2
$\begingroup$

In my QFT lecture notes, it is written that the Lorentz group elements can be written as \begin{equation*} \Lambda = e^{i\vec{\theta}\cdot\vec{J} + i\vec{\eta}\cdot\vec{K}} \end{equation*}

where $\Big\{\vec{J}, \vec{K}\Big\}$ are the generators of the Lorentz algebra.

Ater this, they write that Weyl spinors transform under a representation of the Lorentz group, as \begin{equation*} \phi' = e^{i\frac{\vec{\sigma}}{2}\cdot\left(\vec{\theta} - i\vec{\eta}\right)}\phi \end{equation*}

Here, as $\Big\{\frac{\vec{\sigma}}{2}, -i\frac{\vec{\sigma}}{2}\Big\}$ indeed satisfies the Lorentz algebra commutation relations, it is indeed a representation of the Lorentz algebra $\Big\{\vec{J}, \vec{K}\Big\}$. However, not all representations of Lie algebras lead to a representation of the Lie group by exponentiation. So, for the case of the Weyl spinor, how can we show that the transformation rule is indeed a representation of the Lorentz group?

$\endgroup$
1
  • $\begingroup$ Elementary Weyl spinors transform either in the (1/2,0) or (0,1/2) projective representation of the restricted Lorentz group or in the true (1/2,0) or (0,1/2) representation of $\text{SL}(2,\mathbb C)$. $\endgroup$
    – DanielC
    Oct 1, 2021 at 9:36

1 Answer 1

2
$\begingroup$

TL;DR: To discuss non-projective group representations of spinors we need to go to the universal covering group.

In detail:

  1. First define a (left) Weyl spinor $\phi$ to transform in the defining group representation of $SL(2,\mathbb{C})$, which is the double cover of the restricted Lorentz group $SO^+(1,3;\mathbb{R})$.

  2. Only thereafter, we should identify the corresponding Lie algebra $sl(2,\mathbb{C})\cong so(1,3;\mathbb{R})$, the Lie algebra representation, and their 6 generators of boosts and rotations.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.