# Correct transformation of left-handed Weyl spinor

In the book "Matthew D. Schwartz, Quantum Field Theory and the Standard Model", page 164, it says that a left-handed spinor transforms as

$$\psi_L \rightarrow e^{\frac{1}{2}(i\vec{\theta} - \vec{\beta})\vec{\sigma}} \psi_L.\tag{10.39}$$

In the book "Peskin, Schröder, An Introduction To Quantum Field Theory", page 44, it says that a lefthanded spinor transforms as

$$\psi_L \rightarrow e^{-\frac{1}{2}(i\vec{\theta} + \vec{\beta})\vec{\sigma}} \psi_L.\tag{3.37}$$

In the book "Anthony Duncan, The Conceptual Framework of Quantum Field Theory", page 76, it says that a spinor in the $$(1/2,0)$$ representation (I assume that is a lefthanded spinor) transforms as

$$\psi_L \rightarrow e^{-\frac{1}{2}(i\vec{\theta} - \vec{\beta})\vec{\sigma}} \psi_L.$$

All of them are different. Which transformation is correct?

If all of them are, why the different signs?

• Unfortunate fact: Different authors have different conventions. – Qmechanic Sep 26 '19 at 7:19
• All signs here are solely a matter of convention. For example, switching between thinking of rotations/boosts as active, vs. passive, produces a sign flip. There is no problem as long as you are consistent in your conventions. – knzhou Sep 26 '19 at 7:19
• Unfortunately, spinors are a subject that, within intro QFT, are almost 100% made of conventions. That is, in a typical introductory chapter on spinors with 100 equations in it, over half are really definitions of conventions, not deductions. As such, you shouldn't expect any equations from any book to match any other book. – knzhou Sep 26 '19 at 7:20
• There must be a convention though for when a spinor is called lefthanded (regardless of the signs involved in the representations above). – Thomas Sep 26 '19 at 10:56

As stated in some of the comments above, signs are a matter of convention, but they are not arbitrary ! Consistency is the key.

Since $$e^{\pm i \frac{1}{2} \vec{\phi} \cdot \vec{\sigma} + \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}}$$ and $$e^{\pm i \frac{1}{2} \vec{\phi} \cdot \vec{\sigma} - \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}}$$ (same sign for the rotation generators but a sign difference in the boost generators) are unitarily inequivalent, only one of them can be the transformation for a lefthanded spinor.

By consulting the sources I considered the most trustworthy (in the sense that they are quite explicit and comprehensive in the statement of their assumptions and conventions, and seem to use them consistently throughout) - among them:

• 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)

• Peskin Schröder, An Introduction to QFT

• Srednicki, Quantum Field Theory

I came to the conclusion that the following transformation is commonly agreed upon to be a active rotation (counterclockwise) with angle $$\phi$$ around the axis $$\hat{\phi}$$ and active boost with rapidity $$\vec{\beta}$$ for a lefthanded spinor:

$$M_L = e^{-i \frac{1}{2} \vec{\phi} \cdot \vec{\sigma} - \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}}$$

In the case of "Anthony Duncan, The Conceptual Framework of Quantum Field Theory", he

• first talks about passive rotations and boosts (page 110, formula 5.9, 5.10)
• but then later seems to use a different conventions:

$${\Lambda^\mu}_\nu = {\delta^\mu}_\nu + {\omega^\mu}_\nu$$ for infinitesimal transformations (page 174, right after formula 7.19) and

$$\vec{\phi} = (\omega_{23},\omega_{31},\omega_{12})$$, $$\vec{\beta} = (\omega_{10},\omega_{20},\omega_{30})$$ (page 175, right after formula 7.35)

which together implie that he is using passive rotations but active boosts.

The latter is then consistent with

$$M_L = e^{-i \frac{1}{2} \vec{\phi} \cdot \vec{\sigma} + \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}}$$ (page 176, formula 7.36)