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I have read, that if you have a Dirac spinor \begin{equation} \psi = \begin{pmatrix} \phi_R\\ \phi_L \end{pmatrix} \end{equation}

that you can apply a Lorentz boost along the $z$-direction with rapidity $y$ like this:

\begin{equation} \phi_R\rightarrow e^{-\frac{1}{2}\sigma_zy}\phi_R; \quad\phi_L\rightarrow e^{+\frac{1}{2}\sigma_zy}\phi_L \end{equation}

and a general boost like this: \begin{equation} \phi_R\rightarrow e^{-\frac{1}{2}\mathbf{\hat n}\cdot{\bf\sigma}}\phi_R; \quad\phi_L\rightarrow e^{+\frac{1}{2}\mathbf{\hat n}\cdot{\bf\sigma}}\phi_L \end{equation}

Why is this the correct way to transform the spinor? Also are opposite signs in the exponentials just convention or do they have deeper meaning?

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    $\begingroup$ These transformations come from the way the spinor representation is defined. You do not have "a Dirac spinor" if it doesn't transform like this. The sign are of course related to the parity ${}_{R/L}$. I'm not quite sure what you want to know. $\endgroup$
    – ACuriousMind
    May 25, 2015 at 13:55
  • $\begingroup$ Ok, why do the parities have different signs and how do you choose which one is positive and negative? $\endgroup$
    – Luka Milic
    May 25, 2015 at 14:19
  • $\begingroup$ Sorry I meant why do you use oppositely signed exponentials to transform the left and right handed spinors $\endgroup$
    – Luka Milic
    May 30, 2015 at 21:23

2 Answers 2

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You may wish to look at the derivation of Lorentz transformations for Dirac spinors, say in the book by Itzykson, Zuber. They are derived from the condition of relativistic covariance of the Dirac equation. The generators of Lorentz transformations are $[\gamma^\mu,\gamma^\nu]$ (up to a constant factor). This expression gives the different signs for boosts for the right-handed and left-handed parts of the Dirac spinor in the chiral representation.

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Comments to the question (v3):

  1. Recall that the restricted Lorentz group $$\tag{1} SO^+(3,1)~\cong~ SL(2,\mathbb{C})/\mathbb{Z}_2$$ is locally isomorphic to the Lie group of complex $2\times 2$ matrices with unit determinant, cf. e.g. my Phys.SE answer here. The Lie group of 3D rotations $$\tag{2} SO(3)~\cong~ SU(2)/\mathbb{Z}_2$$ is a subgroup thereof.

  2. The left-handed Weyl spinor representation is the fundamental representation of $SL(2,\mathbb{C})$. A left-handed Weyl spinor $\psi^{\alpha}_L$ (with upper indices) transforms as

$$\tag{3} \psi_L~\longrightarrow~\psi^{\prime}_L ~=~g \psi_L, \qquad g~\in~SL(2,\mathbb{C}). $$

  1. The right-handed Weyl spinor representation is mathematically speaking the complex conjugate representation. This means a right-handed Weyl spinor $\psi^{\dot{\alpha}}_R$ (with upper indices) transforms as $$\tag{4} \psi^{\bullet}_R~\longrightarrow~\psi^{\prime\bullet}_R ~=~g^{\ast} \psi^{\bullet}_R, \qquad g~\in~SL(2,\mathbb{C}), $$ where $g^{\ast}$ denotes the complex conjugate $2\times 2$ matrix.

  2. Phycisists often lower the index of the right-handed Weyl spinor $$\tag{5}\psi_{\dot{\alpha}}^R~=~\varepsilon_{\dot{\alpha}\dot{\beta}}\psi^{\dot{\beta}}_R$$ with the Levi-Civita tensor $\varepsilon_{\dot{\alpha}\dot{\beta}}$. The right-handed Weyl spinor $\psi_{\dot{\alpha}}^R$ (with lower indices) transforms as the Hermitian conjugate representation $$\tag{6}\psi_{\bullet}^R~\longrightarrow~ \psi^{\prime R}_{\bullet} ~=~(g^{-1})^{\dagger} \psi_{\bullet}^R, \qquad g~\in~SL(2,\mathbb{C}), $$ due to special properties of $2\times 2$ matrices.

  3. In summary, the difference between the transformations (3) and (6) is rendered moot for compact 3D rotations [since they are implemented by unitary $2\times 2$ matrices, cf. eq. (2)], but induces an important relative minus sign in the non-compact Lorentz boosts transformation rule between the left-handed and the right-handed Weyl spinors.

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  • $\begingroup$ Comment about notation: In this answer, all spinors are implicitly understood as column vectors. Note that often in physics, the right-handed spinors are implicitly assumed to be row vectors. $\endgroup$
    – Qmechanic
    Jun 7, 2015 at 20:12

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