Given the matrices
$\gamma_{k}=\begin{bmatrix} O & -i\sigma_{k}\\ +i\sigma_{k}& O \end{bmatrix}$
where $\sigma_{k}$ is the $k^{th}$ Pauli matrix
$\gamma_{4}=\begin{bmatrix} I^{2} &0 \\ 0 & -I^{2} \end{bmatrix}$
$\gamma_{5}=\gamma_{1}.\gamma_{2}.\gamma_{3}.\gamma_{4}$
The anticommutator rule is defined by $\left [ x,y \right ]=xy+yx$
Show that the anticommutator relation $\gamma_{u}.\gamma_{v}+\gamma_{v}.\gamma_{u}=2\delta _{u v}I$ is satisfied for all $u,v=1,2,3,4$ where $I$ is the $4 \times 4$ identity matrix.
What I have defined is
\[ScriptCapitalO] = {{0, 0}, {0, 0}};
Subscript[\[Gamma], 1] = {{\[ScriptCapitalO], -I PauliMatrix[1]},
{I PauliMatrix[1], \[ScriptCapitalO]}};
Subscript[\[Gamma], 2] = {{\[ScriptCapitalO], -I PauliMatrix[2]},
{I PauliMatrix[2], \[ScriptCapitalO]}};
Subscript[\[Gamma], 3] = {{\[ScriptCapitalO], -I PauliMatrix[3]},
{I PauliMatrix[3], \[ScriptCapitalO]}};
Subscript[\[Gamma], 4] = {{IdentityMatrix[2], \[ScriptCapitalO]},
{\[ScriptCapitalO], -IdentityMatrix[2]}};
Subscript[\[Gamma], 5] =
Subscript[\[Gamma], 1].Subscript[\[Gamma], 2].Subscript[\[Gamma],3].Subscript[\[Gamma], 4];
One way to do this would be to show the identity holds individually. But this would be tedious. Can someone help me with a more efficient and general way to this?
ArrayFlatten[]
, BTW? $\endgroup$ – J. M.'s ennui♦ Apr 8 '17 at 11:56