diff --git a/src/external/libGapIntegrals/GapIntegrals.pdf b/src/external/libGapIntegrals/GapIntegrals.pdf index ee36ca2c..5a13b18c 100644 Binary files a/src/external/libGapIntegrals/GapIntegrals.pdf and b/src/external/libGapIntegrals/GapIntegrals.pdf differ diff --git a/src/external/libGapIntegrals/GapIntegrals.tex b/src/external/libGapIntegrals/GapIntegrals.tex index 29c1180d..2dafd92b 100644 --- a/src/external/libGapIntegrals/GapIntegrals.tex +++ b/src/external/libGapIntegrals/GapIntegrals.tex @@ -84,7 +84,7 @@ In order not to do too many unnecessary function calls during the final numerica \frac{\partial f}{\partial E} = -\frac{1}{k_{\mathrm B}T}\frac{\exp(E/k_{\mathrm B}T)}{\left(1+\exp(E/k_{\mathrm B}T)\right)^2} = -\frac{1}{4k_{\mathrm B}T} \frac{1}{\cosh^2\left(E/2k_{\mathrm B}T\right)}. \end{equation} Using (\ref{derivative}) and doing the substitution $E'^2 = E^2-\Delta^2(\varphi,T)$, equation (\ref{eq:int_phi}) can be written as -\begin{equation} +\begin{equation}\label{eq:bmw_2d} \begin{split} I(T) & = 1 - \frac{1}{4\pi k_{\mathrm B}T}\int_0^{2\pi}\int_{\Delta(\varphi,T)}^{\infty}\frac{1}{\cosh^2\left(E/2k_{\mathrm B}T\right)}\frac{E}{\sqrt{E^2-\Delta^2(\varphi,T)}}\mathrm{d}E\mathrm{d}\varphi \\ & = 1 - \frac{1}{4\pi k_{\mathrm B}T}\int_0^{2\pi}\int_{0}^{\infty}\frac{1}{\cosh^2\left(\sqrt{E'^2+\Delta^2(\varphi,T)}/2k_{\mathrm B}T\right)}\mathrm{d}E'\mathrm{d}\varphi\,. @@ -210,28 +210,55 @@ within the semi-classical model assuming a cylindrical Fermi surface and a mixed \subsection*{Gap Integrals for $\bm{\theta}$-, and $\bm{(\theta, \varphi)}$-dependent Gaps}% -\noindent For the most general case for which the gap-symmetry is $(E,\varphi,\theta)$-dependent, the integral to be calculate takes the form +First some general formulae as found in Ref.\,\cite{Prozorov}. It assumes an anisotropic response which can be classified in 3 directions ($a$, $b$, and $c$). -\begin{equation}\label{eq:int_phi_theta} - I = 1 + \frac{1}{2\pi} \int_0^\pi \sin(\theta)\, \mathrm{d}\theta \int_0^{2\pi} \mathrm{d}\varphi \int_{\Delta(\varphi,\theta,T)}^\infty \mathrm{d}E \left\{ \left(\frac{\partial f}{\partial E}\right) \frac{E}{\sqrt{E^2 - \Delta^2(\varphi, \theta, T)}} \right\} +\noindent For the case of a 2D Fermi surface (cylindrical symmetry): + +\begin{equation}\label{eq:n_anisotrope_2D} +n_{aa \atop bb}(T) = 1 - \frac{1}{2\pi k_{\rm B} T} \int_0^{2\pi} \mathrm{d}\varphi\, {\cos^2(\varphi) \atop \sin^2(\varphi)} \underbrace{\int_0^\infty \mathrm{d}\varepsilon\, \left\{ \cosh\left[\frac{\sqrt{\varepsilon^2 + \Delta^2}}{2 k_{\rm B}T}\right]\right\}^{-2}}_{= G(\Delta(\varphi), T)} \end{equation} +\noindent For the case of a 3D Fermi surface: -\subsubsection*{Gap Integrals for $\bm{\theta}$-dpendent Gaps}% +\begin{eqnarray} + n_{aa \atop bb}(T) &=& 1 - \frac{3}{4\pi k_{\rm B} T} \int_0^1 \mathrm{d}z\, (1-z^2) \int_0^{2\pi} \mathrm{d}\varphi\, {\cos^2(\varphi) \atop \sin^2(\varphi)} \cdot G(\Delta(z,\varphi), T) \label{eq:n_anisotrope_3D_aabb} \\ + n_{cc}(T) &=& 1 - \frac{3}{2\pi k_{\rm B} T} \int_0^1 \mathrm{d}z\, z^2 \int_0^{2\pi} \mathrm{d}\varphi\, {\cos^2(\varphi) \atop \sin^2(\varphi)} \cdot G(\Delta(z,\varphi), T) \label{eq:n_anisotrope_3D_cc} +\end{eqnarray} -\noindent In case the gap-symmetry is only depending on $(E,\theta)$ the $\varphi$-integral can be carried out and Eq.(\ref{eq:int_phi_theta}) simplifies to +\noindent The ``powder averaged'' superfluid density is then defined as -\begin{equation}\label{eq:int_theta} - I(T) = 1 + \int_0^\pi \sin(\theta)\, \mathrm{d}\theta \int_{\Delta(\varphi,\theta,T)}^\infty \mathrm{d}E \left\{ \left(\frac{\partial f}{\partial E}\right) \frac{E}{\sqrt{E^2 - \Delta^2(\varphi, \theta, T)}} \right\} +\begin{equation} + n_{\rm S} = \frac{1}{3}\cdot \left[ \sqrt{n_{aa} n_{bb}} + \sqrt{n_{aa} n_{cc}} + \sqrt{n_{bb} n_{cc}} \right] \end{equation} -\noindent Following the same simplification steps as for Eq.(\ref{eq:int_phi}) we find +\subsubsection*{Isotropic s-Wave Gap} -\begin{equation}\label{eq:int_tilde_theta} - \tilde{I}(T) = 1 - \frac{\pi E_c}{4 k_{\rm B}T}\, \int_0^1 \mathrm{d}x \sin(\pi x) \int_0^1 \mathrm{d}F\, \frac{1}{\cosh^2\left(\sqrt{(E_c\cdot F)^2 + \Delta^2(\pi\cdot x, T)}/(2 k_{\rm B} T)\right)} +\noindent For the 2D/3D case this means that $\Delta$ is just a constant. + +\noindent For the 2D case it follows + +\begin{equation} + n_{aa \atop bb}(T) = 1 - \frac{1}{2 k_{\rm B} T} \cdot G(\Delta, T) = n_{\rm S}(T). \end{equation} -\noindent where $x = \theta / \pi$, and $F = E^\prime / E_c$. +\noindent This is the same as Eq.(\ref{eq:bmw_2d}), assuming a $\Delta \neq f(\varphi)$. + +\vspace{5mm} + +\noindent The 3D case for $\Delta \neq f(\theta, \varphi)$: + +\noindent The variable transformation $z = \cos(\theta)$ leads to $\mathrm{d}z = -\sin(\theta)\,\mathrm{d}\theta$, $z=0 \to \theta=\pi/2$, $z=1 \to \theta=0$, and hence to + +\begin{eqnarray*} + n_{aa \atop bb}(T) &=& 1 - \frac{3}{4\pi k_{\rm B} T} \underbrace{\int_0^{\pi/2} \mathrm{d}\theta \, \sin^3(\theta)}_{= 2/3} \, \underbrace{\int_0^{2\pi} \mathrm{d}\varphi {\cos^2(\varphi) \atop \sin^2(\varphi)}}_{=\pi} \cdot G(\Delta, T) \\ + &=& 1 - \frac{1}{2 k_{\rm B} T} \cdot G(\Delta, T). \\ + n_{cc}(T) &=& 1 - \frac{3}{2\pi k_{\rm B} T} \underbrace{\int_0^{\pi/2} \mathrm{d}\theta \, \cos^2(\theta)\sin(\theta)}_{=1/3} \, \underbrace{\int_0^{2\pi} \mathrm{d}\varphi {\cos^2(\varphi) \atop \sin^2(\varphi)}}_{=\pi} \cdot G(\Delta, T) \\ + &=& 1 - \frac{1}{2 k_{\rm B} T} \cdot G(\Delta, T). +\end{eqnarray*} + +\noindent And hence + +$$ n_{\rm S}(T) = 1- \frac{1}{2 k_{\rm B} T} \cdot G(\Delta, T). $$ \subsection*{License} The \gapint library is free software; you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation \cite{GPL}; either version 2 of the License, or (at your option) any later version.