0
$\begingroup$

There are some definitions and properties for Pauli matrices and their combinations:

$$ \varepsilon^{\alpha \beta } = \varepsilon^{\dot {\alpha} \dot {\beta} } = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}_{\alpha \beta }, \quad \varepsilon_{\dot {\alpha} \dot {\beta}} = \varepsilon_{\alpha \beta} = -\varepsilon^{\alpha \beta }, $$ $$ (\sigma^{\mu})_{\alpha \dot {\alpha} } = (\hat {\mathbf E} , \hat {\mathbf \sigma})^{\mu}_{\alpha \dot {\alpha}}, \quad (\tilde {\sigma}^{\mu})^{\dot {\beta } \beta } = \varepsilon^{\alpha \beta}\varepsilon^{\dot {\alpha }\dot {\beta }}(\sigma^{\mu})_{\alpha \dot {\alpha}} = (\hat {\mathbf E}, -\hat {\mathbf \sigma})^{\mu , \dot {\beta }\beta }, $$ $$ (\sigma^{\mu \nu})_{\alpha \beta} = -\frac{1}{4}\left( (\sigma^{\mu} \tilde {\sigma}^{\nu})_{\alpha \beta } - (\sigma^{\nu} \tilde {\sigma}^{\mu})_{\alpha \beta }\right), \quad (\tilde {\sigma}^{\mu \nu})_{\dot {\alpha }\dot {\beta }} = -\frac{1}{4}\left( (\tilde{\sigma}^{\mu} \sigma^{\nu})_{\dot {\alpha} \dot {\beta} } - (\tilde {\sigma}^{\nu} \sigma^{\mu})_{\dot {\alpha} \dot {\beta} }\right), $$ $$ (\tilde {\sigma}^{\mu})^{\dot {\alpha }\alpha}(\sigma_{\mu})_{\beta \dot {\beta }} = 2\delta^{\dot {\alpha}}_{\dot {\beta}}\delta^{\alpha}_{\beta}. $$ How to show, that $$ (\sigma^{\alpha \beta})_{a b }(\tilde {\sigma}^{\mu \nu})_{\dot {c} \dot {d}}g_{\alpha \mu} = 0? $$ (It helps to show, that spinor irreducible representation of the generators of Lorentz group expands on two spinor subgroups).

My attempt.

I tried to show this, but only got following. $$ (\sigma^{\alpha \beta})_{a b }(\tilde {\sigma}^{\mu \nu})_{\dot {c} \dot {d}}g_{\alpha \mu} = \frac{1}{16}\left( (\sigma^{\alpha})_{a \dot {n}}(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\tilde {\sigma}^{\mu})_{\dot {c}}^{\quad m }(\sigma^{\nu })_{m \dot {d}} - (\sigma^{\alpha})_{a \dot {n}}(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\tilde {\sigma}^{\nu})_{\dot {c}}^{ \quad m}(\sigma^{\mu })_{m \dot {d}}\right) $$ $$ + \frac{1}{16}\left(-(\sigma^{\beta })_{a \dot {n}}(\tilde {\sigma}^{\alpha })^{ \dot {n}}_{\quad b}(\tilde {\sigma}^{\mu})_{\dot {c}}^{\quad m}(\sigma^{\nu })_{m \dot {d}} + (\sigma^{\beta })_{a \dot {n}}(\tilde {\sigma}^{\alpha })^{ \dot {n}}_{\quad b}(\tilde {\sigma}^{\nu})_{\dot {c}}^{\quad m }(\sigma^{\mu })_{m \dot {d}} \right)g_{\alpha \mu} $$ After that I transformed each summand like $$ (\sigma^{\alpha})_{a \dot {n}}(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\tilde {\sigma}^{\mu})_{\dot {c}}^{\quad m }(\sigma^{\nu })_{m \dot {d}}g_{\alpha \mu} = (\sigma^{\alpha})_{a \dot {n}}(\tilde {\sigma}_{\alpha})_{\dot {c}}^{\quad m }(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\sigma^{\nu })_{m \dot {d}} = $$ $$ = \varepsilon_{\dot {c}\dot {\gamma}}(\sigma^{\alpha})_{a \dot {n}}(\tilde {\sigma}_{\alpha})^{\dot {\gamma} m }(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\sigma^{\nu })_{m \dot {d}} = 2\varepsilon_{\dot {c}\dot {\gamma }}\delta^{\dot {\gamma}}_{\dot {m}}\delta^{n}_{a}(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\sigma^{\nu })_{m \dot {d}} = $$ $$ 2\varepsilon_{\dot {c}\dot {m}}(\tilde {\sigma}^{\beta })^{ \dot {m}}_{\quad b}(\sigma^{\nu })_{a \dot {d}} = 2(\sigma^{\beta } )_{b \dot {c}}(\sigma^{\nu })_{a \dot {d}}. $$ Finally, I got $$ (\sigma^{\alpha \beta})_{a b }(\tilde {\sigma}^{\mu \nu})_{\dot {c} \dot {d}}g_{\alpha \mu} = \frac{1}{8}\left( (\sigma^{\beta})_{b\dot {c}}(\sigma^{\nu})_{a\dot {d}} + (\sigma^{\beta })_{b \dot {d}}(\sigma^{\nu})_{a \dot {c}} + (\sigma^{\beta })_{a \dot {c}}(\sigma^{\nu})_{b \dot {d}} + (\sigma^{\beta})_{a \dot {d}}(\sigma^{\nu})_{b \dot {c}}\right). $$ What to do next?

$\endgroup$
  • $\begingroup$ Why the downvoted? $\endgroup$ – Abhimanyu Pallavi Sudhir Aug 28 '13 at 15:47
  • $\begingroup$ @DImension10AbhimanyuPS . What did you mean? $\endgroup$ – user8817 Aug 28 '13 at 15:56
  • $\begingroup$ I meant "Why the downvote?". I can see there's a downvote on this answer. Just click on the vote count... Oh, wait, you don't have 1000 rep, so you can't see up-down vote counts....... $\endgroup$ – Abhimanyu Pallavi Sudhir Aug 28 '13 at 16:32
1
$\begingroup$

Note, that, on the third line, the $\beta$ indice of $\sigma^{\mu\nu}$ and the $\dot \beta$ indice of $\tilde \sigma^{\mu\nu}$ must be raised for indice coherence. Same error for the following lines.

The formula you want to demonstrate is certainly false. Take $\beta = \nu = 0$, and noting that $\sigma^{\alpha0}= -\frac{1}{2}\sigma^\alpha,\tilde \sigma^{\mu0}= \frac{1}{2}\sigma^\mu $, if the formula was exact, it would imply (with $g_{11}=g_{22}=g_{33}$), and in short tensorial notation :

$-\frac{1}{4} \vec \sigma \otimes \vec \sigma = 0\tag{1}$

which is obviously false.

$\endgroup$
  • $\begingroup$ I don't understand, why indices must be raised. Why it must be? Maybe it is easy to lower these indices by $\varepsilon$ matrice, and it isn't be an error. Why not? $\endgroup$ – user8817 Aug 28 '13 at 16:55
  • 2
    $\begingroup$ @PhysiXxx : With your definitions, in the second line, of $\sigma^{\mu\nu}$ and $\tilde \sigma^{\mu\nu}$, you have to raise the $\beta, \dot \beta$ indices. If you don't want, you have to insert $\epsilon$ matrices, with lower indices, for coherence, but I think it will bring unnecessary complexity, at least, in this specific question. So, it is better to raise the indices. $\endgroup$ – Trimok Aug 28 '13 at 17:04
  • $\begingroup$ By the way, this result (non-zero convolution) is strange. According to this, an algebra of two irreducible spinor tensors (generators) of the Lorentz group isn't splitted. $\endgroup$ – user8817 Aug 28 '13 at 17:37

Your Answer

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