Velocity Operator and Zitterbewegung

We will work for a while in the Heisenberg representation in which the operators depend on time and we can see some of the general behavior of electrons. If we work in a state of definite energy, the time dependence of the operators is very simple, just the usual exponentials.

The operator for velocity in the x direction can be computed from the commutator with the Hamiltonian.

\dot{x}={i\over\hbar}[H,x]={i\over\hbar}ic[\gamma_4\gamma_jp_j,x]=ic\gamma_4\gamma_1 \\
v_j=ic\gamma_4\gamma_j \\

The velocity operator then is \bgroup\color{black}$v_j=ic\gamma_4\gamma_j$\egroup.

Its not hard to compute that the velocity eigenvalues (any component) are \bgroup\color{black}$\pm c$\egroup.

ic\gamma_4\gamma_1=ic\pmatrix{1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 ...
...2-1)^2=0 \\
(\lambda^2-1)=0 \\
\lambda=\pm 1 \\
v_x=\pm c \\

Thus, if we measure the velocity component in any direction, we should either get plus or minus \bgroup\color{black}$c$\egroup. This seems quite surprising, but we should note that a component of the velocity operator does not commute with momentum, the Hamiltonian, or even the other components of the velocity operator. If the electron were massless, velocity operators would commute with momentum. (In more speculative theories of particles, electrons are actually thought to be massless, getting an effective mass from interactions with particles present in the vacuum state.)

The states of definite momentum are not eigenstates of velocity for a massive electron. The velocity eigenstates mix positive and ``negative energy'' states equally.

\pmatrix{0 & 0 & 0 & 1 \cr 0 & 0 & 1 & 0 \cr 0 & 1 & 0 & 0 \cr...
...1\cr 0\cr 0\cr -1\cr}\qquad
u=\pmatrix{0\cr 1\cr -1\cr 0\cr} \\

Thus, while momentum is a constant of the motion for a free electron and behaves as it did in NR Quantum Mechanics, velocity behaves very strangely in the Dirac theory, even for a free electron. Some further study of this effect is in order to see if there are physical consequences and what is different about the Dirac theory in this regard.

We may get the differential equation for the velocity of a free electron by computing the derivative of velocity. We attempt to write the derivative in terms of the constants of the motion \bgroup\color{black}$E$\egroup and \bgroup\color{black}$\vec{p}$\egroup.


This is a differential equation for the Heisenberg operator \bgroup\color{black}$v_j$\egroup which we may solve.

\begin{displaymath}\bgroup\color{black} v_j(t)=c^2p_j/E+(v_j(0)-c^2p_j/E)e^{-2iEt/\hbar} \egroup\end{displaymath}

To check, differentiate the above

\begin{displaymath}\bgroup\color{black} \dot{v}_j(t)=-{2iE\over\hbar}(v_j(0)-c^2...
...\hbar}={i\over\hbar}(-2Ev_j(0)+2c^2p_j) e^{-2iEt/\hbar} \egroup\end{displaymath}

and compare it to the original derivative.

={i\over\hbar}(-2Ev_j(0)+2c^2p_j)e^{-2iEt/\hbar} \\

This checks so the solution for the velocity as a function of time is correct.

\begin{displaymath}\bgroup\color{black} v_j(t)=c^2p_j/E+(v_j(0)-c^2p_j/E)e^{-2iEt/\hbar} \egroup\end{displaymath}

There is a steady motion in the direction of the momentum with the correct magnitude \bgroup\color{black}$\beta c$\egroup. There are also very rapid oscillations with some amplitude. Since the energy includes \bgroup\color{black}$mc^2$\egroup, the period of these oscillations is at most \bgroup\color{black}${2\pi\hbar\over 2mc^2}={\pi\hbar c\over mc^2c}={(3.14)(197....
...c)}=1200F/c={1200\times 10^{-13}\over 3\times 10^{10}}
=4\times 10^{-21}$\egroup seconds. This very rapid oscillation is known as Zitterbewegung. Obviously, we would see the same kind of oscillation in the position if we integrate the above solution for the velocity. This very rapid motion of the electron means we cannot localize the electron extremely well and gives rise to the Darwin term. This operator analysis is not sufficient to fully understand the effect of Zitterbewegung but it illustrates the behavior.

Jim Branson 2013-04-22