A Superconductor Is Not a Bose-Einstein Condensate of Cooper Pairs — Epoche B2
Beyond Simple Condensation: The True Nature of Superconductivity A slogan you meet early in solid-state physics runs like this [1] : below a critical temperature $T_c$ certain metals lose all electrical resistance because their electrons bind into Cooper pairs [2] , and a Cooper pair, carrying integer spin, is a boson, so the superconductor is simply the Bose-Einstein condensate (BEC) of those bosons. The slogan is not wrong that a single macroscopic quantum state forms. But it quietly smuggles in a false analogy: it pictures superconductivity as a gas of ready-made particles dropping into one state [3] , when what actually happens is that the electron system rebuilds itself from the ground up . This note assembles the argument piece by piece — what a normal metal's electrons look like, what a genuine BEC requires, why electrons pair, and what the pairing does to the ground state and its excitations. The decisive event, we shall see, is the opening of an energy gap in a reconstructed many-body state, and its fingerprint is written plainly in the heat capacity. The normal metal: a gapless Fermi sea Electrons are fermions, so the Pauli exclusion principle forbids two of them from occupying the same quantum state. At absolute zero the electrons in a metal therefore stack one per state, filling every level up to a sharp ceiling, the Fermi energy $E_F$; this filled block of states is the Fermi sea . At finite temperature the occupation of a state of energy $\varepsilon$ is smeared over a width $\sim k_B T$ by the Fermi-Dirac distribution, $$ \bar{n}(\varepsilon) = \frac{1}{e^{(\varepsilon - E_F)/k_B T} + 1}, $$ where $k_B$ is Boltzmann's constant and $T$ the temperature; $\bar{n}$ runs from $1$ (states well below $E_F$, full) to $0$ (states well above, empty). The property that matters here is that the spectrum is gapless : an electron sitting just below $E_F$ can be lifted to a state just above it at arbitrarily small cost, so the available excitation energies run continuously down to zero. This is exactly why a normal metal's electronic heat capacity is linear in temperature, $C_n = \gamma T$, with $\gamma \simeq \tfrac{\pi^2}{3}N(0)k_B^2$ the Sommerfeld coefficient and $N(0)$ the density of electronic states at the Fermi level: only the electrons within about $k_B T$ of $E_F$ can be thermally excited, and their number grows in proportion to $T$. Keep this link in mind — gapless spectrum implies a linear heat capacity , the Sommerfeld result derived in Kittel's Introduction to Solid State Physics — because superconductivity breaks it. What a genuine Bose-Einstein condensate requires Bosons obey no exclusion rule, so any number of them may share one state. Cool an ideal gas of $N$ such bosons below a critical temperature and a macroscopic fraction of them collapses into the single lowest-energy state. The number $N_0$ in that ground state follows $$ \frac{N_0}{N} = 1 - \left(\frac{T}{T_c}\right)^{3/2}, $$ so at $T = 0$ every particle sits in one state ($N_0 = N$) and the fraction falls to zero at $T_c$. The transition temperature itself is fixed by the number density $n$ and the particle mass $m_B$ alone, $$ k_B T_c = \frac{2\pi\hbar^2}{m_B}\left(\frac{n}{\zeta(3/2)}\right)^{2/3}, \qquad \zeta(3/2) \approx 2.612, $$ where $\hbar$ is the reduced Planck constant and $\zeta(3/2)$ is a numerical constant from the sum over states. It is worth putting this to work, because the result is a warning. Liquid $^4$He has $n \approx 2.2\times10^{28}\,\mathrm{m^{-3}}$ and $m_B = 6.65\times10^{-27}\,$kg, which gives $T_c \approx 3.1\,$K; the observed superfluid transition sits at $2.17\,$K, and the condensate fraction at $T=0$ is not $100\%$ but about $8\%$, because the helium atoms interact strongly. The clean realisations of the formula are instead dilute alkali gases, some $10^5$ times more rarefied, which condense in the nanokelvin range. Three assumptions are built into this textbook BEC, and superconductivity will overturn all three: the bosons are (i) pre-formed , existing as stable particles whether or not they condense; (ii) dilute and effectively point-like; and (iii) merely rearranged by condensation — the single-particle spectrum stays gapless, and the ideal condensate's heat capacity follows a soft power law, $C \propto T^{3/2}$, with no activation barrier of any kind. Nothing here reconstructs the particles themselves. Why the Fermi sea is unstable: Cooper pairing Electrons repel through the Coulomb force, so binding them seems hopeless. The lattice supplies the loophole. A moving electron drags the positive ions towards it, and because the heavy ions relax slowly it leaves a fleeting wake of excess positive charge that attracts a second electron. This retarded, phonon-mediated attraction is feeble, yet in 1956 Cooper showed it is decisive. Add two electrons just above a filled Fermi sea; because every state below $E_F$ is Pauli-blocked, the pair can only scatter within a thin shell of width $\sim \hbar\omega_D$ above the Fermi surface, and in that squeezed phase space a bound state with positive binding energy $E_b$ appears for any attraction, however weak: $$ E_b \approx 2\hbar\omega_D \exp\!\left(-\frac{2}{N(0)V}\right), $$ where $\hbar\omega_D$ is the Debye energy set by the lattice vibrations, $V$ measures the strength of the attraction, and $N(0)$ is again the density of states at $E_F$. The precise value is unimportant; the existence of a bound state is everything. The normal Fermi sea is unstable — it can always lower its energy by pairing. The favoured pair carries zero total momentum and opposite spins, $(\mathbf{k}\uparrow, -\mathbf{k}\downarrow)$, a spin singlet. The reconstructed ground state and the energy gap A single Cooper pair is only the seed. Because pairing lowers the energy everywhere on the Fermi surface, the whole surface pairs at once and self-consistently, producing an entirely new ground state in which each $(\mathbf{k}\uparrow, -\mathbf{k}\downarrow)$ mode is a coh