You’ve probably heard of superconductivity - isn’t it the phenomenon where a material has zero resistance, usually at very low temperatures? It sounds simple, but it actually requires the full force of quantum mechanics to describe it, and this is what Bardeen, Cooper and Schreiffer did in their Nobel-prize winning work, 60 years ago. In this two part series, I’ll aim to describe the physics behind superconductors, which will involve interesting detours into solid state physics, electrodynamics and quantum mechanics.
A recurring motif in condensed matter systems like superconductors is fundamental degrees of freedom (like electrons, photons, etc.) conspiring to form emergent phenomena when viewed at a slightly larger scale. These phemonena can be described as being initiated by objects with different properties to the original, fundamental ones, and these objects are called quasiparticles. An easy-to-visualise example is of an electron hole: imagine a lattice of atoms. This has room for electrons to live, but the absence of an electron in its designated spot creates a hole. As it turns out, we can ascribe particle-like properties to this hole, regarding it as a fundamental object in its own right. Indeed, this was Dirac’s original interpretation of the antiparticle solutions to his famous equation.
Wavefunctions and quantum particles
Superconductors also involve an emergent phenomenon, but I must first digress to introduce some key aspects of quantum mechanics. I will emphasise the conceptual features and aim to keep mathematical notation to a minimum. As you might be aware, quantum mechanical objects do not have a definite position. Rather, they are described by a wavefunction which is a function of space and time, denoted $\psi(\mathbf r, t)$. At a given time, $\psi(\mathbf r)$ specifies a complex number at each point in space such that the probability of finding the particle in some infinitesimal volume $\mathrm dV$ is $\lvert\psi(\mathbf r)\rvert^2\mathrm dV$, where $\lvert\cdot\rvert^2$ denotes the magnitude squared. As opposed to this probability, the value of the wavefunction is not itself measurable. So why bother working with $\psi$ and not just with $\lvert\psi\rvert^2$ directly? Because, similar to light waves, when multiple wavefunctions (for instance, corresponding to two electron beams passing through different slits) coalesce, the wavefunctions $\psi_1$ and $\psi_2$ themselves add constructively and destructively (and not the probabilities $\lvert\psi_1\rvert^2$ and $\lvert\psi_2\rvert^2$) . A similar thing happens with classical electromagnetic waves: the (classical) waves add as $A=A_1+A_2$ - their individual intensities on a screen, given by $\lvert A_1\rvert^2$ and $\lvert A_2\rvert^2$, do not add! This is the reason electrons display the same band-like pattern on a screen as light rays during the double-slit experiment. Finally (although we will not need it here), the way these spatial probabilities change over time is governed by the famous Schrödinger equation.
Now for deep reasons, there are two categories of fundamental particles, bosons and fermions. Qualitatively, the difference is that no two fermions can have the same quantum state, while as many bosons can pile up into a given quantum state as they’d like. If these sound familiar, it’s because they describe the Pauli exclusion principle and the Bose-Einstein condensate respectively. Famous fundamental fermions include the electron, neutrinos, and all 6 quarks while photons and gluons are fundamental bosons.
For the same deep reasons as above, a composite object containing an odd number of fermions is still fermionic, such as a proton, which contains two up quarks and a down quark and similarly for the neutron. Indeed, that the Pauli exclusion principle continues to act for the proton and neutron is responsible for the structure of almost all visible matter in the universe! However when an object is composed of an even number of fermions actually acts as a boson. Why doesn’t the Pauli exclusion principle fundamentally prevent this? See here.
The atomic lattice
If you’ve digested this, it’s time to move on to the superconductor setup. It’s not very difficult - we just need a lattice of metal atoms. Assume for now that the metal ions are fixed in place, and the electrons don’t interact with anything. As simple as it sounds, the finite-size lattice actually forces electrons to take on discrete values of momentum. In fact, this is not too hard to visualise: quantum mechanics tells us that particles have a wavefunction. The momentum of a wave is given by how rapidly it is changing, or in other words its frequency. But imagine (or try!) attaching a string, another wave-producing object, between two beams at a distance of $L$. If you pluck the string at different places, you’ll find that only certain wavelengths $\lambda_n$ persist after a few seconds. This is the case when $\lambda_n=\frac{2L}{n}$ - that is, wavelength can only take on a set of discrete values (or else the vibration dies out by destructive interference). Since frequency is just speed divided by wavelength, the possible frequencies form a discrete set: $f_n=\frac{vn}{2L}$. Analysing the Schrödinger equation for an electron in the finite, periodic lattice uses essentially the same argument to conclude that the electron’s momentum also takes on a set of discrete values. We say that momentum is quantized - but instead of $v$ in the previous formula, we have a factor of $2\pi\hbar$, where $\hbar$ is Planck’s constant, signalling the presence of quantum mechanics. So $p_n=\frac{n\pi\hbar}{L}$. We see that the larger the dimensions of the lattice are, the less spaced out the possible momenta. One more thing - our system has three finite dimensions, so the possible discrete momenta themselves form a 3-dimensional lattice, known as the reciprocal lattice. The sites of this lattice are the obvious generalisation of the one-dimensional case: $p=(p_x, p_y, p_z)$ are the three components of momentum in different directions. So for instance, $p_{1,0,0}=\left(\frac{\pi\hbar}{L}, 0, 0\right)$, $p_{0,1,0}=\left( 0,\frac{\pi\hbar}{L}, 0\right)$, $p_{43, 9, 17} = \left(\frac{43\pi\hbar}{L}, \frac{9\pi\hbar}{L}, \frac{17\pi\hbar}{L}\right)$, and so on. Hopefully you’ll be able to keep track of whether I’m talking about the physical metal lattice or the momentum lattice.
Why is this important? Remember what we said about fermions: they can’t have the same quantum state, one aspect of which is momentum. So suppose each electron in the lattice is granted a certain amount of kinetic energy - they can’t all take the maximum allowed momentum (according to $K=\frac{p_x^2}{2m}+\frac{p_y^2}{2m}+\frac{p_z^2}{2m}$), because this would violate the Pauli exclusion principle! Instead, in order to settle into the lowest energy state while obeying the exlucion principle, the electrons must fill up the possible momenta one by one, so one electron has $p=p_{1,0,0}$, one $p=p_{0,1,0}$ and another $p=p_{0,0,1}$. These three electrons have the minimum kinetic energy. The next least energetic electrons will have $p=p_{2,0,0}$, and so on.
If there are $n$ electrons at 0K, then the filled momentum states will occupy a spherical shape in the momentum lattice, so as to minimise the total energy for stability. This minimum energy state is called the ground state, the bounding sphere is called the Fermi surface, and the corresponding energy is (imaginatively) called the Fermi energy. The entire setup is called the perfect metallic state or the free electron gas. Remember that the Fermi surface isn’t a physical surface containing the electrons! It’s just meant to denote what momenta the electrons can attain - as an analogy, the Fermi surface for a 1D string might be $[0, \frac{5v}{2L}]$. If you know what a Fourier transform is, it’s just an interval in Fourier (momentum) space.
Suppose now that a spurt of additional energy is provided to the lattice (the real lattice, not the momentum lattice!). The electron states deep within the Fermi surface can’t be excited - they’d have to occupy states which are already occupied, and you know that electrons don’t like to share lattice sites. Instead, only electrons close to the Fermi surface can use the additional energy to escape to unoccupied zones - these electrons are the ones with a high momentum and kinetic energy already. Similarly, instead of providing additional energy externally, these electrons could escape with the help of thermal fluctuations, provided that the system was at a non-zero temperature $T$. Heuristically, these thermal fluctuations can provide an energy of $k_BT$ ($k_B$ is the ubiquitous Boltzmann constant, which roughly serves as a conversion factor between temperature and energy) to the electrons at the edge of the Fermi surface.
Adding interactions
At this stage we can improve the model by allowing electrons to interact with the ions. The electrons at the Fermi level have a high momentum, and as they travel through the lattice, the positive ions are attracted to them, though their acceleration is slow because of their large mass. So a fast moving electron creates a slow “ripple” of increased positive charge density behind it. This positive charge can then attract another electron in the lattice, even if it’s far away from the original electron. So electrons, which ordinarily repel each other due to their like charge, can find themselves weakly attracting each other! The full quantum analysis yields the same result, with only one difference: these vibrational ripples in the ion lattice become quasiparticles (remember those?), known as phonons. We say that phonons mediate a weak, long-range attractive force between electrons in a lattice. They form what’s known as a Cooper pair, which is a boson since it is composed of two fermionic electrons. It has a charge of $2e$.
What happens when two particles are conjoined by an attractive force? Well, think of a classical hydrogen atom: unlike free electrons, which can whizz around and be fired chaotically in beams, electrons bound to a nucleus (which, in this case, is just a singular proton) are much more sedate - we need to provide some energy to ionise the electron first. What’s more, the mass of this hydrogen atom is less than the sum of the individual masses of the proton and the electron! Here’s why: the total energy of the atomic system is: the rest energy of the proton ($m_pc^2$, by Einstein’s relation), the rest energy of the electron ($m_ec^2$) and also the energy due to the attractive force. The first two terms are the same as in the respective free particles, but this third contributor is negative - because we need to supply additional (positive) energy to get rid of it. Analogously in the case of electron-phonon attraction, the two electrons are bound, so their total energy decreases slightly.
We now have an idea of what a single Cooper pair looks like, but what happens when we examine the entire system at once, will all electrons subject to attractive forces by the ion lattice? This is what Bardeen, Schreiffer and Cooper analysed in 1957, and they were able to find an approximate wavefunction that describes the behaviour of all the electrons in the system. Provided that temperatures are low, we’ve described its qualitative features above: low-momentum electrons deep within the Fermi surface are unbound, and pretty much act like normal electrons. We can even guess why - it’s because there aren’t any close states with larger momentum for them to get excited to, so they can’t really interact via phonons. Close to the Fermi surface, however, we typically have occupied Cooper pairs, although some slots may be empty (strictly speaking these two exist as a superposition). The final kicker is that the interaction has completely changed the dynamics of excited states of the system. No longer do states with more energy look like faster electrons whizzing around. It’s not even the Cooper pairs that get excited and behave as a free particle. Instead, there are Boguliobons, quasiparticles that don’t have a definite charge, but are a mixture of electrons and holes.
How does all of this lead to superconductivity then? Find out in the next post.