Dirac adjoints are typically introduced in an ad hoc manner to replace the Hermitian adjoints which are not spinors. I ask how this can be cast in a more refined way in the language of representation theory.

I hope this is the right section as the question is about Lie groups and representations.

First and foremost, in this post, I'll be dealing with Dirac and Weyl spinor (not spinor fields) representations of the Lorentz algebra. Also, for simplicity, I'll use the chiral representation later on. Given a Dirac spinor $ \Psi $, physics books typically introduce the Dirac adjoint $ \overline{\Psi}=\Psi^\dagger\gamma^0 $ because the Hermitian adjoint "does not transform like a spinor," and in fact, under a Lorentz transformation $$ \Psi\rightarrow D(\Lambda)\Psi\implies\Psi^\dagger\rightarrow \Psi^\dagger D(\Lambda) \quad \text{and} \quad \overline{\Psi}\rightarrow\overline{\Psi}\gamma^0 D(\Lambda)\gamma^0=[D(\Lambda)]^{-1} $$

Now, I don't see any compelling motivation for the statement that the Hermitian adjoint "is not a spinor" and the Dirac adjoint is, other than requiring the Lorentz invariance of $ \overline{\Psi}\Psi $. I'd better understand this if it were cast properly in the language of representation theory.

First of all, I would translate the last statement to the following: there is (resp. there isn't) a (spinorial) representation of the Lorentz algebra that acts as above. So, let's go straight to the core of the question. If I have $ \Psi\in(1/2,0)\oplus(0,1/2) $, then what representation does $ \overline{\Psi} $ belong to? I'm not sure this is the conjugate representation, which as I know should transform with the complex conjugate matrix, not the inverse, which are different in the present case: this is a finite-dimensional representation, so due to the non-compactness of the Lorentz group it can't be unitary. Is this the dual representation (the transpose only depends on the fact we write the adjoint as a row/column spinor)? In such case, what do we label the dual representation of $ (1/2,0)\oplus(0,1/2) $?

  • 4
    $\begingroup$ Not really a research level question, unfortunately. The details are in many standard references, e.g., these notes by Trautman. $\endgroup$ Feb 12 at 11:51
  • $\begingroup$ You might get a more sympathetic resonse on physics.stackexchange.com $\endgroup$
    – iSeeker
    2 days ago


Your Answer

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