A superconductor is a material which at low temperature has zero resistance. That is, it can support DC currents without a voltage drop. This is in spite of containing impurities. How do the electrons avoid scattering off the impurities? The answer is that at some critical temperature they begin to pair up into bosons (called ``Cooper pairs"). These bosons all occupy a single quantum state (the "Bose Condensate"). It acts as a wave, instead of a particle, and simply coherently flows around the impurities. Phenomonologically, this means that as one lowers the temperature the resistivity suddenly drops to zero when pairs form. The pairs act as a short circuit. Of course, right at the critical temperature, there are very few pairs so the maximum current they can carry is finite. This leads to the concept of "critical current". The critical current grows as temperature is lowered. There are really three questions that comes to mind:

1. What is the glue holding the electrons together? [A more detailed version of this question, is ``How can we predict which materials will become superconducting, and what the transition temperature will be?"]
2. Why does a condensate lead to zero resistance?
3. What is the effective description of the superconductor?

Glue

The glue which holds electrons together is ``phonons". This is sometimes known as the soft-mattress argument. Imagine two people in a bed. The weight of one person deforms the mattress, which attracts the other person. For electrons (which repel each-other via the Coulomb interaction) it is really the finite frequence, "retarded" version of this argument that applies. One person gets into a bed with a "memory foam mattress". They make a dent. They then get up. Once they leave another person gets in bed. They roll into the dimple left by the first. Thus each electron is attracted to where the other one "used to be." It turns out that at zero temperature an infinitesmal interaction suffices to drive a superconducting transition. Unfortunately, the transition temperature is exponentially sensitive to the interactions, and it is almost impossible to reliably calculate transition temparatures. There are two immediate consequences of the phonon mechanism. First, since the phonon properties are modified by changing the mass of the nuclei, there is an "isotope effect". You can change the transition temperature of a superconductor by substituting different isotopes. This was one of the main clues for a phononic mechanism. The second consequence is that metals with stronger electron-phonon coupling will have higher transition temperatures. Typically those same metals will have lower normal-state resistivity, as the electrons are then more sensitive to lattice defects. Thus we have the perverse rule of thumb that bad metals make good superconductors.

Zero resistivity of a Bose condensate

The zero resistivity of a Bose condensate comes from the fact that it is a single quantum state. Think of a plane wave $\psi=e^{i k x}$. It carries a current, without any potential drop. Imagine I add some scattering potential near the origin. Far to the left of the impurity, the wavefunction will be modified to $\psi=e^{i k x}+ R e^{-i k x}$, and far to the right it will be $\psi=T e^{i k x}$. The coefficients $T$ and $R$ stand for transmitted and reflected. As long as $T\neq 0$ there is a current -- and still there is no potential drop.

In a typical system of electrons (with many states), what happens is you get interference between all of the various eigenstates, so that in the absence of a potential there is no net current. If somehow you only have one mode, then you can get zero resistance.

Effective description

A typical phenomological model of a superconductor involves writing a ``Schrodinger equation" for the macroscopically occupied mode: \begin{equation} \label{gp} (i \partial_t-q \phi)\psi= -\frac{1}{2 m}(\nabla-i q {\bf A})^2\psi - a^2 \psi+b^2 |\psi|^2 \psi, \end{equation} where the coefficients $a$ and $b$ are phenomonological, and depend on temperature. The mass is $m=2m_e$ and the charge is $q=-2 e$, corresponding to electron pairs. This differs from the regular Schrodinger equation in that it includes a ``self-interaction" term, which in some manner captures the repulsion of the electrons.

Meisner Effect and Vortices

One interesting consequence of Eq.~(\ref{gp}) is that a superconductor will expell magnetic fields. For example imagine putting a chunk of aluminum in a solenoid. Cool it down to 1.2K (the superconducting temperature). You will find that the magnetic field is zero in the aluminum -- and a little bigger around it (to account for conservation of magnetic flux). This is known as the Meisner effect. In your homework you will derive this effect.

Actually, there are two types of superconductors. Instead of expelling flux, type-II superconductors sequester it in narrow "flux tubes," also known as vortex lines. These form triangular arrays. One can derive this property from Eq.~(\ref{gp}).

Josephson Effects

We now know enough about superconductivity to do some modeling. Lets imagine a device consisting of two superconductors separated by a thin oxide tunnel barrier, and no electromagnetic fields. Thinking of Eq.~(\ref{gp}), one would expect that the ground state would have $\psi$ completely uniform. On the other hand, if we made the tunnel barrier very thick, the wavefunctions on the two sides of the barrier should be independent. Thus a good low-energy description should have a uniform $\psi(r)=\psi_L$ on the left, and $\psi(r)=\psi_R$ on the right. If there were no tunneling, we would write the equation of motion as \begin{eqnarray} i\partial_t\psi_L&=& \gamma_L(|\psi_L|^2-\rho_L) \psi_L\\ i\partial_t\psi_R&=& \gamma_R(|\psi_R|^2-\rho_R) \psi_R, \end{eqnarray} Where the $\gamma$'s and $\rho$'s are just new ways of writing the $a$'s and $b$'s. Thinking of this like Ammonia, the way to add tunneling should be \begin{eqnarray} i\partial_t\psi_L&=& \gamma_L(|\psi_L|^2-\rho_L) \psi_L-J \psi_R\\ i\partial_t\psi_R&=& \gamma_R(|\psi_R|^2-\rho_R) \psi_R-J \psi_L. \end{eqnarray} For simplicity, lets assume the two superconductors are identical ($\gamma_L=\gamma_R=\gamma$ and $\rho_L=\rho_R=\rho$. If we linearize these equations we find that this isolated device can show oscillations. These can be detected using microwave spectroscopy. Complete the Josephson Oscillation worksheet. Next we will want to put this device (known as a Josephson junction) in a circuit. Before we think about leads, lets firs think about applying a voltage to the left or right. Going back to Eq.~(\ref{gp}), these voltages appear as \begin{eqnarray} i\partial_t\psi_L&=& \gamma(|\psi_L|^2-\rho) \psi_L-J \psi_R+q V_R \psi_R\\ i\partial_t\psi_R&=& \gamma(|\psi_R|^2-\rho) \psi_R-J \psi_L+q V_L\psi_L. \end{eqnarray} As with the Josephson oscillations, this is more readily understood by writing \begin{eqnarray} \psi_L&=& f_L e^{i \theta_L}\\ \psi_R&=& f_R e^{i \theta_R}. \end{eqnarray} In terms of this polar representation \begin{eqnarray}\label{cur} \partial_t \left(|f_L|^2-|f_R|^2\right)= I/q =J f_R f_L \sin(\theta_L-\theta_R)\\ \partial_t \left(\theta_L-\theta_R\right)=\gamma \left(f_R^2-f_L^2\right) + \left(V_R-V_L\right) +\cdots \end{eqnarray} where in the second term we have omitted a term of order $J$, which is typically small. The first equation defines the current in terms of the rate of change of the number of cooper pairs in each superconductor. The second shows that a potential difference winds the phase. This relationship is used in metrology devices to define the Volt. Now imagine we add leads, and drive a current through this device. Assuming there is no build-up of charge, we expect $f_R=F_L$ to be constant. The equation for the current, Eq.~(\ref{cur}) tells us that $\theta_L-\theta_R$ will be constant. Plugging that into the second equation, we see $V_R-V_L=0$. Thus this device carries current with no voltage drop. It is a superconductor. Interestingly, there is a maximum current the device can handle: $I_C = q J f_R f_L$. For larger current we need to model the electrons which are not paired. They act like a shunt resistor, carrying the remainder of the current.

Unconventional Superconductors

I must finally mention that there are "unconventional superconductors" whose behavior are not consistent with the phonon mechanism. These include Cuprates, Iron Pnictides, and Heavy Fermion Superconductors. These are areas of current research.