first, we use Cartesian coordinate to define the matrix representation of the Mirror, 180 degree Rotation and Parity.

M_x=    \left(  \begin{array}{ccc}  -1 & 0 & 0 \\  0 & 1 & 0 \\  0 & 0 & 1  \end{array}  \right)

R_x=    \left(  \begin{array}{ccc}  1 & 0 & 0 \\  0 & -1 & 0 \\  0 & 0 & -1  \end{array}  \right)

P=    \left(  \begin{array}{ccc}  -1 & 0 & 0 \\  0 & -1 & 0 \\  0 & 0 & -1  \end{array}  \right)

i guess the reader can think the other matrix.

the group table is:

\left(  \begin{array}{cccccccc}  i & M_x & M_y & M_z & R_x & R_x & R_z & P \\  M_x & i & R_z & R_x & P & M_z & M_y & R_x \\  M_y & R_z & i & R_x & M_z & P & M_x & R_x \\  M_z & R_x & R_x & i & M_y & M_x & P & R_z \\  R_x & P & M_z & M_y & i & R_z & R_x & M_x \\  R_x & M_z & P & M_x & R_z & i & R_x & M_y \\  R_z & M_y & M_x & P & R_x & R_x & i & M_z \\  P & R_x & R_x & R_x & M_x & M_y & M_z & i  \end{array}  \right)

where i is identity matrix. we can see it is a group of order 8. there is a sub-group of (i, R_x, R_y, R_z) . since the group table are diagonal symmetry, each element from its own class.

since a rotational group can be represented by 2X2 matrix instead of 3X3 by introducing complex number. those matrix is Pauli’s spinor.

\sigma_x=    \left(  \begin{array}{cc}  0 & -1 \\  1 & 0  \end{array}  \right)

the transform from 3X3 matrix to 2X2 matrix is that…… then we can map the mirror matrix and parity into 2X2 matrix.

Advertisements