To kick off SMTM, I’ll look at a topic that I never really understood when I was in graduate school: the Lamb shift.

The Lamb Shift: what it is

In its most commonly-discussed form, the Lamb shift is a small effect.  In fact, it’s a very small effect (which is probably why I never bothered to learn it in the first place).  The Lamb shift is a miniscule change in (some of) the energy levels of the hydrogen atom relative to where it seems like they should be.  For example, the binding energy of an electron to the hydrogen nucleus (a proton) is about $13.6$ electron volts.  The Lamb shift is a phenomenon that changes this energy level by about $4\times10^{-6}$ eV, or about $0.00003$%.  But the existence of this shift was a serious puzzle to physicists in the 1940s and 50s, and its final resolution provided a beautiful piece of physics that helped spur the development of quantum electrodynamics, one of the most spectacularly successful scientific theories in history.

The essence of the Lamb shift can be stated like this: it is the energy of interaction between hydrogen and empty space.

The dominant contribution, of course, to the energy of the hydrogen atom is the interaction of the electron with the proton it’s orbiting.  If you want a really quick way to derive the energy of the hydrogen atom, all you need to remember is that the size of the electron cloud around the proton has some characteristic size $a$.  Confining the electron to within this cloud costs some kinetic energy $\sim \hbar^2 /m a^2$, and it buys you some energy of attraction, $\sim -e^2/a$ (here I’m being too lazy to write out the $4\pi\epsilon_0$s that come in SI units).  So the total energy is something like $E \sim \hbar^2/ma^2 - e^2/a$.  If you minimize $E$ with respect to $a$ (take the derivative and set it to zero), you’ll find that

$a \sim \hbar^2/me^2$

and

$E \sim -me^4/\hbar^2 \sim -e^2/a$.

The constant $a \approx 0.5$ Angstroms is the Bohr radius, which is the typical size of the hydrogen atom (and, roughly speaking, any atom).  The energy $-e^2/a \equiv R$ is the Rydberg energy.

The Lamb shift comes from the way this balanced state between electron and proton is influenced by the slight, random buffetings from the vacuum itself.

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The hydrogen atom

There are two players in this story: the hydrogen atom, and empty space.  I’ll describe the former first, since the latter (paradoxically) is considerably harder.

In reality, the Lamb shift is most easily observed in excited states of hydrogen (the P states, see Footnote 1 at the bottom), but for the purpose of this discussion it’s easiest to think about the ground state.  In terms of the electron probability cloud, the ground state of the hydrogenic electron looks like this:

It has a peak right at the middle of the atom, and it falls of exponentially.

I know there is a lot of trickiness associated with whether to think about an electron as a particle or as a wave (my own favorite take is here — in short, an electron is a particle that surfs on a wave), but for this discussion it’s easiest to think of an electron as a point object that just happens to arrange itself in space according to the probability density plotted above.

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The vacuum

There is a lot going on in empty space.  If this crazy idea is completely new to you, I would (humbly) suggest reading my post on the Casimir effect.  The upshot of it is that all of space is filled by endlessly boiling quantum fields, and one of these, the electromagnetic field, is responsible for conveying electromagnetic forces.  As a result of its indelible boiling, however, the electromagnetic field can push on charged objects, like our electron, even when there are no other charged objects around to seemingly initiate the pushing.

To get a better description of the electromagnetic field in vacuum, it will be helpful to imagine that our electron sits inside a large metal box with size $L$.  Inside this metal box are lots of randomly-arising electromagnetic waves (“virtual photons”).  Something like this:

When dealing with quantum fields, a good rule of thumb is to expect that, in vacuum, every possible oscillatory mode will be occupied by one quantum of energy.  In this case, it means that for every possible vector $k = 2\pi / \lambda$, where $\lambda = 2L, 2L/2, 2L/3, 2L/4, ...$ is a permissible photon wavelength, there will be roughly one photon present inside the box.  This photon has an energy $E_k = \hbar c k$, where $c$ is the speed of light.  One can estimate the typical magnitude of the electric field the photon creates, $|\vec{\mathcal{E}_k}|$, by remembering that $|\vec{\mathcal{E}_k}|^2$ gives the energy density of an electric field.  Since the photon fills the whole box, this means that $|\vec{\mathcal{E}_k}|^2 \times L^3 \sim E_k$, or $|\mathcal{E}_k|^2 \sim \hbar c k/L^3$.  This electric field oscillates with a frequency $ck$.

So now the stage is set.  The hydrogen atom sits inside a “large box” (which we’ll do away with later), and inside the box with is a huge mess of random electric fields that can push on the electron.  Now we should figure out how all this pushing affects the hydrogen energy.

[By the way, you may be bothered by the fact that all these randomly-arising photons seem to endow the interior of the box with an infinite amount of energy.  If that is the case, then there's nothing much I can say except that you and I are in the same club, with only speculation to assuage our uneasiness.]

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How the vacuum pushes on hydrogen

The essence of the Lamb shift is that the random electric fields push on the electron, and in doing so they move it slightly further away from the proton, on average, than it would otherwise be.  Another way to say this is that the distribution of the electron’s position gets blurred over some particular (small) length scale $\delta r$.  In particular, the sharp peak in the distribution near the center of the atom should get slightly rounded, like this:

The “smearing length” δr is greatly exaggerated in this picture.

The resulting shift of the electron distribution away from the center lowers the interaction energy of the electron to the proton.  To estimate the amount of energy that the electron loses, you can think that in those moments where the electron happens to approach be within a distance $\delta r$ of the nucleus, it frequently finds itself getting pushed outward by an amount $\delta r$.  As a result of this outward push it loses an energy $e^2/\delta r$.  This means that the Lamb shift energy

$\Delta E \sim (e^2/\delta r) \times [\text{fraction of time the electron spends within } \delta r \text{ of the nucleus}]$

$\Delta E \sim (e^2/\delta r) \times (\delta r)^3/a^3$

$\Delta E \sim e^2 (\delta r)^2/a^3$

Now all that’s left is to estimate $\delta r$.

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The trick here is to realize that all of those photons within the metal “box” are independently shaking the electron, and each push is in a random direction.  So if some photon with wave vector $k_1$ produces, by itself, a displacement of the electron $(\delta r)_{k_1}$, then the total displacement $(\delta r)$ satisfies

$(\delta r)^2 = (\delta r)^2_{k_1} + (\delta r)^2_{k_2} + (\delta r)^2_{k_3} ...$

[This is a general rule of statistics: independently-contributing things add together in quadrature.]

A photon is essentially just an electric field that keeps reversing sign.

In our case, each $(\delta r)_k$ comes from the influence of a photon, which has electric field $\vec{\mathcal{E}_k}$.  The simplest way to estimate $(\delta r)_k$ is to imagine that the electric field $\vec{\mathcal{E}_k}$ pushes on the electron for a time $\tau \sim 1/kc$ (the period of its oscillation), after which it reverses its direction (as shown on the right).  During that time, the acceleration of the electron is something like $|\vec{A}| \sim |\vec{F}|/m$, where $\vec{F} = e\vec{\mathcal{E}_k}$ is the force of the electric field pushing on the electron, and its net displacement $(\delta r)_k \sim |\vec{A}| \tau^2$.  This means

$(\delta r)_k \sim e |\vec{\mathcal{E}_k}|/mk^2 c^2$.

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Now we should just add up all the $(\delta r)^2_k$s.  Since there is a very large number of photons with a nearly continuous range of energies (due to the very large size of the confining “box”), we can replace a sum over all $k$‘s with an integral: $\sum_k (\delta r)^2_k\rightarrow L^3 \int d^3 k (\delta r)^2_k$.  Inserting the expressions for $(\delta r)_k$ and $|\vec{\mathcal{E}_k}|^2$ gives the following:

$(\delta r)^2 \sim (e^2 \hbar/m^2 c^3) \int (1/k) dk$.

You can notice that the size $L$ of the box drops out of the expression, which is good because the box was completely fictitious anyway.

The only remaining thing to figure out is what to do with the integral $\int(1/k) dk$, which is technically equal to infinity.  In physics, however, when you get an infinite answer, it means that you forgot to stop counting things that shouldn’t actually count.  In this case, we should stop counting photons whose wavelength is either too short or too long to affect the electron.  On the long wavelength side, we should stop counting photons when their wavelength gets bigger than the size of the atom, $a$.  Such long wavelength photons make an electric field that oscillates so slowly that by the time it has changed sign, the electron has likely moved to a completely different part of the atom, and the net effect is zero.  One the short wavelength side, we should stop counting photons when their wavelength gets shorter than the Compton wavelength.  Such photons are super-energetic, with energy larger than $mc^2$, which means their energy is so high that they don’t push around electrons anymore: they spontaneously create new electrons from the vacuum.  [In essence, it doesn't make sense to talk about electron position at length scales shorter than the Compton wavelength.]

Using these two wavelengths as the upper and lower cutoffs of the integral gives $\int (1/k) dk = ln(1/\alpha)$.  Here, $\alpha = e^2/\hbar c \approx 1/137$ is the much-celebrated fine structure constant.

[It is perhaps worth pausing to note, as so many before me have done, what a strange and interesting object the fine structure constant is.  $\alpha$ contains only the most fundamental constants of electricity, quantum mechanics, and relativity, ($e, \hbar$, and $c$), and they combine to produce exactly one dimensionless number.  How strange that this number should be as large as 137.  As a general rule, fundamental numbers produced by the universe are usually close to $1$.]

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Now we have all the pieces necessary to assemble a result for the Lamb shift.  And actually, if you like the fine structure constant, then you’ll love the final answer.  It looks like this:

$\Delta E/R \sim \alpha^3 \ln(1/\alpha)$.

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At the beginning of this post I mentioned that the Lamb shift is very small — only about 1/500,000 of the energy of hydrogen (the Rydberg energy, $R$).   Now, if you want to know why the Lamb shift is so small, the best answer I have is that the Lamb shift is proportional to $\alpha^3$, and in our universe the fine structure constant $\alpha$ just happens to be a small number.

It’s interesting to note that if we somehow lived in a universe where $\alpha$ was not so small, then the Lamb shift would get big, and those random fluctuations of the quantum field would get large enough to completely knock the electron off the nucleus.  This would be a universe without atoms, and consequently, without you and me.

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Footnotes

Probability clouds for different hydrogen states. Different states along the same row are supposed to have the same energy, but the Lamb shift splits the S states from the P and D states.

1)  You might notice that the Lamb shift appeared only because the electron probability cloud had a peak at the center of the atom.  If it didn’t have a peak — say, if it went to zero near the center — then there would be no Lamb shift.  This is in fact exactly how the Lamb shift was discovered.  Certain excited states of the hydrogen atom have a peak near the center and others go to zero.  So while normal quantum mechanics predicts that, say, the 2S and 2P states (shown to the right) have the same energy, in fact the Lamb shift makes a small difference between them.  This difference can be observed as a faint radio wave microwave signal from interstellar hydrogen.

2)  It’s probably worth noting that if you increased the fine structure constant $\alpha$, you would have run into bigger problems long before you started fussing about the Lamb shift.

3) Also, while $\alpha$ is a small number in our normal world, it’s not hard to imagine synthetic worlds (the interior of certain materials) where $\alpha$ is not a small number.  For example, graphene is a material inside of which there is an effective speed of light that is 300 times smaller than $c$.  This makes $\alpha$ very close to 1, and the sort of effects discussed in this post get very real.  This is part of why there has been so much ado about graphene among physicists: it’s quite an exciting (and frustrating) playground for people like me.

UPDATE:

4)  I just came across this video of Freeman Dyson (one of my personal favorite physicists) explaining the Lamb shift and some of the history behind it.  His conceptual summary of it starts at 2:43.

1. April 25, 2013 4:27 am

I really enjoyed your post. There was a lot that I missed, but a lot that I took in. I would suggest for people that don’t have the physics experience that you have, to define every variable. You don’t necessarily have to explain every single one, but it would be nice what every variable stood for in your equations.

Thanks again. Good post, and good blog.

April 25, 2013 8:24 am

Sorry. I hate when people do that. I think what I missed is that $\hbar$ is Planck’s constant (divided by $2\pi$), $m$ is the electron mass, and $e$ is the electron charge.