The Quantum Nature of Superconductivity — Epoche B2
Beyond Resistance: How the Superconducting State Admits Magnetic Flux A superconductor is usually introduced twice over: it carries current with zero resistivity, $\rho = 0$, and it expels magnetic field from its interior, $\mathbf{B} = 0$. Both statements are true, and the second is the deeper one, because flux expulsion is a quantum effect with no classical counterpart. But taken together they leave a reader with a false picture — of a material that simply refuses magnetism, and refuses it harder the better a superconductor it is. The opposite is closer to the truth. Every superconductor gives up on flux expulsion at some field, and the materials that survive the largest fields are precisely the ones that stop expelling flux at the smallest . They do not resist the field; they let it in, in indivisible quantum lumps, and carry current around each one. This essay works out where the line falls, why it falls there, and what the field looks like once it is inside; the material is standard, and Tinkham's and Kittel's texts treat it at full length [1] . The Meissner state has a price, and the price is finite Expelling a field is not free. To keep $\mathbf{B}=0$ in its interior, a superconductor must run screening currents that cancel the applied field $B_a$ throughout its bulk, and excluding a field from a volume costs the magnetic energy that field would otherwise have held there. Per unit volume that cost is $$ u_{\text{field}} = \frac{B_a^{2}}{2\mu_0}, $$ where $\mu_0 = 4\pi\times10^{-7}\ \mathrm{H\,m^{-1}}$ is the permeability of free space and $B_a$ is the applied flux density in tesla. Against this the superconductor has only one asset: the energy it saved by condensing in the first place. Writing $f_n$ and $f_s$ for the free-energy densities of the normal and superconducting states at the same temperature, the condensation energy density $f_n - f_s$ is a fixed property of the material, and it is conventional to quote it as a field, the thermodynamic critical field $B_c$: $$ f_n - f_s \equiv \frac{B_c^{2}}{2\mu_0}. $$ The comparison is now immediate. Superconductivity survives only while the saving exceeds the cost [2] , that is while $B_a \lt B_c$; above $B_c$ the material is better off normal and the transition is thrown. The numbers are small. Lead has $B_c(0) = 80.3\,$mT, so its condensation energy density is $(0.0803)^2/(2\mu_0) \approx 2.6\ \mathrm{kJ\,m^{-3}}$; aluminium, with $B_c(0) = 10.5\,$mT, saves only about $44\ \mathrm{J\,m^{-3}}$. For comparison, a hospital scanner works at $1.5$ or $3\,$T — twenty to forty times $B_c$ for lead. If flux expulsion were the whole story, no superconductor could be used to build a magnet at all. Two lengths, not one The escape route appears once we notice that the superconducting state is characterised by two independent lengths, and that nothing forces them to be similar. The first is the penetration depth $\lambda$, the distance a magnetic field soaks into the surface before the screening currents have cancelled it, $B(x) = B(0)e^{-x/\lambda}$. It is set by how much supercurrent a given field can drive, $$ \lambda = \sqrt{\frac{m}{\mu_0\, n_s e^{2}}}, $$ with $m$ the carrier mass, $e$ the elementary charge and $n_s$ the density of superconducting carriers. For a metal with $n_s \sim 3\times10^{28}\ \mathrm{m^{-3}}$ this gives $\lambda \approx 30\,$nm. The second is the coherence length $\xi$, and it measures something quite different: the shortest distance over which the density of the condensate can change. The superconducting state is described by an order parameter $\Psi$, one complex number per point of the sample, whose squared modulus $|\Psi|^{2}$ is the local density of Cooper pairs. Bending $|\Psi|$ costs energy, so it cannot be switched off abruptly; forced to zero at some point, it recovers to its full value over a distance $\xi$. In the weak-coupling theory $$ \xi_0 = \frac{\hbar v_F}{\pi \Delta}, $$ where $\hbar$ is the reduced Planck constant, $v_F$ the Fermi velocity of the electrons and $\Delta$ the superconducting energy gap — the minimum energy needed to break a pair. Aluminium has $v_F \approx 2.0\times10^{6}\ \mathrm{m\,s^{-1}}$ and $\Delta \approx 0.18\,$meV $= 2.9\times10^{-23}\,$J, which puts $\xi_0$ at a couple of micrometres; the tabulated value is $1.6\ \mu$m. That is roughly four thousand aluminium lattice spacings. A Cooper pair is an enormous, floppy object, and of order a million other pair centres lie inside the volume of any one of them. So $\lambda$ and $\xi$ have nothing to do with each other: one is fixed by carrier density, the other by the gap and the Fermi velocity. In aluminium $\xi \gg \lambda$; in a dirty alloy, where electron scattering shortens $\xi$ without shortening $\lambda$, the inequality can reverse completely. Their ratio is the Ginzburg–Landau parameter $$ \kappa = \frac{\lambda}{\xi}, $$ and everything that follows is decided by which side of a single number $\kappa$ falls on. The sign of a surface Suppose a sample is partly normal and partly superconducting, with a flat wall between the two regions. Follow the energy across that wall. On the superconducting side the order parameter cannot rise instantly — it needs a distance $\xi$ — so a slab of thickness $\xi$ is deprived of its full condensation energy, at a cost $\xi\, B_c^{2}/2\mu_0$ per unit area. But the field does not stop instantly either: it leaks a distance $\lambda$ into the superconducting side, so a slab of thickness $\lambda$ escapes the cost of exclusion, a saving of $\lambda\, B_c^{2}/2\mu_0$ per unit area. The wall's surface energy is the difference, $$ \sigma_{ns} \sim \frac{B_c^{2}}{2\mu_0}\,(\xi - \lambda), $$ and its sign is the whole story. If $\xi \gt \lambda$ the surface energy is positive: interfaces are expensive, the sample makes as few as possible, and it stays fully field-free until $B_c$ and then goes normal in one step. This is a type-I superconductor. If $\lambda \gt \xi$ the surface energy is nega