A Cooper Pair Is a Correlation in Momentum Space, Not a Molecule — Epoche B2
Beyond Simple Pairs: The Quantum Reality of Superconductivity The first picture most of us are given of superconductivity is a chemical one [1] : two electrons, glued together by the lattice, form a Cooper pair [2] , and the pair is a little molecule that drifts through the crystal without friction. The picture is memorable, and it gets one thing right — electrons really do pair. Everything else about it is misleading, and the reason is worth spelling out precisely. A Cooper pair is not a compact object bound in real space; it is a correlation in momentum space , an agreement about which pair modes carry amplitude, spread over a region of the crystal a million times wider than a chemical bond. This note builds that claim from the Fermi sea upwards, and ends by computing the size of a pair from nothing but the width of the shell of momentum states that pairing disturbs. The Fermi sea and the sharpness of its surface A metal is a rigid lattice of positive ions in a sea of mobile conduction electrons. Electrons are fermions, so the Pauli principle allows one per single-particle state, and at zero temperature they fill every state up to the Fermi energy $E_F$ — a few electronvolts, $11.7\,\text{eV}$ in aluminium. The occupation of a state whose energy is $\xi_{\mathbf{k}}=\varepsilon_{\mathbf{k}}-E_F$, measured from the Fermi level, is the Fermi-Dirac function $$ f(\xi_{\mathbf{k}}) = \frac{1}{e^{\xi_{\mathbf{k}}/k_B T} + 1} \;\xrightarrow[\,T\to 0\,]{}\; \begin{cases} 1, & \xi_{\mathbf{k}}\lt 0,\\ 0, & \xi_{\mathbf{k}}\gt 0,\end{cases} $$ where $k_B$ is Boltzmann's constant and $T$ the temperature. At $T=0$ this is a perfect step: the boundary between filled and empty states — the Fermi surface , a sphere of radius $k_F$ in momentum space — is infinitely sharp. Two further quantities will be needed. $N(0)$, the density of single-particle states per unit energy per spin at the Fermi level, measures how thickly states are packed there. And $v_F=\hbar k_F/m$ is the Fermi velocity, about $2\times10^{6}\,\text{m}\,\text{s}^{-1}$ in aluminium: the speed of the electrons that do all the work. The aluminium figures used throughout — $E_F$, $v_F$, $k_F$, the Debye temperature and $T_c$ — are the tabulated values in Kittel [3] . The lattice supplies the attraction that pairing needs. A passing electron drags the heavy ions towards it; they respond sluggishly, and the lingering excess of positive charge draws in a second electron. The attraction is weak and delayed, and it acts only between electrons whose energies lie within the Debye energy $\hbar\omega_D$ of each other — the largest energy the lattice vibrations can carry, $37\,\text{meV}$ in aluminium. Cooper's 1956 result is that against a filled Fermi sea even an arbitrarily weak attraction of this kind binds a pair of electrons with opposite momenta and opposite spins, $(\mathbf{k}\uparrow,-\mathbf{k}\downarrow)$. Because that is true everywhere on the Fermi surface, the surface does not bind one pair; it reorganises all at once. What the reorganisation actually is The Bardeen-Cooper-Schrieffer ground state expresses that reorganisation as a statement about pair modes [4] . For each $\mathbf{k}$ the mode $(\mathbf{k}\uparrow,-\mathbf{k}\downarrow)$ is either empty or doubly occupied, and the state is a coherent superposition of both possibilities: $$ |\text{BCS}\rangle = \prod_{\mathbf{k}}\left(u_{\mathbf{k}} + v_{\mathbf{k}}\,c^{\dagger}_{\mathbf{k}\uparrow}c^{\dagger}_{-\mathbf{k}\downarrow}\right)|0\rangle, \qquad |u_{\mathbf{k}}|^2 + |v_{\mathbf{k}}|^2 = 1. $$ Here $|0\rangle$ is the state with no electrons, $c^{\dagger}_{\mathbf{k}\uparrow}$ creates an electron of momentum $\hbar\mathbf{k}$ and spin up, $|v_{\mathbf{k}}|^2$ is the probability that the mode is occupied and $|u_{\mathbf{k}}|^2$ that it is empty. Notice what this expression does not contain: any reference to where an electron is. It assigns amplitudes to momenta. Whatever a Cooper pair is, it is built here. Solving the pairing problem self-consistently produces an energy gap $\Delta$, which in the weak-coupling limit is $$ \Delta = 2\hbar\omega_D\,\exp\!\left(-\frac{1}{N(0)V}\right), $$ with $V>0$ the strength of the attraction and $N(0)V$ the dimensionless coupling. For aluminium $\Delta(0)\approx0.18\,\text{meV}$, consistent with the BCS relation $\Delta(0)\approx1.76\,k_BT_c$ and $T_c\approx1.2\,\text{K}$. The elementary excitations are not electrons but Bogoliubov quasiparticles — coherent mixtures of an added electron and a removed one — and their energies are $$ E_{\mathbf{k}} = \sqrt{\xi_{\mathbf{k}}^{\,2} + \Delta^{2}} \;\ge\; \Delta . $$ Minimising the energy of $|\text{BCS}\rangle$ fixes the amplitudes in terms of these two quantities: $$ |v_{\mathbf{k}}|^2 = \frac{1}{2}\left(1 - \frac{\xi_{\mathbf{k}}}{E_{\mathbf{k}}}\right) = \frac{1}{2}\left(1 - \frac{\xi_{\mathbf{k}}}{\sqrt{\xi_{\mathbf{k}}^{\,2}+\Delta^{2}}}\right). $$ Read this against the step function above. Deep below the Fermi level, $\xi_{\mathbf{k}}\ll-\Delta$, it returns $1$; far above, $\xi_{\mathbf{k}}\gg\Delta$, it returns $0$; exactly at the Fermi level it returns $\tfrac12$. Between those extremes the step has been rounded off, and the rounding is confined to a band of order $\Delta$ on either side: at $\xi_{\mathbf{k}}=\pm\Delta$ the occupation is $\tfrac12(1\mp1/\sqrt2)$, that is $0.85$ and $0.15$, so essentially the whole crossover is done within $\pm2\Delta$ (Fig. 1). Everything outside that band is untouched. The superconducting transition rearranges nothing except a skin on the Fermi surface. Two curves on the same axes. The blue curve is the zero-temperature Fermi-Dirac occupation of a normal metal: exactly one below the Fermi level, dropping vertically to exactly zero above it. The amber curve is the BCS pair occupation, which equals one half at the Fermi level and rounds the step off over a band of order the gap on either side, reaching 0.85 at minus one gap and 0.15 at plus one gap. 1 ½ 0