Zitterbewegung

Zitterbewegung is the rapidly oscillating term in the free Dirac position operator in its usual representation. It connects positive and negative energy sectors and disappears from its expectation in a state confined to either free sector. The calculation is an operator and interference result; it does not, by itself, establish a literal microscopic trajectory of an isolated electron.

Required background. The Dirac Hamiltonian as an Operator gives the spectral projectors and velocity; Dirac Negative-Energy Solutions explains the sector interpretation. Helpful background. Localization Problems discusses relativistic position observables.

Heisenberg evolution of the velocity

Retain $c,\hbar$ and take the free massive Hamiltonian $H=c\boldsymbol\alpha\cdot\mathbf P+\beta mc^2$ on all of space. The following identities hold on a suitable common invariant core, such as Schwartz spinors. The free gap makes $H^{-1}$ bounded for $m>0$. Since

$$\dot X_i=\frac{i}{\hbar}[H,X_i]=c\alpha_i,\qquad \{H,\alpha_i\}=2cP_i,$$

the velocity matrix satisfies

$$\dot\alpha_i=\frac{2i}{\hbar}(cP_i-\alpha_iH).$$

Define the oscillatory part $A_i=\alpha_i(0)-cP_iH^{-1}$. It anticommutes with $H$, because both terms have the same anticommutator $2cP_i$. As $\mathbf P,H$ are constants of the free motion, the exact solution is

$$\alpha_i(t)=cP_iH^{-1}+A_i e^{-2iHt/\hbar}.$$

The operator order is deliberate: $A_i$ does not commute with $H$. Differentiating this expression reproduces the equation above.

Position is a drift plus an oscillatory term

Integrating $\dot X_i=c\alpha_i(t)$ yields

$$\begin{aligned} X_i(t)={}&X_i(0)+c^2P_iH^{-1}t\\ &+\frac{i\hbar c}{2}A_iH^{-1} \left(e^{-2iHt/\hbar}-I\right). \end{aligned}$$

At fixed momentum the phase frequency has magnitude $2E_{\mathbf p}/\hbar$. Near rest this is $2mc^2/\hbar$, and the coefficient of the oscillatory part has the length scale $\hbar/(mc)$. These scales follow from the energy gap; they are not a universal measured oscillation amplitude for every packet. Amplitudes also depend on its sector coherence and on the chosen position observable.

The last term is Hermitian even though its factors do not individually make that obvious. Anticommutation implies $A_iH^{-1}=-H^{-1}A_i$ and moves the exponential to its inverse when it passes $A_i$. Keeping the factor order avoids an apparent sign or Hermiticity contradiction.

Why a single energy sector has no free interference term

With $P_\pm$ the free spectral projectors, anticommutation gives

$$P_\pm A_iP_\pm=0,\qquad A_i=P_+A_iP_-+P_-A_iP_+.$$

Thus the oscillatory expectation contains only coherence between opposite sectors. For $P_+\psi=\psi$ it vanishes, and the drift velocity becomes $c^2\mathbf P/E_{\mathbf P}$; for the negative sector the sign reverses.

Projecting a state and transforming a position operator are different operations. A unitary change of representation must transform observables as well as states. Position operators adapted to a positive-energy subspace have their own momentum dependence. One cannot keep the same component expression for position after changing representation and infer that a physical observable changed.

A simple internal-spinor check and packet dephasing

At fixed momentum $\mathbf p=0$, choose the unit internal spinor $\zeta=(1,0,1,0)^T/\sqrt2$ in the Dirac basis. It has equal positive and negative energy amplitudes. Since $H=\beta mc^2$,

$$\zeta^\dagger c\alpha_z(t)\zeta =c\cos(2mc^2t/\hbar).$$

This is a velocity-matrix check at one momentum. A plane wave does not have a localized position expectation, so it must not be drawn as a measured particle trajectory. A sufficiently narrow packet around zero momentum with the corresponding coherence has an approximately sinusoidal contribution while its mode phases remain aligned. Integrating the displayed cosine gives the characteristic coefficient $\hbar/(2mc)$.

For a generic packet with momentum width $\Delta p\ll mc$ near rest, $2E_{\mathbf p}=2mc^2+\mathbf p^2/m+\cdots$. the general phase-spread estimate is $t_{\rm dephase}\sim\hbar/\Delta(2E)$. If the spread in $\mathbf p^2$ is of order $(\Delta p)^2$, as for a near-rest Gaussian, this gives the order-of-magnitude time scale

$$t_{\rm dephase}\sim\frac{m\hbar}{(\Delta p)^2}.$$

An angular spread at nearly fixed $|\mathbf p|$ need not produce this energy spread. The precise damping envelope depends on the packet, its spin preparation, and spatial overlap of the sectors. A monochromatic calculation has no such dephasing and should not be treated as the generic long-time behavior of a localized packet.

What quantum simulations demonstrate

Trapped-ion experiments can engineer a two-level internal system and motional coordinate with an effective one-dimensional Dirac Hamiltonian. Gerritsma et al. (2010) observed the corresponding simulated zitterbewegung dynamics. The effective velocity, mass, position, and initial sector mixture are defined by the simulator’s mapping. This tests controlled Dirac-like interference; it is not a direct observation of a free electron moving along the formal trajectory of the first-quantized position operator.

For physical charged relativistic particles, statements about localization, pair creation, and detector signals require the field and measurement description. In particular, the exact Dirac current bound prevents this oscillation from implying superluminal signal transport.

Exercises

1. Prove $\{H,A_i\}=0$ and show that $A_i$ sends an energy eigenvector to the opposite energy subspace, unless it annihilates it.

Solution

$\{H,cP_iH^{-1}\}=2cP_i=\{H,\alpha_i\}$, so the difference anticommutes. If $H\psi=E\psi$, then $H A_i\psi=-E A_i\psi$. This gives the off-diagonal projector structure used above.

2. Differentiate the oscillatory position term and recover the oscillatory velocity, retaining the displayed factor order.

Solution

The derivative is $(i\hbar c/2)A_iH^{-1}(-2iH/\hbar)e^{-2iHt/\hbar} =cA_i e^{-2iHt/\hbar}$. Adding the drift $c^2P_iH^{-1}$ gives $c\alpha_i(t)$.

3. Can an incoherent mixture of positive and negative energy packets have a nonzero oscillatory expectation merely because both are occupied?

Solution

Not if its density operator is block diagonal in the free energy sectors. The oscillatory operator is off diagonal, so its trace against that state vanishes. Occupation of both sectors and coherence between them are different conditions.

References

• J. D. Bjorken and S. D. Drell, Relativistic Quantum Mechanics, McGraw–Hill, 1964 — Heisenberg velocity and zitterbewegung.
• R. Gerritsma, G. Kirchmair, F. Zähringer, E. Solano, R. Blatt, and C. F. Roos, “Quantum Simulation of the Dirac Equation,” Nature 463, 68–71, 2010, doi:10.1038/nature08688 — a trapped-ion simulation and its observable mapping.
• B. Thaller, The Dirac Equation, Springer, 1992 — spectral-sector and position-operator interpretation.