QMC

问题 9/27

The density compressibility. Defi:

$$\begin{equation}\tag{1} \kappa = \frac{\beta}{2L^2} \lgroup\langle(\sum_in_i)^2\rangle -\langle\sum_in_i\rangle^2\rgroup \end{equation}$$
and by definition,
$$\begin{equation}\tag{2} n_i = c_{i\uparrow}^{\dagger}c_{i\uparrow}+c_{i\downarrow}^{\dagger}c_{i\downarrow} \end{equation}$$
first item--average over sum's square:
$$\begin{equation}\tag{3} \sum_i\sum_j \langle (c_{i1}^{\dagger}c_{i1}c_{j1}^{\dagger}c_{j1} +c_{i1}^{\dagger}c_{i1}c_{j2}^{\dagger}c_{j2} +c_{i2}^{\dagger}c_{i2}c_{j1}^{\dagger}c_{j1} +c_{i2}^{\dagger}c_{i2}c_{j2}^{\dagger}c_{j2}) \rangle \end{equation}$$
use wick's theorem,
$$\begin{split} \langle c_{i1}^{\dagger}c_{i1}c_{j1}^{\dagger}c_{j1}\rangle&=\langle c_{i1}^{\dagger}c_{i1}\rangle\langle c_{j1}^{\dagger}c_{j1}\rangle+\langle c_{i1}^{\dagger}c_{j1}\rangle\langle c_{i1} c_{j1}^{\dagger}\rangle\\ &=\langle n_{i1}\rangle\langle n_{j1}\rangle + (\delta_{ij}-g_{ji,1})g_{ij,1} \\\\ \langle c_{i1}^{\dagger}c_{i1}c_{j2}^{\dagger}c_{j2}\rangle&=\langle c_{i1}^{\dagger}c_{i1}\rangle\langle c_{j2}^{\dagger}c_{j2}\rangle+\langle c_{i1}^{\dagger}c_{j2}\rangle\langle c_{i1} c_{j2}^{\dagger}\rangle\\ &=\langle n_{i1}\rangle\langle n_{j2}\rangle\\\\ \end{split}$$
so the first term can be like,
$$\langle n_{i1}\rangle\langle n_{j1}\rangle + (\delta_{ij}-g_{ji,1})g_{ij,1}+\langle n_{i1}\rangle\langle n_{j2}\rangle+\langle n_{i2}\rangle\langle n_{j2}\rangle + (\delta_{ij}-g_{ji,2})g_{ij,2}+\langle n_{i2}\rangle\langle n_{j1}\rangle$$
which is
$$\sum_{i,j}\langle n_i\rangle\langle n_j\rangle+(\delta_{ij}-g_{ji,1})g_{ij,1}+(\delta_{ij}-g_{ji,2})g_{ij,2}$$
and the second item is,
$$(\sum_i\langle n_i\rangle)^2=\sum_{i,j}\langle n_i\rangle\langle n_j\rangle$$
so by the Defi of $\kappa$,
$$\kappa = (\delta_{ij}-g_{ji,1})g_{ij,1}+(\delta_{ij}-g_{ji,2})g_{ij,2}$$

问题 9/26

Wick 定理

$$\begin{split} \langle n_{i\uparrow}n_{i\downarrow}\rangle =\langle c_{i\uparrow}^{\dagger}c_{i\uparrow} c_{i\downarrow}^{\dagger}c_{i\downarrow}\rangle =\langle n_{i\uparrow}\rangle\langle n_{i\downarrow}\rangle + \langle c_{i\uparrow}^{\dagger}c_{i\downarrow}\rangle\langle c_{i\uparrow} c_{i\downarrow}^{\dagger}\rangle \end{split}$$
why
$$\langle c_{i\uparrow}^{\dagger}c_{i\downarrow}\rangle\langle c_{i\uparrow} c_{i\downarrow}^{\dagger}\rangle=?~ 0$$

问题 9/18

接下来，我写一些我们要测的物理量的格林函数表达式。也摘录下程序中的几段。麻烦学长检查一下表达式...

1 Kinetic energy

• spin channel:
  $$\begin{split} E_{kin} &= -\frac{t}{N_0}\sum_{i,j,\alpha}\langle c^{\dagger}_{i\alpha}c_{j\alpha}+h.c.\rangle\\ &= \frac{t}{N_0}\sum_{i,j}(g_{ij,\uparrow}+g_{i,j,\downarrow}) \end{split}$$
  对应的代码

  do i = 1,Nsite
     do j = 1,Nsite
        g_hop = g_hop - T_hop*free_con(i,j)*( gfun_2nd(j,i,+1)+gfun_2nd(j,i,-1) )
     end do
  end do

• charge channel
  不区分spin向上向下，因此是

    do i=1,Nsite
      do j=1,Nsite
        g_hopup = g_hopup - gfun_up_2nd(i,j)*free_con(i,j)*T_hop*N0*2
      end do
    end do

2 on-site energy

• spin channel
  $$\begin{split} E_{int} & = \frac{U}{N_0}\sum_{i}\langle(n_{i\uparrow}-\frac{1}{2})(n_{i\downarrow}-\frac{1}{2})\rangle\\ & =\frac{U}{N_0}\sum_i[\langle n_{i\uparrow}n_{i\downarrow}\rangle - \frac{1}{2}\langle n_{i\uparrow}+n_{i\downarrow}\rangle+\frac{1}{4}] \end{split}$$
  其中$\langle n_{i\uparrow}\rangle=1-g_{ii,\uparrow},~~\langle n_{i\downarrow}\rangle=1-g_{ii,\downarrow},~~ \langle n_{i\uparrow}n_{i\downarrow}\rangle=(1-g_{ii,\uparrow})(1-g_{ii,\downarrow})$
  因此占据能的表达式为：
  $$\begin{split} E_{int} = \frac{1}{2N_0}\sum_i[2g_{ii,\uparrow}g_{ii,\downarrow}-g_{ii,\uparrow}-g_{ii,\downarrow}+\frac{1}{2}] \end{split}$$
  对应的代码是：

  do i=1,Nsite
     g_onsite = g_onsite + U/2.d0*( 2.d0*gfun_2nd(i,i,+1)*gfun_2nd(i,i,-1)  &
          - gfun_2nd(i,i,+1) - gfun_2nd(i,i,-1) + 1.d0/2.d0 )
  end do

• charge channel
  $$\begin{split} E_{int} & = \frac{U}{2N_0}\sum_i\langle (n_{i\uparrow}+n_{i\downarrow}-1)^2\rangle\\ & = \frac{U}{N_0}\sum_i\langle n_{i\uparrow}n_{i\downarrow}\rangle\\ & = \frac{U}{N_0}\sum_i(1-g_{ii})^2 \end{split}$$
  对应代码：

    do i=1,Nsite
      g_onsite = g_onsite + U*( (1.d0-gfun_up_2nd(i,i))**2.d0 )    &
                 *Nfavor*(Nfavor-1)/2.d0
    end do

3 spin-spin correlation

• spin channel
  定义算符$S_{\alpha\beta,i} = c_{i,\alpha}^{\dagger}c_{i,\beta} - \frac{1}{2N} n_i\delta_{\alpha\beta}$
  对$N=2$自旋1/2粒子，令$\alpha,\beta=~\uparrow,\downarrow$。那么有
  $$\begin{split} & S_{\uparrow\uparrow,i} = \frac{1}{2}(c_{i\uparrow}^{\dagger}c_{i\uparrow}-c_{i\downarrow}^{\dagger}c_{i\downarrow})=S^{z} = - S_{\downarrow\downarrow,i}\\ & S_{\uparrow\downarrow,i} = c_{i\uparrow}^{\dagger}c_{i\downarrow} = S_i^+ \\ & S_{\downarrow\uparrow,i} = c_{i\downarrow}^{\dagger}c_{i\uparrow} = S_i^- \end{split}$$
  求出一些形如$\langle S_{\alpha\beta,i}S_{\beta\alpha,j}\rangle$的spin-spin correlation的关系：
  $\mathcal{local~moments}:~\langle m_i^2\rangle=4\langle S_i^zS_i^z\rangle$
  $$\begin{split} Scor\_11:~~ \langle S_{i\uparrow\uparrow}S_{j\uparrow\uparrow}\rangle&=\langle S_i^zS_j^z\rangle\\ &=\frac{1}{4}(\langle c_{i\uparrow}^{\dagger}c_{i\uparrow}c_{j\uparrow}^{\dagger}c_{j\uparrow}+ c_{i\downarrow}^{\dagger}c_{i\downarrow}c_{j\downarrow}^{\dagger}c_{j\downarrow}-c_{i\uparrow}^{\dagger}c_{i\uparrow}c_{j\downarrow}^{\dagger}c_{j\downarrow}- c_{i\downarrow}^{\dagger}c_{i\downarrow}c_{j\uparrow}^{\dagger}c_{j\uparrow}\rangle)\\ & = \frac{1}{4}[(g_{ii}^{\uparrow}-g_{ii}^{\downarrow})+(\delta_{ji}-g_{ji}^{\uparrow})g_{ij}^{\uparrow}+(\delta_{ji}-g_{ji}^{\downarrow})g_{ij}^{\downarrow}] \end{split}$$
  $\mathcal{spin-spin~correlation}:~c_{\alpha\beta}(r)=\langle c_{i+r,\downarrow}^{\dagger}c_{i+r,\uparrow}c_{i\uparrow}^{\dagger}c_{i\downarrow}\rangle=\langle S^-_jS^+_i\rangle$
  $$\begin{split} Scor\_21:~~\langle S_i^-S_j^+\rangle &= \langle S_{\downarrow\uparrow,i} S_{\uparrow\downarrow,j} \rangle\\ &= \langle c_{i\downarrow}^{\dagger}c_{i\uparrow} \rangle \langle c_{j\uparrow}^{\dagger}c_{j\downarrow}\rangle + \langle c_{i\downarrow}^{\dagger}c_{j\downarrow}\rangle\langle c_{i\uparrow}c_{j\uparrow}^{\dagger}\rangle\\ &= (\delta_{ij}-g_{ji}^{\downarrow})g_{ij}^{\uparrow} \end{split}$$
  对应程序中的代码：

    do i=1,Nsite
     do j=1,Nsite
        delta = 0.d0
        if (i == j) delta = 1.d0
        Scor_11(i,j) = Scor_11(i,j) +  (gfun_2nd(i,i,+1) - gfun_2nd(i,i,-1)) * &
             (gfun_2nd(j,j,+1) - gfun_2nd(j,j,-1)) &
             + ( delta-gfun_2nd(j,i,+1) ) * gfun_2nd(i,j,+1)  &
             + ( delta-gfun_2nd(j,i,-1) ) * gfun_2nd(i,j,-1)
        Scor_22(i,j) = Scor_11(i,j)
        Scor_12(i,j) = Scor_12(i,j) + ( delta-gfun_2nd(j,i,+1) ) * gfun_2nd(i,j,-1)
        Scor_21(i,j) = Scor_21(i,j) + ( delta-gfun_2nd(j,i,-1) ) * gfun_2nd(i,j,+1)
     end do
  end do

• charge channel
  $\langle S_{i\uparrow\uparrow}S_{j\uparrow\uparrow}\rangle$简化为：
  $$\langle S_{i\uparrow\uparrow}S_{j\uparrow\uparrow}\rangle=\frac{1}{2}(\delta_{ji}-g_{ji})g_{ij}$$
  那么对所有自旋关联求和，即
  $$\langle\sum_{\alpha\beta}S_{\alpha\beta,i}S_{\beta\alpha,j}\rangle=3(\delta_{ij}-g_{ji})g_{ij}$$
  代码：

    do i=1,Nsite
    do j=1,Nsite
      delta = 0.d0
      if(i.eq.j) delta = 1.d0
      Scor(j,i) = Scor(j,i) + (delta-gfun_up_2nd(j,i))*gfun_up_2nd(i,j) *3
      ' 注: *3 没有在原程序中，因为他计算的只是Scor_12. '
      ' 注： 指标可能有点错误，应该是Scor(i,j) = Scor(i,j)+ ... '
    end do   
    end do

4 m(0),m(Q)

• spin channel
  按照$m(0),m(Q)$的定义，可知:
  $$\begin{split} m(0) &= \langle\sum_{ij}S_i^+S_j^-\rangle/N=\sum_{ij}\langle S_{\uparrow\downarrow,i} S_{\downarrow\uparrow,j} \rangle /N\\ &= \sum_{ij}(\delta_{ij}-g_{ji}^{\uparrow})g_{ij}^{\downarrow}/N\\ \\ m(Q) &= \langle\sum_{ij}(-1)^{i+j}S_i^+S_j^-\rangle = \sum_{ij}(-1)^{i+j}\langle S_{\uparrow\downarrow,i} S_{\downarrow\uparrow,j} \rangle/N\\ &= \sum_{ij}(-1)^{i+j}(\delta_{ij}-g_{ji}^{\uparrow})g_{ij}^{\downarrow}/N \end{split}$$
  所以代码具体为：

  FM = 0.d0
  call mn_lieb(pn)
  AF = 0.d0
  do i = 1, Nsite
     do j = 1, Nsite
        FM = FM + Scor_12(i,j)
        AF = AF + pn(i)*pn(j)*Scor_12(i,j)
     end do
  end do

• charge channel
  在陈国的code中，由于他的Scor的定义不同,

'm(0)'
    do i=1,Nsite
    do j=1,Nsite
      sus = sus + Scor(i,j)
    end do
    end do
'm(Q)'
    call mn_lieb_(pn)
    do i=1,Nsite
    do j=1,Nsite
      afm  = afm + pn(i)*pn(j)*Scor(i,j)
    end do
    end do

计算结果对比

===== System Parameters
total core number 3
Row,Col 3 3
Nsite 27
2N 2
U 4.00000000000000
beta 38.0000000000000
interval 1900
p0 10
===== ===== ====== ======

items	charge channel	spin channel
local magnetic moment (with err)	0.789 $\pm$ 0.003	0.787 $\pm$ 0.001
Kinetic energy (with err)	-1.084 $\pm$ 0.002	-1.084 $\pm$ 0.003
Total energy (with bar)	-0.664 $\pm$ 0.005	-1.659 $\pm$ 0.001 + U/4
m(0) (with err)	0.63 $\pm$ 0.01	0.81 $\pm$ 0.04 *HERE
m(Q) (with err)	2.23 $\pm$ 0.02	2.90 $\pm$ 0.09 *HERE

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问题 9/5

• 1
  HS 分解：
  $$exp(-\Delta\tau U\sum_i(n_{i,\uparrow}-\frac{1}{2})(n_{i,\downarrow}-\frac{1}{2}))=C\sum_{s_1,...,s_N=\pm1}exp(\lambda\sum_is_i(n_{i,\uparrow}-n_{i,\downarrow}))$$
  其中$\lambda=cosh(\Delta\tau U/2)$。
  Assaad文章里说上式，对一个给定的辅助场$\{s_i\}$，破坏了SU(2)-spin symmetry, i.e. $c_{j,\sigma}\rightarrow[exp(i\phi\vec{e}\vec{\sigma}/2)]_{\sigma,s'}c_{j,s'}$是为什么呢？如果把这个变换作用在$(n_{i,\uparrow}-n_{i,\downarrow})$上

问题 9/3

• 1
  $\hat{c}_i^{\dagger}\sigma\hat{c}_i/2$就是$\hat{s}_i^z$。
  那$\langle (\hat{c}_i^{\dagger}\sigma\hat{c}_i/2)^2\rangle=\langle (\hat{s}_i^z)^2\rangle$和由$M_i$的定义里$M_i^2=\langle\hat{s}_i^z\rangle \langle\hat{s}_i^z\rangle$都是对一个基态的本征值，所以是相同的？

  然后$m(0)$里是对所有格点求和，设原胞数是$n$，那$m(0)=\frac{1}{n*(L_x*L_y/a^2-1)}\sum_{i,j}\langle S_i^+S_j^-\rangle$是个双粒子算符；而$M=\sum_iM_i$只是个单粒子算符啊，怎么相等的。$\frac{1}{n}\sum_{i,j}\langle S_i^+S_j^-\rangle=22\sum_i M_i$

• 2
  $M_i\rightarrow\sum_R\langle S_R^z(i)\rangle$的对应是说i格点的自旋完全由处于vacancy引入的零能态$\vert\psi_0\rangle$的电子决定，那不在零能态的电子没有贡献吗？

• 3
  S 对称性。$S=P·D$里$P=R·K$的定义，$R: \hat{c}_i\rightarrow (-1)\hat{d}^{\dagger}_i$，求出$PHP^{\dagger}$是