
Nuclear, particle & frontier orientation · Interactive lesson
Cooper pairing and 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.
UPIV-15.05 · Lesson 5 of 9 in this unit. Position is not a completion score.Nuclear, particle & frontier orientation. These numbered units provide nuclear, particle and frontier orientation, not a substitute for a complete specialist course. Check your university’s required depth and prerequisites.
Build the model
However weak the attraction, the Fermi sea cannot survive it.
The key idea
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 minimum energy of each quasiparticle released when one breaks, so breaking a pair costs 2Δ.
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.
The complete physical model and assumptions
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 sea is no longer frozen and states within ħωD below EF pair as well, doubling the phase space. 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, on the evidence of a near-zero isotope effect, do not set the pairing scale in the cuprates.
1A 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 bound-state condition then sets the integral of dξ/(2ξ − E) from 0 to ħωD equal to 1/N(0)V; that integral 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.
2Retardation 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 one per cent of the sea is involved.
3Self-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), where 1.13 is 2 exp(0.5772)/π = 1.134 and 0.5772 is Euler's constant. Divide the two: the exponential and ħωD both cancel, leaving 2Δ(0)/kB Tc = 4/1.134 = 3.53, a pure number with nothing fitted in it.
4Three 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 tunnelling measurements of such thresholds won him a share of the 1973 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 it jumps by 1.43 times the normal-state electronic heat capacity. 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.
5The 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 with microwaves of frequency ν and flat voltage steps appear at exactly V = nhν/2e, which is how the SI volt has been realised since 2019, with KJ = 483597.8484 GHz/V fixed by definition.
6Where 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 large isotope shift, and there the live issue is independent replication, not mechanism.
Key equations
Read the units, assumptions and limits before substituting.
Cooper's bound pair
Eb = 2ħωD exp(−2/N(0)V)
It has no Taylor series at V = 0, so no order of perturbation theory finds it: any attraction binds.
Eb = pair binding energy (J); N(0) = states per spin per unit volume at EF (J⁻¹ m⁻³); V = attraction strength (J m³); ħωD ≈ kB ΘDBCS gap at T = 0
Δ(0) = 2ħωD exp(−1/N(0)V)
Half the exponent of one lone pair, because states within ħωD below EF now pair as well, doubling the phase space — not a slipped factor.
Same N(0)V; Δ in joules, quoted in meV. Weak coupling means N(0)V ≲ 0.3.The parameter-free ratio
2Δ(0) = 3.53 kB Tc
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.Isotope effect
Tc ∝ M(−α), with α = 1/2
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.Flux quantum
Φ = n Φ₀, Φ₀ = h/2e = 2.068 × 10⁻¹⁵ Wb
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.Josephson relations
I = Ic sin φdφ/dt = 2eV/ħ
They turn a dc voltage into a frequency, which is why the SI volt has been realised through 2e/h since 2019.
Ic = critical current (A); φ = gauge-invariant phase difference (rad); V in volts. f = 2eV/h = 483.598 MHz per µV.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. 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.
Live results
- CRITICAL 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
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Debye temperature Θ at 200 u: 275 K; Ion mass M: 200 u; Coupling N(0)V: 0.28
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Students picture a Cooper pair as two electrons orbiting one another like a small molecule, then cannot see how it survives Coulomb repulsion or squeezes through a lattice. The pairing is in momentum space, not real space: the partners are k↑ and −k↓, and the correlation extends over the coherence length ξ₀ = ħvF/πΔ — about 1.6 µm in aluminium, thousands of lattice spacings, and still 38 nm, over a hundred spacings, in niobium. The volume of one pair holds the centres of order a million others, so pairs are not separate objects that occasionally collide; they overlap completely and share one phase, and that overlap is exactly what makes the condensate rigid. The repulsion is dodged in time, by retardation, not in space by bringing the partners close together.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Niobium 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.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- 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.
Answer. Predicted 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.
MediumExample 2
Mercury 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.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- 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 about three parts in ten 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.
Answer. Tc(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.
HardExample 3
HgBa₂Ca₂Cu₃O₈₊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.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- 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.
Answer. 2Δ/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.
Quick check
Select an answer and explain it before checking.
Review & continue
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 minimum energy of each quasiparticle released when one breaks, so breaking a pair costs 2Δ.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.