I am interested in calculating the dynamics of a wavepacket obeying the Dirac equation, eventually with an applied Electric field, but I am stuck on the case without an applied field. I eventually want to study the $m\to0$ limit, and so want to be careful not to assume that the velocity of a wavepacket is defined by its momentum.

Abusing notation, define the center of charge of some wavepacket as: $$ x^{j}(t)\equiv \int d^{3}x \psi^{\dagger}(\vec{x}, t)\psi(\vec{x}, t)x^{j}=-\int d^{3}x \left(\overline{\psi}\gamma^{0}\psi\right) x^{j} $$ here $\psi^{\dagger}\psi=-\overline{\psi}\gamma^{0}\psi$ is the charge density. Below, I show that $\partial_{t}x^{j}(t)=\int d^{3}x \overline{\psi}\gamma^{j}\psi$, as we would expect from current conservation $\partial_{\mu}\overline{\psi}\gamma^{\mu}\psi=0$. I also expect that $\partial_{t}^{2} x^{j}(t)$ should vanish, but that's not what I get, and my question is how to make sense of the acceleration of the center-of-charge.

Working in the $(-+++)$ convention, with $\{\gamma^{\mu}, \gamma^{\nu}\}=2\eta^{\mu\nu}$, $(\gamma^{\mu})^{\dagger}=\gamma^{0}\gamma^{\mu}\gamma^{0}$, and $\overline{\psi}=-\psi^{\dagger}\gamma^{0}=\psi^{\dagger}\gamma_{0}$, the Dirac Lagrangian is $$ \mathcal{L}=-i\overline{\psi}(\gamma^{\mu}\partial_{\mu}-m)\psi $$ This gives us the Dirac equation: $$ \gamma^{\mu}\partial_{\mu}\psi-m\psi=0\implies \partial_{0}\psi=\gamma^{0}\gamma^{i}\partial_{i}\psi-m\gamma^{0}\psi $$ and its conjugate: $$ -(\partial_{\mu}\overline{\psi})\gamma^{\mu}-m\overline{\psi}=0\implies \partial_{0}\psi^{\dagger}=-(\partial_{i}\psi^{\dagger})\gamma^{i}\gamma^{0}+m\psi^{\dagger}\gamma^{0} $$ where Latin letters $i, j$ are summed over spatial indices.

We can quickly calculate: $$ \partial_{t}x^{j}(t) =\int d^{3}x\left[\left(-(\partial_{i}\psi^{\dagger})\gamma^{i}\gamma^{0}+m\psi^{\dagger}\gamma^{0}\right)\psi+\psi^{\dagger}\left(\gamma^{0}\gamma^{i}\partial_{i}\psi-m\gamma^{0}\psi\right)\right]x^{j} $$ $$ =\int d^{3}x~\partial_{i}\left(\psi^{\dagger}\gamma^{0}\gamma^{i}\psi\right)x^{j}=\int d^{3}x \psi^{\dagger}\gamma^{0}\gamma^{j}\psi=\int d^{3}x\overline{\psi} \gamma^{j}\psi $$ which is exactly what we expect from the Noether conservation equation $\partial_{\mu}\overline{\psi}\gamma^{\mu}\psi=0$.

Now, if I calculate the second derivative, I should expect it to vanish. Instead, I get: $$ \partial_{t}^{2}x^{j}=\int d^{3}x\left[(\partial_{t}\psi^{\dagger})\gamma^{0}\gamma^{j}\psi+\psi^{\dagger}\gamma^{0}\gamma^{j}\partial_{t}\psi\right] $$ $$ =\int d^{3}x\left[\left(-(\partial_{i}\psi^{\dagger})\gamma^{i}\gamma^{0}+m\psi^{\dagger}\gamma^{0}\right)\gamma^{0}\gamma^{j}\psi + \psi^{\dagger}\gamma^{0}\gamma^{j}\left(\gamma^{0}\gamma^{i}\partial_{i}\psi-m\gamma^{0}\psi\right)\right] $$ $$ =\int d^{3}x\left[(\partial_{i}\psi^{\dagger})\gamma^{i}\gamma^{j}\psi+\psi^{\dagger}\gamma^{j}\gamma^{i}\partial_{i}\psi-2m\psi^{\dagger}\gamma^{j}\psi\right] $$ Is this calculation incorrect? If not, how am I to make sense of this final equation? Is there something wrong with studying the center-of-change and expecting it to behave like a center-of-mass? What about if $m=0$, or if we have just a single (classical) Weyl fermion?

  • $\begingroup$ Zitterbewegung? $\endgroup$ – AccidentalFourierTransform Jun 1 '18 at 20:28
  • $\begingroup$ Oooh this is it! Thank you! I'll also try to see what happens with just a single Weyl Fermion. $\endgroup$ – MDM Jun 1 '18 at 20:45

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