Is Zitterbewegung an artefact of single-particle theory? I have seen a number of articles on Zitterbewegung claiming searches for it such as this one: http://arxiv.org/abs/0810.2186. Others such as the so-called ZBW interpretation by Hestenes seemingly propose to explain electron spin as a consequence of ZBW.
According to Itzykson and Zuber p.62, Zitterbewegung is an artefact of considering a single-particle theory. It has been pointed out in this question's replies that it is not a physical phenomenon: What was missing in Dirac's argument to come up with the modern interpretation of the positron?. 
How does the upgrade to many-particle theory solve the issue?
 A: See this relevant quotation and reference; ''  Seeing that the fact that an old interest in ZB has recently been rekindled by the investigations on spintronic, graphene, and superconducting systems, etc., in this paper we present a quantum-field-theory investigation on ZB and obtain the conclusion that, the ZB of an electron arises from the influence of virtual electron-positron pairs (or vacuum fluctuations) on the electron.  '' https://arxiv.org/ftp/quant-ph/papers/0612/0612090.pdf
A: ZBW doesn't exist on Pauli equation, which describes a non-relativistic particle with spin one-half. Thus spin exists without ZBW contrary to Hestenes claim.
Neither ZBW is an artifact of a single-particle wavefunction theory, because ZBW continue present beyond the one-particle theory. In wavefunction theory, the ZBW follows from the coupling of the positive and negative energy branches. ZBW also exist in the quantum field theory where is reinterpreted as the effect from the coupling of the electrons and positrons
A: The Zitterbewegung is more of a relic of the early Dirac equation days. It does not exist in the standard position, velocity and acceleration operators of the single particle field, only in alternatively derived versions. These alternative versions were developed because people thought the standard operators were wrong. In fact they didn't understand the standard operators. The standard method is using:
$\frac{\partial {\cal \tilde{O}}}{dt} ~=~ \frac{i}{\hbar}\!\!\left[~\tilde{H},\tilde{O}~\right]$
Where the misunderstanding comes from is easy to see in the modern Chiral representation. We will show that the standard operators are correct. If we define a position, a velocity and an acceleration operator for the Dirac field then the (averaged) position and velocity and acceleration are given by:

Position, Velocity and Acceleration operators applied on the Dirac field:
$\vec{x}_{avg} ~=~ \frac{1}{2mc}\int dx^3 ~~ \psi^* \vec{X}~\psi  ~~~~~~~~~ (\vec{X}: \mbox{position operator}) $
$\vec{v}_{avg} ~=~ \frac{1}{2mc}\int dx^3 ~~ \psi^* \vec{V}~\psi \,~~~~~~~~~ (\vec{V}: \mbox{velocity operator}) $
$\vec{a}_{avg} ~=~ \frac{1}{2mc}\int dx^3 ~~ \psi^* \vec{A}~\psi \,~~~~~~~~~ (\vec{A}: \mbox{acceleration operator}) $

Velocity operator
Now $\vec{X}$ is simply the position $\vec{x}$ of each point of the wavefunction. The velocity operator can be derived by commutating with the Hamiltonian.
$\tilde{V}^i\psi\ =\  \frac{i}{\hbar}\left[~\tilde{H},\tilde{X}^i~\right]\psi\ =\ c \left( \begin{array}{cc} -\sigma^i & 0 \\ 0 & \sigma^i \end{array} \right)\psi$
This velocity operator is in fact totally correct but it was thought to be erroneous in the early days because people misunderstood it to mean that the electron can only move with $\pm\,c$, and therefor it must be wrong, they thought. 
What they were actually expecting was something like the $\vec{v}=\vec{p}/m$ as they got in non relativistic theories, but they found something which only contained $\pm\,c$. However, if we evaluate the expression for $\vec{v}_{avg}$ then we get.
$\vec{v}_{avg} ~=~ \frac{c}{2mc}\int dx^3 ~~ \psi^* \left( \begin{array}{cc} -\sigma^i & 0 \\ 0 & \sigma^i \end{array} \right)\psi  ~~=~~ \frac{c}{2mc}\int dx^3 ~~ \bar{\psi} \gamma^i \psi  ~~=~~ \frac{c}{2mc}\int dx^3 ~ j^i$
This is an integral over the current density, or the momentum with the appropriate units. Now the momentum $\vec{p}$ is a factor $\gamma$ larger as the velocity $\vec{v}$ but the integral over the Lorentz contracted field compensates this so we end up with the velocity of the particle as required! The velocity operator is perfectly fine.
The other big misunderstanding was that the x,y and z-components of the velocity operator do not commute while they do so in the non-relativistic theory and therefor the operator must be wrong, they thought. You can still find this quoted in many textbooks.
But as you see the expression derives the velocity from the momentum and as we know the momentum components (the boost components) should not commute. In fact they should commute just like in the velocity operator. Again the operator behaves exactly in the right way, and it doesn't show a zitter-bewegung at all

Acceleration operator
We'll briefly handle the standard acceleration operator as well and show that there is no zitterbewung and that the result transforms in the right way under Lorentz transform. It can be actually be shown that it transforms like the Lorentz Force
$\psi^*\tilde{A}^i\psi   ~~=~~ \frac{i}{m}\frac{d\vec{p}}{dt} ~~\mbox{transforms like:}~~ \frac{iq}{m}\left(\vec{v}\times\vec{B} ~+~ \vec{E}\right)$
Because $\psi^*\tilde{A}^i\psi $ gives rise to two terms which transform like the electron's magnetization and polarization. The construction which therefor transforms like the Lorentz force is thus actually.
$\psi^*\tilde{A}^i\psi  ~~\mbox{transforms like:}~~ \frac{iq}{m}\left(-\vec{v}\times\mu_o\vec{M} ~+~ \frac{1}{\epsilon_o}\vec{P}\right)$
If you note that $\vec{v}\times\vec{M}~\propto~\vec{p}\times\vec{j}_A$ then you can recognize the two terms in the standard acceleration operator which is.
$\psi^*\tilde{A}^i\psi   ~~=~~ c~\bar{\psi}\left[\,\gamma^i\gamma^5\times(\partial_i-i\frac{e}{\hbar}\!A_i) ~\right]\psi~+~ \frac{imc^3}{\hbar}~\bar{\psi}\gamma^0\gamma^i\psi$
The acceleration is zero in a plane-wave in the absence of a B or E field. In this case the electron field has its own inherent M and P values and the two terms cancel each other. If the inherent M and P values change because of external B and E fields (by addition) then the electron accelerates.

Chiral representation
Now what about the c in the velocity operator. This behavior of the propagator is easy to understand in the modern chiral representation and the propagator of the field. In principle all fields are massless and propagate with c. Due to coupling however propagators can have any speed between +c and -c. The electron has two such massless components.
$\psi~~=~~\left(\begin{array}{c}\psi_L\\ \psi_R \end{array}\right)$
So, these two components do move at the speed of light. In the rest frame they move exactly opposite to each other and the combined speed is zero. The big difference with the zitterbewegung is that they both happen at the same time. There is no overall alternating net velocity.
Now the time evolution in the restframe is.
$e^{-Ht}\left(\begin{array}{c}\psi_L\\ \psi_R \end{array}\right) ~~=~~ 
\left(\begin{array}{c}\psi_L\cos(mt)-i\psi_R\sin(mt)\\ \psi_R\cos(mt)-i\psi_L\sin(mt) \end{array}\right)$
So, you see the $\psi_L$ and $\psi_R$ alternating but is there a zitterbewegung of $\psi$ or the individual components $\psi_L$ and $\psi_R$? The answer is: NO for electrons and NO for positrons. This is because these are exactly the only two solutions of the Dirac equation which do not show a zitterbewegung. The reason for this is.
electron at rest: $\psi_L=+\psi_R$
positron at rest: $\psi_L=-\psi_R$
The other "exotic" states where $\psi_L\neq\pm\psi_R$ at rest do show a zitterbewegung, for instance $\psi_L=i\psi_R$ or $\psi_L=\sigma_z\psi_R$. This is actually the reason why these states are not allowed. They would radiate away electromagnetic energy with the frequency corresponding to their mass. 

Hans.
