Understanding Fierz rearrangement identity

Question

I'm trying to get a better grasp of the Fierz rearrangement identity for 2-component spinors (Equation 2.20 I'll be using the Van der Waerden notionation used in the given link) $$ \chi_\alpha (\xi \eta) + \xi_\alpha (\eta \chi) + \eta_\alpha (\chi \xi) = 0. $$

Now I can easily prove that this is true by brute force using the anti-commutation relations of the spinors and expanding the dot product.

However I was looking for a more elegant reason of why this should be true. I believe there should be some group theoretical reason as to why this identity occurs rather than a useful coincidence.

This is in hopes that I may be able to generalize the relation to different spinor representations (Dirac, Majorana) and higher dimensions.

This post imported from StackExchange Physics at 2015-06-15 18:11 (UTC), posted by SE-user Peter Anderson

Answer 1

One way to see the expression vanishes (maybe the one you already know how to do?) is to use the fact that there can not exist a 2*2*2 totally antisymmetric matrix. First, write everything using indices:

$$\begin{align} A &= \chi_\alpha (\xi \eta) + \xi_\alpha (\eta \chi) + \eta_\alpha (\chi \xi) \\ &= \chi_\alpha \xi_\beta\eta_\gamma\varepsilon^{\beta\gamma} + \xi_\alpha\eta_\beta\chi_\gamma\varepsilon^{\beta\gamma} + \eta_\alpha\chi_\beta\xi_\gamma\varepsilon^{\beta\gamma} \\ &= \varepsilon^{\beta\gamma}(\chi_\alpha\xi_\beta\eta_\gamma +\xi_\alpha\eta_\beta\chi_\gamma+\eta_\alpha\chi_\beta\xi_\gamma) \end{align}$$ Then use the antisymmetry of the epsilon tensor and the anticommutivity of spinors to get $$\begin{align} A_\alpha &= \tfrac12\varepsilon^{\beta\gamma}( \chi_\alpha\xi_\beta\eta_\gamma - \chi_\alpha\xi_\gamma\eta_\beta +\xi_\alpha\eta_\beta\chi_\gamma-\xi_\alpha\eta_\gamma\chi_\beta +\eta_\alpha\chi_\beta\xi_\gamma-\eta_\alpha\chi_\gamma\xi_\beta) = \tfrac12\varepsilon^{\beta\gamma}( \chi_\alpha\xi_\beta\eta_\gamma - \chi_\alpha\xi_\gamma\eta_\beta +\chi_\gamma\xi_\alpha\eta_\beta-\chi_\beta\xi_\alpha\eta_\gamma +\chi_\beta\xi_\gamma\eta_\alpha-\chi_\gamma\xi_\beta\eta_\alpha) \end{align}$$ This can be written as $$ A_\alpha = \varepsilon^{\beta\gamma}A_{\alpha\beta\gamma}$$ where $A_{\alpha\beta\gamma}=A_{[\alpha\beta\gamma]} =3\chi_{[\alpha}\xi_\beta\eta_{\gamma]}$ is totally antisymmetric and thus must be zero.

This can generalize to other spinors, but the expressions scale as the dimensions!

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Maybe the more algorithmic way is to use Clifford algebra (sigma / gamma matrix) identities to get all of the terms into a standard form. For example, move the free index onto the $\chi$ spinor. You can do this with 2-component spinors using the identity $$\psi_\alpha \chi_\beta = \tfrac12 \varepsilon_{\alpha\beta} (\psi\chi) - \tfrac12 \sigma^{ab}_{\alpha\beta}(\psi\sigma_{ab}\chi) $$ A couple of simplifications then reduce the expression for $A_\alpha$ to zero.

Once again, similar things work with other spinors, but as they are higher dimensional spin representations, the expressions are in general larger.

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I talked a bit more about Fierz identites and Majorana Fierz identities in some older questions. Just as in those posts I'm using the conventions in Kuzenko and Buchbinder and recommend you have a look at Generalized Fierz Identities and references within.

This post imported from StackExchange Physics at 2015-06-15 18:11 (UTC), posted by SE-user Simon

Comments

Thanks, the second method is actually exactly what I was looking for! (by the way $\sigma_{\alpha\beta}^{ab}$ threw me off for a bit =p I'm used to seeing ${{\sigma^{ab}}_\alpha}^\beta$) One thing I'd like to add to this answer, and please correct me if I'm wrong, is how to get that identity (or to make sure you have accounted for everything in the identity) using group representation. On the LHS you have (1/2,0)x(1/2,0) so on the RHS you should have (0,0) + (1,0) which is exactly what you put up to some constants. Which is useful when you don't have a complete list of identities.

This post imported from StackExchange Physics at 2015-06-15 18:11 (UTC), posted by SE-user Peter Anderson