1
$\begingroup$

There have been several questions with good answers in physics.stackexchange about the motivation of the complexification of the Lorentz Lie algebra, basically as a mathematically nice way to deal with the situation generated by the non-existence of finite dimensional unitary representations of the Lorentz group.

But I'm interested in a clear outline about how to link this with the need of the Dirac spinor in QFT and how the complexification prompts one to go from a 2-spinor to a 4-spinor.

I think this is related to the need to go from the spin structure (Spin group) to the spin-c structure (Spin-c group with complex reresentation) in 4-dimensional Minkowski space with charged spin 1/2 particles carrying a unitary representation but I'm not sure exactly how.

$\endgroup$
2

1 Answer 1

0
$\begingroup$

The complexification of the Lorentz group is not connected to the fact that there are no finite dimensional unitary representations of the Lorentz group. The Dirac spinor representation is not unitary!

Furthermore, there is no need, a priori, for Dirac spinors in QFT. If one demands invariance under the orthochronous Lorentz group, the Dirac spinor is reducible, and we may instead consider Weyl fermions. Dirac Spinors are irreducible if one enlarges the $Spin$-group to a $Pin$-group containing spatial reflections or time-reflections.

For charged particles, the spinors fields should take value in $\mathbb{C} \otimes V$, where $V$ is a representation space of the $Spin$ group and the $\mathbb{C}$-factor describes charge. This is an example of a $Spin^c$-structure, but note that there is no need to do this unless the particle is charged, and even if it is charged, we don't have to consider a general $Spin^c$ structure.

$\endgroup$
17
  • $\begingroup$ I didn't write that the Dirac spinor representation was unitary, the infinite-dimensional linear representations of the Lorentz($SO^+(1,3)$) and Poincare groups are. I hope you can see that complexification is related to the role of classifying more easily these infinite-dimensional representations. My question was referring to charged fermions in its respect to Dirac spinors, yes, as they are quite important in QFT and SM. I didn't understand what you meant about not considering a general $Spin^c$ structure in your more helpful in relation to my specific question third paragraph. $\endgroup$
    – bonif
    Oct 19, 2018 at 16:39
  • 1
    $\begingroup$ you were asking about the connection between complexification and going from 2- to 4-spinors, and there is none. In particular, Weyl fermions describe charged fermions. The 4-spinor is important (irrespective of charge) if Lorentz transformations outside the identity component of the Lorentz group are symmetries. $\endgroup$ Oct 19, 2018 at 16:52
  • 1
    $\begingroup$ I don't quite know what these infinite-dimensional unitary representations of the Poincare group are supposed to do in a question about spinors (they are finite-dimensional non-unitary representations) and i was not talking about them. And the lack of unitary finite-dimensional representation is more connected to the fact that the Lorentz group is non-compact. How this comes about is a different question. So please do not mix these things. $\endgroup$ Oct 30, 2018 at 12:14
  • 1
    $\begingroup$ Defining particle representations and field representations is not a priori connected. Weinberg discusses this very neatly. $\endgroup$ Oct 31, 2018 at 9:54
  • 1
    $\begingroup$ I think there are some things mixed up here. Let's put it this way: in the electroweak sector, there are no dirac particles. Fundamentally, one has doublets and singlets of weyl fermions, only in some regime the theory is described by dirac spinors (namely, if the discrete symmetries are approximately conserved). In this case, one goes from the Spin group to a Pin group, as is nicely explained in the link you provided. The Lie algebra is an approximation of the Lie group close to the identity. The elements constituting the difference between Spin and Pin are very far from the identity. $\endgroup$ May 12, 2019 at 18:36

Your Answer

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

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