Dirac adjoint

In quantum field theory, the Dirac adjoint defines the dual operation of a Dirac spinor. The Dirac adjoint is motivated by the need to form well-behaved, measurable quantities out of Dirac spinors, replacing the usual role of the Hermitian adjoint.

Possibly to avoid confusion with the usual Hermitian adjoint, some textbooks do not provide a name for the Dirac adjoint but simply call it " ψ -bar".

Definition
Let $$\psi$$ be a Dirac spinor. Then its Dirac adjoint is defined as


 * $$\bar\psi \equiv \psi^\dagger \gamma^0$$

where $$\psi^\dagger$$ denotes the Hermitian adjoint of the spinor $$\psi$$, and $$\gamma^0$$ is the time-like gamma matrix.

Spinors under Lorentz transformations
The Lorentz group of special relativity is not compact, therefore spinor representations of Lorentz transformations are generally not unitary. That is, if $$\lambda$$ is a projective representation of some Lorentz transformation,


 * $$\psi \mapsto \lambda \psi$$,

then, in general,


 * $$\lambda^\dagger \ne \lambda^{-1}$$.

The Hermitian adjoint of a spinor transforms according to


 * $$\psi^\dagger \mapsto \psi^\dagger \lambda^\dagger$$.

Therefore, $$\psi^\dagger\psi$$ is not a Lorentz scalar and $$\psi^\dagger\gamma^\mu\psi$$ is not even Hermitian.

Dirac adjoints, in contrast, transform according to


 * $$\bar\psi \mapsto \left(\lambda \psi\right)^\dagger \gamma^0$$.

Using the identity $$\gamma^0 \lambda^\dagger \gamma^0 = \lambda^{-1}$$, the transformation reduces to


 * $$\bar\psi \mapsto \bar\psi \lambda^{-1}$$,

Thus, $$\bar\psi\psi$$ transforms as a Lorentz scalar and $$\bar\psi\gamma^\mu\psi$$ as a four-vector.

Usage
Using the Dirac adjoint, the probability four-current J for a spin-1/2 particle field can be written as


 * $$J^\mu = c \bar\psi \gamma^\mu \psi$$

where c is the speed of light and the components of J represent the probability density ρ and the probability 3-current j :


 * $$\boldsymbol J = (c \rho, \boldsymbol j)$$.

Taking and using the relation for gamma matrices


 * $$\left(\gamma^0\right)^2 = I$$,

the probability density becomes


 * $$\rho = \psi^\dagger \psi$$.