Free Energy vs. Renormalize Paramaters
p-wave
cal
\begin{align}
\delta^p(\vec{q}, z) =& \mathrm{Arg}\left[ \frac{Mk_{n^2}}{2}\frac{1}{R}
\left(
\frac{1}{4\pi}\cdot \frac{2R}{k_n^2 v}
+ \tilde{z}\cdot\frac{1}{4\pi}
+ \frac{2 R}{M k_n^2}\Pi_r(\vec{q},z)
\right)
\right] \\
=& \mathrm{Arg}\left[
\frac{1}{4\pi}\cdot \frac{2R}{k_n^2 v}
+ \tilde{z}\cdot\frac{1}{4\pi}
+ \frac{2 R}{M k_n^2}\Pi_r(\vec{q},z + \mathrm{i}0^+)
\right]
\end{align}
其中 \(\tilde{z}=z/E_n\) , \(E_n = k_n^2/(2M)\) , \(k_n^3 = 6\pi^2n\) , \(n = N/V\)
\begin{align} \frac{2 R}{M k_n^2}\Pi_r(\vec{q},z) =& (k_n R)\cdot\Pi_r \cdot \frac{2}{Mk_n^3}\\ =& \tilde{R}\left[ \frac{2}{Mk_n^3}\left( -\frac{M}{V} \right)\sum_{\vec{k}}1 - \tilde{z}E_n \frac{M^2}{V}\frac{2}{Mk_n^3}\sum_{\vec{k}}\frac{1}{k^2} + \frac{2}{Mk_n^3}\Pi^{l=1}(\vec{q},z) \right] \\ =& \tilde{R}\left[ -\frac{1}{\pi^2}\int \mathrm{d}\tilde{k}\cdot \tilde{k}^2 -\tilde{z} \frac{1}{2\pi^2}\int \mathrm{d}\tilde{k} +\tilde{\Pi}^{l=1} \right] \end{align}其中 \(\tilde{R} = k_nR\) , \(\tilde{k} = k/k_n\)
\begin{align} \tilde{\Pi}^{l=1} = &\frac{2}{Mk_n^3}\Pi^{l=1}(\vec{q},\omega) \\ =& \frac{2}{Mk_n^3}\frac{1}{V}\frac{V}{(2\pi)^3}\int \mathrm{d}\tilde{k} \left[ k^2 \cdot 4\pi |Y_{lm}(\hat{k})|^2 \frac{1+n(\xi_{\vec{k}+\vec{q}/2}) + n(\xi_{-\vec{k}+\vec{q}/2})} {\xi_{\vec{k}+\vec{q}/2} + \xi_{-\vec{k}+\vec{q}/2} - \omega} \right] \\ =& \frac{2}{\pi^2}\int \mathrm{d}\tilde{k}\cdot\tilde{k}^4\left[ \frac{1+n(\xi_{\vec{k}+\vec{q}/2}) + n(\xi_{-\vec{k}+\vec{q}/2})} {\tilde{\xi}_{\vec{k}+\vec{q}/2} + \tilde{\xi}_{-\vec{k}+\vec{q}/2} - \tilde{\omega}} \right] \end{align}其中 \(\tilde{\xi} = \xi/E_n\) , \(\tilde{\omega} = \omega/E_n\) , \(n(\xi) = \frac{1}{e^{\beta \xi}-1}\)
最终
\begin{align} \frac{\tilde{\Omega}}{N E_n} =& \frac{1}{N E_n} \frac{V}{(2\pi^3)}\int \mathrm{d}^3\vec{q} \cdot \int \frac{\mathrm{d}\omega}{\pi}\cdot \frac{1}{e^{\beta\omega}-1} \delta^p \\ =& \frac{3}{\pi} \int \mathrm{d}\tilde{q}\cdot \tilde{q}^2 \int_{-\infty}^{+\infty}\mathrm{d}\tilde{\omega} \cdot \frac{1}{e^{\tilde{\beta}\tilde{\omega}}-1} \tilde{\delta}^p(\vec{q},z) \end{align}其中 \(\tilde{\beta} = \beta E_n\) . 得自由能
\begin{align} \frac{F}{NE_n} = \frac{\tilde{\Omega}}{N E_n} -\frac{\mu}{E_n} \end{align}即
\begin{align} f(\tilde{\mu}, \tilde{R}) = \tilde{\Omega}'(\tilde{\mu}, \tilde{R})-\tilde{\mu} \end{align}其中 \(\tilde{\mu} = \mu/E_n\).
\(\mu\) 由
\begin{align} N = - \frac{\partial\Omega}{\partial \mu} \end{align}决定.
以 \(\varepsilon\) 为单位
若以某一能量 \(\varepsilon\) 为单位, 对应的长度单位 \(k_{\varepsilon} = \sqrt{2M\varepsilon}\) , 密度单位 \(n_{\varepsilon} = k_{\varepsilon}^3/(6\pi^2)\) , 那么
\begin{align} \frac{\Omega}{N \varepsilon} = & \frac{n_{\varepsilon}}{n}\int \mathrm{d}\tilde{q}\cdot \tilde{q}^2 \int_{-\infty}^{+\infty}\mathrm{d}\tilde{\omega} \cdot\frac{3}{\pi}\cdot \frac{1}{e^{\tilde{\beta}\tilde{\omega}}-1} \tilde{\delta}^p(\vec{q},z) \\ = & \frac{n_{\varepsilon}}{n}\int \mathrm{d}\tilde{q}\cdot \tilde{q}^2 \int_{-\infty}^{+\infty}\mathrm{d}\tilde{\omega} \cdot f(\tilde{q}, \tilde{\omega}, \tilde{\mu}, \tilde{\beta}) \end{align}其中
\begin{align} f(\tilde{q}, \tilde{\omega}, \tilde{\mu}, \tilde{\beta}) = \frac{3}{\pi}\cdot \frac{1}{e^{\tilde{\beta}\tilde{\omega}}-1} \tilde{\delta}^p(\vec{q},z) \end{align}而
\begin{align} \frac{n}{n_{\varepsilon}} =& - \frac{1}{n_{\varepsilon}V} \frac{\partial\Omega}{\partial\mu} =- \frac{1}{n_{\varepsilon}V} \frac{\partial\Omega/\mu}{\partial\tilde{\mu}}\\ =& - \frac{1}{n_{\varepsilon}V} \frac{\partial}{\partial\tilde{\mu}}\left[ V n_{\varepsilon} \int \mathrm{d}\tilde{q}\cdot \tilde{q}^2 \int_{-\infty}^{+\infty}\mathrm{d}\tilde{\omega} \cdot f(\tilde{q}, \tilde{\omega}, \tilde{\mu}, \tilde{\beta}) \right] \\ =& - \frac{\partial}{\partial\tilde{\mu}}\left[ \int \mathrm{d}\tilde{q}\cdot \tilde{q}^2 \int_{-\infty}^{+\infty}\mathrm{d}\tilde{\omega} \cdot f(\tilde{q}, \tilde{\omega}, \tilde{\mu}, \tilde{\beta}) \right] \end{align}所以最终要求的为
\begin{align} \frac{\Delta F}{NE_n} =& \frac{\Omega}{NE_n} + \frac{\mu}{E_n} \\ =&\frac{\Omega}{N\varepsilon}\left( \frac{\varepsilon}{E_n} \right) + \tilde{\mu} \left( \frac{\varepsilon}{E_n} \right) \\ =& \left( \frac{n_{\varepsilon}}{n} \right)^{5/3} \int\mathrm{d}\tilde{q}\cdot \tilde{q}^2 \int_{-\infty}^{+\infty}\mathrm{d}\tilde{\omega} \cdot f(\tilde{q}, \tilde{\omega}, \tilde{\mu}, \tilde{\beta}) + \tilde{\mu} \left( \frac{n_{\varepsilon}}{n} \right)^{2/3} \\ \end{align}横坐标为
\begin{align} \frac{2R}{k_n^2v} = \frac{2R}{k_{\varepsilon v}}\cdot \left( \frac{n_{\varepsilon}}{n} \right)^{2/3} \end{align}result
code
计算 \(\Delta F\)
from matplotlib import pyplot as plt |
计算 \(T_{C}\)
N = 100 |
def tm(omega, q, rkv, mu): |