Here, I will demonstrate an astonishing fact (and thus an astonishing lack of understanding):

It is well known from the uncertainty principle that an electron cannot be at rest.

However, consider the relativistic limit, in which we treat the electron as a spinor. Then, by solving the free equation of motion, we get \begin{align} (i\gamma^\mu \partial_\mu -m)\Psi&=0\\ \partial_0\Psi&=im\gamma^0\Psi \end{align} Which yields, if we explicitly write out the left and right-handed components as $\Psi=\begin{bmatrix}\zeta_1&\zeta_2\end{bmatrix}$, $\partial_0^2\zeta_i=m^2\zeta_i$ thus yielding $$\Psi(x,t)=e^{\pm im t}\Psi(x,0)$$ And hence we are free to talk about things such as "an electron at rest".

Why do we have to wait until quantum field theory to discuss electrons at rest? Why does it make sense in the relativistic context?


closed as unclear what you're asking by Ben Crowell, Kyle Kanos, ACuriousMind, Brandon Enright, JamalS Nov 1 '14 at 14:50

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ And hence we are free to talk about things such as "an electron at rest". Huh? $\endgroup$ – Ben Crowell Nov 1 '14 at 2:45

The Dirac equation is not the relativistic limit of the Schrodinger equation, it is the high energy completion of it (for a spin 1/2 particle). Therefore, it explains higher energy electrons as well as low energy ones. Also in the above by taking, $i \partial_\mu \Psi = m \gamma_0 \Psi$, you assumed that $\vec{p}=0 $, so of course you found the electron is at rest.


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