# Why do pairing and condensation occur simultaneously in superconductors?

Bose condensation in the weak-coupling limit (such as BCS superconductors) and the strong-coupling limit (like BECs of atoms) can be unified into a single framework by the BEC-BCS crossover. One key feature distinguishing the BCS side from BEC is that the pairing of fermions near the Fermi surface into Cooper pairs and the condensation of those Cooper pairs happens simultaneously at the same temperature.

1. Is there a simple argument to understand why this is so?

2. For a usual BCS-type interaction, at what interaction strength does this break down? Is there a known expression for the difference, as a function of interaction strength, between the temperature at which pairs become bound and $T_c$ for them to condense?

1) On the Bose side pairs form when the temperature is lower than the binding energy, $k_BT<B$. Note that this is not a sharp phase transition, and there is no order parameter associated with it. Pairs condense at the Einstein temperature $$T_c = \frac{2\pi\hbar^2}{mk_B} \left( \frac{n}{\zeta(3/2)}\right)^{2/3}$$ where $n$ is the density of pairs, and $m\simeq 2m_a$ is the mass of a pair. This is a sharp phase transition (Bose condensation). The pair formation and BEC temperatures become equal when $T_c\sim B$.

Note that parametrically, $T_c$ is on the order of the would-be Fermi energy $E_F=k_F^2/(2m_a)$ of the fermionic constituents. Here, $k_F$ is defined via the density of the gas, $n_F=k_F^3/(3\pi^2)$. The pair density is $n=n_F/2$. This means that the crossover estimate $T_c\sim B$ is roughly consistent with the dimensional estimate $T_F\sim B$.

2) Coming from the BCS side the interaction is weak, and in 3d there are no bound states in vacuum. This means that Cooper pairs form at the BCS transition temperature $$T_c = \frac{8e^\gamma E_F}{(4e)^{1/3}e^2\pi}\exp\left(-\frac{\pi}{2k_F|a|} \right)$$ where $a$ is the scattering length. The crossover is reached when $T_c$ is of order $E_F$, corresponding to $a\to \infty$.

3) We know that the crossover is smooth (no other phase transitions intervene). This has been established experimentally, by numerical simulation, and is consistent with simple many-body theories. At $a\to\infty$ there is some discussion about a possible "pseudo-gap" phase above $T_c$. This is hard to settle, because there is no completely sharp definition of what a pseudo-gap is.

• Thanks Thomas, this is helpful. Maybe I can refine my question to this: In the BCS limit and just above $T_c$, is there still some (possibly small) number of fermions in the low-energy part of the thermal distribution that form noncondensed Cooper pairs? Do these pairs exist but are generally ignored because they don't contribute much to thermodynamic properties? And if not, why not? – Rococo Aug 2 '17 at 2:09
• One final way of phrasing the same question: if you sit at 1.1 $T_c$ and move from the BEC to BCS limit, does the number of atoms in non-condensed bound pairs just smoothly asymptote towards zero, or does it actually become precisely zero at some finite interaction strength? – Rococo Aug 2 '17 at 2:11
• There is no sharp transition, and not even a sharp definition of an operator that would correspond to the "fraction of pre-formed pairs". Having said this, there is no reason to expect any pre-formed pairs on the BCS side. There are no bound states in vacuum, and the BCS wvae function does not predict pre-formed pairs. – Thomas Aug 2 '17 at 6:46
• Thanks Thomas. There is still something not clicking here for me, but I might just need to sit with this. I don't see how the statement "There are no bound states in vacuum" is relevant, since the ground state is a filled Fermi sea instead of a vacuum. That's why pairing is possible with an arbitrarily weak attraction, right? So why wouldn't this happen above $T_c$ too? – Rococo Aug 3 '17 at 13:59
• Well, I'm just asking what the possible reasons are for expecting bound states. 1) Bound states already exist in vacuum (not the case on the BCS) side, 2) we have a many-body wave function, and it predicts pair correlatons above $T_c$. Possibility 2) is of course an option, but deep on the BCS side the BCS variational wave function is reliable, and it does not predict pair correlations above $T_c$. (Roughly, it is "too simple". It has only one variational parameter related to pair correlations, and that parameter goes to zero at $T_c$). – Thomas Aug 3 '17 at 15:53