1. Introduction

Because photon is massless, its energy E (= h̄ω) is originally proportional to its momentum p (= h̄k):

This feature can be described from another point of view. Because Maxwell’s wave equation is of the second order of the time coordinate t, it has a time-reversal symmetry in the absence of the static magnetic field, so its dispersion as a function of the wave number k is symmetric:

However, this logic has a pitfall. It was shown by the transmission line theory that the dispersion curves can be linear if we have two modes and they coincide in the zone center as illustrated in Fig. 1 [13, 14, 15, 16]. Here, the two dispersion curves are symmetric with each other about the origin of k, so Eq. (3) is satisfied in spite of their finite slopes at k = 0. The coincidence of the two bands was realized by accidental degeneracy with appropriate choice of the sample structure.

 

Fig. 1 The linear dispersion in the zone center realized by two modes that are symmetric with each other about the origin of k.

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On the other hand, I recently reformulated the same problem based on the tight-binding approximation and group theory, and found that the necessary and sufficient condition for this phenomenon is that the two modes have different spatial symmetries for k = 0 and have the same symmetry for k ≠ 0 [17, 18]. The two bands are mixed and repel each other for k ≠ 0 because they have the same symmetry, while they cross each other for k = 0 because their symmetries are different. The balance of these crossing and anti-crossing behaviors results in the linear dispersion relation in the vicinity of the Γ point of the Brillouin zone. This property can actually be materialized, for example, by one-dimensional periodic metamaterials composed of metallic unit structures with the C_2v (rectangular) spatial symmetry [17]. Thus the effective mass m^* vanishes again due to the constraint imposed by the mode symmetry.

Now, the extension of this property to higher dimensions is quite an interesting problem, since the isotropic linear dispersion, or Dirac cone, in two and three dimensions brings about many exotic phenomena like unidirectional edge states [19, 20, 21], Zitterbewegung [22], and pseudo-diffusive transmission [23, 24]. In addition, it brings in a new prospect to the photon-dispersion engineering and may find applications in photonic and electromagnetic wave generation and propagation devices.

The presence of the Dirac cone on the Brillouin zone boundary was known for two-dimensional photonic crystals [25], and so far, those photonic crystals have been examined theoretically to predict the above-mentioned characteristics [19, 20, 21, 22, 23, 24]. Quite recently, Chan and his collaborators showed by the multiple scattering theory that the Dirac cone can also be materialized in the zone center of a two-dimensional photonic crystal using the accidental degeneracy of a doubly degenerate mode and a non-degenerate mode [26].

In this paper, I present another way to materialize the isotropic Dirac cone in the Brillouin zone center of two- and three-dimensional metamaterials. To the best of my knowledge, this is the first theoretical prediction of the presence of the Dirac cone in the three-dimensional periodic structure. This paper is organized as follows. In Section 2, the secular equation of the electromagnetic field, which is derived based on the resonant states of each unit structure of a two-dimensional square-lattice metamaterial, is solved to show the presence of the isotropic Dirac cone. In Section 3, the theory is extended to three dimensions. We deal with the case of accidental degeneracy of a triply degenerate mode and a non-degenerate mode in the cubic lattice and show the presence of the Dirac cone that is isotropic in all directions. Lengthy calculations to derive and solve the secular equation for the square lattice are given in Appendix A. Those for the cubic lattice and detailed analysis of electromagnetic transfer integrals for the T_1u and A_1g modes are given in Appendix B.

2. Isotropic Dirac cone in two dimensions

Maxwell’s wave equation for the magnetic field H(r,t) is given by

The essential point to realize the Dirac cone in higher dimensions is mixing and repulsion of modes for all k except the Γ point. As will be proved in the following, this condition can be fulfilled by accidental degeneracy of a doubly degenerate E mode and a non-degenerate A₁ mode on the Γ point in the square lattice of unit structures with the C_4v (regular square) symmetry. Figure 2 illustrates an example of such metamaterials, which was analyzed in Ref. [18] and the presence of a resonant state of the E symmetry was shown by numerical calculation. In general, there are resonant states of all possible symmetries of the C_4v point group.

 

Fig. 2 Left: Metallic unit structure with the C_4v (regular square) symmetry. Right: Square array of the unit metallic structure on a uniform dielectric-slab waveguide with a back electrode.

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For a single unit structure described by dielectric constant ε_s(r), we assume a doubly degenerate E resonant state and a non-degenerate A₁ state, and denote their magnetic field distributions by H⁽¹⁾ and H⁽²⁾ for the former and H⁽⁰⁾ for the latter. Thus, these magnetic fields satisfy the following eigenvalue equations:

We know from the group theory that the two eigenfunctions (magnetic field) of the E state (H⁽¹⁾ and H⁽²⁾) have the spatial symmetry as the x and y coordinates, respectively, while the eigenfunction of the A₁ state (H⁽⁰⁾) has the symmetry of the z coordinate [27]. When there are no other resonant states in the relevant frequency range, we can form the Bloch functions by linear combinations of these resonant states. Then, on the k_x axis, where the parity with respect to the y coordinate is a good quantum number, H⁽¹⁾ and H⁽⁰⁾ have the same parity so that their Bloch functions are mixed and repel each other. Similarly, the Bloch functions made of H⁽²⁾ and H⁽⁰⁾ are mixed and repel each other on the k_y axis.

In the following, I show that the mixing and repulsion take place for all two-dimensional directions to create the isotropic Dirac cone. Here we should note that the mixing does not take place on the Γ point, which is another necessary condition for the linear dispersion, since the Bloch functions of the Γ point have the full symmetry of the basis functions, and so the Bloch functions made of the E and A₁ modes have different symmetries.

Thus I solve Maxwell’s wave equation by the tight-binding approximation to show these characters. This approximation is justified because the electromagnetic waves are well localized in each unit structure of metallic metamaterials [17, 18]. I describe the total magnetic field, according to the prescription of the tight-binding approximation in solid state physics, as a linear combination of H⁽⁰⁾, H⁽¹⁾, and H⁽²⁾ localized in each unit structure with a phase factor that is dependent on the wave vector k to obtain a Bloch function.

Then, we can derive the secular equation of the electromagnetic eigenvalue problem, whose details are described in Appendix A. The problem is reduced to the diagonalization of a Hermitian matrix of the following form:

Now, we assume accidental degeneracy of the A₁ and E mode frequencies in the zone center and denote the degenerate frequency by ω_Γ. Such accidental degeneracy is generally possible by adjusting sample parameters. For example, it was attained by adjusting the lattice constant in Ref. [17]. By diagonalizing matrix B to the second order of k_x and k_y, we obtain the following dispersion relations:

 

Fig. 3 Isotropic Dirac cone in two dimensions realized by accidental degeneracy of a doubly degenerate E mode and a non-degenerate A₁ mode. The dashed square is the third parabolic band. The origin of the frequency ω is shifted to the degenerate frequency ω_Γ.

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3. Isotropic Dirac cone in three dimensions

In three dimensions, a similar situation can be realized by combining a T_1u mode and an A_1g mode in the simple cubic lattice of unit structures of the O_h symmetry, that is, the symmetry of a regular cube. Such a metamaterial is illustrated in Fig. 4. The T_1u mode is triply degenerate and its three wave functions behave like the x, y, and z coordinates, respectively, whereas the A_1g mode behaves like x² + y² + z² [27]. On each axis in the k space, the latter Bloch wave function has the same parity as one of the former three, so they are mixed and repel each other. Thus we can expect the generation of the Dirac cone that is isotropic in all directions.

 

Fig. 4 Left: Metallic unit structure with the O_h (regular cube) symmetry. Right: Simple-cubic array of the unit metallic structures.

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We can derive the secular equation of the electromagnetic field in a similar manner as described in the previous section. In the present case, the Hermitian matrix to be diagonalized is

In this paper, we examined two cases, that is, the accidental degeneracy of the E and A₁ mode of the square lattice and that of the T_1u and A_1g mode of the simple cubic lattice. Because the Dirac cone is realized by the balance of crossing and anti-crossing properties due to particular combinations of spatial symmetries of resonant states, we may be able to find still more different mode combinations to create the Dirac cone, which remains as an interesting future work. Numerical studies for the practical design of metamaterials to achieve the accidental degeneracy as described in this paper are also required in future.

4. Conclusion

It was shown by analytical calculation based on the tight-binding approximation and group theory that the isotropic Dirac cone in the Brillouin zone center can be created in two- and three-dimensional periodic metamaterials by accidental degeneracy of two modes. In the case of two dimensions, accidental degeneracy of a doubly degenerate E mode and a non-degenerate A₁ mode of the square lattice of the C_4v symmetry was examined and the presence of the isotropic Dirac cone, or a pair of modes whose dispersion is linear to k, in the Brillouin zone center was shown. For three dimensions, accidental degeneracy of a triply degenerate T_1u mode and a non-degenerate A_1g mode was examined and the presence of the Dirac cone in the zone center that was isotropic in all directions was shown. To the best of my knowledge, this is the first theoretical prediction of the presence of the Dirac cone in the three-dimensional periodic structure.

The essential point for creating the Dirac cone is the crossing and anti-crossing properties caused by particular combinations of spatial symmetries of resonant states localized in a single unit structure of the metamaterial. That is, the two relevant modes must have different symmetries for k = 0 so that they are not mixed and cross each other when the accidental degeneracy takes place, while the two modes must have the same symmetry for k ≠ 0 so that they are mixed and repel each other. The balance of these crossing and anti-crossing behaviors results in the linear dispersion in the vicinity of the Brillouin zone center.

To prove these properties, the dispersion relation of the periodic metamaterial was analytically calculated for small k by tight-binding approximation with the localized resonance states in the single unit structure as basis functions. The mutual relations among different electromagnetic transfer integrals were derived based on the group theory and used to solve the secular equation.

Appendix

A. E and A₁ modes of two-dimensional square lattice

An essential part of the theory is evaluation of the electromagnetic transfer integral. In two dimensions, it is defined by

(25)$$L_{lm}^{\left(ij\right)}\equiv \frac{1}{V}{\int }_{V}d\mathbf{r}{\mathbf{H}}^{\left(i\right)*}\left(\mathbf{r}\right)\cdot 𝒧{\mathbf{H}}^{\left(j\right)}\left(\mathbf{r}-{\mathbf{r}}_{lm}\right),$$

where H⁽ⁱ⁾(r) (i = 0, 1, 2) are the magnetic fields of the resonant states of the E symmetry (i = 1, 2) and the A₁ symmetry (i = 0), V is the volume on which the periodic boundary condition is imposed, 𝒧 is the operator that defines the eigenvalue problem of the periodic metamaterial given in Eq. (5), and r_lm are the lattice vectors of the square lattice given by

(26)$${\mathbf{r}}_{lm}=l\left(\begin{array}{c}a\\ 0\\ 0\end{array}\right)+m\left(\begin{array}{c}0\\ a\\ 0\end{array}\right),$$

where a is the lattice constant, and l and m are integers.

Because of the high symmetry of the unit structure (C_4v) and the square lattice, we can derive various relations among $L_{lm}^{\left(ij\right)}$. For the two eigenfunctions of the E mode, relations among their transfer integrals are given in Eqs. (14) – (18) of Ref. [18]. For the nearest neighbor lattice points, they are

(27)$$L_{0,0}^{\left(11\right)}=L_{0,0}^{\left(22\right)}\equiv \frac{{\omega }_1^2}{c^2}+M_1,$$

(28)$$L_{\pm 1,0}^{\left(11\right)}=L_{0,\pm 1}^{\left(22\right)}\equiv M_1^{\prime },$$

(29)$$L_{0,\pm 1}^{\left(11\right)}=L_{\pm 1,0}^{\left(22\right)}\equiv M_1^{″},$$

where ω₁ is the resonance frequency of the E mode. In these equations, I use notations of M different from those in Ref. [18] for the sake of convenience. All other $L_{lm}^{\left(ij\right)}$ for the nearest neighbor lattice points are equal to zero. As for the A₁ mode, it is invariant by any symmetry operation of the C_4v point group. Using this property, we can easily prove

(30)$$L_{0,0}^{\left(00\right)}\equiv \frac{{\omega }_0^2}{c^2}+M_0,$$

(31)$$L_{\pm 1,0}^{\left(00\right)}=L_{0,\pm 1}^{\left(00\right)}\equiv M_0^{\prime },$$

(32)$$\pm L_{\pm 1,0}^{\left(01\right)}=\mp L_{\pm 1,0}^{\left(10\right)*}=\pm L_{0,\pm 1}^{\left(02\right)}=\mp L_{0,\pm 1}^{\left(20\right)*}\equiv M_2,$$

where ω₀ is the resonance frequency of the A₁ mode.

According to the prescription of the tight-binding approximation, the Bloch eigenfunction is expressed by the linear combination of the three basis functions over all unit structures in V with k-dependent phase factors:

(33)$${\mathbf{H}}_{\mathbf{k}}\left(\mathbf{r}\right)=\frac{1}{V}\sum _{l,m}{e}^{i\mathbf{k}\cdot {\mathbf{r}}_{lm}}\sum _{i=0}^2{A}_i{\mathbf{H}}^{\left(i\right)}\left(\mathbf{r}-{\mathbf{r}}_{lm}\right).$$

Following the same procedure as described in Ref. [18], the eigenvalue equation,

(34)$$𝒧{\mathbf{H}}_k\left(\mathbf{r}\right)=\frac{{\omega }_k^2}{c^2}{\mathbf{H}}_k\left(\mathbf{r}\right),$$

leads to the secular equation

(35)$$\left|\text{B}-\frac{{\omega }_{\mathbf{k}}^2}{c^2}\text{I}\right|=0,$$

where I is the rank-3 unit matrix. Elements of matrix B are defined by

(36)$$B_{ij}=\sum _{lm}{e}^{ia\left(l{k}_x+m{k}_y\right)}{L}_{lm}^{\left(ij\right)}.$$

When we keep the nearest neighbor terms, we obtain Eqs. (10) – (14).

Now the next task is to solve the secular equation, Eq. (35), analytically. In the following we denote

(37)$$\xi =\frac{{\omega }_{\mathbf{k}}^2}{c^2},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\xi }_0=\frac{{\omega }_0^2}{c^2},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\xi }_1=\frac{{\omega }_1^2}{c^2}.$$

Then, the secular equation reduces to

(38)$${\xi }^3+b_2{\xi }^2+b_1\xi +b_0=0,$$

where

(39)$$\begin{array}{lll}{b}_2\hfill & =\hfill & -\left(B_{00}+B_{11}+B_{22}\right)\hfill \\ \hfill & =\hfill & -\left\{{\xi }_0+2{\xi }_1+M_0+2M_1+2\left(M_0^{\prime }+M_1^{\prime }+M_1^{″}\right)\left(\text{cos}{k}_{x}a+\text{cos}{k}_{y}a\right)\right\},\hfill \end{array}$$

(40)$$\begin{array}{lll}{b}_1\hfill & =\hfill & B_{00}{B}_{11}+B_{11}{B}_{22}+B_{22}{B}_{00}-{\left|B_{01}\right|}^2-{\left|B_{12}\right|}^2-{\left|B_{20}\right|}^2\hfill \\ \hfill & =\hfill & \left({\xi }_1+M_1\right)\left(2{\xi }_0+{\xi }_1+2M_0+M_1\right)\hfill \\ \hfill & \hfill & +2\left\{\left({\xi }_1+M_1\right)\left(2M_0^{\prime }+M_1^{\prime }+M_1^{″}\right)+\left({\xi }_0+M_0\right)\left(M_1^{\prime }+M_1^{″}\right)\right\}\left(\text{cos}{k}_{x}a+\text{cos}{k}_{y}a\right)\hfill \\ \hfill & \hfill & +4\left\{M_1^{\prime }{M}_1^{″}+M_0^{\prime }\left(M_1^{\prime }+M_1^{″}\right)\right\}\left({\text{cos}}^2{k}_{x}a+{\text{cos}}^2{k}_{y}a\right)\hfill \\ \hfill & \hfill & +4\left\{\left(M_1^{\prime 2}+M_1^{″2}\right)+2M_0^{\prime }\left(M_1^{\prime }+M_1^{″}\right)\right\}\text{cos}{k}_{x}a\text{cos}{k}_{y}a\hfill \\ \hfill & \hfill & -4{\left|M_2\right|}^2\left({\text{sin}}^2{k}_{x}a+{\text{sin}}^2{k}_{y}a\right),\hfill \end{array}$$

(41)$$\begin{array}{lll}{b}_0\hfill & =\hfill & -2\text{Re}\left\{B_{01}{B}_{12}{B}_{20}\right\}-B_{00}{B}_{11}{B}_{22}+\left(B_{00}{\left|B_{12}\right|}^2+B_{11}{\left|B_{20}\right|}^2+B_{22}{\left|B_{01}\right|}^2\right)\hfill \\ \hfill & =\hfill & -\{\left({\xi }_0+M_0\right){\left({\xi }_1+M_1\right)}^2\hfill \\ \hfill & \hfill & +2\left\{M_0^{\prime }{\left({\xi }_1+M_1\right)}^2+\left({\xi }_0+M_0\right)\left({\xi }_1+M_1\right)\left(M_1^{\prime }+M_1^{″}\right)\right\}\left(\text{cos}{k}_{x}a+\text{cos}{k}_{y}a\right)\hfill \\ \hfill & \hfill & +4\left\{\left({\xi }_0+M_0\right)M_1^{\prime }{M}_1^{″}+M_0^{\prime }\left({\xi }_1+M_1\right)\left(M_1^{\prime }+M_1^{″}\right)\right\}\left({\text{cos}}^2{k}_{x}a+{\text{cos}}^2{k}_{y}a\right)\hfill \\ \hfill & \hfill & +4\left\{\left({\xi }_0+M_0\right)\left(M_1^{\prime 2}+M_1^{″2}\right)+2M_0^{\prime }\left({\xi }_1+M_1\right)\left(M_1^{\prime }+M_1^{″}\right)\right\}\text{cos}{k}_{x}a\text{cos}{k}_{y}a\hfill \\ \hfill & \hfill & +8M_0^{\prime }\left(M_1^{\prime 2}+M_1^{″2}+M_1^{\prime }{M}_1^{″}\right)\left({\text{cos}}^2{k}_{x}a\text{cos}{k}_{y}a+\text{cos}{k}_{x}a{\text{cos}}^2{k}_{y}a\right)\hfill \\ \hfill & \hfill & +8M_0^{\prime }{M}_1^{\prime }{M}_1^{″}\left({\text{cos}}^3{k}_{x}a+{\text{cos}}^3{k}_{y}a\right)\}.\hfill \end{array}$$

For the Γ point (k = 0) of the Brillouin zone, Eq. (38) can be solved easily. Its three solutions are

(42)$$\xi =\{\begin{array}{l}{\xi }_0+M_0+4M_0^{\prime }\equiv {\xi }_{\mathrm{\Gamma }}^{\left(0\right)},\hfill \\ {\xi }_1+M_1+2\left(M_1^{\prime }+M_1^{″}\right)\equiv {\xi }_{{}^{\mathrm{\Gamma }}}^{\left(1\right)}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(\text{double}\hspace{0.17em}\text{root}\right).\hfill \end{array}$$

To solve Eq. (38) for general k, we change the unknown variable from ξ to

(43)$$\eta =\xi +\frac{b_2}{3}.$$

Then, Eq. (38) is reduced to

(44)$${\eta }^3+p\eta +q=0,$$

where

(45)$$p=b_1-\frac{b_2^2}{3},$$

(46)$$q=b_0-\frac{b_1{b}_2}{3}+\frac{2b_2^3}{27}.$$

To evaluate the k dependence of the solution in the vicinity of the Γ point, we calculate p and q to the third order of k_x and k_y. The results are

(47)$$p=-\frac{1}{3}{\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)}^2+\left\{\frac{1}{3}\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)\left(M_1^{\prime }+M_1^{″}-2M_0^{\prime }\right)-4{\left|M_2\right|}^2\right\}{k}^2{a}^2,$$

(48)$$\begin{array}{lll}q\hfill & =\hfill & \frac{2}{27}{\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)}^3\hfill \\ \hfill & \hfill & -\frac{1}{9}\left\{{\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)}^2\left(M_1^{\prime }+M_1^{″}-2M_0^{\prime }\right)-12\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right){\left|M_2\right|}^2\right\}{k}^2{a}^2,\hfill \end{array}$$

where $k=\sqrt{k_x^2+k_y^2}$.

We consider the case of accidental degeneracy of the E and A₁ modes at the Γ point in the following. So, we assume

(49)$${\xi }_{\mathrm{\Gamma }}^{\left(1\right)}={\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\equiv {\xi }_{\mathrm{\Gamma }}.$$

Then, from Eqs. (39), (47), and (48),

(50)$$b_2=-3{\xi }_{\mathrm{\Gamma }}+{Mk}^2{a}^2,$$

(51)$$p=-4{\left|M_2\right|}^2{k}^2{a}^2,$$

(52)$$q=0,$$

where

(53)$$M=M_0^{\prime }+M_1^{\prime }+M_1^{″}.$$

Thus, we obtain

(54)$$\xi =\{\begin{array}{l}{\xi }_{\mathrm{\Gamma }}\pm 2\left|M_2\right|ka-{Mk}^2{a}^2/3,\hfill \\ {\xi }_{\mathrm{\Gamma }}-{Mk}^2{a}^2/3.\hfill \end{array}$$

In the vicinity of the Γ point where $ck\ll {\omega }_{\mathrm{\Gamma }}\equiv c\sqrt{{\xi }_{\mathrm{\Gamma }}}$, we obtain

(55)$$\begin{array}{lll}{\omega }_{\mathbf{k}}\hfill & =\hfill & c\sqrt{\xi }\hfill \\ \hfill & =\hfill & \{\begin{array}{l}{\omega }_{\mathrm{\Gamma }}\pm \left|M_2\right|a{c}^{2}k/{\omega }_{\mathrm{\Gamma }}-{Ma}^2{c}^2{k}^2/6{\omega }_{\mathrm{\Gamma }},\hfill \\ {\omega }_{\mathrm{\Gamma }}-{Ma}^2{c}^2{k}^2/6{\omega }_{\mathrm{\Gamma }}.\hfill \end{array}\hfill \end{array}$$

B. T₁ and A_1g modes of simple cubic lattice

In three dimensions, we define L by

(56)$$L_{lmn}^{\left(ij\right)}\equiv \frac{1}{V}{\int }_{V}d\mathbf{r}{\mathbf{H}}^{\left(i\right)*}\left(\mathbf{r}\right)\cdot 𝒧{\mathbf{H}}^{\left(j\right)}\left(\mathbf{r}-{\mathbf{r}}_{lmn}\right),$$

where r_lmn are lattice vectors of the simple cubic lattice:

(57)$${\mathbf{r}}_{lm}=l\left(\begin{array}{c}a\\ 0\\ 0\end{array}\right)+m\left(\begin{array}{c}0\\ a\\ 0\end{array}\right)+n\left(\begin{array}{c}0\\ 0\\ a\end{array}\right).$$

A polynomial representation of the T_1u mode is given by {x, y, z} [27]. So, we can choose three eigenfunctions of the T_1u symmetry, H⁽¹⁾(r), H⁽²⁾(r), and H⁽³⁾(r), such that they are transformed as the following three functions when any R ∈ O_h, whose elements are illustrated in Fig. 5, is operated.

(58)$$f_1\left(\mathbf{r}\right)=x,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{f}_2\left(\mathbf{r}\right)=y,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\text{and}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{f}_3\left(\mathbf{r}\right)=z.$$

By definition, they are transformed as

(59)$$R{f}_i\left(\mathbf{r}\right)=f_i\left(R^{-1}\mathbf{r}\right).$$

So, for example,

(60)$${\sigma }_x{f}_1\left(\mathbf{r}\right)=-x=-f_1\left(\mathbf{r}\right),$$

(61)$${\sigma }_x{f}_2\left(\mathbf{r}\right)=y=f_2\left(\mathbf{r}\right),$$

(62)$${\sigma }_x{f}_3\left(\mathbf{r}\right)=z=f_3\left(\mathbf{r}\right).$$

We write these three equations in a matrix form:

(63)$${\sigma }_x=\left(\begin{array}{c}{f}_1\\ f_2\\ f_3\end{array}\right)=\left(\begin{array}{ccc}-1& 0& 0\\ 0,& 1& 0\\ 0,& 0& 1\end{array}\right)\left(\begin{array}{c}{f}_1\\ f_2\\ f_3\end{array}\right).$$

 

Fig. 5 Symmetry operations for the O_h point group

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In this way, we can introduce the matrix representation of R ∈ O_h:

(64)$$\begin{array}{llll}{\sigma }_x:\hfill & \left(\begin{array}{ccc}-1,& 0,& 0\\ 0,& 1,& 0\\ 0,& 0,& 1\end{array}\right),\hfill & C_{4x}:\hfill & \left(\begin{array}{ccc}1,& 0,& 0\\ 0,& 0,& 1\\ 0,& -1,& 0\end{array}\right),\hfill \end{array}$$

(65)$$\begin{array}{llll}{\sigma }_y:\hfill & \left(\begin{array}{ccc}1,& 0,& 0\\ 0,& -1,& 0\\ 0,& 0,& 1\end{array}\right),\hfill & C_{4y}:\hfill & \left(\begin{array}{ccc}0,& 0,& -1\\ 0,& 1,& 0\\ 1,& 0,& 0\end{array}\right),\hfill \end{array}$$

(66)$$\begin{array}{llll}{\sigma }_z:\hfill & \left(\begin{array}{ccc}1,& 0,& 0\\ 0,& 1,& 0\\ 0,& 0,& -1\end{array}\right),\hfill & C_{4z}:\hfill & \left(\begin{array}{ccc}0,& 1,& 0\\ -1,& 0,& 0\\ 0,& 0,& 1\end{array}\right),\hfill \end{array}$$

(67)$$\begin{array}{llll}{C}_{4x}^{-1}:\hfill & \left(\begin{array}{ccc}1,& 0,& 0\\ 0,& 0,& -1\\ 0,& 1,& 0\end{array}\right),\hfill & C_{2x}:\hfill & \left(\begin{array}{ccc}1,& 0,& 0\\ 0,& -1,& 0\\ 0,& 0,& -1\end{array}\right),\hfill \end{array}$$

(68)$$\begin{array}{llll}{C}_{4y}^{-1}:\hfill & \left(\begin{array}{ccc}0,& 0,& 1\\ 0,& 1,& 0\\ -1,& 0,& 0\end{array}\right),\hfill & C_{2y}:\hfill & \left(\begin{array}{ccc}-1,& 0,& 0\\ 0,& 1,& 0\\ 0,& 0,& -1\end{array}\right),\hfill \end{array}$$

(69)$$\begin{array}{llll}{C}_{4z}^{-1}:\hfill & \left(\begin{array}{ccc}0,& -1,& 0\\ 1,& 0,& 0\\ 0,& 0,& 1\end{array}\right),\hfill & C_{2z}:\hfill & \left(\begin{array}{ccc}-1,& 0,& 0\\ 0,& -1,& 0\\ 0,& 0,& 1\end{array}\right),\hfill \end{array}$$

(70)$$\begin{array}{llll}{C}_{2\left(110\right)}:\hfill & \left(\begin{array}{ccc}0,& 1,& 0\\ 1,& 0,& 0\\ 0,& 0,& -1\end{array}\right),\hfill & C_{2\left(1\overline{1}0\right)}:\hfill & \left(\begin{array}{ccc}0,& -1,& 0\\ -1,& 0,& 0\\ 0,& 0,& -1\end{array}\right),\hfill \end{array}$$

(71)$$\begin{array}{llll}{C}_{2\left(011\right)}:\hfill & \left(\begin{array}{ccc}-1,& 0,& 0\\ 0,& 0,& 1\\ 0,& 1,& 0\end{array}\right),\hfill & C_{2\left(01\overline{1}\right)}:\hfill & \left(\begin{array}{ccc}-1,& 0,& 0\\ 0,& 0,& -1\\ 0,& -1,& 0\end{array}\right),\hfill \end{array}$$

(72)$$\begin{array}{llll}{C}_{2\left(101\right)}:\hfill & \left(\begin{array}{ccc}0,& 0,& 1\\ 0,& -1,& 0\\ 1,& 0,& 0\end{array}\right),\hfill & C_{2\left(\overline{1}01\right)}:\hfill & \left(\begin{array}{ccc}0,& 0,& -1\\ 0,& -1,& 0\\ -1,& 0,& 0\end{array}\right),\hfill \end{array}$$

(73)$$\begin{array}{llll}{C}_{3\left(111\right)}:\hfill & \left(\begin{array}{ccc}0,& 1,& 0\\ 0,& 0,& 1\\ 1,& 0,& 0\end{array}\right),\hfill & C_{3\left(111\right)}^{-1}:\hfill & \left(\begin{array}{ccc}0,& 0,& 1\\ 1,& 0,& 0\\ 0,& 1,& 0\end{array}\right),\hfill \end{array}$$

(74)$$\begin{array}{llll}{C}_{3\left(\overline{1}11\right)}:\hfill & \left(\begin{array}{ccc}0,& 0,& -1\\ -1,& 0,& 0\\ 0,& 1,& 0\end{array}\right),\hfill & C_{3\left(\overline{1}11\right)}^{-1}:\hfill & \left(\begin{array}{ccc}0,& -1,& 0\\ 0,& 0,& 1\\ -1,& 0,& 0\end{array}\right),\hfill \end{array}$$

(75)$$\begin{array}{llll}{C}_{3\left(1\overline{1}1\right)}:\hfill & \left(\begin{array}{ccc}0,& 0,& 1\\ -1,& 0,& 0\\ 0,& -1,& 0\end{array}\right),\hfill & C_{3\left(1\overline{1}1\right)}^{-1}:\hfill & \left(\begin{array}{ccc}0,& -1,& 0\\ 0,& 0,& -1\\ 1,& 0,& 0\end{array}\right),\hfill \end{array}$$

(76)$$\begin{array}{llll}{C}_{3\left(11\overline{1}\right)}:\hfill & \left(\begin{array}{ccc}0,& 0,& -1\\ 1,& 0,& 0\\ 0,& -1,& 0\end{array}\right),\hfill & C_{3\left(11\overline{1}\right)}^{-1}:\hfill & \left(\begin{array}{ccc}0,& 1,& 0\\ 0,& 0,& -1\\ -1,& 0,& 0\end{array}\right),\hfill \end{array}$$

(77)$$\begin{array}{llll}I:\hfill & \left(\begin{array}{ccc}-1,& 0,& 0\\ 0,& -1,& 0\\ 0,& 0,& -1\end{array}\right),\hfill & E:\hfill & \left(\begin{array}{ccc}1,& 0,& 0\\ 0,& 1,& 0\\ 0,& 0,& 1\end{array}\right),\hfill \end{array}$$

(78)$$\begin{array}{llll}I{C}_{4x}:\hfill & \left(\begin{array}{ccc}-1,& 0,& 0\\ 0,& 0,& -1\\ 0,& 1,& 0\end{array}\right),\hfill & I{C}_{4x}^{-1}:\hfill & \left(\begin{array}{ccc}-1,& 0,& 0\\ 0,& 0,& 1\\ 0,& -1,& 0\end{array}\right),\hfill \end{array}$$

(79)$$\begin{array}{llll}I{C}_{4y}:\hfill & \left(\begin{array}{ccc}0,& 0,& 1\\ 0,& -1,& 0\\ -1,& 0,& 0\end{array}\right),\hfill & I{C}_{4y}^{-1}:\hfill & \left(\begin{array}{ccc}0,& 0,& -1\\ 0,& -1,& 0\\ 1,& 0,& 0\end{array}\right),\hfill \end{array}$$

(80)$$\begin{array}{llll}I{C}_{4z}:\hfill & \left(\begin{array}{ccc}0,& -1,& 0\\ 1,& 0,& 0\\ 0,& 0,& -1\end{array}\right),\hfill & I{C}_{4z}^{-1}:\hfill & \left(\begin{array}{ccc}0,& 1,& 0\\ -1,& 0,& 0\\ 0,& 0,& -1\end{array}\right),\hfill \end{array}$$

(81)$$\begin{array}{llll}{\sigma }_{dx}^{\prime }:\hfill & \left(\begin{array}{ccc}1,& 0,& 0\\ 0,& 0,& 1\\ 0,& 1,& 0\end{array}\right),\hfill & {\sigma }_{dx}^{″}:\hfill & \left(\begin{array}{ccc}1,& 0,& 0\\ 0,& 0,& -1\\ 0,& -1,& 0\end{array}\right),\hfill \end{array}$$

(82)$$\begin{array}{llll}{\sigma }_{dy}^{\prime }:\hfill & \left(\begin{array}{ccc}0,& 0,& 1\\ 0,& 1,& 0\\ 1,& 0,& 0\end{array}\right),\hfill & {\sigma }_{dy}^{″}:\hfill & \left(\begin{array}{ccc}0,& 0,& -1\\ 0,& 1,& 0\\ -1,& 0,& 0\end{array}\right),\hfill \end{array}$$

(83)$$\begin{array}{llll}{\sigma }_{dz}^{\prime }:\hfill & \left(\begin{array}{ccc}0,& 1,& 0\\ 1,& 0,& 0\\ 0,& 0,& 1\end{array}\right),\hfill & {\sigma }_{dz}^{″}:\hfill & \left(\begin{array}{ccc}0,& -1,& 0\\ -1,& 0,& 0\\ 0,& 0,& 1\end{array}\right),\hfill \end{array}$$

(84)$$\begin{array}{llll}I{C}_{3\left(111\right)}:\hfill & \left(\begin{array}{ccc}0,& -1,& 0\\ 0,& 0,& -1\\ -1,& 0,& 0\end{array}\right),\hfill & I{C}_{3\left(111\right)}^{-1}:\hfill & \left(\begin{array}{ccc}0,& 0,& -1\\ -1,& 0,& 0\\ 0,& -1,& 0\end{array}\right),\hfill \end{array}$$

(85)$$\begin{array}{llll}I{C}_{3\left(\overline{1}11\right)}:\hfill & \left(\begin{array}{ccc}0,& 0,& 1\\ 1,& 0,& 0\\ 0,& -1,& 0\end{array}\right),\hfill & I{C}_{3\left(\overline{1}11\right)}^{-1}:\hfill & \left(\begin{array}{ccc}0,& 1,& 0\\ 0,& 0,& -1\\ 1,& 0,& 0\end{array}\right),\hfill \end{array}$$

(86)$$\begin{array}{llll}I{C}_{3\left(1\overline{1}1\right)}:\hfill & \left(\begin{array}{ccc}0,& 0,& -1\\ 1,& 0,& 0\\ 0,& 1,& 0\end{array}\right),\hfill & I{C}_{3\left(1\overline{1}1\right)}^{-1}:\hfill & \left(\begin{array}{ccc}0,& 1,& 0\\ 0,& 0,& 1\\ -1,& 0,& 0\end{array}\right),\hfill \end{array}$$

(87)$$\begin{array}{llll}I{C}_{3\left(11\overline{1}\right)}:\hfill & \left(\begin{array}{ccc}0,& 0,& 1\\ -1,& 0,& 0\\ 0,& 1,& 0\end{array}\right),\hfill & I{C}_{3\left(11\overline{1}\right)}^{-1}:\hfill & \left(\begin{array}{ccc}0,& -1,& 0\\ 0,& 0,& 1\\ 1,& 0,& 0\end{array}\right),\hfill \end{array}$$

It is easily confirmed that this matrix representation gives the correct values of characters for the T_1u representation in Table 1, which are the trace (sum of the diagonal elements) of the above matrices. Because {H⁽ⁱ⁾} (i = 1, 2, 3) are also a T_1u representation, they transform like {f_i}. But we should note that there is a difference due to the vector nature of the former. Their transformation is defined as

(88)$$\left[R{\mathbf{H}}^{\left(i\right)}\right]\left(\mathbf{r}\right)\equiv R{\mathbf{H}}^{\left(i\right)}\left(R^{-1}\mathbf{r}\right).$$

Table 1. Character table for the A_1g and T_1u modes of the O_h point group

View Table

Let us proceed to the derivation of relations among $L_{lnm}^{\left(ij\right)}$ introduced by Eq. (56). As the first example, let us change the variable of integration from r to r′ = C_4yr and evaluate $L_{000}^{\left(11\right)}$. Since C_4y does not change the size of volume elements, we have

(89)$$\begin{array}{lll}V{L}_{000}^{\left(11\right)}\hfill & =\hfill & {\int }_{V}d{r}^{\prime }{\mathbf{H}}^{\left(1\right)*}\left(C_{4y}^{-1}{r}^{\prime }\right)\cdot \left[𝒧{\mathbf{H}}^{\left(1\right)}\left(C_{4y}^{-1}{r}^{\prime }\right)\right]\hfill \\ \hfill & =\hfill & {\int }_{V}d{r}^{\prime }\left[C_{4y}^{-1}{C}_{4y}{\mathbf{H}}^{\left(1\right)*}\left(C_{4y}^{-1}{r}^{\prime }\right)\right]\cdot \left[C_{4y}^{-1}{C}_{4y}𝒧C_{4y}^{-1}{C}_{4y}{\mathbf{H}}^{\left(1\right)}\left(C_{4y}^{-1}{r}^{\prime }\right)\right],\hfill \end{array}$$

where we introduced three pairs of $C_{4y}^{-1}{C}_{4y}\left(\equiv E\right)$ for later use. Since $C_{4y}^{-1}$ does not change the value of inner products, we obtain

(90)$$V{L}_{000}^{\left(11\right)}={\int }_{V}d{r}^{\prime }\left[C_{4y}{\mathbf{H}}^{\left(1\right)*}\left(C_{4y}^{-1}{r}^{\prime }\right)\right]\cdot \left[{𝒧}^{\prime }{C}_{4y}{\mathbf{H}}^{\left(1\right)}\left(C_{4y}^{-1}{r}^{\prime }\right)\right],$$

where 𝒧′ is defined as

(91)$${𝒧}^{\prime }=C_{4y}𝒧C_{4y}^{-1}.$$

It is an operator in the r′ coordinate system equivalent to 𝒧 in the r coordinate system. Substituting the second relation of Eq. (65) and Eq. (88), we obtain

(92)$$\begin{array}{lll}V{L}_{00}^{\left(11\right)}\hfill & =\hfill & {\int }_{V}d{r}^{\prime }\left[-{\mathbf{H}}^{\left(3\right)*}\left(r^{\prime }\right)\right]\cdot {𝒧}^{\prime }\left[-{\mathbf{H}}^{\left(3\right)}{\left(\mathbf{r}\right)}^{\prime }\right]\hfill \\ \hfill & =\hfill & V{L}_{000}^{\left(33\right)}.\hfill \end{array}$$

Similarly, by changing the variable from r to r′ = C_4zr and using the second relation of Eq. (66), we obtain

(93)$$L_{000}^{\left(11\right)}=L_{000}^{\left(22\right)}.$$

So, we denote them by

(94)$$L_{000}^{\left(11\right)}=L_{000}^{\left(22\right)}=L_{000}^{\left(33\right)}\equiv \frac{{\omega }_1^2}{c^2}+M_1,$$

where the first term on the right-hand side is the original value for an isolated unit structure.

Next, let us examine $L_{000}^{\left(12\right)}$. By changing the variable of integration from r to r′ = C_4xr and using the second relation of Eq. (64), we obtain

(95)$$\begin{array}{lll}V{L}_{000}^{\left(12\right)}\hfill & =\hfill & {\int }_{V}d{r}^{\prime }\left[C_{4x}{\mathbf{H}}^{\left(1\right)*}\left(C_{4x}^{-1}{r}^{\prime }\right)\right]\cdot \left[{𝒧}^{\prime }{C}_{4x}{\mathbf{H}}^{\left(2\right)}\left(C_{4x}^{-1}{r}^{\prime }\right)\right]\hfill \\ \hfill & =\hfill & {\int }_{V}d{r}^{\prime }{\mathbf{H}}^{\left(1\right)*}\left(r^{\prime }\right)\cdot {𝒧}^{\prime }{\mathbf{H}}^{\left(3\right)}\left(r^{\prime }\right)\hfill \\ \hfill & =\hfill & V{L}_{000}^{\left(13\right)}.\hfill \end{array}$$

Note that ${𝒧}^{\prime }=C_{4x}𝒧C_{4x}^{-1}$ for this case. On the other hand, by changing the variable of integration from r to $r^{\prime }=C_{4x}^{-1}\mathbf{r}$ and using the first relation of Eq. (67), we obtain

(96)$$\begin{array}{lll}V{L}_{000}^{\left(12\right)}\hfill & =\hfill & {\int }_{V}d{r}^{\prime }\left[C_{4x}^{-1}{\mathbf{H}}^{\left(1\right)*}\left(C_{4x}{r}^{\prime }\right)\right]\cdot \left[{𝒧}^{\prime }{C}_{4x}^{-1}{\mathbf{H}}^{\left(2\right)}\left(C_{4x}{r}^{\prime }\right)\right]\hfill \\ \hfill & =\hfill & {\int }_{V}d{r}^{\prime }\left[C_{4x}^{-1}{\mathbf{H}}^{\left(1\right)*}\right]\left(r^{\prime }\right)\cdot \left[{𝒧}^{\prime }{C}_{4x}^{-1}{\mathbf{H}}^{\left(2\right)}\right]\left(r^{\prime }\right)\hfill \\ \hfill & =\hfill & {\int }_{V}d{r}^{\prime }{\mathbf{H}}^{\left(1\right)*}\left(r^{\prime }\right)\cdot {𝒧}^{\prime }\left[-{\mathbf{H}}^{\left(3\right)}\left(r^{\prime }\right)\right]\hfill \\ \hfill & =\hfill & -V{L}_{000}^{\left(13\right)}.\hfill \end{array}$$

So, from Eqs. (95) and (96), we obtain

(97)$$L_{000}^{\left(12\right)}=L_{000}^{\left(13\right)}=0.$$

By similar calculations, we also obtain

(98)$$L_{000}^{\left(21\right)}=L_{000}^{\left(31\right)}=L_{000}^{\left(23\right)}=L_{000}^{\left(32\right)}=0.$$

Let us proceed to the evaluation of $L_{100}^{\left(11\right)}$. Using r′ = C_4yr,

(99)$$\begin{array}{lll}V{L}_{100}^{\left(11\right)}\hfill & =\hfill & {\int }_{V}d{r}^{\prime }\left[C_{4y}{\mathbf{H}}^{\left(1\right)*}\right]\left(r^{\prime }\right)\cdot {𝒧}^{\prime }\left[C_{4y}{\mathbf{H}}^{\left(1\right)}\right]\left(r^{\prime }-C_{4y}{\mathbf{r}}_{100}\right)\hfill \\ \hfill & =\hfill & {\int }_{V}d{r}^{\prime }\left[-{\mathbf{H}}^{\left(3\right)*}\left(r^{\prime }\right)\right]\cdot {𝒧}^{\prime }\left[-{\mathbf{H}}^{\left(3\right)}\left(r^{\prime }-{\mathbf{r}}_{00,-1}\right)\right]\hfill \\ \hfill & =\hfill & V{L}_{00,-1}^{\left(33\right)}.\hfill \end{array}$$

On the other hand, by changing the variable of integration from r to $r^{\prime }=C_{4y}^{-1}\mathbf{r}$, we obtain

(100)$$\begin{array}{lll}V{L}_{100}^{\left(11\right)}\hfill & =\hfill & {\int }_{V}d{r}^{\prime }\left[C_{4y}^{-1}{\mathbf{H}}^{\left(1\right)*}\right]\left(r^{\prime }\right)\cdot {𝒧}^{\prime }\left[C_{4y}^{-1}{\mathbf{H}}^{\left(1\right)}\right]\left(r^{\prime }-C_{4y}^{-1}{\mathbf{r}}_{100}\right)\hfill \\ \hfill & =\hfill & V{L}_{001}^{\left(33\right)}.\hfill \end{array}$$

These two relations and similarly calculations for C_4z, $C_{4z}^{-1}$, and C_2y lead to

(101)$$L_{\pm 1,00}^{\left(11\right)}=L_{0,\pm 1,0}^{\left(22\right)}=L_{00,\pm 1}^{\left(33\right)}\equiv M_1^{\prime }.$$

Application of all other R ∈ O_h gives the same result as Eq. (101). As for $L_{010}^{\left(11\right)}$, we may repeat similar calculations and obtain

(102)$$L_{0,\pm 1,0}^{\left(11\right)}=L_{00,\pm 1}^{\left(11\right)}=L_{\pm 1,00}^{\left(22\right)}=L_{00,\pm 1}^{\left(22\right)}=L_{\pm 1,00}^{\left(33\right)}=L_{0,\pm 1,0}^{\left(33\right)}\equiv M_1^{″}.$$

Next, let us examine $L_{100}^{\left(12\right)}$. Using r′ = C_4xr,

(103)$$\begin{array}{lll}V{L}_{100}^{\left(12\right)}\hfill & =\hfill & {\int }_{V}d{r}^{\prime }\left[C_{4x}{\mathbf{H}}^{\left(1\right)*}\right]\left(r^{\prime }\right)\cdot {𝒧}^{\prime }\left[C_{4x}{\mathbf{H}}^{\left(2\right)}\right]\left(r^{\prime }-C_{4x}{\mathbf{r}}_{100}\right)\hfill \\ \hfill & =\hfill & {\int }_{V}d{r}^{\prime }{\mathbf{H}}^{\left(1\right)*}\left(r^{\prime }\right)\cdot {𝒧}^{\prime }{\mathbf{H}}^{\left(3\right)}\left(r^{\prime }-{\mathbf{r}}_{100}\right)\hfill \\ \hfill & =\hfill & V{L}_{100}^{\left(13\right)}.\hfill \end{array}$$

On the other hand, by changing the variable of integration from r to $r^{\prime }=C_{4x}^{-1}\mathbf{r}$, we obtain

(104)$$\begin{array}{lll}V{L}_{100}^{\left(12\right)}\hfill & =\hfill & {\int }_{V}d{r}^{\prime }\left[C_{4x}^{-1}{\mathbf{H}}^{\left(1\right)*}\right]\left(r^{\prime }\right)\cdot {𝒧}^{\prime }\left[C_{4x}^{-1}{\mathbf{H}}^{\left(2\right)}\right]\left(r^{\prime }-C_{4x}^{-1}{\mathbf{r}}_{100}\right)\hfill \\ \hfill & =\hfill & -V{L}_{100}^{\left(13\right)}.\hfill \end{array}$$

So, we obtain

(105)$$L_{100}^{\left(12\right)}=L_{100}^{\left(13\right)}=0.$$

This equation and similar calculations for all other R ∈ O_h lead to

(106)$$\begin{array}{lll}{L}_{\pm 1,00}^{\left(12\right)}\hfill & =\hfill & L_{0,\pm 1,0}^{\left(12\right)}=L_{\pm 1,00}^{\left(21\right)}=L_{0,\pm 1,0}^{\left(21\right)}=L_{\pm 1,00}^{\left(13\right)}=L_{00,\pm 1}^{\left(13\right)}\hfill \\ \hfill & =\hfill & L_{\pm 1,00}^{\left(31\right)}=L_{00,\pm 1}^{\left(31\right)}=L_{0,\pm 1,0}^{\left(23\right)}=L_{00,\pm 1}^{\left(23\right)}=L_{0,\pm 1,0}^{\left(32\right)}=L_{00,\pm 1}^{\left(32\right)}\hfill \\ \hfill & =\hfill & 0,\hfill \end{array}$$

where we used the following relation:

(107)$$L_{lmn}^{\left(ij\right)}=L_{-l,-m,-n}^{\left(ji\right)*}.$$

As for $L_{001}^{\left(12\right)}$, we change the variable of integration from r to r′ = σ_xr and obtain

(108)$$\begin{array}{lll}V{L}_{001}^{\left(12\right)}\hfill & =\hfill & {\int }_{V}d{r}^{\prime }\left[{\sigma }_x{\mathbf{H}}^{\left(1\right)*}\right]\left(r^{\prime }\right)\cdot {𝒧}^{\prime }\left[{\sigma }_x{\mathbf{H}}^{\left(2\right)}\right]\left(r^{\prime }-{\sigma }_x{\mathbf{r}}_{001}\right)\hfill \\ \hfill & =\hfill & {\int }_{V}d{r}^{\prime }\left[-{\mathbf{H}}^{\left(1\right)*}\left(r^{\prime }\right)\right]\cdot {𝒧}^{\prime }{\mathbf{H}}^{\left(2\right)}\left(r^{\prime }-{\mathbf{r}}_{001}\right)\hfill \\ \hfill & =\hfill & -V{L}_{001}^{\left(12\right)}.\hfill \end{array}$$

So, we obtain $L_{001}^{\left(12\right)}=0$. Similar calculations lead to

(109)$$L_{00,\pm 1}^{\left(12\right)}=L_{00,\pm 1}^{\left(21\right)}=L_{0,\pm 1,0}^{\left(13\right)}=L_{0,\pm 1,0}^{\left(31\right)}=L_{\pm 1,00}^{\left(23\right)}=L_{\pm 1,00}^{\left(32\right)}=0.$$

Equations (94), (97), (98), (101), (102), (106), and (109) give all relations among the three functions of the T_1u mode.

Now let us proceed to the A_1g mode. Its eigen function H⁽⁰⁾ is invariant when any R ∈ O_h is applied as can be found in Table 1:

(110)$$\left[R{\mathbf{H}}^{\left(0\right)}\right]\left(\mathbf{r}\right)={\mathbf{H}}^{\left(0\right)}\left(\mathbf{r}\right).$$

First, let us denote $L_{000}^{\left(00\right)}$ by

(111)$$L_{000}^{\left(00\right)}\equiv \frac{{\omega }_0^2}{c^2}+M_0.$$

Next, let us examine $L_{100}^{\left(00\right)}$. By changing the variable of integration from r to r′ = σ_xr, we obtain

(112)$$\begin{array}{lll}V{L}_{100}^{\left(00\right)}\hfill & =\hfill & {\int }_{V}d{r}^{\prime }{\mathbf{H}}^{\left(0\right)*}\left(r^{\prime }\right)\cdot {𝒧}^{\prime }{\mathbf{H}}^{\left(0\right)}\left(r^{\prime }-{\sigma }_x{\mathbf{r}}_{100}\right)\hfill \\ \hfill & =\hfill & V{L}_{-1,00}^{\left(00\right)}.\hfill \end{array}$$

In addition to this equation, by applying C₃₍₁₁₁₎, $C_{3\left(111\right)}^{-1}$, σ_y, and σ_z, we obtain

(113)$$L_{\pm 1,00}^{\left(00\right)}=L_{0,\pm 1,0}^{\left(00\right)}=L_{00,\pm 1}^{\left(00\right)}\equiv M_0^{\prime }.$$

By using Eq. (107), we can prove

(114)$$M_0^{\prime *}=M_0^{\prime },$$

so M′₀ is a real number.

As for cross terms between the T_1u and A_1g modes, let us start with $L_{000}^{\left(01\right)}$. By changing the variable of integration from r to r′ = σ_xr, we can easily prove

(115)$$L_{000}^{\left(01\right)}=0.$$

Similar calculations lead to

(116)$$L_{000}^{\left(01\right)}=L_{000}^{\left(10\right)}=L_{000}^{\left(02\right)}=L_{000}^{\left(20\right)}=L_{000}^{\left(03\right)}=L_{000}^{\left(30\right)}=0.$$

Also using σ_x, we obtain

(117)$$L_{010}^{\left(01\right)}=0.$$

Similar calculations in this case result in

(118)$$\begin{array}{l}{L}_{0,\pm 1,0}^{\left(01\right)}=L_{00,\pm 1}^{\left(01\right)}=L_{0,\pm 1,0}^{\left(10\right)}=L_{00,\pm 1}^{\left(10\right)}=L_{\pm 1,00}^{\left(02\right)}=L_{00,\pm 1}^{\left(02\right)}\\ =L_{\pm 1,00}^{\left(20\right)}=L_{00,\pm 1}^{\left(20\right)}=L_{\pm 1,00}^{\left(03\right)}=L_{0,\pm 1,0}^{\left(03\right)}=L_{\pm 1,00}^{\left(30\right)}=L_{0,\pm 1,0}^{\left(30\right)}\\ =0.\end{array}$$

Next, let us examine $L_{100}^{\left(01\right)}$. By changing the variable of integration from r to r′ = σ_xr, we have

(119)$$\begin{array}{lll}V{L}_{100}^{\left(01\right)}\hfill & =\hfill & {\int }_{V}d{r}^{\prime }\left[{\sigma }_x{\mathbf{H}}^{\left(0\right)*}\right]\left(r^{\prime }\right)\cdot {𝒧}^{\prime }\left[{\sigma }_x{\mathbf{H}}^{\left(1\right)}\right]\left(r^{\prime }-{\sigma }_x{\mathbf{r}}_{100}\right)\hfill \\ \hfill & =\hfill & {\int }_{V}d{r}^{\prime }{\mathbf{H}}^{\left(0\right)*}\left(r^{\prime }\right)\cdot {𝒧}^{\prime }\left[-{\mathbf{H}}^{\left(1\right)}\left(r^{\prime }-{\mathbf{r}}_{-1,00}\right)\right]\hfill \\ \hfill & =\hfill & -V{L}_{-1,00}^{\left(01\right)}.\hfill \end{array}$$

Similarly, by using C₃₍₁₁₁₎ and $C_{3\left(111\right)}^{-1}$, and further applying σ_y and σ_z, we obtain

(120)$$\pm L_{\pm 1,00}^{\left(01\right)}=\pm L_{0,\pm 1,0}^{\left(02\right)}=\pm L_{00,\pm 1}^{\left(03\right)}\equiv M_2.$$

Then, from Eq. (107), we also obtain

(121)$$\mp L_{\pm 1,00}^{\left(10\right)}=\mp L_{0,\pm 1,0}^{\left(20\right)}=\mp L_{00,\pm 1}^{\left(30\right)}\equiv M_2^{*}.$$

In summary, we have the following relations:

(122)$$L_{000}^{\left(00\right)}\equiv \frac{{\omega }_0^2}{c^2}+M_0,$$

(123)$$L_{\pm 1,00}^{\left(00\right)}=L_{0,\pm 1,0}^{\left(00\right)}=L_{00,\pm 1}^{\left(00\right)}\equiv M_0^{\prime },$$

(124)$$L_{000}^{\left(11\right)}=L_{000}^{\left(22\right)}=L_{000}^{\left(33\right)}\equiv \frac{{\omega }_1^2}{c^2}+M_1,$$

(125)$$L_{\pm 1,00}^{\left(11\right)}=L_{0,\pm 1,0}^{\left(22\right)}=L_{00,\pm 1}^{\left(33\right)}\equiv M_1^{\prime },$$

(126)$$L_{0,\pm 1,0}^{\left(11\right)}=L_{00,\pm 1}^{\left(11\right)}=L_{\pm 1,00}^{\left(22\right)}=L_{00,\pm 1}^{\left(22\right)}=L_{\pm 1,00}^{\left(33\right)}=L_{0,\pm 1,0}^{\left(33\right)}\equiv M_1^{″},$$

(127)$$\pm L_{\pm 1,00}^{\left(01\right)}=\mp L_{\pm 1,00}^{\left(10\right)*}=\pm L_{0,\pm 1,0}^{\left(02\right)}=\mp L_{0,\pm 1,0}^{\left(20\right)*}=\pm L_{00,\pm 1}^{\left(03\right)}=\mp L_{00,\pm 1}^{\left(30\right)*}\equiv M_2.$$

All other $L_{lmn}^{\left(ij\right)}$ for the nearest neighbor lattice points are equal to zero. So, the number of independent elements among 112 $L_{lmn}^{\left(ij\right)}$ is six.

The Bloch function in the present case is given by a linear combination of the eigenfunctions of A_1g mode (H⁽⁰⁾) and T_1u mode (H⁽¹⁾, H⁽²⁾, H⁽³⁾):

(128)$${\mathbf{H}}_{\mathbf{k}}\left(\mathbf{r}\right)=\frac{1}{V}\sum _{l,m,n}{e}^{i\mathbf{k}\cdot {\mathbf{r}}_{lmn}}\sum _{i=0}^3{A}_i{\mathbf{H}}^{\left(i\right)}\left(\mathbf{r}-{\mathbf{r}}_{lmn}\right).$$

Then, the secular equation is derived as before by keeping the nearest neighbor terms. The Hermitian matrix B that should be diagonalized is now a 4 × 4 matrix. Its elements are given by

(129)$$B_{ij}=\sum _{lmn}{e}^{ia\left(l{k}_x+m{k}_y+n{k}_z\right)}{L}_{lmn}^{\left(ij\right)}.$$

When we keep the nearest neighbor terms, we obtain Eqs. (17) – (23). We denote

(130)$$\xi =\frac{{\omega }_{\mathbf{k}}^2}{c^2},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\xi }_0=\frac{{\omega }_0^2}{c^2},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\xi }_1=\frac{{\omega }_1^2}{c^2}$$

as before. Then the secular equation reduces to

(131)$${\xi }^4+b_3{\xi }^3+b_2{\xi }^2+b_1\xi +b_0=0,$$

where

(132)$$b_3=-\left(B_{00}+B_{11}+B_{22}+B_{33}\right),$$

(133)$$\begin{array}{l}{b}_2=B_{00}{B}_{11}+B_{00}{B}_{22}+B_{00}{B}_{33}+B_{11}{B}_{22}+B_{11}{B}_{33}+B_{22}{B}_{33}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-\left({\left|B_{01}\right|}^2+{\left|B_{02}\right|}^2+{\left|B_{03}\right|}^2\right),\end{array}$$

(134)$$\begin{array}{l}{b}_1=-\left(B_{00}{B}_{11}{B}_{22}+B_{11}{B}_{22}{B}_{33}+B_{22}{B}_{33}{B}_{00}+B_{33}{B}_{00}{B}_{11}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+{\left|B_{01}\right|}^2\left(B_{22}+B_{33}\right)+{\left|B_{02}\right|}^2\left(B_{11}+B_{33}\right)+{\left|B_{03}\right|}^2\left(B_{11}+B_{22}\right),\end{array}$$

(135)$$b_0=B_{00}{B}_{11}{B}_{22}{B}_{33}-\left({\left|B_{01}\right|}^2{B}_{22}{B}_{33}+{\left|B_{02}\right|}^2{B}_{11}{B}_{33}+{\left|B_{03}\right|}^2{B}_{11}{B}_{22}\right).$$

For the Γ point (k = 0) of the Brillouin zone, the secular equation, Eq. (131), can be solved easily because the off-diagonal elements of matrix B are all vanishing. Its four solutions are

(136)$$\xi =\{\begin{array}{l}{\xi }_0+M_0+6M_0^{\prime }\equiv {\xi }_{\mathrm{\Gamma }}^{\left(0\right)},\hfill \\ {\xi }_1+M_1+2M_1^{\prime }+4M_1^{″}\equiv {\xi }_{\mathrm{\Gamma }}^{\left(1\right)}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(\text{triple}\hspace{0.17em}\text{root}\right).\hfill \end{array}$$

To solve Eq. (131) for general k, we change the unknown variable from ξ to

(137)$$\eta =\xi +\frac{b_3}{4}.$$

Then, Eq. (131) is reduced to

(138)$${\eta }^4+p{\eta }^2+q\eta +r=0,$$

where

(139)$$p=b_2-\frac{3b_3^2}{8},$$

(140)$$q=b_1-\frac{b_2{b}_3}{2}+\frac{b_3^3}{8},$$

(141)$$r=b_0-\frac{b_1{b}_3}{4}+\frac{b_2{b}_3^2}{16}-\frac{3b_3^4}{256}.$$

To evaluate the k dependence of the solution in the vicinity of the Γ point, p, q, and r were evaluated to the second, third, and fourth order of k_x, k_y and k_z, respectively. The results are

(142)$$\begin{array}{l}p=-\frac{3}{8}{\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)}^2\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+\frac{1}{4}\left\{\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)\left(M_1^{\prime }+2M_1^{″}-3M_0^{\prime }\right)-16{\left|M_2\right|}^2\right\}{k}^2{a}^2\end{array}$$

(143)$$\begin{array}{l}q=\frac{1}{8}{\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)}^3\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-\frac{1}{8}\left\{{\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)}^2\left(M_1^{\prime }+2M_1^{″}-3M_0^{\prime }\right)-16\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right){\left|M_2\right|}^2\right\}{k}^2{a}^2\end{array}$$

(144)$$\begin{array}{l}r=-\frac{3}{256}{\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)}^4\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+\left\{\frac{1}{64}{\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)}^3\left(M_1^{\prime }+2M_1^{″}-3M_0^{\prime }\right)-\frac{1}{4}{\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)}^2{\left|M_2\right|}^2\right\}{k}^2{a}^2\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+\{\frac{1}{128}{\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)}^2\left[8\left(M_1^{\prime }+2M_1^{″}\right)M_0^{\prime }+\left(M_1^{\prime }+2M_1^{″}+M_0^{\prime }\right)\left(7M_1^{\prime }+14M_1^{″}-9M_0^{\prime }\right)\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-\frac{1}{2}\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)\left(M_1^{\prime }+2M_1^{″}+M_0^{\prime }\right){\left|M_2\right|}^2\}{k}^4{a}^4\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+\{-\frac{1}{768}{\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)}^3\left(M_1^{\prime }+2M_1^{″}-3M_0^{\prime }\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+{\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)}^2\left[-\frac{3}{16}\left(2M_1^{\prime }+M_1^{″}\right)M_1^{″}+\frac{1}{12}{\left|M_2\right|}^2\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+2\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)M_1^{″}{\left|M_2\right|}^2\}\left(k_x^4+k_y^4+k_z^4\right)a^4\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+\{-\frac{3}{16}{\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)}^2\left(M_1^{\prime 2}+2M_1^{\prime }{M}_1^{″}+3M_1^{″2}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+2\left({\xi }_{\mathrm{\Gamma }}^{\left(1\right)}-{\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\right)\left(M_1^{\prime }+M_1^{″}\right){\left|M_2\right|}^2\}\left(k_x^2{k}_y^2+k_y^2{k}_z^2+k_z^2{k}_x^2\right)a^4,\end{array}$$

where $k=\sqrt{k_x^2+k_y^2+k_z^2}$.

We consider the case of accidental degeneracy of the T_1u and A_1g modes at the Γ point in the following. So, we assume

(145)$${\xi }_{\mathrm{\Gamma }}^{\left(1\right)}={\xi }_{\mathrm{\Gamma }}^{\left(0\right)}\equiv {\xi }_{\mathrm{\Gamma }}.$$

Then, from Eqs. (142) – (144),

(146)$$p=-4{\left|M_2\right|}^2{k}^2{a}^2,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}q=r=0.$$

Since

(147)$$b_3=-4{\xi }_{\mathrm{\Gamma }}+M^{\prime }{k}^2{a}^2,$$

where M′ = M′₁ + 2M″₁ + M′₀, the solutions of the secular equation, Eq. (131), are

(148)$$\xi =\{\begin{array}{l}{\xi }_{\mathrm{\Gamma }}\pm 2\left|M_2\right|ka-M^{\prime }{k}^2{a}^2/4\hfill \\ {\xi }_{\mathrm{\Gamma }}-M^{\prime }{k}^2{a}^2/4\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(\text{double}\hspace{0.17em}\text{root}\right).\hfill \end{array}$$

So, in the vicinity of the Γ point where $ck\ll {\omega }_{\mathrm{\Gamma }}\equiv c\sqrt{{\xi }_{\mathrm{\Gamma }}}$, we obtain

(149)$${\omega }_{\mathbf{k}}=\{\begin{array}{l}{\omega }_{\mathrm{\Gamma }}\pm \left|M_2\right|a{c}^{2}k/{\omega }_{\mathrm{\Gamma }}-M^{\prime }{a}^2{c}^2{k}^2/8{\omega }_{\mathrm{\Gamma }},\hfill \\ {\omega }_{\mathrm{\Gamma }}-M^{\prime }{a}^2{c}^2{k}^2/8{\omega }_{\mathrm{\Gamma }}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(\text{double}\hspace{0.17em}\text{root}\right).\hfill \end{array}$$

Acknowledgments

This study was supported by a Grant-in-Aid for Scientific Research on Innovative Areas from the Japanese Ministry of Education, Culture, Sports and Technology (Grant number 22109007).

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