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. Since the usual Hermitian adjoint lacks the Lorentz symmetry of the system, the Dirac adjoint must be used instead.

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 ψ be a Dirac spinor. Then its Dirac adjoint is defined as

${\displaystyle {\bar {\psi }}\equiv \psi ^{\dagger }\gamma ^{0}}$

where ψ denotes the Hermitian adjoint of the spinor ψ and γ0 is the time-like gamma matrix.

## Spinors under Lorentz transformations

The Lorentz group of special relativity is not compact, therefore representations of Lorentz transformations in the Dirac spinor space are not unitary. That is, in general,

${\displaystyle \lambda ^{\dagger }\neq \lambda ^{-1}}$

where λ is the corresponding Lorentz transformation that maps spinors:

${\displaystyle \psi \mapsto \lambda \psi }$.

The Hermitian adjoint of spinors transforms according to

${\displaystyle \psi ^{\dagger }\mapsto \psi ^{\dagger }\lambda ^{\dagger }}$.

Therefore, using only the Hermitian adjoint, one finds that ψ ψ is not a Lorentz scalar and ψ γμ ψ is not even Hermitian.

Using the definition, one finds that the Dirac adjoint of spinors transforms according to

${\displaystyle {\bar {\psi }}\mapsto \left(\lambda \psi \right)^{\dagger }\gamma ^{0}}$.

Using the identity γ0 λ γ0 = λ−1, the transformation reduces to

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

which possesses the required Lorentz symmetry for ψ ψ and ψ γμ ψ.

## Usage

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

${\displaystyle J^{\mu }=c{\bar {\psi }}\gamma ^{\mu }\psi }$

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

${\displaystyle {\boldsymbol {J}}=(c\rho ,{\boldsymbol {j}})}$.

Taking μ = 0 and using the relation for gamma matrices

${\displaystyle \left(\gamma ^{0}\right)^{2}=I}$,

the probability density becomes

${\displaystyle \rho =\psi ^{\dagger }\psi }$.