# In what representation do Dirac adjoint spinors lie?

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• qft-El
In summary, the Dirac and Weyl spinor representations of the Lorentz algebra have different adjoints, the hermitian adjoint does not transorm like a spinor and the Dirac adjoint is more useful. There is (resp. there isn't) a (spinorial) representation of the Lorentz algebra that acts as above. The pinor and spinor representations of O(3,1) and SO(3,1), respectively, are derived from the Clifford (geometric) algebra and its even grade subalgebra. Depending on sign conventions, the Clifford algebra for O(3,1) is either isomorphic to \
qft-El
TL;DR Summary
Dirac adjoints are tipically 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 tipically introduce the Dirac adjoint $\overline{\Psi}=\Psi^\dagger\gamma^0$ because the hermitian adjoint "does not transorm like a spinor" and in fact, under a Lorentz transformation
$$\Psi\rightarrow D(\Lambda)\Psi\implies\Psi^\dagger\rightarrow \Psi^\dagger D(\Lambda) \qquad \text{and} \qquad \overline{\Psi}\rightarrow\overline{\Psi}\gamma^0 D(\Lambda)\gamma^0=[D(\Lambda)]^{-1}$$
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)##?

(  I've cleaned up some typos.)
As to the core question. $\overline{\Psi}$ is an element of the dual space. In so far as it is a transformation you must take the Dirac adjoint of the matrix as that is what satisfies the analogue properties:
$$D(\Lambda)^- = D(\Lambda^{-1})=D(\Lambda)^{-1}$$
for a Lorentz group element $\Lambda$ and:
$$D(\lambda)^- = - D(\lambda)^-$$
for $\lambda$ a generator from the Lie algebra of the Lorentz group whereby $\Lambda(s) = e^{s\lambda}$
Here I'm using the superscript dash for Dirac adjoint of a matrix representation of a spinor operator (and s is just a generic parameter).

Understand that the typical matrix adjoint (complex transpose) as it reflects an involution between the underlying vector space and dual space is basis dependent. General linear transformations will not respect the induced hermitian form as, obviously, not all general linear transformations are unitary.

Also understand that the invariant form on the pinor space (direct sum of two spinor rep spaces) is not even symmetric for $O(3,1)$'s pin representation; (using spin for special orthogonal group and pin for general orthogonal group) it is rather a symplectic form.

I will try to outline the "big picture" details. There are various sign conventions which vary from author to author so I'll be explicit when necessary which gets lengthy. The pinor and spinor representations of O(3,1) and SO(3,1) respectively are derived from the Clifford (geometric) algebra and its even grade subalgebra. Depending on sign conventions the Clifford algebra for O(3,1) is either isomorphic to $\mathbb{H}(2)$ the set of 2x2 quaternion matrices (Dirac pinors/spinors) or to $\mathbb{R}(4)$ the set of 4x4 real matrices (Majorana pinors/spinors). In both cases the even subalgebra is isomorphic to $\mathbb{C}(2)$ the algebra of 2x2 complex matrices... but we must be very careful about the meaning of the imaginary unit.

Now you may have notice that Dirac doesn't use quaternions. That is because in both Dirac and Majorana cases we can complexify i.e. introduce a central imaginary unit to extend either algebra to $\mathbb{C}(4)$ the algebra of 2x2 complex matrices.

So our matrix representation of the complexified Clifford algebra will act adjointly on itself (scalar, vector, antisymmetric tensor representations) and act from the left on column vectors (the direct sum of the two spinor types) and will also act from the right on row vectors (the dual pinor space). In the physics Dirac effectively uses the central imaginary unit as the generator of U(1) e-m gauge, i.e. complex conjugation expresses charge conjugation.

So we have two spaces, the pinor space $\mathbb{C}^4$ (column vector rep) and its dual space (row vector rep). The pinor space can also be split into two spinor spaces $\mathbb{C}^2\oplus\mathbb{C}^2$ and their dual spaces are likewise a splitting of the dual pinor representation space.

Now this representation obviously NOT hermitian as we are talking about a finite representation of a non-compact group which won't fit into a compact unitary representation. (This is "cured" by extending to the field theory where we can construct infinite dimensional unitary reps.) So the fact that $\dagger$ adjoint is not invariant under the Lorentz group (nor under space-time reflections of O(3,1) ) means that its meaning in the context of these groups is very much basis dependent, hence arbitrary. The conjugate transpose will not, in general even map one of the spinors to its own dual space. One must construct the adjoint from scratch so-to-speak.

It is very educational to work through the construction for one's self. I think you'll find it worth the effort.

Last edited:
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