Example for group theory 1

How to find the irreducible basis for $\Gamma^{-}_{4}$?

The character table of the point group $O_h$ for $T_{1u}$ ($\Gamma^{-}_{4}$) is give by:

	$E$	$8C_{3}$	$6C’_{2}$	$6C_{4}$	$3C_{2}$	$I$	$6S_{4}$	$8S_{6}$	$3\sigma_{h}$	$6\sigma_{d}$
$T_{1u}$	$3$	$0$	$-1$	$1$	$-1$	$-3$	$-1$	$0$	$1$	$1$

For a given simple cubic lattice with bases located at each lattice point, let us discuss how $\ket{\phi_{1}}$ is written as the irrep. basis of $T_{1u}$. In order to do that, one should use the projection operator:

\[\begin{equation} P^{\mu} = \frac{d_{\mu}}{|G|}\sum_{g \in O_{h}}\chi^{\mu}(g)R(g) \end{equation}\]

where $d_\mu$ is the dimension of $\mu$th irrep, $

G

$ is the order of the group, $\chi^{\mu}(g)$ is the character of $\mu$th irrep, and $R(g)$ is the symmetry operator. The dimension of $T_{1u}$ irrep is given by $\chi^{T_{1u}}(E) = 3$, and the order of the group $

G

$ is $48$.

The next step is to find where the basis at site $1$, $\ket{\phi_{1}}$ maps to for each group element $g$:

• $g = E$.

\[\begin{equation} \chi^{T_{1u}}(E)R(E)\ket{\phi_{1}} = 3\times\ket{\phi_{1}} \end{equation}\]

• $g = 6C’_{2}$.

The $g = 6C’_{2}$ are the rotations about midpoint axes. Referring BCS, one can figure out where does it explicitly map to. For example, a rotation $C’_{2}$ along the axis $[011]$ transforms $(x,y,z)$ to $(-x,z,y)$. Since one has $\ket{\phi_{1}}$ at $(-,-,-)$, acting this operator yields $(+,-,-)$ which corresponds to $\ket{\phi_{2}}$. Doing for all cases:

\[\begin{equation} \chi^{T_{1u}}(6C'_{2})R(6C'_{2})\ket{\phi_{1}} = (-1)\times\left[\ket{\phi_{2}} + \ket{\phi_{4}} + \ket{\phi_{5}} + 3\ket{\phi_{7}}\right] \end{equation}\]

• $g = 6C_{4}$.

\[\begin{equation} \chi^{T_{1u}}(6C_{4})R(6C_{4})\ket{\phi_{1}} = (+1)\times\left[2\ket{\phi_{2}} + 2\ket{\phi_{4}} + 2\ket{\phi_{5}}\right] \end{equation}\]

• $g = 3C_{2}$.

\[\begin{equation} \chi^{T_{1u}}(3C_{2})R(3C_{2})\ket{\phi_{1}} = (-1)\times\left[\ket{\phi_{3}} + \ket{\phi_{6}} + \ket{\phi_{8}}\right] \end{equation}\]

• $g = I$.

\[\begin{equation} \chi^{T_{1u}}(I)R(I)\ket{\phi_{1}} = (-3)\times\ket{\phi_{7}} \end{equation}\]

• $g = 6S_{4}~(C_{4}\times\sigma_{\perp})$.

\[\begin{equation} \chi^{T_{1u}}(6S_{4})R(6S_{4})\ket{\phi_{1}} = (-1)\times\left[2\ket{\phi_{3}} + 2\ket{\phi_{6}} + 2\ket{\phi_{8}}\right] \end{equation}\]

• $g = 3\sigma_{h}$.

\[\begin{equation} \chi^{T_{1u}}(3\sigma_{h})R(3\sigma_{3})\ket{\phi_{1}} = (+1)\times\left[\ket{\phi_{2}} + \ket{\phi_{4}} + \ket{\phi_{5}}\right] \end{equation}\]

• $g = 6\sigma_{d}$.

\[\begin{equation} \chi^{T_{1u}}(6\sigma_{d})R(6\sigma_{d})\ket{\phi_{1}} = (+1)\times\left[\ket{\phi_{3}} + \ket{\phi_{6}} + \ket{\phi_{8}} + 3\ket{\phi_{1}}\right] \end{equation}\]

The $6$ mirror planes for $g = 6\sigma_{d}$ are shown in Fig.~. Finally, summing all them yields:

\[\begin{align} P^{T_{1u}}\ket{\phi_{1}} &= \frac{1}{16}\left[6\ket{\phi_{1}}+2\ket{\phi_{2}}-2\ket{\phi_{3}}+2\ket{\phi_{4}}+2\ket{\phi_{5}}-2\ket{\phi_{6}}-6\ket{\phi_{7}}-2\ket{\phi_{8}}\right] \nonumber \\ &= \frac{1}{8}\left[3\ket{\phi_{1}}+\ket{\phi_{2}}-\ket{\phi_{3}}+\ket{\phi_{4}}+\ket{\phi_{5}}-\ket{\phi_{6}}-3\ket{\phi_{7}}-\ket{\phi_{8}}\right] \end{align}\]

Some cautions. While doing so, all the rotation axes should be the same. If a lattice point maps to outside of the unit cell, one reduces it to the home unit cell by the lattice vector $\mathbf{R}$. However, this generates an additional phase factor $e^{i\mathbf{k}\cdot\mathbf{R}}$ at $\mathbf{k} \neq \mathbf{0}$.