整体思路
\begin{align}
\frac{\Delta F}{NE_n}
= \left[ \tilde{\tilde{\Omega}}_{\mathrm{int}} - \tilde{\Omega}_{0}^{\mathrm{M}}
+ \tilde{\mu} \right]\left( \frac{n_{\varepsilon}}{n} \right)^{2/3}
\end{align}
\begin{align}
\frac{n}{n_{\varepsilon}}
=& - \frac{\partial}{\partial\tilde{\mu}}\left[
\tilde{\tilde{\Omega}}_{\mathrm{int}} + \tilde{\Omega}_{0}^{\mathrm{B}}
\right]
\end{align}
\begin{align}
\tilde{\tilde{\Omega}}_{\mathrm{int}} = \int \mathrm{d}\tilde{q}\cdot \tilde{q}^2
\int_0^{\pi}\sin\theta_q\mathrm{d}\theta_q \int_{-\infty}^{+\infty}\mathrm{d}\tilde{\omega}
\cdot \frac{3}{2\pi}\cdot \frac{1}{e^{\tilde{\beta}\tilde{\omega}}-1}
\delta^p(\vec{q},z)
\end{align}
\begin{align}
\tilde{\Omega}_0^{\mathrm{B}} = \frac{3}{\tilde{\beta}} \int_0^{\infty} \mathrm{d}
\tilde{k} \cdot \tilde{k}^2 \ln \left[1 - e^{- \tilde{\beta} \tilde{\xi}_k}\right]
\end{align}
\begin{align}
\delta^p(\vec{q}, z)
=& \mathrm{Arg}\left[
\frac{1}{4\pi}\cdot \frac{2R}{k_{\varepsilon}^2 v}
+ \tilde{z}\cdot\frac{1}{4\pi}
+ \frac{2 R}{M k_{\varepsilon}^2}\Pi_r(\vec{q},z + \mathrm{i}0^+)
\right]
\end{align}
\begin{align}
\frac{2 R}{M k_{\varepsilon}^2}\Pi_r(\vec{q},z)
=& \frac{2}{\pi^2} \cdot k_{\varepsilon}R\cdot \int \mathrm{d}\tilde{k}\\
& \left\{
\int_0^{2\pi}\mathrm{d}\phi_k \int_{0}^{\pi} \sin\theta \mathrm{d}\theta_k
\cdot \cos^2 \theta_k \left[
1+n(\xi_{\vec{k}+\vec{q}/2}) + n(\xi_{-\vec{k}+\vec{q}/2})
\right]\frac{3}{4\pi}\frac{\tilde{k}^4}
{2\tilde{k}^2 - \tilde{z}}
-\frac{1}{2}\tilde{k}^2
- \frac{1}{4}\tilde{z}
\right\}
\end{align}
而
\begin{align} n(\xi_{\vec{k}+\vec{q}/2}) = \frac{1}{e^{\beta(\tilde{k}^2 + \tilde{q}^2/4 + \tilde{k} \tilde{q} x-2\mu)} -1} \end{align}其中
\begin{align} x = \cos\theta_{kq} =& \sin\theta_k \cos \phi_k \sin \theta_q \cos \phi_q + \sin \theta_k \sin\phi_k \sin \theta_q \sin \phi_q + \cos \theta_k \cos \theta_q \\ =& \sin\theta_k \sin \theta_q \cos(\phi_k-\phi_q) + \cos \theta_k \cos \theta_q \end{align}带 tilde 的都是以 \(\varepsilon\) 或 \(k_{\varepsilon}\) 为单位的无量纲量.
最终画的是 \((1)\) 式. \((1)\) 式左边以 \(E_n\) 为单位, 右边乘上 \(\left( \frac{n_{\varepsilon}}{n} \right)^{2/3}\) 进行单位转换.
将 \((2)~(5)\) 式依次代入可得结果
还有一些关系: \(\tilde{\xi} = \tilde{k}^2 - \tilde{\mu} ,\quad n(\xi) = \frac{1}{e^{\tilde{\beta} \tilde{\xi}}-1}, \quad \tilde{z} = \tilde{\omega} - \frac{\tilde{q}^2}{2} + 2 \tilde{\mu}\)
Fig
fig3-p-nc
