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It is known that the two components spinor $\chi$ is transformed under the $(\frac{1}{2},0)$ representation of lorentz group. This transformation can be written as $$\chi\rightarrow \exp[-\frac{i}{2}\vec\sigma\cdot\vec\theta]\chi\tag{1}$$ for rotation and $$\chi\rightarrow \exp[-\frac{1}{2}\vec\sigma\cdot\vec\zeta]\chi\tag{2}$$ for Lorentz boost.

For rotation, it can be verified that $$\exp[\frac{i}{2}\vec\sigma\cdot\vec\theta]\sigma^\mu\exp[-\frac{i}{2}\vec\sigma\cdot\vec\theta]=\Lambda^\mu_\nu\sigma^\nu,\tag{3}$$ where $\Lambda^\mu_\nu$ describes the corresponding transformation for Lorentz vector. However, it seems impossible to obtain the similar relation $$\exp[\frac{1}{2}\vec\sigma\cdot\vec\zeta]\sigma^\mu\exp[-\frac{1}{2}\vec\sigma\cdot\vec\zeta]=\Lambda^\mu_\nu\sigma^\nu\tag{4}$$ for boost, because $[\sigma^0,\sigma^i]=0$.

Thus, my question is how to obtain the relation $$\exp[\frac{1}{2}\vec\sigma\cdot\vec\zeta]\sigma^\mu\exp[-\frac{1}{2}\vec\sigma\cdot\vec\zeta]=\Lambda^\mu_\nu\sigma^\nu\tag{5}$$ for two components spinor, although this relation can be obtained for Dirac spinor easily due to $[\gamma^0,\gamma^i]\neq 0$.

K. Chang
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  • Related: http://physics.stackexchange.com/q/28505/2451 and links therein. – Qmechanic Aug 29 '16 at 08:43
  • I think I make a mistake about the transformation of boost. The correct form of this transformation reads $\exp[-\frac{1}{2}\vec\sigma\cdot\vec\zeta]\sigma^\mu\exp[-\frac{1}{2}\vec\sigma\cdot\vec\zeta]=\Lambda_\nu^\mu\sigma^\nu$. However, this leads to another question that $\chi^\dagger\chi$ is not invariant under such transformation. – K. Chang Aug 30 '16 at 13:47
  • I think I have solved this question. We must pay attention to the structure of indices. For example, $\chi_\alpha\rightarrow {M_\alpha}^\beta\chi_\beta$ while $\chi^\alpha$ is transformed as $\chi^\alpha\rightarrow {{(M^{-1})^T}^\alpha}_\beta\chi^beta$, where $M$ are the spin-1/2 representation. – K. Chang Sep 02 '16 at 03:05

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