In BCS theory, the superconducting gap is given by solving at different temperatures the integral $$\frac{1}{N(0)V}=\int_0^{\hbar\omega_c}\frac{\tanh\frac{1}{2}\beta(\xi^2+\Delta^2)^{1/2}}{(\xi^2+\Delta^2)^{1/2}} \text{d}\xi$$ In textbooks like Tinkham (2nd edition, page 63) and Phillips (Advanced Solid State Physics, page 246) you can find approximate formulas for certain temperature ranges (typically $T\approx T_C$).

In some other references, such as here and here (for the latter, I couldn't find the arxiv version, sorry), it is mentioned an interpolation formula valid in the whole temperature range, that is $$ \Delta(T)=\Delta_0\tanh (k\sqrt{\frac{T_{C}-T}{T}})$$ with $k=1.74$ or $k=2.2$.

Is there someone who can link a reference to this formula, and how it is obtained?

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/54200/2451 , physics.stackexchange.com/q/122781/2451 and links therein. $\endgroup$
    – Qmechanic
    Jul 3, 2015 at 12:40
  • $\begingroup$ In the post you link there's a numerical method to solve the integral I wrote in the first line; in the comments the interpolation formula I'm looking for is quoted, but without references. References (and the derivation method for this interpolation formula) are actually what I'm asking in this post. $\endgroup$
    – casx
    Jul 3, 2015 at 13:00
  • 1
    $\begingroup$ I'm not sure you can really prove an interpolation formula, but you can numerically show it's correct, up to some minor numerical difference. An interpolation formula is something which pass through all the known points of a curve. For the superconducting gap, one only knows $\Delta\left(T=0\right)=\Delta_{0}$ and $\Delta\left(T=T_{c}\right)=0$. The interpolation formula obviously pass through these points. You might fix the factor $k$ depending on the region you want to focus on. $\endgroup$
    – FraSchelle
    Jul 9, 2015 at 13:49
  • $\begingroup$ I believe that there is a way to find the formula among all the approximations in the following (in French): link:archive.org/details/mq_aslangul_tome3/page/n1179/mode/… $\endgroup$
    – The Tiler
    Feb 13 at 12:37

2 Answers 2


This interpolation formula agrees with both the high- and zero-temperature limits for $\Delta(T)$: \begin{align} \tag{1}\label{eq1} 1-T/T_c \ll 1 &: \Delta(T)\approx 3.06\, k_B T_c\sqrt{1-T/T_c}\\ \tag{2}\label{eq2} T = 0 &: \Delta_0 = 1.764\, k_B T_c. \end{align}

When $T$ is near $T_c$ the argument of $\tanh$ in the interpolation formula is small so we can approximate $\tanh x \approx x$, giving $\Delta(T) \approx k \Delta_0 \sqrt{T_c/T-1}$. Then to recover \eqref{eq1}, we substitute $k = 1.74$, replace $\Delta_0$ using \eqref{eq2}, and use the fact that $\sqrt{T_c/T-1} = \sqrt{1-T/T_c}$ for $T$ near $T_c$.

At zero temperature, the argument of $\tanh$ in the interpolation function is large such that $\tanh x \approx 1$, giving $\Delta(T = 0) = \Delta_0$ which is just \eqref{eq2}.

As for the choice of $k$, for some strong-coupling superconductors, the prefactor in \eqref{eq2} is larger than 1.76.

Not only does the interpolation formula agree at $T=0$ and $T=T_c$, as stated in the comment above, but it also works near $T_c$.


It is just an interpolation formula. However, as Gross (3.11) mentioned in 1980, this formula can be rewritten as $$ \Delta(T)=\Delta(0) \tanh \left( \frac{\pi}{\Delta(0)} \sqrt{a \frac{\delta C}{C} ( \frac{T_{c}}{T}-1} ) \right), $$ where for s-wave superconductors, $Δ(0)$ = $1.76T_{c}$, a = 2/3; for the pure d-wave case $Δ(0)$ = 2.14$T_{c}$, a = 1; and for an s + g wave $Δ(0)$ = 2.77$T_{c}$, a = 2. $\delta C/C$ is specific heat jump.

It looks analitic cause the limits at zero temperature and $T_{c}$, which are fitted, are analitical.


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