Dirac Spinors

A Dirac spinor combines a left Weyl representation and a right Weyl representation in a direct sum. A Dirac mass couples these two chiral components in the equation of motion while preserving the total vector current. This representation structure explains the adjoint and the action of parity; it does not identify upper and lower components with particle and antiparticle states.

Required background. Weyl Spinors fixes the two transformation laws; the Covariant Dirac Equation supplies the equation and current. Helpful background. Free Dirac Spinors supplies normalized modes and spin sums, which are not rederived here.

The direct sum of two chiral representations

Use natural units and the fixed chiral basis. With $\Psi=(\chi_L,\chi_R)^T$, the Lorentz double cover acts as

$$\Psi'(x')=S(A)\Psi(x),\qquad S(A)= \begin{pmatrix} (A^\dagger)^{-1}&0\\ 0&A \end{pmatrix}.$$

The representation is $(1/2,0)\oplus(0,1/2)$: it has $2+2=4$ complex components. A tensor product would instead be $(1/2,1/2)$, the vector representation. Equal component counts do not make these two representations equivalent.

Both chiral subspaces are preserved by every connected Lorentz transformation. Thus a Dirac spinor is reducible as a representation of the connected Lorentz cover. Its massive equation nevertheless couples the two subspaces dynamically. Reducibility of a component representation and decoupling of an equation are different questions.

A mass term connects the two Weyl equations

The chiral gamma blocks turn $(i\gamma^\mu\partial_\mu-m)\Psi=0$ into

$$i\bar\sigma^\mu\partial_\mu\chi_L=m\chi_R,\qquad i\sigma^\mu\partial_\mu\chi_R=m\chi_L.$$

For $m=0$ the equations decouple. For $m\ne0$, setting either component identically to zero forces the other to vanish too. That statement concerns an entire solution, not arbitrary initial data on one time slice: initial data may have only one chiral block, and subsequent evolution then generates the other.

For a positive-frequency plane wave with future $p^\mu=(E,\mathbf p)$, the equations read

$$(E+\boldsymbol\sigma\cdot\mathbf p)\chi_L=m\chi_R, \qquad (E-\boldsymbol\sigma\cdot\mathbf p)\chi_R=m\chi_L.$$

Multiplying the two matrices gives $(E^2-\mathbf p^2)\chi_{L,R}=m^2\chi_{L,R}$. The common mass shell follows, but the first-order relation between the blocks remains necessary.

At rest, a positive-energy mode has $\chi_R=\chi_L$; a negative-energy eigenmode of the rest Hamiltonian has $\chi_R=-\chi_L$. Neither is confined to one chirality. Conversely, a rest spinor with only $\chi_L$ populated is an equal-amplitude superposition of the two energy signs. This is the chiral-basis version of the distinction between component blocks and energy sectors.

Why the adjoint pairs opposite chiralities

Since $\gamma^0_C$ interchanges the blocks,

$$\bar\Psi=(\chi_R^\dagger,\chi_L^\dagger),\qquad \bar\Psi\Psi=\chi_R^\dagger\chi_L+\chi_L^\dagger\chi_R.$$

Each cross term is Lorentz invariant. For example, $\chi_R^\dagger\chi_L$ transforms to $\chi_R^\dagger A^\dagger(A^\dagger)^{-1}\chi_L$. The separate Euclidean norms of the blocks are not scalars under boosts.

The vector current instead adds the two chiral currents:

$$j^\mu =\chi_L^\dagger\bar\sigma^\mu\chi_L +\chi_R^\dagger\sigma^\mu\chi_R.$$

Its density is $j^0=\chi_L^\dagger\chi_L+\chi_R^\dagger\chi_R$. The scalar mass bilinear and the positive probability density therefore use different pairings, even though both contain two spinor factors.

Mass transfers current between chiral components

For free, commuting c-number amplitudes, use the coupled equations and their adjoints to obtain

$$\begin{aligned} \partial_\mu j_L^\mu &=im(\chi_R^\dagger\chi_L-\chi_L^\dagger\chi_R),\\ \partial_\mu j_R^\mu &=-\partial_\mu j_L^\mu . \end{aligned}$$

Their sum is conserved. For packets with vanishing flux at infinity, the integrated left and right weights can change while their sum remains constant. The source term depends on the relative phase of the two chiral components; mass does not imply a universal classical rate of chirality flipping.

The axial current is $j_5^\mu=j_R^\mu-j_L^\mu$ and satisfies

$$\partial_\mu j_5^\mu=2im\,\bar\Psi\gamma^5\Psi.$$

This is a classical free-equation identity. The axial anomaly is an additional quantum effect in suitable interacting field theories; it is not included by this calculation.

A useful time-domain check is the rest Hamiltonian $H=m\gamma^0_C$. Starting with $\Psi(0)=(\chi,0)^T$ gives

$$\Psi(t)= \begin{pmatrix} \cos(mt)\chi\\ -i\sin(mt)\chi \end{pmatrix}.$$

The weights oscillate as $\cos^2(mt)$ and $\sin^2(mt)$. The chosen initial condition contains both energy sectors. It should not be identified with the evolution of an initially positive-energy electron at rest.

Parity and changes of basis

With a conventional intrinsic phase suppressed, the Dirac parity map is $\Psi'(t,\mathbf x)=\gamma^0\Psi(t,-\mathbf x)$. In the chiral basis it exchanges $\chi_L$ and $\chi_R$. Parity lies outside the connected Lorentz group, so this exchange does not contradict preservation of each chirality by $S(A)$. A lone chiral representation does not implement parity within itself in the same way.

For a Dirac-basis column $\Psi_D=(\phi,\eta)^T$, the unitary conversion to this chiral convention is

$$\Psi_C=U\Psi_D,\qquad U=\frac1{\sqrt2} \begin{pmatrix}I&-I\\ I&I\end{pmatrix}.$$

Thus $\chi_L=(\phi-\eta)/\sqrt2$ and $\chi_R=(\phi+\eta)/\sqrt2$. Every gamma matrix and observable must transform with the same $U$: $\gamma^\mu_C=U\gamma^\mu_DU^\dagger$. In particular, “upper components” means different physical projections in the Dirac and chiral bases.

Exercises

1. Check that the displayed $U$ maps $\gamma^0_D=\operatorname{diag}(I,-I)$ to $\gamma^0_C$ and $\gamma^5_D$ to $\operatorname{diag}(-I,I)$.

Solution

Block multiplication gives $U\gamma^0_DU^\dagger=\begin{pmatrix}0&I\\I&0\end{pmatrix}$ and $U\gamma^5_DU^\dagger=\begin{pmatrix}-I&0\\0&I\end{pmatrix}$. Also $UU^\dagger=I$, so the integrated Hilbert norm is unchanged.

2. For the rest evolution above, verify the left-current source term directly with $\chi^\dagger\chi=1$.

Solution

$\chi_R^\dagger\chi_L=i\sin(mt)\cos(mt)$ and $\chi_L^\dagger\chi_R=-i\sin(mt)\cos(mt)$. The source is $-2m\sin(mt)\cos(mt)$, equal to $d\cos^2(mt)/dt$. The right weight changes by the opposite amount.

3. Explain why a massive positive-energy rest spinor has equal left and right weights yet no oscillation between energy sectors.

Solution

Its blocks are equal and share the common phase $e^{-imt}$. It is an energy eigenvector, so the state acquires only that phase. Equal chiral weights do not mean a mixture of positive and negative energies; chirality and energy use different projectors.

References

• H. K. Dreiner, H. E. Haber, and S. P. Martin, “Two-component spinor techniques and Feynman rules for quantum field theory and supersymmetry,” Physics Reports 494, 1–196, 2010, doi:10.1016/j.physrep.2010.05.002; corrected arXiv:0812.1594v6, 2022 — two- and four-component representations and mass terms.
• M. D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press, 2014, doi:10.1017/9781139540940 — Dirac representations and vector and axial currents.
• D. Tong, Quantum Field Theory, University of Cambridge lecture notes, 2006, chapter 4 — chiral components and Dirac dynamics.