Theory of Projectors and Its Application to Molecular Symmetry

Abstract

Projector theory can serve as a powerful tool to perform the symmetric computation of molecular systems. The work of William Harter has long demonstrated the effectiveness of this theory in molecular spectroscopy; however, it seems its usefulness has not been realized by many in the field. We have described this methodology and have considered the D₃ symmetry system and the tetrahedral symmetry of methane molecules as concrete examples where the computed rotation tensors and vibrational wavefunction were derived for some symmetry states.

1. Introduction

In some sense, we have a closed-form solution of the Schrodinger equation for the hydrogen molecule, but, beyond this, there are no closed-form solutions for any other molecule. However, with the aid of numerical techniques, we can solve the molecular problem with a great degree of accuracy. As more and more larger molecules are considered, techniques tend to have limitations. Many clever approaches result in all sorts of approximation schemes, which work well for some systems but not for others. Paousek and Aliev [1] have shown that the general molecular Hamiltonian to is given by,

$$H_i={\displaystyle \sum _{j=1}^{N_e}\left(\frac{{\hslash }^2}{2m_e}{p}_j^2\right)+}{\displaystyle \sum _{i=1}^N\left(\frac{{\hslash }^2}{2m}{P}_i^2\right)}-V,$$

(1)

where $p_i\left({\dot{x}}_j,{\dot{y}}_j, {\dot{z}}_j\right)$ is the momentum of electrons and $P_i\left({\dot{x}}_i,{\dot{y}}_i, {\dot{z}}_i\right)$ is the momentum of the nuclei. It does not lend itself to a description of the stationary states of a polyatomic molecule. Although the Hamiltonian is of a simple form, the numerical integration of the Hamiltonian in terms of the space-fixed coordinates of the atomic nuclei and electrons is extremely difficult, even for the simplest of molecular systems [2,3,4].

A classical description of such a system of atomic nuclei and electrons sometimes reveals translation and rotation motion in space. The atomic nuclei can vibrate around the configuration given by the electronic structure of the molecule, and the electron also can move around the atomic nuclei. As a result, a quantum mechanical description of these motions should yield vibrational, rotational, and electronic energy levels and the corresponding wave function of the Schrodinger equation for this system. The task, however, of getting the information from a full Schrodinger wave equation is almost impossible, therefore, other techniques must be explored [1,2,3,4].

One approach is to develop a classical model of its translation, overall rotation, the vibration of the atomic nuclei, and the electronic motions. This is achieved by replacing the laboratory or the space-fixed coordinates of the atomic nuclei and electrons with new coordinate systems suitable for describing the individual types of motion mentioned above [3,4]. The Hamiltonian is developed in the new coordinates $x_j,y_j, z_j\left(j=1,2,\dots ,N_e\right)$, which are the coordinates of the electrons concerning a moving x,y,z axis system fixed to a molecule. The Euler angles $\theta ,\varphi ,\chi$ define the orientation of the x,y,z axis system with respect to the X, Y, Z axis system. The coordinates $Q_1,Q_2,\dots ,Q_{3N-6}$ are the vibrational displacement vectors of the system atomic nuclei for the x, y, z axis system [1,2,3,4,5]. Next, we make the Born–Oppenheimer approximation (BOA) to obtain a Hamiltonian for the electronic state and a Hamiltonian for vibrational–rotational states for given electronic states. Furthermore, Born–Oppenheimer-like approximations allow for the separation of the vibrational motions from the rotational motion [1,2,3,4,5,6,7,8].

The BOA allows the molecular Hamiltonian to be separated into the electronic part and its rovibrational Hamiltonian. The rovibrational Hamiltonian is given by

$$H_{rv}=\underset{H_r^0}{\underbrace{A{J}_x^2+B{J}_y^2+C{J}_z^2}}+\underset{H_v^0}{\underbrace{{\displaystyle \sum _k{P}_k^2}+\frac{1}{2}{\displaystyle \sum _{k=1}^{2N-6}{\lambda }_k{Q}_k^2}}}+V,$$

(2)

where

$$A=\frac{1}{2I_x}, B=\frac{1}{2I_y}, C=\frac{1}{2I_z},$$

(3)

and $H_r^0$ is the rotational Hamiltonian, and $H_v^0$ is a harmonic oscillator Hamiltonian. The wavefunction for the molecule is given by,

$$\mathsf{\Psi }={\psi }_e\left(r_{j\alpha }^0\right){\psi }_e^0\left(Q_1,Q_2,\dots \right){\psi }_r^0\left(\varphi \theta \chi \right).$$

(4)

Here, we can exploit the spatial symmetry of the molecule in the molecular system. A symmetry operation is a geometrical action that leaves the position of the atoms unchanged. There are five types of geometry operations: Identity (1), reflection (σ), rotation ($C_{\alpha }$), rotation–reflection ($S_{\alpha }$), and inversion ($i$). Each symmetry operation, excluding the identity, is associated with a symmetry element [1,2,3,4,8,9]. The symmetry operation is the actual action, while the symmetry element is the point, line, or plane about which the action occurs. Therefore, our goal in this paper is to illustrate how symmetry analysis can calculate genuine vibrations and classify the spectra according to the point group irreps. Thus, spectral decomposition is a powerful tool that will aid in determining the eigenvectors and the eigenvalues if the symmetry group is known. Harter [3,8] has detailed descriptions of the procedure for such a computation. We will outline how to utilize this method to perform calculations on a tetrahedral molecule. The key in this approach is to determine the projectors associated with the particular group symmetries.

2. Commuting Observable and Symmetry Projectors

Symmetry groups are classified into two categories: Abelian and Non-Abelian. The neat thing about Abelian groups is that we can write all the class operators as a combination of a single set of idempotent projectors. However, this is not the case for Non-Abelian groups. For the very simple fact that every group operator does not commute. All is not lost because we can find a set of mutually commuting operators. We describe how to do this by following the procedure outlined by Harter [3,8]. Moreover, similar approaches have been described in [4,7,9,10,11,12,13,14,15,16,17,18].

A.

Class algebra and all-commuting operator. We create the class structure of the group. This allows us to create communicative algebra. Each class sum commutes with each other and with every operator in the entire group algebra. In other words, the cg’s are mutually commuting with respect to themselves and all-commuting with respect to the whole group. ${\displaystyle \sum _{g=1}^{{}^{°}G}hg{h}^{-1}}=n_g{c}_g$. Harter [3,8] showed that an all-commuting operator C = C_g g that is, one that commutes with all h in the group, is a combination of class sums cg. All commutation (C h = h C) implies,

$$C={\displaystyle \sum _{g=1}^{{}^{°}G}{c}_{g}g}=\frac{1}{{}^{\circ }G}{\displaystyle \sum _{h=1}^{{}^{\circ }G}h\left({\displaystyle \sum _{g=1}^{{}^{\circ }G}{c}_{g}g}\right)}{h}^{-1}={\displaystyle \sum _{g=1}^{{}^{\circ }G}{c}_g{n}_g{c}_g}$$

(5)

B.

Characters and all-commuting projectors. We construct the projectors by solving the minimal equation of the group,

$${\mathbb{P}}^{\alpha }=\frac{{\displaystyle \prod _{\gamma \ne \alpha }\left(c_i-c_g^{\gamma }1\right)}}{{\displaystyle \prod _{\gamma \ne \alpha }\left(c_s^{\alpha }-c_s^{\gamma }\right)}}$$

(6)

where ${\mathbb{P}}^{\alpha }$ are called the All-Commuting Class Idempotent (ACCI), and $c_s^{\alpha }$ and $c_s^{\gamma }$ are eigenvalues from the minimal equation. We can expand the original class operators in terms of all commuting idempotent [3] i.e.,

$$c_j={\displaystyle \sum _{\alpha }{c}_j^{\alpha }{\mathbb{P}}^{\alpha }}$$

(7)

where $c_j^{\alpha }$ is the eigenvalue such that $c_j{\mathbb{P}}^{\alpha }=c_j^{\alpha }{\mathbb{P}}^{\alpha }$. By knowing the character table of the group, (5) can be obtained from,

$$P^{\alpha }={\displaystyle \sum _{m=1}^{{\ell }^{\alpha }}{P}_m^{\alpha }=\frac{{\ell }^{\alpha }}{{}^{\circ }G}}={\displaystyle \sum _{m=1}^{{\ell }^{\alpha }}{D}_{mm}^{\alpha }\left(g\right)g=\frac{{\ell }^{\alpha }}{{}^{\circ }G}{\displaystyle \sum _{m=1}^{{\ell }^{\alpha }}{\chi }^{\alpha }\left(g\right)g}}$$

(8)

Next, we acquire the inverse of (7), which must satisfy the completeness and spectral decomposition relation, i.e.,

$$1={\displaystyle \sum _{\alpha }{\mathbb{P}}^{\alpha }} \mathrm{and} g={\displaystyle \sum _{k=0}^{n=1}{D}^{k_{{}_n}}\left(g\right){\mathbb{P}}^{k_n}}$$

(9)

The way we performed the spectral decomposition will depend on whether the group is Abelian or Non-Abelian. For Abelian grouping, an element $g$ must satisfy a minimal equation ${\mathrm{g}}^n=1$ where there exists $n$ orthogonal idempotents $\left\{p^1,p^2,\dots , p^n\right\}$. We must choose another element $h$ of the group that would yield $\left\{q^1,q^2,\dots , q^n\right\}$. Both sets must satisfy the completeness and the spectral decomposition equation,

$${\displaystyle \sum _{j=1}^n{p}^j=1={\displaystyle \sum _{j=1}^m{q}^j}}$$

(10)

We then multiply (9) by $p^j$, which may result in the splitting of $p^j$ into a sum of operators $p^j{q}^k$,

$$p^j=p^1{q}^1+p^2{q}^2+\dots +p^j{q}^k=\dots +{\mathbb{P}}^2+{\mathbb{P}}^2+\dots +{\mathbb{P}}^r r\le m.$$

(11)

This set of nonzero terms $\left\{{\mathbb{P}}^1,{\mathbb{P}}^2,\dots , {\mathbb{P}}^n\right\}$ will also satisfy orthonormality and completeness relations. Therefore, the spectral decomposition of $g$ and $h$,

$$g={\displaystyle \sum _{j=1}{D}^j\left(g\right){\mathbb{P}}^j} \mathrm{and} h={\displaystyle \sum _{j=1}{D}^j\left(h\right){\mathbb{P}}^j}$$

(12)

C.

Maximal sets of commuting operators (MSOCO). These are the independent operators of the group that can be diagonalized at once. They are not unique since there are numerous sets of operators that compete to be part of a specialized set called an MSOCO (Maximal Set of Commuting Operators or Observables (MSOCO) [3,8]. The number of operators is the rank of the group. As an example, shown in [8], the rank of D₃ turns out to be four. Furthermore, the set diagram of D₃ in Figure 1 shows the class sum algebra of all-commuting operators at the center of the diagram.

The D₃ dimension determines the number of orthogonal irreducible representations (irreps) in the algebra (the irreps A₁, A₂, and E and the three all-commuting projectors $\left\{P^{A_1}, P^{A_2},P^E\right\}$ of the center are uniquely defined; no others exist). A rank of four implies another member of the MSOCO. The rank can be obtained by simply summing down the column of the character table (13). From the first column of the character table, we are able to tell how each irrep will split.

$$\begin{array}{cccc}{\mathrm{D}}_3& \mathrm{I}& \mathit{r},{\mathit{r}}^{\mathbf{2}}& {\mathit{i}}_{\mathbf{1}},{\mathit{i}}_{\mathbf{2}},{\mathit{i}}_{\mathbf{3}}\\ {\mathrm{A}}_1& 1& 1& 1\\ {\mathrm{A}}_2& 1& 1& -1\\ \mathrm{E}& 2& -1& 0\end{array}$$

(13)

The other member of the rank-4 MSOCO is not uniquely chosen; one choice is the operator $i_3$ picked in Figure 1 (other choices are shown in the figure). By choosing this operator to be diagonal, we are selecting a particular way to “split” the $P^E$ idempotent and build a particular set of E-irreps. Furthermore, the number of operators needed depends on the order of group. Thus, in the case of D₃ there are six projectors needed. In general, the number of operators needed is given by,

$${}^{\circ }G={\displaystyle \sum _{\alpha }{\left({\ell }^{\alpha }\right)}^2}$$

(14)

In the case of tetrahedral symmetry, the rank is 10 but only 5 irreps are possible: $A_1, A_2,E,T_1, \mathrm{and} T_2$ and total number operators is 24. However, $T_1$ is not vibrational mode and, thus, only 15 operators are necessary.

D.

Computing irreducible projectors. To compute the irreducible projectors of a non-communicating group we must consider the projectors of its subgroups. The key is finding the non-commutative spectral decomposition of the entire Non-Albelian group. Unlike Abelian, Non-Abelian will consist of nilpotent$({\mathbb{P}}^2=0)$ and idempotent $({\mathbb{P}}^2=\mathbb{P})$ projectors. Nilpotent projectors are important for expanding operators that do not commute. Each of the all-commuting idempotents ${\mathbb{P}}^{\alpha }$ can be split into ${\ell }^{\alpha }$ irreducible idempotents, i.e.,

$${\mathbb{P}}^{\alpha }=P_1^{\alpha }+P_2^{\alpha }+\cdots +P_{{\ell }^{\alpha }}^{\alpha }$$

(15)

$\mathrm{If} {\ell }^{\alpha }=1$, then ${\mathbb{P}}^{\alpha }$ remains un-split. We use the subgroup idempotent to split ${\mathbb{P}}^{\alpha }$. Suppose that $H\subseteq G$ where $\mu$and $\alpha$ are the irreps of $H$and $G$, respectively, such that $\left\{P_1^{\alpha },P_2^{\alpha },\dots ,P_{{\ell }^{\mu } }^{\mu }\right\}$ are projectors of $H$ and $\left\{{\mathbb{P}}^{{\alpha }_1},{\mathbb{P}}^{{\alpha }_2},\dots ,{\mathbb{P}}^{{\alpha }_n}\right\}$ are projectors of $G$, then idempotent of both $H$and $G$satisfy the completeness relation,

$$1={\displaystyle \sum _{k=1}^{{\ell }^{\alpha }}{P}_k^{\mu }} \mathrm{and} 1={\displaystyle \sum _{k=1}^n{P}_k^{\alpha }},$$

(16)

such that:

$$1\cdot 1=\left({\displaystyle \sum _{k=1}^{{\ell }^{\alpha }}{P}_k^{\mu }}\right)\left({\displaystyle \sum _{k=1}^n{P}_k^{\alpha }}\right)={\displaystyle \sum _{k=1}^n{P}_i^{{\alpha }_i}}$$

(17)

Hater [3] refers to this as one times one ($1\cdot 1)$ trick. Thus, a single projector such that ${\ell }^{\alpha }>1$ will split as,

$${\mathbb{P}}^{{\alpha }_{{}_1}}={\mathbb{P}}^{{\alpha }_{{}_1}}1={\mathbb{P}}^{{\alpha }_{{}_1}}{\displaystyle \sum _{k=1}^{{\ell }^{\mu }}{P}_k^{\mu }=P_1^{{\alpha }_1}+P_2^{{\alpha }_1}+\cdots +P_r^{{\alpha }_1}}$$

(18)

Every subgroup will give a different splitting. For octahedral [3], the all-commuting idempotent ${\mathbb{P}}^{T_1}$ splits when multiplied by six idempotents of $D_4$:

$${\mathbb{P}}^{T_1}={\mathbb{P}}^{T_1}1={\mathbb{P}}^{T_1}\left(P_1^E+P_2^E+P^{A_1}+P^{A_2}+P^{B_1}+P^{B_2}\right)=P_1^{T_1}+P_2^{T_1}+0+P_{3 }^{T_1}+0+0$$

(19)

Let’s consider D₃ symmetry a much simpler group with rank 4 and three irreps see (13). Of the three irreps only ${\mathbf{P}}^3= {\mathbf{P}}^E$splits. Since C₂ is a subgroup of D₃ we use it decompose ${\mathbf{P}}^E$ into the irreducible projectors $P_{xx}^E$ and $P_{yy}^E$ shown below,

$$\begin{gathered} {\mathbf{P}}^E{P}^x=P^x {\mathbf{P}}^E=\left(\mathbf{1}+i_3\right)/2 \left(21-r-r^2\right)/3=\left(21-r-r^2-i_1-i_2+2i_3\right)/6={{\mathbf{P}}^E}_{xx}, \\ {\mathbf{P}}^E{\mathbf{P}}^y={\mathbf{P}}^y {\mathbf{P}}^E=\left(\mathbf{1}-i_3\right)/2 \left(21-r-r^2\right)/3=\left(21-r-r^2+i_1+i_2-2i_3\right)/6={{\mathbf{P}}^E}_{yy} . \end{gathered}$$

(20)

Now using the ($1\cdot 1)$ trick with the completeness relation gives,

$$\mathbf{1}={\mathbf{1}}^{.}\mathbf{1}=( P^{A_1}+ P^{A_2}+ {\mathbf{P}}^E)\left(P^x+P^y \right)= P^{A_1}+ {\mathbf{P}}^E+{{\mathbf{P}}^E}_{xx}+{{\mathbf{P}}^E}_{yy}$$

(21)

Next, 1 is wrapped around any operator g of the D₃ algebra to give the following generalized Spectral Decomposition of the form shown in (22),

$$g={\mathbf{1}}^{.}{g}^{.}\mathbf{1}=\left(P^{A_1}+ P^{A_2}+{{\mathbf{P}}^E}_{xx}+{{\mathbf{P}}^E}_{yy} \right){}^{.}{g}^{.}\left( P^{A_1}+ P^{A_2}+{{\mathbf{P}}^E}_{xx}+{{\mathbf{P}}^E}_{yy}\right).$$

(22)

Thus, this result in (21) in which the $P^{A_1}$ and $P^{A_2}$ are unchanged because they are all-commuting,

$$\mathit{g}={\mathbf{1}}^{.}{\mathit{g}}^{.}\mathbf{1}= {\mathit{g}}^{.}{\mathit{P}}^{A_1}+{\mathit{g}}^{.}{\mathit{P}}^{A_2}+{{\mathit{P}}^E}_{xx} {}^{.}{\mathit{g}}^{.}{{\mathit{P}}^E}_{xx} +{{\mathit{P}}^E}_{xx} {}^{.}{\mathit{g}}^{.}{{\mathit{P}}^E}_{yy}+{P^E}_{yy} {}^{.}{g}^{.}{P^E}_{xx}+{P^E}_{yy} {}^{.}{g}^{.}{P^E}_{yy}$$

(23)

The remaining four terms are the E-projectors multiplied by irreps as seen by comparing (24)

$$\begin{gathered} {{\mathbf{P}}^E}_{xx} {}^{.}{g}^{.}{{\mathbf{P}}^E}_{xx}={D^E}_{xx}\left(g\right) {{\mathbf{P}}^E}_{xx} , {{\mathbf{P}}^E}_{xx} {}^{.}{g}^{.}{{\mathbf{P}}^E}_{yy}={D^E}_{xy}\left(g\right) {{\mathbf{P}}^E}_{xy}, \\ {{\mathbf{P}}^E}_{yy} {}^{.}{g}^{.}{{\mathbf{P}}^E}_{xx}={D^E}_{yx}\left(g\right) {{\mathbf{P}}^E}_{yx}, {{\mathbf{P}}^E}_{yy} {}^{.}{g}^{.}{{\mathbf{P}}^E}_{yy}={D^E}_{yy}\left(g\right) {{\mathbf{P}}^E}_{yy}. \end{gathered}$$

(24)

(23) is the form of a generalized non-commutative spectral decomposition of an entire Non-Abelian group. D₃’s decomposition differs from the commutative C₆. D₃ has two nilpotent projectors $P_{xy}^{\mathrm{E}}$ and $P_{yx}^{\mathrm{E}}$ along with four ordinary (idempotent) projectors ${\mathit{P}}^{A_1}$, ${\mathit{P}}^{A_2}$, $P_{xx}^{\mathrm{E}}$, and $P_{yy}^{\mathrm{E}}$. Other Non-Abelian groups have other numbers of these two kinds of P’s.

All commutative groups can be reduced to orthogonal idempotents that satisfy simple orthogonality relations given in [4],

$$P_i P_j= {\mathsf{\delta }}_{ij} P_i=P_j P_i.$$

(25)

A more general formulation spectral decomposition comes from writing (12) as,

$$g=1g1=\left({\displaystyle \sum _{k=1}^{{\ell }^{\alpha }}{P}_k^{\mu }}\right)g\left({\displaystyle \sum _{k=1}^n{P}_k^{\alpha }}\right)={\displaystyle \sum _{\mu }{\displaystyle \sum _m{\displaystyle \sum _n{D}_{mn}^{\mu }\left(g\right)P_{mn’}^n}}}$$

(26)

and (25) replaced by the generalized projector orthonormality relation for non-commutative projectors,

$$P_{jk}^{\mu }{P}_{mn}^{\nu }={\delta }^{\mu \nu }{\delta }_{mn}{P}_{jn}^{\mu }$$

(27)

To obtain generalized projectors, the decomposition (26) must be inverted. The result which will be proved below is the famous Wigner–Weyl projection formula,

$$P_{mn}^{\mu }=\frac{{\ell }^{\mu }}{{}^{\circ }G}{\displaystyle \sum _g{D}_{mn}^{\mu \ast }\left(g\right)}g.$$

(28)

where ^oG = 6 is the order of the group D₃ and ${\ell }^m$ is the m-irrep dimension, (${\ell }^{A_1}=1$, ${\ell }^{A_2}=1$, and ${\ell }^{E_1}=2$). Again, these numbers vary from group to group. For Abelian groups, irrep dimensions are always one (${\ell }^m=1$).

Let us summarize the first set of six irrep projectors which will be used in a band theory model below,

$$\begin{array}{|ccccccc|}\hline & P^{A_1}& P^{A_2}& P_{xx}^E=P_x^{E}1P_x^E& P_{yy}^E=P_y^{E}1P_y^E& P_{xy}^E=D_{xy}^E\left(i_3\right)P_{}^E=P_x^E{i}_3{P}_y^E& P_{yx}^E=D_{yx}^E\left(i_3\right)P_{yx}^E=P_y^E{i}_3{P}_x^E\\ 1& 1& 1& 2& 2& 0& 0\\ r& 1& 1& -1& -1& -1& 1\\ r^2& 1& 1& -1& -1& 1& -1\\ i_1& 1& -1& -1& 1& -1& -1\\ i_2& 1& -1& -1& 1& 1& 1\\ i_3& 1& -1& 2& -2& 0& 0\\ & 1/6& 1/6& 1/6& 1/6& 1/4& 1/4\\ \hline\end{array}$$

(29)

The element $i_2$ is outside of the CSOCO can be used to determine the two non-zero nilpotent projectors P^E_xy and P^E_yx since the idempotents P^E_xx and P^E_yy from (29) are orthogonal. A commuting operator would permit the two projectors to annihilate [8]. The key is to choose any operator such as $r$, $r^2$, $i_1$, or $i_2$ outside of the CSOCO, as shown in Figure 1. This will give all the needed projectors. At this stage, we use the projector to find eigenvalues and eigenvectors,

$$|e_{jk}^{\alpha }A\rangle =\frac{P_{jk}^{\alpha }|A\rangle }{\sqrt{N_k^{\alpha }}}$$

(30)

Suppose that we have the normalized eigenfunction $|e_j^{\alpha }\rangle =P_{jk}^{\alpha } |1\rangle /\sqrt{N_k^{\alpha }}$, then the vector product is $\rangle e_j^{\alpha }|e_j^{\alpha }\rangle =1$ gives,

$$N_k^{\alpha }=\langle 1\left|P_{jk}^{\alpha }{P}_{jk}^{\alpha }\right|1\rangle =\langle 1\left|P_{kk}^{\alpha }\right|1\rangle$$

(31)

Given this and the irrep P-operators in (29), the D₃-symmetry projected states are obtained easily,

$$\begin{array}{|ccccccc|}\hline & |P^{A_1}\rangle & |P^{A_2}\rangle & |P_{xx}^E\rangle & |P_{yy}^E\rangle & |P_{xy}^E\rangle & |P_{yx}^E\rangle \\ 1& 1& 1& 2& 2& 0& 0\\ r& 1& 1& -1& -1& -1& 1\\ r^2& 1& 1& -1& -1& 1& -1\\ i_1& 1& -1& -1& 1& -1& -1\\ i_2& 1& -1& -1& 1& 1& 1\\ i_3& 1& -1& 2& -2& 0& 0\\ & 1/\sqrt{6}& 1/\sqrt{6}& 1/2\sqrt{3}& 1/2\sqrt{3}& 1/2& 1/2\\ \hline\end{array}$$

(32)

The E-matrices ${\mathrm{D}}^{\mathrm{E}}\left(\mathrm{g}\right)=\left(\begin{array}{cc}{\mathrm{D}}_{11}^{\mathrm{E}}& {\mathrm{D}}_{12}^{\mathrm{E}}\\ {\mathrm{D}}_{21}^{\mathrm{E}}& {\mathrm{D}}_{22}^{\mathrm{E}}\end{array}\right)$ we can derive from the projector states for example $D^E\left(r\right)=\left(\begin{array}{cc}\frac{1}{2}& \frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right)$; This comes about by computing the operator $r$ acts on the column element $|P_{11}^E\rangle$ and $|P_{21}^E\rangle$ (i.e., $r|P_{11}^E\rangle$ and $r|P_{21}^E\rangle$),

$$\begin{gathered} r|P_{11}^E\rangle =r\left(2-r-r^2-i_1-i_2+2i_3\right)/2\sqrt{3}=\left(\begin{array}{c}-1\\ 2\\ -1\\ -1\\ 2\\ -1\end{array}\right)/2\sqrt{3} \\ =\frac{-\frac{1}{2}|P_{11}^E\rangle +\frac{\sqrt{3}}{2}|P_{21}^E\rangle }{2\sqrt{3}}, \end{gathered}$$

(33)

$$\begin{gathered} r|P_{21}^E\rangle =r\left(0+r-r^2-i_1+i_2+0i_3\right)/2\sqrt{3}=\left(\begin{array}{c}1\\ 0\\ -1\\ 1\\ 0\\ 1\end{array}\right)/2\sqrt{3} \\ =\frac{-\frac{\sqrt{3}}{2}|P_{11}^E\rangle +\frac{-1}{2}|P_{21}^E\rangle }{2\sqrt{3}}. \end{gathered}$$

(34)

We compute the following products, $D_{11}^E=\langle P_{11}^E\left|r\right|P_{11}^E\rangle$,$D_{21}^E=\langle P_{21}^E\left|r\right|P_{11}^E\rangle$, $D_{12}^E=\langle P_{12}^E\left|r\right|P_{21}^E\rangle$, and $D_{22}^E=\langle P_{12}^E\left|r\right|P_{21}^E\rangle$. In general,

$$\langle P_{ij}^{\alpha }\left|\mathrm{g}\right|P_{kl}^{\beta }\rangle =D_{ik}^{\alpha }\left(\mathrm{g}\right){\delta }^{\alpha \beta }{\delta }_{jl},$$

(35)

and

$$\mathrm{g}|P_{ij}^{\alpha }\rangle ={{\displaystyle \sum }}_{k=1}^{l^{\alpha }}{D}_{ki}^{\alpha }\left(\mathrm{g}\right)|P_{kj}^{\alpha }\rangle$$

(36)

Thus, from (33)–(36) we found all reduce regular representation for our $D_3$ example as shown by (37),

$$\begin{array}{ccccccc}g=& 1& r& r^2& i_1& i_2& i_3\\ D^{A_1}\left(g\right)=& 1& 1& 1& 1& 1& 1\\ D^{A_2}\left(g\right)=& 1& 1& 1& -1& -1& -1\\ D_{x_2 y_2}^E\left(g\right)=& \left(\begin{array}{cc}1& .\\ .& 1\end{array}\right)& \left(\begin{array}{cc}-1/2& -\sqrt{3}/2\\ \sqrt{3}/2& -1/2\end{array}\right)& \left(\begin{array}{cc}-1/2& \sqrt{3}/2\\ -\sqrt{3}/2& -1/2\end{array}\right)& \left(\begin{array}{cc}-1/2& -\sqrt{3}/2\\ -\sqrt{3}/2& 1/2\end{array}\right)& \left(\begin{array}{cc}-1/2& \sqrt{3}/2\\ \sqrt{3}/2& 1/2\end{array}\right)& \left(\begin{array}{cc}1& .\\ .& -1\end{array}\right)\end{array}$$

(37)

Each bra or ket is an operation of P-operator on the “first” state |1$\rangle$. Note the use of conjugation:$P_{m n}^{†}=P_{n m}$.

3. Rovibrational Hamiltonian and Wavefunctions

In (2), we write down the rovibrational Hamiltonian for a molecular system. This is synonymous with that described by Bunker and Jensen [4]. Bunker and Jensen [4] showed that invoking the harmonic oscillator approximation leads to the rigid-rotor harmonic oscillator Hamiltonian for a rotational–vibrational motion. Consequently, our Hamiltonian is given by,

$$H=\frac{1}{2}{\displaystyle \sum _{\alpha }{\mu }_{\alpha \alpha }^e{J}_{\alpha }^2+{\displaystyle \sum _{\alpha }{\displaystyle \sum _i\left[\frac{{\left(p_2^{\alpha }\right)}^{{}^2}}{2}+\frac{{\left({\omega }^{\alpha }{q}_{{}_2}^{\alpha }\right)}^2}{2}\right]}}},$$

(38)

where

$$\frac{\partial H}{\partial p_i^{\alpha }}={\dot{q}}_i^{\alpha }=p_i^{\alpha }$$

(39)

This is then treated as a quantum oscillator, as em-modes are quantized to give quantum field theory. The quantum theory of vibrational modes has the same traditional classical roots as quantum field theory. We will use canonical variables, which are defined by

$$q_j^{\alpha }=\langle e_j^{\alpha }|m|x\rangle , p_j^{\alpha }=\langle e_j^{\alpha }|m|\dot{x}\rangle$$

(40)

Next, we write the canonical variables in ordinary coordinates.

$$X_{lQ}=\langle lQ|x\rangle .$$

(41)

Using the completeness relation gives

$$\begin{array}{l}{X}_{lQ}=\langle lQ|x\rangle ={\displaystyle \sum _{\alpha }{\displaystyle \sum \langle lQ|e_j^{\alpha }\rangle \langle e_j^{\alpha }|m|x\rangle }}\\ ={\displaystyle \sum _{\alpha }{\displaystyle \sum _l\langle lQ|e_j^{\alpha }\rangle q_j^{\alpha }}}\end{array},$$

(42)

where

$$q_j^{\alpha }={\displaystyle \sum _{Q=A_1,E,T_2}{\displaystyle \sum _l\langle e_j^{\alpha }|m|lQ\rangle \langle lQ|x\rangle }}.$$

(43)

We can then write annihilation and creation operators in terms of symmetry coordinates since

$$a_i^{\alpha }=\left[\sqrt{{\omega }^{\alpha }}{q}_i^{\alpha }+\frac{i}{\sqrt{{\omega }^{\alpha }}}{p}_i^{\alpha }\right]/\sqrt{2\hslash },$$

(44)

$$a_i^{{\alpha }^{†}}=\left[\sqrt{{\omega }^{\alpha }}{q}_i^{\alpha }-\frac{i}{\sqrt{{\omega }^{\alpha }}}{p}_i^{\alpha }\right]/\sqrt{2\hslash }.$$

(45)

We can consider how to symmetrize the wavefunction for the rovibrational Hamiltonian. Dijon group [19,20] define a compact form that can be expressed as,

$$|{\mathsf{\Psi }}_{r\nu }^{\mathsf{\Gamma },\gamma }\rangle ={\left[|{\mathsf{\Psi }}_{rot}^{{\mathsf{\Gamma }}_1}\rangle \times |{\mathsf{\Psi }}_{vib}^{{\mathsf{\Gamma }}_2}\rangle \right]}_{\gamma }^{\mathsf{\Gamma }}.$$

(46)

We consider the form for the vibration and rotation as describe by ref. [9,10],

$$|{\mathsf{\Psi }}_{rot}^{{\mathsf{\Gamma }}_1}\rangle ={{\displaystyle \sum }}_{k=-J}^J{s}_{k;\mathsf{\Gamma },\gamma }^J|Jkm\rangle$$

(47)

$$|Jkm\rangle =D_{mk}^J\left(\alpha \beta \gamma \right)/\sqrt{\frac{8{\pi }^2}{2j+1}}$$

(48)

where

$$\begin{array}{l}{D}_{m k}^J\left(g\right)=\sqrt{\left(J-m\right)!\left(J+m\right)!}\sqrt{\left(J-k\right)!\left(J+k\right)!}\times \\ \frac{{\displaystyle \sum _n{\left(-1\right)}^{{}^n}{\left(\cos \frac{\beta }{2}\right)}^{2J+m-k-2n}{\left(sib\frac{\beta }{2}\right)}^{k-m+2n}{e}^{-i\left(m\alpha +k\gamma \right)}}}{\left(J+m-n\right)!\left(k-m+n\right)!n!\left(J-k-n\right)!}\end{array}$$

(49)

and $s_{k;\mathsf{\Gamma },\gamma }^J$ or similarly in spherical harmonics by Wormer [21]

$$|l, m, G, \lambda , i\rangle =\frac{f_{\lambda }}{\left|G\right|}{\displaystyle \sum _{g\in G}{D}^{\left(\lambda \right)}{\left(g^{-1}\right)}_{ij}g{Y}_{lm}}$$

(50)

A.

Rotational Wavefunction

To $|{\mathsf{\Psi }}_{rot}^{{\mathsf{\Gamma }}_1}\rangle ,$we must consider the overall external symmetry $O_3$. This then correlated with the internal symmetry ${\mathsf{\Gamma }}_1$. Knowing ${\mathsf{\Gamma }}_1$ gives a correlation induced between the irrep ${\mathsf{\Gamma }}_1$ and that of $O_3$. The characters of R₃ are given by,

$$Trace{D}^{\ell }\left(\omega 00\right)=\frac{\sin \left(\ell +\frac{1}{2}\right)\omega }{\sin \raisebox{1ex}{$\omega $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$$

(51)

We derive the character of $O_3$ from its outer product in relation to $R_3$ in (51).

$$\begin{array}{l}Trace D\left(\omega 00\right)=\\ \begin{array}{ccccccccccc}{O}_3& \begin{array}{l}1\\ \omega =0^{\circ }\end{array}& \begin{array}{l}r,r^2\\ \omega ={120}^{\circ }\end{array}& \begin{array}{l}{R}^2\\ \omega ={180}^{\circ }\end{array}& \begin{array}{l}R,R^3\\ \omega ={90}^{\circ }\end{array}& \begin{array}{l}i\\ \omega ={180}^{\circ }\end{array}& \begin{array}{l}I\\ \omega =-{180}^{\circ }\end{array}& \begin{array}{l}Ir,I{r}^2\\ \omega ={120}^{\circ }\end{array}& \begin{array}{l}I{R}^2\\ \omega ={180}^{\circ }\end{array}& \begin{array}{l}IR,I{R}^3\\ \omega ={90}^{\circ }\end{array}& \begin{array}{l}Ii\\ \omega ={180}^{\circ }\end{array}\\ 0^{+}& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1\\ 0^{-}& 1& 1& 1& 1& 1& -1& -1& -1& -1& -1\\ 1^{+}& 3& 0& -1& 1& -1& 3& 0& -1& 1& -1\\ 1^{-}& 3& 0& -1& 1& -1& -3& 0& 1& -1& 1\\ 2^{+}& 5& -1& 1& -1& 1& 5& -1& 1& -1& 1\\ 2^{-}& 5& -1& 1& -1& 1& -5& 1& -1& 1& -1\\ 3^{+}& 7& 1& -1& -1& -1& 7& 1& -1& -1& -1\\ 3^{-}& 7& 1& -1& -1& -1& -7& -1& 1& 1& 1\\ 4^{+}& 9& 0& 1& 1& 1& 9& 0& 1& 1& 1\\ 4^{-}& 9& 0& 1& 1& 1& -9& 0& -1& -1& -1\\ 5^{+}& 11& -1& -1& 1& -1& 11& -1& -1& 1& -1\\ 5^{-}& 11& -1& -1& 1& -1& -11& 1& -1& 1& 1\\ 6^{+}& 13& 1& 1& -1& 1& 13& 1& 1& -1& 1\end{array}\end{array}$$

(52)

The frequency $f^{{\mathsf{\Gamma }}_1}\left(J\right)$ of the irreducible representation of ${\mathsf{\Gamma }}_1$subduced to $R_3$ is given by:

$$f^{{\Gamma }_1}=\frac{1}{{}^{\circ }G}{\displaystyle \sum _{classes c_g}{\chi }_g^{{\Gamma }_1^{\ast }}{}^{\circ }{c}_{g}Trace D^{\ell }\left(g\right)}$$

(53)

where ${\mathsf{\Gamma }}_1$ is the irrep label of the point group of the composite rigid body. With ${\mathsf{\Gamma }}_1=T_d$ the Equation (53) and we can use character table (54) to find the induced representation,

$$\overline{)\begin{array}{cccccc}{T}_d& 1& \left(r_1\dots r_1^2\dots \right)& \left(R_1^2\dots \right)& \left(I{R}_1\dots I{R}_3^3\dots \right)& \left(I{i}_1\dots \right)\\ A_1& 1& 1& 1& 1& 1\\ A_2& 1& 1& 1& -1& -1\\ E& 2& -1& 2& 0& 0\\ T_1& 3& 0& -1& 1& -1\\ T_2& 3& 0& -1& -1& 1\end{array}}$$

(54)

Therefore, we give the correlation between $R_3$ and $T_d$ as (55),

$$\begin{array}{cccccc}\begin{array}{l}{T}_d\uparrow O_3\\ S_4\uparrow O_3\end{array}& \begin{array}{l}{A}_1\\ \left\{4\right\}\end{array}& \begin{array}{l}{A}_2\\ \left\{1^4\right\}\end{array}& \begin{array}{l}E\\ \left\{2^2\right\}\end{array}& \begin{array}{l}{T}_1\\ \left\{2,1^2\right\}\end{array}& \begin{array}{l}{T}_2\\ \left\{3,1\right\}\end{array}\\ J^p=0^{+}& 1& .& .& .& .\\ 0^{-}& .& 1& .& .& .\\ 1^{+}& .& .& .& 1& .\\ 1^{-}& .& .& .& .& 1\\ 2^{+}& .& .& 1& .& 1\\ 2^{-}& .& .& 1& 1& .\\ 3^{+}& .& 1& .& 1& 1\\ 3^{-}& 1& .& .& 1& 1\\ 4^{+}& 1& .& 1& 2& 1\\ 4^{-}& .& 1& 1& 1& 2\\ 5^{+}& .& .& 1& 2& 1\\ 5^{-}& .& .& 1& 1& 2\\ 6^{+}& 1& 1& 1& 1& 2\\ 6^{-}& 1& 1& 1& 2& 1\\ 7^{+}& 1& .& 1& 2& 2\\ 7^{-}& .& 1& 1& 2& 2\end{array}$$

(55)

We construct symmetrize rotational wave function as follows

$$|\begin{array}{c}{\mathsf{\Gamma }}_1\\ \gamma \end{array}\rangle =P_{\gamma f}^{{\mathsf{\Gamma }}_1}|\begin{array}{c}J\\ mk\end{array}\rangle /\sqrt{N^{{\mathsf{\Gamma }}_1}}=\frac{1}{\sqrt{N^{{\mathsf{\Gamma }}_1}}}{\displaystyle \sum _{R_l}{D}_{\gamma f}^{{\mathsf{\Gamma }}_1}\left(R_l\right)R_l}|\begin{array}{c}J\\ mk\end{array}\rangle$$

(56)

$$R_l|\begin{array}{c}J\\ m k\end{array}\rangle ={\displaystyle \sum _{R_l}{D}_{n J}^J\left(\alpha \beta \gamma \right)}|\begin{array}{c}J\\ n k\end{array}\rangle$$

(57)

We must compute how J split under a subduction $R_3\supset {\mathsf{\Gamma }}_1$. For instance, if ${\mathsf{\Gamma }}_1={\mathrm{T}}_d$, which is tetrahedral symmetry, and that we the splitting at $J=2$, then the split is $\left(E,T_2\right)$. We build the corresponding wavefunction $\left(|\begin{array}{c}2\\ -2\end{array}\rangle ,|\begin{array}{c}2\\ 1\end{array}\rangle ,|\begin{array}{c}2\\ 0\end{array}\rangle ,|\begin{array}{c}2\\ 1\end{array}\rangle ,|\begin{array}{c}2\\ 2\end{array}\rangle \right)$ and $\left(|\begin{array}{c}E\\ 1\end{array}\rangle ,|\begin{array}{c}E\\ 2\end{array}\rangle ,|\begin{array}{c}{T}_2\\ 1\end{array}\rangle ,|\begin{array}{c}{T}_2\\ 2\end{array}\rangle ,|\begin{array}{c}{T}_2\\ 3\end{array}\rangle \right)$ as outlined by Harter [3] for octahedral symmetry. From (56) we have,

$$|\begin{array}{c}E\\ e\end{array}\rangle =P_{ef}^E|\begin{array}{c}J\\ m k\end{array}\rangle /\sqrt{N^E},$$

(58)

$$|\begin{array}{c}{\mathrm{T}}_2\\ t\end{array}\rangle =P_{tu}^{{\mathsf{\Gamma }}_1}|\begin{array}{c}J\\ m k\end{array}\rangle /\sqrt{N^{{\mathrm{T}}_2}}.$$

(59)

The local symmetry of $C_2=\left\{1,R_3^2\right\}$ has a projector $P^{0_2}$,

$$P^{0_2}|\begin{array}{l}2\\ 2\end{array}\rangle =\frac{1}{2}\left(1+R_3^2\right)|\begin{array}{l}2\\ 2\end{array}\rangle$$

(60)

We need the tetragonal ($T_d\supset D_{2d}\supset C_2$) projectors $P_{ij}^{\alpha }$ of $T_d$ and only those for which $j=0_2.$ Thus, we find the induced representation $D^{0_2}\left(C_2\right)\uparrow T_d$ of $T_d$ by performing a correlation between the character tables of $C_v$ and $T_d$:

$$\overline{)\begin{array}{ccc}{C}_2& 1& R_3^2\\ 0_2& 1& 1\\ 1& 1& -1\end{array}}$$

(61)

Using columns 1 and Ii_I of $T_d$ table from (54) gives,

$$\begin{array}{ccc}{0}_2\left(C_2\right)\uparrow T_d& 0_2& 1_2\\ A_1& 1& .\\ A_2& .& 1\\ E& 1& 1\\ T_1& 1& 2\\ T_2& 2& 1\end{array}$$

(62)

$$\begin{array}{cccccc}{T}_d\uparrow D_{2d}& A_1& A_2& B_1& B_2& E\\ A_1& 1& .& .& .& .\\ A_2& .& .& 1& .& .\\ E& 1& .& 1& .& .\\ T_1& .& .& .& 1& 1\\ T_2& .& 1& .& .& 1\end{array}$$

(63)

It is clear from (62) that $0_2$ correlates with the tetrahedral component $E$ and $T_2$. Therefore, we have ${\mathcal{D}}^E\left(P^{0_2}\right)=\left(\begin{array}{cc}1& .\\ .& 1\end{array}\right)$ and ${\mathcal{D}}^{T_2}\left(P^{0_2}\right)=\left(\begin{array}{ccc}.& .& .\\ .& .& .\\ .& .& 1\end{array}\right)$. This tells us that we must use the third row of the tetragonal D-matrix for $T_2$ symmetry irrep, however, in the case of E symmetry irrep, it said both the first and third could be used. Therefore, consider $D_{2d}$ and repeat correlation with $T_d$ symmetry group, as shown by (63). We find that $A_1$ correlates with E and $A_2$ correlates with $T_2$. Both suggest that the last rows in $E$ and $T_2$ D-matrices are needed (57), depending on the D-function of $R_3$ and symmetry irreps coset leaders. Let’s choose the following coset leaders. We can write each coset leader in terms of its Euler angle rotation. Figure 2. shows the Euler machine designed and built by Harter to perform these operations [3,14]. There are sequenced pivot bearings connecting each frame to its neighbor. One of the three Euler angles $\left(\alpha ,\beta ,\gamma \right)$ is displayed on each bearing’s indicator and dial.

Harter [3] showed that the ordered rotation sequence of lab-based operations orients the body into position relative to the lab, i.e.,

$$R\left(\alpha 00\right)R\left(0\beta 0\right)R\left(00\gamma \right)=R\left(\alpha \beta \gamma \right).$$

(64)

There are steps we must perform to find $R\left(\alpha \beta \gamma \right)$. As shown in Figure 3, $\alpha$ is a rotation around the z-axis, $\beta$ around y-axis, and $\gamma$ rotation around z-axis. The first move is to zero the azimuth angle $R\left(-\varphi 00\right)$ with the z-crank, and, then, zero the polar angle $R\left(0-\theta 0\right)$ with the crank, finally the $\omega$ rotation is given by the z crank $R\left(\omega 00\right)$. To complete this operation, we return $\omega$ to its original position by doing the reserve operation $R\left(\varphi 00\right) R\left(0\theta 0\right)$. This allows us to create the expression between Euler rotations and Darboux rotations,

$$R\left(\varphi 00\right) R\left(0\theta 0\right)R\left(\omega 00\right) R\left(0-\theta 0\right) R\left(-\varphi 00\right)=R\left(\alpha 00\right) R\left(0\beta \beta 0\right)R\left(00\gamma \right)$$

(65)

We obtain the Euler angles from Darboux angles [2] as described by (66):

$$\begin{gathered} cos\left(\alpha +\gamma /2\right)cos\left(\beta /2\right)=cos\left(\omega /2\right), \\ sin\left(\alpha +\gamma /2\right)cos\left(\beta /2\right)=sin\left(\omega /2\right)cos\left(\theta \right), \\ cos\left(\alpha -\gamma /2\right)sin\left(\beta /2\right)=sin\left(\omega /2\right)sin\left(\theta \right)sin\left(\varphi \right), \\ sin\left(\alpha -\gamma /2\right)sin\left(\beta /2\right)=sin\left(\omega /2\right)sin\left(\theta \right)cos\left(\varphi \right). \end{gathered}$$

(66)

Here we find all following for our coset leaders written in their polar form:

$$\begin{gathered} R_1=R\left(-\pi /2\pi /2\pi /2\right),R_1^3=R\left(-\pi /2\pi /2\pi /2\right),R_2=R\left(0\pi /20\right), \\ {R}_2^3=R\left(0-\pi /20\right),R_3^3=R\left(0\pi \pi /2\right), \\ {R}_3=R\left(00\pi /2\right),i_1=R\left(0\pi /4\pi \right),i_2=R\left(\pi \pi /4\pi \right),i_3=R\left(\pi /4\pi /2\pi \right),i_4=R\left(-\pi /4\pi /2\pi \right), \\ {i}_5=R\left(3\pi /2\pi /4\pi \right),i_6=R\left(\pi /2\pi /4\pi \right) \end{gathered}$$

(67)

We could have also us the follow alternative sets of cosets leaders:

$$\begin{gathered} 1=R\left(000\right), R_1^2=R\left(-\pi /2\pi /2\pi \right),R_2^2=R\left(0\pi /2\pi \right),R_3^2=R\left(0\pi /2\pi \right), \\ {r}_1=R\left(-\pi /3\pi /32\pi /3\right), r_2=R\left(\pi /3\pi /3-2\pi /3\right),r_3=R\left(\pi /3-\pi /3-2\pi /3\right), \\ {r}_4=R\left(-\pi /3-\pi /32\pi /3\right),r_1^2=R\left(-\pi /3\pi /3-2\pi /3\right),r_2^2=R\left(\pi /3-\pi /32\pi /3\right),r_3^2 \\ =R\left(\pi /3-\pi /32\pi /3\right), r_4^2=R\left(-\pi /3\pi /3-2\pi /3\right) \end{gathered}$$

(68)

By using (66) we found the Euler representation (56) leading cosets:

$$\begin{gathered} R_1=R\left(-\pi /2\pi /2\pi /2\right),R_1^3=R\left(-\pi /2\pi /2\pi /2\right),R_2=R\left(0\pi /20\right), \\ {R}_2^3=R\left(0-\pi /20\right),R_3^3=R\left(3\pi /40\pi /4\right), \\ {R}_3=R\left(\pi /403\pi /4\right),i_1=R\left(0\pi /2\pi \right),i_2=R\left(3\pi /4\pi /20\right),i_3=R\left(\pi /4\pi /23\pi /4\right),i_4=R\left(3\pi /4\pi /2\pi /4\right), \\ {i}_5=R\left(\pi /2\pi /2\pi /2\right),i_6=R\left(\pi /2\pi /2\pi /2\right), \end{gathered}$$

(69)

Thus, we proceed with the example by finding the representation components of ${\mathfrak{D}}_{m2}^2\left(\alpha \beta \gamma \right)$ for each coset leader by taking only the first column of the ${\mathfrak{D}}_{m2}^2\left(\alpha \beta \gamma \right)$ matrix (49), therefore,

$${\mathfrak{D}}_{m2}^2(\alpha \beta \gamma =(\begin{array}{c}{e}^{-i2\left(\alpha +\gamma \right)}{\left(\frac{1+cos\beta }{2}\right)}^2\\ e^{-i\left(\alpha +2\gamma \right)\left(\frac{1+cos\beta }{2}\right)}sin\beta \\ e^{-i2\gamma }\sqrt{\frac{3}{8}}{\left(sin\beta \right)}^2\\ e^{i\left(\alpha -2\gamma \right)\left(\frac{1-cos\beta }{2}\right)}sin\beta \\ e^{2i\left(\alpha -\gamma \right)}{\left(\frac{1-cos\beta }{2}\right)}^2\end{array}$$

(70)

We evaluate (70) for the coset leaders, and we have

$$\begin{gathered} {\mathfrak{D}}^2(R_3=(\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}, {\mathfrak{D}}^2(R_3^3=(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 1\end{array} , {\mathfrak{D}}^2(R_1=(-\begin{array}{c}1/4\\ -i/2\\ \sqrt{3/8}\\ i/2\\ 1/4\end{array}, {\mathfrak{D}}^2(R_1^3=(-\begin{array}{c}1/4\\ i/2\\ \sqrt{3/8}\\ -i/2\\ 1/4\end{array} \\ {\mathfrak{D}}^2(R_2=(\begin{array}{c}1/4\\ 1/2\\ \sqrt{3/8}\\ 1/2\\ 1/4\end{array}, {\mathfrak{D}}^2(R_2^3=(\begin{array}{c}1/4\\ -1/2\\ \sqrt{3/8}\\ -1/2\\ 1/4\end{array}, {\mathfrak{D}}^2(i_3=(\begin{array}{c}\frac{1}{4}\\ \frac{1-i}{2\sqrt{2}}\\ -i\sqrt{\frac{3}{8}}\\ \frac{-1+i}{2\sqrt{2}}\\ \frac{1}{4}\end{array}, {\mathfrak{D}}^2(i_4=(\begin{array}{c}\frac{1}{4}\\ -\frac{1-i}{2\sqrt{2}}\\ i\sqrt{\frac{3}{8}}\\ \frac{1-i}{2\sqrt{2}}\\ \frac{1}{4}\end{array} \\ {\mathfrak{D}}^2(i_1=(\begin{array}{c}1/4\\ 1/2\\ \sqrt{3/8}\\ 1/2\\ 1/4\end{array}, {\mathfrak{D}}^2(i_2=(\begin{array}{c}1/4\\ -1/2\\ \sqrt{3/8}\\ -1/2\\ 1/4\end{array}, {\mathfrak{D}}^2(i_5=(\begin{array}{c}1/4\\ i/2\\ -\sqrt{3/8}\\ -i/2\\ 1/4\end{array}, {\mathfrak{D}}^2(i_6=(\begin{array}{c}1/4\\ -i/2\\ -\sqrt{3/8}\\ i/2\\ 1/4\end{array} \end{gathered}$$

(71)

Now, we use Equation (54) through (71) to determine our orbital states (72).

$$\begin{array}{l}|\begin{array}{c}{T}_2\\ 1\end{array}\rangle =\left(-|\begin{array}{c}2\\ -1\end{array}\rangle -|\begin{array}{c}2\\ 1\end{array}\rangle \right)/\sqrt{2}\\ |\begin{array}{c}{T}_2\\ 2\end{array}\rangle =i\left(|\begin{array}{c}2\\ -1\end{array}\rangle +|\begin{array}{c}2\\ 2\end{array}\rangle \right)/\sqrt{2}\\ |\begin{array}{c}{T}_2\\ 3\end{array}\rangle =\left(-|\begin{array}{c}2\\ -2\end{array}\rangle +|\begin{array}{c}2\\ 2\end{array}\rangle \right)/\sqrt{2}\\ |\begin{array}{c}E\\ 1\end{array}\rangle =|\begin{array}{c}2\\ 0\end{array}\rangle \\ |\begin{array}{c}E\\ 2\end{array}\rangle =\left(|\begin{array}{c}2\\ -2\end{array}\rangle +|\begin{array}{c}2\\ 2\end{array}\rangle \right)/\sqrt{2}\end{array}$$

(72)

By applying spherical harmonics, we have,

$$\begin{array}{l}{n}_2\langle \theta \varphi |\begin{array}{c}2\\ 2\end{array}\rangle =n_2{Y}_2^2\left(\theta \varphi \right)=\sqrt{\frac{3}{8}}{e}^{2i\varphi }si{n}^2\theta =\sqrt{\frac{3}{8}}{\left(x+iy\right)}^2/r^2\\ n_2\langle \theta \varphi |\begin{array}{c}2\\ 1\end{array}\rangle =n_2{Y}_1^2\left(\theta \varphi \right)=\sqrt{\frac{3}{8}}{e}^{i\varphi }si{n}^2\theta =\sqrt{\frac{3}{8}}\left(x+iy\right)/r^2\\ n_2\langle \theta \varphi |\begin{array}{c}2\\ 0\end{array}\rangle =n_2{Y}_0^2\left(\theta \varphi \right)=\left(3co{s}^2\theta -1\right)/2=\left(2z^2-x^2-y^2\right)/2r^2\\ n_2\langle \theta \varphi |\begin{array}{c}2\\ -1\end{array}\rangle =n_2{Y}_{-1}^2\left(\theta \varphi \right)=-\sqrt{\frac{3}{2}}{e}^{-i\varphi }sin\theta cos\theta =-\sqrt{\frac{3}{2}}\left(x-iy\right)z/r^2\\ n_2\langle \theta \varphi |\begin{array}{c}2\\ -2\end{array}\rangle =n_2{Y}_{-2}^2\left(\theta \varphi \right)=\sqrt{\frac{3}{8}}{e}^{-2i\varphi }si{n}^2\theta =\sqrt{\frac{3}{8}}{\left(x-iy\right)}^2/r^2\end{array}$$

(73)

By combining (72) and (73), we acquire the same result for the rotational tensor given in [3].

B.

Vibrational Wavefunction

Consider the vibration Hamiltonian expanded in terms of raising $a^{†}$ and lowering $a$ operators,

$$H={\displaystyle \sum _{\alpha }{\displaystyle \sum _i\left({\alpha }_i^{{\alpha }^{†}}{\alpha }_i^{\alpha }+\frac{1}{2}\right)}}.$$

(74)

Suppose $\epsilon$ is an eigenstate of H then

$$H|\epsilon \rangle =\epsilon |\epsilon \rangle ,$$

(75)

as such:

$$|\epsilon \rangle =|\cdots n_i^{\alpha }\cdots n_j^{\beta }\cdots \rangle =\cdots {\left(a_i^{{\alpha }^{†}}\right)}^{n_i^{\alpha }}\cdots {\left(a_j^{\beta }\right)}^{n_j^{\beta }}|0\cdots \rangle /\sqrt{N}.$$

(76)

Furthermore, ${\nu }_i$ is the spectroscopy notation for $\mathsf{\Gamma }$, such that in methane ${\nu }_1$ for $A_1$, ${\nu }_2$ for $E$, ${\nu }_3$ for $T_2,$ ${\nu }_4$ for $T_2$. There, E label $\left(2{\nu }_2\right)$ has a triple degenerate:

$$|\cdots n_1^E,n_2^E\cdots \left(2{\nu }_2\right)\rangle =|\cdots 2,0\cdots \left(2{\nu }_2\right)\rangle , |\cdots 1,1\cdots \left(2{\nu }_2\right)\rangle , |\cdots 0,2\cdots \left(2{\nu }_2\right)\rangle .$$

(77)

(77) is exactly the same as Alvarez-Bajo et al. [10], i.e., $|\mathrm{V};\mathrm{v}\rangle$, therefore, we can say

$$|V;{\mathsf{\Psi }}_{vib}^{p, {\mathsf{\Gamma }}_2, {\gamma }_2}\rangle ={\displaystyle \sum _v{D}_v^{p, {\mathsf{\Gamma }}_2, {\gamma }_2}{|V;v\rangle }_p}$$

(78)

4. Concrete Example: Symmetry Analysis of Methane

The chemical formula for methane is CH₄.; there are four hydrogen atoms bonded to one carbon atom. The ground state methane (CH₄) molecule has T_d symmetry, i.e., tetrahedral point group symmetry. Imagine a cube that has four hydrogen atoms placed at four of its corners laying diagonally at equal distances with a carbon atom at the center. These atoms are connected to each other via springs. Thus, the tetrahedral symmetry is shown in Figure 4.

The position of Hydrogen atoms or Y atoms will be represented with symmetry coordinates. Figure 4. shows a set of bases for the Y atoms. We will use these base sets to do the symmetry analysis of XY₄. The vibrational modes will be calculated, and they can be simulated on a computer to understand the dynamics of XY₄ molecules better.

There are twelve bases stated describing the Y atom in XY₄ and three bases stated for the X atom. Bases set of Y atoms in Figure 4. are labeled by set selected symmetry operators in Figure 5,

$$\left\{|A\rangle ,|r_{1}A\rangle ,|r_1^{2}A\rangle ,|R_1^{2}A\rangle ,|r_{4}A\rangle ,|r_2^{2}A\rangle ,|R_2^{2}A\rangle ,|r_{2}A\rangle ,|r_3^{2}A\rangle ,|R_3^{2}A\rangle ,|r_{3}A\rangle ,|r_3^2\rangle \right\}$$

(79)

The symmetry operations for an object with $T_d$ symmetry is given by the following in Figure 6.

Although (a) and (b) make up $T_d$ point symmetry, the symmetry operations of (a) moves the four $Y$ atoms to the same place that the symmetry operations in (b) would. Consequently, the symmetry operation in (a) and (b) of the two cosets of twelve in $T_d$ base vectors of the X atom are,

$$\left\{|C_1\rangle ,|C_2\rangle ,|C_3\rangle \right\}$$

(80)

The eigenvectors are then obtained from the group projectors $P_{jk}^{\alpha }$. When the particular symmetry is known, the eigenvectors or near-eigenvectors $|e_{jk}^{\alpha }A\rangle$ are very easily found,

$$|e_{jk}^{\alpha }A\rangle =\frac{P_{jk}^{\alpha }|A\rangle }{\sqrt{N_k^{\alpha }}}$$

(81)

There exists a local symmetry of $C_{\nu }=\left\{1,\sigma \right\}$ that preserves each coordinate vector $|rA\rangle$ so $\left(\sigma |rA\rangle =|rA\rangle \right)$. It has a projector $P^{0_2}$,

$$P^{0_2}=\frac{1}{2}\left(1+\sigma \right)|A\rangle .$$

(82)

Thus, it spans an induced representation $D^{0_2}\left(C_{\nu }\right)\uparrow T_d$ of $T_d$. The full $T_d$labeled orbit is given by a 12-dimensional induced representation $D^{0_2}\left(C_{\nu }\right)\uparrow T_d$. The character tables of $C_{\nu }$,

$$\overline{)\begin{array}{ccc}{C}_v& 1& \sigma \\ 0_2& 1& 1\\ 1& 1& -1\end{array}}$$

(83)

$$\overline{)\begin{array}{cccccc}{T}_d& 1& \left(r_1\dots r_1^2\dots \right)& \left(R_1^2\dots \right)& \left(I{R}_1\dots I{R}_3^3\dots \right)& \left(I{i}_1\dots \right)\\ A_1& 1& 1& 1& 1& 1\\ A_2& 1& 1& 1& -1& -1\\ E& 2& -1& 2& 0& 0\\ T_1& 3& 0& -1& 1& -1\\ T_2& 3& 0& -1& -1& 1\end{array}}$$

(84)

and $T_d$ and using columns 1 and Ii₁ of $T_d$ table gives (63). However, by considering the sub-group chain $T_d\supset C_{3\nu }\supset C_{\nu }$, we find the symmetry characteristic of the energy levels in $X{Y}_4$ as shown by Figure 5. We start with the $C_{3\nu }$ character table:

$$\begin{array}{cccc}{C}_{3v}& 1& \left(r,r^2\right)& \left({\sigma }_1,{\sigma }_2,{\sigma }_3\right)\\ A_1& 1& 1& 1\\ A_2& 1& 1& -1\\ E& 2& -1& 0\end{array}$$

(85)

Then, we determine the correlation $T_d\supset D_3:T_d\uparrow D_3$ between the $T_d$ character table (60) and the $C_{3\nu }$ character table (85),

$$\begin{array}{cccc}{T}_d\uparrow C_{3v}& A_1& A_2& E\\ A_1& 1& .& .\\ A_2& .& 1& .\\ E& .& .& 1\\ T_1& .& 1& 1\\ T_2& 1& .& 1\end{array}$$

(86)

Next, we obtain the correlation between $C_{3\nu }$ and $C_v$,

$$\begin{array}{ccc}{D}_3\uparrow C_v& 0_2& 1_2\\ A_1& 1& .\\ A_2& .& 1\\ E& 1& 1\end{array}$$

(87)

$$\Rightarrow D^{0_2}\uparrow T_d=A_1\oplus E\oplus T_1\oplus 2T_2.\oplus$$

(88)

From (79)–(88), we construct the following energy level correlation diagram (see Figure 7), which shows symmetry splitting for the subgroup chain $T_d\supset C_{3\nu }\supset C_{\nu }$. This gives the eigenvectors:

$$\begin{array}{l}|e_j^{\alpha }\rangle =P_{ij}^{\alpha }|A\rangle =\frac{l}{{}^{0}G}{\displaystyle \sum _g{D}_{ij}^{{\alpha }^{\ast }}}\left(g\right)|gA\rangle \\ where\begin{array}{cc}& \alpha =A_1\end{array},E,T_1,T_2\end{array},$$

(89)

$$\{\begin{array}{l}|e_1^{A_1}\rangle =\frac{1}{24}{\displaystyle \sum _g{D}_{11}^{A_1}}\left(g\right)|gA\rangle ,\\ |e_j^E\rangle =\frac{1}{24}{\displaystyle \sum _g{D}_{j1}^E}\left(g\right)|gA\rangle , j=1,2,\\ |e_j^{T_1}\rangle =\frac{1}{8}{\displaystyle \sum _g{D}_{j1}^{T_1}}\left(g\right)|gA\rangle , j=1,2,3,\\ |e_j^{T_2}\rangle =\frac{1}{8}{\displaystyle \sum _g{D}_{j1}^{T_2}}\left(g\right)|gA\rangle \begin{array}{cc}, j=1,2,3, & |e_j^{T_2}\rangle =\frac{1}{8}{\displaystyle \sum _g{D}_{j3}^{T_2}}\left(g\right)|gA\rangle , j=1,2,3\end{array}\end{array}$$

(90)

Irrep matrix $D^{{\mathsf{\Gamma }}_1}$ are determined from irrep idempotent $P_j^{{\mathsf{\Gamma }}_1}$ by constructing the normalize element $P_j^{{\mathsf{\Gamma }}_1}\mathrm{g}{P}_j^{{\mathsf{\Gamma }}_1}$. In Appendix A we list the trigonal and tetrahedral bases $3\times 3$ and $2\times 2$ matrices. We demonstrate in Section 2 how to compute $D^{{\mathsf{\Gamma }}_1}$ for $2\times 2$ matrices, however, for more details consult red. 3. In (89) the element of $D^{{\mathsf{\Gamma }}_1}$ are used to find the eigenvectors.

The projected near eigenvectors corresponding to $T_d\supset C_{3v}\supset C_v$ are given below,

$$\begin{array}{|ccccccccccccc|}\hline & |e^{A_1}\rangle & |e^{E_1}\rangle & |e^{E_2}\rangle & |e_{11}^{T_1}\rangle & |e_{12}^{T_1}\rangle & |e_{13}^{T_1}\rangle & |e_{11}^{T_2}\rangle & |e_{12}^{T_2}\rangle & |e_{13}^{T_2}\rangle & |e_{31}^{T_2}\rangle & |e_{32}^{T_2}\rangle & |e_{33}^{T_2}\rangle \\ |1\rangle & 1& 2& 0& 2& 0& 0& 3& 0& 0& 0& 0& 2\\ |r_1\rangle & 1& -1& 1& -1& -3& 0& 3& 0& 0& 0& 3& -1\\ |r_1^2\rangle & 1& -1& -1& -1& 3& 0& 3& 0& 0& 0& -3& -1\\ |R_1^2\rangle & 1& 2& 0& 0& 2& 1& -1& -1& -1& 1& -2& 0\\ |r_4\rangle & 1& -1& 1& 1& -1& 1& -1& 2& 0& 1& 1& 1\\ |r_2^2\rangle & 1& -1& -1& -1& -1& 1& -1& -1& -1& 1& 1& -1\\ |R_2^2\rangle & 1& 2& 0& 0& 2& -1& -1& -1& -1& 1& 2& 0\\ |r_2\rangle & 1& -1& 1& -1& 1& -1& -1& -1& 1& 1& -1& -1\\ |r_3^2\rangle & 1& -1& -1& 1& 1& -1& -1& 2& 0& 1& -1& 1\\ |R_3^2\rangle & 1& 2& 0& -2& 0& 0& -1& 2& 0& 0& 0& -2\\ |r_3\rangle & 1& -1& 1& 1& 3& 0& -1& -1& -1& 0& -3& 1\\ |r_4^2\rangle & 1& -1& -1& 1& -3& 0& -1& -1& -1& 0& 3& 1\\ & 2\sqrt{3}& 2\sqrt{6}& 2\sqrt{2}& 4& 4\sqrt{3}& \sqrt{6}& 6& 2\sqrt{6}& 2\sqrt{2}& \sqrt{6}& 4\sqrt{3}& 4\\ \hline\end{array}$$

(91)

The $T_1$ and $T_2$ eigenvectors are written in the trigonal bases (see Appendix A.4, Appendix A.5 and Appendix A.6) and are related to the subgroup chain $T_d\supseteq D_{2d}\supseteq C_v$ by the unitary transformation,

$$T=\left(\begin{array}{ccc}\frac{1}{\sqrt{2}}& \frac{-1}{\sqrt{2}}& 0\\ \frac{1}{\sqrt{6}}& \frac{1}{\sqrt{6}}& \frac{-2}{\sqrt{6}}\\ \frac{1}{\sqrt{3}}& \frac{1}{\sqrt{3}}& \frac{1}{\sqrt{3}}\end{array}\right)$$

(92)

Applying the transformation to the irrep matrices of the $T_d\supset C_{3\nu }\supset C_{\nu }$ chain gives the irrep matrices for the $T_d\supseteq D_{2d}\supseteq C_v$ chain (see Appendix A.1, Appendix A.2 and Appendix A.3). As a result, we find the eigenvectors in a new bases which is given by,

$$\begin{array}{|ccccccccccccc|}\hline & |e^{A_1}\rangle & |e^{E_1}\rangle & |e^{E_2}\rangle & |e_{11}^{T_1}\rangle & |e_{12}^{T_1}\rangle & |e_{13}^{T_1}\rangle & |e_{11}^{T_2}\rangle & |e_{12}^{T_2}\rangle & |e_{13}^{T_2}\rangle & |e_{31}^{T_2}\rangle & |e_{32}^{T_2}\rangle & |e_{33}^{T_2}\rangle \\ |1\rangle & 1& 2& 0& 1& -1& 0& 1& 0& 0& 0& 0& 1\\ |r_1\rangle & 1& -1& 1& 0& 1& -1& 0& 1& 0& 1& 0& 0\\ |r_1^2\rangle & 1& -1& -1& -1& 0& 1& 0& 0& 1& 0& 1& 0\\ |R_1^2\rangle & 1& 2& 0& 1& 1& 0& 1& 0& 0& 0& 0& -1\\ |r_4\rangle & 1& -1& 1& 0& -1& 1& 0& -1& 0& -1& 0& 0\\ |r_2^2\rangle & 1& -1& -1& -1& 0& -1& 0& 0& -1& 0& -1& 0\\ |R_2^2\rangle & 1& 2& 0& -1& -1& 0& -1& 0& 0& 0& 0& -1\\ |r_2\rangle & 1& -1& 1& 0& 1& 1& 0& 1& 0& -1& 0& 0\\ |r_3^2\rangle & 1& -1& -1& 1& 0& -1& 0& 0& -1& 0& 1& 0\\ |R_3^2\rangle & 1& 2& 0& -1& 1& 0& -1& 0& 0& 0& 0& 1\\ |r_3\rangle & 1& -1& 1& 0& -1& -1& 0& -1& 0& 1& 0& 0\\ |r_4^2\rangle & 1& -1& -1& 1& 0& 1& 0& 0& 1& 0& -1& 0\\ & 2\sqrt{3}& 2\sqrt{6}& 2\sqrt{2}& 2\sqrt{2}& 2\sqrt{2}& 2\sqrt{2}& 2& 2& 2& 2& 2& 2\\ \hline\end{array}$$

(93)

In Figure 8, we display all the genuine vibrations of XY₄. Now, we consider the energy eigenvalues using those bases for XY₄ molecule.

4.1. Solving the Equation of Motion

From the $T_d$ symmetry projection operators, we can construct the eigenstates and use them to partially solve the equation of motion (94),

$$m|\ddot{x}\rangle =-F|x\rangle$$

(94)

The number of elements we need for the force matrix is considerably less than (15)² = 225 due to symmetry considerations. Only 18 elements are needed for the force matrix. Because of the non-genuine vibration of the $T_2$ mode, it will acquire in general (82), which simplifies to a block matrix form (96). To compute the F-matrix we must obtain the component $\langle i\left|F\right|j\rangle$ by computing the product of the projections of each spring on the coordinate axis $x_i=\langle i|x\rangle$ and $x_j=\langle j|x\rangle$ for all spring and summing products.

$$-F_1=\langle 1\left|F\right|1\rangle \langle 1|x\rangle +\langle 1|F|2\rangle \langle 2|x\rangle +\langle 1\left|F\right|3\rangle \langle 3|x\rangle +\dots ,$$

(95)

which implies a force matrix given by,

$$\begin{array}{c}|A\rangle \\ \\ |C_1\rangle \end{array}\left[\begin{array}{ccccccccccccccc}|1\rangle & |2\rangle & |3\rangle & |4\rangle & |5\rangle & |6\rangle & |7\rangle & |8\rangle & |9\rangle & |10\rangle & |11\rangle & |12\rangle & |13\rangle & |14\rangle & |15\rangle \\ k+\frac{j}{3}& \frac{k}{2}+\frac{j}{3}& \frac{k}{2}+\frac{j}{3}& \frac{k}{2}& •& \frac{k}{2}& \frac{k}{2}& \frac{k}{2}& •& •& •& •& \frac{-j}{3}& \frac{-j}{3}& \frac{-j}{3}\\ & & & & & & & & & & & & \frac{4j}{3}& •& •\end{array}\right]$$

(96)

$$\begin{array}{c}|A\rangle \\ \\ |C_1\rangle \end{array}\left[\begin{array}{ccccccccccccccc}|1\rangle & |2\rangle & |3\rangle & |4\rangle & |5\rangle & |6\rangle & |7\rangle & |8\rangle & |9\rangle & |10\rangle & |11\rangle & |12\rangle & |13\rangle & |14\rangle & |15\rangle \\ m& •& •& •& •& •& •& •& •& •& •& •& •& •& •\\ •& •& •& •& •& •& •& •& •& •& •& •& M& •& •\end{array}\right]$$

(97)

Notice in the force matrix, only the first column must be written down. Thus, with our eigenvectors applied to this first row we obtain the eigenfrequencies:

$$\begin{array}{l}{\left({\omega }^{\alpha }\right)}^2=\langle m^{-1}F\rangle =\frac{1}{m}{\displaystyle \sum _g\langle A|}F|gA\rangle D_{ij}^{\alpha }\left(g\right)\\ where\begin{array}{cc}& \alpha =A_1\end{array},E,T_1,T_2\end{array}$$

(98)

$${\mathrm{A}}_1 \mathrm{mode} {\omega }^{A_1}=\sqrt{\frac{4k+j}{m}}$$

(99)

$$\mathrm{\text{E-mode}} {\omega }^E=\sqrt{\frac{2k}{m}}.$$

(100)

T₁ mode has an eigenfrequency of zero corresponding to a pure classical rotation. Since XY₄ is spherically symmetric, the following rotational Hamiltonian can describe $T_1$ modes (see Figure 9),

$$H_r=\frac{J^2}{2I}.$$

(101)

4.2. Non-Genuine Vibration

There are two $T_2$ modes for the H atoms and one for the C atom. Thus, there are six eigenfrequencies since the dimension of $T_2$ is three (see Figure 8). The $T_2$ are not all genuine vibrations, but two vibrational modes are mixed with one translation mode. By using the above $T_2$ eigenvectors and the $|C\rangle$ bases to transform each Q matrix where Q = F or Q = m, we acquire a block diagonal form,

$$\left[\begin{array}{ccc}\langle \begin{array}{c}{T}_2\\ j1\end{array}|Q|\begin{array}{c}{T}_2\\ j1\end{array}\rangle & \langle \begin{array}{c}{T}_2\\ j1\end{array}|Q|\begin{array}{c}{T}_2\\ j3\end{array}\rangle & \langle \begin{array}{c}{T}_2\\ j1\end{array}|Q|C_j\rangle \\ \langle \begin{array}{c}{T}_2\\ j3\end{array}|Q|\begin{array}{c}{T}_2\\ j1\end{array}\rangle & \langle \begin{array}{c}{T}_2\\ j3\end{array}|Q|\begin{array}{c}{T}_2\\ j1\end{array}\rangle & \langle \begin{array}{c}{T}_2\\ j3\end{array}|Q|C_j\rangle \\ \langle C_j|Q|\begin{array}{c}{T}_2\\ j1\end{array}\rangle & \langle C_j|Q|\begin{array}{c}{T}_2\\ j3\end{array}\rangle & \langle C_j|Q|C_j\rangle \end{array}\right]$$

(102)

Note, there are nine eigenvalues and one for each j = 1, 2, 3 three 3 × 3 Γ-matrices, since by symmetry each 3 × 3 matrix is the same. By solving one of them we determine all nine eigenvalues of three triply degenerate levels. (103) gives the matrix element calculation for (102),

$$\langle \begin{array}{c}{T}_2\\ j1\end{array}|Q|\begin{array}{c}{T}_2\\ j1\end{array}\rangle =\langle A|P_{j1}^{T_2}Q{P}_{j1}^{T_2}|A\rangle ={\displaystyle \sum _g\langle A|}Q|gA\rangle D_{11}^{T_2}\left(g\right),$$

(103)

$$\langle \begin{array}{c}{T}_2\\ j3\end{array}|Q|\begin{array}{c}{T}_2\\ j3\end{array}\rangle ={\displaystyle \sum _g\langle A|}Q|gA\rangle D_{33}^{T_2}\left(g\right),$$

(104)

$$\langle \begin{array}{c}{T}_2\\ j1\end{array}|Q|\begin{array}{c}{T}_2\\ j3\end{array}\rangle =\langle \begin{array}{c}{T}_2\\ j3\end{array}|Q|\begin{array}{c}{T}_2\\ j1\end{array}\rangle ={\displaystyle \sum _g\langle A|}Q|gA\rangle D_{13}^{T_2}\left(g\right),$$

(105)

$$\langle \begin{array}{c}{T}_2\\ j3\end{array}|Q|C_j\rangle =\langle \begin{array}{c}{T}_2\\ j1\end{array}|Q|C_j\rangle =\langle A|Q{P}_{1j}^{T_2}|C_j\rangle ,$$

(106)

$$\langle \begin{array}{c}{T}_2\\ j3\end{array}|Q|C_j\rangle =\langle \begin{array}{c}{T}_2\\ j1\end{array}|Q|C_j\rangle =\langle A|Q{P}_{1j}^{T_2}|C_j\rangle ,$$

(107)

$$\langle C_j|Q|C_j\rangle =\langle C_j|Q|C_j\rangle .$$

(108)

First, we consider $Q=F$ in (102) to give the force matrix as

$$\left[\begin{array}{ccc}2\left(\gamma +1/3\right)& \frac{1}{3}& \frac{-2}{3}\\ \frac{1}{3}& \frac{1}{3}& \frac{-2}{3}\\ \frac{-2}{3}& \frac{-2}{3}& \frac{4}{3}\end{array}\right], \gamma =\frac{k}{j}.$$

(109)

Next, by making Q = m in (103) through (107) and Q = M to give,

$$Q=mass\Rightarrow \left[\begin{array}{ccc}m& •& •\\ •& m& •\\ •& •& M\end{array}\right].$$

(110)

As a result, we have the following acceleration matrix,

$$\frac{j}{m}\left[\begin{array}{ccc}2\left(\gamma +1/3\right)& \frac{1}{3}& \frac{-2}{3}\\ \frac{1}{3}& \frac{1}{3}& \frac{-2}{3}\\ \frac{-2}{3}\alpha & \frac{-2}{3}\alpha & \frac{4}{3}\alpha \end{array}\right],\gamma =\frac{k}{j}, \alpha =\frac{m}{M};$$

(111)

We solved the eigenvalue equation,

$$\lambda \left({\lambda }^2-2\left(\gamma +\frac{1}{2}+\frac{2}{3}\alpha \right)\lambda +\frac{2\gamma }{3}\left(4\alpha +1\right)\right)=0.$$

(112)

Solving (112), we have following eigenvalues

$$\begin{array}{l}{\lambda }_0=0;\\ {\lambda }_{\pm }={\Re }_0\pm \sqrt{\left({\Re }_1^2-\frac{2{\Re }^{\prime }}{3}\right)}\end{array},$$

(113)

$$\begin{array}{l}where\begin{array}{cc}& {\omega }_{\pm }\end{array}=\sqrt{\frac{j}{m}{\lambda }_{\pm }}\\ {\Re }_0=\gamma +\frac{2}{3}\alpha +\frac{1}{2}\\ {\Re }_0=\gamma +\frac{2}{3}\alpha +\frac{1}{2}\\ {\Re }^{\prime }=\gamma -\frac{4}{3}\alpha +\frac{1}{6}\end{array}$$

(114)

The corresponding eigenvectors can be written as

$$|{\lambda }_0\rangle =\left[\begin{array}{c}0\\ 2\\ 1\end{array}\right]$$

(115)

$$|{\lambda }_{+}\rangle =\left[\begin{array}{c}\frac{\left({\Re }_1+\sqrt{\Omega }\right)}{4\alpha }\\ -\frac{1}{2\alpha }\\ 1\end{array}\right],$$

(116)

$$|{\lambda }_{-}\rangle =\left[\begin{array}{c}\frac{-\left({\Re }_1+\sqrt{\Omega }\right)}{4\alpha }\\ -\frac{1}{2\alpha }\\ 1\end{array}\right],$$

(117)

$$where\begin{array}{cc}& \Omega =\left({\Re }_1^2-\frac{2{\Re }^{\prime }}{3}\right)\end{array}.$$

(118)

This procedure could be carried out using 2 × 2 matrices instead of 3 × 3 matrices since one of the $T_2$ modes is a pure translation. By excluding that mode, we only need to solve a 2 × 2 matrix. For the translation, both the X atom and the four $Y$ atoms should move by the same amount. As a result, we have,

$$|\begin{array}{c}{T}_2\\ j\end{array}Trans\rangle =2|\begin{array}{c}{T}_2\\ j3\end{array}\rangle +|\begin{array}{c}C\\ j\end{array}\rangle .$$

(119)

We illustrate $|\begin{array}{c}{T}_2\\ j\end{array} Trans\rangle$ in Figure 10 where the translation of H atoms or the Y atoms and X-atom for states that are in the $|C_1\rangle$ direction.

The translational mode and two vibrational modes of $T_2$ are now isolated (see Figure 10). This was achieved by taking linear combinations of the $T_2$modes.

A $T_2$ mode that is orthogonal to (95) must be a genuine vibration, i.e.,

$$|\begin{array}{c}{T}_2\\ j\end{array}vib\rangle =2M|\begin{array}{c}{T}_2\\ j3\end{array}\rangle -4m|\begin{array}{c}C\\ j\end{array}\rangle .$$

(120)

(120) gives the genuine vibration from the non-genuine vibrations. We illustrate (120) pictorially in Figure 11.

We just need to find the overlap matrix between the next $T_2$ mode, namely $|\begin{array}{c}{T}_2\\ j1\end{array}\rangle$, and $|\begin{array}{c}{T}_2\\ j1\end{array} vib\rangle ,$ i.e.,

$$\left(\begin{array}{cc}\langle \begin{array}{c}{T}_2\\ j1\end{array}|Q|\begin{array}{c}{T}_2\\ j\end{array}vib\rangle & \langle \begin{array}{c}{T}_2\\ j1\end{array}|Q|\begin{array}{c}{T}_2\\ j\end{array}vib\rangle \\ \langle \begin{array}{c}{T}_2\\ j\end{array}vib|Q|\begin{array}{c}{T}_2\\ j1\end{array}\rangle & \langle \begin{array}{c}{T}_2\\ j\end{array}vib|Q|\begin{array}{c}{T}_2\\ j\end{array}vib\rangle \end{array}\right)=\frac{j}{m}\left(\begin{array}{cc}\frac{2\left(\gamma +\frac{1}{3}\right)}{3}& \frac{2\left(M+4m\right)}{3}\\ \frac{1}{6M}& \frac{\left(M+4m\right)}{3M}\end{array}\right),$$

(121)

with eigenvalues,

$${\omega }_{\pm }=\sqrt{\frac{j}{m}{\Re }_0\pm \frac{j}{m}\sqrt{\left({\Re }_1^2-\frac{2{\Re }^{\prime }}{3}\right)}},$$

(122)

and the eigensolution

$$|e_j^{T_2}(\pm )\rangle ={\epsilon }_{\pm }|\begin{array}{c}{T}_2\\ j1\end{array}\rangle +|\begin{array}{c}{T}_2\\ j\end{array}vib\rangle ,$$

(123)

$$\mathrm{where} {\epsilon }_{\pm }=6M\left({\Re }_1\pm \sqrt{{\Re }_1^2-\frac{2}{3}{\Re }^{\prime }}\right).$$

(124)

Figure 12 shows how the frequency of vibration is affected as the ratio of the spring coupling increases.

For the CH₄ molecular parameters values $m=1$, $M=12$, $k=0.6$, and $j=2.66$, (113) gives the following acceleration,

$$\left[\begin{array}{cc}2.97& 28.37\\ 0.04& 1.18\end{array}\right].$$

(125)

where we found the eigenvalues to be ${\omega }_{+}=3.44$; ${\omega }_{-}=0.72$ and the eigenvectors of the vibration for the $T_2$ mode are,

$$\begin{array}{l}|e_j^{T_2}(+)\rangle =|\begin{array}{c}{T}_2\\ j1\end{array}\rangle +0.02|\begin{array}{c}{T}_2\\ j\end{array}vib\rangle \\ |e_j^{T_2}(-)\rangle =-|\begin{array}{c}{T}_2\\ j1\end{array}\rangle +0.08|\begin{array}{c}{T}_2\\ j\end{array}vib\rangle \end{array}$$

(126)

Table 1. Frequency comparison.

$\mathit{X}{\mathit{Y}}_4$ Irreps	$\mathbf{Observed} \left(c{m}^{-1}\right)$	$\mathbf{Computed} \left(c{m}^{-1}\right)$
$A_1$	2914.2	3179.8
$E$	1526.0	647.0
$T_1$	3020.3	3023.3
$T_2$	1306.3	1258.1

shows the observed with the computed in units of cm⁻¹. A₁ mode, and two $T_2$modes compare with observed frequency. We obtain computed frequency by using the force constants of $j$, $k$ $5.495 \mathrm{aJ}\dot{{\mathrm{A}}^{-2}}$, and $0.124 \mathrm{aJ}\dot{{\mathrm{A}}^{-2}}$, respectively. However, the E-mode does not compare well.

In the analysis of $X{Y}_4$, we see that only the $T_2$ moves the X atom. This means there is a coupling between the X and Y atoms through the j spring. It is evident that this is the case since the $T_2$ modes repel each other, as shown in Figure 12. In Figure 12, there is a clustering of (E, $T_2$) modes which is more pronounced when the strength of the j spring and mass ratio of X and Y atom increases. For better results with the $X{Y}_4$ molecule, we would need to consider bending forces for a more precise agreement.

5. Conclusions

It is possible to define every molecule according to its symmetry, or lack thereof, which can be expressed in terms of symmetry elements. The Projection Operator Technique utilizes the extended character table which includes each symmetry operation separately. We have demonstrated the power of the projector method to perform molecular calculations. As an example, a CH₄ molecule was considered for its tetrahedral symmetry, and our compute frequencies A₁ and the two T₂′s modes were in range of the observed values. By exploiting a molecule’s symmetry, it is not necessary to solve the Schrödinger equation in detail.