Modern understanding of zitterbewegung? According to my naive understanding,  zitterbewegung is a property of particles obeying the Dirac equation where the expectation value of the position operator has an oscillatory component at frequency $2E/\hbar$, where $E$ is the total relativistic energy. 
More explicitly, the position operator expectation value looks roughly like the following
$$x(t)=x_0+vt+A\mathrm{cos}(2Et/\hbar)$$
Obviously there is no simple way to probe this oscillation, as it occurs on the timescale of the Compton wavelength. But more importantly, this description doesn't really make much sense to me if I think of particles within a Lorentz invariant field theory. Instead it feels like an oddity arising from forcing a semi-classical picture onto the Dirac equation. 
For example, zitterbewegung is explained in terms of position expectation values, but the position operator is not gauge invariant (and has other issues...) . There is also the issue that localizing a particle to its Compton wavelength inevitably creates other particles/antiparticles, so it is not clear whether the expectation value is meaningful anyways.
So what is the modern understanding of zitterbewegung in the usual terms of relativistic quantum field theory?
 A: Many sources describe the QFT whose equation of motion is the Dirac equation, but I'll describe it here in a way that emphasizes general principles more than mathematical detail. Outline:


*

*Sections A through E review the relevant QFT, with emphasis on the role of energy-increasing and energy-decreasing operators and their spatial profiles. (The usual creation/annihilation operators are a special case of this.)

*Section F highlights the assumption that leads to zitterbewegung and why the assumption doesn't hold in relativistic QFT. 

*Section G explains what replaces that assumption in relativistic QFT.
 A. The Dirac equation in QFT 
In QFT, the Dirac equation
$$
\newcommand{\pl}{\partial}
\newcommand{\bfx}{\mathbf{x}}
\newcommand{\bfy}{\mathbf{y}}
\newcommand{\bfX}{\mathbf{X}}
 (i\gamma^\mu\pl_\mu+m)\psi(x) = 0
\tag{1}
$$
is understood to be the equation of motion for a time-dependent field operator in the Heisenberg picture. The spinor field $\psi(x)$ consists of one operator $\psi_k(x)$ for each spacetime point $x$ and for each value of the spinor index $k$. The field operators $\psi_k(x)$ are operators on a Hilbert space, and all of the theory's observables will be expressed in terms of these operators $\psi_k(x)$, as explained below.
(More carefully: $\psi_k(x)$ becomes a well-defined operator when integrated against a smearing function. Calling $\psi_k(x)$ itself an "operator" is a convenient abbreviation, analogous to calling $\delta(x-y)$ a "function.")
The field operators at time $t=0$ satisfy the anticommutation relations
\begin{align}
 \big\{\psi_j(0,\bfx),\,\psi_k(0,\bfy)\big\} &= 0
\\
 \big\{\psi_j(0,\bfx),\,\psi_k^*(0,\bfy)\big\}
  &= \delta^3(\bfx-\bfy)
\tag{2}
\end{align}
with $\{A,B\} := AB+BA$. The asterisk denotes the operator adjoint, and I used the notation $x\equiv (t,\bfx)$ to separate the time coordinate from the others. The equation of motion (1) implicitly expresses the field operators at all other times in terms of those at $t=0$, so equations (1) and (2) together define a spinor's worth of field operators $\psi_k(x)$ for every spacetime point $x$, at least as an abstract algebra (no Hilbert space yet). This all turns out to be Poincaré symmetric, and the symmetry can even be made manifest, but I won't do that here.
To complete the definition of the model, we need to do two more things:


*

*Specify which operators qualify as observables (things that can be measured).

*Construct a representation of the operators as operators on a Hilbert space such that the spectrum condition is satisfied.
After that, we can construct single-electron states and ask how the electron behaves.
 B. Which operators are observables? 
Equations (1)-(2) define a non-commutative algebra $F$ generated by the field operators $\psi_k(x)$ and their adjoints. To specify which operators in $F$ qualify as observables, we can use this rule: A self-adjoint operator in $F$ qualifies as an observable if and only if it is invariant under the transformation
$$
 \psi_k(x)\to e^{i\theta}\psi_k(x)
\tag{3}
$$
for all real numbers $\theta$. (This rule can be motivated by thinking of this QFT as a warm-up for QED, where observables are required to be gauge-invariant.) According to this rule, the simplest observables have the form 
$$
 \psi_j^\dagger(x)\psi_k(y)+\text{adjoint}.
\tag{4}
$$
Let $R$ denote a region of spacetime, and let $F(R)$ denote the set of operators generated by the field operators $\psi_k(x)$ and their adjoints with $x\in R$. If an observable $\Omega$ belongs to $F(R)$, then $\Omega$ is understood to represent something that could be measured in $R$. 
Equations (1)-(2) imply that observables associated with spacelike-separated regions commute with each other.
 C. Energy-increasing and -decreasing operators 
We still need to represent the field operators as things that act on a Hilbert space, and  the Hilbert-space representation should satisfy the spectrum condition (defined below). Such a representation can be constructed using energy-increasing and energy-decreasing operators. I'll start by defining these operators. The usual creation/annihilation operators are a special case of this.
Let $H$ be the time-translation operator (Hamiltonian, aka the total energy operator), defined so that the equation
$$
 i\frac{\pl}{\pl t}\psi(x) = \big[\psi(x),\,H\big]
\tag{5}
$$
is the same as the Dirac equation (1). Any operator $A$ can be translated in time using 
$$
 A(t) := U(-t)AU(t)
 \hskip1cm
 \text{with }\ U(t) := \exp(-iH t).
\tag{6}
$$
Using a Fourier transform over $t$, we can define positive- and negative-frequency parts of $A(t)$. These work as energy-increasing and energy-decreasing operators, where energy is defined by $H$. Let $A^+(t)$ and $A^-(t)$ denote the energy-increasing and -decreasing parts, respectively, so that
$$
 A(t) = A^+(t) + A^-(t).
\tag{7}
$$
The adjoint of the energy-increasing part of an operator is the energy-decreasing part of its adjoint: 
$$
 \big(A^+(t)\big)^*=\big(A^*(t)\big)^-.
\tag{8a}
$$
The first of equations (2) implies that the energy-increasing and -decreasing parts of the field operator $\psi_k(x)$ anticommute with each other at equal time:
$$
 \Big\{\big(\psi_k(t,\bfx)\big)^+,\,\big(\psi_k(t,\bfy)\big)^-\Big\}=0,
\tag{8b}
$$
and similarly for $\psi_k^*(x)$.
The energy-increasing and -decreasing operators $A^\pm(t)$ are not localized in any finite region of spacetime: they do not belong to $F(R)$ for any finite $R$. For example, the operator $(\psi_k(0,\bfx))^+$ involves the operators $\psi_j(0,\bfy)$ for arbitrarily large $|\bfx-\bfy|$. The coefficient of $\psi_j(0,\bfy)$ in $(\psi_k(0,\bfx))^+$ decreases exponentially with increasing $|\bfx-\bfy|$, with a characteristic scale given by the Compton length scale $\hbar/mc$ where $m$ is the electron's mass, so it rapidly approaches zero, but it isn't equal to zero.
 D. The spectrum condition 
We still need to represent the field operators as things that act on a Hilbert space. One of the most important general principles of QFT is that the Hilbert-space representation should satisfy the spectrum condition. This means that we should have
$$
 \langle\Psi| H |\Psi\rangle \geq 0
\tag{9}
$$
for all state-vector $|\Psi\rangle$ in the Hilbert space. (This condition can be expressed in a Lorentz-symmetric way, but I won't do that here.) To construct a Hilbert-space representation of the algebra that satisfies this condition, start with a state-vector $|0\rangle$ that is defined by the condition
$$
 A^-(t)|0\rangle = 0
\tag{10}
$$
for all $A(t)$, and choose the constant term in $H$ so that $H|0\rangle=0$. Acting on this state-vector with the algebra of field operators generates a Hilbert space in which the spectrum condition (9) is satisfied. By construction, the state-vector $|0\rangle$ is the one with the lowest energy.
As far as observables are concerned, this representation is reducible: it contains subspaces that are not mixed with each other by any observables. For example, the state-vectors 
$$
 |0\rangle
 \hskip2cm
 (\psi_k(x))^+|0\rangle
 \hskip2cm
 (\psi^*_k(x))^+|0\rangle
\tag{11}
$$
are not mixed with each other by any observables. This is clear from the fact that they all transform differently under (3). The irreducible subspace relevant to the question will be specified below.
 E. Single-electron states 
The lowest-energy state $|0\rangle$ in the preceding construction is invariant under translations and Lorentz transformations. This state has no particles, by definition.
The question is about single-particle states. Choosing a useful nonperturbative definition of "single-particle state" in a typical QFT is challenging, but in the present case with a linear equation of motion, the usual  textbook approach is sufficient: Each application of the energy-increasing part of a field operator adds a particle to the state. In symbols, the operators $(\psi_k(x))^+$ and $(\psi^*_k(x))^+$ each add one particle to the state.
Observables cannot mix states of the form $(\psi_k(x))^+|0\rangle$ with states of the form $(\psi^*_k(x))^+|0\rangle$, even though both are single-particle states. We can say that the state $(\psi^*_k(x))^+|0\rangle$ has a single electron and that $(\psi_k(x))^+|0\rangle$ has a single positron. Any linear combination of single-electron states is another single-electron state, and likewise for positrons.
 F. From non-QFT to zitterbewegung 
The question asks for the modern perspective on zitterbewegung. Zitterbewegung comes from trying to interpret the $\bfx$ in the Dirac equation (1) as the position observable of a particle. The modern perspective is that that interpretation is a mistake. Zitterbewegung is just one of the consequences of that mistake.
The modern perspective is that QFT is the proper foundation for relativistic quantum physics. In QFT, $\bfx$ isn't an observable at all. It's a parameter, basically a continuous "index" that is used together with the spinor index $k$ to parameterize the field operators $\psi_k(t,\bfx)$. Observables are constructed from the field operators, and they are operators that act on a Hilbert space. The next section looks at the observables that QFT does provide and explores what they can tell us about the behavior of an electron.
 G. How does an electron behave in QFT? 
From now on, only single-electron states will be considered. All such state-vectors are linear combinations of the state-vectors $(\psi^*_k(x))^+|0\rangle$.
As always, whatever questions we ask about the system's behavior should be expressed in terms of the theory's observables. That prevents meaningless questions. 
Let's start with something easy: Pick a direction in space and consider the corresponding momentum operator. The momentum operator is (by definition) the generator of translations along the given direction. It is invariant under (3), so it qualifies as an observable. It clearly maps single-electron states to single-electron states, because it just shifts one of the spatial coordinates in the field operators, so we can use it to say something about the behavior of a single electron. Finally, it commutes with the Hamiltonian, so the electron's momentum is conserved. 
Does that mean that the electron has constant velocity? For the concept of "velocity" to make sense, we would need a velocity observable. And for that to make sense, we would need position observables $\bfX(t)$. (I'm writing boldface $\bfX$ for a set of $3$ operators, one per dimension of space.) But the QFT that we constructed above doesn't have position observables for the electron, at least not in the strict sense that is familiar from nonrelativistic QM. To qualify as position observable for the electron, the operators $\bfX(t)$ would need to satisfy these requirements:


*

*They would need to map single-electron states to single-electron states. Otherwise, we couldn't justify interpreting it as the position (or anything else) of the electron.

*For any finite region $R$ of space at any given time $t$, let $f(\bfx)$ be a function that equals $1$ for $\bfx\in R$ and equals $0$ otherwise. Then if $\bfX(t)$ are the alleged position operators, the projection operator $f(\bfX(t))$ would need to belong to the algebra $F(R)$.
No such operator can exist, because these two requirements contradict each other. The first requirement says that $\bfX$ must be constructed from operators of the form (see (8a)-(8b))
$$
 \big(\psi_k^*(x)\big)^+\big(\psi_j(y)\big)^-
\hskip2cm
 \big(\psi_k(x)\big)^+\big(\psi_j^*(y)\big)^-.
\tag{12}
$$
Operators constructed from these are not localized in any finite region of spacetime, contradicting the second requirement. This argument is only intuitive, but the conclusion is sound: the Reeh-Schlieder theorem implies that relativistic QFT cannot have position observable in the strict sense.
Instead of a strict position observable for the electron, we can use the operators
$$
 \bfX(t) \propto \int d^3x\ \bfx 
  \sum_k\big(\psi_k^*(t,\bfx)\big)^+\big(\psi_k(t,\bfx)\big)^-.
\tag{13}
$$
This satisfies the first requirement above. It doesn't satisfy the second requirement, but it comes close, because the energy-increasing and -decreasing factors in the integrand are almost localized at $\bfx$, except for exponentially-decreasing tails that extend away from $\bfx$. These tails fall off exponentially with characteristic length scale $\hbar/mc$, which might as well be zero in the nonrelativistic approximation, and indeed the operators (13) reduce to the usual strict position observables in the strict nonrelativistic approximation.
We could use the Dirac equation to study the time derivatives of $\bfX(t)$, but since this isn't a strict position observable, we still couldn't conclude draw any strict conclusions about the electron's velocity — simply because "velocity" isn't strictly defined. Again, whatever questions we ask about the system's behavior should be expressed in terms of the theory's observables. Questions that try to ask about the electron's strict position or velocity are not meaningful in realtivistic QFT (or in the real world!), much like the concept of strict simultaneity isn't meaningful in general relativity.
What else can we do with the observables the theory provides? We can also construct quasi-local observables that detect the electron's presence/absence mostly within a given region of spacetime, with exponentially decreasing sensitivity outside the given spatial region. This is closer to how we measure a particle's position in practice anyway, using an array of somewhat-localized detectors to register its presence/absence in each of those little regions of space. However, even in nonrelativistic QM, if we repeatedly measure a particle's position to try to track its motion, it jumps around with an amplitude that increases with finer resolution of the measurements. That's not zitterbewegung, it's just plain-old quantum indeterminism combined with the tendency of the electron's wavepacket to disperse between measurements. This happens already in nonrelativistic QM, where we do have a strict position operator. In relativistic QFT, we can't do any better. 
 Summary 
Does zitterbewegung exist in relativistic QFT? The word zitterbewegung is normally used for a consequence of the assumption that the $\bfx$ in the Dirac equation is a position operator. That assumption is false in relativistic QFT, so relativistic QFT can't have zitterbewegung — not unless we redefine the word to mean something else like "the electron's velocity isn't strictly well-defined." 
Sure, the mathematical circumstances that preclude the existence of a strict position observable are related to the mathematical circumstances that would lead to zitterbewegung if we wrongly assumed that $\bfx$ were a position observable. In that indirect sense, a person could insist that a vestige of zitterbewegung is still present in relativistic QFT, but that would just be playing word games. If we look past the words and focus on the concepts, the messages are simple: 


*

*Zitterbewegung is an artifact of mistaking the parameter $\bfx$ as a position observable.

*Relativistic QFT doesn't have a strict position observable. 
In relatistic QFT, particle-detection observables can't be both perfectly noise-free and perfectly sharply localized in a finite region of spacetime. We can have one or the other, but we can't have both, and nothing about this statement contradicts anything we know from real experiments.
