Specific heat of superconductor

Question

I don't understand how the exponential growth of specific heat of a superconductor is interpreted. Wikipedia says:

Due to the energy gap, the specific heat of the superconductor is suppressed strongly at low temperatures, there being no thermal excitations left. However before reaching the transition temperature, the specific heat of the superconductor becomes even higher than of the normal conductor.

If we have an energy bandgap, shouldn't we need more energy to heat up such a system than to heat up a "normally conductive" system with continuous energy?

Answer 1

A system's specific heat (or heat capacity) is defined to be $C = d\langle E \rangle/dT$, where $\langle E \rangle$ is the system's internal energy. For a gapped quantum system at a temperature far below the energy gap $\Delta E$, we expect the thermal mixture to be almost entirely in the ground state, with only an exponentially small weighting for the excited states. So the expected energy $\langle E \rangle$ is effectively quantized and pinned down to its ground state. The system can only absorb a quantized amount of energy $\Delta E$, so if you start the system in the ground state and put it into thermal contact with a bath at any temperature $T \ll \Delta E$, the system basically doesn't absorb any energy at all until $T \sim \Delta E$, so $d\langle E \rangle = 0$ below that temperature, and so $C$ is exponentially suppressed.

More precisely, if $T \ll \Delta E$, we expect that all excited states above the first one will contribute negligibly, so we can model the system as a two-level system with energies $0$ and $\Delta E$. The expected energy is then $$\langle E \rangle = \frac{0 \times 1 + \Delta E \times e^{-\Delta E/T}}{1 + e^{-\Delta E/T}} = \frac{\Delta E}{1 + e^{\Delta E/T}},$$ and $$C = \frac{(\Delta E/T)^2}{4 \cosh^2(1/2 \times \Delta E/T)} \sim \left(\frac{\Delta E}{T}\right)^2 e^{-\frac{\Delta E}{T}}$$ vanishes exponentially quickly at temperatures far below the energy gap. By contrast, for a gapless system like a metal, the specific heat typically goes to zero much more slowly, usually as a power-law with temperature (e.g. with an electron contribution $C \sim T$ or a phonon contribution $C \sim T^3$).

Answer 2

Previous answers have pointed out why you can expect an exponential behaviour of the specific heat when the system's spectrum is gapped, thus explaining the superconductor's specific heat at small temperatures.

But if I am not wrong, you are confused by the fact that $C_v(T\to T_c^-) > C_v(T\to T_c^+)$ (where $C_v$ is the specific heat and $T_c$ the critical temperature). In words, the specific heat of the superconducting state close to the critical temperature is higher than the specific heat of the normal metallic state close to the critical temperature. You said:

If we have an energy bandgap, shouldn't we need more energy to heat up such a system than to heat up a "normally conductive" system with continuous energy?

but I think this is not the correct interpretation. $C_v = d\left\langle E \right\rangle /dT$ measures the slope of the internal energy vs temperature curve. So what you said is correct: the internal energy in the metallic state is higher than in the superconducting state because the spectral gap is closed, but the rapidity of energy growth with temperature is higher for the superconducting state. When $T \to T_c^-$, the internal energy grows faster with $T$ than in the case $T \to T_c^+$, even though the energy always grows with temperature.

The physical meaning is clear: as $T\to T_c^-$, the energy spectrum changes: the gap $\Delta(T)$ closes very quickly, roughly $\Delta(T) \sim k_BT_c\sqrt{ 1 - \frac{T}{T_c}}$, so the excitation of quasiparticles is easier and easier as $T$ increases. When $T>T_c$ the gap is closed and the spectrum does not change with $T$, thus the energy gain is slower.

Answer 3

Concerning the question's quote (from [1])

"before reaching the transition temperature the specific heat of the superconductor becomes even higher than of the normal conductor"

there is indeed a positive jump $ {\Delta C\equiv C(T_c+0^-)-C(T_c+0^+)} $ in the specific heat at $T_c$. In general ${C = T {dS\over dT}}$, S being the entropy, and ${dS\over dT}$ jumps at $T_c$ due to the jump in ${d\Delta^2\over dT}$, where ${\Delta}$ is the gap.

See (3.56-58) in [2] or (80-84) in [3] for details.

In a nutshell the answer is: beside the gap, mind the gap-induced entropy change. Breaking up Cooper pairs costs and absorbs energy.

[1] https://en.wikipedia.org/wiki/BCS_theory

[2] M. Tinkham Introduction to Superconductivity (2ed., MGH, 1996)

[3] Rafael M. Fernandes, Lecture Notes: BCS theory of superconductivity