In the literature one encounters a lot of different conventions for how left-handed spinor transforms (rotation angle $\phi$, rapidity $\beta$), among them

$M_L = M_{(\frac{1}{2}, 0)} = e^{-i \frac{1}{2} \vec{\theta} \cdot \vec{\sigma} - \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}}$,

$M_L = M_{(\frac{1}{2}, 0)} = e^{-i \frac{1}{2} \vec{\theta} \cdot \vec{\sigma} + \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}}$,

$M_L = M_{(\frac{1}{2}, 0)} = e^{+i \frac{1}{2} \vec{\theta} \cdot \vec{\sigma} + \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}}$

Comparing various sources can be a pain, so I tried to establish a definite convention (as a reference for me), to compare against various sources, but I failed so far to do it consistently.

Specifically I tried to understand the conventions used in the following source:

Dreiner, H. K., Haber, H. E., & Martin, S. P. (2010). Two-component spinor techniques and Feynman rules for quantum field theory and supersymmetry. Physics Reports, 494(1-2), 1–196. doi:10.1016/j.physrep.2010.05.002 (http://arxiv.org/abs/0812.1594)

Is use the version with the $(-1, +1, +1, +1)$ convention that can be obtained from https://www.niu.edu/spmartin/spinors/ (whereas the arxiv version uses the (+1, -1, -1, -1) convention).

They state (page 10) that the transformation of a left-handed spinor ought to be:

$M_L = M_{(\frac{1}{2}, 0)} = e^{-i \frac{1}{2} \vec{\theta} \cdot \vec{\sigma} - \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}}$

I then tried to get the corresponding lorentz-transformation by looking at the transformation of the hermitean matrix $X = \left( \begin{array}{cc} x^0+x^3 & x^1-ix^2\\ x^1+ix^2 & x^0-x^3 \end{array} \right)$

$A: X \rightarrow X' = A X A^\dagger$

To extract the lorentz-transformatin I use

${\Lambda^\mu}_\nu = -\frac{1}{2} \operatorname{tr}[\sigma^\mu A \bar{\sigma}_\nu A^\dagger]$

with $\sigma^\mu = (1, \vec{\sigma})$ and $\bar{\sigma}_\nu = (-1, -\vec{\sigma})$

This is not covered in their paper, but when I do it for

$R = e^{-i\phi\frac{1}{2}\sigma^3} = \left(\begin{array}{cc} e^{-i\frac{1}{2}\phi} & 0 \\ 0 & e^{i\frac{1}{2}\phi} \end{array}\right)$ (rotation around the z-axis)

I get the lorentz transformation

${R^\mu}_\nu = \left( \begin{array}{cc} 1& 0& 0& 0\\ 0& \cos\phi& -\sin\phi & 0 \\ 0& \sin\phi& \cos\phi& 0 \\ 0& 0& 0& 1 \end{array} \right)$

Similar for

$B = e^{-\beta\frac{1}{2}\sigma^3} = \left(\begin{array}{cc} e^{-\frac{1}{2}\beta} & 0 \\ 0 & e^{\frac{1}{2}\beta} \end{array}\right)$ (boost in z-direction)

I get

${B^\mu}_\nu = \left( \begin{array}{cc} \cosh\beta& 0& 0& -\sinh\beta\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -\sinh\beta& 0& 0& \cosh\beta \end{array} \right)$

These correspond to active rotations (counterclockwise) around the z-axis, but passive boost in the z-direction.

But this is in constrast to their statement (page 10) that corresponding infinitesimal transformations are

$\vec{x} \rightarrow \vec{x} + (\vec{\phi} \times \vec{x})$

$\vec{x} \rightarrow \vec{x} + \vec{\beta}t$

which both are active rotations and boosts.

So either:

  • something is wrong in my calculation from $SL(2, \mathbb{C})$ to the Lorentz Group (or may be also there is a hidden convention I got wrong ?)
  • or the cited paper got it wrong (unlikely, since its already in its 5th incarnation, with a list of errata [https://www.niu.edu/spmartin/spinors/])
  • or I am quite generally confused about how exactely spinors should transform, and specifically how the signs are determined.
  • $\begingroup$ Consider to add sources for the various conventions (besides Martin et. al.). $\endgroup$
    – Qmechanic
    Aug 31, 2019 at 8:20
  • 1
    $\begingroup$ A overall difference of sign in the exponential may be due to the user of active vs. passives Transformation. Difference of sign between rotation and boost generator may be due to the use of clockwise vs. countercklockwise rotations... etc. From the inspection of a transformation $M\in SL(2,\mathbb{C})$ alone, it is not possible to deduce all the implicit convention. My point is that certainly different convention are possible. But for now I am mainly interested in the one used by the cited paper. $\endgroup$
    – Thomas
    Sep 1, 2019 at 19:37

1 Answer 1


A wrong assumption was made in the question asked.

An element $A\in SL(2,\mathbb{C})$ that induces the Lorentz transformation $X \rightarrow A X A^\dagger$ is a Spinor transformation for a right-handed Spinor (not left-handed, as claimed).

Rather, a left-handed spinor transformes with an element $A\to SL(2,\mathbb{C})$ that induces the Lorentz transformation $X \rightarrow X^{'} = \left(A^{-1}\right)^\dagger X A^{-1}$.

So everything fits together and is consistent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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