(Σu {)], so a(Σu+ ) = 0, a(Σu{) = 2 ∴ Γ = Σg+ + 2Πg + ∆g + Φg + 2Σu{ + Πu
χ(C2) = a(Σg+ ) – a(Σg{) – a(Σu +) + a(Σu {) Two steps hold the key to producing algebraic equa-
If the point group is C∞v the process is even easier.
Vol. 74 No. 7 July 1997 • Journal of Chemical Education
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Research: Science & Education Consideration of χ (C∞φ) and χ (σv) leads immediately to two equations in two unknowns [a(Σ+) and a(Σ {)] plus n equations in one unknown, where n is the number of different doubly degenerate representations with nonzero coefficients. Consider the example: 2C ∞φ
∞ σv
3 + 6 cos φ + 2 cos 2 φ + 2 cos 3 φ
{1
C∞v E Γ
13
D∞h E Γ
7
2C ∞φ 1 + 2cosφ + 2cos2φ + 2cos3 φ
∞σv 1
2S ∞φ
i
{1 + 2cosφ {7 – 2 cos 2φ + 2 cos3 φ
∞C 2 {1
χ (C∞φ) + χ (S∞φ) = 4 cos φ + 4 cos 3 φ = 2 [a (Σg+ ) + a(Σg {)] + 4a(E1u) cosφ + 4a(E3u) cos3 φ ∴ a (Σg+ ) = 0, a (Σg{) = 0, a (E1u ) ≡ a (Πu) = 1, a (E3u) ≡ a (Φu) = 1
From eq 1 χ (C∞φ) = a(Σ+ ) + a (Σ {) + 2[a(Π) cosφ + a (∆) cos 2φ + a (Φ) cos 3φ]
(10)
χ(C∞φ) – χ(S∞φ) = 2 + 4 cos2φ = 2 [a(Σu+) + a (Σu{)] + 4a (E2u) cos 2φ ∴ a (Σu +) + a (Σu{) = 1, a(E2u ) ≡ a (∆u ) = 1
χ (σv) = a(Σ+ ) – a(Σ {)
(11)
χ (σv) – χ (C 2) = 2 = 2 [a(Σu+) – a(Σu{)] ∴ a (Σu +) – a (Σu{) = 1, a (Σu+) = 1, a (Σu{) = 0
leading immediately to a (Π) = 3, a (∆) = 1, a (Φ) = 1. Equating the constant term in eq 10 to the constant term in the χ (C∞φ) component of the reducible representation and applying eq 11 yields a (Σ+) + a(Σ {) = 3, a(Σ+ ) – a (Σ {) = {1, so that a(Σ+ ) = 1, a(Σ { ) = 2, and Γ = Σ+ + 2Σ { + 3Π + ∆ + Φ. Some more physically realistic applications are presented in the following. Determine the vibrational representation for dicyanoacetylene: N≡C–C≡C–C≡N. Application of standard methods (7) leads to the following representation: D∞h
E
2C ∞φ
Γvib 13 5 + 8 cos φ
∞σv
i
2S ∞φ
∞C 2
5
1
1
1
From the characters of the operations in the reducible representation, χ (C∞φ) + χ(S∞φ ) = 6 + 8 cos φ = 2 [a(Σg+) + a(Σg{)] + 4a(E1u) cosφ ∴ a (Σg+ ) + a (Σg {) = 3, a (E1u ) ≡ a (Πu) = 2 χ (σv) + χ (C2) = 6 = 2 [a(Σg+) – a(Σg{)] ∴ a (Σg +) – a (Σg{) = 3, a (Σg+) = 3, a (Σg{) = 0 χ (C∞φ) – χ (S∞φ) = 4 + 8 cos φ = 2 [a (Σu+) + a(Σu{)] + 4a(E1g ) cosφ ∴ a (Σu+ ) + a (Σu {) = 2, a (E1g) ≡ a (Πg) = 2 χ (σv) – χ (C 2) = 4 = 2 [a(Σu+) – a(Σu{)] ∴ a (Σu+ ) – a (Σu {) = 2, a (Σu +) = 2, a(Σu {) = 0 ∴ Γvib = 3Σg+ + 2Πg + 2Σu + + 2Πu The central atom in a linear X–Y–X molecule is a member of the lanthanide series. Write the following representation of its f orbitals (8) as a sum of irreducible representations:
810
∴ Γ = Σu + + Πu + ∆u + Φu Determine the spatial part of the electronic states arising from the electronic configuration πg 1πg1δg1 (9). The appropriate representation is: D∞h E Γ
2C ∞φ
8 2 + 4 cos 2φ + 2 cos 4φ
∞σv 0
i
2S ∞φ
8 2 + 4 cos 2φ + 2 cos 4φ
∞C 2 0
χ (C∞φ) + χ (S∞φ) = 4 + 8 cos2φ + 4 cos4φ = 2 [a (Σg+ ) + a(Σg {)] + 4a (E2g) cos 2φ + 4a (E4g ) cos4φ ∴ a (Σg+ ) + a (Σg{) = 2, a (E2g) ≡ a (∆g) = 2, a(E4g ) ≡ a(Γg ) = 1 χ (σv) + χ (C2) = 0 = 2 [a (Σg+ ) – a (Σg{)], so a (Σg+) = 1, a (Σg{) = 1 χ (C∞φ ) – χ (S∞φ ) = 0 = 2 [a (Σu+ ) + a (Σu{)] ∴ a (Σu+ ) = 0, a (Σu{) = 0 ∴ Γ = Σg + + Σg{ + 2∆g + Γg This technique requires no mathematical ability beyond the solution of sets of two equations in two unknowns and is much simpler than any technique so far suggested for reducing representations in infinite point groups. Literature Cited 1. 2. 3. 4. 5. 6.
Schäfer, L; Cyvin, S. J. J. Chem. Educ. 1971, 48, 295–296. Strommen, D. P; Lippincott, E. R. J. Chem. Educ. 1972, 49, 341–342. Alvariño, J. M. J. Chem. Educ. 1978, 55, 307–308. Lie, G. C. J. Chem. Educ. 1979, 56, 636–637. Flurry, R. L., Jr. J. Chem. Educ. 1979, 56, 638–640. Kettle, S. F. A. Symmetry and Structure; Wiley: Chichester, 1986; p 323. 7. McGinn, C. J. J. Chem. Educ. 1982, 59, 813. 8. Harris, D. C; Bertolucci, M. D. Symmetry and Spectroscopy; OUP: New York, 1978; p 477. 9. Herzberg, G. Molecular Spectra and Molecular Structure: I. Spectra of Diatomic Molecules, 2nd ed.; Van Nostrand: New York, 1950; p 335.
Journal of Chemical Education • Vol. 74 No. 7 July 1997
