我们定义虚时格林函数如下：
\[\begin{equation}\begin{array}{l}
G_{i j}(\tau, \tau)=\left\langle c_{i}(\tau) c_{j}^{\dagger}(\tau)\right\rangle, \qquad G_{i j}(\tau, 0)=\left\langle c_{i}(\tau) c_{j}^{\dagger}(0)\right\rangle,\\ G_{i j}(0, \tau)=\left\langle T_{\tau} c_{i}(0) c_{j}^{\dagger}(\tau)\right\rangle \equiv-\left\langle c_{j}^{\dagger}(\tau) c_{i}(0)\right\rangle, G_{i j}(0,0)=\left\langle c_{i}(0) c_{j}^{\dagger}(0)\right\rangle
\end{array}\end{equation}\]
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, G_{i j}(0,0)=\left\langle c_{i}(0) c_{j}^{\dagger}(0)\right\rangle
\end{array}\end{equation}\]
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：
\[\begin{equation}\mathbf{B}^{\uparrow}\left(l_{2} \Delta \tau, l_{1} \Delta \tau\right)=\prod_{l=l_{1}+1}^{l_{2}} e^{\alpha \operatorname{Diag}\left(\vec{S}_{l}\right)} e^{-\Delta \tau T}\end{equation}\]
为了方便，我们此处定义带费米算符的：
\[\begin{equation}\mathbf{U}^{\uparrow}\left(l_{2} \Delta \tau, l_{1} \Delta \tau\right)=\prod_{l=l_{1}+1}^{l_{2}} e^{ \hat{\mathbf{c}}^{\dagger} \alpha \operatorname{Diag}\left(\vec{S}_{l}\right)\hat{\mathbf{c}}} e^{-\Delta \tau \hat{\mathbf{c}}^{\dagger}T\hat{\mathbf{c}}}\end{equation}\]
对于某观测量来说，有：
\[\begin{equation}\langle \hat{O}(\tau)\rangle=\frac{\operatorname{Tr}\left[e^{-\beta H} \hat{O}(\tau)\right]}{\operatorname{Tr}\left[e^{-\beta H}\right]}=\sum_{C} \mathrm{P}_{C}\langle \hat{O}(\tau)\rangle_{C}+O\left(\Delta_{\tau}^{2}\right)\end{equation}\]
其中：
\[\begin{equation}\mathrm{P}_{C}=\frac{\operatorname{det}\left(1+B_{C}(\beta, 0)\right)}{\sum_{C} \operatorname{det}\left(1+B_{C}(\beta, 0)\right)}, \quad \langle\hat{O}(\tau)\rangle_{\mathcal{C}}=\frac{\operatorname{Tr}\left\{\hat{U}_{\mathcal{C}}(\beta, \tau) \hat{O} \hat{U}_{\mathcal{C}}(\tau, 0)\right\}}{\operatorname{Tr}\left\{\hat{U}_{\mathcal{C}}(\beta, 0)\right\}}\end{equation}\]
假设该单体算符为：$\hat{O}=\hat{\mathbf{c}}^{\dagger} O \hat{\mathbf{c}}$，则：
\[\begin{equation}\begin{aligned}
\langle\hat{O}\rangle_{\mathcal{C}} &=\left.\frac{\partial}{\partial \eta} \ln \operatorname{Tr}\left[\hat{U}_{\mathcal{C}}(\beta, \tau) e^{\eta \hat{O}} \hat{U}_{\mathcal{C}}(\tau, 0)\right]\right|_{\eta=0} \\
&=\left.\frac{\partial}{\partial \eta} \ln \operatorname{Det}\left[\mathbf{1}+B_{\mathcal{C}}(\beta, \tau) e^{\eta O} B_{\mathcal{C}}(\tau, 0)\right]\right|_{\eta=0} \\
&=\left.\frac{\partial}{\partial \eta} \operatorname{Tr} \ln \left[\mathbf{1}+B_{\mathcal{C}}(\beta, \tau) e^{\eta O} B_{\mathcal{C}}(\tau, 0)\right]\right|_{\eta=0} \\
&=\operatorname{Tr}\left[B_{\mathcal{C}}(\tau, 0)\left(1+B_{\mathcal{C}}(\beta, 0)\right)^{-1} B_{\mathcal{C}}(\beta, \tau) O\right] \\
&=\operatorname{Tr}\left[\left(1-\left(1+B_{\mathcal{C}}(\tau, 0) B_{\mathcal{C}}(\beta, \tau)\right)^{-1}\right) O\right]
\end{aligned}\end{equation}\]
\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}\end{equation}\]
最后一个等号用到了Sherman-Morrison公式：
\[\begin{equation}
\mathbf{U}\left(\mathbf{I}_{k}+\mathbf{V U}\right)^{-1} \mathbf{V} =\mathbf{I}-(\mathbf{I}+\mathbf{U V})^{-1}
\end{equation}\]
\mathbf{U}\left(\mathbf{I}_{k}+\mathbf{V U}\right)^{-1} \mathbf{V} =\mathbf{I}-(\mathbf{I}+\mathbf{U V})^{-1}
\end{equation}\]
$\mathbf{U}$ 是 $N \times k$ 的矩阵， $\mathbf{V}$ 是 $k \times N$ 的矩阵。证明见附录。
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：
\[\begin{equation}G(\tau, \tau)=[\mathbf{1}+B(\tau, 0) B(\beta, \tau)]^{-1}\end{equation}\]
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：
\[\begin{equation}B_{C}\left(\tau_{1}, \tau_{2}\right) G_{C}\left(\tau_{2}, \tau_{2}\right) B_{C}^{-1}\left(\tau_{1}, \tau_{2}\right)=G_{C}\left(\tau_{1}, \tau_{1}\right)\end{equation}\]
对于非等时格林函数
\[\begin{equation}G_{C}\left(\tau_{1}, \tau_{2}\right)_{x, y}=\left\langle T c_{x}\left(\tau_{1}\right) c_{y}^{\dagger}\left(\tau_{2}\right)\right\rangle_{C}=\left\{\begin{array}{c}
\left\langle c_{x}\left(\tau_{1}\right) c_{y}^{\dagger}\left(\tau_{2}\right)\right\rangle_{C} \text { if } \tau_{1} \geq \tau_{2} \\
-\left\langle c_{y}^{\dagger}\left(\tau_{2}\right) c_{x}\left(\tau_{1}\right)\right\rangle_{C} \text { if } \tau_{1}<\tau_{2}
\end{array}\right.\end{equation}\]
\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.\end{equation}\]
我们这里首先证明一个qm常考题，设
\[\begin{equation}c_x(\tau)=e^{\tau \mathbf{c}^{\dagger} A c} c_{x} e^{-\tau c^{\dagger} A c}\end{equation}\]
求一次导得：
\[\begin{equation}\frac{\partial c_x(\tau)}{\partial \tau}=e^{\tau c^{\dagger} A c}\left[\boldsymbol{c}^{\dagger} A \boldsymbol{c}, c_{x}\right] e^{-\tau \boldsymbol{c}^{\dagger} A \boldsymbol{c}}=-\sum_{z} A_{x, z} c_{z}(\tau)\end{equation}\]
同理n阶导的结果可知。于是根据泰勒展开：
\[\begin{equation} c_x(\tau)=c_x - \tau \left(A\boldsymbol{c}(\tau)\right)_x + \tau^2 \left(A^2\boldsymbol{c}(\tau)\right)_x+ \cdots = \left(e^{-\tau A} c\right)_{x}\end{equation}\]
取 $\tau=1$，有：
\[\begin{equation}e^{ c^{\dagger} A c} c_{x} e^{- c^{\dagger} A c}=\left(e^{-A} c\right)_{x}\end{equation}\]
类似的，我们有：
\[\begin{equation}e^{ c^{\dagger} A c} c^{\dagger}_{x} e^{- c^{\dagger} A c}=\left(\boldsymbol{c}^{\dagger} e^{A}\right)_{x}\end{equation}\]
由此我们知道：
\[\begin{equation}\begin{array}{l}
U_{C}^{-1}\left(\tau_{1}, \tau_{2}\right) c_{x} U_{C}\left(\tau_{1}, \tau_{2}\right)=\left(B_{C}\left(\tau_{1}, \tau_{2}\right) c\right)_{x} \\
U_{C}^{-1}\left(\tau_{1}, \tau_{2}\right) c_{x}^{\dagger} U_{C}\left(\tau_{1}, \tau_{2}\right)=\left(c^{\dagger} B_{C}^{-1}\left(\tau_{1}, \tau_{2}\right)\right)_{x}
\end{array}\end{equation}\]
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}\end{equation}\]
于是我们有：
\[\begin{equation}\begin{aligned}
\left\langle c_{x}\left(\tau_{1}\right) c_{y}^{\dagger}\left(\tau_{2}\right)\right\rangle_{C} &=\frac{\operatorname{Tr}\left[U_{C}\left(\beta, \tau_{1}\right) c_{x} U_{C}\left(\tau_{1}, \tau_{2}\right) c_{y}^{\dagger} U_{C}\left(\tau_{2}, 0\right)\right]}{\operatorname{Tr}\left[U_{C}(\beta, 0)\right]} \\
&=\frac{\operatorname{Tr}\left[U_{C}\left(\beta, \tau_{2}\right) U_{C}^{-1}\left(\tau_{1}, \tau_{2}\right) c_{x} U_{C}\left(\tau_{1}, \tau_{2}\right) c_{y}^{\dagger} U_{C}\left(\tau_{2}, 0\right)\right]}{\operatorname{Tr}\left[U_{C}(\beta, 0)\right]}\\
&=\sum_z B_{C}\left(\tau_{1}, \tau_{2}\right)_{xz} \frac{\operatorname{Tr}\left[ U_{C}\left(\beta, \tau_{2}\right) c_{z} c_{y}^{\dagger} U_{C}\left(\tau_{2}, 0\right) \right]}{\operatorname{Tr}\left[U_{C}(\beta, 0)\right]}\\
&=\left[B_{C}\left(\tau_{1}, \tau_{2}\right) G_{C}\left(\tau_{2}, \tau_{2}\right)\right]_{x, y} \qquad \qquad \qquad \text{当} \; \tau_{1}>\tau_{2}
\end{aligned}\end{equation}\]
\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}\end{equation}\]
类似的，对于 $\tau_{2}>\tau_{1}$的情况
\[\begin{equation}G_{C}\left(\tau_{1}, \tau_{2}\right)_{x, y}=-\left\langle c_{y}^{\dagger}\left(\tau_{2}\right) c_{x}\left(\tau_{1}\right)\right\rangle_{C}=-\left[\left(1-G_{C}\left(\tau_{1}, \tau_{1}\right)\right) B_{C}^{-1}\left(\tau_{2}, \tau_{1}\right)\right]_{x, y}\end{equation}\]
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。
在前文我们已经知道，对于某个辅助场的构型，对应的费米子部分的权重为：
\[\begin{equation}
\prod_{\sigma=\uparrow \downarrow} \operatorname{Det}\left[\mathbf{I}+ \mathbf{B}_{C}^{\sigma}(\beta, 0)\right]
\end{equation}\]
\prod_{\sigma=\uparrow \downarrow} \operatorname{Det}\left[\mathbf{I}+ \mathbf{B}_{C}^{\sigma}(\beta, 0)\right]
\end{equation}\]
因为中间的各种矩阵都是关于自旋上下块的，所以我们可以分开考虑自旋上下的情况，最后再把上和下的权重乘起来。
根据定义，我们知道某个辅助场构型对应的权重为 $\operatorname{det}[\mathbf{I}+\mathbf{B}(\beta, \tau) \mathbf{B}(\tau, 0)]$ ，而更新了$\tau$层（ $\tau =l \Delta \tau$ ）的辅助场后的权重为：
\[\begin{equation}
\begin{aligned}
& \operatorname{det}[\mathbf{I}+\mathbf{B}(\beta, \tau) e^{V\left(\vec{S}^{\prime}_{l}\right)} e^{-\Delta \tau T} \mathbf{B}(\tau -\Delta \tau, 0)]\\
= & \operatorname{det}[\mathbf{I}+\mathbf{B}(\beta, \tau) e^{V\left(\vec{S}^{\prime}_{l}\right)} e^{-V\left(\vec{S}_{l}\right)} \mathbf{B}(\tau , 0)]\\
=&\operatorname{det}[\mathbf{I}+\mathbf{B}(\beta, \tau) (\mathbf{I}+\mathbf{\Delta}) \mathbf{B}(\tau , 0)]
\end{aligned}
\end{equation}\]
\begin{aligned}
& \operatorname{det}[\mathbf{I}+\mathbf{B}(\beta, \tau) e^{V\left(\vec{S}^{\prime}_{l}\right)} e^{-\Delta \tau T} \mathbf{B}(\tau -\Delta \tau, 0)]\\
= & \operatorname{det}[\mathbf{I}+\mathbf{B}(\beta, \tau) e^{V\left(\vec{S}^{\prime}_{l}\right)} e^{-V\left(\vec{S}_{l}\right)} \mathbf{B}(\tau , 0)]\\
=&\operatorname{det}[\mathbf{I}+\mathbf{B}(\beta, \tau) (\mathbf{I}+\mathbf{\Delta}) \mathbf{B}(\tau , 0)]
\end{aligned}
\end{equation}\]
其中， $\mathbf{\Delta}=e^{V\left(\vec{S}^{\prime}_{l}\right)} e^{-V\left(\vec{S}_{l}\right)}-\mathbf{I}$ 。我们注意到，对于诸如Hubbard等on-site的相互作用的形式， $\mathbf{\Delta}=e^{V\left(\vec{S}^{\prime}_{l}\right)-V\left(\vec{S}_{l}\right)} -\mathbf{I}$ 。当改变 $L \times L \times L_{\tau}$ 中的一个site的辅助场 $s_{i,l}$ 的值的时候， $\mathbf{\Delta}$ 矩阵只有一个元素非零，且为 $e^{\pm \alpha \Delta s_{i,l}}-1$ 。其中的 $\pm$ 对应自旋上下。
因此我们在计算更新前后的接受概率（权重的比值）时，有：
\[\begin{equation}
\begin{aligned}
& \frac{\operatorname{det}[\mathbf{I}+\mathbf{B}(\beta, \tau)(\mathbf{I}+\Delta) \mathbf{B}(\tau, 0)]}{\operatorname{det}[\mathbf{I}+\mathbf{B}(\beta, \tau) \mathbf{B}(\tau, 0)]} \\
=& \frac{\operatorname{det}[\mathbf{I}+\mathbf{B}(\beta, 0)+\mathbf{B}(\beta, \tau) \Delta \mathbf{B}(\tau, 0)]}{\operatorname{det}[\mathbf{I}+\mathbf{B}(\beta, 0)]} \\
=& \operatorname{det}\left[\mathbf{I}+\mathbf{B}(\beta, \tau) \Delta \mathbf{B}(\tau, 0)(\mathbf{I}+\mathbf{B}(\beta, 0))^{-1}\right] \\
=& \operatorname{det}\left[\mathbf{I}+\Delta \mathbf{B}(\tau, 0)(\mathbf{I}+\mathbf{B}(\beta, 0))^{-1} \mathbf{B}(\beta, \tau)\right] \\
=& \operatorname{det}\left[\mathbf{I}+\boldsymbol{\Delta}\left(\mathbf{I}-(\mathbf{I}+\mathbf{B}(\tau, 0) \mathbf{B}(\beta, \tau))^{-1}\right)\right] \\
=& \operatorname{det}[\mathbf{I}+\mathbf{\Delta}(\mathbf{I}-\mathbf{G}(\tau, \tau))]
\end{aligned}
\end{equation}\]
\begin{aligned}
& \frac{\operatorname{det}[\mathbf{I}+\mathbf{B}(\beta, \tau)(\mathbf{I}+\Delta) \mathbf{B}(\tau, 0)]}{\operatorname{det}[\mathbf{I}+\mathbf{B}(\beta, \tau) \mathbf{B}(\tau, 0)]} \\
=& \frac{\operatorname{det}[\mathbf{I}+\mathbf{B}(\beta, 0)+\mathbf{B}(\beta, \tau) \Delta \mathbf{B}(\tau, 0)]}{\operatorname{det}[\mathbf{I}+\mathbf{B}(\beta, 0)]} \\
=& \operatorname{det}\left[\mathbf{I}+\mathbf{B}(\beta, \tau) \Delta \mathbf{B}(\tau, 0)(\mathbf{I}+\mathbf{B}(\beta, 0))^{-1}\right] \\
=& \operatorname{det}\left[\mathbf{I}+\Delta \mathbf{B}(\tau, 0)(\mathbf{I}+\mathbf{B}(\beta, 0))^{-1} \mathbf{B}(\beta, \tau)\right] \\
=& \operatorname{det}\left[\mathbf{I}+\boldsymbol{\Delta}\left(\mathbf{I}-(\mathbf{I}+\mathbf{B}(\tau, 0) \mathbf{B}(\beta, \tau))^{-1}\right)\right] \\
=& \operatorname{det}[\mathbf{I}+\mathbf{\Delta}(\mathbf{I}-\mathbf{G}(\tau, \tau))]
\end{aligned}
\end{equation}\]
其中第四个等号用到了Sherman-Morrison公式：
\[\begin{equation}
(\mathbf{I}+\mathbf{U V})^{-1}=\mathbf{I}-\mathbf{U}\left(\mathbf{I}_{k}+\mathbf{V U}\right)^{-1} \mathbf{V}
\end{equation}\]
(\mathbf{I}+\mathbf{U V})^{-1}=\mathbf{I}-\mathbf{U}\left(\mathbf{I}_{k}+\mathbf{V U}\right)^{-1} \mathbf{V}
\end{equation}\]
$\mathbf{U}$ 是 $N \times k$ 的矩阵， $\mathbf{V}$ 是 $k \times N$ 的矩阵。证明见附录。
因为此处的 $\mathbf{\Delta}$ 矩阵只有一个元素非零，则接受概率实际上可以通过如下方式简单计算：
\[\begin{equation}
\begin{aligned}
r=&\operatorname{det}[\mathbf{I}_{N \times N}+\mathbf{\Delta}_{N \times N}(\mathbf{I}-\mathbf{G}(\tau, \tau))_{N \times N}] \\
=& \operatorname{det}[\mathbf{I}_{N \times N}+\mathbf{\Delta}_{N \times 1}(\mathbf{I}-\mathbf{G}(\tau, \tau))_{1 \times N}]\\
=& \operatorname{det}[\mathbf{I}_{1 \times 1}+(\mathbf{I}-\mathbf{G}(\tau, \tau))_{1 \times N} \mathbf{\Delta}_{N \times 1}]\\
=& \operatorname{det}[\mathbf{I}_{1 \times 1}+(\mathbf{I}-\mathbf{G}(\tau, \tau))_{1 \times 1} \mathbf{\Delta}_{1 \times 1}]\\
=& 1+\mathbf{\Delta}_{i i}\left(1-\mathbf{G}_{i i}\right)
\end{aligned}
\end{equation}\]
\begin{aligned}
r=&\operatorname{det}[\mathbf{I}_{N \times N}+\mathbf{\Delta}_{N \times N}(\mathbf{I}-\mathbf{G}(\tau, \tau))_{N \times N}] \\
=& \operatorname{det}[\mathbf{I}_{N \times N}+\mathbf{\Delta}_{N \times 1}(\mathbf{I}-\mathbf{G}(\tau, \tau))_{1 \times N}]\\
=& \operatorname{det}[\mathbf{I}_{1 \times 1}+(\mathbf{I}-\mathbf{G}(\tau, \tau))_{1 \times N} \mathbf{\Delta}_{N \times 1}]\\
=& \operatorname{det}[\mathbf{I}_{1 \times 1}+(\mathbf{I}-\mathbf{G}(\tau, \tau))_{1 \times 1} \mathbf{\Delta}_{1 \times 1}]\\
=& 1+\mathbf{\Delta}_{i i}\left(1-\mathbf{G}_{i i}\right)
\end{aligned}
\end{equation}\]
其中第二个等号利用了公式
\[\begin{equation}
\det(I_{m}+AB)=\det(I_{n}+BA)
\label{eq1}
\end{equation}\]
\det(I_{m}+AB)=\det(I_{n}+BA)
\label{eq1}
\end{equation}\]
，证明见附录：
\[\begin{equation}
\det(\mathbf{I}_{N}+\mathbf{U V})=\det(\mathbf{I}_{k}+\mathbf{ VU})
\end{equation}\]
\det(\mathbf{I}_{N}+\mathbf{U V})=\det(\mathbf{I}_{k}+\mathbf{ VU})
\end{equation}\]
比如自旋上部分的权重比就为 $1+(e^{\alpha \Delta s_{i,l}}-1)(1-G^{\uparrow}(\tau,\tau)_{i,i})$ ，自旋下部分 $1+(e^{-\alpha \Delta s_{i,l}}-1)(1-G^{\downarrow}(\tau,\tau)_{i,i})$ 。我们看到，因为Hubbard的特殊形式，导致我们其实在有对应等时格林函数的值的时候，不需要计算一个巨大矩阵的乘法再加上行列式，只需要计算一个数，即能得到接受概率。
而这样的方式需要我们随时更新的等时格林函数的数值。根据定义：
\[\begin{equation}
\begin{aligned}
\mathbf{G}^{\prime}(\tau, \tau) &=[\mathbf{I}+(\mathbf{I}+\boldsymbol{\Delta}) \mathbf{B}(\tau, 0) \mathbf{B}(\beta, \tau)]^{-1} \\
&=\mathbf{G}(\tau, \tau)[(\mathbf{I}+(\mathbf{I}+\boldsymbol{\Delta}) \mathbf{B}(\tau, 0) \mathbf{B}(\beta, \tau)) \mathbf{G}(\tau, \tau)]^{-1} \\
&=\mathbf{G}(\tau, \tau)\left[\left(\mathbf{I}+(\mathbf{I}+\boldsymbol{\Delta})\left(\mathbf{G}^{-1}(\tau, \tau)-\mathbf{I}\right)\right) \mathbf{G}(\tau, \tau)\right]^{-1} \\
&=\mathbf{G}[\mathbf{I}+\boldsymbol{\Delta}(\mathbf{I}-\mathbf{G})]^{-1}
\end{aligned}
\end{equation}\]
\begin{aligned}
\mathbf{G}^{\prime}(\tau, \tau) &=[\mathbf{I}+(\mathbf{I}+\boldsymbol{\Delta}) \mathbf{B}(\tau, 0) \mathbf{B}(\beta, \tau)]^{-1} \\
&=\mathbf{G}(\tau, \tau)[(\mathbf{I}+(\mathbf{I}+\boldsymbol{\Delta}) \mathbf{B}(\tau, 0) \mathbf{B}(\beta, \tau)) \mathbf{G}(\tau, \tau)]^{-1} \\
&=\mathbf{G}(\tau, \tau)\left[\left(\mathbf{I}+(\mathbf{I}+\boldsymbol{\Delta})\left(\mathbf{G}^{-1}(\tau, \tau)-\mathbf{I}\right)\right) \mathbf{G}(\tau, \tau)\right]^{-1} \\
&=\mathbf{G}[\mathbf{I}+\boldsymbol{\Delta}(\mathbf{I}-\mathbf{G})]^{-1}
\end{aligned}
\end{equation}\]
同样的利用Sherman-Morrison公式：
\[\begin{equation}
(\mathbf{I}+\mathbf{U V})^{-1}=\mathbf{I}-\mathbf{U}\left(\mathbf{I}_{k}+\mathbf{V U}\right)^{-1} \mathbf{V}
\end{equation}\]
(\mathbf{I}+\mathbf{U V})^{-1}=\mathbf{I}-\mathbf{U}\left(\mathbf{I}_{k}+\mathbf{V U}\right)^{-1} \mathbf{V}
\end{equation}\]
我们有：
\[\begin{equation}
\begin{aligned}
\mathbf{G}^{\prime}(\tau, \tau) &=\mathbf{G}_{N \times N
}(\mathbf{I}_{N \times N}+\boldsymbol{\Delta}_{N \times N} (\mathbf{I}-\mathbf{G})_{N \times N})^{-1} \\
&=\mathbf{G}_{N \times N}(\mathbf{I}_{N \times N}+\boldsymbol{\Delta}_{N \times 1} (\mathbf{I}-\mathbf{G})_{1 \times N})^{-1} \\
&\left.=\mathbf{G}_{N \times N}\left[\mathbf{I}_{N \times N}-\boldsymbol{\Delta}_{N \times 1}\left(\mathbf{I}_{1 \times 1}+(\mathbf{I}-\mathbf{G})_{1 \times N} \boldsymbol{\Delta}_{N \times 1}\right)^{-1} (\mathbf{I}-\mathbf{G})_{1 \times N}\right)\right] \\
&=\mathbf{G}_{N \times N}-\mathbf{G}_{N \times N} \boldsymbol{\Delta}_{N \times 1} \; \frac{1}{r}\; (\mathbf{I}-\mathbf{G})_{1 \times N} \\
&=\mathbf{G}_{N \times N}-\mathbf{G}_{N \times 1} \boldsymbol{\Delta}_{1 \times 1} \; \frac{1}{r}\; (\mathbf{I}-\mathbf{G})_{1 \times N}\\
&=\mathbf{G}_{N \times N}-\mathbf{G}(:,i) \boldsymbol{\Delta}_{ii} \; \frac{1}{r}\; (\mathbf{I}-\mathbf{G})(i,:)
\end{aligned}
\end{equation}\]
\begin{aligned}
\mathbf{G}^{\prime}(\tau, \tau) &=\mathbf{G}_{N \times N
}(\mathbf{I}_{N \times N}+\boldsymbol{\Delta}_{N \times N} (\mathbf{I}-\mathbf{G})_{N \times N})^{-1} \\
&=\mathbf{G}_{N \times N}(\mathbf{I}_{N \times N}+\boldsymbol{\Delta}_{N \times 1} (\mathbf{I}-\mathbf{G})_{1 \times N})^{-1} \\
&\left.=\mathbf{G}_{N \times N}\left[\mathbf{I}_{N \times N}-\boldsymbol{\Delta}_{N \times 1}\left(\mathbf{I}_{1 \times 1}+(\mathbf{I}-\mathbf{G})_{1 \times N} \boldsymbol{\Delta}_{N \times 1}\right)^{-1} (\mathbf{I}-\mathbf{G})_{1 \times N}\right)\right] \\
&=\mathbf{G}_{N \times N}-\mathbf{G}_{N \times N} \boldsymbol{\Delta}_{N \times 1} \; \frac{1}{r}\; (\mathbf{I}-\mathbf{G})_{1 \times N} \\
&=\mathbf{G}_{N \times N}-\mathbf{G}_{N \times 1} \boldsymbol{\Delta}_{1 \times 1} \; \frac{1}{r}\; (\mathbf{I}-\mathbf{G})_{1 \times N}\\
&=\mathbf{G}_{N \times N}-\mathbf{G}(:,i) \boldsymbol{\Delta}_{ii} \; \frac{1}{r}\; (\mathbf{I}-\mathbf{G})(i,:)
\end{aligned}
\end{equation}\]
在更新完某一虚时层的辅助场之后，我们转而尝试更新下一层的辅助场，此时将得到新虚时位置的格林函数，从格林函数的定义我们知道：
\[\begin{equation}
\mathbf{G}(\tau+\Delta \tau, \tau+\Delta \tau)=\mathbf{B}(\tau+\Delta \tau, \tau) \mathbf{G}(\tau, \tau) \mathbf{B}^{-1}(\tau+\Delta \tau, \tau)
\end{equation}\]
\mathbf{G}(\tau+\Delta \tau, \tau+\Delta \tau)=\mathbf{B}(\tau+\Delta \tau, \tau) \mathbf{G}(\tau, \tau) \mathbf{B}^{-1}(\tau+\Delta \tau, \tau)
\end{equation}\]
类似的，我们有：
\[\begin{equation}
\mathbf{G}(\tau-\Delta \tau, \tau-\Delta \tau)=\mathbf{B}^{-1}(\tau, \tau-\Delta \tau) \mathbf{G}(\tau, \tau) \mathbf{B}(\tau, \tau -\Delta \tau)
\end{equation}\]
\mathbf{G}(\tau-\Delta \tau, \tau-\Delta \tau)=\mathbf{B}^{-1}(\tau, \tau-\Delta \tau) \mathbf{G}(\tau, \tau) \mathbf{B}(\tau, \tau -\Delta \tau)
\end{equation}\]
我们实际计算中，经常从第1层到第 $L_\tau$ 层的辅助场逐层逐个尝试更新，在每一层更新的时候，计算该层的格林函数 $G^{\uparrow / \downarrow}(\tau,\tau)$ 和相应的更新的权重 $r^{\uparrow} r^{\downarrow}$ 。然后在将要进行下一层更新的时候，用传递的方式得到下一层的格林函数 $\mathbf{G}(\tau+\Delta \tau, \tau+\Delta \tau)$ ，即我们只需知道该虚时层所对应的格林函数，而非时刻知道所有的等时格林函数的值。在进行了从第1层到第 $L_\tau$ 层的sweep之后，我们从第 $L_\tau$ 层到第1层再反过来一遍，中间的行为是类似的。
附录
Theorem 1
\[\begin{equation}
\det(I_{m}+AB)=\det(I_{n}+BA)
\end{equation}\]
\det(I_{m}+AB)=\det(I_{n}+BA)
\end{equation}\]
其中，$A$ 为 $m \times n$ 的矩阵，$B$ 为 $n \times m$ 的矩阵，$I_{m}$ 为 $m \times m$ 的单位阵.
Proof:
\[\begin{equation}\begin{aligned}
&\left|I_{m}+A B\right|=\left|\begin{array}{cc}
I_{m}+A B & 0 \\
B & I_{n}
\end{array}\right|=\left|\begin{array}{ccc}
I_{m} & -A \\
0 & I_{n}
\end{array}\right|\left|\begin{array}{ccc}
I_{m}+A B & 0 \\
B & I_{n}
\end{array}\right|\\
&=\left|\begin{array}{cc}
I_{m} & -A \\
B & I_{n}
\end{array}\right|=\left|\begin{array}{cc}
I_{m} & 0 \\
-B & I_{n}
\end{array}\right|\left|\begin{array}{cc}
I_{m} & -A \\
B & I_{n}
\end{array}\right|=\left|\begin{array}{cc}
I_{m} & -A \\
0 & I_{n}+B A
\end{array}\right|\\
&=\left|I_{n}+B A\right|
\end{aligned}
\label{eqeq}
\end{equation}\]
&\left|I_{m}+A B\right|=\left|\begin{array}{cc}
I_{m}+A B & 0 \\
B & I_{n}
\end{array}\right|=\left|\begin{array}{ccc}
I_{m} & -A \\
0 & I_{n}
\end{array}\right|\left|\begin{array}{ccc}
I_{m}+A B & 0 \\
B & I_{n}
\end{array}\right|\\
&=\left|\begin{array}{cc}
I_{m} & -A \\
B & I_{n}
\end{array}\right|=\left|\begin{array}{cc}
I_{m} & 0 \\
-B & I_{n}
\end{array}\right|\left|\begin{array}{cc}
I_{m} & -A \\
B & I_{n}
\end{array}\right|=\left|\begin{array}{cc}
I_{m} & -A \\
0 & I_{n}+B A
\end{array}\right|\\
&=\left|I_{n}+B A\right|
\end{aligned}
\label{eqeq}
\end{equation}\]
Theorem 2
Sherman-Morrison 公式:
\[\begin{equation}
(\mathbf{A}+\mathbf{U V})^{-1}=\mathbf{A}^{-1}-\mathbf{A}^{-1} \mathbf{U}\left(\mathbf{I}_{k}+\mathbf{V A^{-1} U}\right)^{-1} \mathbf{V}\mathbf{A}^{-1}\end{equation}\]
(\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}\end{equation}\]
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。
Proof:
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：$\left(\mathbf{A}+\mathbf{U V}\right) x=b$ ：
\[\begin{equation}\left[\begin{array}{cc}
\mathbf{A} & \mathbf{U} \\
\mathbf{V} & -\mathbf{I_{k}}
\end{array}\right]\left[\begin{array}{l}
x \\
\xi
\end{array}\right]=\left[\begin{array}{l}
b \\
0
\end{array}\right]\end{equation}\]
\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]\end{equation}\]
进行分解：
\[\begin{equation}\left[\begin{array}{cc}
\mathbf{A} & \mathbf{U} \\
\mathbf{V} & -\mathbf{I_{k}}
\end{array}\right]=\left[\begin{array}{cc}
\mathbf{I_{N}} & 0 \\
\mathbf{V} \mathbf{A}^{-1} & \mathbf{I_{k}}
\end{array}\right]\left[\begin{array}{cc}
\mathbf{A} & \mathbf{U} \\
0 & -\mathbf{I_{k}}-\mathbf{V} \mathbf{A}^{-1} \mathbf{U}
\end{array}\right]\end{equation}\]
\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]\end{equation}\]
利用
\[\begin{equation}\label{eq6}
\left[\begin{array}{cc}
\mathbf{I_{N}} & 0 \\
\mathbf{B} & \mathbf{I_{k}}
\end{array}\right]^{-1}=\left[\begin{array}{cc}
\mathbf{I_{N}} & 0 \\
-\mathbf{B} & \mathbf{I_{k}}
\end{array}\right]
\end{equation}\]
\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]
\end{equation}\]
\[\begin{equation}\left[\begin{array}{cc}
\mathbf{A} & \mathbf{U} \\
0 & -\mathbf{I_{k}}-\mathbf{V} \mathbf{A}^{-1} \mathbf{U}
\end{array}\right]\left[\begin{array}{l}
x \\
\xi
\end{array}\right]=\left[\begin{array}{cc}
\mathbf{I_{N}} & 0 \\
-\mathbf{V} \mathbf{A}^{-1} & \mathbf{I_{k}}
\end{array}\right]\left[\begin{array}{l}
b \\
0
\end{array}\right]\end{equation}\]
\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]\end{equation}\]
我们有：
\[\begin{equation}
\mathbf{A}x+\mathbf{U}\xi=b \qquad \left(-\mathbf{I_{k}}-\mathbf{V} \mathbf{A}^{-1} \mathbf{U}\right)\xi =-\mathbf{V} \mathbf{A}^{-1}b
\end{equation}\]
\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
\end{equation}\]
则：
\[\begin{equation}
x=\mathbf{A}^{-1} \left(b-\mathbf{U}\xi\right) \qquad \xi =-\left(-\mathbf{I_{k}}-\mathbf{V} \mathbf{A}^{-1} \mathbf{U}\right)^{-1}\mathbf{V} \mathbf{A}^{-1}b
\end{equation}\]
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
\end{equation}\]
则：
\[\begin{equation}
x=\left[\mathbf{A}^{-1}-\mathbf{A}^{-1} \mathbf{U}\left(\mathbf{I}_{k}+\mathbf{V A^{-1} U}\right)^{-1} \mathbf{V}\mathbf{A}^{-1}\right]b
\end{equation}\]
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
\end{equation}\]
或：
\[\begin{equation}(\mathbf{A}+\mathbf{U V})^{-1}=\mathbf{A}^{-1}-\mathbf{A}^{-1} \mathbf{U}\left(\mathbf{I}_{k}+\mathbf{V A^{-1} U}\right)^{-1} \mathbf{V}\mathbf{A}^{-1}\end{equation}\]
因为 $\left(\mathbf{A}+\mathbf{U V}\right) x=b$.
接下来只需令 $\mathbf{A}=\mathbf{I_N}$，即有：
\[\begin{equation}
(\mathbf{I}+\mathbf{U V})^{-1}=\mathbf{I}-\mathbf{U}\left(\mathbf{I}_{k}+\mathbf{V U}\right)^{-1} \mathbf{V}
\label{eq:70}
\end{equation}\]
(\mathbf{I}+\mathbf{U V})^{-1}=\mathbf{I}-\mathbf{U}\left(\mathbf{I}_{k}+\mathbf{V U}\right)^{-1} \mathbf{V}
\label{eq:70}
\end{equation}\]