The Darwin term, a correction to the non-relativistic hydrogen Hamiltonian due to the zitterbewegung of the electron, is given by $$H_{Darwin}=\frac{e^2\hbar^2}{8m^2c^2\epsilon_{0}}\delta^3(\boldsymbol{r}).$$ Due to the Dirac delta, only states where the electron has a finite probability of being found in the nucleus (s-states) are affected by this term. What is the physical justification for why only s-states are affected? Do electrons in other states not possess zitterbewegung?

Edit: According to links such as this (towards the bottom), the zitterbewegung of the electron is caused by electromagnetic zero-point fluctuations. If that physical interpretation is indeed correct, shouldn't all electron states be affected?

  • $\begingroup$ The Darwin term accounts for relativistic terms that occur at small r , where the electron's momentum becomes comparable to $m_ec$. For electrons with $l=0$ there is a Darwin terms proportional to $|\psi (r=0)|^2 $ that has no classical analogue. (see Cohen-Tannoudji, Diu, Laloe, Quantum Mechanics, vol II, chapter XII). $\endgroup$ – porphyrin Jul 5 '16 at 7:27

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