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