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Pierre-Francois Loos 2022-09-21 11:49:50 +02:00
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@ -673,6 +673,8 @@ It is then a trivial matter to integrate out exactly the $n_k$ variables, leadin
\\ \\
+ \sum_{p=1}^{\infty} \sum_{I_1 \notin \cD_0, \hdots , I_p \notin \cD_{p-1}} + \sum_{p=1}^{\infty} \sum_{I_1 \notin \cD_0, \hdots , I_p \notin \cD_{p-1}}
\qty[ \prod_{k=0}^{p-1} \mel{ I_k }{ P_k \qty( H-E \Id )^{-1} P_k (-H)(1-P_k) }{ I_{k+1} } ] \qty[ \prod_{k=0}^{p-1} \mel{ I_k }{ P_k \qty( H-E \Id )^{-1} P_k (-H)(1-P_k) }{ I_{k+1} } ]
\\
\times
\mel{ I_p }{ P_p \qty( H-E \Id)^{-1} P_p }{ I_N } \mel{ I_p }{ P_p \qty( H-E \Id)^{-1} P_p }{ I_N }
\end{multline} \end{multline}
As an illustration, Appendix \ref{app:A} reports the exact derivation of this formula in the case of a two-state system. As an illustration, Appendix \ref{app:A} reports the exact derivation of this formula in the case of a two-state system.
@ -688,8 +690,9 @@ In fact, there is a more direct way to derive the same equation by resorting to
where $H_0$ is some arbitrary reference Hamiltonian, we have the Dyson equation where $H_0$ is some arbitrary reference Hamiltonian, we have the Dyson equation
\be \be
\label{eq:GE} \label{eq:GE}
G^E_{ij}= \titou{G^E_{0,ij}} + \sum_{kl} G^{E}_{0,ik} (H_0-H)_{kl} G^E_{lj}. G^E_{ij} = G^E_{0,ij} + \sum_{kl} G^{E}_{0,ik} (H_0-H)_{kl} G^E_{lj},
\ee \ee
with $G^E_{0,ij} = \mel{i}{ \qty( H_0-E \Id )^{-1} }{j}$.
Let us choose $H_0$ such that $\mel{ i }{ H_0 }{ j } = \mel{ i }{ P_i H P_i }{ j }$ for all $i$ and $j$. Let us choose $H_0$ such that $\mel{ i }{ H_0 }{ j } = \mel{ i }{ P_i H P_i }{ j }$ for all $i$ and $j$.
Then, The Dyson equation \eqref{eq:GE} becomes Then, The Dyson equation \eqref{eq:GE} becomes
\begin{multline} \begin{multline}
@ -794,7 +797,7 @@ Thus, from a practical point of view, a trade-off has to be found between the \t
Let us consider the one-dimensional Hubbard Hamiltonian for a chain of $N$ sites Let us consider the one-dimensional Hubbard Hamiltonian for a chain of $N$ sites
\be \be
\hat{H}= -t \sum_{\expval{ i j } \sigma} \hat{a}^+_{i\sigma} \hat{a}_{j\sigma} H= -t \sum_{\expval{ i j } \sigma} \hat{a}^+_{i\sigma} \hat{a}_{j\sigma}
+ U \sum_i \hat{n}_{i\uparrow} \hat{n}_{i\downarrow}, + U \sum_i \hat{n}_{i\uparrow} \hat{n}_{i\downarrow},
\ee \ee
where $\langle i j\rangle$ denotes the summation over two neighboring sites, $\hat{a}_{i\sigma}$ ($\hat{a}_{i\sigma}$) is the fermionic creation (annihilation) operator of an spin-$\sigma$ electron (with $\sigma$ = $\uparrow$ or $\downarrow$) on site $i$, $\hat{n}_{i\sigma} = \hat{a}^+_{i\sigma} \hat{a}_{i\sigma}$ the number operator, $t$ the hopping amplitude, and $U$ the on-site Coulomb repulsion. where $\langle i j\rangle$ denotes the summation over two neighboring sites, $\hat{a}_{i\sigma}$ ($\hat{a}_{i\sigma}$) is the fermionic creation (annihilation) operator of an spin-$\sigma$ electron (with $\sigma$ = $\uparrow$ or $\downarrow$) on site $i$, $\hat{n}_{i\sigma} = \hat{a}^+_{i\sigma} \hat{a}_{i\sigma}$ the number operator, $t$ the hopping amplitude, and $U$ the on-site Coulomb repulsion.
@ -822,17 +825,19 @@ The parameters $\alpha$ and $\beta$ are optimized by minimizing the variational
As discussed above, the efficiency of the method depends on the choice of states forming each domain. As discussed above, the efficiency of the method depends on the choice of states forming each domain.
As a general guiding principle, it is advantageous to build domains associated with a large average trapping time in order to integrate out the most important part of the Green's matrix. As a general guiding principle, it is advantageous to build domains associated with a large average trapping time in order to integrate out the most important part of the Green's matrix.
Here, as a first illustration of the method, we shall consider the large-$U$ regime of the Hubbard model where the construction of such domains is rather natural. Here, as a first illustration of the method, we shall consider the large-$U$ regime of the Hubbard model where the construction of such domains is rather natural.
At large $U$ and half-filling, the Hubbard model approaches the Heisenberg limit where only the $2^N$ states with no double occupancy, $n_D(n)=0$, have a significant weight in the wave function. Indeed, at large $U$ and half-filling, the Hubbard model approaches the Heisenberg limit where only the $2^N$ states with no double occupancy, $n_D(n)=0$, have a significant weight in the wave function.
The contribution of the other states vanishes as $U$ increases with a rate increasing sharply with $n_D(n)$. The contribution of the other states vanishes as $U$ increases with a rate increasing sharply with $n_D(n)$.
In addition, for a given number of double occupations, configurations with large values of $n_A(n)$ are favored due to their high kinetic energy (electrons move more easily). In addition, for a given number of double occupations, configurations with large values of $n_A(n)$ are favored due to their high kinetic energy.
Therefore, we build domains associated with small $n_D$ and large $n_A$ in a hierarchical way as described below. Therefore, we build domains associated with small $n_D$ and large $n_A$ in a hierarchical way as described below.
For simplicity and decreasing the number of diagonalizations to perform, we shall consider only one non-trivial domain called here the main domain and denoted as $\cD$.
This domain will be chosen common to all states belonging to it, that is For simplicity and reducing the number of diagonalizations to perform, we shall consider only one non-trivial domain called here the main domain and denoted as $\cD$.
This domain will be chosen common to all states belonging to it, that is,
\be \be
\cD_i= \cD \qq{for} \ket{i} \in \cD. \cD_i= \cD \qq{for} \ket{i} \in \cD.
\ee \ee
For the other states, we choose a single-state domain For the other states, we choose a single-state domain as
\be \be
\cD_i= \qty{ \ket{i} } \qq{for} \ket{i} \notin \cD. \cD_i= \qty{ \ket{i} } \qq{for} \ket{i} \notin \cD.
\ee \ee
@ -840,15 +845,15 @@ To define $\cD$, let us introduce the following set of states
\be \be
\titou{\cS_{ij} = \qty{ \ket{n} : n_D(n)=i \land n_A(n)=j }}. \titou{\cS_{ij} = \qty{ \ket{n} : n_D(n)=i \land n_A(n)=j }}.
\ee \ee
$\cD$ is defined as the set of states having up to $n_D^\text{max}$ double occupations and, for each state with a number of double occupations equal to $m$, a number of nearest-neighbor antiparallel pairs between $n_A^\text{min}(m)$ and $n_A^\text{max}(m)$. which means that $\cD$ contains the set of states having up to $n_D^\text{max}$ double occupations and, for each state with a number of double occupations equal to $m$, a number of nearest-neighbor antiparallel pairs between $n_A^\text{min}(m)$ and $n_A^\text{max}(m)$.
Here, $n_A^\text{max}(m)$ will not be varied and taken to be the maximum possible for a given $m$, $n_A^\text{max}(m)= \max(N-1-2m,0)$. Here, $n_A^\text{max}(m)$ will not be varied and taken to be the maximum possible for a given $m$, $n_A^\text{max}(m)= \max(N-1-2m,0)$.
Using these definitions, the main domain is taken as the union of some elementary domains Using these definitions, the main domain is taken as the union of some elementary domains
\be \be
\cD = \cup_{n_D=0}^{n_D^\text{max}}\cD(n_D,n_A^{\rm min}(n_D)) \cD = \bigcup_{n_D=0}^{n_D^\text{max}}\cD(n_D,n_A^{\rm min}(n_D))
\ee \ee
where the elementary domain is defined as where the elementary domain is defined as
\be \be
\cD(n_D,n_A^\text{min}(n_D))=\cup_{ n_A^\text{min}(n_D) \leq j \leq n_A^{\rm max}(n_D)} \cS_{n_D j} \cD(n_D,n_A^\text{min}(n_D))=\bigcup_{ n_A^\text{min}(n_D) \leq j \leq n_A^{\rm max}(n_D)} \cS_{n_D j}
\ee \ee
The two quantities defining the main domain are thus $n_D^\text{max}$ and $n_A^\text{min}(m)$. The two quantities defining the main domain are thus $n_D^\text{max}$ and $n_A^\text{min}(m)$.
To give an illustrative example, let us consider the 4-site case. There are 6 possible elementary domains To give an illustrative example, let us consider the 4-site case. There are 6 possible elementary domains
@ -903,87 +908,90 @@ and
Let us begin with a small chain of 4 sites with $U=12$. Let us begin with a small chain of 4 sites with $U=12$.
From now on, we shall take $t=1$. From now on, we shall take $t=1$.
The size of the linear space is ${\binom{4}{2}}^2= 36$ and the ground-state energy obtained by exact diagonalization is $E_0=-0.768068...$. The two variational parameters of the trial vector have been optimized and fixed at the values of $\alpha=1.292$, and $\beta=0.552$ with a variational energy of $E_\text{T}=-0.495361...$. In what follows The size of the linear space is ${\binom{4}{2}}^2= 36$ and the ground-state energy obtained by exact diagonalization is $E_0=-0.768068...$.
$|I_0\rangle$ will be systematically chosen as one of the two N\'eel states, {\it e.g.} $|I_0\rangle =|\uparrow,\downarrow, \uparrow,...\rangle$. The two variational parameters of the trial vector have been optimized and fixed at the values of $\alpha=1.292$, and $\beta=0.552$ with a variational energy of $E_\text{T}=-0.495361...$.
In what follows $|I_0\rangle$ will be systematically chosen as one of the two N\'eel states, {\it e.g.} $|I_0\rangle =|\uparrow,\downarrow, \uparrow,...\rangle$.
Figure \ref{fig1} shows the convergence of $H_p$ as a function of $p$ for different values of the reference energy $E$. Figure \ref{fig1} shows the convergence of $H_p$ as a function of $p$ for different values of the reference energy $E$.
We consider the simplest case where a single-state domain is associated to each state. We consider the simplest case where a single-state domain is associated to each state.
Five different values of $E$ have been chosen, namely $E=-1.6,-1.2,-1,-0.9$, and Five different values of $E$ have been chosen, namely $E=-1.6,-1.2,-1,-0.9$, and $E=-0.8$.
$E=-0.8$. Only $H_0$ is computed analytically ($p_{ex}=0$). At the scale of the figure error bars are too small to be perceptible. Only $H_0$ is computed analytically ($p_{ex}=0$). At the scale of the figure error bars are too small to be perceptible.
When $E$ is far from the exact value of $E_0=-0.768...$ the convergence is very rapid and only a few terms of the $p$-expansion are necessary. When $E$ is far from the exact value of $E_0=-0.768...$ the convergence is very rapid and only a few terms of the $p$-expansion are necessary.
In constrast, when $E$ approaches the exact energy, In contrast, when $E$ approaches the exact energy, a slower convergence is observed, as expected from the divergence of the matrix elements of the Green's matrix at $E=E_0$ where the expansion does not converge at all.
a slower convergence is observed, as expected from the divergence of the matrix elements of the Note the oscillations of the curves as a function of $p$ due to a parity effect specific to this system.
Green's matrix at $E=E_0$ where the expansion does not converge at all. Note the oscillations of the curves as a function of $p$ due to a In practice, it is not too much of a problem since a smoothly convergent behavior is nevertheless observed for the even- and odd-parity curves.
parity effect specific to this system. In practice, it is not too much of a problem since The ratio, $\cE_{QMC}(E,p_{ex}=1,p_{max})$ as a function of $E$ is presented in figure \ref{fig2}. Here, $p_{max}$ is taken sufficiently large so that the convergence at large $p$ is reached.
a smoothly convergent behavior is nevertheless observed for the even- and odd-parity curves. The values of $E$ are $-0.780$, $-0.790$, $-0,785$, $-0,780$, and $-0.775$. For smaller $E$ the curve is extrapolated using the two-component expression. The estimate of the energy obtained from $\cE(E)=E$ is $-0.76807(5)$ in full agreement with the exact value of $-0.768068...$.
The ratio, $\cE_{QMC}(E,p_{ex}=1,p_{max})$ as a function of $E$ is presented in figure \ref{fig2}. Here, $p_{max}$ is taken sufficiently large
so that the convergence at large $p$ is reached. The values of $E$ are $-0.780$, $-0.790$, $-0,785$, $-0,780$, and $-0.775$. For smaller $E$ the curve is extrapolated using
the two-component expression. The estimate of the energy obtained from $\cE(E)=E$ is $-0.76807(5)$ in full agreement with the exact value of $-0.768068...$.
%%% FIG 1 %%%
\begin{figure}[h!] \begin{figure}[h!]
\includegraphics[width=\columnwidth]{fig1} \includegraphics[width=\columnwidth]{fig1}
\caption{1D-Hubbard model, $N=4$, $U=12$. $H_p$ as a function of $p$ for $E=-1.6,-1.2,-1.,-0.9,-0.8$. $H_0$ is \caption{One-dimensional Hubbard model for $N=4$ and $U=12$.
computed analytically and $H_p$ (p > 0) by Monte Carlo. Error bars are smaller than the symbol size.} $H_p$ as a function of $p$ for $E=-1.6$, $-1.2$, $-1$, $-0.9$, and $-0.8$.
$H_0$ is computed analytically and $H_p$ ($p > 0$) by Monte Carlo.
Error bars are smaller than the symbol size.}
\label{fig1} \label{fig1}
\end{figure} \end{figure}
%%% FIG 2 %%%
\begin{figure}[h!] \begin{figure}[h!]
\includegraphics[width=\columnwidth]{fig2} \includegraphics[width=\columnwidth]{fig2}
\caption{1D-Hubbard model, $N=4$ and $U=12$. $\mathcal{E}(E)$ as a function of $E$. \caption{One-dimensional Hubbard model for $N=4$ and $U=12$.
The horizontal and vertical lines are at $\mathcal{E}(E_0)=E_0$ and $E=E_0$, respectively. $\cE(E)$ as a function of $E$.
$E_0$ is the exact energy of -0.768068.... The dotted line is the two-component extrapolation. The horizontal and vertical lines are at $\cE(E_0)=E_0$ and $E=E_0$, respectively.
$E_0 = -0.768068\ldots$ is the exact energy.
The dotted line is the \titou{two-component} extrapolation.
Error bars are smaller than the symbol size.} Error bars are smaller than the symbol size.}
\label{fig2} \label{fig2}
\end{figure} \end{figure}
Table \ref{tab1} illustrates the dependence of the Monte Carlo results upon the choice of the domain. The reference energy is $E=-1$. Table \ref{tab1} illustrates the dependence of the Monte Carlo results upon the choice of the domain.
The reference energy is $E=-1$.
The first column indicates the various domains consisting of the union of some elementary domains as explained above. The first column indicates the various domains consisting of the union of some elementary domains as explained above.
The first line of the table gives the results when using a minimal single-state domain for all states, and the last The first line of the table gives the results when using a minimal single-state domain for all states, and the last one for the maximal domain containing the full linear space.
one for the maximal domain containing the full linear space. The size of the various domains is given in column 2, the average trapping time The size of the various domains is given in column 2, the average trapping time for the state $|I_0\rangle$ in the third column, and an estimate of the speed of convergence of the $p$-expansion for the energy in the fourth column.
for the state $|I_0\rangle$ in the third column, and an estimate of the speed of convergence of the $p$-expansion for the energy in the fourth column. To quantify the rate of convergence, we report the quantity, $p_{conv}$, defined as the smallest value of $p$ for which the energy is converged with six decimal places.
To quantify the rate of convergence, we report the quantity, $p_{conv}$, defined as The smaller $p_{conv}$, the better the convergence is.
the smallest value of $p$ for which the energy is converged with six decimal places. The smaller $p_{conv}$, the better the convergence is.
Although this is a rough estimate, it is sufficient here for our purpose. Although this is a rough estimate, it is sufficient here for our purpose.
As clearly seen, the speed of convergence is directly related to the magnitude of $\bar{t}_{I_0}$. The As clearly seen, the speed of convergence is directly related to the magnitude of $\bar{t}_{I_0}$.
longer the stochastic trajectories remain trapped within the domain, the better the convergence. Of course, when the domain is chosen to be the full space, The longer the stochastic trajectories remain trapped within the domain, the better the convergence.
the average trapping time becomes infinite. Let us emphasize that the rate of convergence has no reason to be related to the size of the domain. Of course, when the domain is chosen to be the full space, the average trapping time becomes infinite.
For example, the domain ${\cal D}(0,3) \cup {\cal D}(1,0)$ has a trapping time for the N\'eel state of 6.2, while Let us emphasize that the rate of convergence has no reason to be related to the size of the domain.
the domain ${\cal D}(0,3) \cup {\cal D}(1,1)$ having almost the same number of states (28 states), has an average trapping time about 6 times longer. Finally, the last column gives the energy obtained for For example, the domain ${\cal D}(0,3) \cup {\cal D}(1,0)$ has a trapping time for the N\'eel state of 6.2, while the domain ${\cal D}(0,3) \cup {\cal D}(1,1)$ having almost the same number of states (28 states), has an average trapping time about 6 times longer.
$E=-1$. The energy is expected to be independent of the domain and to converge to a common value, which is indeed the case here. Finally, the last column gives the energy obtained for $E=-1$.
The energy is expected to be independent of the domain and to converge to a common value, which is indeed the case here.
The exact value, $\cE(E=-1)=-0.75272390...$, can be found at the last row of the Table for the case of a domain corresponding to the full space. The exact value, $\cE(E=-1)=-0.75272390...$, can be found at the last row of the Table for the case of a domain corresponding to the full space.
In sharp contrast, the statistical error depends strongly on the type of domains used. As expected, the largest error of $3 \times 10^{-5}$ is obtained in the case of In sharp contrast, the statistical error depends strongly on the type of domains used.
a single-state domain for all states. The smallest statistical error is obtained for the "best" domain having the largest average As expected, the largest error of $3 \times 10^{-5}$ is obtained in the case of a single-state domain for all states. The smallest statistical error is obtained for the "best" domain having the largest average trapping time.
trapping time. Using this domain leads to a reduction in the statistical error as large as about three orders of magnitude, nicely illustrating the Using this domain leads to a reduction in the statistical error as large as about three orders of magnitude, nicely illustrating the critical importance of the domains employed.
critical importance of the domains employed.
\begin{table}[h!] \begin{table}[h!]
\centering \centering
\caption{$N$=4, $U$=12, $E$=-1, $\alpha=1.292$, $\beta=0.552$,$p_{ex}=4$. Simulation with 20 independent blocks and $10^5$ stochastic paths \caption{$N=4$, $U=12$, $E=-1$, $\alpha=1.292$, $\beta=0.552$, $p_{ex}=4$.
starting from the N\'eel state. $\bar{t}_{I_0}$ Simulation with 20 independent blocks and $10^5$ stochastic paths starting from the N\'eel state.
is the average trapping time for the $\bar{t}_{I_0}$ is the average trapping time for the N\'eel state.
N\'eel state. $p_{\rm conv}$ is a measure of the convergence of $\cE_{QMC}(p)$ as a function of $p$, see text.} $p_{\rm conv}$ is a measure of the convergence of $\cE_{QMC}(p)$ as a function of $p$, see text.}
\label{tab1} \label{tab1}
\begin{ruledtabular} \begin{ruledtabular}
\begin{tabular}{lcccl} \begin{tabular}{lrrrl}
Domain & Size & $\bar{t}_{I_0}$ & $p_{\rm conv}$ & $\;\;\;\;\;\;\cE_{QMC}$ \\ Domain & Size & $\bar{t}_{I_0}$ & $p_{\rm conv}$ & $\;\;\;\;\;\;\cE_{QMC}$ \\
\hline \hline
Single & 1 & 0.026 & 88 &$\;\;\;\;$-0.75276(3)\\ Single & 1 & 0.026 & 88 &$-0.75276(3)$\\
${\cal D}(0,3)$ & 2 & 2.1 & 110 &$\;\;\;\;$-0.75276(3)\\ $\cD(0,3)$ & 2 & 2.1 & 110 &$-0.75276(3)$\\
${\cal D}(0,2)$ & 4 & 2.1 & 106 &$\;\;\;\;$-0.75275(2)\\ $\cD(0,2)$ & 4 & 2.1 & 106 &$-0.75275(2)$\\
${\cal D}(0,1)$ & 6 & 2.1& 82 &$\;\;\;\;$-0.75274(3)\\ $\cD(0,1)$ & 6 & 2.1& 82 &$-0.75274(3)$\\
${\cal D}(0,3)$ $\cup$ ${\cal D}(1,1)$ &14 &4.0& 60 & $\;\;\;\;$-0.75270(2)\\ $\cD(0,3)\cup\cD(1,1)$ &14 &4.0& 60 & $-0.75270(2)$\\
${\cal D}(0,3)$ $\cup$ ${\cal D}(1,0)$ &26 &6.2& 45 & $\;\;\;\;$-0.752730(7) \\ $\cD(0,3)\cup\cD(1,0)$ &26 &6.2& 45 & $-0.752730(7)$ \\
${\cal D}(0,2)$ $\cup$ ${\cal D}(1,1)$ &16 &10.1 & 36 &$\;\;\;\;$-0.75269(1)\\ $\cD(0,2)\cup\cD(1,1)$ &16 &10.1 & 36 &$-0.75269(1)$\\
${\cal D}(0,2)$ $\cup$ ${\cal D}(1,0)$ &28 &34.7 & 14&$\;\;\;\;$-0.7527240(6)\\ $\cD(0,2)\cup\cD(1,0)$ &28 &34.7 & 14&$-0.7527240(6)$\\
${\cal D}(0,1)$ $\cup$ ${\cal D}(1,1)$ &18 & 10.1 & 28 &$\;\;\;\;$-0.75269(1)\\ $\cD(0,1)\cup\cD(1,1)$ &18 & 10.1 & 28 &$-0.75269(1)$\\
${\cal D}(0,1)$ $\cup$ ${\cal D}(1,0)$ &30 & 108.7 & 11&$\;\;\;\;$-0.75272400(5) \\ $\cD(0,1)\cup\cD(1,0)$ &30 & 108.7 & 11&$-0.75272400(5)$ \\
${\cal D}(0,3)$ $\cup$ ${\cal D}(1,1)$ $\cup$ ${\cal D}$(2,0) &20 & 4.1 & 47 &$\;\;\;\;$-0.75271(2)\\ $\cD(0,3)\cup\cD(1,1)\cup\cD$(2,0) &20 & 4.1 & 47 &$-0.75271(2)$\\
${\cal D}(0,3)$ $\cup$ ${\cal D}(1,0)$ $\cup$ ${\cal D}$(2,0) &32 & 6.5 & 39 &$\;\;\;\;$-0.752725(3)\\ $\cD(0,3)\cup\cD(1,0)\cup\cD$(2,0) &32 & 6.5 & 39 &$-0.752725(3)$\\
${\cal D}(0,2)$ $\cup$ ${\cal D}(1,1)$ $\cup$ ${\cal D}$(2,0) &22 & 10.8 & 30 &$\;\;\;\;$-0.75270(1)\\ $\cD(0,2)\cup\cD(1,1)\cup\cD$(2,0) &22 & 10.8 & 30 &$-0.75270(1)$\\
${\cal D}(0,2)$ $\cup$ ${\cal D}(1,0)$ $\cup$ ${\cal D}$(2,0) &34 & 52.5 & 13&$\;\;\;\;$-0.7527236(2)\\ $\cD(0,2)\cup\cD(1,0)\cup\cD$(2,0) &34 & 52.5 & 13&$-0.7527236(2)$\\
${\cal D}(0,1)$ $\cup$ ${\cal D}(1,1)$ $\cup$ ${\cal D}$(2,0) & 24 & 10.8 & 26&$\;\;\;\;$-0.75270(1)\\ $\cD(0,1)\cup\cD(1,1)\cup\cD$(2,0) & 24 & 10.8 & 26&$-0.75270(1)$\\
${\cal D}(0,1)$ $\cup$ ${\cal D}(1,0)$ $\cup$ ${\cal D}$(2,0) & 36 & $\infty$&1&$\;\;\;\;$-0.75272390\\ $\cD(0,1)\cup\cD(1,0)\cup\cD$(2,0) & 36 & $\infty$&1&$-0.75272390$\\
\end{tabular} \end{tabular}
\end{ruledtabular} \end{ruledtabular}
\end{table} \end{table}
@ -1027,10 +1035,10 @@ laptop. Of course, it will also be particularly interesting to take advantage of
All these aspects will be considered in a forthcoming work. All these aspects will be considered in a forthcoming work.
\begin{table} \begin{table}
\caption{$N=4$, $U=12$, and $E=-1$. Dependence of the statistical error on the energy with the number of $p$-components calculated \caption{$N=4$, $U=12$, and $E=-1$.
analytically. Same simulation as for Table \ref{tab1}. Results are presented when a single-state domain Dependence of the statistical error on the energy with the number of $p$-components calculated analytically.
is used for all states and when Same simulation as for Table \ref{tab1}.
${\cal D}(0,1) \cup {\cal D}(1,0)$ is chosen as main domain.} Results are presented when a single-state domain is used for all states and when $\cD(0,1) \cup \cD(1,0)$ is chosen as main domain.}
\label{tab2} \label{tab2}
\begin{ruledtabular} \begin{ruledtabular}
\begin{tabular}{lcc} \begin{tabular}{lcc}
@ -1051,8 +1059,10 @@ $8$ & $2.2 \times10^{-5}$ &$ 0.05 \times 10^{-8}$\\
\begin{table} \begin{table}
\caption{$N=4$, $U=12$, $\alpha=1.292$, $\beta=0.552$. Main domain = ${\cal D}(0,1) \cup {\cal D}(1,0)$. Simulation with 20 independent blocks and $10^6$ paths. \caption{$N=4$, $U=12$, $\alpha=1.292$, $\beta=0.552$, and $p_{ex}=4$.
$p_{ex}=4$. The various fits are done with the five values of $E$} Main domain = $\cD(0,1) \cup \cD(1,0)$.
Simulation with 20 independent blocks and $10^6$ paths.
The various fits are done with the five values of $E$}
\label{tab3} \label{tab3}
\begin{ruledtabular} \begin{ruledtabular}
\begin{tabular}{lc} \begin{tabular}{lc}
@ -1072,11 +1082,11 @@ $E_0$ exact & -0.768068...\\
\end{table} \end{table}
\begin{table} \begin{table}
\caption{$N$=4, Domain ${\cal D}(0,1) \cup {\cal D}(1,0)$} \caption{$N=4$, Domain $\cD(0,1) \cup \cD(1,0)$}
\label{tab4} \label{tab4}
\begin{ruledtabular} \begin{ruledtabular}
\begin{tabular}{cccccc} \begin{tabular}{cccccc}
$U$ & $\alpha,\beta$ & $E_{var}$ & $E_{ex}$ & $\bar{t}_{I_0}$ \\ $U$ & $\alpha,\beta$ & $E_\text{T}$ & $E_{ex}$ & $\bar{t}_{I_0}$ \\
\hline \hline
8 & 0.908,\;0.520 & -0.770342... &-1.117172... & 33.5\\ 8 & 0.908,\;0.520 & -0.770342... &-1.117172... & 33.5\\
10 & 1.116,\;0.539 & -0.604162... &-0.911497... & 63.3\\ 10 & 1.116,\;0.539 & -0.604162... &-0.911497... & 63.3\\