Physical significance of left- and right-handed spinors

Question

We know that the $(1/2,0)$ and $(0,1/2)$ representations of Lorentz group represent the left- and right-handed spinors respectively. What’s the reason behind this nomenclature? What they represent physically?

Answer 1

You will probably see other answers (and links) explaining chirality in terms of representations of Lorentz groups generators as matrices and the spinors as columns acted on by these matrices. But in this answer I will present a purely algebraic explanation without any reference to matrix/column representation.

The whole business of the "physical significance of left- and right-handed spinors" boils down to the identity of how the pseudoscalar $\gamma_0\gamma_1\gamma_2\gamma_3$ acts on the spinor: $$ \gamma_0\gamma_1\gamma_2\gamma_3 \psi_{L} = - i\psi_{L},\\ \gamma_0\gamma_1\gamma_2\gamma_3 \psi_{R} = + i\psi_{R}, $$ where $\psi_{L}$ and $\psi_{R}$ are the left-handed spinor and right-handed spinor, respectively. Note that above identity is an algebraic identity, which is not explicitly dependent on the representation.

The reduction of the pseudoscalar $\gamma_0\gamma_1\gamma_2\gamma_3$ to negative imaginary number $-i$ for left-handed spinor and to positive imaginary number $i$ for right-handed spinor is actually all you need in terms of how left-handed and right-handed spinors behave differently under Lorentz $Spin(1,3)$ (or $SO(1,3)$ for vectors) transformations. Of course, left-handed and right-handed spinors behave differently under electroweak transformations, but the electroweak chirality would be a subject for a different post.

Specifically, the Lorentz algebra $spin(1,3)$ comprises 6 independent generators (using the "anti-hermician" version preferred by mathematicians) $$ \{\gamma_1\gamma_2, \gamma_2\gamma_3, \gamma_3\gamma_1,\gamma_0\gamma_3,\gamma_0\gamma_1,\gamma_0\gamma_2\}. $$

The mentioned identity $\gamma_0\gamma_1\gamma_2\gamma_3 \psi_{L/R} = \mp i\psi_{L/R}$ makes 3 of the generators degenerate for a given chirality. For example: $$ \gamma_0\gamma_1 \psi_{L/R} = -(\gamma_2\gamma_3)^2 (\gamma_0\gamma_1)\psi_{L/R}= -(\gamma_2\gamma_3) (\gamma_0\gamma_1\gamma_2\gamma_3)\psi_{L/R} = \pm i\gamma_2\gamma_3\psi_{L/R}, $$ (the first equality is resulted from $(\gamma_2\gamma_3)^2 = -1$) whereby real Lorentz boost ($\gamma_0\gamma_1$) is translated into imaginary rotation ($i\gamma_2\gamma_3$) and vice-versa: $$ e^{\theta\gamma_0\gamma_1} \psi_{L/R} = e^{\pm i\theta\gamma_2\gamma_3}\psi_{L/R}. $$

Therefore the original Lorentz algebra generators are turned into $$ \{\gamma_1\gamma_2, \gamma_2\gamma_3, \gamma_3\gamma_1,i\gamma_1\gamma_2, i\gamma_2\gamma_3, i\gamma_3\gamma_1\} $$ and $$ \{\gamma_1\gamma_2, \gamma_2\gamma_3, \gamma_3\gamma_1,-i\gamma_1\gamma_2, -i\gamma_2\gamma_3, -i\gamma_3\gamma_1\} $$ for left-handed spinor $\psi_{L}$ and right-handed spinor $\psi_{R}$, respectively. Now you only have 3 independent complexified $SU(2)$ generators ($\{\gamma_1\gamma_2, \gamma_2\gamma_3, \gamma_3\gamma_1\} \sim su(2)$) for each chirality. This is all the $(1/2,0)$ and $(0,1/2)$ "representations" business is about.