T-signature or T-simplex, and signature or simplex

Let us now consider operator $\hat{\cal{O}}$ which is even or odd with respect to one of the six antilinear, $\hat{\cal{Z}}^2$= $\hat{\cal{E}}$, operators (see Sec. 3.1.2), and simultaneously even or odd with respect to one of the six linear, $\hat{\cal{X}}^2$= $\bar{\cal{E}}$= $-\hat{\cal{E}}$, operators (see Sec. 3.1.3). In such a case, simplification of the single-particle basis is possible only for pairs of $\hat{\cal{Z}}$ and $\hat{\cal{X}}$ operators which correspond to two different Cartesian directions. Indeed, focusing our attention on signatures, $\hat{\cal{R}}_{k}^T$ commutes with $\hat{\cal{R}}_{k}$, and therefore (being antilinear) it flips the $\hat{\cal{R}}_{k}$ signature quantum number. Therefore, the eigenstates of $\hat{\cal{R}}_{k}$ cannot be eigenstates of $\hat{\cal{R}}_{k}^T$. We may then only work either in the basis of eigenstates of $\hat{\cal{R}}_{k}^T$, Sec. 3.1.2, or in the basis of eigenstates of $\hat{\cal{R}}_{k}$, Sec. 3.1.3. On the other hand, for l$\neq$k $\hat{\cal{R}}_{l}^T$ anticommutes with $\hat{\cal{R}}_{k}$, and therefore, it conserves the $\hat{\cal{R}}_{k}$ signature quantum number. Hence, the eigenstates of $\hat{\cal{R}}_{k}$ can be rendered the eigenstates of $\hat{\cal{R}}_{l}^T$ by a suitable choice of phases.

It is easy to check that after multiplying eigenstates listed in Table 4 by the following phase factors:

 

$$\Phi_{lk} \exp\left\{i\zeta{\textstyle\frac{\pi}{2}}(N_l+1)\right\}, \mbox{~~~~~~~~~~~~~~~~~~for~} k<l,$$ = (27)

$$\Phi_{lk} \exp\left\{i\zeta{\textstyle\frac{\pi}{4}}+ i\zeta{\textstyle\frac{\pi}{2}}(N_l+N_k+1)\right\}, \mbox{~~for~} l<k,$$ = (28)

one obtains the basis states,

$$\vert\mbox{{\boldmath {$n$ }}}\,\zeta\rangle_{lk}=\Phi_{lk}\vert\mbox{{\boldmath {$n$ }}}\,\zeta\rangle_{k},$$ (29)

which simultaneously fulfill Eqs. (9), (14), and (19), i.e.,

 

$$\hat{\cal{T}}\vert\mbox{{\boldmath {$n$ }}}\,\zeta\rangle_{lk} \zeta\vert\mbox{{\boldmath {$n$ }}}\,-\!\zeta\rangle_{lk},$$ = (30)

$$\hat{\cal{Z}}_l\vert\mbox{{\boldmath {$n$ }}}\,\zeta\rangle_{lk} \vert\mbox{{\boldmath {$n$ }}}\,\zeta\rangle_{lk},$$ = (31)

$$\hat{\cal{X}}_k\vert\mbox{{\boldmath {$n$ }}}\,\zeta\rangle_{lk} i\zeta\vert\mbox{{\boldmath {$n$ }}}\,\zeta\rangle_{lk}.$$ = (32)

Here, $\hat{\cal{Z}}_l$ stands for $\hat{\cal{R}}_{l}^T$ or $\hat{\cal{S}}_{l}^T$, and $\hat{\cal{X}}_k$ stands for $\hat{\cal{R}}_{k}$ or $\hat{\cal{S}}_{k}$. In Eqs. (27), symbols (N_x,N_y,N_z) refer to (n_x,n_y,n_z) for the $\hat{\cal{S}}_{l}^T$ or $\hat{\cal{S}}_{k}$operators, and to (n_y+n_z,n_x+n_z,n_x+n_y) for the $\hat{\cal{R}}_{l}^T$ or $\hat{\cal{R}}_{k}$ operators. Moreover, a circular ordering of Cartesian directions is assumed, i.e., x<y<z<x, in order to define conditions l<k and k<l.

For operators $\hat{\cal{O}}$ even or odd simultaneously with respect to $\hat{\cal{Z}}_l$ and $\hat{\cal{X}}_k$, see Eqs. (15) and (18), bases defined by Eq. (29) allow for a very simple forms of matrices ${\cal{O}}$. Combining conditions (17) and (21) one obtains block diagonal and real matrix elements, e.g. for $\epsilon_Z$=+1,

$$\left(\begin{array}{cc} A & 0 \\ 0 & B \end{array} \rig... ...array}{cc} 0 & Y \\ \widetilde{Y} & 0 \end{array} \right),$$ (33)

for $\epsilon_X$=+1 and $\epsilon_X$=-1, respectively, with A and B real symmetric, and Y arbitrary real matrix.