1
$\begingroup$

In the set of SUSY notes I'm following, the Pauli operator is given as: ${(\sigma^\mu)}_{\alpha\dot{\alpha}} = (I_2, \sigma^1, \sigma^2, \sigma^3)$.

The antisymmetric tensor that lowers and raises indices is defined as: $\epsilon^{\alpha\beta} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$

When we define the barred Pauli operator by applying this twice,

${(\bar{\sigma}^\mu)}^{\dot{\alpha}\alpha} = \epsilon^{\alpha\beta}\epsilon^{\dot{\alpha}\dot{\beta}} {(\sigma^\mu)}_{\beta\dot{\beta}} = (I_2, -\sigma^1, -\sigma^2, -\sigma^3)$.

I really struggle with Spinor indices, and keep trying to think of them as matrices and which I'm sure is incredibly wrong. For example, acting on this for ${(\bar{\sigma}^0)}^{\dot{\alpha}\alpha} = \epsilon^{\alpha\beta}\epsilon^{\dot{\alpha}\dot{\beta}} {(\sigma^0)}_{\beta\dot{\beta}}$, I have written

$\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}^{\alpha\beta}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}^{\dot{\alpha}\dot{\beta}}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}_{\beta\dot{\beta}} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}^{\dot{\alpha}\alpha} = (-I_2)^{\dot{\alpha}\alpha}$,

whereas looking at the notes I should not have a minus sign. How does applying the epsilon tensor leave the first element of the Pauli operator ($I_2$) intact but applying a minus sign to the rest of the barred tensor? Or is this just a convention that we define?

Apologies if there are any errors in the TeX, still learning! Any help on this area would be great, really think I'm doing this wrong by trying to write things out in matrices and contract Spinor indices.

$\endgroup$
1
$\begingroup$

Multiplying matrices is not correct here. You need to work these things out using the definition, $(\bar{\sigma}^\mu)^{\dot{\alpha}\alpha}=\epsilon^{\alpha\beta}\epsilon^{\dot{\alpha}\dot{\beta}}(\sigma^\mu)_{\beta\dot{\beta}}$. We have, for example, $$(\bar{\sigma}^0)^{11}=\epsilon^{12}\epsilon^{12}(\sigma^0)_{22}=(\sigma^0)_{22}=1,$$ while $$(\bar{\sigma}^1)^{12}=\epsilon^{12}\epsilon^{21}(\sigma^1)_{21}=-(\sigma^1)_{21}=-1.$$ This shows how the negative sign is obtained in $\bar{\sigma}^1$ but not in $\bar{\sigma}^0$. The other cases can be worked out similarly.

$\endgroup$
0
0
$\begingroup$

I know this post is old, but I just wanted to remark that it is possible to use matrix multiplication here, provided one is somewhat careful. The equation

$(\bar{\sigma}^\mu)^{\dot{\alpha}\alpha} = \epsilon^{\alpha\beta} \epsilon^{\dot{\alpha}\dot{\beta}} (\sigma^\mu)_{\beta\dot{\beta}} = -\epsilon^{\alpha\beta} (\sigma^\mu)_{\beta\dot{\beta}} \,\epsilon^{\beta \alpha}$

can be written using matrix multiplication as

$(\bar{\sigma}^\mu)^T = -\epsilon (\sigma^\mu) \epsilon$

or, equivalently,

$\bar{\sigma}^\mu = -\epsilon (\sigma^\mu)^T \epsilon$

since $(A^T)^T = A$, $(ABC)^T = C^T B^T A^T$ and $\epsilon^T = -\epsilon$ ($\epsilon$ is skew-symmetric) for any square matrices $A$, $B$ and $C$ of the same size.

A short calculation shows that, if

$A = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$

then

$-\epsilon A^T \epsilon = \left( \begin{array}{cc} d & -b \\ -c & a \end{array} \right)$.

Thus $\bar{\sigma}^0 = I$ and $\bar{\sigma}^i = - \sigma^i$ using matrix notation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.