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Let $\chi$ be a left-handed Weyl spinor transforming as $$\delta\chi=\frac{1}{2}\omega_{\mu\nu}\sigma^{\mu\nu}\chi.$$ In my lecture notes it is explicitly stated that complex conjugation interchanges the representations which spinors transform under, i.e. we would should expect that $\chi^*$ transform as a right-handed spinor; $$\delta\chi^*=\frac{1}{2}\omega_{\mu\nu}\bar\sigma^{\mu\nu}\chi^*$$ where

$$\sigma^{\mu\nu}=\frac{1}{4}(\sigma^\mu\bar\sigma^\nu-\sigma^\nu\bar\sigma^\mu)$$ and $$\bar\sigma^{\mu\nu}=\frac{1}{4}(\bar\sigma^\mu\sigma^\nu-\bar\sigma^\nu\sigma^\mu).$$

I have the identity $$\bar\sigma^2(\sigma^{\mu\nu})^*\sigma^2=-\bar\sigma^{\mu\nu}.$$

Using this I can show that the quantity $-\bar\sigma^2\chi^*$ transforms as a right handed spinor (under the assumption that the variation of the spinor under lorentz transformation commutes with complex conjugation, i.e that $$(\delta\chi)^*=\delta\chi^*,$$ which I'm not totally sure is true but I think it is based on other notes I've seen. We obtain,

$$\delta(-\bar\sigma^2\chi^*)=-\bar\sigma^2(\delta\chi)^*=\frac{1}{2}\omega_{\mu\nu}\bar\sigma^{\mu\nu}(-\bar\sigma^2\chi^*).$$

However, I have also shown that $\chi^*$ does not transform as a right handed spinor. In particular, we have

$$\delta(\chi)^*=\frac{1}{2}\omega_{\mu\nu}\bar\sigma^2\bar\sigma^{\mu\nu}\sigma^2\chi^*.$$

In what sense then does complex conjugation interchange the representations under which spinors transform?

Edit: I've just noticed that by using $\bar\sigma^2=-\sigma^2$ and the fact that the Pauli matricies are Hermitian I can rewrite this as

$$\delta(\chi^*)=\frac{1}{2}(-\omega_{\mu\nu})\sigma^{2\dagger}\bar\sigma^{\mu\nu}\sigma^2\chi^*$$ which I guess in just the transformation to a new basis since we've applied a similarity transformation to the generators and doing the lorentz transformation in the "opposite" direction. Is this correct?

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  • $\begingroup$ Which lecture notes? $\endgroup$ – Qmechanic Oct 21 '15 at 18:05
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Remember the index structure of all the objects involved, namely $(\sigma^{\mu\nu})_{\alpha}{}^\beta$ and $( {\overline\sigma}^{\mu\nu} )^{\dot \alpha}{}_{\dot \beta}$. Then, the transformation of right and left-handed spinors is $$ \delta \chi_{\alpha} = \frac{1}{2} \omega_{\mu\nu} (\sigma^{\mu\nu})_{\alpha}{}^\beta \chi_\beta $$ and $$ \delta \psi^{\dot \alpha} = \frac{1}{2} \omega_{\mu\nu}( {\overline\sigma}^{\mu\nu} )^{\dot \alpha}{}_{\dot \beta} \psi^{\dot \beta} $$ Importantly, the right-handed spinor has an index raised as opposed to the left-handed spinor. Recall also that spinor indices are raised and lowered using ${\varepsilon}^{\alpha\beta}$ which is the 2-index Levi-Civita tensor which is nothing but $i \sigma^2$.

Now, if I take a complex conjugation of the first equation and use all the formulae that you have written down in the question, we do end up getting the second equation as long as we keep track of the change of index structure.

I will let you work out the exact details to show that this is true.

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  • $\begingroup$ So using your fact that $i(\sigma^2)^{ab}=\epsilon^{ab}$ I was able to show $(\delta\chi_\alpha)^*=\frac{1}{2}\omega_{\mu\nu}\bar\sigma^{\mu\nu}_{\dot\alpha\dot\beta}(\chi^*)^{\dot\beta}$. Is this okay even though one of the spinor indicies isn't raised? $\endgroup$ – Okazaki Oct 21 '15 at 18:31
  • $\begingroup$ Ya. You can just raise the ${\dot \alpha}$ index on both sides and get precisely the correct form. $\endgroup$ – Prahar Oct 21 '15 at 18:32
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Prahar has already given a good answer. Here we will instead focus on the pertinent Lie group (as opposed to the Lie algebra and its generators).

  1. The Lie group $SL(2,\mathbb{C})$ is the double cover of the restricted Lorentz group $SO^+(3,1)$, cf. e.g. this Phys.SE post.

  2. The fundamental/defining representation $V\cong\mathbb{C}^2$ of the Lie group $SL(2,\mathbb{C})$ transforms as $$\tag{1} \psi^{\prime\alpha} ~=~g^{\alpha}{}_{\beta}~ \psi^{\beta}, \qquad g~\in~SL(2,\mathbb{C}). $$

  3. The representation quartet mechanism. For each group representation (1), it is possible to define three sister representations of the same dimension: $$\tag{2} \psi^{\prime}_{\alpha} ~=~((g^{-1})^t)_{\alpha}{}^{\beta}~ \psi_{\beta},\qquad\text{(dual/contragredient repr.)}$$ $$\tag{3} \psi^{\prime\dot{\alpha}} ~=~\overline{g^{\dot{\alpha}}{}_{\dot{\beta}}}~ \psi^{\dot{\beta}},\qquad\text{(complex conjugate repr.)}$$ $$\tag{4} \psi^{\prime}_{\dot{\alpha}} ~=~((g^{-1})^{\dagger})_{\dot{\alpha}}{}^{\dot{\beta}}~ \psi_{\dot{\beta}},\qquad\text{(dual/contragred. complex conj. repr.)}$$

  4. By raising and lowering indices with the 2-dimensional Levi-Civita symbol (which happens to be equal to $\pm i$ times the 2nd Pauli matrix), it is possible to show that the two un-dotted representations (1) and (2) are isomorphic/equivalent. They are collectively known as the left Weyl spinor representation.

  5. Similarly, the two dotted representations (3) and (4) are isomorphic/equivalent (via similarity transformations with the dotted Levi-Civita symbol), and collectively known as the right Weyl spinor representation.

  6. In summary, the representation quartet mechanism produce two inequivalent 2-dimensional representations, the left and right Weyl spinor representations.

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