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Methane Symmetry Operations - Symmetry Properties of Vibrational Functions

6.   Symmetry Properties of Vibrational Functions

Vibrational symmetry coordinates for CH4 can be written as

$$\begin{eqnarray*} S_1 &=& 4^{-1/2} (\delta r_1 + \delta r_2 + \delta r_3 + \delta r_4)\\ S_{2a}&=&12^{-1/2} (2\delta\alpha_{12} + 2\delta\alpha_{34} - \delta\alpha_{13} - \delta\alpha_{24} - \delta\alpha_{23} - \delta\alpha_{14})\\ S_{2b}&=& 4^{-1/2} (\delta\alpha_{13} + \delta\alpha_{24} - \delta\alpha_{23} - \delta\alpha_{14})\\ S_{3x}&=& 4^{-1/2} (+\delta r_1 - \delta r_2 + \delta r_3 - \delta r_4)\\ S_{3y}&=& 4^{-1/2} (-\delta r_1 + \delta r_2 + \delta r_3 - \delta r_4)\\ S_{3z}&=& 4^{-1/2} (+\delta r_1 + \delta r_2 - \delta r_3 - \delta r_4)\\ S_{4x}&=& 2^{-1/2} (\delta\alpha_{13} - \delta\alpha_{24})\\ S_{4y}&=& 2^{-1/2} (\delta\alpha_{23} - \delta\alpha_{14})\\ S_{4z}&=& 2^{-1/2} (\delta\alpha_{12} - \delta\alpha_{34}) ~ , \end{eqnarray*}$$

(eq. 20)

where δri indicates a change in the C−Hi bond length, and δαij indicates a change in the Hi−C−Hj angle. Unfortunately, the positions of signs and subscripts in (eq. 20) depend on how numbered hydrogens are associated with the corners of the cube in Fig. 1. Equations (20) are thus written differently in a discussion of the methane force field [26].

While an intuitive application of the permutation-inversion operations on the right side of (eq. 20) will lead to correct results, it is perhaps safer to recall that transformation properties of the displacement vectors di have already been fixed, so that transformation properties of δri and δαij should be determined by noting that they are expressed in terms of the displacement vectors d, i = 1, 2, 3, 4, and dC by the equations

$$\begin{eqnarray*} \delta r_i &=& (\mbox{$a$}_i/a) \cdot (\mbox{$d$}_i-\mbox{$d$}_{\rm C})\\ \delta\alpha_{ij} &=&- \, {\scriptstyle{1\over4}} \sqrt{2} \, \left[ \frac{(\mbox{$a$}_i/a) \cdot (\mbox{$d$}_i-\mbox{$d$}_{\rm C})}{a} + \frac{(\mbox{$a$}_j/a) \cdot (\mbox{$d$}_j-\mbox{$d$}_{\rm C})}{a} \right]^{\phantom I}\\ &~&- \, {\scriptstyle{3\over4}} \sqrt{2} \, \left[ \frac{(\mbox{$a$}_i/a) \cdot (\mbox{$d$}_j-\mbox{$d$}_{\rm C})}{a} + \frac{(\mbox{$a$}_j/a) \cdot (\mbox{$d$}_i-\mbox{$d$}_{\rm C})}{a}\right]^{\phantom I} ~ , \end{eqnarray*}$$

(eq. 21)

where the ai represent vectors from the equilibrium C position to the equilibrium Hi positions and a = | ai | . The ai do not change under the symmetry operations. Application of the substitutions indicated in (eq. 12) and (eq. 16) and use of the relationships in (eq. 13) and (eq. 17) between the equilibrium positions leads immediately to the vibrational symmetry species indicated in Table 8.

Table 8. Symmetry species of the vibrational coordinates of (eq. 20)
 

  S1 S2a S2b S3x S3y S3z S4x S4y S4z
Td A1 Ea Eb F2x F2y F2z F2x F2y F2z
D2d A1 A1 B1 Ex Ey B2 Ex Ey B2


The symmetry species of the lowest vibrational state of the ground electronic state of methane is, of course, A1. The symmetry species of vibrational levels obtained by exciting one quantum of a given vibration are the same as the species of the corresponding vibrational coordinate. The symmetry species of levels obtained by multiple excitation of a single vibration are given by the symmetrized products [Γυ] [27]. They can be obtained from Table X−l3 of Ref. [4] (after correcting a small error in the results for even υ), and are reproduced here in Table 9. The symmetry species Γ(υ1, υ2, υ3, υ4) of levels obtained by multiply exciting several vibrations can be obtained by first determining the species for multiple excitation of each individual vibration, and then determining all possible direct products, as given in Table 10, between those sets of species [4]. Symbolically,

$${\mit\Gamma}(v_1, v_2, v_3, v_4) = [A_1^{v_1}] \times [E^{v_2}] \times [F_2^{v_3}] \times [F_2^{v_4}] ~ , $$

(eq. 22)

where a slight simplification is achieved by noting that [A1υ1] ≡ A1.

Table 9. Symmetry species [Γυ] for multiple excitation of a single degenerate vibration in CH4

Species corresponding to the symmetrized products [F1υ] can be obtained from the second half of this table by exchanging the subscripts 1 and 2 when υ is odd.

  υ [Eυ]
6p p(A1 + A2 + 2E) + A1
6p + 1 p(A1 + A2 + 2E) + E
6p + 2 p(A1 + A2 + 2E) + A1 + E
6p + 3 p(A1 + A2 + 2E) + A1 + A2 + E
6p + 4 p(A1 + A2 + 2E) + A1 + 2E
6p + 5 p(A1 + A2 + 2E) + A1 + A2 + 2E
υ [F2υ]
12p (3p2 + p)  Γ + pΓ1 + A1 
12p + 1 (3p2 + p)  Γ − pΓ2 + F2 
12p + 2 (3p2 + 2p)  Γ + pΓ1 + A1 + E + F2 
12p + 3 (3p2 + 2p)  Γ − pΓ2 + A1 + F1 + 2F2
12p + 4 (3p2 + 3p)  Γ + pΓ1 + 2A1 + 2E + F1 + 2F2 
12p + 5 (3p2 + 3p)  Γ − pΓ2 + A1 + E + 2F1 + 4F2 
12p + 6 (3p2 + 4p)  Γ + pΓ1 + 3A1 + A2 + 3E + 2F1 + 4F2 
12p + 7 (3p2 + 4p)  Γ − pΓ2 + 2A1 + 2E + 4F1 + 6F2 
12p + 8 (3p2 + 5p)  Γ + pΓ1 + 4A1 + A2 + 5E + 4F1 + 6F2 
12p + 9 (3p2 + 5p)  Γ − pΓ2 + 3A1 + A2 + 3E + 6F1 + 9F2 
12p + 10 (3p2 + 6p)  Γ + pΓ1 + 5A1 + 2A2 + 7E + 6F1 + 9F2 
12p + 11  (3p2 + 6p)  Γ − pΓ2 + 4A1 + A2 + 5E + 9F1 + 12F2 

Γ = A1 + A2 + 2E + 3F1 + 3F2 
Γ1 = 2A1A2 + E − 3F1 
Γ2 = 2A2A1 + E − 3F2 


Table 10. Direct product table for the point group Td

Antisymmetrized products are enclosed in curly brackets { }
 
  A1 A2 E F1 F2
A1 A1 A2 E F1 F2
A2 A2 A1 E F2 F1
E E E A1 + {A2} + E F1 + F2 F1 + F2
F1 F1 F2 F1 + F2 A1 + E + {F1} + F2 A2 + E + F1 + F2
F2 F2 F1 F1 + F2 A2 + E + F1 + F2 A1 + E + {F1} + F2

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Created September 20, 2016, Updated June 2, 2021