Dirac Dynamics in Electromagnetic Fields

The Dirac equation in a prescribed electromagnetic field evolves a four-component amplitude while conserving its positive norm. It gives exact operator identities for velocity, Lorentz force, and electrical work, but those identities do not automatically reduce to a classical trajectory. Stationary backgrounds support spectral problems; time-dependent backgrounds can mix instantaneous energy sectors. Neither calculation alone defines a vacuum pair-production probability.

Required background. Minimal Coupling derives the Hamiltonian, Hamiltonian Form specifies its free operator setting, and Gauge Covariance relates time-dependent gauge descriptions.

The background Hamiltonian and its domain

Retain SI-normalized potentials, $\hbar,c$, and signed charge $q$. The Hamiltonian derived at the coupling owner is

$$H(t)=c\boldsymbol\alpha\cdot\boldsymbol\pi(t) +\beta mc^2+q\Phi(t), \qquad \boldsymbol\pi=\mathbf p-q\mathbf A.$$

Real potentials make the formal expression Hermitian. To obtain a self-adjoint operator, its spatial domain must also be fixed. For example, bounded real $\Phi,\mathbf A$ are bounded Hermitian perturbations of the free operator on $H^1(\mathbb R^3,\mathbb C^4)$, so that domain gives a self-adjoint realization. Sufficient regularity in time then gives a unitary evolution family.

Singular Coulomb sources, boundaries, or unbounded gauges require the corresponding domain analysis rather than this bounded-perturbation shortcut. The identities below are derived on smooth spinors where the products and commutators are meaningful, with appropriate decay or boundary conditions when taking expectations.

The density and current remain $\rho=\psi^\dagger\psi$ and $\mathbf j=c\psi^\dagger\boldsymbol\alpha\psi$. The real electromagnetic terms cancel between the equation and its adjoint, giving the same continuity equation as in the free case. Norm conservation does not require the external field to preserve free-particle energy projectors.

Exact velocity and the kinetic Lorentz force

The Heisenberg velocity is

$$\dot{\mathbf X} =\frac{i}{\hbar}[H,\mathbf X] =c\boldsymbol\alpha.$$

It is not $\boldsymbol\pi/m$ as an operator on the full Dirac space. The relation to a classical group velocity requires a suitable energy sector and packet approximation.

Use $[\pi_i,\pi_j]=iq\hbar\epsilon_{ijk}B_k$ and $[\pi_i,f]=-i\hbar\partial_i f$. Including the explicit time dependence of $\boldsymbol\pi$ gives

$$\begin{aligned} \dot\pi_i &=\frac{i}{\hbar}[H,\pi_i]-q\partial_tA_i\\ &=qE_i+qc\,\epsilon_{ijk}\alpha_jB_k. \end{aligned}$$

Thus $\dot{\boldsymbol\pi}=q(\mathbf E+\dot{\mathbf X}\times\mathbf B)$. In this representation the alpha matrices commute with the multiplication operators $B_k(\mathbf X,t)$, so the displayed magnetic term is Hermitian. For more general velocity operators one must check the ordering before using the same notation.

Taking an expectation does not replace $\langle\alpha_jB_k(\mathbf X)\rangle$ by $\langle\alpha_j\rangle B_k(\langle\mathbf X\rangle)$ without an approximation. Spatial spread and spin–position correlations can matter.

Mechanical energy and electrical work

Define the gauge-covariant mechanical-energy operator

$$\mathcal K=H-q\Phi =c\boldsymbol\alpha\cdot\boldsymbol\pi+\beta mc^2.$$

It is not the positive square root of a classical kinetic-energy expression; it acts on both Dirac energy sectors. Directly,

$$\frac{d\mathcal K}{dt} =\frac{i}{\hbar}[H,\mathcal K] +\partial_t\mathcal K =qc\,\boldsymbol\alpha\cdot\mathbf E.$$

The magnetic field does no work in this identity. For a static background, $\langle H\rangle$ is conserved, but the mechanical energy can change against the electrostatic potential. For a time-dependent background,

$$\frac{d\langle H\rangle}{dt} =\left\langle q\partial_t\Phi-qc\boldsymbol\alpha\cdot\partial_t\mathbf A \right\rangle.$$

The background is prescribed; the equation does not evolve the source supplying this energy. Including backreaction requires an additional field or source dynamics.

Stationary states and the danger of squaring too early

For static potentials, a stationary state obeys $H\psi_E=E\psi_E$. The kinetic operator has the exact square

$$\mathcal K^2 =m^2c^4+c^2\boldsymbol\pi^2 -q\hbar c^2\boldsymbol\Sigma\cdot\mathbf B.$$

The mass cross terms vanish by alpha–beta anticommutation, while the magnetic term comes from the noncommuting kinetic momenta. For a purely magnetic background with $\Phi=0$, an energy eigenstate therefore satisfies the corresponding equation with eigenvalue $E^2$. The original first-order equation still selects the allowed spinor components and energy sign.

For spatially varying $\Phi$, however, $\mathcal K\psi_E=(E-q\Phi)\psi_E$ cannot be squared as commuting algebra. Instead

$$\mathcal K^2\psi_E =\left[(E-q\Phi)^2 -iq\hbar c\,\boldsymbol\alpha\cdot\mathbf E\right]\psi_E,$$

where $\mathbf E=-\nabla\Phi$ in this static setting. The extra term follows from $[\mathcal K,E-q\Phi] =i\hbar c\,\boldsymbol\alpha\cdot\nabla(q\Phi)$. This second-order equation is an energy-dependent consequence of the first-order eigenproblem, not an ordinary scalar Schrödinger Hamiltonian.

Keeping the electric commutator is essential in relativistic bound-state reductions. Dropping it would erase terms responsible for spin and gradient corrections in the low-energy expansion.

A uniform electric field gives a finite mode equation

Choose temporal gauge for $\mathbf E=E_0\hat{\mathbf z}$: $\Phi=0$ and $\mathbf A=-E_0t\,\hat{\mathbf z}$. Spatial translation symmetry allows $\psi(t,\mathbf x)=e^{i\mathbf p\cdot\mathbf x/\hbar}w_{\mathbf p}(t)$, where $\mathbf p$ is the conserved canonical label. The four-spinor obeys

$$i\hbar\dot w_{\mathbf p} =\left[ c\boldsymbol\alpha\cdot (\mathbf p+qE_0t\,\hat{\mathbf z}) +\beta mc^2 \right]w_{\mathbf p}.$$

At each time this matrix is Hermitian. For $m>0$ it has two eigenvalues of each sign,

$$E_\pm(t)=\pm\sqrt{ m^2c^4+c^2|\mathbf p+qE_0t\,\hat{\mathbf z}|^2}.$$

The exact mode norm is conserved. The instantaneous eigenvectors vary, so initial projection onto one instantaneous sector does not automatically keep the solution there. An adiabatic estimate compares $\hbar\|\dot H_{\mathbf p}\|$ with the squared instantaneous gap, together with the duration and mode geometry. A finite gap alone does not prove adiabatic following.

This finite matrix problem is a useful mode solver. Interpreting mode conversion as particle production adds the quantized matter field, initial vacuum or occupation data, and suitable final modes. The norm of a lower component alone is not a produced-particle number.

Exercises

1. Derive the electric part of $\dot{\boldsymbol\pi}$ and explain why differentiating only the operator commutator is insufficient in a time-dependent gauge.

Solution

$(i/\hbar)[q\Phi,\pi_i]=-q\partial_i\Phi$. The explicit derivative of $-qA_i$ adds $-q\partial_tA_i$, giving $qE_i$. Without that explicit derivative, the temporal-gauge uniform electric field would incorrectly produce no electric force.

2. Check the mechanical work identity for a purely static magnetic field with $\Phi=0$.

Solution

$\mathbf E=0$, $\mathcal K=H$, and $H$ has no explicit time dependence. Its expectation is conserved, even though the kinetic momentum can turn under the magnetic force.

3. Why can solving the squared stationary Dirac equation produce candidates that are not solutions of the original first-order equation?

Solution

The squared equation is necessary but loses first-order component relations and the chosen energy branch. Candidate modes must be substituted back into $H\psi_E=E\psi_E$, with the same boundary conditions and operator domain.

References

• J. D. Bjorken and S. D. Drell, Relativistic Quantum Mechanics, McGraw–Hill, 1964 — external-field dynamics and stationary reductions.
• W. Greiner, Relativistic Quantum Mechanics: Wave Equations, 3rd ed., Springer, 2000, doi:10.1007/978-3-662-04275-5 — electromagnetic backgrounds and spinor solutions.
• B. Thaller, The Dirac Equation, Springer, 1992 — self-adjoint Dirac operators and external fields.