In BCS-BEC crossover, we will get two self-consistent equations about the gap $\Delta$ and the chemical potential $\mu$. But when I compute the particle density: $n=\beta^{-1}\sum_k tr(G\frac{\partial G^{-1}}{\partial \mu})$ my result is $n=-\sum_k\frac{\zeta_k}{E_k}\tanh(\beta E_k/2)$, how can I get the right result $n=\sum_k(1-\frac{\zeta_k}{E_k}\tanh(\beta E_k/2))$. Who can give me a calculation in detail.
Here: $G^{-1}=\begin{pmatrix}-\partial_\tau-\frac{p^2}{2m}+\mu & \Delta\\ \bar\Delta&-\partial_\tau+\frac{p^2}{2m}-\mu\end{pmatrix}$, $\tau=it$ is the imaginary time, $\beta=1/T$ and $T$ is the temperature, $\zeta_k=\frac{p^2}{2m}-\mu$, $E_k=\sqrt{\zeta_k^2+|\Delta|^2}$.
This is my calculation: $n=\beta^{-1} tr\left(G\frac{\partial G^{-1}}{\partial \mu}\right)=\beta^{-1}tr\left[G\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right]\\=\beta^{-1}\sum_{k}tr\left[\begin{pmatrix}i\omega_n-\zeta_k&\Delta\\\bar\Delta&i\omega_n+\zeta_k\end{pmatrix}^{-1}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right]\\=\beta^{-1}\sum_k\frac{-2\zeta_k}{\omega_n^2+E_k^2}=\sum_{\bf k}\frac{1}{2\pi i}\oint \frac{-2\zeta_k}{-z^2+E_k^2}\frac{dz}{\exp\{\beta z\}+1}=-\sum_{\bf k}\frac{\zeta_k}{E_k}\tanh{\frac{\beta E_k}{2}}$
• You should first give us all your notation details. What are $\tau$, $\beta$, ... and explain us how you get a negative particle density ! Clearly, what you gave is $G^{-1}$ if I define $G$ as the Green's function, something you didn't do, too ... – FraSchelle May 31 '17 at 3:20
• The problem is that you have thrown away the $i\omega_n$ term in the numerator. You should keep it. Including the convergence term $e^{i\omega_n}$, this gives an additional $1$. – Adam May 31 '17 at 10:27