The Darwin Term

We get a correction at the origin from the Dirac equation.

$$\bgroup\color{black}H_D={\pi e^2\hbar^2\over{2m^2_ec^2}}\delta^3(\vec r)\egroup$$

When we take the expectation value of this, we get the probability for the electron and proton to be at the same point.

$$\bgroup\color{black}\left<\psi\left\vert H_D\right\vert\psi\r... ...2\hbar^2\over{2m^2_ec^2}} \left\vert\psi(0)\right\vert^2\egroup$$

Now, $\psi(0)=0$ for $\ell>0$ and $\psi(0)={1\over\sqrt{4\pi}}2\left({z\over{na_0}}\right)^{3/2}$ for $\ell=0$, so

$$\bgroup\color{black} \left<H_D\right>_{n00}={4e^2\hbar^2\over... ...2c^2\over{2n^3a_0m^2c^2\hbar^2}} = {2nE^2_n\over{mc^2}} \egroup$$

This is the same as $\ell=0$ term that we got for the spin orbit correction. This actually replaces the $\ell=0$ term in the spin-orbit correction (which should be zero) making the formula correct!