超导
库珀对
$$\hat{Z}_{\boldsymbol{k}}^\dagger=\hat{c}_{\boldsymbol{k},\uparrow}^\dagger\hat{c}_{-\boldsymbol{k},\downarrow}^\dagger$$
相互作用哈密顿量
$$\hat{H}^{(2)}=\frac{1}{2V}\sum_{\boldsymbol{q,k,k’}}\sum_{\sigma,\sigma’}U(\boldsymbol{q})\hat{c}_{\boldsymbol{k+q},\sigma}^\dagger\hat{c}_{\boldsymbol{k’-q},\sigma’}^\dagger\hat{c}_{\boldsymbol{k’},\sigma’}\hat{c}_{\boldsymbol{k},\sigma}$$
简化图景$\boldsymbol{k}=-\boldsymbol{k’},\sigma=-\sigma’,U(\boldsymbol{q})=U<0$，引入$\displaystyle\hat{\Delta}\equiv \frac{U}{V}\sum_{\boldsymbol{k}}\hat{Z}_{\boldsymbol{k}}$
$$\hat{H}^{(2)}=\frac{U}{2V}\sum_{\boldsymbol{q,k}}\sum_{\sigma}\hat{c}_{\boldsymbol{k+q},\sigma}^\dagger\hat{c}_{\boldsymbol{-k-q},-\sigma}^\dagger\hat{c}_{\boldsymbol{-k},-\sigma}\hat{c}_{\boldsymbol{k},\sigma}\\
\qquad\\
=\frac{U}{V}\left(\sum_{\boldsymbol{k}}\hat{Z}_{\boldsymbol{k}}^\dagger\right)\left(\sum_{\boldsymbol{k}}\hat{Z}_{\boldsymbol{k}}\right)=\frac{V}{U}\hat{\Delta}^\dagger\hat{\Delta}$$
无相互作用费米气体$\text{NIF}$基态
$$\ket{g}=\prod_{|\boldsymbol{k}|<k_F}\hat{c}_{\boldsymbol{k},\uparrow}^\dagger\hat{c}_{-\boldsymbol{k},\downarrow}^\dagger\ket{0}
=\prod_{|\boldsymbol{k}|<k_F}\hat{Z}_{\boldsymbol{k}}^\dagger\ket{0}$$
引入$\ket{g_\boldsymbol{k}}=(u_\boldsymbol{k}+v_\boldsymbol{k}\hat{Z}_{\boldsymbol{k}}^\dagger)\ket{0}=u_\boldsymbol{k}\ket{0}+v_\boldsymbol{k}\ket{2}$，$u_\boldsymbol{k}^*u_\boldsymbol{k}+v_{\boldsymbol{k}}^*v_{\boldsymbol{k}}=1$，取试探波函数为
$$\ket{g_S}=\prod_{\boldsymbol{k}}\ket{g_\boldsymbol{k}}$$
在$\ket{2},\ket{0}$基下有
$$\hat{Z}_{\boldsymbol{k}}^\dagger=\begin{pmatrix}
0&1\\0&0
\end{pmatrix}\qquad \hat{Z}_{\boldsymbol{k}}=\begin{pmatrix}
0&0\\1&0
\end{pmatrix}\qquad \hat{n}_\boldsymbol{k}+\hat{n}_\boldsymbol{-k}=\begin{pmatrix}
2&0\\0&0
\end{pmatrix}$$
巨正则系综一阶哈密顿量
$$\hat{H}^{(1)}=\sum_{\boldsymbol{k},\sigma}E(\boldsymbol{k})\hat{c}_{\boldsymbol{k},\sigma}^\dagger\hat{c}_{\boldsymbol{k},\sigma}-\mu\hat{N}=\sum_{\boldsymbol{k},\sigma}\varepsilon_\boldsymbol{k}\hat{n}_{\boldsymbol{k},\sigma}$$
总动能
$$E_K=\bra{g_S}\hat{H}^{(1)}\ket{g_S}=2\sum_{\boldsymbol{k}}\varepsilon_\boldsymbol{k}|v_\boldsymbol{k}|^2$$
引入$\displaystyle\Delta=\langle \hat{\Delta}\rangle=\frac{U}{V}\sum_{\boldsymbol{k}}u_\boldsymbol{k}^*v_\boldsymbol{k}$，由于$\displaystyle \sum_{\boldsymbol{k,k’}}\langle \hat{\Delta}^\dagger\hat{\Delta}\rangle$贡献主要来自$\boldsymbol{k}\neq \boldsymbol{k’}$，势能为
$$E_P=\frac{V}{U}\langle{\hat{\Delta}^\dagger\hat{\Delta}}\rangle\approx \frac{V}{U}|\Delta|^2\\
\qquad\\
=\frac{U}{V}\left|\sum_{\boldsymbol{k}}u_\boldsymbol{k}^*v_\boldsymbol{k}\right|^2$$
总能量对$u_\boldsymbol{k},v_\boldsymbol{k}$变分为
$$E=2\sum_{\boldsymbol{k}}\varepsilon_\boldsymbol{k}|v_\boldsymbol{k}|^2+\frac{U}{V}\left|\sum_{\boldsymbol{k}}u_\boldsymbol{k}^*v_\boldsymbol{k}\right|^2\\
\qquad\\
\delta E=2\varepsilon_\boldsymbol{k}v_\boldsymbol{k}^*\delta v_\boldsymbol{k}+\Delta^*u_\boldsymbol{k}^*\delta v_\boldsymbol{k}+\Delta v_\boldsymbol{k}^*\delta u_\boldsymbol{k}\\
\qquad\\ \require{enclose}
\overset{u_\boldsymbol{k}^*\delta u_\boldsymbol{k}+v_\boldsymbol{k}^*\delta v_\boldsymbol{k}=0}{=}(-2\varepsilon_\boldsymbol{k}u_\boldsymbol{k}^*-\Delta^*u_\boldsymbol{k}^{*2}/ v_\boldsymbol{k}^*+\Delta v_\boldsymbol{k}^*)\delta u_\boldsymbol{k}\\
\qquad\\
\frac{\delta E}{\delta u_\boldsymbol{k}}=0\Rightarrow 2\varepsilon_\boldsymbol{k}u_\boldsymbol{k}^*v_\boldsymbol{k}\overset{\enclose{downdiagonalstrike}{\Delta^*}}{=}\Delta(|v_\boldsymbol{k}|^2-|u_\boldsymbol{k}|^2)\\
\qquad\\
4|u_\boldsymbol{k}|^2|v_\boldsymbol{k}|^2=\frac{|\Delta|^2}{\varepsilon_\boldsymbol{k}^2+|\Delta|^2}$$
由非超导$\Delta=0$的$\text{NIF}$基态知此二次方程根中$|v_\boldsymbol{k}|^2$取负号
$$
|u_\boldsymbol{k}|^2,|v_\boldsymbol{k}|^2=\frac{1}{2}\left(1\pm \frac{\varepsilon_\boldsymbol{k}}{\sqrt{\varepsilon_\boldsymbol{k}^2+|\Delta|^2}}\right)\\
\qquad\\
u_\boldsymbol{k}^*v_\boldsymbol{k}=-\frac{1}{2}\frac{\Delta}{\sqrt{\varepsilon_\boldsymbol{k}^2+|\Delta|^2}}\\
\qquad\\
\Delta=\frac{U}{V}\sum_{\boldsymbol{k}}u_\boldsymbol{k}^*v_\boldsymbol{k}=\frac{U}{V}\sum_{\boldsymbol{k}}-\frac{1}{2}\frac{\Delta}{\sqrt{\varepsilon_\boldsymbol{k}^2+|\Delta|^2}}\\
\qquad\\
=\frac{|U|}{2}\int\frac{\dd\boldsymbol{k}}{(2\pi)^3}\frac{\Delta}{\sqrt{\varepsilon_\boldsymbol{k}^2+|\Delta|^2}}=\frac{|U|}{2}\int\frac{\Delta\nu(\varepsilon)\dd\varepsilon}{\sqrt{\varepsilon^2+|\Delta|^2}}$$
能量$\varepsilon$对应的态密度$\nu(\varepsilon)$与$\text{NIF}$接近，截止能量$E_{\text{cut}}>>|\Delta|$
$$\frac{|U|}{2}\int\frac{\nu(\varepsilon)\dd\varepsilon}{\sqrt{\varepsilon^2+|\Delta|^2}}\approx \frac{|U|\nu(0)}{2}\int_{-E_{\text{cut}}}^{E_{\text{cut}}}\frac{\dd\varepsilon}{\sqrt{\varepsilon^2+|\Delta|^2}}\\
\qquad\\
=|U|\nu(0)\ln{\frac{2E_{\text{cut}}}{|\Delta|}}=1\\
\qquad\\
|\Delta|=2E_{\text{cut}}\exp\left\{-\frac{1}{|U|\nu(0)} \right\}
$$
$$$$