Cooper Pair Instability

Derivation

Cooper Pair Instability

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Statement

Two electrons added just above a filled Fermi sea, interacting through an arbitrarily weak attractive potential, form a bound state whose energy lies below twice the Fermi energy. Modelling the attraction as a constant \( -V \) acting only within a shell of width \( \hbar\omega_D \) above \( \epsilon_F \), the binding energy of the most favourable pair (zero centre-of-mass momentum, spin singlet) is \[ \Delta \;=\; 2\hbar\omega_D\,\exp\!\left(-\frac{2}{N(0)\,V}\right), \] where \( N(0) \) is the single-spin density of states at the Fermi level. Because \( \Delta>0 \) for every \( V>0 \), the normal Fermi sea is unstable against pairing.

Why it matters

This is Cooper's 1956 argument, the seed crystal of BCS theory. It shows that the free-electron ground state assumed by Sommerfeld is not the true ground state of a metal with any net attraction between electrons: an infinitesimal attraction already binds pairs, so the system must reorganise. The exponential and its essential singularity at \( V=0 \) explain at a stroke why superconductivity is invisible to perturbation theory and why transition temperatures are so much smaller than the Debye scale.

The same \( e^{-1/\lambda} \) structure recurs across physics wherever a filled sea plus a marginal interaction meets a constant density of states at threshold: the Kondo effect, the BCS gap, chiral symmetry breaking. Cooper's calculation is the cleanest place to see the mechanism naked.

Assumptions

A filled, inert Fermi sea.The two added electrons are forbidden the states \( k<k_F \) by Pauli exclusion, but the sea electrons are otherwise frozen. Drop this and you have a genuine many-body problem; the restriction \( k>k_F \) is exactly what makes the density of states finite at the lower limit and forces binding.

Zero centre-of-mass momentum, spin singlet.Electrons paired as \( (\mathbf{k}\uparrow,\,-\mathbf{k}\downarrow) \). If the pair carries net momentum \( \mathbf{q}\neq 0 \), the shell of allowed states shrinks and the binding weakens, so \( \mathbf{q}=0 \) is optimal.

Separable contact interaction.\( V_{\mathbf{k}\mathbf{k}'}=-V \) for both states in the shell \( \epsilon_F<\epsilon_k,\epsilon_{k'}<\epsilon_F+\hbar\omega_D \), else zero. Drop the constant form and the gap equation no longer decouples; the log integral becomes a kernel eigenvalue problem.

Constant density of states across the shell.\( N(\epsilon)\approx N(0) \) over the thin phonon shell \( \hbar\omega_D\ll\epsilon_F \). If \( N(\epsilon) \) varies appreciably (narrow bands, van Hove singularity at \( \epsilon_F \)) the prefactor and exponent shift, though binding survives.

Retardation replaced by a sharp cutoff.The phonon-mediated attraction is dynamical; here it is idealised as instantaneous within \( |\xi|<\hbar\omega_D \). Dropping the cutoff removes the ultraviolet regulator and the log integral diverges.

Derivation

\[ \psi(\mathbf{r}_1,\mathbf{r}_2)=\left(\sum_{k>k_F} g_k\, e^{\,i\mathbf{k}\cdot(\mathbf{r}_1-\mathbf{r}_2)}\right)\frac{|\!\uparrow\downarrow\rangle-|\!\downarrow\uparrow\rangle}{\sqrt2} \]

Two-electron ansatz with total momentum zero; the orbital part is symmetric so the spin part is the antisymmetric singlet. The restriction \( k>k_F \) enforces exclusion by the frozen sea. B

\[ \left[-\frac{\hbar^2}{2m}\left(\nabla_1^2+\nabla_2^2\right)+V(\mathbf{r}_1-\mathbf{r}_2)\right]\psi=E\,\psi \]

The stationary two-body Schrödinger equation for the added pair, energies measured from the bottom of the band. A

\[ \left(2\epsilon_k-E\right)g_k=-\sum_{k'>k_F} V_{kk'}\,g_{k'},\qquad \epsilon_k=\frac{\hbar^2k^2}{2m} \]

Project onto plane wave \( e^{i\mathbf{k}\cdot(\mathbf{r}_1-\mathbf{r}_2)} \): the Laplacians give \( 2\epsilon_k \) (each electron contributes \( \hbar^2k^2/2m \)) and \( V \) couples \( g_k \) to all \( g_{k'} \) through its Fourier component \( V_{kk'} \). B

\[ \left(2\epsilon_k-E\right)g_k=V\sum_{k'>k_F} g_{k'}\equiv V\,C \]

Insert the constant kernel \( V_{kk'}=-V \) inside the shell. The right side no longer depends on \( k \); call the common sum \( C \). This is where the separable model earns its keep. A

\[ g_k=\frac{V\,C}{2\epsilon_k-E} \]

Solve algebraically for each amplitude. For a bound state \( E<2\epsilon_k \) throughout the shell, so \( g_k \) has a fixed sign — a coherent, in-phase superposition. A

\[ 1=V\sum_{k>k_F}\frac{1}{2\epsilon_k-E} \]

Sum step 5 over \( k \) and divide by \( C=\sum_k g_k \) (non-zero for a real pair). This self-consistency condition fixes \( E \). B

\[ 1=V\!\int_{\epsilon_F}^{\epsilon_F+\hbar\omega_D}\frac{N(\epsilon)\,d\epsilon}{2\epsilon-E} \]

Replace the sum by an energy integral, \( \sum_k\to\int N(\epsilon)\,d\epsilon \), with \( N(\epsilon) \) the single-spin density of orbital states per unit volume (each pair labelled by one \( \mathbf{k} \)). Limits are the phonon shell. B

\[ 1=N(0)V\!\int_{\epsilon_F}^{\epsilon_F+\hbar\omega_D}\frac{d\epsilon}{2\epsilon-E}=\frac{N(0)V}{2}\,\ln\!\left(\frac{2\epsilon_F+2\hbar\omega_D-E}{2\epsilon_F-E}\right) \]

Pull the slowly varying \( N(\epsilon)\approx N(0) \) out of the thin shell and integrate \( \int d\epsilon/(2\epsilon-E)=\tfrac12\ln(2\epsilon-E) \). Legal because \( \hbar\omega_D\ll\epsilon_F \). B

\[ \Delta\equiv 2\epsilon_F-E>0:\qquad 1=\frac{N(0)V}{2}\,\ln\!\left(\frac{2\hbar\omega_D+\Delta}{\Delta}\right) \]

Define the binding energy \( \Delta \) as the depth of \( E \) below the two-particle continuum edge \( 2\epsilon_F \). Rewrite the logarithm in terms of \( \Delta \). A solution with \( \Delta>0 \) is a true bound state. A

\[ \text{weak coupling }N(0)V\ll1\Rightarrow\Delta\ll\hbar\omega_D:\quad \frac{2}{N(0)V}=\ln\!\frac{2\hbar\omega_D}{\Delta} \]

Small \( N(0)V \) forces the logarithm large, hence \( \Delta\ll\hbar\omega_D \); drop \( \Delta \) beside \( 2\hbar\omega_D \) in the numerator. Solve for \( \Delta \). C

Result

\[ \boxed{\;\Delta \;=\; 2\hbar\omega_D\,\exp\!\left(-\dfrac{2}{N(0)V}\right)\;>\;0\quad\text{for all }V>0\;} \]

Reading. The added pair sits an energy \( \Delta \) below the top of two free electrons at the Fermi surface, so the normal metal can always lower its energy by pairing — it is unstable. The binding is non-analytic in \( V \): every term of its Taylor series about \( V=0 \) vanishes, so no order of perturbation theory ever produces it. The Fermi sea is indispensable: without it the lower integration limit would go to zero and a weak attraction in 3D would not bind. Note \( \Delta\propto\hbar\omega_D \), tying pairing to the lattice — the origin of the isotope effect.

Units check. \( \hbar\omega_D \) is an energy. The coupling \( N(0)V \) is dimensionless: \( N(0) \) has units \( (\text{energy}\cdot\text{volume})^{-1} \) and the contact strength \( V \) has units \( \text{energy}\cdot\text{volume} \), so the exponent is a pure number. Hence \( [\Delta]=\text{energy} \). Good.

Limiting cases

Weak coupling \( N(0)V\to0^+ \): \( \Delta\to0 \) faster than any power of \( V \); binding is exponentially small but strictly positive.
Strong coupling \( N(0)V\gtrsim1 \): keep the full step-9 form, \( \Delta=2\hbar\omega_D/\!\left(e^{2/N(0)V}-1\right) \), which stays finite and approaches \( N(0)V\,\hbar\omega_D \) as \( V \) grows.
No Fermi sea (\( k_F\to0 \)): lower limit \( \to0 \), the log integral no longer diverges at threshold, and in 3D a finite \( V \) is needed to bind — Cooper's effect disappears.
Heavy-isotope limit: \( \omega_D\propto M^{-1/2} \Rightarrow \Delta\propto M^{-1/2} \), the observed isotope shift of \( T_c \).

Breaks when

The attraction is overwhelmed by Coulomb repulsion. If the net \( V_{kk'} \) is repulsive in the relevant channel, \( \Delta \) has no real positive solution and the sea is stable; the Morel–Anderson pseudopotential \( \mu^\ast \) must be subtracted before this formula applies.
The single-pair picture fails at finite density. Cooper treats one pair against a rigid sea. Once many pairs form they overlap and screen one another; the correct ground state is the BCS coherent state, and \( \Delta \) is replaced by the self-consistent gap (same exponent, factor-2 differences in prefactor).
Density of states varies across the shell. A van Hove singularity or narrow band at \( \epsilon_F \) violates \( N(\epsilon)\approx N(0) \); the log integral must be redone and the simple exponential is only leading order.
Strong retardation / large \( \hbar\omega_D/\epsilon_F \). The instantaneous-cutoff model breaks; Eliashberg theory with the full frequency-dependent phonon kernel is required.

Failure modes

Forgetting the sign in the exponent's factor of 2. With single-spin \( N(0) \) the exponent is \( -2/N(0)V \); with total (both-spin) DOS \( N_{\text{tot}}=2N(0) \) it is \( -1/N_{\text{tot}}V \). Mixing the two doubles or halves \( \Delta \).
Dropping the \( k>k_F \) restriction. Integrating from \( \epsilon=0 \) rather than \( \epsilon_F \) destroys the mechanism; students then wrongly conclude weak attraction cannot bind in 3D.
Trying to expand \( \Delta \) in powers of \( V \). The essential singularity means the perturbation series is identically zero. Attempting a Taylor result is not just inaccurate, it is qualitatively wrong.
Using \( 2\epsilon_k=\hbar^2k^2/2m \). The pair kinetic energy is \( 2\epsilon_k=\hbar^2k^2/m \); halving it misplaces the continuum edge.
Confusing binding energy \( \Delta \) with the BCS gap. Cooper's \( \Delta \) is the single-pair binding; the thermodynamic gap \( \Delta_{\text{BCS}} \) is a distinct (though similarly exponential) quantity.
Reading \( E<2\epsilon_F \) as \( E<0 \). The pair energy is negative relative to the Fermi surface, not relative to the band bottom.

Discussion

The physics lives in the lower limit of the step-7 integral. In a 3D vacuum the two-body density of states \( \propto\sqrt{\epsilon} \) vanishes at threshold, so \( \int d\epsilon\,\sqrt{\epsilon}/(2\epsilon-E) \) converges and a minimum attraction is needed to bind. The Fermi sea replaces that threshold behaviour with a constant \( N(0) \) starting at \( \epsilon_F \); the integral then diverges logarithmically as \( E\to2\epsilon_F \), and any \( V>0 \) can satisfy the self-consistency. Pairing is therefore a Fermi-surface phenomenon, effectively one-dimensional in the radial energy variable.

The coherent, single-sign amplitude \( g_k \) of step 5 is the real-space signature of a pair: all the plane waves in the shell add in phase, giving a wavefunction that is spatially compact on the scale of the coherence length \( \xi\sim\hbar v_F/\Delta \). Because \( \Delta \) is exponentially small, \( \xi \) is large — hundreds of nanometres — so Cooper pairs are enormous and heavily overlapping, which is precisely why the collective BCS treatment, not a dilute bosonic gas, is the right sequel.

The essential singularity \( e^{-2/N(0)V} \) is the hallmark of a marginally relevant coupling in the renormalisation-group sense: the pairing interaction grows logarithmically under scaling toward the Fermi surface, and the scale at which it reaches order unity is exactly \( \Delta \). This is the same structure that produces the Kondo temperature and dynamical mass generation, and it is why mean-field theory, which sums the leading logarithms, is qualitatively correct here despite being uncontrolled by any small parameter.

Common misconceptions. The attraction need not overcome the bare Coulomb repulsion pointwise — retardation lets the phonon-mediated attraction act after the fast Coulomb repulsion has passed, and it is the net effective \( V \) that enters. Also, Cooper pairing does not mean two electrons orbit like a hydrogen-like molecule; the pair is a momentum-space correlation spread over many lattice sites.

Worked examples

\[ \Delta=2\hbar\omega_D\,e^{-2/N(0)V},\qquad \hbar\omega_D=25\ \text{meV},\quad N(0)V=0.30 \]

Typical weak-coupling metal (Debye temperature \( \sim290\ \text{K} \), so \( \hbar\omega_D=k_B\cdot290\,\text{K}\approx25\ \text{meV} \)). Symbols first, numbers next. A

\[ \frac{2}{N(0)V}=\frac{2}{0.30}=6.67,\qquad e^{-6.67}=1.27\times10^{-3} \]

Evaluate the dimensionless exponent, then exponentiate. A

\[ \Delta=2(25\ \text{meV})(1.27\times10^{-3})=6.4\times10^{-2}\ \text{meV} \]

Multiply through; \( 2\times25=50\ \text{meV} \) prefactor. A

\[ \Delta\approx0.064\ \text{meV}\;\approx\;k_B\times0.74\ \text{K} \]

Reading. A four-order-of-magnitude suppression below \( \hbar\omega_D \): the binding energy, and hence \( T_c \), is tiny compared with the phonon scale — exactly the puzzle the exponential resolves. Converting, \( 0.064\ \text{meV}/0.0862\ \text{meV K}^{-1}=0.74\ \text{K} \).

Units check. meV \( \times \) dimensionless = meV; division by \( k_B \) gives kelvin.

\[ \text{Given }\hbar\omega_D=25\ \text{meV},\ \Delta=0.50\ \text{meV};\quad \text{find }N(0)V\text{ and }\frac{\delta\Delta}{\Delta}\text{ for }V\to1.1V \]

Invert the result to extract the coupling, then test exponential sensitivity. B

\[ N(0)V=\frac{2}{\ln\!\left(2\hbar\omega_D/\Delta\right)}=\frac{2}{\ln(50/0.50)}=\frac{2}{\ln100}=\frac{2}{4.605}=0.434 \]

Solve step-10 form for \( N(0)V \). A

\[ \frac{\Delta'}{\Delta}=\exp\!\left[\frac{2}{N(0)V}-\frac{2}{1.1\,N(0)V}\right]=\exp\!\left[4.605\left(1-\tfrac{1}{1.1}\right)\right]=e^{0.419} \]

Ratio of the two exponentials; the prefactor cancels. A 10% rise in \( V \) enters through the exponent. B

\[ N(0)V=0.43,\qquad \frac{\Delta'}{\Delta}=e^{0.42}\approx1.52 \]

Reading. A mere 10% increase in the coupling raises the binding energy by 52%. The exponential makes \( \Delta \) (and \( T_c \)) hypersensitive to \( N(0)V \) — why small material changes swing transition temperatures so violently.

Units check. \( N(0)V \) dimensionless; the ratio \( \Delta'/\Delta \) dimensionless. Consistent.

Problems

(A) Set up the pair equation. Starting from the ansatz, derive \( (2\epsilon_k-E)g_k=-\sum_{k'}V_{kk'}g_{k'} \) and, for \( V_{kk'}=-V \), reduce it to the self-consistency condition \( 1=V\sum_k(2\epsilon_k-E)^{-1} \).
Solution
Insert \( \psi=\sum_k g_k e^{i\mathbf{k}\cdot\mathbf{r}} \) (relative coordinate \( \mathbf{r} \)) into \( [-(\hbar^2/m)\nabla^2+V(\mathbf{r})]\psi=E\psi \). The Laplacian gives \( (\hbar^2k^2/m)g_k=2\epsilon_k g_k \). Projecting onto \( e^{-i\mathbf{k}\cdot\mathbf{r}} \) and using \( V(\mathbf{r})=\sum_{kk'}V_{kk'} \) yields \( (2\epsilon_k-E)g_k=-\sum_{k'}V_{kk'}g_{k'} \). With \( V_{kk'}=-V \), the right side is \( V\sum_{k'}g_{k'}=VC \), so \( g_k=VC/(2\epsilon_k-E) \). Summing over \( k \) and cancelling \( C \): \( 1=V\sum_k(2\epsilon_k-E)^{-1} \).
(A) Binding energy. A metal has \( \hbar\omega_D=20\ \text{meV} \) and \( N(0)V=0.25 \). Compute \( \Delta \) in meV and in kelvin.
Solution
\( \Delta=2\hbar\omega_D e^{-2/N(0)V}=2(20)\,e^{-8}\ \text{meV} \). \( e^{-8}=3.35\times10^{-4} \), so \( \Delta=40\times3.35\times10^{-4}=1.34\times10^{-2}\ \text{meV}=0.0134\ \text{meV} \). In temperature, \( 0.0134/0.0862=0.155\ \text{K} \).
(B) Non-analyticity. Show that \( f(V)=e^{-2/N(0)V} \) has \( f(0^+)=0 \) with every derivative \( f^{(n)}(0^+)=0 \), so its Maclaurin series is identically zero. Explain what this says about perturbation theory.
Solution
Let \( x=N(0)V\to0^+ \). \( f=e^{-2/x} \). Each derivative is \( e^{-2/x} \) times a polynomial in \( 1/x \); since \( e^{-2/x} \) decays faster than any power of \( x \) grows, \( \lim_{x\to0^+}(1/x)^m e^{-2/x}=0 \) for all \( m \). Hence \( f^{(n)}(0^+)=0 \) for every \( n \), and the Taylor series \( \sum f^{(n)}(0)V^n/n!\equiv0\neq f(V) \). Physically, the pairing energy is invisible order-by-order in \( V \): no finite-order Feynman-diagram sum in the bare coupling produces superconductivity; one must resum the leading logarithms.
(B) Isotope effect. Since \( \Delta\propto\hbar\omega_D \) and \( \omega_D\propto M^{-1/2} \) for ionic mass \( M \), find the fractional change in \( \Delta \) when \( M \) increases by 8% (e.g. \( ^{200}\text{Hg}\to^{216}\text{Hg} \)-like), assuming \( N(0)V \) is fixed.
Solution
\( \Delta\propto M^{-1/2} \), so \( \delta\Delta/\Delta=-\tfrac12\,\delta M/M=-\tfrac12(0.08)=-0.04 \). The binding energy (and \( T_c \)) falls by about 4%. This \( T_c\propto M^{-1/2} \) scaling is the classic isotope-effect fingerprint of phonon-mediated pairing.
(C) Finite-momentum pair. For a pair with centre-of-mass momentum \( \hbar\mathbf{q} \), the two members occupy \( \mathbf{k}+\tfrac{\mathbf q}{2} \) and \( -\mathbf{k}+\tfrac{\mathbf q}{2} \). Argue qualitatively (via the shell of allowed states) why \( \mathbf{q}=0 \) maximises the binding, and estimate the momentum scale over which pairing survives.
Solution
Both members must lie in the attractive shell \( \epsilon_F<\epsilon<\epsilon_F+\hbar\omega_D \). For \( \mathbf q=0 \) the pair \( (\mathbf k,-\mathbf k) \) can use the entire spherical shell — maximal phase space, hence the largest \( \sum_k(2\epsilon_k-E)^{-1} \) and deepest binding. For \( \mathbf q\neq0 \) the condition that both \( \mathbf k+\mathbf q/2 \) and \( -\mathbf k+\mathbf q/2 \) sit in the shell is met only in a lens-shaped region whose volume shrinks as \( q \) grows, weakening \( \Delta \). Binding is lost once the pair kinetic energy \( \sim\hbar v_F q \) exceeds the binding \( \Delta \), i.e. for \( q\gtrsim\Delta/\hbar v_F\sim1/\xi \), the inverse coherence length. Thus only near-zero-momentum pairs bind, and they do so over a length scale \( \xi=\hbar v_F/\Delta \).