Relativistic Correction and the Fine Structure of Hydrogen

Wait — we just finished building up perturbation theory, and now we’re… back to hydrogen?!

Yep. That’s exactly right.

Time to revisit hydrogen one more time. And the reason we can revisit it is precisely because we just learned perturbation theory.

Here’s the deal: the hydrogen atom, as we solved it, already has a perturbation baked in. It’s not the clean story we’ve been telling ourselves. It needs a tiny correction because of “certain factors.”

And those certain factors are:

1. The relativistic effect (relativistic correction)
2. Spin-orbit coupling (LS coupling)
3. Lamb shift

Number 1!!!! We have to think about the relativistic effect!!!

What does this mean? Well — the electron is actually zipping around at roughly $c/136$. About 1/100 the speed of light. So the kinetic energy is not just $p^2/2m$…

Which means we need to tack a correction onto the $E_n^{0}$ we got from the clean hydrogen model.

Number 2. Spin-orbit coupling (LS coupling).

What’s this one? OK so — when the electron orbits the nucleus, it generates a magnetic field $B$. But the nucleus has spin too. And there’s this tiny little error that comes from the interaction between the electron’s orbital angular momentum and the nucleus’s spin.

Now — the version of hydrogen that takes both of these (relativistic + LS) into account is called the fine structure!!!

And the size of this energy correction? About 1 part in 10,000???????????? Yeah, roughly $10^{-4}$, lol.

(Quick aside — I actually asked my professor about this.

“Professor… are there any examples where humans figuring out a ~1/10000 level energy correction led to engineering advances for humanity?”

And the professor goes: “………what could humans possibly do with 1/10000 of an energy correction………… hahahahahahaha

There’s something you need to really sit with, about why we study science. You have to let go of the idea that a theory is good only if it’s useful somewhere, and bad if it isn’t. We’re people who study science. Not engineering, right?

Just by thinking about stuff like this, we discover a 1/10000 energy error — and by exactly that much, human thought has expanded.”

Man… hearing that gave me literal goosebumps hahahahahaha. Amazing. Amazing. Exactly. Exactly. Exactly. I am a student who studies science!!)

(Though — fair warning — LS coupling has actually become a hot research topic lately. Scientists keep finding weird new materials — like, you know, the kind where the surface is a conductor but the interior is an insulator. Turns out the secret behind a lot of that stuff is hidden in LS coupling. For the details, Google it — go go go.)

And then there’s one more factor, the Lamb shift.

Where does this error come from? It’s a small correction from the interaction between photons and electrons. (Related to the quantization of the electromagnetic field, apparently.) The size of this correction? About 1 in 100,000………… heh heh.

But — this one isn’t something you can cover in undergrad QM. My professor basically said: if you’re going to major in this, or go to grad school, that’s when you learn it properly.

Oh, and yet another one — the hyperfine structure (hyperfine structure splitting). Think of it as an error that pops up like you’ve got two little magnets — the electron’s spin and the nucleus’s spin — talking to each other.

OK. So for now, we’re going to focus on the fine structure. That is: relativistic effect + LS coupling.

1. The relativistic correction

Like I said — the electron’s kinetic energy is not actually

$$\dfrac{1}{2}mv^{2}.$$

Nope.

Remember back in “Modern Physics I Studied #5” — as the speed creeps toward $c$, the momentum $p$ is better described by

$$p \;=\; \frac{m_{0}v}{\sqrt{1-\dfrac{v^{2}}{c^{2}}}}$$

So relativistically, the total energy is

(energy from moving) + (rest mass energy)

Which means the kinetic energy (the “energy from moving” part) = total energy − rest mass energy:

$$\text{kinetic energy} \;=\; \text{total energy} \;-\; \text{rest mass energy} \\ K \;=\; \sqrt{p^{2}c^{2} \;+\; m_{0}^{2}c^{4}} \;-\; m_{0}c^{2} \\ \;=\; c^{2}\sqrt{\frac{p^{2}}{c^{2}} \;+\; m_{0}^{2}} \;-\; m_{0}c^{2} \\ \;=\; m_{0}c^{2}\sqrt{\frac{p^{2}}{m_{0}^{2}c^{2}} \;+\; 1} \;-\; m_{0}c^{2} \\ \;=\; m_{0}c^{2}\!\left(\sqrt{\frac{p^{2}}{m_{0}^{2}c^{2}} \;+\; 1} \;-\; 1\right)$$$$=\; m_{0}c^{2}\!\left\{\!\left(1 \;+\; X^{2}\right)^{\!\frac{1}{2}} \;-\; 1 \;\right\} \quad \text{now Taylor expand the front piece} \\ =\; m_{0}c^{2}\!\left\{1+\frac{1}{2}X^{2} \;-\frac{1}{8}\!\left(X^{2}\right)^{\!2} \;+\; \cdots \;-\; 1 \;\right\} \\ =\; m_{0}c^{2}\!\left\{\frac{1}{2}\!\left(\frac{p}{m_{0}c}\right)^{\!2} \;-\; \frac{1}{8}\!\left(\!\left(\frac{p}{m_{0}c}\right)^{\!2}\right)^{\!2} \;+\; \cdots\right\} \\ =\; \frac{p^{2}}{2m_{0}} \;-\; \frac{p^{4}}{8m_{0}^{3}c^{2}} \;+\; \cdots \\ \text{the blue is the 0th order — that's the classical kinetic energy,} \\ \text{the red is the 1st order correction — the biggest piece of the relativistic energy correction.}$$$$\text{Ohhh — so *that* guy is literally} \\ \text{'the perturbed Hamiltonian!!!'} \\ H_{relative}^{'} \;=\; -\frac{p^{4}}{8m_{0}^{3}c^{2}} \\ \text{Computing the 1st-order energy correction for a perturbation like this — we've done this a thousand times.} \\ E_{r}^{1} \;=\; \left\langle\psi_{n,l,m}^{0}\!\left|H_{r}^{'}\right|\!\psi_{n,l,m}^{0}\right\rangle \quad \text{and since this is hydrogen, we just plug in the hydrogen wave function, right?} \\ =\; \left\langle\psi_{n,l,m}^{0}\!\left|-\frac{p^{4}}{8m_{0}^{3}c^{2}}\right|\!\psi_{n,l,m}^{0}\right\rangle \;=\; -\frac{1}{8m_{0}^{3}c^{2}}\left\langle\psi_{n,l,m}^{0}\!\left|p^{4}\right|\!\psi_{n,l,m}^{0}\right\rangle \\ \text{And because } p \text{ is Hermitian,} \\ =-\frac{1}{8m_{0}^{3}c^{2}}\left\langle p^{2}\psi_{n,l,m}^{0}\,\middle|\,p^{2}\psi_{n,l,m}^{0}\right\rangle$$$$p^{2}\psi_{n,l,m}^{0} \text{ — we can pull this straight out of the time-independent Schrödinger equation.} \\ \frac{p^{2}}{2m}\psi_{n,l,m}^{0} \;+\; V\psi_{n,l,m}^{0} \;=\; E\psi_{n,l,m}^{0} \\ \therefore \quad p^{2}\psi_{n,l,m}^{0} \;=\; 2m\!\left(E-V\right)\psi_{n,l,m}^{0}$$$$E_{r}^{1} \;=\; -\frac{1}{8m_{0}^{3}c^{2}}\left\langle 2m\!\left(E-V\right)\psi_{n,l,m}^{0}\,\middle|\,2m\!\left(E-V\right)\psi_{n,l,m}^{0}\right\rangle \\ \;=\; -\;\frac{1}{2m_{0}c^{2}}\!\left\langle\!\left(E-V\right)^{2}\right\rangle \;=\; -\;\frac{1}{2m_{0}^{3}c^{2}}\!\left\langle E^{2} \;-\; 2EV \;+\; V^{2}\right\rangle \\ =\; -\;\frac{1}{2m_{0}c^{2}}\!\left[\left\langle E^{2}\right\rangle -2\left\langle EV\right\rangle +\left\langle V^{2}\right\rangle\right] \\ =\; -\;\frac{1}{2m_{0}c^{2}}\!\left[E^{2} \;-2E\!\left\langle V\right\rangle +\left\langle V^{2}\right\rangle\right]$$$$\text{And in hydrogen, } V \;=\; -\frac{1}{4\pi\epsilon_{0}}\frac{e^{2}}{r}, \\ \text{so the things we actually have to compute are} \\ \left\langle\frac{1}{r}\right\rangle \;\text{ and }\; \left\langle\frac{1}{r^{2}}\right\rangle. \\ \text{THAT'S why we grinded through those proofs so hard in the last post!!!!><}$$$$\left\langle\frac{1}{r}\right\rangle \;=\; \frac{1}{n^{2}a} \quad // \quad \left\langle\frac{1}{r^{2}}\right\rangle \;=\; \frac{1}{\!\left(\ell \;+\; \dfrac{1}{2}\right)n^{3}a^{2}}$$

Those were the results!!!!!! Plug ’em in!!!

$$E_{r}^{1} \;=\; -\;\frac{1}{2m_{0}c^{2}}\!\left[E^{2} \;-2E\!\left\langle V\right\rangle +\left\langle V^{2}\right\rangle\right] \\ =\; -\;\frac{1}{2m_{0}c^{2}}\!\left[E_{n}^{2} \;-2E_{n}\!\left(-\frac{e^{2}}{4\pi\epsilon_{0}}\right)\!\left(\frac{1}{n^{2}a}\right) +\left(-\frac{e^{2}}{4\pi\epsilon_{0}}\right)^{\!2}\!\left(\frac{1}{\!\left(\ell \;+\; \frac{1}{2}\right)n^{3}a^{2}}\right)\right] \\ \therefore \quad E_{r}^{1} \;=\; -\frac{E_{n}^{2}}{2m_{0}c^{2}}\!\left[\frac{4n}{\ell \;+\; \dfrac{1}{2}}-3\right]$$

Look at that $E_{n}^{2}$ — yep, the correction really is on the order of $10^{-4}$.

And one more thing. Notice what $E_{r}^{1}$ is telling us: the original $E_n$ had zero dependence on angular momentum — but this correction? It depends on $\ell$. The angular momentum is suddenly in the game.

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Originally written in Korean on my Naver blog (2015-12). Translated to English for gdpark.blog.