Bilinear Covariants

Spinor components depend on a basis; bilinears organize their content into quantities with definite spacetime transformation laws. In four dimensions, scalar, pseudoscalar, vector, axial-vector, and antisymmetric tensor bilinears account for the sixteen independent matrix structures. The factor of $i$ in the pseudoscalar and the order of $\gamma^5$ in the axial vector are consequential conventions.

Required background. The Covariant Dirac Equation fixes spinor covariance; Gamma-Matrix Conventions fixes the matrix signs; Free Dirac Spinors supplies normalized on-shell examples. Use $\hbar=c=1$ below.

The sixteen bilinear structures

For a commuting Dirac wavefunction, define

Name	Definition	Independent components
Scalar	$S=\bar\psi\psi$	1
Pseudoscalar	$P=i\bar\psi\gamma^5\psi$	1
Vector	$V^\mu=\bar\psi\gamma^\mu\psi$	4
Axial vector	$A^\mu=\bar\psi\gamma^\mu\gamma^5\psi$	4
Tensor	$T^{\mu\nu}=\bar\psi\sigma^{\mu\nu}\psi=-T^{\nu\mu}$	6

The symbols $S,P$ in this table label bilinears, not the Lorentz-spinor transformation matrix or the parity operator. Off-diagonal transition bilinears $\bar\psi\Gamma\chi$ use the same matrix structures but need not be real.

Why is this list complete? Complex $4\times4$ matrices form a sixteen-dimensional space, and these $1+1+4+4+6$ matrices are linearly independent. The trace pairings establish that independence. In addition to the scalar and vector traces in the convention ledger,

$$\begin{aligned} \operatorname{tr}[(i\gamma^5)^2]&=-4,\\ \operatorname{tr}(\gamma^\mu\gamma^5\gamma^\nu\gamma^5) &=-4\eta^{\mu\nu},\\ \operatorname{tr}(\sigma^{\mu\nu}\sigma^{\rho\sigma}) &=4(\eta^{\mu\rho}\eta^{\nu\sigma} -\eta^{\mu\sigma}\eta^{\nu\rho}). \end{aligned}$$

Pairings between the different sectors vanish. Each sector’s trace pairing is nondegenerate, so no nontrivial linear combination vanishes. This completeness underlies expansions of interaction vertices and spin density matrices. It does not mean the sixteen bilinear values of a single spinor are freely independent numbers; they obey additional nonlinear relations.

Lorentz transformation of the bilinears

For a proper orthochronous transformation, $\psi'(x')=S_\Lambda\psi(x)$ and $\bar\psi'(x')=\bar\psi(x)S_\Lambda^{-1}$. Thus

$$\bar\psi'(x')\Gamma\psi'(x') =\bar\psi(x)S_\Lambda^{-1}\Gamma S_\Lambda\psi(x).$$

The gamma intertwiner gives the vector law directly: $V'^\mu(x')=\Lambda^\mu{}_\nu V^\nu(x)$. Taking the commutator of two transformed gamma matrices gives the tensor law,

$$T'^{\mu\nu}(x') =\Lambda^\mu{}_\rho\Lambda^\nu{}_\sigma T^{\rho\sigma}(x).$$

The oriented four-gamma product defining $\gamma^5$ is unchanged under the proper group. Hence $P$ is a scalar and $A^\mu$ a vector under that group. Their “pseudo” and “axial” character becomes visible under orientation-reversing transformations such as parity. Likewise $S$ is a scalar, while $\psi^\dagger\psi=V^0$ is only one component of a vector, not a Lorentz scalar.

Reality uses the Dirac adjoint

Define the Dirac adjoint of a matrix by $\overline\Gamma=\gamma^0\Gamma^\dagger\gamma^0$. Then

$$(\bar\psi\Gamma\chi)^*=\bar\chi\,\overline\Gamma\,\psi.$$

For $\Gamma=I,i\gamma^5,\gamma^\mu,\gamma^\mu\gamma^5,\sigma^{\mu\nu}$ one has $\overline\Gamma=\Gamma$. Therefore all five diagonal bilinear types in the table are real for commuting wavefunctions. Bare $\gamma^5$ instead obeys $\overline{\gamma^5}=-\gamma^5$, so $\bar\psi\gamma^5\psi$ is purely imaginary. This explains the factor $i$ in $P$.

No extra $i$ belongs in the axial vector with this convention. Also $\gamma^5\gamma^\mu=-\gamma^\mu\gamma^5$, so reversing that order reverses the axial bilinear. Tables from different sources must be translated at their definitions.

Parity distinguishes polar and axial objects

For the standard spinor parity action $\psi'(t,\mathbf x)=\eta_P\gamma^0\psi(t,-\mathbf x)$ with $|\eta_P|=1$, the diagonal bilinear transforms through $\Gamma\mapsto\gamma^0\Gamma\gamma^0$. The intrinsic phase cancels. Comparing the transformed quantity at $(t,\mathbf x)$ with the original at $(t,-\mathbf x)$ gives

Component	Parity sign
$S$	$+$
$P$	$-$
$V^0$, $\mathbf V$	$+$, $-$
$A^0$, $\mathbf A$	$-$, $+$
$T^{0i}$, $T^{ij}$	$-$, $+$

For example, $\gamma^0\gamma^5\gamma^0=-\gamma^5$ gives the pseudoscalar sign; the extra spatial-vector sign cancels it for $\gamma^i\gamma^5$, giving an even spatial axial vector. These are transformation properties. A dynamics containing an external parity-breaking background need not have parity as a symmetry.

Spin information in an on-shell state

Take a massive $u(p)$ state normalized by $\bar uu=2m$. If its rest two-spinor satisfies $\chi^\dagger\boldsymbol\sigma\chi=\mathbf n$, $|\mathbf n|=1$, define the polarization four-vector

$$s^0=\frac{\mathbf p\cdot\mathbf n}{m},\qquad \mathbf s=\mathbf n+ \frac{\mathbf p(\mathbf p\cdot\mathbf n)}{m(E+m)}.$$

It obeys $p\cdot s=0$ and $s^2=-1$. Direct evaluation at rest, followed by Lorentz covariance, gives

$$\bar u\gamma^\mu u=2p^\mu,\qquad \bar u\gamma^\mu\gamma^5u=2m s^\mu.$$

The vector current is independent of the spin orientation, while the axial bilinear carries it. At rest $s^\mu=(0,\mathbf n)$ and $T^{ij}=2m\epsilon^{ijk}n_k$, $T^{0i}=0$. Also $i\bar u\gamma^5u=0$: inserting the on-shell slash on either side and anticommuting it through $\gamma^5$ gives the negative of the same bilinear for $m\ne0$. This diagonal result does not make all transition pseudoscalars vanish.

The polarization vector with $1/m$ has no direct massless rest-frame limit. Use helicity and chiral spinors for that case rather than substituting $m=0$ into this massive parameterization.

Exercises

1. For a rest spin-up state $u=\sqrt{2m}(1,0,0,0)^T$, compute $S,V^\mu,A^\mu$, and $T^{12}$.

Solution

$S=2m$, $V^\mu=(2m,0,0,0)$, $A^\mu=(0,0,0,2m)$, and $T^{12}=2m$. The current is future timelike while the polarization bilinear is spacelike. They encode different information.

2. If $M=aI+b\,i\gamma^5$ is an arbitrary matrix in this two-dimensional subspace, recover $a,b$ by traces.

Solution

$a=\operatorname{tr}M/4$ and $b=-\operatorname{tr}(i\gamma^5M)/4$. The minus sign follows from $(i\gamma^5)^2=-I$, not from a different spacetime metric convention.

3. Show $p\cdot s=0$ using the displayed boosted polarization.

Solution

Writing $d=\mathbf p\cdot\mathbf n$,

$$p\cdot s=d\left[\frac Em-1- \frac{\mathbf p^2}{m(E+m)}\right]=0,$$

because $\mathbf p^2=(E-m)(E+m)$. The result constrains polarization to the rest-frame spatial subspace transported with the particle.

References

• C. Itzykson and J.-B. Zuber, Quantum Field Theory, McGraw–Hill, 1980 — bilinear classification and polarized spinors.
• M. D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press, 2014, doi:10.1017/9781139540940 — gamma-matrix traces and spinor matrix elements.
• B. Thaller, The Dirac Equation, Springer, 1992 — spinor transformations and observables.