我们定义虚时格林函数如下：
\[ \begin{array}{l} G_{i j}(\tau, \tau)=\left\langle c_{i}(\tau) c_{j}^{\dagger}(\tau)\right\rangle, \qquad G_{i j}(\tau, 0)=\left\langle c_{i}(\tau) c_{j}^{\dagger}(0)\right\rangle,\\ G_{i j}(0, \tau)=\left\langle T_{\tau} c_{i}(0) c_{j}^{\dagger}(\tau)\right\rangle \equiv-\left\langle c_{j}^{\dagger}(\tau) c_{i}(0)\right\rangle, \qquad G_{i j}(0,0)=\left\langle c_{i}(0) c_{j}^{\dagger}(0)\right\rangle \end{array} \]注意，不同地方的格林函数的定义可能与此处相差一个负号。
此前我们已经定义了诸如：
\[ \mathbf{B}^{\uparrow}\left(l_{2} \Delta \tau, l_{1} \Delta \tau\right)=\prod_{l=l_{1}+1}^{l_{2}} e^{\alpha \operatorname{Diag}\left(\vec{S}_{l}\right)} e^{-\Delta \tau T} \]为了方便，我们此处定义带费米算符的：
\[ \mathbf{U}^{\uparrow}\left(l_{2} \Delta \tau, l_{1} \Delta \tau\right)=\prod_{l=l_{1}+1}^{l_{2}} e^{ \hat{\mathbf{c}}^{\dagger} \alpha \operatorname{Diag}\left(\vec{S}_{l}\right)\hat{\mathbf{c}}} e^{-\Delta \tau \hat{\mathbf{c}}^{\dagger}T\hat{\mathbf{c}}} \]对于某观测量来说，有：
\[ \langle \hat{O}(\tau)\rangle=\frac{\operatorname{Tr}\left[e^{-\beta H} \hat{O}(\tau)\right]}{\operatorname{Tr}\left[e^{-\beta H}\right]}=\sum_{C} \mathrm{P}_{C}\langle \hat{O}(\tau)\rangle_{C}+O\left(\Delta_{\tau}^{2}\right) \]其中：
\[ \mathrm{P}_{C}=\frac{\operatorname{det}\left(1+B_{C}(\beta, 0)\right)}{\sum_{C} \operatorname{det}\left(1+B_{C}(\beta, 0)\right)}, \quad \langle\hat{O}(\tau)\rangle_{\mathcal{C}}=\frac{\operatorname{Tr}\left\{\hat{U}_{\mathcal{C}}(\beta, \tau) \hat{O} \hat{U}_{\mathcal{C}}(\tau, 0)\right\}}{\operatorname{Tr}\left\{\hat{U}_{\mathcal{C}}(\beta, 0)\right\}} \]假设该单体算符为： \(\hat{O}=\hat{\mathbf{c}}^{\dagger} O \hat{\mathbf{c}}\) ，则：
\[ \begin{aligned} \langle\hat{O}\rangle_{\mathcal{C}} &=\left.\frac{\partial}{\partial \eta} \ln \operatorname{Tr}\left[\hat{U}_{\mathcal{C}}(\beta, \tau) e^{\eta \hat{O}} \hat{U}_{\mathcal{C}}(\tau, 0)\right]\right|_{\eta=0} \\ &=\left.\frac{\partial}{\partial \eta} \ln \operatorname{Det}\left[\mathbf{1}+B_{\mathcal{C}}(\beta, \tau) e^{\eta O} B_{\mathcal{C}}(\tau, 0)\right]\right|_{\eta=0} \\ &=\left.\frac{\partial}{\partial \eta} \operatorname{Tr} \ln \left[\mathbf{1}+B_{\mathcal{C}}(\beta, \tau) e^{\eta O} B_{\mathcal{C}}(\tau, 0)\right]\right|_{\eta=0} \\ &=\operatorname{Tr}\left[B_{\mathcal{C}}(\tau, 0)\left(1+B_{\mathcal{C}}(\beta, 0)\right)^{-1} B_{\mathcal{C}}(\beta, \tau) O\right] \\ &=\operatorname{Tr}\left[\left(1-\left(1+B_{\mathcal{C}}(\tau, 0) B_{\mathcal{C}}(\beta, \tau)\right)^{-1}\right) O\right] \end{aligned} \]最后一个等号用到了Sherman-Morrison公式：
\[ \mathbf{U}\left(\mathbf{I}_{k}+\mathbf{V U}\right)^{-1} \mathbf{V} =\mathbf{I}-(\mathbf{I}+\mathbf{U V})^{-1} \]其中， \(\mathbf{U}\) 是 \(N \times k\) 的矩阵， \(\mathbf{V}\) 是 \(k \times N\) 的矩阵。证明见文末。
对于等时格林函数 \(\left\langle\hat{c}_{i} \hat{c}_{j}^{\dagger}\right\rangle\) 来说， \(\hat{O}=\delta_{i j}-\hat{c}_{j}^{\dagger} \hat{c}_{i}\) 。有：
\[ \mathbf{G}(\tau, \tau)=[\mathbf{1}+\mathbf{B}(\tau, 0) \mathbf{B}(\beta, \tau)]^{-1} \]顺便，因为有 \(A B^{-1} C= \left(C^{-1} B A^{-1}\right)^{-1}\) ，根据等时格林函数的定义我们能很容易的验证有：
\[ B_{C}\left(\tau_{1}, \tau_{2}\right) G_{C}\left(\tau_{2}, \tau_{2}\right) B_{C}^{-1}\left(\tau_{1}, \tau_{2}\right)=G_{C}\left(\tau_{1}, \tau_{1}\right) \]对于非等时格林函数：
\[ G_{C}\left(\tau_{1}, \tau_{2}\right)_{x, y}=\left\langle T c_{x}\left(\tau_{1}\right) c_{y}^{\dagger}\left(\tau_{2}\right)\right\rangle_{C}=\left\{\begin{array}{c} \left\langle c_{x}\left(\tau_{1}\right) c_{y}^{\dagger}\left(\tau_{2}\right)\right\rangle_{C} \text { if } \tau_{1} \geq \tau_{2} \\ -\left\langle c_{y}^{\dagger}\left(\tau_{2}\right) c_{x}\left(\tau_{1}\right)\right\rangle_{C} \text { if } \tau_{1}<\tau_{2} \end{array}\right. \]我们这里首先证明一个qm常考题，设
\[ c_x(\tau)=e^{\tau \mathbf{c}^{\dagger} A c} c_{x} e^{-\tau c^{\dagger} A c} \]求一次导得：
\[ \frac{\partial c_x(\tau)}{\partial \tau}=e^{\tau c^{\dagger} A c}\left[\boldsymbol{c}^{\dagger} A \boldsymbol{c}, c_{x}\right] e^{-\tau \boldsymbol{c}^{\dagger} A \boldsymbol{c}}=-\sum_{z} A_{x, z} c_{z}(\tau) \]同理n阶导的结果可知。于是根据泰勒展开：
\(c_x(\tau)=c_x - \tau \left(A\boldsymbol{c}(\tau)\right)_x + \tau^2 \left(A^2\boldsymbol{c}(\tau)\right)_x+ \cdots = \left(e^{-\tau A} c\right)_{x}\)
取 \(\tau=1\) ，有：
\[ e^{ c^{\dagger} A c} c_{x} e^{- c^{\dagger} A c}=\left(e^{-A} c\right)_{x} \]类似的，我们有：
\[ e^{ c^{\dagger} A c} c^{\dagger}_{x} e^{- c^{\dagger} A c}=\left(\boldsymbol{c}^{\dagger} e^{A}\right)_{x} \]由此我们知道：
\[ \begin{array}{l} U_{C}^{-1}\left(\tau_{1}, \tau_{2}\right) c_{x} U_{C}\left(\tau_{1}, \tau_{2}\right)=\left(B_{C}\left(\tau_{1}, \tau_{2}\right) c\right)_{x} \\ U_{C}^{-1}\left(\tau_{1}, \tau_{2}\right) c_{x}^{\dagger} U_{C}\left(\tau_{1}, \tau_{2}\right)=\left(c^{\dagger} B_{C}^{-1}\left(\tau_{1}, \tau_{2}\right)\right)_{x} \end{array} \]于是我们有：
\[ \begin{aligned} \left\langle c_{x}\left(\tau_{1}\right) c_{y}^{\dagger}\left(\tau_{2}\right)\right\rangle_{C} &=\frac{\operatorname{Tr}\left[U_{C}\left(\beta, \tau_{1}\right) c_{x} U_{C}\left(\tau_{1}, \tau_{2}\right) c_{y}^{\dagger} U_{C}\left(\tau_{2}, 0\right)\right]}{\operatorname{Tr}\left[U_{C}(\beta, 0)\right]} \\ &=\frac{\operatorname{Tr}\left[U_{C}\left(\beta, \tau_{2}\right) U_{C}^{-1}\left(\tau_{1}, \tau_{2}\right) c_{x} U_{C}\left(\tau_{1}, \tau_{2}\right) c_{y}^{\dagger} U_{C}\left(\tau_{2}, 0\right)\right]}{\operatorname{Tr}\left[U_{C}(\beta, 0)\right]}\\ &=\sum_z B_{C}\left(\tau_{1}, \tau_{2}\right)_{xz} \frac{\operatorname{Tr}\left[ U_{C}\left(\beta, \tau_{2}\right) c_{z} c_{y}^{\dagger} U_{C}\left(\tau_{2}, 0\right) \right]}{\operatorname{Tr}\left[U_{C}(\beta, 0)\right]}\\ &=\left[B_{C}\left(\tau_{1}, \tau_{2}\right) G_{C}\left(\tau_{2}, \tau_{2}\right)\right]_{x, y} \qquad \qquad \qquad \text{当} \; \tau_{1}>\tau_{2} \end{aligned} \]类似的，对于 \(\tau_{2}>\tau_{1}\) 的情况
\[ G_{C}\left(\tau_{1}, \tau_{2}\right)_{x, y}=-\left\langle c_{y}^{\dagger}\left(\tau_{2}\right) c_{x}\left(\tau_{1}\right)\right\rangle_{C}=-\left[\left(1-G_{C}\left(\tau_{1}, \tau_{1}\right)\right) B_{C}^{-1}\left(\tau_{2}, \tau_{1}\right)\right]_{x, y} \]而对于其他物理量，如 \(\left\langle c_{x}^{\dagger}\left(\tau_{1}\right) c_{x}\left(\tau_{1}\right) c_{y}^{\dagger}\left(\tau_{2}\right) c_{y}\left(\tau_{2}\right)\right\rangle\) ，我们通过Wick定理，转化为两点格林函数乘积的加减，同样能通过二点格林函数的值来得到。
附录
随手抄的一个Sherman-Morrison 公式的证明：
\[ \label{eq7} (\mathbf{A}+\mathbf{U V})^{-1}=\mathbf{A}^{-1}-\mathbf{A}^{-1} \mathbf{U}\left(\mathbf{I}_{k}+\mathbf{V A^{-1} U}\right)^{-1} \mathbf{V}\mathbf{A}^{-1} \]其中 \(\mathbf{A}\) 是 \(N \times N\) 的矩阵， \(\mathbf{U}\) 是 \(N \times k\) 的矩阵， \(\mathbf{V}\) 是 \(k \times N\) 的矩阵。
证明如下：
令 \(\xi=\mathbf{V} x\) 且有 \(\left(\mathbf{A}+\mathbf{U V}\right) x=b\) ，其中 \(x\) 和 \(b\) 是 \(N \times 1\) 的向量， 而 \(\xi\) 是 \(k \times 1\) 的向量.
写下如下形式： \(\left(\mathbf{A}+\mathbf{U V}\right) x=b\) ：
\[ \left[\begin{array}{cc} \mathbf{A} & \mathbf{U} \\ \mathbf{V} & -\mathbf{I_{k}} \end{array}\right]\left[\begin{array}{l} x \\ \xi \end{array}\right]=\left[\begin{array}{l} b \\ 0 \end{array}\right] \]进行分解：
\[ \left[\begin{array}{cc} \mathbf{A} & \mathbf{U} \\ \mathbf{V} & -\mathbf{I_{k}} \end{array}\right]=\left[\begin{array}{cc} \mathbf{I_{N}} & 0 \\ \mathbf{V} \mathbf{A}^{-1} & \mathbf{I_{k}} \end{array}\right]\left[\begin{array}{cc} \mathbf{A} & \mathbf{U} \\ 0 & -\mathbf{I_{k}}-\mathbf{V} \mathbf{A}^{-1} \mathbf{U} \end{array}\right] \]利用
\[ \label{eq6} \left[\begin{array}{cc} \mathbf{I_{N}} & 0 \\ \mathbf{B} & \mathbf{I_{k}} \end{array}\right]^{-1}=\left[\begin{array}{cc} \mathbf{I_{N}} & 0 \\ -\mathbf{B} & \mathbf{I_{k}} \end{array}\right] \]\[ \left[\begin{array}{cc} \mathbf{A} & \mathbf{U} \\ 0 & -\mathbf{I_{k}}-\mathbf{V} \mathbf{A}^{-1} \mathbf{U} \end{array}\right]\left[\begin{array}{l} x \\ \xi \end{array}\right]=\left[\begin{array}{cc} \mathbf{I_{N}} & 0 \\ -\mathbf{V} \mathbf{A}^{-1} & \mathbf{I_{k}} \end{array}\right]\left[\begin{array}{l} b \\ 0 \end{array}\right] \]我们有：
\[ \mathbf{A}x+\mathbf{U}\xi=b \qquad \left(-\mathbf{I_{k}}-\mathbf{V} \mathbf{A}^{-1} \mathbf{U}\right)\xi =-\mathbf{V} \mathbf{A}^{-1}b \]则：
\[ x=\mathbf{A}^{-1} \left(b-\mathbf{U}\xi\right) \qquad \xi =-\left(-\mathbf{I_{k}}-\mathbf{V} \mathbf{A}^{-1} \mathbf{U}\right)^{-1}\mathbf{V} \mathbf{A}^{-1}b \]则：
\[ x=\left[\mathbf{A}^{-1}-\mathbf{A}^{-1} \mathbf{U}\left(\mathbf{I}_{k}+\mathbf{V A^{-1} U}\right)^{-1} \mathbf{V}\mathbf{A}^{-1}\right]b \]或：
\[ (\mathbf{A}+\mathbf{U V})^{-1}=\mathbf{A}^{-1}-\mathbf{A}^{-1} \mathbf{U}\left(\mathbf{I}_{k}+\mathbf{V A^{-1} U}\right)^{-1} \mathbf{V}\mathbf{A}^{-1} \]因为 \(\left(\mathbf{A}+\mathbf{U V}\right) x=b\)
接下来只需令 \(\mathbf{A}=\mathbf{I_N}\) ，即有：
\[ (\mathbf{I}+\mathbf{U V})^{-1}=\mathbf{I}-\mathbf{U}\left(\mathbf{I}_{k}+\mathbf{V U}\right)^{-1} \mathbf{V} \label{eq:70} \]（写回答那种小引用怎么搞啊…还没学会…不加ref了，问就是全抄师祖的。我们不生产知识，我们只是祖师爷们知识的搬运工。）