Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
the short answer is yes. Zitterbewegung is due to the "small components" of the dirac 4-spinor (the last two components in a Dirac rep. where gamma-0 is diagonal) oscillating with a frequency $2mc^2$, which is entirely due to the little bit of antiparticle mixed up in the particle wavefunction for a nonrelativistic motion. It doesn't appear in the correct second quantized theory, or rather, it is resolved by using Feynman propagators and doing QED. Nevertheless, it is an interesting way to understand certain QED effects heuristically from a single particle picture, so I hesitate to be glib. –  Ron Maimon May 20 '12 at 20:16
Can you see the mistake on Itzykson and Zuber p.62? They mix electrons and positrons in (2-58) in order to construct a spatial Gaussian waveform. However, what should have been done is simply mixing electrons with positive and negative momenta. The latter won't give a Zitterbewegung. –  Hans de Vries May 21 '12 at 4:53
Note: If you mix correctly normalized positive en negative electron momentum eigen states then you'll be able to get one chiral component Gaussian shaped while the other won't be Gaussian but will be the first order Gaussian Hermite function, if I'm correct here working "without pen and paper" –  Hans de Vries May 21 '12 at 6:58
Ignore the note above, its more complicated.... –  Hans de Vries May 21 '12 at 7:51
I need some time to study your very detailed reply and understand it but when you talk about a mistake in the book, do you mean the book is wrong or it's demonstrating how the ambiguity arises? The authors say there is an inconsistency in the calculation but it wasn't clear to me where. –  Whelp May 21 '12 at 12:51
show 1 more comment

2 Answers

up vote 7 down vote accepted

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.


share|improve this answer
Very thorough answer, thanks a lot. –  Whelp May 22 '12 at 13:13
Now just compute the average of $v^2$ and you will obtain $c^2$, which correspond to a Dirac particle moving at $\pm c$ –  juanrga Mar 21 at 1:26
add comment

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

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.