The Cooper Instability

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Hello! Welcome to our next lesson on the microscopic theory of superconductivity.

In our previous lesson, we used the phenomenological Ginzburg-Landau theory to understand the behavior of superconductors, defining the coherence length $ξ$ and penetration depth $λ$ , and using their ratio $κ$ to distinguish between Type I and Type II materials. While powerful, GL theory doesn't explain why superconductivity occurs. It begins with the assumption of a macroscopic wavefunction but doesn't explain its origin.

Today, we take the first crucial step towards a microscopic understanding. We will address the fundamental question: How can two electrons, which normally repel each other, form the bound pairs (Cooper pairs) that constitute the superconducting condensate? We will demonstrate the Cooper instability, a remarkable result showing that even an infinitesimally weak attractive force between electrons near the Fermi surface is enough to create a bound state, destabilizing the normal metallic state (the Fermi sea).

1. The Origin of Attraction: Electron-Phonon Coupling

The first puzzle is to find a source for an attractive interaction between electrons in a solid. The bare Coulomb interaction is, of course, repulsive. However, electrons in a crystal are not in a vacuum; they are immersed in a lattice of positively charged ions. The interaction between an electron and the lattice vibrations (phonons) can lead to an effective attraction between two electrons.

The physical picture, first proposed by Herbert Fröhlich, is as follows:

An electron moves through the crystal, attracting the nearby positive ions. This creates a slight, localized distortion of the lattice—a region with a higher density of positive charge.
This distortion persists for a short time, on the order of the inverse Debye frequency ( $\sim ω_{D}^{- 1}$ ), which is the characteristic timescale for lattice vibrations.
If a second electron passes through this region before the lattice has relaxed, it will be attracted to the excess positive charge left in the wake of the first electron.

Effectively, the two electrons have interacted attractively, with a phonon acting as the mediator of the force. This is analogous to how force-carrying particles (like photons) mediate fundamental interactions in quantum field theory.

Mod-01 Lec-30 Cooper Pairs

This video provides a good qualitative description of this phonon-mediated attraction, which overcomes the screened Coulomb repulsion, and also introduces the Feynman diagram for the process.

Please watch from 27:27 to 32:45. Focus on the physical picture of lattice polarization and the competition between the screened Coulomb repulsion and the phonon-mediated attraction. Pay attention to the Feynman diagram presented, which depicts one electron emitting a virtual phonon that is later absorbed by another.

The net interaction between two electrons is a sum of the repulsive (screened) Coulomb potential and this attractive phonon-mediated potential. For superconductivity to occur, the attractive part must win, at least for electrons near the Fermi surface. While the resulting net attraction, $V$ , is typically very weak, Leon Cooper showed in 1956 that this is all that is needed.

2. The Cooper Problem

Cooper's brilliant insight was to simplify the overwhelmingly complex many-body problem into a tractable two-body problem. Instead of trying to solve for all $\sim 1 0^{23}$ electrons at once, he asked:

What happens if we add two interacting electrons to the top of a filled Fermi sea at absolute zero?

The key assumptions are:

The Fermi sea itself is "inert": all states with energy $E < E_{F}$ are filled. Due to the Pauli exclusion principle, the two extra electrons cannot scatter into these occupied states. They can only scatter into empty states with $E > E_{F}$ .
The two electrons interact via a simple, attractive potential $V_{k, q} = - V_{0}$ (where $V_{0} > 0$ ) if their energies $E_{k}$ and $E_{q}$ are within a thin shell above the Fermi energy, i.e., $E_{F} < E_{k}, E_{q} < E_{F} + ℏ ω_{D}$ . Outside this shell, the interaction is zero. The cutoff $ℏ ω_{D}$ is physically motivated by the fact that the attraction is mediated by phonons, whose maximum energy is the Debye energy.
To maximize the interaction, the two electrons form a pair with zero total momentum. This means if one electron has momentum $k$ , the other has $- k$ .

Our goal is to solve the Schrödinger equation for this pair and show that their total energy is less than $2 E_{F}$ , indicating the formation of a bound state.

Cooper Pair Formation and Phonon-Mediated Attraction — This illustration depicts the core concept of the Cooper problem. Two electrons with opposite momentum ($+\mathbf{k}, -\mathbf{k}$) and spin ($\uparrow, \downarrow$), existing just outside the filled Fermi sea, interact via an effective attractive potential $U$ (mediated by lattice phonons) to form a bound state, the Cooper pair.

3. Derivation of the Bound State Energy

Let's outline the derivation of the Cooper pair binding energy. We are looking for a two-particle stationary state $Ψ (r_{1}, r_{2})$ that is an eigenstate of the Schrödinger equation:

(- \frac{ℏ ^{2}}{2 m} \nabla_{1}^{2} - \frac{ℏ ^{2}}{2 m} \nabla_{2}^{2} + V (r_{1} - r_{2})) Ψ (r_{1}, r_{2}) = E Ψ (r_{1}, r_{2})

We construct the wavefunction as a superposition of plane-wave states for a pair with zero total momentum. To ensure the total wavefunction is antisymmetric as required for fermions, we pair a symmetric spatial part with an antisymmetric spin part (a spin singlet).

The BCS Theory of Superconductivity

The following text provides a concise mathematical derivation showing that an attractive potential leads to a bound state. It follows the exact logic we've just set up.

Please read the section titled 'To see that pairs of electrons can form bound states...' on page 2. Follow the derivation from Eq. (1) to the final result for the energy E. Pay close attention to how the sum over states is converted into an integral over energy, and how the assumption of a constant density of states n(E_F) simplifies the problem.

Let's summarize the key steps and the result from the derivation:

The pair wavefunction is written as $Ψ (r_{1}, r_{2}) = \sum_{k} g_{k} e^{i k \cdot (r_{1} - r_{2})}$ (ignoring the spin part for now), where the sum is over unoccupied states $∣ k ∣ > k_{F}$ .
Plugging this into the Schrödinger equation leads to an equation for the coefficients $g_{q}$ :
$(E - 2 ϵ_{q}) g_{q} = \sum_{k} V_{k, q} g_{k}$
where $ϵ_{q} = ℏ^{2} q^{2} /2 m$ is the single-particle energy.
Using the simplified potential $V_{k, q} = - V_{0}$ within the shell $E_{F} < ϵ_{k}, ϵ_{q} < E_{F} + ℏ ω_{D}$ , the equation becomes an integral equation.
Assuming the density of states (DOS), $N (ϵ)$ , is roughly constant and equal to $N (E_{F})$ in this narrow shell, we can solve the integral. This leads to the condition:
$1 = - V_{0} \int_{E_{F}}^{E_{F} + ℏ ω_{D}} \frac{N ( ϵ )}{E - 2 ϵ} d ϵ \approx V_{0} N (E_{F}) \int_{E_{F}}^{E_{F} + ℏ ω_{D}} \frac{d ϵ}{2 ϵ - E}$
Solving this for the total energy $E$ of the pair yields:
$E \approx 2 E_{F} - 2ℏ ω_{D} exp (- \frac{2}{N ( E _{F} ) V _{0}})$

This result is the heart of the Cooper instability.

A Bound State Exists: The total energy $E$ is less than $2 E_{F}$ (the energy of two non-interacting electrons at the Fermi surface) for any attractive potential $V_{0} > 0$ .
Binding Energy: The energy reduction, known as the binding energy $Δ_{B} = 2 E_{F} - E$ , is
$Δ_{B} = 2ℏ ω_{D} e^{- 2/ N (E_{F}) V_{0}}$
Non-Perturbative Nature: Notice that the binding energy is non-analytic at $V_{0} = 0$ . You cannot obtain this result using standard perturbation theory by expanding in powers of $V_{0}$ . This is a hallmark of many-body phenomena like superconductivity.

4. The Crucial Role of the Fermi Sea

A perceptive question to ask is: why does this work? In your quantum mechanics courses, you learned that in three dimensions, a weak, short-range attractive potential is generally not strong enough to form a bound state. You need the potential to have a minimum depth and range. How does an arbitrarily weak attraction succeed here?

The answer lies in the presence of the Fermi sea. Although the electrons in the sea are "inert" spectators in Cooper's model, their existence fundamentally changes the problem through the Pauli principle. By forbidding the two interacting electrons from scattering into states with $E < E_{F}$ , the Fermi sea restricts their available phase space. They can only scatter into states on or near the Fermi surface. This restriction of momentum space effectively reduces the dimensionality of the problem.

20. Fermi gases, BEC-BCS crossover

Prof. Wolfgang Ketterle gives a brilliant and intuitive explanation for why the 3D Cooper problem effectively behaves like a 2D problem, where an arbitrarily weak attraction is always sufficient to form a bound state.

Watch from 15:57 to 29:30. The first part (until 25:40) reviews why bound states always form in 1D and 2D but not 3D. The crucial part for us is the second half (from 25:40), where he explains how Cooper's problem of two electrons on top of a Fermi sea leads to the same mathematical structure (a logarithmic divergence) as a two-dimensional system, thus guaranteeing a bound state.

As the video explains, the key is that the integral for the binding energy involves the density of states. Because the interaction is restricted to a narrow shell around $E_{F}$ , the 3D density of states $N (E) \propto E$ is approximated as a constant, $N (E_{F})$ . An integral with a constant density of states is mathematically characteristic of a 2D system.

So, is Cooper pairing a two-body or many-body effect? The answer is subtle: it is a two-body binding problem that is only possible because of the many-body environment (the Fermi sea), which provides the necessary kinematic constraints via Pauli blocking.

Test your understanding!

Suppose you have two different metals, A and B. They have identical Debye frequencies ( $ω_{D}$ ) and the same effective attractive potential ( $V_{0}$ ). However, Metal A has a much larger density of states at the Fermi level, $N_{A} (E_{F})$ , than Metal B, $N_{B} (E_{F})$ . Which metal would you expect to have a larger Cooper pair binding energy, and why?

Show answer

The binding energy is given by $Δ_{B} = 2ℏ ω_{D} exp (- \frac{2}{N ( E _{F} ) V _{0}})$ . Since the exponent is negative, a larger value of the term $N (E_{F}) V_{0}$ in the denominator will make the exponent less negative (i.e., closer to zero), leading to a larger value for $Δ_{B}$ . Therefore, Metal A, with the larger density of states $N_{A} (E_{F})$ , will have a larger Cooper pair binding energy. This makes physical sense: a higher density of states means there are more available final states for the pair to scatter into, which enhances the binding effect.

Conclusion

Today, we've unraveled the first piece of the microscopic puzzle of superconductivity. By considering the simplified Cooper problem, we've seen how the normal metallic state, the Fermi sea, is inherently unstable in the presence of any net attractive interaction between electrons.

Key Takeaways:

Electron-phonon coupling can mediate an effective attractive interaction between electrons, which can overcome their screened Coulomb repulsion.
The Cooper problem considers two electrons interacting above a filled, inert Fermi sea.
Due to Pauli blocking, scattering is restricted to a thin energy shell above $E_{F}$ , which kinematically resembles a 2D system.
In this situation, an arbitrarily weak attractive potential is sufficient to form a bound state, a Cooper pair, with energy $E < 2 E_{F}$ .
The binding energy is non-perturbative and depends exponentially on the inverse of the interaction strength and the density of states at the Fermi level.

Preview of the Next Lesson

The Cooper problem is a monumental step, but it only describes two electrons. The real system has $\sim 1 0^{23}$ interacting electrons. How do we generalize this pairing concept to all of them?

In our next lesson, we will dive into the full Bardeen-Cooper-Schrieffer (BCS) theory. We will see how to construct a new many-body ground state wavefunction for the superconductor, where all electrons are paired up. This will lead us to the famous BCS gap equation, which self-consistently determines the superconducting energy gap—a quantity directly measurable in experiments like tunneling spectroscopy.