考虑一个多体态

$$\psi(z_1,z_2\dots) = \frac{1}{\sqrt{N!}}\sum_{P}(-1)^P\phi_{P(m_1)}(z_1)\phi_{P(m_2)}(z_2)\dots$$

以及CF波函数

$$\psi_{CF}(\{z_i,\eta_i\}) = \frac{1}{\sqrt{N!}}\exp(\sum_i \frac{z_i\bar\eta_i}{2l_b^2})\sum_{P'}(-1)^{P'}\varphi_{P(m_1)}(z_1,\eta_1)\dots$$

对于前两个 $\Lambda$ level 填满的情况

$$\varphi_{m}(z,\eta)= \exp(\frac{z\bar\eta}{2l_b^2}) \left\{ \begin{aligned} &z^{m}, &0\leq m<N/2\\ &\frac{l_B}{\sqrt \gamma}(2mz^{m-1}-\frac{\bar\eta}{l^2}z^{m}), & N/2\leq m<N \end{aligned} \right.$$

Bergman 空间这个波函数的归一化系数为

$$A = \frac{2\pi l_B l_b}{\sqrt{2^{m+2}\pi^2 m! l_b^2l^{2(m+1)}}} = \frac{l_B}{\sqrt{2^m m!}l^{m+1}}$$

其中 $\frac{1}{l^2} = \frac{1}{l_B^2}-\frac{1}{l_b^2},\,\gamma = \frac{1}{q},\, \tilde\nu = \frac{q}{2q+1}$。

那么overlap就是

$$\begin{aligned} &\int\prod_i\mu_B(z_i)\psi^*\int \prod_id\mu_b(\eta_i)\prod_{i<j}(\eta_i-\eta_j)^2\psi_{CF}(\{z_i,\eta_i\})\\ =&\sum_\alpha C_\alpha^* \int\prod_i d\mu_b(\eta_i)\prod_{i<j}(\eta_i-\eta_j)^2\\ &\frac{1}{N!}\sum_{P,P'}(-1)^{P+P'} \int\prod_i \left[\exp(\frac{z_i\bar\eta_i}{2l_b^2})d\mu_B(z_i)\right]\phi_{m_{P(1)})}^*(z_1)\varphi_{m_{P'(1)}}(z_1,\eta_1)\dots \\ =&\sum_\alpha C_\alpha^* \int\prod_i d\mu_b(\eta_i)\prod_{i<j}(\eta_i-\eta_j)^2\\ &\frac{1}{N!}\sum_{P,P'}(-1)^{P+P'} A_{P(1),P'(1)}(\eta_1)A_{P(2),P'(2)}(\eta_2)\dots \\ \end{aligned}$$

这里有一个trick，整个表达式实际上对于 $\eta_i$ 是轮换对称的，前面 $\eta_i$ 和 $z_i$ 同时出现，让我忽略了这一点。那么，那么可以考虑将 $\eta_i$ 的指标换成 $P(i)$ ，那么

$$\begin{aligned} =&\frac{1}{N!}\sum_{P,P'}(-1)^{P+P'}\sum_\alpha C_\alpha^* \int\prod_i d\mu_b(\eta_{P(i)})\prod_{i<j}(\eta_{P(i)}-\eta_{P(j)})^2\\ &A_{P(1),P'(1)}(\eta_{P(1)})A_{P(2),P'(2)}(\eta_{P(2)})\dots \\ =&\frac{1}{N!}\sum_{P}(-1)^{P}\sum_\alpha C_\alpha^* \int\prod_i d\mu_b(\eta_{P(i)})\prod_{i<j}(\eta_{P(i)}-\eta_{P(j)})^2\\ &\det(A_{P(i) j})\dots \\ =&\sum_\alpha C_\alpha^* \int\prod_i d\mu_b(\eta_i)\prod_{i<j}(\eta_{i}-\eta_{j})^2\det(A_{ij}) \end{aligned}$$

其中

$$\begin{aligned} A_{ij}=A_{i,j}(\eta_i)= & \int d\mu_B(z)\exp(\frac{z\bar\eta_i}{2l_b^2})\phi_{m_i}^*(z)\varphi_{m_j}(z,\eta_i)\\ =& \left\{ \begin{aligned} &\int d\mu_B(z) \exp(\frac{z\bar\eta_i}{2l_b^2}){\bar z}^{m_i} z^{m_j}, &0\leq j<N/2\\ &\int d\mu_B(z) \exp(\frac{z\bar\eta}{2l_b^2}){\bar z}^{m_i}(2m_jz^{m_j-1}-\frac{\bar\eta_i}{l^2}z^{m_j}), & N/2\leq j<N \end{aligned} \right. \end{aligned}$$

$d\mu_B(z) = \frac{d^2z}{2\pi l_B^2}\exp(-|z|^2/2l_B^2)$

$A_{ij}(\eta_i)$ 的计算

$0\leq m_j<N/2$

$$\begin{aligned} &\int d\mu_B(z) \exp(\frac{z\bar\eta_i}{2l_b^2}){\bar z}^{m_i} z^{m_j} = \sum_{k}\frac{1}{k!}\left(\frac{\bar\eta}{2l_b^2}\right)^{k}\int d\mu_B(z)z^{m_j+k}\bar z^{m_i} \\ = &\sum_k\frac{1}{k!}\left(\frac{\bar\eta}{2l_b^2}\right)^{k}(2l_B^2)^{m_i} m_i!\delta_{m_i,m_j+k} \\ =& \frac{1}{(m_i-m_j)!}\left(\frac{\bar\eta}{2l_b^2}\right)^{m_i-m_j}(2l_B^2)^{m_i} m_i! \end{aligned}$$

这里 $m_i\geq m_j$。$\phi = z^m$ 和 $\varphi = z^m \exp(\frac{z\eta}{2l_b^2})$ 的归一化系数分别为 $\frac{1}{\sqrt{2^m m!}l_B^m}$ 和 $\frac{l_B}{\sqrt{2^m m!}l^{m+1}}$ 从而

$$A_{ij}=\frac{\sqrt{2^{m_j}m_i!}}{\sqrt{2^{m_i}m_j!}}\frac{1}{(m_i-m_j)!}(\frac{\bar\eta}{l_b})^{m_i-m_j}\frac{(l_B)^{m_i+1}}{(l_b)^{m_i-m_j}}\frac{1}{l^{m_j+1}}$$

$N/2\leq m_j<N$

$$\begin{aligned} &\int d\mu_B(z) \exp(\frac{z\bar\eta}{2l_b^2}){\bar z}^{m_i}(2m_jz^{m_j-1}-\frac{\bar\eta}{l^2}z^{m_j}) \\ =&\sum_k \frac{1}{k!}\left(\frac{\bar\eta}{2l_b^2}\right)^k\int d\mu_B(z) {\bar z}^{m_i}(2m_jz^{m_j-1+k}-\frac{\bar\eta}{l^2}z^{m_j+k}) \\ =&\sum_k\frac{1}{k!}\left(\frac{\bar\eta}{2l_b^2}\right)^k (2l_B^2)^{m_i}m_i!(2n\delta_{m_i,m_j-1+k}-\frac{\bar\eta}{l^2}\delta_{m_i,m_j+k}) \\ =&\frac{1}{(m_i-m_j+1)!}\left(\frac{\bar\eta}{2l_b^2}\right)^{m_i-m_j+1} (2l_B^2)^{m_i}m_j! 2m_j-\frac{1}{(m_i-m_j)!}\left(\frac{\bar\eta}{2l_b^2}\right)^{m_i-m_j} (2l_B^2)^{m_i}m_j!\frac{\bar\eta}{l^2} \\ =&\frac{1}{(m_i-m_j)!}\left(\frac{\bar\eta}{2l_b^2}\right)^{m_i-m_j+1} (2l_B^2)^{m_i}m_i!\left(\frac{2m_j}{m_i-m_j+1}-\frac{2l_b^2}{l^2}\right) \end{aligned}$$

这里 $m_i\geq m_j$, 当 $m_i=m_j-1$时，只保留第一项，当 $m_i<m_j-1$ 时为零。 考虑归一化因子 $\frac{1}{\sqrt{2^{m_i} m_i!}l_B^{m_i}}\frac{l_B}{\sqrt{2^{m_j} m_j!}l^{m_j+1}}\frac{l_B}{\sqrt{\gamma}}$

$$A_{ij} = \frac{1}{2\sqrt{\gamma}}\sqrt{\frac{2^{m_j}m_i!}{2^{m_i}m_j!}}\frac{1}{(m_i-m_j)!}\left(\frac{\bar\eta}{l_b}\right)^{m_i-m_j+1}\frac{l_B^{2m_i+2}}{l^{m_j+1}l_b^{m_i-m_j+1}}\left(\frac{2m_j}{m_i-m_j+1}-\frac{2l_b^2}{l^2}\right)$$

两个CF波函数的内积（算符期望值）

如果考虑算符的期望值，需要计算

$$\begin{aligned} &\hat H \psi_{CF}(\{z_i,\eta_i\}) \\ =&\sum_{n_1n_2n_1'n_2'}H_{n_1n_2n_1'n_2'}a_{n_1}^{\dagger}a_{n_2}^{\dagger}a_{n_1'}a_{n_2'}\frac{1}{\sqrt{N!}}\sum_P(-1)^P \,\varphi_{m_{P(1)}}(z_1,\bar\eta_1)\dots \end{aligned}$$

这个不是很好算，考虑将CF波函数用电子的朗道能级波函数展开 LLL 的

$$z^m\exp(\frac{z\bar\eta}{2l_b^2}) = \sum_n z^{m+n}(\frac{\bar\eta}{2l_b^2})^n\frac{1}{n!}$$

n lambda level, k landau level

$$\int d\mu_B(z) z^{m-k}L_k^{m-k}(\frac{|z|^2}{2l_B^2}) \exp(\frac{z\bar \eta}{2l_b^2})\left(2l_B^2\partial_z+(\frac{l_B^2}{l_b^2}-1)\bar\eta\right)^n f(z)$$

Appendix

$$\begin{aligned} &(a^\dagger)^n \exp(\frac{\bar\eta z}{2l_b^2})f(z) \\ = &\frac{1}{\sqrt{\gamma^n}l_B^n}(2l_B^2\partial_z-\bar\eta)^nf(z) \exp(\frac{\bar\eta z}{2l_b^2})\\ = &\frac{1}{\sqrt{\gamma^n}l_B^n}\sum_{k=0}^n(-\bar\eta)^{n-k}(2l_B^2\partial_z)^kf(z) \exp(\frac{\bar\eta z}{2l_b^2}) \\ = &\frac{1}{\sqrt{\gamma^n}l_B^n}\sum_{k=0}^n(-\bar\eta)^{n-k}(2l_B^2)^k\exp(\frac{\bar\eta z}{2l_b^2})(\partial_z+\frac{\bar\eta}{2l_b^2})^k f(z) \\ = &\frac{1}{\sqrt{\gamma^n}l_B^n}\exp(\frac {z\bar \eta}{2l_b^2})\left(2l_B^2\partial_z+(\frac{l_B^2}{l_b^2}-1)\bar\eta\right)^n f(z) \end{aligned}$$

overlap和上面同理

$$\begin{aligned} &\int\prod_i\mu_B(z_i)\int \prod_id\mu_b(\eta_i)\mu_b(\eta_i')\prod_{i<j}(\bar\eta_i-\bar\eta_j)^2(\eta_i'-\eta_j')^2\psi_{CF}(\{\bar z_i,\eta_i\})\psi_{CF}(\{z_i,\bar\eta_i'\})\\ =&\int \prod_id\mu_b(\eta_i)\mu_b(\eta_i')\prod_{i<j}(\bar\eta_i-\bar\eta_j)^2(\eta_i'-\eta_j')^2\\ &\frac{1}{N!}\sum_{P,P'}(-1)^{P+P'} \int\prod_i \left[\exp(\frac{\bar z_i\eta_i+z_i\bar\eta_i'}{2l_b^2})d\mu_B(z_i)\right]\varphi_{m_{P(1)}}^*(z_1,\eta_1)\varphi_{m_{P'(1)}}(z_1,\bar\eta_1')\dots \\ =&\int \prod_id\mu_b(\eta_i)\mu_b(\eta_i')\prod_{i<j}(\bar\eta_i-\bar\eta_j)^2(\eta_i'-\eta_j')^2\\ &\frac{1}{N!}\sum_{P,P'}(-1)^{P+P'} A_{m_{P(1)},m_{P'(1)}}(\eta_1,\bar\eta_1')\dots \\ =&\int \prod_id\mu_b(\eta_i)\mu_b(\eta_i')\prod_{i<j}(\bar\eta_i-\bar\eta_j)^2(\eta_i'-\eta_j')^2\\ &\frac{1}{N!}\sum_{P,P'}(-1)^{P+P'} A_{m_{P(1)},m_{P'(1)}}(\eta_{P(1)},\bar\eta_{P'(1)}')\dots \\ =&\int \prod_id\mu_b(\eta_i)\mu_b(\eta_i')\prod_{i<j}(\bar\eta_i-\bar\eta_j)^2(\eta_i'-\eta_j')^2\frac{1}{N!}\sum_{P'}(-1)^{P'}\det(A_{i,P'(j)}) \\ =&\int \prod_id\mu_b(\eta_i)\mu_b(\eta_i')\prod_{i<j}(\bar\eta_i-\bar\eta_j)^2(\eta_i'-\eta_j')^2 \det(A_{ij}(\eta_i,\bar\eta_j')) \\ \end{aligned}$$

其中 $$ \begin{aligned} &A_{ij} = A_{ij}(\eta_i,\bar \eta_j’) \ =&\eta_i^n(\bar\eta_j’)^{n’}\int d\mu_B(z)\exp\left(\frac{\bar z\eta_i+z\bar\eta_j’}{2l_b^2}\right)

\left(1+\frac{2l_B^2l_b^2}{l_B^2-l_b^2}\frac{\partial_{\bar z}}{\eta_i}\right)^n\left(\frac{\bar z}{l_B}\right)^{m}

\left(1+\frac{2l_B^2l_b^2}{l_B^2-l_b^2}\frac{\partial_z}{\bar\eta_j’}\right)^{n’}\left(\frac{z}{l_B}\right)^{m’} \end{aligned} $$ 令 $2l_Bl_b/(l_B^2-l_b^2) = \alpha$，（$l_B^2/l_b^2=2\tilde\nu$）那么

$$\begin{aligned} A_{ij} =& \eta_i^n(\bar\eta_j')^{n'}\sum_{k=0}^n\sum_{k'=0}^{n'} &\frac{n!}{k!(n-k)!}\frac{\alpha^k}{(\eta_i/l_b)^k}\frac{m!}{(m-k)!} \\ &&\frac{n'!}{k'!(n'-k')!}\frac{\alpha^{k'}}{(\bar \eta_j'/l_b)^{k'}}\frac{m'!}{(m'-k')!}\\ &&\int d\mu_B(z)\left(\frac{\bar z}{l_B}\right)^{m-k}\left(\frac{z}{l_B}\right)^{m'-k'} \end{aligned}$$

其中

$$\int d\mu_B(z)\left(\frac{\bar z}{l_B}\right)^{m-k}\left(\frac{z}{l_B}\right)^{m'-k'} = \delta_{m-k,m'-k'}2^{m-k}(m-k)!$$

那么，

$$A_{ij} = \sum_{k=\max(0,m'-m-n')}^{\min(n',m'-m)}...$$

下面处理 $\int \prod_id\mu_b(\eta_i)\mu_b(\eta_i')\prod_{i<j}(\bar\eta_i-\bar\eta_j)^2(\eta_i'-\eta_j')^2 \det(A_{ij}(\eta_i,\bar\eta_j'))$

我们令

$$\Delta(\bar\eta) = \prod_{i<j}(\bar\eta_i-\bar\eta_j)^2,\,\Delta(\eta') = \prod_{i<j}(\eta_i'-\eta_j')^2$$

令

$$\det(A_{ij}(\eta_i,\bar\eta_j')) = \exp(\sum\frac{\eta_i\bar\eta_i'}{2l_b^2})f(\eta,\bar\eta') = \exp(\frac{\eta\bar\eta'}{2l_b^2})f(\eta,\bar\eta')$$

那么积分可以写成

$$\int d\mu_b(\eta)d\mu_b(\eta')\Delta(\bar\eta)\Delta(\eta')\exp(\eta\bar\eta'/2l_b^2)f(\eta,\bar\eta')$$

我们先对 $\eta'$ 积分，得到

$$\begin{aligned} &\int d\mu_b(\eta)\Delta(\bar\eta)N_{-}[\Delta(\eta')f(\eta,2l_b^2\partial_{\eta'})]_{\eta'\rightarrow\eta} \\ =& \int d\mu_b(\eta)\Delta(\bar\eta)f(\eta,2l_b^2\partial_{\eta})\Delta(\eta) \end{aligned}$$

这里 $f(\eta,2l_b^2\partial_\eta)$ 中的 $2l_b^2\partial_\eta$ 只作用在右侧的 $\Delta(\eta)$ 上面，我们将 $f$ 展开

$$f(\eta,2l_b^2\partial_\eta) = \sum_m c_m(\eta)(2l_b^2\partial_\eta)^m = \sum_mc_m(\{\eta_i\})(\prod_i2l_b^2\partial_{\eta_i})^m$$

那么积分就等于

$$\begin{aligned} &\sum_m\int d\mu_b(\eta)\Delta(\bar\eta) c_m(\eta)(2l_b^2\partial_\eta)^m \Delta(\eta) \\ = &\sum_m\frac{1}{2\pi l_b^2}\int d^2\eta\exp(-\frac{\eta\bar\eta}{2l_b^2})\Phi(\eta,\bar\eta)(2l_b^2\partial_\eta)\Psi(\eta) \\ = &\sum_m\frac{1}{2\pi l_b^2}\int d^2\eta (2l_b^2\partial_\eta)\left(\exp(-\frac{\eta\bar\eta}{2l_b^2})\Phi(\eta,\bar\eta)\Psi(\eta)\right)-\int d^2\eta\Psi(\eta)(2l_b^2\partial_\eta)\exp(-\frac{\eta\bar\eta}{2l_b^2})\Phi(\eta,\bar\eta)\\ = &\sum_m\int d\mu_b(\eta) \Psi(\eta)(\bar\eta-2l_b^2\partial_\eta)\Phi(\eta,\bar\eta) \\ =&\sum_m\int d\mu_b(\eta) \left[(\bar\eta-2l_b^2\partial_\eta)\Phi(\eta,\bar\eta)\right]\left[(2l_b^2\partial_\eta)^{m-1}\Delta(\eta)\right] \\ =&\sum_m\int d\mu_b(\eta) \left[(\bar\eta-2l_b^2\partial_\eta)^m\Phi(\eta,\bar\eta)\right]\Delta(\eta) \\ =&\sum_m\int d\mu_b(\eta) \left[(\bar\eta-2l_b^2\partial_\eta)^mc_m(\eta)\right]|\Delta(\eta)|^2 \\ =&\int d\mu_b(\eta)|\Delta(\eta)|^2f(\eta,\bar\eta-2l_b^2\partial_\eta) \end{aligned}$$

即

$$\int d\mu_b(\eta)d\mu_b(\eta')\Delta(\bar\eta)\Delta(\eta')\exp(\eta\bar\eta'/2l_b^2)f(\eta,\bar\eta') = \int d\mu_b(\eta)|\Delta(\eta)|^2f(\eta,\bar\eta-2l_b^2\partial_\eta)$$

这里的 $\partial_\eta$ 作用到 $f$ 中的 $\eta$ 上面。下面求 $f(\eta,\bar\eta-2l_b^2\partial_\eta)$，只需要将 $f(\eta,\bar\eta')$ 的 $\bar\eta'$ 写到 $\eta$ 的左边，然后将 $\bar\eta'$ 替换成 $\bar\eta-2l_b^2\partial_\eta$ $$ \begin{aligned} f(\eta,\bar\eta’) &=\exp(-\frac{\eta\bar\eta’}{2l_b^2})\det(A_{ij}(\eta_i,\bar\eta_j’)) \

\end{aligned} $$ 由于 $\bar\eta'$ 含有求导项目，这个行列式存在交叉项，并不好处理。然而除了可以让 $\eta,\eta'$ 分别绑定行列指标，也可以让它们都绑定行指标，这样的积分结果不变，此时(注意下面的f和上面的f并不是相等的，但是积分是相等的) $$ \begin{aligned} f(\eta,\bar\eta’) &=\exp(-\frac{\eta\bar\eta’}{2l_b^2})\det(A_{ij}(\eta_i,\bar\eta_i’)) \

\end{aligned}

$$$$

\begin{aligned} f(\eta,\bar\eta’) &=\exp(-\frac{\eta\bar\eta’}{2l_b^2})\det(A_{ij}(\eta_i,\bar\eta_i’)) \ &=\sum_P(-1)^P \exp(-\eta_i\bar\eta_i’/2l_b^2)A_{iP(j)}(\eta_i,\bar\eta_i’)\cdots\ &=\sum_P(-1)^P \tilde A_{iP(j)}(\eta_i,\bar\eta_i’)\dots \ &=\det(\tilde{A}_{ij}(\eta_i,\bar\eta_i’)) \end{aligned} $$

其中

$$\begin{aligned} \tilde A_{ij}(\eta_i,\bar\eta_i') &=\exp(-\frac{\eta_i\bar\eta_i'}{2l_b^2})A_{ij}(\eta_i,\bar\eta_i') \\ &=\exp(-\frac{\eta_i\bar\eta_i'}{2l_b^2}) \sum_{k=0}^n\sum_{k'=0}^{n'} \frac{n!}{k!(n-k)!}\frac{n'!}{k'!(n'-k')!}\frac{m!}{(m-k)!}\frac{m'!}{(m'-k')!}\alpha^{k+k'} \\ &\left(\frac{\eta_i}{l_b}\right)^{n-k}\left(\frac{\bar\eta_i'}{l_b}\right)^{n'-k'}\delta_{m-k,m'-k'}2^{m-k}(m-k)! \\ &=\exp(-\frac{\eta_i\bar\eta_i'}{2l_b^2})\sum_k F(k)\left(\frac{\bar\eta_i'}{l_b}\right)^{n'-k+m-m'}\left(\frac{\eta_i}{l_b}\right)^{n-k} \\ &=\sum_q\frac{1}{q!}\left(-\frac{\eta_i\bar\eta_i'}{2l_b^2}\right)^q \sum_k F(k)\left(\frac{\bar\eta_j'}{l_b}\right)^{n'-k+m-m'}\left(\frac{\eta_i}{l_b}\right)^{n-k}\\ &=\sum_{q}\sum_k\frac{1}{q!}\frac{1}{(-2)^q}F(k)\left(\frac{\bar\eta'_i}{l_b}\right)^{n'-k+m-m'+q}\left(\frac{\eta_i}{l_b}\right)^{n-k+q} \end{aligned}$$

我们将 $\bar\eta_i'$ 替换成 $\bar\eta_i-2l_b^2\partial_{\eta_i}$，需要指出的是，由于 $\eta'$ 和 $\eta$ 都是和行指标绑定，所以在求行列式的过程中不会出现跨行的求导项，所以我们可以直接在矩阵元内部直接完成替换和求导。上式就等于

$$\begin{aligned} &\sum_{q}\sum_k\frac{1}{q!}\frac{1}{(-2)^q}F(k)\left(\frac{\bar\eta_i-2l_b^2\partial_{\eta_i}}{l_b}\right)^{n'-k+m-m'+q}\left(\frac{\eta_i}{l_b}\right)^{n-k+q} \\ &=\sum_q\sum_k\frac{1}{q!}\frac{1}{(-2)^q}F(k)\sum_t\left(\frac{\bar\eta_i}{l_b}\right)^{n'-k+m-m'+q-t}\left(-2l_b\partial_{\eta_i}\right)^{t}\left(\frac{\eta_i}{l_b}\right)^{n-k+q} \\ &=\sum_q\sum_k\frac{1}{q!}\frac{1}{(-2)^q}F(k)\sum_t\frac{(n'-k+m -m'+q)!}{t!(n'-k+m -m'+q-t)!}\left(\frac{\bar\eta_i}{l_b}\right)^{n'-k+m-m'+q-t}\left(-2\right)^{t}\frac{(n-k+q)!}{(n-k+q-t)!}\left(\frac{\eta_i}{l_b}\right)^{n-k+q-t} \\ \end{aligned}$$

其中 $q$ 的求和范围是 $0,\infty$，$k$ 便利 $\Lambda$ level, t依赖于 $n'-k+m-m'+q$，这个矩阵元是一个无穷求和，计算量比较大。

另一方面，将 $\bar\eta_i'$ 替换成 $\bar\eta_i-2l_b^2\partial_{\eta_i}$的操作实际上可以等价为一个指数微分算符

$$\exp(-2l_b^2\partial_{\eta_i}\partial_{\bar\eta_i})\tilde A_{ij}(\eta_i,\bar\eta_i)$$

证明：

$$\begin{aligned} &\tilde A_{ij}(\eta_i,\bar\eta_i-2l_b^2\partial_{\eta_i}) \\ = &\sum_{\alpha\beta}c_{\alpha\beta}(\bar\eta_i-2l_b^2\partial_{\eta_i})^\beta\eta_i^\alpha \\ = &\sum_{\alpha\beta}\sum_{k}c_{\alpha\beta} \begin{pmatrix} \beta\\ k \end{pmatrix} \bar\eta_i^{\beta-k}(-2l_b^2\partial_{\eta_i})^{k}\eta_i^{\alpha} \\ =&\sum_{\alpha\beta}\sum_{k}\frac{\beta!}{(\beta-k)!}\frac{\alpha!}{(\alpha-k)!}\frac{(-2l_b^2)^k}{k!}c_{\alpha\beta}\bar\eta_i^{\beta-k}\eta_i^{\alpha-k} \\ =&\sum_{k}\frac{(-2l_b^2\partial_{\bar\eta_i}\partial_{\eta_i})^k}{k!}\sum_{\alpha\beta}c_{\alpha\beta}\eta_i^{\alpha}\bar\eta_i^{\beta} \\ =&\exp(-2l_b^2\partial_{\eta_i}\partial_{\bar\eta_i})\tilde A_{ij}(\eta_i,\bar\eta_i) = \mathcal B_{ij}(\eta_i,\bar\eta_i) \end{aligned}$$

这等价于

$$\tilde A_{ij}(\eta,\bar\eta) = \int \frac{d^2\eta'}{2\pi l_b^2}\exp(-\frac{|\eta'-\eta|^2}{2l_b^2})\mathcal B_{ij}(\eta',\bar\eta')$$

因为根据上面的指数微分，关系可以得到

$$\tilde A_{ij}(\eta,\bar\eta) = \exp(2l_b^2\partial_{\eta}\partial_{\bar\eta})\mathcal B_{ij}(\eta,\bar\eta)$$

而根据

$$\exp(-2\partial_x\partial_y)[e^{Axy} x^a x^b] = \exp(\frac{A}{1+2A}xy)\sum_{s=0}^{\min(a,b)} \begin{pmatrix} a\\ s \end{pmatrix} \begin{pmatrix} b\\ s \end{pmatrix} s!(-2)^s (1+2A)^{-(a+b+1-s)} x^{a-s} y^{b-s}$$

可以通过生成函数证明，我们考虑一个生成函数

$$\exp(t\partial_x\partial_y)[e^{Axy+ux+vy}]$$

这个生成函数取 $t=2$，并对 $u, v$ 求偏导之后取 $u=v=0$ 即可得到上面的结果。生成函数本身可以这样求，我们令 $F(t) = \exp(t D)[e^{Axy+ux+vy}]$，其中 $D=\partial_x\partial_y$ 容易发现

$$\partial_t F(t) = DF(t)$$

可以设

$$F(t) = \exp(A(t)xy+u(t)x+v(t)y+d(t))$$

那么

$$\begin{aligned} &\partial_t F(t) = \left(A'(t)xy+u'(t)x+v'(t)y+d'(t)\right)F(t) \\ &DF(t) = \left[A(t)+(A(t)x+v(t))(A(t)y+u(t))\right]F(t) \end{aligned}$$

对比系数可以发现

$$\begin{cases} A'(t) = A^2(t)\\ u'(t) = A(t)u(t) \\ v'(t) = A(t)v(t) \\ d'(t)= A(t)+u(t)v(t) \end{cases}$$

初始条件为 $A(0)=A, \, u(0) = u,\,v(0) = v,\, d(0) = 1$，容易解得

$$\begin{cases} A(t) &= \frac{A}{1-At}\\ u(t) &= \frac{u}{1-At}\\ v(t) &= \frac{v}{1-At} \\ d(t) &= -\ln|At-1|+\frac{uv t}{1-At} \end{cases}$$

那么就得到了 $F(t)$

$$F(t) = \frac{1}{|1-At|}\exp(\frac{Axy+ux+vy}{1-At}) = \exp(t\partial_x\partial_y)[e^{Axy+ux+vy}]$$

然后根据 $$ \begin{aligned} &\exp(-2\partial_x\partial_y)[e^{Axy}x^ay^b] \ =&\partial_u^a\partial_v^bF(-2) |{u=v=0}\ =&\frac{1}{|2A+1|}\exp\left(\frac{Axy}{1+2A}\right)\sum{r=0}^{\min(a,b)}\frac{a!}{(a-r)!}\frac{b!}{(b-r)!}\frac{1}{r!}\left(\frac{-2}{1+2A}\right)^r\left(\frac{x}{1+2A}\right)^{a-r}\left(\frac{y}{1+2A}\right)^{b-r}

\end{aligned}

$$然而对于这里的$$

\tilde{A}_{ij}(\eta_i,\bar\eta_i) = \exp\left(-\frac{\eta_i\bar\eta_i}{2l_b^2}\right)\times多项式 $$ 对应于 $A=-1/2$，可以发现这是上面表达式的奇异点，公式失效。

下面做法，可以多保留一个高斯因子，转化成有限项求和，我们从下面表达式出发 $$ \begin{aligned} &A_{ij} = A_{ij}(\eta_i,\bar \eta_i’) \ =&\eta_i^n(\bar\eta_i’)^{n’}\int d\mu_B(z)\exp\left(\frac{\bar z\eta_i+z\bar\eta_i’}{2l_b^2}\right)

\left(1+\alpha\frac{\partial_{\bar z/l_B}}{\eta_i/l_b}\right)^n\left(\frac{\bar z}{l_B}\right)^{m}

\left(1+\alpha\frac{\partial_{z/l_B}}{\bar\eta_i’/l_b}\right)^{n’}\left(\frac{z}{l_B}\right)^{m’} \

=&\eta_i^n(\bar\eta_i’)^{n’}\sum_{kk’}^{nn’} \begin{pmatrix} n\ k \end{pmatrix} \begin{pmatrix} n’\ k’ \end{pmatrix} \alpha^{k+k’} \frac{l_b^{k+k’}}{\eta_i^k\bar\eta_i’^{k’}} \int d\mu_B(z)\exp\left(\frac{\bar z\eta_i+z\bar\eta_i’}{2l_b^2}\right)

\left(\partial_{\bar z/l_B}\right)^k \left(\frac{\bar z}{l_B}\right)^{m}

\left(\partial_{z/l_B}\right)^{k’}\left(\frac{z}{l_B}\right)^{m’}\ =&\eta_i^n(\bar\eta_i’)^{n’}\sum_{kk’}^{nn’} \begin{pmatrix} n\ k \end{pmatrix} \begin{pmatrix} n’\ k’ \end{pmatrix} \alpha^{k+k’} \frac{l_b^{k+k’}}{\eta_i^k\bar\eta_i’^{k’}} \frac{m!}{(m-k)!}\frac{m’!}{(m’-k’)!} \int d\mu_B(z)\exp\left(\frac{\bar z\eta_i+z\bar\eta_i’}{2l_b^2}\right)

\left(\frac{\bar z}{l_B}\right)^{m-k} \left(\frac{z}{l_B}\right)^{m’-k’}\ \end{aligned}

$$由于$$

\begin{aligned} &\int d\mu_B(z)\exp(J\bar z+Kz)\bar z^a z^b\ =&\partial_J^a\partial_K^b\exp(2l_B^2JK) \ =& \partial_J^a(2l_B^2J)^b\exp(2l_B^2JK) \ = &(2l_B^2)^{a+b}\exp(2l_B^2JK)\sum_{\gamma=0}^{\min(a,b)} \begin{pmatrix}a \ \gamma\end{pmatrix} \begin{pmatrix}b \ \gamma\end{pmatrix} \gamma!(2l_B^2)^{-\gamma}J^{b-\gamma}K^{a-\gamma} \end{aligned}

$$那么$$

\begin{aligned} &\int d\mu_B(z)\exp\left(\frac{\bar z\eta_i+z\bar\eta_i’}{2l_b^2}\right) \left(\frac{\bar z}{l_B}\right)^{a} \left(\frac{z}{l_B}\right)^{b} \ = &\exp(\rho\frac{\eta_i\bar\eta_i’}{2l_b^2})\sum_{\gamma=0}^{\min(a,b)} \begin{pmatrix}a\ \gamma\end{pmatrix} \begin{pmatrix}b\ \gamma\end{pmatrix} \gamma!2^{\gamma}\rho^{(a+b-2\gamma)/2}\left(\frac{\eta_i}{l_b}\right)^{b-\gamma}\left(\frac{\bar\eta_i’}{l_b}\right)^{a-\gamma} \end{aligned} $$ 其中 $\rho = l_B^2/l_b^2$，从而 $$ \begin{aligned} &\tilde A_{ij}(\eta_i,\bar\eta_i’) \ =&\exp\left((\rho-1)\frac{\eta_i\bar\eta_i’}{2l_b^2}\right) \eta_i^n(\bar\eta_i’)^{n’}\sum_{kk’}^{nn’} \begin{pmatrix} n\ k \end{pmatrix} \begin{pmatrix} n’\ k’ \end{pmatrix} \alpha^{k+k’} \frac{l_b^{k+k’}}{\eta_i^k\bar\eta_i’^{k’}} \frac{m!}{(m-k)!}\frac{m’!}{(m’-k’)!}\

&\sum_{\gamma=0}^{\min(a,b)} \begin{pmatrix}a\ \gamma\end{pmatrix} \begin{pmatrix}b\ \gamma\end{pmatrix} \gamma!2^{\gamma}\rho^{(a+b-2\gamma)/2}\left(\frac{\eta_i}{l_b}\right)^{b-\gamma}\left(\frac{\bar\eta_i’}{l_b}\right)^{a-\gamma}\ \end{aligned} $$ 这里的 $a=m-k,\,b = m'-k'$，这个结果多保留了一个高斯因子，对应的 $A = \frac{\rho-1}{2}\neq -\frac{1}{2}$。从而

$$\begin{aligned} &\exp\left(-2l_b^2\partial_{\eta_i}\partial_{\bar\eta_i}\right)\tilde{A}_{ij}(\eta_i,\bar\eta_i) \\ =&\exp\left(-2l_b^2\partial_{\eta_i}\partial_{\bar\eta_i}\right)\exp((\rho-1)\eta_i\bar\eta_i/2l_b^2)\sum_{\alpha\beta}c_{\alpha\beta}(\frac{\eta_i}{l_b})^\alpha(\frac{\bar\eta_i}{l_b})^\beta \\ =&\frac{1}{\rho}\exp\left(\frac{\rho-1}{\rho}\frac{\eta_i\bar\eta_i}{2l_b^2}\right)\sum_{\alpha\beta}c_{\alpha\beta}\sum_{r=0}^{\min(\alpha,\beta)}\\ &\frac{\alpha!}{(\alpha-r)!}\frac{\beta!}{(\beta-r)!}\frac{1}{r!}\left(\frac{-2}{\rho}\right)^r\left(\frac{\eta_i}{\rho l_b}\right)^{a-r}\left(\frac{\bar\eta_i}{\rho l_b}\right)^{b-r} \end{aligned}$$

我们令 $\mathcal B_{ij}(\eta_i,\bar\eta_i) =\exp\left(-2l_b^2\partial_{\eta_i}\partial_{\bar\eta_i}\right)\tilde{A}_{ij}(\eta_i,\bar\eta_i)=\exp\left(-2l_b^2\partial_{\eta_i}\partial_{\bar\eta_i}\right)[\exp(-\frac{\eta_i\bar\eta_i}{2l_b^2})A_{ij}(\eta_i,\bar\eta_i)]$，那么overlap就是

$$\int \prod_{i}d\mu_b(\eta_i)|\Delta(\eta)|^2\det(\mathcal B_{ij}(\eta_i,\bar\eta_i))$$

$$\int \prod_id\mu_b(\eta_i)\mu_b(\eta_i')\Delta(\bar\eta)\Delta(\eta') \det(A_{ij}(\eta_i,\bar\eta_i'))$$

整个过程就是：插入一个 $\exp(+\frac{\eta_i\bar\eta_i'}{2l_b^2})\exp(-\frac{\eta_i\bar\eta_i'}{2l_b^2})$， 对 $\eta'$ 积分，这等价于将 $\bar\eta'$ 移动到 $\eta'$ 的左边，然后替换成 $2l_b^2\partial_{\eta'}$，求导完之后替换成 $\eta$。这里面涉及到对 $\Delta(\eta')$ 的求导，利用分部积分可以将 $\Delta(\eta')$ 移动到求导算符左边，等最终等价与将 $\bar\eta'$ 替换成 $\eta-$ 。将 $\bar\eta_i'$ 替换0成 $\bar\eta_i-2l_b^2\partial_{\eta_i}$的操作实际上可以等价为一个指数微分算符$\exp(-2l_b^2\partial_{\eta_i}\partial_{\bar\eta_i})\tilde A_{ij}(\eta_i,\bar\eta_i)$

考虑到一个 $\Lambda$ level 内的填充数 $M$ 应该远大于 $\Lambda$ level 的数量 N，那么可以得到矩阵元的计算复杂度为 $\frac{3}{2}O(M^2N^2)$，矩阵的维度为 $MN$，得到整个矩阵的复杂度就是$O(M^4N^4)$，这个复杂度高于行列式本身的复杂度，是leading order项。粒子数就是 $MN=N_p$ ，整体复杂度就是 $O(N_p^4)$。如果每次更新一个坐标，那么只需要更新一行，$O(N_p^3)$

另一方面，我们其实可以将 $\tilde A_{ij}(\eta_i,\bar\eta_i)$ 用连带拉盖尔多项式展开。考虑到

$$L_n^m(x) = \sum_{\gamma=0}^n\frac{(n+m)!}{(n-\gamma)!(m+\gamma)!}\frac{(-x)^\gamma}{\gamma!} = \sum_{\gamma=0}^n \begin{pmatrix} n+m\\ n-\gamma \end{pmatrix} \frac{(-x)^\gamma}{\gamma!}$$

那么 $$ \sum_{\gamma=0}^{\min(a,b)} \begin{pmatrix} a\ \gamma \end{pmatrix}

\begin{pmatrix} b\ \gamma \end{pmatrix}

\gamma!

z^\gamma

=z^{\min(a,b)}\min(a,b)! L_{\min(a,b)}^{|a-b|}(-\frac{1}{z}) $$ 那么 当 $m'-k'<m-k$

$$\begin{aligned} \tilde A_{ij} =&\exp\left((\rho-1)\frac{\eta_i\bar\eta_i}{2l_b^2}\right) \eta_i^n(\bar\eta_i)^{n'}\sum_{kk'}^{nn'} \begin{pmatrix} n\\ k \end{pmatrix} \begin{pmatrix} n'\\ k' \end{pmatrix} \alpha^{k+k'} \frac{l_b^{k+k'}}{\eta_i^k\bar\eta_i^{k'}} \frac{m!}{(m-k)!}\frac{m'!}{(m'-k')!}\\ &2^{m'-k'} (m'-k')!\left(\sqrt{\rho}\frac{\bar\eta_i}{l_b}\right)^{|m-m'+k'-k|}L_{m'-k'}^{m-m'+k'-k}\left(-\frac{\rho |\eta|^2}{2l_b^2}\right) \end{aligned}$$

当 $m'-k'>m-k$

$$\begin{aligned} \tilde A_{ij} =&\exp\left((\rho-1)\frac{\eta_i\bar\eta_i}{2l_b^2}\right) \eta_i^n(\bar\eta_i)^{n'}\sum_{kk'}^{nn'} \begin{pmatrix} n\\ k \end{pmatrix} \begin{pmatrix} n'\\ k' \end{pmatrix} \alpha^{k+k'} \frac{l_b^{k+k'}}{\eta_i^k\bar\eta_i^{k'}} \frac{m!}{(m-k)!}\frac{m'!}{(m'-k')!}\\ &2^{m-k}(m-k)!\left(\sqrt{\rho}\frac{\eta_i}{l_b}\right)^{|m-m'+k'-k|}L_{m-k}^{m-m'+k'-k}\left(-\frac{\rho |\eta|^2}{2l_b^2}\right) \end{aligned}$$

全部代入的结果 $$ \begin{aligned} \mathcal B_{ij}(\eta_i,\bar\eta_i) = &\exp\left(-2l_b^2\partial_{\eta_i}\partial_{\bar\eta_i}\right)\tilde A_{ij}(\eta_i,\bar\eta_i) \ =&\exp\left(-2l_b^2\partial_{\eta_i}\partial_{\bar\eta_i}\right)\exp\left((\rho-1)\frac{\eta_i\bar\eta_i}{2l_b^2}\right) \sum_{kk’}^{nn’} \begin{pmatrix} n\ k \end{pmatrix} \begin{pmatrix} n’\ k’ \end{pmatrix} \alpha^{k+k’}

\frac{m!}{(m-k)!}\frac{m’!}{(m’-k’)!}\

&\sum_{\gamma=0}^{\min(a,b)} \begin{pmatrix}a\ \gamma\end{pmatrix} \begin{pmatrix}b\ \gamma\end{pmatrix} \gamma!2^{\gamma}\rho^{(a+b-2\gamma)/2}\left(\frac{\eta_i}{l_b}\right)^{b+n-k-\gamma}\left(\frac{\bar\eta_i}{l_b}\right)^{a+n’-k’-\gamma}\ =&\frac{1}{\rho}\exp\left(\frac{\rho-1}{\rho}\frac{\eta_i\bar\eta_i}{2l_b^2}\right) \sum_{kk’}^{nn’} \begin{pmatrix} n\ k \end{pmatrix} \begin{pmatrix} n’\ k’ \end{pmatrix} \alpha^{k+k’}

\frac{m!}{(m-k)!}\frac{m’!}{(m’-k’)!}\

&\sum_{\gamma=0}^{\min(a,b)} \begin{pmatrix}a\ \gamma\end{pmatrix} \begin{pmatrix}b\ \gamma\end{pmatrix} \gamma!2^{\gamma}\rho^{(a+b-2\gamma)/2} \sum_{r=0}^{\min(p,q)}\ &\frac{p!}{(p-r)!}\frac{q!}{(q-r)!}\frac{1}{r!}\left(\frac{-2}{\rho}\right)^r\left(\frac{\eta_i}{\rho l_b}\right)^{p-r}\left(\frac{\bar\eta_i}{\rho l_b}\right)^{q-r}

\end{aligned} $$ $p=b+n-k-\gamma,\quad q=a+n'-k'-\gamma,\quad a = m-k,\quad b = m'-k'$ $i = (n,m)$ and $j = (n',m')$ $\alpha=\sqrt{\rho}/(\rho-1), \,0<\rho<1$

当 $n_{max} = 0$ 时

$$\begin{aligned} B_{ij} = &\frac{1}{\rho}\exp\left(\frac{\rho-1}{\rho}\frac{\eta_i\bar\eta_i}{2l_b^2}\right)\sum_{\gamma=0}^{\min(m,m')} \begin{pmatrix}m\\ \gamma\end{pmatrix} \begin{pmatrix}m'\\ \gamma\end{pmatrix} \gamma!2^{\gamma}\rho^{(m+m'-2\gamma)/2} \sum_{r=0}^{\min(m'-\gamma,m-\gamma)}\\ &\frac{(m'-\gamma)!}{(m'-\gamma-r)!}\frac{(m-\gamma)!}{(m-\gamma-r)!}\frac{1}{r!}\left(\frac{-2}{\rho}\right)^r\left(\frac{\eta_i}{\rho l_b}\right)^{m'-\gamma-r}\left(\frac{\bar\eta_i}{\rho l_b}\right)^{m-\gamma-r} \\ \end{aligned}$$

令

$$s=r+\gamma$$

那么 $\gamma\leq s \leq \min(m,m')-\gamma$

$$\sum_{\gamma=0}^{\min(m,m')}\sum_{s=\gamma}^{\min(m,m')} F(\gamma,s) = \sum_{s=0}^{\min(m,m')}\sum_{\gamma=0}^s F(\gamma,s)$$

关于 $\gamma$ 的部分为

$$\sum_{\gamma=0}^s \frac{2^\gamma\rho^{-\gamma}}{\gamma!(m-\gamma)!(m'-\gamma)!}\frac{(m'-\gamma)!(m-\gamma)!(-2)^{-\gamma}}{(s-\gamma)!\rho^{-\gamma}} = \sum_{\gamma=0}^s \frac{(-1)^\gamma}{(s-\gamma)!\gamma!} = \frac{1}{s!}\delta_{0,s}$$

从而

$$B_{ij}=\rho^{-1-\frac{m+m'}{2}}\exp\left(\frac{\rho-1}{\rho}\frac{|\eta_i|^2}{2l_b^2}\right)\left(\frac{\eta_i}{l_b}\right)^{m'}\left(\frac{\bar\eta_i}{l_b}\right)^{m}$$

那么

$$\det(B) \propto \prod_i \bar\eta_i^i \prod_{i<j}(\eta_i-\eta_j)$$

$$B_{ij}(\eta_i,\bar\eta_i) = \exp(-2l_b^2\partial_{\eta_i}\partial_{\bar\eta_i})\exp(-\frac{\bar\eta_i\eta_i}{2l_b^2})\int d\mu_B(z)\exp(\frac{\bar z\eta_i+z\bar\eta_i}{2l_b^2})\varphi_i(\bar z,\eta_i)\varphi_j(z,\bar\eta_i)$$

那么我们可以考虑

$$\begin{aligned} \exp(-2\partial_\eta\partial_{\bar\eta})f(\eta,\bar\eta) = \exp(-\frac{1}{2}\nabla^2)f(x,y) = \exp(-\frac 1 2\partial_x^2-\frac 1 2\partial_y^2)f(x,y) \end{aligned}$$

这里将指数微分拆成了独立的两个部分，根据

$$\exp(-\frac{C^2}{2}) = \frac{1}{\sqrt{2\pi}}\int_R \mathrm e^{-s^2/2}\mathrm e^{isC} ds$$

那么

$$\begin{aligned} &\exp\left(-\frac{\partial_x^2+\partial_y^2}{2}\right) \\ = &\frac{1}{2\pi}\int_{\mathbb R^2} \mathrm d^2 \xi\,\mathrm{e}^{-|\xi|^2/2}\mathrm{e}^{i\xi_x\partial_x+i\xi_y\partial_y} \\ = &\frac{1}{2\pi}\int_{\mathbb R^2} \mathrm d^2 \xi\,\mathrm{e}^{-|\xi|^2/2}\mathrm{e}^{i\xi_x\partial_x+i\xi_y\partial_y} \\ =&\frac{1}{2\pi}\int_{\mathbb R^2} \mathrm d^2 \xi\,\mathrm{e}^{-|\xi|^2/2}\mathrm{e}^{i\xi\partial_\eta+i\bar\xi\partial_{\bar\eta}} \end{aligned}$$

那么

$$\exp(-2\partial_\eta\partial_{\bar\eta})f(\eta,\bar\eta) = \frac{1}{2\pi}\int_{\mathbb R^2} \mathrm d^2 \xi\,\mathrm{e}^{-|\xi|^2/2}f(\eta+i\xi,\bar\eta+i\bar\xi) = \frac{1}{2\pi}\int_{\mathbb R^2} \mathrm d^2 \xi\,\mathrm{e}^{-|\xi|^2/2}f(\eta+\xi,\bar\eta-\bar\xi)$$

那么

$$\begin{aligned} &\exp(-2l_b^2\partial_{\eta}\partial_{\bar\eta})\exp(-\frac{\bar\eta\eta}{2l_b^2})\int d\mu_B(z)\exp(\frac{\bar z\eta+z\bar\eta}{2l_b^2})\varphi_i(\bar z,\eta)\varphi_j(z,\bar\eta) \\ =&\int\mathrm d\mu_B(z)\int\frac{\mathrm d^2\xi}{2\pi l_b^2}\,\mathrm e^{-|\xi|^2/2l_b^2}\mathrm e^{-(|\eta|^2-|\xi|^2-\bar\xi\eta+\bar\eta\xi)/2l_b^2}\mathrm e^{(\bar z\eta+z\bar\eta+\bar z\xi-z\bar\xi)/2l_b^2}\varphi_i(\bar z,\eta+\xi)\varphi_j(z,\bar\eta-\bar\xi) \\ =&\int\mathrm d\mu_B(z)\mathrm e^{(-|\eta|^2+\bar z\eta+z\bar\eta)/2l_b^2}\int\frac{\mathrm d^2\xi}{2\pi l_b^2}\,\mathrm e^{(\bar\xi(\eta-z)+\xi(\bar z-\bar\eta))/2l_b^2}\varphi_i(\bar z,\eta+\xi)\varphi_j(z,\bar\eta-\bar\xi) \end{aligned}$$

考虑到

$$\varphi_{nm}(z,\bar\eta) = (2l_B^2\partial_z+(\frac{l_B^2}{l_b^2}-1)\bar\eta)^n z^m = (2l_B^2\partial_z-\alpha\bar\eta)^nz^m = \partial_s^n\mathrm{e}^{s(2l_B^2\partial_z-\alpha\bar\eta)}z^m|_{s=0}$$

$$\partial_s^n\mathrm{e}^{s(2l_B^2\partial_z-\alpha\bar\eta)}z^m = \partial_s^n \mathrm e^{-s\alpha\bar\eta}\mathrm e^{2sl_B^2\partial_z}z^m = \partial_s^n\mathrm e^{-s\alpha\bar\eta}(z+2sl_B^2)^m$$

那么我们可以将 $\varphi$ 替换成对应的生成函数，然后再对参数求导，即

$$\begin{aligned} &\varphi_i(\bar z,\eta+\xi) \rightarrow \mathrm e^{-\alpha\bar s(\eta+\xi)}(\bar z+2\bar sl_B^2)^{m_i} \\ &\varphi_j(z,\bar\eta-\bar\xi) \rightarrow \mathrm e^{-\alpha s(\bar \eta-\bar\xi)}(z+2sl_B^2)^{m_j} \end{aligned}$$

$\alpha = 1-l_B^2/l_b^2$，我们可以发现被积函数对 $\xi$ 的依赖都跑到了指数上面，$\xi$ 依赖的积分结果为

$$\int\frac{\mathrm d^2\xi}{2\pi l_b^2}\exp\left(\frac{\xi(\bar z-\bar\eta-2l_b^2\bar s\alpha)-\bar\xi (z-\eta-2l_b^2s\alpha)}{2l_b^2}\right) = 2\pi l_b^2\delta^{(2)}(z-\eta-2l_b^2s\alpha)$$

z的积分就是做个替换 $z\rightarrow \eta+2l_b^2\alpha s$，结果为

$$\begin{aligned} \frac{l_b^2}{l_B^2}\mathrm e^{-|\eta|^2/2l^2}\exp\left(-\frac{2l_b^4\alpha^2|s|^2}{l_B^2}-\alpha\frac{l_b^2}{l_B^2}(\bar s\eta+\bar\eta s))\right)(\bar\eta+2l_b^2\bar s)^{m_i}(\eta+2l_b^2s)^{m_j} \end{aligned}$$

可以将模方项写成积分，利用前面用到的变换

$$\exp(-\frac{\bar A A}{2}) = \exp(-\frac{1}{2}(A_x^2+A_y^2)) = \frac{1}{2\pi}\int_{\mathbb R^2}e^{-|w|^2/2}e^{i(w\cdot A)} d^2 w = \frac{1}{2\pi}\int_{\mathbb R^2}e^{-|w|^2/2}e^{\frac i 2(\bar wA+w\bar A)} d^2 w = \frac{1}{2\pi}\int_{\mathbb R^2}e^{-|w|^2/2}e^{\frac 1 2(-\bar wA+w\bar A)} d^2 w$$

所以

$$\exp(-\frac{2l_b^4\alpha^2|s|^2}{l_B^2}) = \int \frac{d^2 w}{2\pi l_B^2}e^{-|w|/2l_B^2}e^{l_b^2\alpha s\bar w/l_B^2}e^{-l_b^2\alpha\bar s w/l_B^2}$$

代入，关于 s 和 $\bar s$ 的指数项结果为

$$\mathrm{\exp}\left[\frac{l_b^2}{l_B^2}(\bar w-\bar\eta)\alpha s\right],\quad\mathrm{\exp}\left[-\frac{l_b^2}{l_B^2}(w+\eta)\alpha \bar s\right]$$

乘上后面的多项式，可以发现就是 $\varphi$ 的生成函数，即

$$\mathrm{\exp}\left[\frac{l_b^2}{l_B^2}(\bar w-\bar\eta)\alpha s\right](\eta+2l_b^2 s)^{m_j}\Leftrightarrow (2 l_b^2\partial_\eta+\frac{l_b^2}{l_B^2}(\bar w-\bar\eta)\alpha)^{n_i}\eta^{m_j} =$$

或者说

$$\begin{aligned} &\partial_s^{n_j}\mathrm{\exp}\left[\frac{l_b^2}{l_B^2}(\bar w-\bar\eta)\alpha s\right](\eta+2l_b^2 s)^{m_j}|_{s=0} \\ =& \partial_{s}^{n_j}\mathrm{exp}\left[-\frac{l_b^2}{l_B^2}(\bar\eta-\bar w)\alpha s+2l_b^2s\partial_\eta\right]\eta^{m_j} \\ = &(-\frac{l_b^2}{l_B^2}(\bar\eta-\bar w)\alpha +2l_b^2\partial_\eta)^{n_j}\eta^{m_j} \\ = &\left(\frac{l_b^2}{l_B^2}\right)^{n_i}\left(2l_B^2\partial_\eta-\alpha(\bar\eta-\bar w)\right)^{n_j}\eta^{m_j} \\ = &\left(\frac{l_b^2}{l_B^2}\right)^{n_j}\varphi_{n_jm_j}(\eta,\bar\eta-\bar w) \end{aligned}$$

同理

$$\partial_s^{n_j}\mathrm{\exp}\left[-\frac{l_b^2}{l_B^2}(w+\eta)\alpha s\right](\bar\eta+2l_b^2 s)^{m_i}|_{s=0} = \left(\frac{l_b^2}{l_B^2}\right)^{n_i}\varphi_{n_im_i}(\bar \eta, \eta+w)$$

因此

$$B_{ij}(\eta_i,\bar\eta_i) = \left(\frac{l_b^2}{l_B^2}\right)^{n_i+n_j+1}\mathrm e^{-|\eta|^2/2l^2}\int d\mu_B(w) \varphi_{n_im_i}(\bar\eta_i,\eta_i+w)\varphi_{n_jm_j}(\eta_i,\bar\eta_i-\bar w)$$

所以 $$

$$$$

\begin{aligned} &\frac{1}{N!}\sum_{P}\det(B_{ij}(\eta_P(i),\bar\eta_P(i))) \ =&\frac{1}{N!}\prod_{i,j}\left(\frac{l_b^2}{l_B^2}\right)^{n_i+n_j+1}\prod_i e^{-|\eta_i|/2l^2}\int \prod_i d\mu_B(w_i) \det(\varphi_{j}(\bar\eta_i,\eta_i+w_i))\det(\varphi_{j}(\eta_i,\bar\eta_i-\bar w_i )) \end{aligned} $$ 可以令 $U_{ji} = \varphi_j(\bar\eta_i,\eta_i+w_i)$, $V_{ji} = \varphi_j(\eta_i, \bar\eta_i-\bar w_i)$

考虑算符 $\rho_q = \exp(iq\cdot r) = \exp(i\bar q z/2+iq\bar z/2)$，对应的矩阵为

$$B_{ij}^{\rho_q}(\eta_i,\bar\eta_i) = \left(\frac{l_b^2}{l_B^2}\right)^{n_i+n_j+1}\mathrm e^{-|\eta|^2/2l^2}\mathrm e^{\frac i 2(q\bar\eta+\bar q \eta)}\int d\mu_B(w) \varphi_{n_im_i}(\bar\eta_i,\eta_i+w-iql_B^2)\varphi_{n_jm_j}(\eta_i,\bar\eta_i-\bar w-i\bar ql_B^2)$$

相当于 $w$ 平移了 $-iql_B^2$，然后前面多了一个相位。可以令

$$U_{ji}^{\rho_q} = e^{\frac i 2 q\bar\eta_i}\varphi_j(\bar\eta_i,\eta_i+w_i-iql_B^2),\quad V_{ji}^{\rho_q} = e^{\frac i 2\bar q\eta_i}\varphi_j(\eta_i,\bar\eta_i-\bar w -i\bar q l_B^2)$$

下面来看一下全排列平均之后算符期望值的表达式，由于

$$\braket{\Psi|\rho(\boldsymbol q)|\Psi} = \int d\mu_b(\eta)\prod_{i<j}|\eta_i-\eta_j|^4 \det(\mathcal B)\mathrm{Tr}(\mathcal B^{-1}\mathcal B^{\rho(q)})$$

所以有一个问题在于求 $B^{-1}$，保留积分结构求逆是困难的，所以实际上不如利用这个表达式

$$\braket{\Psi|\rho(\boldsymbol q)|\Psi} = \int d\mu_b(\eta)\prod_{i<j}|\eta_i-\eta_j|^4 \sum_{ij}\mathrm{Cof}_{ij}(\mathcal B)\mathcal B_{ij}^{\rho(q)}$$

这里的

$$\begin{aligned} &\sum_{ij}\mathrm{Cof}_{ij}(\mathcal B)\mathcal B_{ij}^{\rho(q)} \\ =&\sum_{i}\det(\mathcal B^{i}) \end{aligned}$$

其中

$$\mathcal B^{i}_{kl} = \begin{cases} B_{kl},\quad k\neq i\\ B_{kl}^{\rho_q},\quad k=i \end{cases}$$

$\mathcal B^i$ 的意思是将 $B$ 的第 i 行替换成 $B^{\rho_q}$ 的第 i 行，而 $B$ 和 $B^{\rho_q}$ 每个矩阵元的积分结构都是相似的，我们可以只考虑积分的核，即

$$B_{kl}\rightarrow U_{kk}V_{lk},\quad B^{\rho_q}_{kl}\rightarrow U^{\rho_q}_{kk}V^{\rho_q}_{lk}$$

$B^{i}_{kl}$ 实际上就是当 $k=i$ 的时候，$B^{i}_{{k=i},l}\rightarrow U_{kk}^{\rho_q}V_{lk}^{\rho_q}$，那么

$$\det(\mathcal B^{i}_{kl}) \leftrightarrow U_{11}U_{22}\cdots U_{ii}^{\rho_q}\dots \begin{vmatrix} V_{11}\dots V_{1i}^{\rho_q}\dots V_{1N} \\ ......V_{2i}^{\rho_q}....... \\ ................. \\ \\ ......V_{Ni}^{\rho_q}....... \\ \end{vmatrix}$$

后面的行列式在交换 $\eta_i$ 的指标的时候（也就是交换两列直接的列指标），并不一定会产生一个负号，因为任意一列和 $k=i$ 列交换的时候会产生一个新的行列式。但是这个新的行列式实际上可以是其他 $\det(B^{j}_{kl})$ 进行列交换的结果，也就是说单独一个 $\det(\mathcal B^{i}_{kl})$ 在对称话的时候没有办法写成 $\det(U^i)\det(V^i)$ ，但是整个 $\sum_i\det(B_{kl}^i)$ 对称化的结果一定是 $\sum_i \det(U^i)\det(V^i)$。 所以

$$\begin{aligned} &\mathcal S\sum_{ij}\mathrm{Cof}_{ij}(\mathcal B)\mathcal B_{ij}^{\rho(q)} \\ =&\frac{1}{N!}\prod_{i,j}\left(\frac{l_b^2}{l_B^2}\right)^{n_i+n_j+1}\prod_i e^{-|\eta_i|/2l^2}\int \prod_i d\mu_B(w_i) \sum_i|U^i(\bar\eta_,\eta+w)||V^i(\eta,\bar\eta-\bar w )| \end{aligned}$$

其中

$$U^i_{kl} = \begin{cases} U_{kl},l\neq i\\ U^{\rho_q}_{kl},l=i \end{cases}\qquad V^i_{kl} = \begin{cases} V_{kl},l\neq i\\ V^{\rho_q}_{kl},l=i \end{cases}$$

含义就是替换第 l 列为算符的矩阵元，这里行指标是轨道指标，列指标是坐标指标。 例子：

$$T_1= \widetilde F_1(x_1)F_2(x_2) \begin{vmatrix} \widetilde G_1(x_1)&G_1(x_2)\\ \widetilde G_2(x_1)&G_2(x_2) \end{vmatrix}, \quad T_2= F_1(x_1)\widetilde F_2(x_2) \begin{vmatrix} G_1(x_1)&\widetilde G_1(x_2)\\ G_2(x_1)&\widetilde G_2(x_2) \end{vmatrix}.$$

$T_1+T_2$ 整体对称化。结果是

$$\begin{aligned} &\frac12 \begin{vmatrix} \widetilde F_1(x_1)&F_1(x_2)\\ \widetilde F_2(x_1)&F_2(x_2) \end{vmatrix} \begin{vmatrix} \widetilde G_1(x_1)&G_1(x_2)\\ \widetilde G_2(x_1)&G_2(x_2) \end{vmatrix} \\ + &\frac12 \begin{vmatrix} F_1(x_1)&\widetilde F_1(x_2)\\ F_2(x_1)&\widetilde F_2(x_2) \end{vmatrix} \begin{vmatrix} G_1(x_1)&\widetilde G_1(x_2)\\ G_2(x_1)&\widetilde G_2(x_2) \end{vmatrix}. \end{aligned}$$

考虑算符 $\rho_q = \exp(iq\cdot (r_1-r_2))$，

$$\sum_{a\neq b}\det F^{(a:+q,b:-q)}\det G^{(a:+,b:-)}$$

这里 $F^{(a:+q,b:-q)}$ 含义就是替换 $a$ 列的为 $+q$ 的矩阵元，替换 $b$ 列为 $-q$ 的矩阵元。

列替换行列式的快速计算

$$\det A^{(a\leftarrow u, b\leftarrow v)} = \det A \begin{vmatrix} (A^{-1}u)_a &(A^{-1}v)_a \\ (A^{-1}u)_b &(A^{-1}v)_b \end{vmatrix}$$

也就是求解

$$Ay = u, Az = v$$

可观测量的计算

单体算符

$$O = \sum O(z_n,\bar z_n)$$

$$\begin{aligned} \braket{\Psi|O|\Psi} =& \sum_n\int \prod_id\mu_b(\eta_i)\mu_b(\eta_i')\prod_{i<j}(\bar\eta_i-\bar\eta_j)^2(\eta_i'-\eta_j')^2\\ &\frac{1}{N!}\sum_{P,P'}(-1)^{P+P'}A_{P(1)P'(1)}(\eta_{P(1)},\bar\eta'_{P(1)})\dots A^O_{P(n)P'(n)}(\eta_{P(n)},\bar\eta'_{P(n)})\dots \\ =&\sum_n\int \prod_id\mu_b(\eta_i)\mu_b(\eta_i')\prod_{i<j}(\bar\eta_i-\bar\eta_j)^2(\eta_i'-\eta_j')^2\\ &\frac{1}{N!}\sum_{P}(-1)^{P}\sum_{j}(-1)^{P(n)+j}M_{P(n)j} A^O_{P(n)j} \\ =&\int \prod_id\mu_b(\eta_i)\mu_b(\eta_i')\prod_{i<j}(\bar\eta_i-\bar\eta_j)^2(\eta_i'-\eta_j')^2\\ &\frac{1}{N!}\sum_{P}(-1)^{P}\sum_n\sum_{j}(-1)^{P(n)+j}M_{P(n)j} A^O_{P(n)j} \\ =&\int \prod_id\mu_b(\eta_i)\mu_b(\eta_i')\prod_{i<j}(\bar\eta_i-\bar\eta_j)^2(\eta_i'-\eta_j')^2\\ &\frac{1}{N!}\sum_{P}(-1)^{P}(-1)^{P}\sum_n\sum_{j}(-1)^{n+j}M_{nj} A^O_{nj} \\ =&\int \prod_id\mu_b(\eta_i)\mu_b(\eta_i')\prod_{i<j}(\bar\eta_i-\bar\eta_j)^2(\eta_i'-\eta_j')^2\sum_{ij}(-1)^{i+j}M_{ij} A^O_{ij} \\ =&\int \prod_id\mu_b(\eta_i)\prod_{i<j}|\bar\eta_i-\bar\eta_j|^4\sum_{ij}(-1)^{i+j}\mathcal M_{ij} \mathcal{B}^O_{ij} \end{aligned}$$

其中 $(-1)^{i+j}\mathcal M_{ij}$ 是 $\mathcal{B}_{ij}$ 的代数余子式，$\mathcal B_{ij}^O$ 是算符 $O$ 对应的矩阵元

$$\mathcal B_{ij}^O(\eta_i,\bar\eta_i) = \exp(-2l_b^2\partial_{\eta_i}\partial_{\bar\eta_i})\tilde{A}^O_{ij}(\eta_i,\bar\eta_i)$$

这里直接计算密度算符 $\rho(\boldsymbol q)=\sum \exp(i\boldsymbol q\cdot \boldsymbol r)$ 的矩阵元。

$$\begin{aligned} &\tilde A_{ij}^O(\eta_i,\bar\eta_i) \\ =&\exp(-\frac{\eta_i\bar\eta_i}{2l_b^2})A_{ij}^O(\eta_i,\bar\eta_i)\\ =&\exp(-\frac{\eta_i\bar\eta_i}{2l_b^2})\eta_i^n(\bar\eta_i)^{n'}\sum_{kk'}^{nn'} \begin{pmatrix} n\\ k \end{pmatrix} \begin{pmatrix} n'\\ k' \end{pmatrix} \alpha^{k+k'} \frac{l_b^{k+k'}}{\eta_i^k\bar\eta_i^{k'}} \frac{m!}{(m-k)!}\frac{m'!}{(m'-k')!} \\ &\int d\mu_B(z)\exp\left(\frac{\bar z\eta_i+z\bar\eta_i}{2l_b^2}\right) \left(\frac{\bar z}{l_B}\right)^{m-k} \exp(iq\cdot \boldsymbol r_i) \left(\frac{z}{l_B}\right)^{m'-k'}\\ \end{aligned}$$

由于

$$\exp\left(\frac{\bar z\eta_i+z\bar\eta_i'}{2l_b^2}\right)\exp(iq\cdot \boldsymbol r_i) = \exp\left(\frac{\bar z(\eta_i+l_b^2(iq_x-q_y))+z(\bar\eta_i+l_b^2(iq_x+q_y))}{2l_b^2}\right)$$

那么 $$ \begin{aligned} =&\exp\left((\rho-1)\frac{\eta_i\bar\eta_i}{2l_b^2}\right) \eta_i^n(\bar\eta_i)^{n’}\sum_{kk’}^{nn’} \begin{pmatrix} n\ k \end{pmatrix} \begin{pmatrix} n’\ k’ \end{pmatrix} \alpha^{k+k’} \frac{l_b^{k+k’}}{\eta_i^k\bar\eta_i^{k’}} \frac{m!}{(m-k)!}\frac{m’!}{(m’-k’)!}\

&\sum_{\gamma=0}^{\min(a,b)} \begin{pmatrix}a\ \gamma\end{pmatrix} \begin{pmatrix}b\ \gamma\end{pmatrix} \gamma!2^{\gamma}\rho^{(a+b-2\gamma)/2}\left(\frac{\eta_i+l_b^2(iq_x-q_y)}{l_b}\right)^{b-\gamma}\left(\frac{\bar\eta_i+l_b^2(iq_x+q_y)}{l_b}\right)^{a-\gamma}\ \end{aligned} $$ 如此我们便得到了 $\rho(\boldsymbol q)=\sum \exp(iq\cdot r_i)$ 的期望值

$$\braket{\Psi|\rho(\boldsymbol q)|\Psi} = \int d\mu_b(\eta)\prod_{i<j}|\eta_i-\eta_j|^4 \sum_{ij}\mathrm{Cof}_{ij}(\mathcal B)\mathcal B_{ij}^{\rho(q)}$$

其中 $\mathrm{Cof}_{ij}(\mathcal B)$ 是 $\mathcal B$ 的代数余子式，如果 $\mathcal B$ 存在逆，那么

$$\mathrm{Cof}_{ij}(\mathcal B) = \det(\mathcal B)(\mathcal B^{-1})_{ji}$$

那么

$$\sum_{ij}\det(\mathcal B)(\mathcal B^{-1})_{ji}\mathcal B^{\rho(q)}_{ij} = \det(\mathcal B)\mathrm{Tr}(\mathcal B^{-1}\mathcal B^{\rho(q)})$$

$$\braket{\Psi|\rho(\boldsymbol q)|\Psi} = \int d\mu_b(\eta)\prod_{i<j}|\eta_i-\eta_j|^4 \det(\mathcal B)\mathrm{Tr}(\mathcal B^{-1}\mathcal B^{\rho(q)})$$

$$\braket{\rho(\boldsymbol q)} = \frac{\braket{\Psi|\rho(\boldsymbol q)|\Psi}}{\braket{\Psi|\Psi}}$$

两体算符

需要计算 $$ \begin{aligned} &\tilde A_{ij,kl}^O(\eta_i,\eta_j, \bar\eta_i,\bar\eta_j) \ =&\exp(-\frac{\eta_i\bar\eta_i}{2l_b^2}-\frac{\eta_j\bar\eta_j}{2l_b^2})

A_{ij,kl}^O(\eta_i,\eta_j, \bar\eta_i,\bar\eta_j) \

=&\exp(-\frac{\eta_i\bar\eta_i}{2l_b^2}-\frac{\eta_j\bar\eta_j}{2l_b^2})

\int d\mu_B(z_1)d\mu_B(z_2)

\varphi_i(\bar z_1,\eta_i)\varphi_j(\bar z_2,\eta_j)O(\boldsymbol z_1,\boldsymbol z_2)\varphi_k(z_1,\bar\eta_i)\varphi_l(z_2,\bar\eta_j)

\end{aligned}

$$那么$$

\mathcal B^O_{ij,kl}(\eta_i,\eta_j, \bar\eta_i,\bar\eta_j) = \exp(-2l_b^2\partial_{\eta_i}\partial_{\bar\eta_i})\exp(-2l_b^2\partial_{\eta_j}\partial_{\bar\eta_j})\tilde A_{ij,kl}^O(\eta_i,\eta_j, \bar\eta_i,\bar\eta_j)

$$那么$$

\braket{\Psi|O|\Psi}= \frac{1}{2}\int d\mu_b(\eta)|\Delta(\eta)|^2\sum_{ij,kl}\mathrm{Cof}{ij,kl}^{(2)}[\mathcal B]\mathcal B{ij,kl}^O(\eta_i,\eta_j, \bar\eta_i,\bar\eta_j)

$$其中$$

\mathrm{Cof}_{ij,kl}^{(2)}[\mathcal B] $$ 是 $\mathcal B$ 的二阶代数余子式。 考虑算符 $O(z_1,z_2) = \exp(iq\cdot(r_1-r_2))$，那么可以直接拆成两个独立的部分

$$\begin{aligned} &\tilde A^{O}_{ij,kl} \\ =&\tilde A^{\rho(q)}_{ik}(\eta_i,\bar\eta_i)\tilde A^{\rho(-q)}_{jl}(\eta_j,\bar\eta_j) \end{aligned}$$

同样的，

$$\mathcal B^O_{ij,kl}(\eta_i,\eta_j, \bar\eta_i,\bar\eta_j) = \mathcal B_{ik}^{\rho(q)}(\eta_i,\bar\eta_i)\mathcal B^{\rho(-q)}_{jl}(\eta_j,\bar\eta_j)$$

这里 $i\neq j, \,k\neq l$ ，如果 $\mathcal B$ 存在逆，那么

$$\mathrm{Cof}_{ij,kl}^{(2)}(\mathcal B) = \det(\mathcal B)\left[(\mathcal B^{-1})_{ki}\mathcal B^{-1})_{lj}-(\mathcal B^{-1})_{kj}(\mathcal B^{-1})_{li}\right]$$

所以

$$\sum_{ij,kl}\mathrm{Cof}_{ij,kl}^{(2)}[\mathcal B]\mathcal B_{ij,kl}^O(\eta_i,\eta_j, \bar\eta_i,\bar\eta_j) = \det(\mathcal B)(\mathrm{Tr}[\mathcal B^{-1}\mathcal B^{\rho(q)}]\mathrm{Tr}[\mathcal B^{-1}\mathcal B^{\rho(-q)}]-\mathrm{Tr}[\mathcal B^{-1}\mathcal B^{\rho(-q)}\mathcal B^{-1}\mathcal B^{\rho(q)}])$$

因此

$$\begin{aligned} &\braket{\Psi|:\rho(-q)\rho(q):|\Psi}\\ = &\int d\mu_b(\eta)|\Delta(\eta)|^2\det(\mathcal B)(\mathrm{Tr}[\mathcal B^{-1}\mathcal B^{\rho(q)}]\mathrm{Tr}[\mathcal B^{-1}\mathcal B^{\rho(-q)}]-\mathrm{Tr}[\mathcal B^{-1}\mathcal B^{\rho(-q)}\mathcal B^{-1}\mathcal B^{\rho(q)}]) \end{aligned}$$

而

$$\braket{\Psi|:\rho(-q)\rho(q):|\Psi} = \braket{\Psi|\rho(-q)\rho(q)|\Psi}-N\braket{\Psi|\Psi}$$

前面讨论过计算出 $\mathcal B$ 的复杂度占据主导的，这里一系列矩阵相乘，求逆，求迹的过程消耗的时间不是主要的。

由密度算符计算单体算符和相互作用能量

单体算符

对于任意一个单体算符，可以进行傅里叶展开

$$o(\boldsymbol r) = \int \frac{d^2 q}{(2\pi)^2}\exp(iq\cdot \boldsymbol r)o(\boldsymbol q)$$

那么

$$O = \sum o(\boldsymbol r_i) = \int \frac{d^2q}{(2\pi)^2}\rho(\boldsymbol q)o(q)$$

相互作用能

由于相互作用具有平移不变性，也可以用密度算符展开：

$$\begin{aligned} &\hat H_{int} \\ = &\frac{1}{2}\sum_{i\neq j}V(\boldsymbol r_i-\boldsymbol r_j) \\ = &\frac{1}{2}\sum_{i\neq j}\int \frac{d^2q}{(2\pi)^2}V(q)\exp(iq(r_i-r_j)) \\ = &\frac{1}{2}\int\frac{d^2q}{(2\pi)^2}V(q)[\rho(-\boldsymbol q)\rho(\boldsymbol q)-N] \end{aligned}$$

那就需要计算 $\braket{\Psi|\rho(-q)\rho(q)|\Psi}$

结构因子

就是 $\braket{\rho(-q)\rho(q)}$

对关联函数 $\braket{:\rho(r)\rho(0):}$

对静态结构因子进行傅里叶变换可以得到

直接采样计算单体算符和两体算符

单体算符

单体算符和密度算符的处理一样，只需要计算

$$\begin{aligned} &\tilde A_{ij}^O(\eta_i,\bar\eta_i) \\ =&\exp(-\frac{\eta_i\bar\eta_i}{2l_b^2})A_{ij}^O(\eta_i,\bar\eta_i) \\ =&\exp(-\frac{\eta_i\bar\eta_i}{2l_b^2})\int d\mu_B(z)\exp(\frac{\bar z_i\eta_i+z_i\bar\eta_i}{l_b^2})\varphi_i(\bar z,\eta_i)O(z,\bar z)\varphi_j(z,\bar\eta_i) \end{aligned}$$

两体算符

需要计算 $$ \begin{aligned} &\tilde A_{ij,kl}^O(\eta_i,\eta_j, \bar\eta_i,\bar\eta_j) \ =&\exp(-\frac{\eta_i\bar\eta_i}{2l_b^2}-\frac{\eta_j\bar\eta_j}{2l_b^2})

A_{ij,kl}^O(\eta_i,\eta_j, \bar\eta_i,\bar\eta_j) \

=&\exp(-\frac{\eta_i\bar\eta_i}{2l_b^2}-\frac{\eta_j\bar\eta_j}{2l_b^2})

\int d\mu_B(z_1)d\mu_B(z_2)

\varphi_i(\bar z_1,\eta_i)\varphi_j(\bar z_2,\eta_j)O(\boldsymbol z_1,\boldsymbol z_2)\varphi_k(z_1,\bar\eta_i)\varphi_l(z_2,\bar\eta_j)

\end{aligned}

$$那么$$

\mathcal B^O_{ij,kl}(\eta_i,\eta_j, \bar\eta_i,\bar\eta_j) = \exp(-2l_b^2\partial_{\eta_i}\partial_{\bar\eta_i})\exp(-2l_b^2\partial_{\eta_j}\partial_{\bar\eta_j})\tilde A_{ij,kl}^O(\eta_i,\eta_j, \bar\eta_i,\bar\eta_j)

$$总能量就是$$

\mathcal E = \frac{1}{2}\int d\mu_b(\eta)|\Delta(\eta)|^2\sum_{ij,kl}\mathrm{Cof}{ij,kl}^{(2)}[\mathcal B]\mathcal B{ij,kl}^O(\eta_i,\eta_j, \bar\eta_i,\bar\eta_j)

$$其中$$

\mathrm{Cof}_{ij,kl}^{(2)}[\mathcal B] $$ 是 $\mathcal B$ 的二阶代数余子式。

$$\int \prod_i d^2 \eta_id^2\xi_i \exp(-\frac{|\xi_i|^2+|\eta|^2}{2l_B^2})\prod_{i<j}|\eta_i-\eta_j|^4\det(U)\det(V)$$

$\xi$ 可以用正态分布生成，$\eta$ 则通过Metropolis采样得到。我的结果显示存在多个 $\Lambda$ level 的时候依然有符号问题。下面是一个参数下的结果： p=2, M = 20, interval = 50*N, measurement_samples=20000 p是 $\nu=p/(2p+1)$ 中的p，M是一个 lambda level内轨道数，interval 是采样间隔，measurement_samples是有效样本数。

在这个参数下，每次运行之后、平均之后的det(U)det(V)的数量级、正负号甚至都会发生变化，这表明符号问题引起的误差已经淹没掉真实值了。我们知道当 $\xi = 0$ 的时候，后面两个行列式是正定的，但是对于以 $\sigma = l_B$ 分布的 $\xi$ 而言，$\det(U)\det(V)$ 的正负是完全随机的。所以我想看一下，这个过程中发生了什么。下面一张图展示了正号比例随着 $\xi$ 方差的变化，也就是将 $\exp(-\frac{|\xi_i|^2}{2l_B^2})$ 替换成 $\exp(-\frac{|\xi_i|^2}{2(\lambda l_B)^2})$ ![[626b780e9099120d1d6f5ea9c14e9ce4.png]] 横轴是 $\lambda$ ，纵轴是 $\det(U)\det(V)$ 正号比例。 可以看到，当 $\lambda = 1$ 的时候，$\det(U)\det(V)$ 50% 是 ‘+’， 50% 是 ‘-’。

可以通过下面的例子看出符号问题的来源

$$U_{ij} = \varphi_i(\bar\eta_j,\eta_j+\xi_j), V_{ij} = \varphi_i(\eta_j,\bar\eta_j-\xi_j)$$

考虑每个 $\Lambda-level$ 只填充一个CF，即 $M=1$。一共 $p$ 个 $\Lambda$ level。 那么

$$U_{ij} = (\eta_j+\xi_j)^{n_i}, V_{ij} = (\bar\eta_j-\bar\xi_j)^{n_i}$$

从而

$$\det(U)\det(V) = \prod_{i<j}(\eta_i+\xi_i-\eta_j-\xi_j)(\bar\eta_i-\bar\xi_i-\bar\eta_j+\bar\xi_j)$$

当 $p = 2$ 的时候，只有两个粒子

$$\det(U)\det(V) = (\eta_1+\xi_1-\eta_2-\xi_2)(\bar\eta_1-\bar\xi_1-\bar\eta_2+\bar\xi_2)$$

对 $\xi$ 平均之后，可以直接丢掉 $\xi_i$ 的奇次项

$$\braket{\det(U)\det(V)}_\xi = |\eta_1-\eta_2|^2-\braket{|\xi_1|^2+|\xi_2|^2}_{\xi} = |\eta_1-\eta_2|^2-4\sigma^2$$

$\sigma$ 是 $\xi$ 分布的方差。可以看出，即便把 $\xi$ 的积分积掉，剩下的 $\eta$ 积分还会存在符号问题，只要采样得到的 $\eta$ 之间的间距小于 $2\sigma$ ，上面就给出负数。对于这样只有两个粒子的系统，采样得到的大部分结果其实都是正的，但是当系统变大的时候正负比例会逐渐平衡。

我的尝试： 通过在分布函数里面添加额外的Jastrow因子 $|\eta_i-\eta_j|^2$ 增加采样的 $\eta$ 间距，但好像没什么效果。可能还是需要解析层面让正负号配对一下。$