University Physics IV · Frontiers of Modern Physics · 15.5
Cooper Pairing & the BCS Gap
Superconductivity stops being a list of strange facts once you accept one thing: a filled Fermi sea is unstable to any attraction at all. Here is the gap that follows, the three measurements that read it off, and the line where the theory stops working.
Build the model
Connect the measurement to the mechanism.
Ordinary metals are stable because the Fermi sea is the lowest-energy filling of the available states. Cooper's 1956 calculation broke that: add two electrons above a filled sea with any attraction, however feeble, and they bind — the sea blocks every state below EF, so the available phase space is a thin shell rather than a volume, and the energy integral diverges logarithmically. The binding energy comes out as 2ħωD exp(−2/N(0)V), a function with no Taylor series at V = 0, which is why no order of perturbation theory around the normal metal could ever have found it.
BCS then made the instability self-consistent: every pair forms at once, k↑ with −k↓, all sharing a single phase, and the spectrum opens a gap Δ(0) = 2ħωD exp(−1/N(0)V) — half the exponent of the single-pair problem, because the pairs now condense against each other. The attraction itself is retarded: an electron leaves an overscreened positive wake in the ion lattice that persists for about one Debye period, by which time the electron is hundreds of lattice spacings away, so the second electron feels the wake without feeling the first electron's Coulomb repulsion. Three costs come with the model.
The ħωD ceiling caps Tc at a few per cent of the Debye temperature. The derivation assumes N(0)V is small. And it assumes phonons — which the cuprates do not use.
- Simple definition
- A Cooper pair is the bound state formed by two electrons of opposite momentum and spin just outside a filled Fermi sea, and the BCS gap Δ is the energy needed to break one — half of it per quasiparticle released.
- Example
- Tin superconducts below 3.72 K, so weak-coupling BCS gives 2Δ(0) = 3.53 kB Tc = 1.13 meV, and a tin–oxide–tin tunnel junction indeed passes no quasiparticle current until the bias reaches 1.13 mV.
It has no Taylor series at V = 0, so no order of perturbation theory finds it: any attraction binds.
N(0) = states per spin per unit volume at EF (J⁻¹ m⁻³); V = attraction strength (J m³); ħωD ≈ kB ΘD
Half the exponent of one lone pair, because every pair condenses at once — self-consistency, not a slipped factor.
Same N(0)V; Δ in joules, quoted in meV. Weak coupling means N(0)V ≲ 0.3.
One measured Tc predicts the tunnelling threshold with nothing fitted, so it is the theory's sharpest test.
Tc in K, gap in J. Sn: 3.72 K → 2Δ(0) = 1.13 meV. Nb returns 3.89 and Pb 4.3.
The mass in the answer is the lattice's, not the electron's — that is the evidence for a phonon glue.
M = ion mass in u. Hg: 4.185 K at 199.5 u → 4.146 K at 203.4 u, giving α = 0.48.
The 2 in the denominator is the carrier's charge, read straight off a ring: pairs, not single electrons.
n an integer; Φ the flux in Wb threading a superconducting ring. Measured in 1961, not assumed.
They turn a dc voltage into a frequency, which is why the SI volt has been realised through 2e/h since 2019.
φ = gauge-invariant phase difference (rad); V in volts. f = 2eV/h = 483.598 MHz per µV.
A filled sea is what makes any attraction enough
Cooper's 1956 question was deliberately small: freeze a filled Fermi sea, add exactly two electrons above it at k↑ and −k↓, and let them attract weakly. In free space a 3-D attraction must exceed a threshold before it binds, because the density of low-energy states vanishes as √E and there is nothing to sum. The sea removes that. Every state below EF is occupied, so the two extra electrons may only scatter within a shell of thickness ħωD above EF, where the density of states is essentially the constant N(0). The self-energy sum then runs as the integral of dξ/(2ξ − E) from 0 to ħωD, which is a logarithm, and a logarithm can be made to equal 1/N(0)V no matter how small V is. Out comes a bound state for every non-zero attraction, with binding energy 2ħωD exp(−2/N(0)V). The normal metal is not a metastable minimum; it is not a minimum at all.
Retardation is what makes electrons attract
An electron moving through the lattice pulls the positive ions towards its track. The ions are some 10⁵ times heavier, so they arrive late: the lattice needs roughly one Debye period, 2π/ωD ≈ 1.7 × 10⁻¹³ s for ΘD = 275 K, to build the distortion. In that time an electron travelling at vF ≈ 10⁶ m s⁻¹ has gone about 170 nm — hundreds of lattice spacings. A second electron arriving later sees a region of excess positive charge and is attracted to it, while the first electron is far too distant for the screened Coulomb repulsion, which reaches only about 0.1 nm, to matter. Repulsion and attraction are separated in time, not in strength. The exchange only works for electrons within ħωD of EF — 23.7 meV in niobium, against EF ≈ 5 eV, so under half a per cent of the sea is involved.
Self-consistency turns one pair into a gap
Cooper froze the sea; BCS let every electron pair at once and demanded consistency. The condition is 1 = N(0)V × ∫ dξ/√(ξ² + Δ²) taken from 0 to ħωD, which is N(0)V⋅arcsinh(ħωD/Δ), and for ħωD ≫ Δ it inverts to Δ(0) = 2ħωD exp(−1/N(0)V) — the same exponential as the single pair but with half the exponent, because the pairs are now condensing against a sea that is itself paired. The excitation spectrum becomes Eₖ = √(ξₖ² + Δ²), so the cheapest excitation costs Δ and breaking a pair costs 2Δ. Repeating the integral at finite T, with a tanh(E/2kBT) factor, gives kB Tc = 1.13 ħωD exp(−1/N(0)V). Divide the two: the exponential and ħωD both cancel, leaving 2Δ(0)/kB Tc = 4/1.13 = 3.53, a pure number with nothing fitted in it.
Three independent ways to read the gap off an experiment
Tunnelling is the most direct. In a normal–insulator–superconductor junction no quasiparticle current flows until eV = Δ; in a symmetric superconductor–insulator–superconductor junction a quasiparticle must be made on both sides, so the threshold is eV = 2Δ. For tin that is 1.13 mV, and Giaever's 1960 measurement of exactly this won a share of the Nobel Prize. Heat capacity is the second: below about Tc/2 the electronic term is exponentially activated, C ∝ exp(−Δ/kBT), so ln C against 1/T is a straight line of slope −Δ/kB, and at Tc there is a jump of ΔC/γTc = 1.43. Third, far-infrared absorption switches on when a photon can break a pair, at ħω = 2Δ — 1.13 meV for tin, which is 274 GHz. Three different apparatus, one number.
The pair charge is measured, not assumed
Nothing above proves the carrier has charge 2e; two 1961 experiments did. Deaver and Fairbank and, independently, Doll and Näbauer found the flux trapped in a superconducting ring quantised in units of h/2e = 2.068 × 10⁻¹⁵ Wb — half the h/e a single-electron condensate would give. Josephson added the phase in 1962. Link two superconductors weakly and they share a gauge-invariant phase difference φ, with I = Ic sin φ flowing at zero voltage; hold a voltage across the link and the phase winds, dφ/dt = 2eV/ħ, so the current oscillates at f = 2eV/h = 483.598 MHz per microvolt. A 25 µV bias gives 12.09 GHz. Irradiate the junction and the steps lock to the drive, which is how the SI volt has been realised since 2019, with KJ = 483597.8484 GHz/V fixed by definition.
Where BCS stops, and what replaced it
Put a cuprate into the formulae and they break. Hg-1223 superconducts at 133 K; with ΘD ≈ 400 K the weak-coupling expression demands N(0)V = 0.82, far outside the N(0)V ≪ 1 that produced the exponential in the first place. Its measured 2Δ/kB Tc is 5.2, not 3.53, and its isotope exponent at optimal doping is about 0.02 rather than 0.5, so the lattice is not setting the scale. Phase-sensitive tricrystal experiments and ARPES nodes give a d(x²−y²) order parameter that changes sign around the Fermi surface, where BCS gives an isotropic s-wave gap. Pairs survive — flux is still quantised in h/2e — but the glue does not. The pressurised hydrides run the other way: H₃S at 203 K and 155 GPa behaves like strong-coupling BCS, with a full isotope shift, and there the live issue is independent replication, not mechanism.
Change one variable at a time
Make the relationship visible.
Hold the mass at 200 u and drag N(0)V from 0.20 to 0.40: Tc climbs from 2.1 K to 25.5 K, twelvefold out of a doubled coupling. Then take the ion mass from 150 u to 250 u at fixed N(0)V and Tc drops 22.5 per cent — the square root of the mass ratio, which is the isotope effect.
CRITICAL TEMPERATURE Tc8.74 K
DEBYE TEMPERATURE Θ275 K
GAP 2Δ(0) = 3.53 kB Tc2.658 meV
Tc AS A FRACTION OF Θ0.0318
Live interpretationCRITICAL TEMPERATURE Tc: 8.74 K. DEBYE TEMPERATURE Θ: 275 K. GAP 2Δ(0) = 3.53 kB Tc: 2.658 meV. Tc AS A FRACTION OF Θ: 0.0318
Catch the common trap
Explain before calculating.
Tin has Tc = 3.72 K and follows weak-coupling BCS. In a symmetric tin–oxide–tin tunnel junction held at 0.3 K, at what bias voltage does quasiparticle current switch on?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyNiobium superconducts below Tc = 9.25 K. Use weak-coupling BCS to predict Δ(0) in meV, then compare with the tunnelling value Δ(0) = 1.55 meV and say what the comparison means.
- Weak coupling fixes the ratio with nothing fitted: 2Δ(0) = 3.53 kB Tc, and kB = 8.617 × 10⁻⁵ eV K⁻¹.
- 2Δ(0) = 3.53 × 8.617 × 10⁻⁵ eV K⁻¹ × 9.25 K = 2.81 × 10⁻³ eV = 2.81 meV, so the predicted Δ(0) = 1.41 meV.
- Now run the measurement back through the ratio: kB Tc = 8.617 × 10⁻⁵ × 9.25 = 0.797 meV, and 2 × 1.55 meV ÷ 0.797 meV = 3.89.
- 3.89 rather than 3.53 is not pairing failing. It is niobium sitting at the strong-coupling edge, N(0)V ≈ 0.28, where the retardation corrections BCS discarded start to count — lead, at 4.3, is further out still.
AnswerPredicted Δ(0) = 1.41 meV against 1.55 meV measured, a ratio 2Δ(0)/kB Tc = 3.89. Weak coupling underestimates the gap, as it does for every strong-coupling metal.
MediumMercury of mean isotope mass 199.5 u has Tc = 4.185 K. Predict Tc for the 203.4 u sample; given the measured 4.146 K, extract the isotope exponent α; then say what α ≈ 0.02 in optimally doped YBCO would mean.
- BCS lets the ion mass in through the Debye frequency alone: ωD ∝ √(K/M), so Tc ∝ ωD ∝ M(−1/2).
- Predicted: Tc = 4.185 K × √(199.5/203.4) = 4.185 × 0.9904 = 4.145 K, against 4.146 K measured — agreement to one part in four thousand.
- Extract α from the two points: α = −ln(4.146/4.185) ÷ ln(203.4/199.5) = 0.009363 ÷ 0.019360 = 0.484.
- α = 0.48 says the lattice sets the energy scale. No electron mass appears in Tc, so a purely electronic mechanism would give α = 0.
- YBCO's α ≈ 0.02 says the opposite: swapping ¹⁶O for ¹⁸O moves the oxygen mass 12.5 per cent and barely moves 92 K, so whatever pairs the cuprates is not simple phonon exchange.
AnswerTc(203.4 u) = 4.145 K predicted against 4.146 K measured, giving α = 0.48 — textbook BCS. YBCO's α ≈ 0.02 rules the same mechanism out.
HardHgBa₂Ca₂Cu₃O₈₊δ has Tc = 133 K, a tunnelling gap 2Δ ≈ 60 meV, and ΘD ≈ 400 K. Test whether weak-coupling BCS can account for it: find 2Δ/kB Tc, then find the N(0)V that Tc = 1.13 ΘD exp(−1/N(0)V) would demand.
- Gap ratio first: kB Tc = 8.617 × 10⁻⁵ eV K⁻¹ × 133 K = 11.46 meV, so 2Δ/kB Tc = 60 ÷ 11.46 = 5.24.
- That is 1.5 times the weak-coupling 3.53, and strong-coupling corrections only push the ratio up, so the coupling cannot be small.
- Now the coupling: Tc ÷ (1.13 ΘD) = 133 ÷ (1.13 × 400) = 133 ÷ 452 = 0.2942, so 1/N(0)V = −ln(0.2942) = 1.223 and N(0)V = 0.82.
- 0.82 is not weak coupling. The exponential was derived by assuming N(0)V ≪ 1, so the number is a demonstration that the formula does not apply here, not a measurement of a coupling.
- Two further facts close the case: the isotope exponent near optimal doping is about 0.02 rather than 0.5, and tricrystal phase-sensitive experiments give a d(x²−y²) gap with nodes where BCS gives an isotropic s-wave gap.
Answer2Δ/kB Tc = 5.24 and the demanded N(0)V = 0.82. Both sit outside weak-coupling BCS, and the near-zero isotope exponent plus d-wave symmetry say the mechanism itself differs.