Why do we ignore the proton when computing relativistic correction in the Hydrogen atom? I have read a number of textbooks (e.g. Sakurai), and they all seem to say that the unperturbed Hamiltonian of hydrogen is:
$$
H_0 = \frac{p^2 }{2m_e} - \frac{e^2}{r} \tag{1}
$$
and the relativistic correction is given by:
$$
T = \sqrt{p^2c^2 + m_e^2c^4} - m_e c^2 \approx \frac{p^2}{2m_e} - \frac{p^4}{8m_ec^3}
$$
and hence the perturbed Hamiltonian is:
$$
H = H_0 + H_p 
$$
where $H_p$ is the perturbation given by:
$$
H_p = - \frac{p^4}{8m_ec^3}.
$$
I really cannot bring myself to agree with (1), isn't the Hamiltonian of the unperturbed hydrogen:
$$
H_0 = \frac{p^2 }{2\mu} + \frac{e^2}{r} \tag{2}
$$
where $\mu$ is the reduced mass, and thus the perturbation should be:
$$
H_p = - \frac{p^4}{8 \mu c^3}.
$$
Why are the textbooks ignoring the mass of the proton? My guess is that they are doing an approximation, but is this approximation really valid?
 A: Well, the difference between $m_e$ and $\mu$ is really small, so I suspect most places just assume $\mu \approx m_e$.
Remember, since the electron is around a 1000 times less massive than the proton, $$\frac{\mu}{m_e} = \frac{m_p}{m_p+m_e} \approx 0.9995,$$ which is pretty darn close to 1! :)
EDIT: @EmilioPisanty has brought up an important point that I didn't consider: while the correction due to the mass of the proton is tiny, the relativistic correction is even tinier. (1 part in $10^7$, as pointed out.) As a result, it doesn't make sense to speak of the perturbed Hamiltonian as
$$\hat{H} = \frac{\hat{p}^2}{2 m_e} + V(r) - \frac{\hat{p}^4}{8 m_e^3 c^2},$$ since we're ignoring a term of the order $\sim 10^{-3}$ but considering one of the order $\sim 10^{-7}$. When taking the relativistic corrections into account, the correct method would be to use the reduced mass $\mu$ everywhere to avoid this.
However, practically, it doesn't seem to be a problem since (at least to first order) the shift in energy due to this perturbation depends on the mass only through $E_n$:
$$\Delta E_{nlm} = E_n \frac{\alpha^2}{n^2}\left( \frac{n}{l + 1/2} -\frac{3}{4}\right),$$
where $\alpha = \frac{e^2}{4\pi\epsilon_0 \hbar c} \approx \frac{1}{137}$ is the fine-structure constant, and $n,l,m$ represent the usual quantum numbers.

EDIT: An interesting side-note is that it is important to consider $\mu$ when we're dealing with "hydrogen-like" systems where one of the masses isn't so much larger than the other. One example is for positronium, a system consisting of an electron and a positron. If you used the "naive" Hamiltonian given above with $m_e$ instead of $\mu$, you might think that the energy spectrum of positronium is the same as that of the Hydrogen atom, but it isn't! In fact, since in this case $\mu = m_e/2$, the frequencies of the spectral lines are less than half of those for the corresponding Hydrogen lines.
