The Lamb and Retherford Experiment; Quantum Field Theory

The prediction was that two levels coincide exactly. A prediction of zero is the most fragile kind a theory can make, because any measured difference at all is infinitely far from it.

🎯 Why this matters

This is why the sharpest tests of a theory are its zeros. Agreement to three decimal places in a large quantity can survive a wrong theory; a splitting that is supposed not to exist cannot.

§5.1 built a theory. This section is about the measurement that proved it was not enough — and it is a measurement on hydrogen, the simplest atom there is, made with a magnet and a radio transmitter.

5.2 A degeneracy that should have been exact

📐 Physics you need first — the hydrogen levels, in three layers

Layer 1: Schrödinger. In a $-1/r$ potential the energy depends only on the principal quantum number: $E_n = -13.6/n^2$ eV. Everything with the same $n$ is degenerate, however the electron is orbiting. That degeneracy is an accident of the $1/r$ shape and is enormous — the $n = 2$ shell holds 2S and 2P states at identical energy.

Layer 2: Dirac. Solve the relativistic equation instead and expand in α. The answer, Eq. (5.23), depends on $n$ and on $j$ , the total electronic angular momentum, but not on the orbital momentum $L$ :

E_{n,j} = -\htmlClass{t-ryd}{\frac{Rhc}{n^2}} \left[1 + \htmlClass{t-a}{\frac{\alpha^2}{n}} \left(\htmlClass{t-j}{\frac{1}{j + 1/2}} - \frac{3}{4n}\right)\right]

(5.23)

The Dirac energy levels of hydrogen, expanded in α. Read what it depends on — and, more importantly, what it does not.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

That partial lifting of the degeneracy is the fine structure — spin–orbit coupling, tens of μeV. It splits 2P₃/₂ away from 2P₁/₂. But 2S₁/₂ and 2P₁/₂ have the same $n$ and the same $j$ and different $L$ , so Dirac says they are still exactly degenerate.

Layer 3 is what this section is about, and it is not in the Dirac equation at all.

Notation: $nL_j$ , with the spin multiplicity dropped because for one electron it is always 2. Since $s = 1/2$ , every $L$ gives $j = L \pm 1/2$ — except S waves, where only $j = 1/2$ exists.

⚠️ Erratum — the 2S₁/₂–2P₁/₂ gap in Fig. 5.1

Figure 5.1 labels the Lamb shift as 6.3 μeV. The measured value, Eq. (5.25), is 1057.8 MHz — and $h \times 1057.8\ \text{MHz} = 4.37$ μeV. 6.3 μeV would be 1523 MHz.

The other number on the same figure is right: the 2P₃/₂–2P₁/₂ fine-structure splitting of 45.2 μeV is 10.9 GHz, which checks out. Only the Lamb-shift label is wrong. Confirmed on the page render.

The idea

The shift is a thousand times smaller than the fine structure, so it cannot be seen in an optical spectrum. Lamb and Retherford did something else entirely: they drove transitions between the levels with a radio-frequency field and measured the frequency, which is where the precision is.

The trick that makes it possible is that 2S₁/₂ is metastable. It cannot decay to 1S₁/₂ — that would need $\Delta l = 0$ , and a photon carries one unit — and the 2S → 2P route is hopeless because the spontaneous rate scales as the cube of the energy difference, and this difference is a millionth of an optical one.

Why the 2S level lives 100 μs instead of 1.6 ns

h = 4.135667696e-15          # eV s
E_lyman = 10.20              # eV, 2S -> 1S
E_lamb  = h * 1057.8e6       # eV, 2S -> 2P

print(f"  2S -> 1S  Lyman-alpha:  {E_lyman:.2f} eV   forbidden anyway"
      f" (needs dl = 1, and both are S)")
print(f"  2S -> 2P  Lamb shift : {E_lamb:.2e} eV   allowed, but the rate goes as (dE)^3")

r = E_lyman / E_lamb
print(f"\n  ratio of energies      = {r:.2e}")
print(f"  ratio of rates (cubed) = {r**3:.2e}")
print("  so a transition that would take nanoseconds instead takes ~1e5 s of level-lifetime")
print("  -> the 2S state survives the 0.8 m flight through the apparatus, and nothing else does")

prints

  2S -> 1S  Lyman-alpha:  10.20 eV   forbidden anyway (needs dl = 1, and both are S)
2S -> 2P  Lamb shift : 4.37e-06 eV   allowed, but the rate goes as (dE)^3

ratio of energies      = 2.33e+06
ratio of rates (cubed) = 1.27e+19
so a transition that would take nanoseconds instead takes ~1e5 s of level-lifetime
-> the 2S state survives the 0.8 m flight through the apparatus, and nothing else does

🔬 Experiment card — Lamb and Retherford, Columbia 1947

Apparatus

A beam of atomic hydrogen in an adjustable magnetic field, crossing a radio-frequency cavity and ending on a tungsten plate. Six stages, each solving one problem — see the diagram and its callouts below.

What is measured

An electric current, in picoamps. The radio frequency is fixed, the magnetic field is swept, and the current is watched for dips: a dip means the RF has come into resonance with a 2S → 2P₃/₂ transition, the metastable atoms have been pumped into a state that decays immediately, and they no longer reach the detector.

The result

Three families of resonances, one per reachable 2P₃/₂ Zeeman level, each linear in B. Extrapolated to zero field they meet at a single frequency — and it is not where the Dirac equation puts it.

What it proved

\Delta E(2S_{1/2} - 2P_{1/2}) = 1057.8 \pm 0.1

MHz, where the theory of the day said exactly zero. Together with Kusch and Foley’s

g_e \neq 2

in the same year, it forced the construction of quantum field theory.

🛠️ Fig. 5.3 — the apparatus, and the problem each stage solves

Click a numbered marker for what that piece does.

Redrawn from Bettini Fig. 5.3. Read it as five separate problems solved in series: make the atoms, excite them without light, keep only the metastable ones, drive the transition, and detect an excited atom against 10⁸ ground-state ones.

Why a magnetic field at all

2S₁ᐟ₂ m = +1/2
2S₁ᐟ₂ m = −1/2
2P₁ᐟ₂ m = +1/2
2P₁ᐟ₂ m = −1/2

Fig. 5.2, computed rather than traced: ν = (Lamb shift) + g_J m (μ_B/h) B, with μ_B/h = 13.996 GHz/T, g_J = 2 for 2S₁ᐟ₂ and 2/3 for 2P₁ᐟ₂. The zero-field gap between the solid and dashed pairs IS the Lamb shift. Watch the m = −1/2 branch: it crosses the two 2P₁ᐟ₂ levels at 0.057 and 0.113 T, mixes with them, and stops being metastable — which is precisely how the experiment throws it away.

💡 What this really says — the field is not a probe, it is a filter and a ruler

It is easy to read the magnetic field as “something to split the levels with”. It is doing two jobs, and both are essential.

A filter. The 2S₁ᐟ₂ m = −1/2 branch slides down into the 2P₁ᐟ₂ levels and mixes with them, which destroys the metastability that the whole experiment depends on. Ten thousand times shorter lifetime, 0.08 mm of flight instead of 0.8 m: it never reaches the detector. The field selects one spin state by ruining the other.

A ruler. Sweeping B moves the levels through resonance with a fixed RF frequency, so a frequency measurement — the most precise kind there is — is converted into a field sweep and back. That is why the data in the next figure are lines rather than points: each line is one transition traced out across the field.

The result

measured — with the Lamb shift
Dirac prediction — no shift
data

Fig. 5.4, redrawn. Three transitions from 2S₁ᐟ₂(m = +1/2) to the three reachable 2P₃ᐟ₂ Zeeman levels, each linear in B with a slope set by g_J and m. The measurement is the INTERCEPT: extrapolate the three solid lines to zero field and they meet at 9.89 GHz, while the Dirac prediction (dashed) meets at 10.95 GHz. Points are simulated at the published scatter; the lines are computed from the Zeeman slopes.

\Delta E\left(2S_{1/2} - 2P_{1/2}\right) = \htmlClass{t-v}{1057.8 \pm 0.1}\ \text{MHz}

(5.25)

Bettini p. 191, the 1952 value. The Dirac equation predicts exactly zero.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

5.3 What had to change

The reading that fails is the one where the proton’s electric field is given, a fixed external potential that the electron moves in. The field is itself a quantum system, made of photons, and the electron interacts with it — including with the part of it that the electron itself produced.

Bettini introduces Feynman diagrams here, three sections before deriving anything about them, purely as a way to picture that. So will we.

Fig. 5.5 — an electron scattering off a nucleus

Click a vertex or an internal line.

Time runs to the right, space is vertical. The nucleus is drawn as a horizontal line because it is far too heavy to move during the interaction. Click the vertices and the photon.

Now three things that a classical picture has no room for at all.

Fig. 5.6 — self-energy

Click a vertex or an internal line.

An electron in vacuum emits and reabsorbs its own photon.

Fig. 5.7 — vacuum polarization

Click a vertex or an internal line.

A photon becomes an e⁺e⁻ pair and back. §5.8 is what this does to α.

Fig. 5.8 — both at once

Click a vertex or an internal line.

The exchanged photon of Fig. 5.5, with a pair loop on it.

⚠️ These processes do not conserve energy — and that is allowed

An electron emitting a photon it has not got the energy for is, taken literally, a violation. The escape is that the violation is unobservable:

\htmlClass{t-de}{\Delta E}\,\htmlClass{t-dt}{\Delta t} \gtrsim \hbar

The licence for every virtual process in this chapter. It is not a statement that energy conservation is suspended — it is a statement about what can be measured.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

To detect an energy discrepancy $\Delta E$ you must measure to better than $\Delta E$ , and that takes at least a time $\hbar/\Delta E$ . If the discrepancy is repaid faster than that, no measurement can catch it, and quantum mechanics permits what cannot be detected.

This is the same inequality that gave a resonance its width in §4.1 and that fixed the range of a force as $R = \hbar/mc$ in §1.10. Three uses, one relation — it is worth recognising on sight.

⚙️ Engineer’s bridge — a diagram is a netlist, and the amplitude is a product along it

Press what does this diagram MEAN? on any diagram above. The picture resolves into an algebraic expression: one factor of √α per vertex, one propagator per internal line, multiplied together.

That is a netlist. Vertices are components with a gain; internal lines are transfer functions between them; the response is the product along the path. An electrical engineer reading a signal-flow graph does exactly this arithmetic.

Two places the analogy has to be extended, and both are the quantum part:

you must sum over every diagram, not evaluate one — every topology connecting the same external legs contributes, and they add as complex amplitudes before anything is squared;
the sum is over amplitudes, so the observable, a probability, is the squared modulus of the total. Interference between diagrams is physical, and §5.6 shows a case where it is the whole measurement.

Where it breaks entirely: in a circuit the internal nodes have measurable voltages. Here they do not — an internal line is a virtual particle, off the mass shell, and no experiment can observe it. The netlist is a bookkeeping device for something that is not made of parts.

Renormalization, in one paragraph

Computed naively, the self-energy of Fig. 5.6 is infinite — exactly as the self-energy of a point charge is infinite in classical electromagnetism. Bethe’s observation, a month after the experiment, is that it does not matter: the bare mass is not observable, only the mass including the self-interaction is, and one may subtract an infinite constant to define the measured value. That is renormalization .

After the subtraction a free electron’s self-interaction contributes nothing, by construction. But a bound electron is different. Quantum fluctuations smear it over a small region, so it is no longer a point charge, and its binding energy is slightly less than a point charge’s would be. The effect is larger for a state with $L = 0$ — because only an S-wave electron has appreciable probability of being at the nucleus.

That is the Lamb shift: 2S₁ᐟ₂ is pushed up relative to 2P₁ᐟ₂ because the S-wave electron spends time where the field is strongest.

What the three quantum-field effects do, and where each is measured in this book

effect	diagram	what it does	measured in
self-energy	Fig. 5.6	Smears the electron over a small region, so a bound one is no longer a point charge. Divergent, and absorbed into the measured mass by renormalization — after which it still shifts BOUND levels, and more for S waves than for P.	§5.2, the Lamb shift
vacuum polarization	Fig. 5.7	Pairs appear around a charge and orient, screening it. The charge you measure therefore depends on how close you get — which is the running of α.	§5.8
vertex correction	Fig. 5.39(b)	Modifies how the electron couples to an external field, and hence its magnetic moment. Schwinger computed α/2π = 0.00116 in 1948 and it matched Kusch and Foley to every measured digit.	§5.9

All three are the same physics — the electron interacting with its own field — appearing in three different observables. <strong>Two of them were measured in 1947, within months of each other</strong>, and together they are why quantum electrodynamics exists.

💡 What this really says — the particle count is not fixed

The deepest consequence is easy to state and hard to absorb. In quantum mechanics, $|\psi|^2$ is the probability of finding the electron: one electron, present throughout. Fig. 5.9’s hydrogen atom does not work that way. Sometimes it holds two electrons and a positron. Sometimes three and two.

So ψ cannot be a wave function for a fixed particle. It has to be a field, carrying operators that create and destroy electrons and positrons — and the photon field alongside it does the same for photons. That is what “quantum field theory” means, and it is forced on us by a 1057.8 MHz measurement on hydrogen.

The reinterpretation is already visible in Fig. 5.5: the incoming electron is destroyed at the vertex and the outgoing one created. Whether they are the same electron is not a question the theory can be asked.

🔑 If you remember only three things

An energy difference measured as a frequency is measured better than anything else. Turning the splitting into a radio tone is what bought the precision that made the discrepancy undeniable.
Four microelectronvolts out of 13.6 eV decided it. One part in three million is the whole discrepancy, which is why the experiment had to be built around a null rather than a reading.
The repair changed what a state is. Fixing the number meant allowing the particle count itself to fluctuate, which is a heavier price than any new constant would have been.

Where this goes next

§5.4–5.5 makes the diagrams quantitative: where the propagator comes from, what a vertex is worth, and why the perturbative series converges.
§5.6 shows that the internal line of Fig. 5.5 can run backwards in time — and that this forces antiparticles to exist.
§5.8 is Fig. 5.7 taken seriously: vacuum polarization makes α depend on the momentum transfer.
§5.9 is the other 1947 measurement, the electron magnetic moment, now known to 0.13 parts per trillion.

✅ Check yourself — the Lamb shift and why fields must be quantised

0/6 answered · 0 correct

1.What exactly did the Dirac equation predict that Lamb and Retherford falsified?
2.Why does the apparatus need a magnetic field, given that the quantity being measured exists at zero field?
3.The detector is a tungsten plate. Why does that count only the atoms of interest?
4.The processes in Figs. 5.6–5.8 do not conserve energy. Why is that permitted?
5.Press 'what does this diagram MEAN?' on Fig. 5.5. What is the netlist view telling you?
6.Why is the Lamb shift bigger for $2S_{1/2}$ than for $2P_{1/2}$ ?