Matrix representation of rank-2 spinors

[grzz]

TL;DR

Relating spinors m^AB and m^BA.

Assume that all spinors are 2-component ones.
If spinor m^AB = n^Ak^B, what are the matrices representing spinors m^AB and m^BA?

 

[anuttarasammyak]

I have read the matrix representation in Landau and Lifshitz.

 

[julian]

Write

\begin{align*}
n^A &=
\begin{pmatrix}
n^0\\
n^1
\end{pmatrix},
&
k^A &=
\begin{pmatrix}
k^0\\
k^1
\end{pmatrix}.
\end{align*}

If

\begin{align*}
m^{AB} = n^A k^B,
\end{align*}

then the matrix representing $m^{AB}$, with row index $A$ and column index$B$, is

\begin{align*}
[m^{AB}]
&=
\begin{pmatrix}
m^{00} & m^{01}\\
m^{10} & m^{11}
\end{pmatrix} \\
&=
\begin{pmatrix}
n^0 k^0 & n^0 k^1\\
n^1 k^0 & n^1 k^1
\end{pmatrix}.
\end{align*}

This is just the outer product: $m = nk^T$.

Now $m^{BA}$ means the spinor obtained by interchanging the two indices. Since

\begin{align*}
m^{BA}=n^B k^A,
\end{align*}

its components, displayed again with row index $A$ and column index $B$, are

\begin{align*}
[m^{BA}]
&=
\begin{pmatrix}
m^{00} & m^{10}\\
m^{01} & m^{11}
\end{pmatrix} \\
&=
\begin{pmatrix}
n^0 k^0 & n^1 k^0\\
n^0 k^1 & n^1 k^1
\end{pmatrix}.
\end{align*}

Thus,

\begin{align*}
[m^{BA}] = [m^{AB}]^T.
\end{align*}