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?
Answer
Define DiracMatrix
:
DiracMatrix[k_] /; k == 1 || k == 2 || k == 3 :=
ArrayFlatten[{{0, - I PauliMatrix[k]}, {I PauliMatrix[k], 0}}]
DiracMatrix[4] := ArrayFlatten[{{ IdentityMatrix[2], 0}, {0, -IdentityMatrix[2]}}]
DiracMatrix[5] := Dot @@ Table[DiracMatrix[k], {k, 4}]
To prove the identity we could check e.g.
And @@ Flatten @
Table[ DiracMatrix[i].DiracMatrix[k] + DiracMatrix[k].DiracMatrix[i]
== 2 KroneckerDelta[i, k] IdentityMatrix[4], {i, 4}, {k, 4}]
True
sometimes one would prefer a visual test, something like e.g.
L[i_, k_] := DiracMatrix[i].DiracMatrix[k] + DiracMatrix[k].DiracMatrix[i] -
2 KroneckerDelta[i, k] IdentityMatrix[4]
Table[ L[i, k] // MatrixForm, {i, 4}, {k, 4}] // MatrixForm
Comments
Post a Comment