Model 部分
Hamiltonian 为
\begin{align*} H =& \sum_{k,\sigma}\varepsilon_{k,\sigma} a_{k,\sigma}^{\dagger} a_{k,\sigma} + \sum_k (\varepsilon_{k,b}+\delta)b_k^{\dagger}b_k \\ & + \frac{\lambda}{\sqrt{\Omega}} \sum_{p,q}(b_{p+q}^{\dagger}a_{p,1}a_{q,2}+\mathrm{H.c.}) \end{align*}前两项是原子和分子态的能量, 最后一项是态 1, 2 和分子态之间的耦合项. 其中
\begin{align*} \varepsilon_{k,\sigma} =& \frac{\hbar^2 k^2}{2m} - \mu_{\sigma} \\ \varepsilon_{k,b} =& \frac{\hbar^2 k^2}{4m} - \mu_1 -\mu_2 \\ \mu_1 =& \mu_3 = \mu ,\quad \mu_2 =0 \end{align*}将耦合部分简记为
\begin{align*} \Lambda = \sum_{p,q}(b_{p+q}^{\dagger}a_{p,1}a_{q,2}+\mathrm{H.c.}) \end{align*}微扰为
\begin{align*} V = \sum_k(a_{k,2}^{\dagger}a_{k,3} e^{-\mathrm{i} \omega't}+a_{k,3}^{\dagger}a_{k,2}e^{\mathrm{i}\omega't}) \end{align*}Sum rules 部分
假设
假设系统的基态为 1 和 3 态的费米海
\begin{align*} |GS\rangle = |F \rangle = \prod_{k< k_F} a_{k,1}^{\dagger} a_{k,3}^{\dagger} |0\rangle \end{align*}Transition rate form 3 to 2
\begin{align*}
I(\omega) =& 2\pi \sum_f |\langle GS |V |f \rangle|^2 \delta(\omega+\mu -E_f +E_0) \\
=& 2 \sum_f |\langle GS |V |f \rangle|^2 \mathrm{Im}\frac{1}{\omega +\mu -E_f +E_0 -\mathrm{i}0^{ +}} \\
\propto &\mathrm{Im} \sum_f \langle GS |V \frac{1}{\omega +\mu -E_f +E_0 -\mathrm{i}0^{ +}} |f \rangle\langle f| V^{\dagger} |GS\rangle \\
=& \mathrm{Im} \langle GS |V \frac{1}{\omega +\mu -H +E_0 -\mathrm{i}0^{ +}} V^{\dagger} |GS\rangle \\
=& \mathrm{Im} \langle GS |V \frac{1}{\omega - \bar{H}} V^{\dagger} |GS\rangle
\end{align*}
其中
\begin{align*} \bar{H} = H -E_0 -\mu \end{align*}第二个等号用到了 Dirac Identity
\begin{align*} \frac{1}{x - x_0 \pm \mathrm{i}0^+} = \mathcal{P}\frac{1}{x-x_0} \mp \mathrm{i}\pi\delta(x-x_0) \end{align*}零温公式推导
考虑中间态 \(|n\rangle\) 有两种
\begin{align*} |q \rangle =& a_{q,2}^{\dagger} a_{q,3}| F\rangle \quad(q< k_F)\\ |p,q \rangle =& b_{p+q}^{\dagger} a_{p,1} a_{q_3}|F \rangle \quad(p,q< k_F) \end{align*}耦合作用到中间态上的结果为
\begin{align*} \Lambda |q \rangle =& \sum_p |p,q\rangle \\ \Lambda |p,q\rangle =& |q \rangle \end{align*}作用是在两种中间态之间切换, 则 \(\bar{H}\) 在空间中的矩阵元为
\begin{align*} \langle q |\bar{H} |q \rangle =& \varepsilon_{q,2} -\varepsilon_{q,3} - \mu = 0 \\ \langle p,q|\bar{H} |p,q \rangle =& \delta + \varepsilon_{p+q,b} - \varepsilon_{p,1}-\varepsilon_{q,3}-\mu \\ \langle q| \bar{H}|p,q\rangle =& \langle p,q |\bar{H} | q\rangle = \lambda/\sqrt{\Omega} \end{align*}Transition rate form 3 to 2 在上述空间中为
\begin{align*} I(\omega) \propto \mathrm{Im}\sum_{n,n'} \langle GS| V | n \rangle \langle n| \frac{1}{\omega-\bar{H}} | n'\rangle \langle n' V^{\dagger}|GS\rangle \end{align*}将耦合 \(\lambda/\sqrt{\Omega}\Lambda\) 看作小量,即 \(\bar{H}= \bar{H_0}+\lambda/\sqrt{\Omega}\Lambda\) , 可做如下展开
\begin{align*} \frac{1}{\omega-\bar{H}} &=\frac{1}{\omega - \bar{H_0}-\lambda/\sqrt{\Omega}\Lambda}\\ =& \frac{1}{\omega - \bar{H_0}}\cdot \left( \frac{1}{1- \frac{\lambda}{\sqrt{\Omega}}\frac{1}{\omega-\bar{H_0}}\Lambda} \right) \\ =& \frac{1}{\omega - \bar{H_0}}\cdot \sum_m \left( \frac{\lambda}{\sqrt{\Omega}}\frac{1}{\omega-\bar{H_0}}\Lambda \right)^m \\ \end{align*}而且将微扰 \(V\) 的表达式代入有
\begin{align*} \langle GS| V |n\rangle = \sum_q e^{\mathrm{i}\omega't}\delta_{n,q}\\ \langle n'| V |GS\rangle = \sum_{q'} e^{-\mathrm{i}\omega't}\delta_{n',q'}\\ \end{align*}所以
\begin{align*} I(\omega) \propto& \mathrm{Im}\sum_{q,q'} \langle q| \frac{1}{\omega-\bar{H}} | q'\rangle \\ =&\mathrm{Im}\sum_{q,q'} \langle q|\frac{1}{\omega - \bar{H_0}}\cdot \sum_m \left( \frac{\lambda}{\sqrt{\Omega}}\frac{1}{\omega-\bar{H_0}}\Lambda \right)^m | q'\rangle \\ \end{align*}由前面得出的耦合 \(\Lambda\) 作用在中间态上的结果可知, 耦合作用奇数次会由于正交性使结果为 \(0\) ( \(\bar{H_0}\) 为本征态), 所以 \(m\) 只能取偶数. 而且不同的 \(|q\rangle\) 也是正交的, 所以
\begin{align*} I(\omega) \propto& \mathrm{Im}\sum_q \langle q|\frac{1}{\omega - \bar{H_0}}\cdot \sum_m \left( \frac{\lambda}{\sqrt{\Omega}}\frac{1}{\omega-\bar{H_0}}\Lambda \right)^{2m} | q\rangle \\ =& \mathrm{Im}\sum_q \langle q|\frac{1}{\omega - \bar{H_0}}\cdot \sum_m \left( \frac{\lambda^2}{\Omega}\frac{1}{\omega-\bar{H_0}}\Lambda\frac{1}{\omega-\bar{H_0}}\Lambda \right)^m | q\rangle \\ =& \mathrm{Im}\sum_q \langle q|\frac{1}{\omega - \bar{H_0}}\cdot \frac{1}{1- \frac{\lambda^2}{\Omega}\frac{1}{\omega-\bar{H_0}}\Lambda\frac{1}{\omega-\bar{H_0}}\Lambda } | q\rangle \\ =& \mathrm{Im}\sum_q \langle q|\frac{1}{\frac{1}{\omega - \bar{H_0}}- \frac{\lambda^2}{\Omega}\Lambda\frac{1}{\omega-\bar{H_0}}\Lambda } | q\rangle \\ =& \mathrm{Im}\sum_{q< q_F}\frac{1}{\omega+ \mu +\varepsilon_{q,3}-\varepsilon_{q,2}-\lambda^2 \theta(q,\omega)} \end{align*}其中
\begin{align*} \theta(q,\omega) = \frac{1}{\Omega} \sum_{p< k_F} \frac{1}{\omega+ \mu + \varepsilon_{q,3}+\varepsilon_{p,1} - \varepsilon_{p+q,b}-\delta} \end{align*}零温数值
化简公式
\begin{align*}
\mu + \varepsilon_{q,3} -\varepsilon_{q,2} = 0
\end{align*}
所以
\begin{align*} I(\omega) \propto & \mathrm{Im} \sum_{q< q_{F}} \frac{1}{\omega -\lambda^2 \theta(q,\omega)} \\ =& \sum_{q< k_F} \frac{\lambda^2\mathrm{Im}\theta(q,\omega)}{\left[ \omega - \lambda^2 \mathrm{Re}\theta(q,\omega) \right]^2 + \left[ \lambda^2 \mathrm{Im}\theta(q,\omega) \right]^2} \end{align*}而
\begin{align*} \theta(q,\omega) =& \frac{1}{\Omega}\sum_{p< k_F} \frac{1}{\omega + \frac{\hbar^2}{4m}\left( p-q\right)^2-\delta} \end{align*}无量纲化
对以下量进行无量纲化:
\begin{align*} \tilde{\omega} = \frac{\omega}{\varepsilon_F} ;\quad \tilde{\delta} = \frac{\delta}{\varepsilon_F} ;\quad \tilde{p} = \frac{p}{k_F} ;\quad \\ \tilde{q} = \frac{q}{k_F}; \quad \tilde{\lambda}^2 = \frac{\lambda^2}{\Omega \varepsilon_F^2} \end{align*}原式就变为
\begin{align*} \theta(\tilde{q},\tilde{\omega}) =& \frac{1}{\Omega \varepsilon_F} \sum_{\tilde{p}< 1} \frac{1}{\tilde{\omega} + \frac{1}{2}(\tilde{p}-\tilde{q})^2 -\tilde{\delta} } \\ =&\frac{1}{\Omega\varepsilon_F}(B+ \mathrm{i}A) \end{align*}其中
\begin{align*} A =& \pi \sum_{\tilde{p}< 1} \delta\left( \tilde{\omega}-\tilde{\delta} + \frac{1}{2}(\tilde{p}-\tilde{q})^2 \right) \\ B =& \sum_{\tilde{p} < 1} \mathcal{P}\frac{1}{\tilde{\omega}-\tilde{\delta} + \frac{1}{2}(\tilde{p}-\tilde{q})^2 } \end{align*}Transition rate 就记为:
\begin{align*} \varepsilon_F I(\tilde{\omega}) =\sum_{\tilde{q}< 1} \frac{ \tilde{\lambda}^2A}{\left( \tilde{\omega} -\tilde{\lambda}^2 B\right)^2+\tilde{\lambda}^4 A^2} \end{align*}数值处理
Dirac Delta 函数和主值分别用其极限形式:
\begin{align*} \delta(x) =& \frac{1}{\pi} \lim_{b\to 0}\frac{b}{b^2+x^2} \\ \mathcal{P}\frac{1}{x} =& \lim_{b\to 0}\frac{x}{b^2+x^2} \end{align*}计算中 \(b=0.0001\) .
求和近似为从 \(0\) 到 \(1\) 的积分.
数值取值为
\begin{align*} \tilde{\lambda} = 1 \end{align*}结果
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结果如图
图中从上到下, \(\tilde{\delta}\) 在区间 \([8,-8]\) 每间隔 \(0.5\) 取一个值.
上图的峰一些尖. 如果 Dirac delta 函数中的参数 \(b\) 取得更大一些, 图就会平滑一些.
下图是 \(b=0.5\) 时的结果. 为了直观,线的间距也由 \(100\) 改为 \(10\) .
eps矢量图: fig4a.eps
零温解析
推导回顾
原公式为:
\begin{align*} I(\omega) \propto \mathrm{Im}\sum_{q< q_F}\frac{1}{\omega+ \mu +\varepsilon_{q,3}-\varepsilon_{q,2}-\lambda^2 \theta(q,\omega)} \end{align*}其中
\begin{align*} \theta(q,\omega) = \frac{1}{\Omega} \sum_{p< k_F} \frac{1}{\omega+ \mu + \varepsilon_{q,3}+\varepsilon_{p,1} - \varepsilon_{p+q,b}-\delta} \end{align*}进行化简, 首先在这个系统中有如下关系
\begin{align*} \mu + \varepsilon_{q,3} -\varepsilon_{q,2} = 0 \end{align*}所以原式可以化为:
\begin{align*} I(\omega) \propto \sum_{q< k_F} \frac{\lambda^2\mathrm{Im}\theta(q,\omega)}{\left[ \omega - \lambda^2 \mathrm{Re}\theta(q,\omega) \right]^2 + \left[ \lambda^2 \mathrm{Im}\theta(q,\omega) \right]^2} \end{align*}其中
\begin{align*} \theta(q,\omega) =& \frac{1}{\Omega}\sum_{p< k_F} \frac{1}{\omega + \frac{\hbar^2}{4m}\left( p-q\right)^2-\delta} \end{align*}对以下量进行无量纲化:
\begin{align*} \tilde{\omega} = \frac{\omega}{\varepsilon_F} ;\quad \tilde{\delta} = \frac{\delta}{\varepsilon_F} ;\quad \tilde{p} = \frac{p}{k_F} ;\quad \\ \tilde{q} = \frac{q}{k_F}; \quad \tilde{\lambda}^2 = \frac{\lambda^2}{\Omega \varepsilon_F^2} \end{align*}原式就变为
\begin{align} \varepsilon_F I(\tilde{\omega}) =\sum_{\tilde{q}< 1} \frac{ \tilde{\lambda}^2A}{\left( \tilde{\omega} -\tilde{\lambda}^2 B\right)^2+\tilde{\lambda}^4 A^2} \end{align}其中
\begin{align} A =& \pi \sum_{\tilde{p}< 1} \delta\left( \tilde{\omega}-\tilde{\delta} + \frac{1}{2}(\tilde{p}-\tilde{q})^2 \right) \\ B =& \sum_{\tilde{p} < 1} \mathcal{P}\frac{1}{\tilde{\omega}-\tilde{\delta} + \frac{1}{2}(\tilde{p}-\tilde{q})^2 } \end{align}解析计算
求和化积分
\begin{align*} A \approx& \pi \frac{Vk^3_F}{(2\pi)^3} \int_{|\tilde{p}| < 1}\mathrm{d}^3\tilde{p} \cdot \delta\left( \tilde{\omega}-\tilde{\delta} + \frac{1}{2}(\tilde{p}-\tilde{q})^2 \right) =\pi \frac{Vk^3_F}{(2\pi)^3} a\\ B \approx& \frac{Vk^3_F}{(2\pi)^3} \int_{|\tilde{p}| < 1}\mathrm{d}^3\tilde{p} \cdot\mathcal{P}\frac{1}{\tilde{\omega}-\tilde{\delta} + \frac{1}{2}(\tilde{p}-\tilde{q})^2 } = \frac{Vk^3_F}{(2\pi)^3} b \end{align*}解析的计算积分 \(a\) 和 \(b\)
\begin{align*} a =&2\pi \int_{-1}^{1}\mathrm{d}x \int_0^1\mathrm{d}\tilde{p} \cdot \delta\left( \tilde{\omega}-\tilde{\delta} + \frac{\tilde{p}^2}{2} +\frac{\tilde{q}^2}{2} -\tilde{p}\tilde{q}x \right) \tilde{p}^2\\ b =&2\pi\int_{-1}^{1}\mathrm{d}x \int_0^1\mathrm{d}\tilde{p} \cdot\mathcal{P}\frac{1}{\tilde{\omega}-\tilde{\delta} + \frac{\tilde{p}^2}{2} +\frac{\tilde{q}^2}{2} -\tilde{p}\tilde{q}x }\tilde{p}^2 \end{align*}积分 \(a\) 是一个对 \(\delta\) 函数在 \([0,1]\) 区间内的积分. 所以要对 \(\delta\) 函数内的 部分分情况讨论.
如果 \(\tilde{\omega}-\tilde{\delta} + \frac{\tilde{p}^2}{2} +\frac{\tilde{q}^2}{2} -\tilde{p}\tilde{q}x = 0\) 没有根, 那么积分 \(a=0\) . 如果有根, 记两个根为 \(r_1, r_2\) , 那么积分 \(a\) 化为
\begin{align*} a =&2\pi\int_{-1}^{1}\mathrm{d}x \int_0^1\mathrm{d}\tilde{p} \cdot \delta\left[ (\tilde{p}-r_1)(\tilde{p}-r_2) \right] \tilde{p}^2 \\ =&2\pi\frac{1}{|r_1-r_2|} \int_{-1}^{1}\mathrm{d}x \int_0^1\mathrm{d}\tilde{p} \cdot \left[ \delta(\tilde{p}-r_1) +\delta(\tilde{p}-r_2) \right] \tilde{p}^2 \end{align*}积分 \(b\) 是一个对主值的积分,也需要分情况讨论.
如果 \(\tilde{\omega}-\tilde{\delta} +\frac{\tilde{p}^2}{2} +\frac{\tilde{q}^2}{2} -\tilde{p}\tilde{q}x =0\) 没有根, 那么它就可以当做普通的积分
\begin{align*} b =&\int_0^1\mathrm{d}\tilde{p} \cdot \frac{1}{\tilde{\omega}-\tilde{\delta} + \frac{1}{2}(\tilde{p}-\tilde{q})^2 } \\ =&\frac{1}{\tilde{\omega}-\tilde{\delta}} \int_0^1 \mathrm{d}\tilde{p} \cdot \frac{1}{1+\left( \frac{\tilde{p}- \tilde{q}}{ \sqrt{2(\tilde{\omega} -\tilde{\delta})} } \right)^2} \\ =&\frac{1}{\tilde{\omega}-\tilde{\delta}} \cdot \left[ \arctan(t) \right]_{t==\frac{-\tilde{q}}{ \sqrt{2(\tilde{\omega} -\tilde{\delta})} }} ^{t=\frac{1-\tilde{q}}{ \sqrt{2(\tilde{\omega} -\tilde{\delta})} }} \end{align*}如有两个根 \(r_1, r_2\) 那么
\begin{align*} b =&\int_0^1\mathrm{d}\tilde{p} \cdot \mathcal{P} \frac{1}{(\tilde{p}-r_1)(\tilde{p}-r_2)} \\ =&\frac{1}{r_1-r_2}\int_0^1\mathrm{d}\tilde{p} \cdot\left[ \mathcal{P} \frac{1}{\tilde{p}-r_1}-\mathcal{P}\frac{1}{\tilde{p}-r_2}\right] \end{align*}两个根 \(r_1, r_2\) 是否在积分区间 \([0,1]\) 之间也会对积分 \(a, b\) 的结果有影响.
根据根的情况讨论积分结果
\begin{align}
\tilde{\omega}-\tilde{\delta} +\frac{1}{2}(\tilde{p}-\tilde{q})^2 = 0
\end{align}
如果有根, 记 \(r_1 = \tilde{q} + \sqrt{2(\tilde{\delta}-\tilde{\omega})}, r_1 = \tilde{q} + \sqrt{2(\tilde{\delta}-\tilde{\omega})}\) . 当 \(\tilde{p}, \tilde{\omega}\) 取不同的值时, 根的情况如图
no root 时
\begin{align*} a =&0 \\ b = &\frac{1}{\tilde{\omega}-\tilde{\delta}} \cdot \left[ \arctan(t) \right]_{t==\frac{-\tilde{q}}{ \sqrt{2(\tilde{\omega} -\tilde{\delta})a} }} ^{t=\frac{1-\tilde{q}}{ \sqrt{2(\tilde{\omega} -\tilde{\delta})} }} \end{align*}\(r_1, r_2 \in [0,1]\) 时
\begin{align*} a =& \frac{1}{ \sqrt{2(\tilde{\delta}-\tilde{\omega})} } \\ b =& \frac{1}{r_1-r_2} \ln \frac{ (1-r_1)r_2 }{ (1-r_2)r_1 } \end{align*}\(r_1\in [0,1]\) 和 \(r_2 \in [0,1]\) 时
\begin{align*} a =& \frac{1}{2 \sqrt{2(\tilde{\delta}-\tilde{\omega})} } \\ b =& \frac{1}{r_1-r_2} \ln \frac{ -(1-r_1)r_2 }{ (1-r_2)r_1 } \end{align*}\(r_1, r_2 \notin [0,1]\) 时
\begin{align*} a =&0 \\ b =& \frac{1}{r_1-r_2} \ln \frac{ (1-r_1)r_2 }{ (1-r_2)r_1 } \end{align*}代回原式
\begin{align*}
\varepsilon_F I(\tilde{\omega}) =&\sum_{\tilde{q}< 1} \frac{ \tilde{\lambda}^2A}{\left(
\tilde{\omega} -\tilde{\lambda}^2 B\right)^2+\tilde{\lambda}^4 A^2} \\
\approx& \int_0^1 \mathrm{d}\tilde{q}\cdot \frac{ \tilde{\lambda}^2 \pi A}{\left(
\frac{(2\pi)^3}{Vk_F}\tilde{\omega} -\tilde{\lambda}^2 B\right)^2+\tilde{\lambda}^4 \pi^2 A^2}
\end{align*}
积分和结果
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有限温公式推导
2 粒子的松原函数(有自能修正)
\begin{align*} G_2(k, \mathrm{i} \omega_2) =& \frac{1}{G_2^0(k,\mathrm{i}\omega_2)^{-1} -\lambda^2 \Sigma (k, \mathrm{i}\omega_2)} \\ =& \frac{1}{\mathrm{i}\omega_2 - \epsilon_{k,2} -\lambda^2 \Sigma (k, \mathrm{i}\omega_2)} \end{align*}3 粒子的松原函数(无相互作用)
\begin{align*} G_3^0(k,\mathrm{i}\omega_3) = \frac{1}{\mathrm{i}\omega_3 - \epsilon_{k,3}} \end{align*}代入总的松原函数
\begin{align*} R(k , \mathrm{i}\omega_n) =& \frac{1}{\beta}\sum_{\omega_2} G_3(k,\mathrm{i}(\omega_2-\omega_n)) G_2(k,\mathrm{i}\omega_2)\\ =& \frac{1}{\beta} \sum_{\omega_2} \frac{1}{\mathrm{i}(\omega_2 -\omega_n) - \epsilon_{k,3}}\cdot \frac{1}{\mathrm{i}\omega_2 - \epsilon_{k,2} -\lambda^2 \Sigma (k, \mathrm{i}\omega_2)} \\ =&\frac{1}{\beta} \sum_{\omega_2}\left( \frac{1}{\mathrm{i}(\omega_2 -\omega_n) - \epsilon_{k,3}} + \frac{-1}{\mathrm{i}\omega_2 - \epsilon_{k,2} -\lambda^2 \Sigma (k, \mathrm{i}\omega_2)} \right)\cdot \frac{1}{\mathrm{i}\omega_n +\epsilon_{k,3}-\epsilon_{k,2}-\lambda^2 \Sigma (k, \mathrm{i}\omega_2)} \end{align*}接下来单独计算第一项
\begin{align*} \frac{1}{\beta} \sum_{\omega_2} \frac{1}{\mathrm{i}(\omega_2 -\omega_n) - \epsilon_{k,3}}\cdot \frac{1}{\mathrm{i}\omega_n +\epsilon_{k,3}-\epsilon_{k,2}-\lambda^2 \Sigma (k, \mathrm{i}\omega_2)} \end{align*}利用环路积分计算此项. 积分
\begin{align*} \oint_{R\to\infty}\frac{1}{\mathrm{i}(\omega_2 -\omega_n) - \epsilon_{k,3}}\cdot \frac{1}{\mathrm{i}\omega_n +\epsilon_{k,3}-\epsilon_{k,2}-\lambda^2 \Sigma (k, \mathrm{i}\omega_2)} \cdot \frac{1}{e^{\mathrm{i}\beta\omega_2} +1}\mathrm{d}\omega_2 = 0 \end{align*}积分的留数为:
第三项对应的留数
\begin{align*} R_3 = \frac{1}{\beta} \sum_{\omega_2} \frac{1}{\mathrm{i}(\omega_2 -\omega_n) - \epsilon_{k,3}}\cdot \frac{1}{\mathrm{i}\omega_n +\epsilon_{k,3}-\epsilon_{k,2}-\lambda^2 \Sigma (k, \mathrm{i}\omega_2)} \end{align*}其中 \(\omega_2 = (2m+1)\pi/\beta\) . 就是要求的总的松原函数 \(R(k,\mathrm{i}\omega)\) .
第一项对应的留数
\begin{align*} R_1 =& \frac{1}{\mathrm{i}\omega_n +\epsilon_{k,3}-\epsilon_{k,2}-\lambda^2 \Sigma (k, \mathrm{i}\omega_n +\epsilon_{k,3})} \cdot \frac{1}{e^{\beta(\mathrm{i}\omega_n + \epsilon_{k,3})} +1} \\ =& \frac{ f(\epsilon_{k,3}) }{\mathrm{i}\omega_n +\epsilon_{k,3}-\epsilon_{k,2}-\lambda^2 \Sigma (k, \mathrm{i}\omega_n +\epsilon_{k,3})} \end{align*}有限温及零温的最终重复结果
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参考文献
Junjun Xu, Qiang Gu, and Erich J. Mueller Phys. Rev. A 88, 023604 (2013)