# Calculating Darwin Term for Hydrogen

The standard classical approximation of the Darwin term in hydrogen's fine structure goes something like this. We approximate the 'smeared' potential due to the electron's fluctuating position as $V(\vec{r}+\vec{\epsilon})$, and then Taylor expand this to get $$V(\vec{r}+\vec{\epsilon})-U(\vec{r})\approx\vec{\epsilon}\cdot\nabla U(\vec{r})+\frac{1}{2}(\vec{\epsilon}\cdot\nabla)^{2}U(\vec{r}).$$ Then it is claimed that by the isotropy of the fluctuations of the electron's position, when we take the expectation values of all the terms, the first term vanishes and the second term evaluates to $$\frac{1}{6}\langle\vec{\epsilon}\rangle^{2}\nabla^{2}U(\vec{r}).$$ I can't really figure out how exactly to go about calculating the expectation values of both these terms. It makes sense that if the fluctuations are isotropic, then $\langle\vec{\epsilon}\rangle$ vanishes, but then why doesn't $\langle\vec{\epsilon}^{2}\rangle$ vanish as well?

• look up random processes – jim Jul 20 '17 at 21:45

Where does that $\frac{1}{6}$ factor in your second expression come from??
The expectation value of Epsilon is related to the heisemberg inequality, isolating $\Delta$x and imposing that the velocity of the electron is almost c you get that the smearing is about the order of compton wavelenght for the electron.
$E\approx pc\rightarrow p\approx\frac{E}{c}\Rightarrow p^2\approx \left(\frac{\frac{1}{2}mc^2}{c}\right)^2=\frac{1}{4}m^2c^2\\ \Delta p=\sqrt[]{\langle p^2 \rangle-\langle p\rangle^2}=\sqrt[]{\langle p^2\rangle}\approx\frac{1}{2}mc \Rightarrow \Delta x\geq \frac{\hslash}{mc}=\lambda_c$