4

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?

Qmechanic
  • 201,751
D. Soul
  • 694

1 Answers1

4

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.

Philip
  • 11,393
  • When solving for radial wave functions of the hydrogen atom, we always include the mass of the proton and hence utilize the reduced mass to solve for the radial wavefunctions.

    Does this mean we can technically ignore the proton term when solving the radial equation? That is:

    $$ H_0 = \frac{\hat{P}_p}{2m_p} + \frac{\hat{P}_e}{2m_e} + V(r) \approx \frac{\hat{P}_e}{2m_e} + V(r) $$

     – D. Soul Jul 02 '20 at 05:57
  • Yes, but since as you have said above, $\mu \approx m_e$, this means that technically I am able to reduce the relative coordinate portion to:

    $$ \frac{P^2_{rel}}{2 m_e} + V(r) $$

    This would imply that my radial wavefunctions would all contain $m_e$ instead of $\mu$?

     – D. Soul Jul 02 '20 at 06:06
  • Yes, that's what people do, generally. They use the Bohr radius, which uses $m_e$, not $\mu$. – Philip Jul 02 '20 at 06:08
  • Ah okay thanks! Because my professor has taught us to use $\mu$ so perhaps is just peoples preference – D. Soul Jul 02 '20 at 06:09
  • Thanks! You really helped! – D. Soul Jul 02 '20 at 06:14
  • 3
    Yes, the proton-mass effects are weak, but the relativistic effects are weaker (one part in $10^7$). This answer's argument fails completely - it's only appealing if you don't think about the details at all. – Emilio Pisanty Jul 02 '20 at 07:37
  • @EmilioPisanty Oh I think I see what you mean: are you saying that since the relativistic correction is so much smaller, it doesn't make sense to approximate the reduced mass with the electron's mass, since it produces a much higher order error to the overall Hamiltonian than the relativistic correction corrects for? – Philip Jul 02 '20 at 07:55
  • 1
    @EmilioPisanty Is it possible for you to elaborate more? – D. Soul Jul 02 '20 at 09:20
  • @EmilioPisanty I've edited my answer a little to take into account what I think was your objection with it. Let me know if you think that it's still wrong/misleading, and I'll remove it. – Philip Jul 02 '20 at 10:24
  • 1
    The first paragraph is still misleading, but the answer as a whole is now OK, I guess. The point is that it depends on what you want to do. If you want to explain how fine-structure works, as in a textbook, you can use any hamiltonian (i.e. either mass) which will allow you to do that in a clear and simple way. If the goal is a quantitatively accurate description of the spectroscopy down to fine-structure level, then you must include all effects that contribute at that level of importance or higher, and that includes the proton-mass effects. – Emilio Pisanty Jul 02 '20 at 11:45