Parity

Spatial inversion acts on a Dirac wavefunction through both its argument and its spinor components. The matrix $\gamma^0$ compensates the sign change of spatial derivatives, making the transformed spinor solve the equation in the parity-transformed electromagnetic background. This covariance becomes a symmetry of one fixed problem only when its background and domain are also invariant. Parity owns the general unitary-operator and selection-rule framework; this page develops the relativistic equation maps.

Required background. Gamma-Matrix Conventions fixes the matrices and projectors, and Minimal Coupling fixes the electromagnetic signs. Helpful background. Parity in Quantum Mechanics supplies the operator interpretation; Hydrogen Fine Structure Revisited supplies central-field spinor harmonics.

Scalar and spinor inversion maps

Write $x_P=(ct,-\mathbf x)$. For a complex scalar solution, define

$$\phi_P(t,\mathbf x)=\eta_P\phi(t,-\mathbf x), \qquad |\eta_P|=1.$$

For a Dirac solution, use

$$\psi_P(t,\mathbf x) =\eta_P\beta\psi(t,-\mathbf x), \qquad \beta=\gamma^0.$$

These maps are complex-linear: they do not conjugate amplitudes or reverse charge. On Dirac initial data, the spatial reflection and the unitary matrix $\beta$ preserve $\int d^3x\,\psi^\dagger\psi$. We choose $\eta_P=1$ for the ordinary Dirac parity operator $\Pi=\beta R$, with $(R\psi)(\mathbf x)=\psi(-\mathbf x)$. Then $\Pi^2=I$ and its eigenvalues are $\pm1$.

A general phase gives $\Pi_{\eta}^2=\eta_P^2I$. Requiring the square to be exactly $I$ restricts that convention to $\eta_P=\pm1$. Extra conditions on the field, such as a fixed Majorana reality structure, can require a different phase convention; the coordinate inversion alone does not settle every internal phase.

The scalar Klein–Gordon operator has two spatial derivatives, so reflection preserves its free form directly. For the first-order Dirac equation, the required identities are

$$\beta\gamma^0\beta=\gamma^0,\qquad \beta\gamma^i\beta=-\gamma^i.$$

They compensate the reflected spatial derivatives. A componentwise reflection $\psi(t,\mathbf x)\mapsto\psi(t,-\mathbf x)$ without $\beta$ does not do so. Tong’s Dirac lectures discuss this need to extend beyond the connected Lorentz group.

Electromagnetic backgrounds and covariance

For a real electromagnetic background, define

$$\begin{aligned} \Phi_P(t,\mathbf x)&=\Phi(t,-\mathbf x),\\ \mathbf A_P(t,\mathbf x)&=-\mathbf A(t,-\mathbf x). \end{aligned}$$

The charge parameter $q$ is unchanged. The fields transform as

$$\mathbf E_P(t,\mathbf x)=-\mathbf E(t,-\mathbf x), \qquad \mathbf B_P(t,\mathbf x)=+\mathbf B(t,-\mathbf x).$$

Thus electric field is polar and magnetic field is axial. For the covariant scalar equation, the temporal derivative transforms evenly and the spatial covariant derivatives oddly, leaving their Lorentz contraction unchanged.

The spinor calculation is especially transparent in Hamiltonian form. Since $R\mathbf pR^{-1}=-\mathbf p$ and $\beta\alpha_i\beta=-\alpha_i$,

$$\begin{aligned} \Pi H_q[\Phi,\mathbf A](t)\Pi^{-1} ={}&c\boldsymbol\alpha\cdot \bigl(\mathbf p+q\mathbf A(t,-\mathbf x)\bigr)\\ &+\beta mc^2+q\Phi(t,-\mathbf x)\\ ={}&H_q[\Phi_P,\mathbf A_P](t). \end{aligned}$$

Because $\Pi$ is time independent, $\psi_P$ obeys the transformed Schrödinger equation with no extra time-derivative term. This is an intertwining relation between two external-field problems.

For a fixed static background to have parity as a symmetry in this gauge, require $\Phi_P=\Phi$, $\mathbf A_P=\mathbf A$, and a parity-invariant operator domain. More generally the transformed potentials may be gauge equivalent to the originals; then the symmetry operator must include the compensating gauge phase described on Gauge Covariance. Comparing field strengths or potentials without keeping this distinction can give an incorrect symmetry verdict.

For example, a central electrostatic potential is parity symmetric. A uniform magnetic field also respects parity: in the symmetric gauge $\mathbf A=\mathbf B\times\mathbf x/2$, the parity-transformed potential equals the original. A fixed nonzero uniform electric field is instead mapped to the opposite field and is not parity symmetric.

Spin, chirality, and currents

The spin matrix is axial: $\beta\boldsymbol\Sigma\beta=\boldsymbol\Sigma$. Meanwhile inversion reverses momentum. Hence helicity changes sign,

$$\Pi\frac{\boldsymbol\Sigma\cdot\mathbf p}{2|\mathbf p|} \Pi^{-1} =-\frac{\boldsymbol\Sigma\cdot\mathbf p}{2|\mathbf p|}.$$

Since $\beta\gamma^5\beta=-\gamma^5$, parity also exchanges the chiral projectors,

$$\beta P_L=P_R\beta,\qquad \beta P_R=P_L\beta.$$

In the chiral basis, $\gamma^0$ interchanges the left and right Weyl columns. A single Weyl representation alone therefore does not admit this parity action as a map within the same component space. A Dirac representation contains both. That representation-theoretic fact does not prove that arbitrary interactions respect parity.

The Dirac probability density and current obey

$$\begin{aligned} \rho_P(t,\mathbf x)&=\rho(t,-\mathbf x),\\ \mathbf j_P(t,\mathbf x)&=-\mathbf j(t,-\mathbf x). \end{aligned}$$

The complete scalar, pseudoscalar, vector, axial-vector, and tensor parity table belongs to Bilinear Covariants. For example, its axial spatial current is even while its axial time component is odd. Those transformation laws remain meaningful even in a background that breaks parity.

Parity of a central-field Dirac eigenstate

Use the separated state

$$\Psi_{\kappa m_j} =\frac1r \begin{pmatrix} G(r)\Omega_{\kappa m_j}\\ iF(r)\Omega_{-\kappa m_j} \end{pmatrix}.$$

The upper harmonic has orbital label $\ell$ and inversion sign $(-1)^\ell$. The lower harmonic has $\ell'=2j-\ell=\ell\pm1$, so it has the opposite orbital sign. The lower block of $\beta$ contributes another minus sign. Both components therefore transform with the same total parity:

$$\Pi\Psi_{\kappa m_j} =(-1)^\ell\Psi_{\kappa m_j}.$$

The Dirac state has a definite parity although its two components have different orbital angular momenta. For the point Coulomb problem, $2S_{1/2}$ and $2P_{1/2}$ have opposite parities despite their equal energies. A parity-even perturbation cannot mix them, but parity alone does not require their diagonal energy shifts to be equal.

Exercises

A parity-breaking electric field. Take $\Phi=-E_0z$, $\mathbf A=0$. Find the transformed potential and field. Does the charge reverse?

Solution

The transformed potential is $\Phi_P=+E_0z$, so $\mathbf E_P=-E_0\hat{\mathbf z}$. The charge is still $q$. The transformed state solves a different background problem; this is covariance, not a symmetry of the fixed field.

Upper and lower orbital parity. For $\kappa=-2$, identify $\ell,j,\ell'$ and the total parity. Repeat for $\kappa=+2$.

Solution

For $\kappa=-2$, $\ell=1$, $j=3/2$, and $\ell'=2$; the total parity is negative. For $\kappa=+2$, $\ell=2$, $j=3/2$, and $\ell'=1$; the total parity is positive. In each case the lower $\beta$ sign compensates its opposite orbital parity.

Chiral projectors. Starting with $P_L=(I-\gamma^5)/2$, derive the parity exchange and explain why it leaves the Dirac mass coupling consistent.

Solution

Anticommutation gives $\beta P_L=(I+\gamma^5)\beta/2=P_R\beta$. Parity interchanges left and right components. The real scalar mass term couples those components symmetrically, so it is invariant under that exchange. This argument assumes the ordinary scalar Dirac mass, not an arbitrary pseudoscalar background.

References

• Bjorken, James D., and Sidney D. Drell. Relativistic Quantum Mechanics. McGraw–Hill (1964).
• Thaller, Bernd. The Dirac Equation. Springer (1992). doi:10.1007/978-3-662-02753-0.
• Tong, David. Quantum Field Theory. University of Cambridge lecture notes (2006), chapter 4. The Dirac Equation.