Variational Monte Carlo methods

Given a hamiltonian $H$ and a trial wave function $\Psi_T$, the variational principle states that the expectation value of $\langle H \rangle$, defined through

 
$$E[H]= \langle H \rangle = \frac{\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R})H(\boldsymbol{R})\Psi_T(\boldsymbol{R})} {\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R})\Psi_T(\boldsymbol{R})},$$

 
is an upper bound to the ground state energy $E_0$ of the hamiltonian $H$, that is

 
$$E_0 \le \langle H \rangle .$$

 
In general, the integrals involved in the calculation of various expectation values are multi-dimensional ones. Traditional integration methods such as the Gauss-Legendre will not be adequate for say the computation of the energy of a many-body system.

The trial wave function can be expanded in the eigenstates of the hamiltonian since they form a complete set, viz.,

 
$$\Psi_T(\boldsymbol{R})=\sum_i a_i\Psi_i(\boldsymbol{R}),$$

 
and assuming the set of eigenfunctions to be normalized one obtains

 
$$\frac{\sum_{nm}a^*_ma_n \int d\boldsymbol{R}\Psi^{\ast}_m(\boldsymbol{R})H(\boldsymbol{R})\Psi_n(\boldsymbol{R})} {\sum_{nm}a^*_ma_n \int d\boldsymbol{R}\Psi^{\ast}_m(\boldsymbol{R})\Psi_n(\boldsymbol{R})} =\frac{\sum_{n}a^2_n E_n} {\sum_{n}a^2_n} \ge E_0,$$

 
where we used that $H(\boldsymbol{R})\Psi_n(\boldsymbol{R})=E_n\Psi_n(\boldsymbol{R})$. In general, the integrals involved in the calculation of various expectation values are multi-dimensional ones. The variational principle yields the lowest state of a given symmetry.

In most cases, a wave function has only small values in large parts of configuration space, and a straightforward procedure which uses homogenously distributed random points in configuration space will most likely lead to poor results. This may suggest that some kind of importance sampling combined with e.g., the Metropolis algorithm may be a more efficient way of obtaining the ground state energy. The hope is then that those regions of configurations space where the wave function assumes appreciable values are sampled more efficiently.

The tedious part in a VMC calculation is the search for the variational minimum. A good knowledge of the system is required in order to carry out reasonable VMC calculations. This is not always the case, and often VMC calculations serve rather as the starting point for so-called diffusion Monte Carlo calculations (DMC). DMC is a way of solving exactly the many-body Schroedinger equation by means of a stochastic procedure. A good guess on the binding energy and its wave function is however necessary. A carefully performed VMC calculation can aid in this context.

• Construct first a trial wave function $\psi_T(\boldsymbol{R},\boldsymbol{\alpha})$, for a many-body system consisting of $N$ particles located at positions

$\boldsymbol{R}=(\boldsymbol{R}_1,\dots ,\boldsymbol{R}_N)$. The trial wave function depends on $\alpha$ variational parameters $\boldsymbol{\alpha}=(\alpha_1,\dots ,\alpha_M)$.

• Then we evaluate the expectation value of the hamiltonian $H$

 
$$E[H]=\langle H \rangle = \frac{\int d\boldsymbol{R}\Psi^{\ast}_{T}(\boldsymbol{R},\boldsymbol{\alpha})H(\boldsymbol{R})\Psi_{T}(\boldsymbol{R},\boldsymbol{\alpha})} {\int d\boldsymbol{R}\Psi^{\ast}_{T}(\boldsymbol{R},\boldsymbol{\alpha})\Psi_{T}(\boldsymbol{R},\boldsymbol{\alpha})}.$$

• Thereafter we vary $\alpha$ according to some minimization algorithm and return to the first step.

Choose a trial wave function $\psi_T(\boldsymbol{R})$.

 
$$P(\boldsymbol{R})= \frac{\left|\psi_T(\boldsymbol{R})\right|^2}{\int \left|\psi_T(\boldsymbol{R})\right|^2d\boldsymbol{R}}.$$

 
This is our new probability distribution function (PDF). The approximation to the expectation value of the Hamiltonian is now

 
$$E[H(\boldsymbol{\alpha})] = \frac{\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R},\boldsymbol{\alpha})H(\boldsymbol{R})\Psi_T(\boldsymbol{R},\boldsymbol{\alpha})} {\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R},\boldsymbol{\alpha})\Psi_T(\boldsymbol{R},\boldsymbol{\alpha})}.$$

Define a new quantity

 
$$E_L(\boldsymbol{R},\boldsymbol{\alpha})=\frac{1}{\psi_T(\boldsymbol{R},\boldsymbol{\alpha})}H\psi_T(\boldsymbol{R},\boldsymbol{\alpha}), \tag{1}$$

 
called the local energy, which, together with our trial PDF yields

 
$$E[H(\boldsymbol{\alpha})]=\int P(\boldsymbol{R})E_L(\boldsymbol{R}) d\boldsymbol{R}\approx \frac{1}{N}\sum_{i=1}^NP(\boldsymbol{R_i},\boldsymbol{\alpha})E_L(\boldsymbol{R_i},\boldsymbol{\alpha}) \tag{2}$$

 
with $N$ being the number of Monte Carlo samples.

The Algorithm for performing a variational Monte Carlo calculations runs thus as this

• Initialisation: Fix the number of Monte Carlo steps. Choose an initial $\boldsymbol{R}$ and variational parameters $\alpha$ and calculate $\left|\psi_T^{\alpha}(\boldsymbol{R})\right|^2$.
• Initialise the energy and the variance and start the Monte Carlo calculation.

• Calculate a trial position $\boldsymbol{R}_p=\boldsymbol{R}+r*step$ where $r$ is a random variable $r \in [0,1]$.
• Metropolis algorithm to accept or reject this move $w = P(\boldsymbol{R}_p)/P(\boldsymbol{R})$.
• If the step is accepted, then we set $\boldsymbol{R}=\boldsymbol{R}_p$.
• Update averages
• Finish and compute final averages.

Observe that the jumping in space is governed by the variable step. This is Called brute-force sampling. Need importance sampling to get more relevant sampling, see lectures below.

The radial Schroedinger equation for the hydrogen atom can be written as

 
$$-\frac{\hbar^2}{2m}\frac{\partial^2 u(r)}{\partial r^2}- \left(\frac{ke^2}{r}-\frac{\hbar^2l(l+1)}{2mr^2}\right)u(r)=Eu(r),$$

 
or with dimensionless variables

 
$$-\frac{1}{2}\frac{\partial^2 u(\rho)}{\partial \rho^2}- \frac{u(\rho)}{\rho}+\frac{l(l+1)}{2\rho^2}u(\rho)-\lambda u(\rho)=0, \tag{3}$$

 
with the hamiltonian

 
$$H=-\frac{1}{2}\frac{\partial^2 }{\partial \rho^2}- \frac{1}{\rho}+\frac{l(l+1)}{2\rho^2}.$$

 
Use variational parameter $\alpha$ in the trial wave function

 
$$u_T^{\alpha}(\rho)=\alpha\rho e^{-\alpha\rho}. \tag{4}$$

Inserting this wave function into the expression for the local energy $E_L$ gives

 
$$E_L(\rho)=-\frac{1}{\rho}- \frac{\alpha}{2}\left(\alpha-\frac{2}{\rho}\right).$$

 
A simple variational Monte Carlo calculation results in

We note that at $\alpha=1$ we obtain the exact result, and the variance is zero, as it should. The reason is that we then have the exact wave function, and the action of the hamiltionan on the wave function

 
$$H\psi = \mathrm{constant}\times \psi,$$

 
yields just a constant. The integral which defines various expectation values involving moments of the hamiltonian becomes then

 
$$\langle H^n \rangle = \frac{\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R})H^n(\boldsymbol{R})\Psi_T(\boldsymbol{R})} {\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R})\Psi_T(\boldsymbol{R})}= \mathrm{constant}\times\frac{\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R})\Psi_T(\boldsymbol{R})} {\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R})\Psi_T(\boldsymbol{R})}=\mathrm{constant}.$$

 
This gives an important information: the exact wave function leads to zero variance! Variation is then performed by minimizing both the energy and the variance.

For boson in a harmonic oscillator-like trap we will use is a spherical (S) or an elliptical (E) harmonic trap in one, two and finally three dimensions, with the latter given by

 
$$\begin{equation} V_{ext}(\mathbf{r}) = \Bigg\{ \begin{array}{ll} \frac{1}{2}m\omega_{ho}^2r^2 & (S)\\ \strut \frac{1}{2}m[\omega_{ho}^2(x^2+y^2) + \omega_z^2z^2] & (E) \tag{5} \end{array} \end{equation}$$

 
where (S) stands for symmetric and

 
$$\begin{equation} \hat{H} = \sum_i^N \left( \frac{-\hbar^2}{2m} { \bigtriangledown }_{i}^2 + V_{ext}({\bf{r}}_i)\right) + \sum_{i < j}^{N} V_{int}({\bf{r}}_i,{\bf{r}}_j), \tag{6} \end{equation}$$

 
as the two-body Hamiltonian of the system.

We will represent the inter-boson interaction by a pairwise, repulsive potential

 
$$\begin{equation} V_{int}(|\mathbf{r}_i-\mathbf{r}_j|) = \Bigg\{ \begin{array}{ll} \infty & {|\mathbf{r}_i-\mathbf{r}_j|} \leq {a}\\ 0 & {|\mathbf{r}_i-\mathbf{r}_j|} > {a} \end{array} \tag{7} \end{equation}$$

 
where $a$ is the so-called hard-core diameter of the bosons. Clearly, $V_{int}(|\mathbf{r}_i-\mathbf{r}_j|)$ is zero if the bosons are separated by a distance $|\mathbf{r}_i-\mathbf{r}_j|$ greater than $a$ but infinite if they attempt to come within a distance $|\mathbf{r}_i-\mathbf{r}_j| \leq a$.

Our trial wave function for the ground state with $N$ atoms is given by

 
$$\begin{equation} \Psi_T(\mathbf{R})=\Psi_T(\mathbf{r}_1, \mathbf{r}_2, \dots \mathbf{r}_N,\alpha,\beta)=\prod_i g(\alpha,\beta,\mathbf{r}_i)\prod_{i < j}f(a,|\mathbf{r}_i-\mathbf{r}_j|), \tag{8} \end{equation}$$

 
where $\alpha$ and $\beta$ are variational parameters. The single-particle wave function is proportional to the harmonic oscillator function for the ground state

 
$$\begin{equation} g(\alpha,\beta,\mathbf{r}_i)= \exp{[-\alpha(x_i^2+y_i^2+\beta z_i^2)]}. \tag{9} \end{equation}$$

For spherical traps we have $\beta = 1$ and for non-interacting bosons ($a=0$) we have $\alpha = 1/2a_{ho}^2$. The correlation wave function is

 
$$\begin{equation} f(a,|\mathbf{r}_i-\mathbf{r}_j|)=\Bigg\{ \begin{array}{ll} 0 & {|\mathbf{r}_i-\mathbf{r}_j|} \leq {a}\\ (1-\frac{a}{|\mathbf{r}_i-\mathbf{r}_j|}) & {|\mathbf{r}_i-\mathbf{r}_j|} > {a}. \end{array} \tag{10} \end{equation}$$

The helium atom consists of two electrons and a nucleus with charge $Z=2$. The contribution to the potential energy due to the attraction from the nucleus is

 
$$-\frac{2ke^2}{r_1}-\frac{2ke^2}{r_2},$$

 
and if we add the repulsion arising from the two interacting electrons, we obtain the potential energy

 
$$V(r_1, r_2)=-\frac{2ke^2}{r_1}-\frac{2ke^2}{r_2}+ \frac{ke^2}{r_{12}},$$

 
with the electrons separated at a distance $r_{12}=|\boldsymbol{r}_1-\boldsymbol{r}_2|$.

The hamiltonian becomes then

 
$$\hat{H}=-\frac{\hbar^2\nabla_1^2}{2m}-\frac{\hbar^2\nabla_2^2}{2m} -\frac{2ke^2}{r_1}-\frac{2ke^2}{r_2}+ \frac{ke^2}{r_{12}},$$

 
and Schroedingers equation reads

 
$$\hat{H}\psi=E\psi.$$

 
All observables are evaluated with respect to the probability distribution

 
$$P(\boldsymbol{R})= \frac{\left|\psi_T(\boldsymbol{R})\right|^2}{\int \left|\psi_T(\boldsymbol{R})\right|^2d\boldsymbol{R}}.$$

 
generated by the trial wave function. The trial wave function must approximate an exact eigenstate in order that accurate results are to be obtained.

Choice of trial wave function for Helium: Assume $r_1 \rightarrow 0$.

 
$$E_L(\boldsymbol{R})=\frac{1}{\psi_T(\boldsymbol{R})}H\psi_T(\boldsymbol{R})= \frac{1}{\psi_T(\boldsymbol{R})}\left(-\frac{1}{2}\nabla^2_1 -\frac{Z}{r_1}\right)\psi_T(\boldsymbol{R}) + \mathrm{finite \hspace{0.1cm}terms}.$$

 
$$E_L(R)= \frac{1}{\mathbf{R}_T(r_1)}\left(-\frac{1}{2}\frac{d^2}{dr_1^2}- \frac{1}{r_1}\frac{d}{dr_1} -\frac{Z}{r_1}\right)\mathbf{R}_T(r_1) + \mathrm{finite\hspace{0.1cm} terms}$$

 
For small values of $r_1$, the terms which dominate are

 
$$\lim_{r_1 \rightarrow 0}E_L(R)= \frac{1}{\mathbf{R}_T(r_1)}\left(- \frac{1}{r_1}\frac{d}{dr_1} -\frac{Z}{r_1}\right)\mathbf{R}_T(r_1),$$

 
since the second derivative does not diverge due to the finiteness of $\Psi$ at the origin.

This results in

 
$$\frac{1}{\mathbf{R}_T(r_1)}\frac{d \mathbf{R}_T(r_1)}{dr_1}=-Z,$$

 
and

 
$$\mathbf{R}_T(r_1)\propto e^{-Zr_1}.$$

 
A similar condition applies to electron 2 as well. For orbital momenta $l > 0$ we have

 
$$\frac{1}{\mathbf{R}_T(r)}\frac{d \mathbf{R}_T(r)}{dr}=-\frac{Z}{l+1}.$$

 
Similarly, studying the case $r_{12}\rightarrow 0$ we can write a possible trial wave function as

 
$$\psi_T(\boldsymbol{R})=e^{-\alpha(r_1+r_2)}e^{\beta r_{12}}. \tag{11}$$

 
The last equation can be generalized to

 
$$\psi_T(\boldsymbol{R})=\phi(\boldsymbol{r}_1)\phi(\boldsymbol{r}_2)\dots\phi(\boldsymbol{r}_N) \prod_{i < j}f(r_{ij}),$$

 
for a system with $N$ electrons or particles.

During the development of our code we need to make several checks. It is also very instructive to compute a closed form expression for the local energy. Since our wave function is rather simple it is straightforward to find an analytic expressions. Consider first the case of the simple helium function

 
$$\Psi_T(\boldsymbol{r}_1,\boldsymbol{r}_2) = e^{-\alpha(r_1+r_2)}$$

 
The local energy is for this case

 
$$E_{L1} = \left(\alpha-Z\right)\left(\frac{1}{r_1}+\frac{1}{r_2}\right)+\frac{1}{r_{12}}-\alpha^2$$

 
which gives an expectation value for the local energy given by

 
$$\langle E_{L1} \rangle = \alpha^2-2\alpha\left(Z-\frac{5}{16}\right)$$

With closed form formulae we can speed up the computation of the correlation. In our case we write it as

 
$$\Psi_C= \exp{\left\{\sum_{i < j}\frac{ar_{ij}}{1+\beta r_{ij}}\right\}},$$

 
which means that the gradient needed for the so-called quantum force and local energy can be calculated analytically. This will speed up your code since the computation of the correlation part and the Slater determinant are the most time consuming parts in your code.

We will refer to this correlation function as $\Psi_C$ or the linear Pade-Jastrow.

We can test this by computing the local energy for our helium wave function

 
$$\psi_{T}(\boldsymbol{r}_1,\boldsymbol{r}_2) = \exp{\left(-\alpha(r_1+r_2)\right)} \exp{\left(\frac{r_{12}}{2(1+\beta r_{12})}\right)},$$

 
with $\alpha$ and $\beta$ as variational parameters.

The local energy is for this case

 
$$E_{L2} = E_{L1}+\frac{1}{2(1+\beta r_{12})^2}\left\{\frac{\alpha(r_1+r_2)}{r_{12}}(1-\frac{\boldsymbol{r}_1\boldsymbol{r}_2}{r_1r_2})-\frac{1}{2(1+\beta r_{12})^2}-\frac{2}{r_{12}}+\frac{2\beta}{1+\beta r_{12}}\right\}$$

 
It is very useful to test your code against these expressions. It means also that you don't need to compute a derivative numerically as discussed in the code example below.

For the computation of various derivatives with different types of wave functions, you will find it useful to use python with symbolic python, that is sympy, see online manual. Using sympy allows you autogenerate both Latex code as well c++, python or Fortran codes. Here you will find some simple examples. We choose the $2s$ hydrogen-orbital (not normalized) as an example

 
$$\phi_{2s}(\boldsymbol{r}) = (Zr - 2)\exp{-(\frac{1}{2}Zr)},$$

 
with $ r^2 = x^2 + y^2 + z^2$.

from sympy import symbols, diff, exp, sqrt
x, y, z, Z = symbols('x y z Z')
r = sqrt(x*x + y*y + z*z)
r
phi = (Z*r - 2)*exp(-Z*r/2)
phi
diff(phi, x)

This doesn't look very nice, but sympy provides several functions that allow for improving and simplifying the output.

We can improve our output by factorizing and substituting expressions

from sympy import symbols, diff, exp, sqrt, factor, Symbol, printing
x, y, z, Z = symbols('x y z Z')
r = sqrt(x*x + y*y + z*z)
phi = (Z*r - 2)*exp(-Z*r/2)
R = Symbol('r') #Creates a symbolic equivalent of r
#print latex and c++ code
print printing.latex(diff(phi, x).factor().subs(r, R))
print printing.ccode(diff(phi, x).factor().subs(r, R))

We can in turn look at second derivatives

from sympy import symbols, diff, exp, sqrt, factor, Symbol, printing
x, y, z, Z = symbols('x y z Z')
r = sqrt(x*x + y*y + z*z)
phi = (Z*r - 2)*exp(-Z*r/2)
R = Symbol('r') #Creates a symbolic equivalent of r
(diff(diff(phi, x), x) + diff(diff(phi, y), y) + diff(diff(phi, z), z)).factor().subs(r, R)
# Collect the Z values
(diff(diff(phi, x), x) + diff(diff(phi, y), y) +diff(diff(phi, z), z)).factor().collect(Z).subs(r, R)
# Factorize also the r**2 terms
(diff(diff(phi, x), x) + diff(diff(phi, y), y) + diff(diff(phi, z), z)).factor().collect(Z).subs(r, R).subs(r**2, R**2).factor()
print printing.ccode((diff(diff(phi, x), x) + diff(diff(phi, y), y) + diff(diff(phi, z), z)).factor().collect(Z).subs(r, R).subs(r**2, R**2).factor())

With some practice this allows one to be able to check one's own calculation and translate automatically into code lines.

The Metropolis algorithm , see the original article (see also the FYS3150 lectures) was invented by Metropolis et. al and is often simply called the Metropolis algorithm. It is a method to sample a normalized probability distribution by a stochastic process. We define $\mathbf{P}_i^{(n)}$ to be the probability for finding the system in the state $i$ at step $n$. The algorithm is then

• Sample a possible new state $j$ with some probability $T_{i\rightarrow j}$.
• Accept the new state $j$ with probability $A_{i \rightarrow j}$ and use it as the next sample. With probability $1-A_{i\rightarrow j}$ the move is rejected and the original state $i$ is used again as a sample.

We wish to derive the required properties of $T$ and $A$ such that $\mathbf{P}_i^{(n\rightarrow \infty)} \rightarrow p_i$ so that starting from any distribution, the method converges to the correct distribution. Note that the description here is for a discrete probability distribution. Replacing probabilities $p_i$ with expressions like $p(x_i)dx_i$ will take all of these over to the corresponding continuum expressions.

The dynamical equation for $\mathbf{P}_i^{(n)}$ can be written directly from the description above. The probability of being in the state $i$ at step $n$ is given by the probability of being in any state $j$ at the previous step, and making an accepted transition to $i$ added to the probability of being in the state $i$, making a transition to any state $j$ and rejecting the move:

 
$$\mathbf{P}^{(n)}_i = \sum_j \left [ \mathbf{P}^{(n-1)}_jT_{j\rightarrow i} A_{j\rightarrow i} +\mathbf{P}^{(n-1)}_iT_{i\rightarrow j}\left ( 1- A_{i\rightarrow j} \right) \right ] \,.$$

 
Since the probability of making some transition must be 1, $\sum_j T_{i\rightarrow j} = 1$, and the above equation becomes

 
$$\mathbf{P}^{(n)}_i = \mathbf{P}^{(n-1)}_i + \sum_j \left [ \mathbf{P}^{(n-1)}_jT_{j\rightarrow i} A_{j\rightarrow i} -\mathbf{P}^{(n-1)}_iT_{i\rightarrow j}A_{i\rightarrow j} \right ] \,.$$

For large $n$ we require that $\mathbf{P}^{(n\rightarrow \infty)}_i = p_i$, the desired probability distribution. Taking this limit, gives the balance requirement

 
$$\sum_j \left [ p_jT_{j\rightarrow i} A_{j\rightarrow i} -p_iT_{i\rightarrow j}A_{i\rightarrow j} \right ] = 0 \,.$$

 
The balance requirement is very weak. Typically the much stronger detailed balance requirement is enforced, that is rather than the sum being set to zero, we set each term separately to zero and use this to determine the acceptance probabilities. Rearranging, the result is

 
$$\frac{ A_{j\rightarrow i}}{A_{i\rightarrow j}} = \frac{p_iT_{i\rightarrow j}}{ p_jT_{j\rightarrow i}} \,.$$

The Metropolis choice is to maximize the $A$ values, that is

 
$$A_{j \rightarrow i} = \min \left ( 1, \frac{p_iT_{i\rightarrow j}}{ p_jT_{j\rightarrow i}}\right ).$$

 
Other choices are possible, but they all correspond to multilplying $A_{i\rightarrow j}$ and $A_{j\rightarrow i}$ by the same constant smaller than unity.\footnote{The penalty function method uses just such a factor to compensate for $p_i$ that are evaluated stochastically and are therefore noisy.}

Having chosen the acceptance probabilities, we have guaranteed that if the $\mathbf{P}_i^{(n)}$ has equilibrated, that is if it is equal to $p_i$, it will remain equilibrated. Next we need to find the circumstances for convergence to equilibrium.

The dynamical equation can be written as

 
$$\mathbf{P}^{(n)}_i = \sum_j M_{ij}\mathbf{P}^{(n-1)}_j$$

 
with the matrix $M$ given by

 
$$M_{ij} = \delta_{ij}\left [ 1 -\sum_k T_{i\rightarrow k} A_{i \rightarrow k} \right ] + T_{j\rightarrow i} A_{j\rightarrow i} \,.$$

 
Summing over $i$ shows that $\sum_i M_{ij} = 1$, and since $\sum_k T_{i\rightarrow k} = 1$, and $A_{i \rightarrow k} \leq 1$, the elements of the matrix satisfy $M_{ij} \geq 0$. The matrix $M$ is therefore a stochastic matrix.

The Metropolis method is simply the power method for computing the right eigenvector of $M$ with the largest magnitude eigenvalue. By construction, the correct probability distribution is a right eigenvector with eigenvalue 1. Therefore, for the Metropolis method to converge to this result, we must show that $M$ has only one eigenvalue with this magnitude, and all other eigenvalues are smaller.

We need to replace the brute force Metropolis algorithm with a walk in coordinate space biased by the trial wave function. This approach is based on the Fokker-Planck equation and the Langevin equation for generating a trajectory in coordinate space. The link between the Fokker-Planck equation and the Langevin equations are explained, only partly, in the slides below. An excellent reference on topics like Brownian motion, Markov chains, the Fokker-Planck equation and the Langevin equation is the text by Van Kampen Here we will focus first on the implementation part first.

For a diffusion process characterized by a time-dependent probability density $P(x,t)$ in one dimension the Fokker-Planck equation reads (for one particle /walker)

 
$$\frac{\partial P}{\partial t} = D\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x} -F\right)P(x,t),$$

 
where $F$ is a drift term and $D$ is the diffusion coefficient.

The new positions in coordinate space are given as the solutions of the Langevin equation using Euler's method, namely, we go from the Langevin equation

 
$$\frac{\partial x(t)}{\partial t} = DF(x(t)) +\eta,$$

 
with $\eta$ a random variable, yielding a new position

 
$$y = x+DF(x)\Delta t +\xi\sqrt{\Delta t},$$

 
where $\xi$ is gaussian random variable and $\Delta t$ is a chosen time step. The quantity $D$ is, in atomic units, equal to $1/2$ and comes from the factor $1/2$ in the kinetic energy operator. Note that $\Delta t$ is to be viewed as a parameter. Values of $\Delta t \in [0.001,0.01]$ yield in general rather stable values of the ground state energy.

The process of isotropic diffusion characterized by a time-dependent probability density $P(\mathbf{x},t)$ obeys (as an approximation) the so-called Fokker-Planck equation

 
$$\frac{\partial P}{\partial t} = \sum_i D\frac{\partial }{\partial \mathbf{x_i}}\left(\frac{\partial }{\partial \mathbf{x_i}} -\mathbf{F_i}\right)P(\mathbf{x},t),$$

 
where $\mathbf{F_i}$ is the $i^{th}$ component of the drift term (drift velocity) caused by an external potential, and $D$ is the diffusion coefficient. The convergence to a stationary probability density can be obtained by setting the left hand side to zero. The resulting equation will be satisfied if and only if all the terms of the sum are equal zero,

 
$$\frac{\partial^2 P}{\partial {\mathbf{x_i}^2}} = P\frac{\partial}{\partial {\mathbf{x_i}}}\mathbf{F_i} + \mathbf{F_i}\frac{\partial}{\partial {\mathbf{x_i}}}P.$$

The drift vector should be of the form $\mathbf{F} = g(\mathbf{x}) \frac{\partial P}{\partial \mathbf{x}}$. Then,

 
$$\frac{\partial^2 P}{\partial {\mathbf{x_i}^2}} = P\frac{\partial g}{\partial P}\left( \frac{\partial P}{\partial {\mathbf{x}_i}} \right)^2 + P g \frac{\partial ^2 P}{\partial {\mathbf{x}_i^2}} + g \left( \frac{\partial P}{\partial {\mathbf{x}_i}} \right)^2.$$

 
The condition of stationary density means that the left hand side equals zero. In other words, the terms containing first and second derivatives have to cancel each other. It is possible only if $g = \frac{1}{P}$, which yields

 
$$\mathbf{F} = 2\frac{1}{\Psi_T}\nabla\Psi_T,$$

 
which is known as the so-called quantum force. This term is responsible for pushing the walker towards regions of configuration space where the trial wave function is large, increasing the efficiency of the simulation in contrast to the Metropolis algorithm where the walker has the same probability of moving in every direction.

The Fokker-Planck equation yields a (the solution to the equation) transition probability given by the Green's function

 
$$G(y,x,\Delta t) = \frac{1}{(4\pi D\Delta t)^{3N/2}} \exp{\left(-(y-x-D\Delta t F(x))^2/4D\Delta t\right)}$$

 
which in turn means that our brute force Metropolis algorithm

 
$$A(y,x) = \mathrm{min}(1,q(y,x))),$$

 
with $q(y,x) = |\Psi_T(y)|^2/|\Psi_T(x)|^2$ is now replaced by the Metropolis-Hastings algorithm as well as Hasting's article,

 
$$q(y,x) = \frac{G(x,y,\Delta t)|\Psi_T(y)|^2}{G(y,x,\Delta t)|\Psi_T(x)|^2}$$

The full code is this link. Here we include only the parts pertaining to the computation of the quantum force and the Metropolis update. The program is a modfication of our previous c++ program discussed previously. Here we display only the part from the vmcsolver.cpp file. Note the usage of the function GaussianDeviate.

void VMCSolver::runMonteCarloIntegration()
{
  rOld = zeros<mat>(nParticles, nDimensions);
  rNew = zeros<mat>(nParticles, nDimensions);
  QForceOld = zeros<mat>(nParticles, nDimensions);
  QForceNew = zeros<mat>(nParticles, nDimensions);

  double waveFunctionOld = 0;
  double waveFunctionNew = 0;

  double energySum = 0;
  double energySquaredSum = 0;

  double deltaE;

  // initial trial positions
  for(int i = 0; i < nParticles; i++) {
    for(int j = 0; j < nDimensions; j++) {
      rOld(i,j) = GaussianDeviate(&idum)*sqrt(timestep);
    }
  }
  rNew = rOld;

The general derivative formula of the Jastrow factor is (the subscript $C$ stands for Correlation)

 
$$\frac{1}{\Psi_C}\frac{\partial \Psi_C}{\partial x_k} = \sum_{i=1}^{k-1}\frac{\partial g_{ik}}{\partial x_k} + \sum_{i=k+1}^{N}\frac{\partial g_{ki}}{\partial x_k}$$

 
However, with our written in way which can be reused later as

 
$$\Psi_C=\prod_{i < j}g(r_{ij})= \exp{\left\{\sum_{i < j}f(r_{ij})\right\}},$$

 
the gradient needed for the quantum force and local energy is easy to compute. The function $f(r_{ij})$ will depends on the system under study. In the equations below we will keep this general form.

In the Metropolis/Hasting algorithm, the acceptance ratio determines the probability for a particle to be accepted at a new position. The ratio of the trial wave functions evaluated at the new and current positions is given by ($OB$ for the onebody part)

 
$$R \equiv \frac{\Psi_{T}^{new}}{\Psi_{T}^{old}} = \frac{\Psi_{OB}^{new}}{\Psi_{OB}^{old}}\frac{\Psi_{C}^{new}}{\Psi_{C}^{old}}$$

 
Here $\Psi_{OB}$ is our onebody part (Slater determinant or product of boson single-particle states) while $\Psi_{C}$ is our correlation function, or Jastrow factor. We need to optimize the $\nabla \Psi_T / \Psi_T$ ratio and the second derivative as well, that is the $\mathbf{\nabla}^2 \Psi_T/\Psi_T$ ratio. The first is needed when we compute the so-called quantum force in importance sampling. The second is needed when we compute the kinetic energy term of the local energy.

 
$$\frac{\mathbf{\mathbf{\nabla}} \Psi}{\Psi} = \frac{\mathbf{\nabla} (\Psi_{OB} \, \Psi_{C})}{\Psi_{OB} \, \Psi_{C}} = \frac{ \Psi_C \mathbf{\nabla} \Psi_{OB} + \Psi_{OB} \mathbf{\nabla} \Psi_{C}}{\Psi_{OB} \Psi_{C}} = \frac{\mathbf{\nabla} \Psi_{OB}}{\Psi_{OB}} + \frac{\mathbf{\nabla} \Psi_C}{ \Psi_C}$$

The expectation value of the kinetic energy expressed in atomic units for electron $i$ is

 
$$\langle \hat{K}_i \rangle = -\frac{1}{2}\frac{\langle\Psi|\mathbf{\nabla}_{i}^2|\Psi \rangle}{\langle\Psi|\Psi \rangle},$$

 
$$\hat{K}_i = -\frac{1}{2}\frac{\mathbf{\nabla}_{i}^{2} \Psi}{\Psi}.$$

The second derivative which enters the definition of the local energy is

 
$$\frac{\mathbf{\nabla}^2 \Psi}{\Psi}=\frac{\mathbf{\nabla}^2 \Psi_{OB}}{\Psi_{OB}} + \frac{\mathbf{\nabla}^2 \Psi_C}{ \Psi_C} + 2 \frac{\mathbf{\nabla} \Psi_{OB}}{\Psi_{OB}}\cdot\frac{\mathbf{\nabla} \Psi_C}{ \Psi_C}$$

 
We discuss here how to calculate these quantities in an optimal way,

We have defined the correlated function as

 
$$\Psi_C=\prod_{i < j}g(r_{ij})=\prod_{i < j}^Ng(r_{ij})= \prod_{i=1}^N\prod_{j=i+1}^Ng(r_{ij}),$$

 
with $r_{ij}=|\mathbf{r}_i-\mathbf{r}_j|=\sqrt{(x_i-x_j)^2+(y_i-y_j)^2+(z_i-z_j)^2}$ in three dimensions or $r_{ij}=|\mathbf{r}_i-\mathbf{r}_j|=\sqrt{(x_i-x_j)^2+(y_i-y_j)^2}$ if we work with two-dimensional systems.

In our particular case we have

 
$$\Psi_C=\prod_{i < j}g(r_{ij})=\exp{\left\{\sum_{i < j}f(r_{ij})\right\}}.$$

The total number of different relative distances $r_{ij}$ is $N(N-1)/2$. In a matrix storage format, the relative distances form a strictly upper triangular matrix

 
$$\mathbf{r} \equiv \begin{pmatrix} 0 & r_{1,2} & r_{1,3} & \cdots & r_{1,N} \\ \vdots & 0 & r_{2,3} & \cdots & r_{2,N} \\ \vdots & \vdots & 0 & \ddots & \vdots \\ \vdots & \vdots & \vdots & \ddots & r_{N-1,N} \\ 0 & 0 & 0 & \cdots & 0 \end{pmatrix}.$$

 
This applies to $\mathbf{g} = \mathbf{g}(r_{ij})$ as well.

In our algorithm we will move one particle at the time, say the $kth$-particle. This sampling will be seen to be particularly efficient when we are going to compute a Slater determinant.

We have that the ratio between Jastrow factors $R_C$ is given by

 
$$R_{C} = \frac{\Psi_{C}^\mathrm{new}}{\Psi_{C}^\mathrm{cur}} = \prod_{i=1}^{k-1}\frac{g_{ik}^\mathrm{new}}{g_{ik}^\mathrm{cur}} \prod_{i=k+1}^{N}\frac{ g_{ki}^\mathrm{new}} {g_{ki}^\mathrm{cur}}.$$

 
For the Pade-Jastrow form

 
$$R_{C} = \frac{\Psi_{C}^\mathrm{new}}{\Psi_{C}^\mathrm{cur}} = \frac{\exp{U_{new}}}{\exp{U_{cur}}} = \exp{\Delta U},$$

 
where

 
$$\Delta U = \sum_{i=1}^{k-1}\big(f_{ik}^\mathrm{new}-f_{ik}^\mathrm{cur}\big) + \sum_{i=k+1}^{N}\big(f_{ki}^\mathrm{new}-f_{ki}^\mathrm{cur}\big)$$

One needs to develop a special algorithm that runs only through the elements of the upper triangular matrix $\mathbf{g}$ and have $k$ as an index.

The expression to be derived in the following is of interest when computing the quantum force and the kinetic energy. It has the form

 
$$\frac{\mathbf{\nabla}_i\Psi_C}{\Psi_C} = \frac{1}{\Psi_C}\frac{\partial \Psi_C}{\partial x_i},$$

 
for all dimensions and with $i$ running over all particles.

For the first derivative only $N-1$ terms survive the ratio because the $g$-terms that are not differentiated cancel with their corresponding ones in the denominator. Then,

 
$$\frac{1}{\Psi_C}\frac{\partial \Psi_C}{\partial x_k} = \sum_{i=1}^{k-1}\frac{1}{g_{ik}}\frac{\partial g_{ik}}{\partial x_k} + \sum_{i=k+1}^{N}\frac{1}{g_{ki}}\frac{\partial g_{ki}}{\partial x_k}.$$

 
An equivalent equation is obtained for the exponential form after replacing $g_{ij}$ by $\exp(f_{ij})$, yielding:

 
$$\frac{1}{\Psi_C}\frac{\partial \Psi_C}{\partial x_k} = \sum_{i=1}^{k-1}\frac{\partial g_{ik}}{\partial x_k} + \sum_{i=k+1}^{N}\frac{\partial g_{ki}}{\partial x_k},$$

 
with both expressions scaling as $\mathcal{O}(N)$.

Using the identity

 
$$\frac{\partial}{\partial x_i}g_{ij} = -\frac{\partial}{\partial x_j}g_{ij},$$

 
we get expressions where all the derivatives acting on the particle are represented by the second index of $g$:

 
$$\frac{1}{\Psi_C}\frac{\partial \Psi_C}{\partial x_k} = \sum_{i=1}^{k-1}\frac{1}{g_{ik}}\frac{\partial g_{ik}}{\partial x_k} -\sum_{i=k+1}^{N}\frac{1}{g_{ki}}\frac{\partial g_{ki}}{\partial x_i},$$

 
and for the exponential case:

 
$$\frac{1}{\Psi_C}\frac{\partial \Psi_C}{\partial x_k} = \sum_{i=1}^{k-1}\frac{\partial g_{ik}}{\partial x_k} -\sum_{i=k+1}^{N}\frac{\partial g_{ki}}{\partial x_i}.$$

For correlation forms depending only on the scalar distances $r_{ij}$ we can use the chain rule. Noting that

 
$$\frac{\partial g_{ij}}{\partial x_j} = \frac{\partial g_{ij}}{\partial r_{ij}} \frac{\partial r_{ij}}{\partial x_j} = \frac{x_j - x_i}{r_{ij}} \frac{\partial g_{ij}}{\partial r_{ij}},$$

 
we arrive at

 
$$\frac{1}{\Psi_C}\frac{\partial \Psi_C}{\partial x_k} = \sum_{i=1}^{k-1}\frac{1}{g_{ik}} \frac{\mathbf{r_{ik}}}{r_{ik}} \frac{\partial g_{ik}}{\partial r_{ik}} -\sum_{i=k+1}^{N}\frac{1}{g_{ki}}\frac{\mathbf{r_{ki}}}{r_{ki}}\frac{\partial g_{ki}}{\partial r_{ki}}.$$

Note that for the Pade-Jastrow form we can set $g_{ij} \equiv g(r_{ij}) = e^{f(r_{ij})} = e^{f_{ij}}$ and

 
$$\frac{\partial g_{ij}}{\partial r_{ij}} = g_{ij} \frac{\partial f_{ij}}{\partial r_{ij}}.$$

 
Therefore,

 
$$\frac{1}{\Psi_{C}}\frac{\partial \Psi_{C}}{\partial x_k} = \sum_{i=1}^{k-1}\frac{\mathbf{r_{ik}}}{r_{ik}}\frac{\partial f_{ik}}{\partial r_{ik}} -\sum_{i=k+1}^{N}\frac{\mathbf{r_{ki}}}{r_{ki}}\frac{\partial f_{ki}}{\partial r_{ki}},$$

 
where

 
$$\mathbf{r}_{ij} = |\mathbf{r}_j - \mathbf{r}_i| = (x_j - x_i)\mathbf{e}_1 + (y_j - y_i)\mathbf{e}_2 + (z_j - z_i)\mathbf{e}_3$$

 
is the relative distance.

The second derivative of the Jastrow factor divided by the Jastrow factor (the way it enters the kinetic energy) is

 
$$\left[\frac{\mathbf{\nabla}^2 \Psi_C}{\Psi_C}\right]_x =\ 2\sum_{k=1}^{N} \sum_{i=1}^{k-1}\frac{\partial^2 g_{ik}}{\partial x_k^2}\ +\ \sum_{k=1}^N \left( \sum_{i=1}^{k-1}\frac{\partial g_{ik}}{\partial x_k} - \sum_{i=k+1}^{N}\frac{\partial g_{ki}}{\partial x_i} \right)^2$$

But we have a simple form for the function, namely

 
$$\Psi_{C}=\prod_{i < j}\exp{f(r_{ij})},$$

 
and it is easy to see that for particle $k$ we have

 
$$\frac{\mathbf{\nabla}^2_k \Psi_C}{\Psi_C }= \sum_{ij\ne k}\frac{(\mathbf{r}_k-\mathbf{r}_i)(\mathbf{r}_k-\mathbf{r}_j)}{r_{ki}r_{kj}}f'(r_{ki})f'(r_{kj})+ \sum_{j\ne k}\left( f''(r_{kj})+\frac{2}{r_{kj}}f'(r_{kj})\right)$$

• Monte Carlo simulations can be treated as computer experiments
• The results can be analysed with the same statistical tools as we would use analysing experimental data.
• As in all experiments, we are looking for expectation values and an estimate of how accurate they are, i.e., possible sources for errors.

A very good article which explains blocking is H. Flyvbjerg and H. G. Petersen, Error estimates on averages of correlated data, Journal of Chemical Physics 91, 461-466 (1989).

• As in other experiments, Monte Carlo experiments have two classes of errors:

• Statistical errors
• Systematical errors
• Statistical errors can be estimated using standard tools from statistics
• Systematical errors are method specific and must be treated differently from case to case. (In VMC a common source is the step length or time step in importance sampling)

The probability distribution function (PDF) is a function $p(x)$ on the domain which, in the discrete case, gives us the probability or relative frequency with which these values of $X$ occur:

 
$$p(x) = \mathrm{prob}(X=x)$$

 
In the continuous case, the PDF does not directly depict the actual probability. Instead we define the probability for the stochastic variable to assume any value on an infinitesimal interval around $x$ to be $p(x)dx$. The continuous function $p(x)$ then gives us the density of the probability rather than the probability itself. The probability for a stochastic variable to assume any value on a non-infinitesimal interval $[a,\,b]$ is then just the integral:

 
$$\mathrm{prob}(a\leq X\leq b) = \int_a^b p(x)dx$$

 
Qualitatively speaking, a stochastic variable represents the values of numbers chosen as if by chance from some specified PDF so that the selection of a large set of these numbers reproduces this PDF.

Also of interest to us is the cumulative probability distribution function (CDF), $P(x)$, which is just the probability for a stochastic variable $X$ to assume any value less than $x$:

 
$$P(x)=\mathrm{Prob(}X\leq x\mathrm{)} =\int_{-\infty}^x p(x^{\prime})dx^{\prime}$$

 
The relation between a CDF and its corresponding PDF is then:

 
$$p(x) = \frac{d}{dx}P(x)$$

A particularly useful class of special expectation values are the moments. The $n$-th moment of the PDF $p$ is defined as follows:

 
$$\langle x^n\rangle \equiv \int\! x^n p(x)\,dx$$

 
The zero-th moment $\langle 1\rangle$ is just the normalization condition of $p$. The first moment, $\langle x\rangle$, is called the mean of $p$ and often denoted by the letter $\mu$:

 
$$\langle x\rangle = \mu \equiv \int\! x p(x)\,dx$$

A special version of the moments is the set of central moments, the n-th central moment defined as:

 
$$\langle (x-\langle x \rangle )^n\rangle \equiv \int\! (x-\langle x\rangle)^n p(x)\,dx$$

 
The zero-th and first central moments are both trivial, equal $1$ and $0$, respectively. But the second central moment, known as the variance of $p$, is of particular interest. For the stochastic variable $X$, the variance is denoted as $\sigma^2_X$ or $\mathrm{var}(X)$:

 
$$\begin{align} \sigma^2_X\ \ =\ \ \mathrm{var}(X) & = \langle (x-\langle x\rangle)^2\rangle = \int\! (x-\langle x\rangle)^2 p(x)\,dx \tag{12}\\ & = \int\! \left(x^2 - 2 x \langle x\rangle^{2} + \langle x\rangle^2\right)p(x)\,dx \tag{13}\\ & = \langle x^2\rangle - 2 \langle x\rangle\langle x\rangle + \langle x\rangle^2 \tag{14}\\ & = \langle x^2\rangle - \langle x\rangle^2 \tag{15} \end{align}$$

 
The square root of the variance, $\sigma =\sqrt{\langle (x-\langle x\rangle)^2\rangle}$ is called the standard deviation of $p$. It is clearly just the RMS (root-mean-square) value of the deviation of the PDF from its mean value, interpreted qualitatively as the spread of $p$ around its mean.

Another important quantity is the so called covariance, a variant of the above defined variance. Consider again the set $\{X_i\}$ of $n$ stochastic variables (not necessarily uncorrelated) with the multivariate PDF $P(x_1,\dots,x_n)$. The covariance of two of the stochastic variables, $X_i$ and $X_j$, is defined as follows:

 
$$\begin{align} \mathrm{cov}(X_i,\,X_j) &\equiv \langle (x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle \nonumber\\ &= \int\!\cdots\!\int\!(x_i-\langle x_i \rangle)(x_j-\langle x_j \rangle)\, P(x_1,\dots,x_n)\,dx_1\dots dx_n \tag{16} \end{align}$$

 
with

 
$$\langle x_i\rangle = \int\!\cdots\!\int\!x_i\,P(x_1,\dots,x_n)\,dx_1\dots dx_n$$

If we consider the above covariance as a matrix $C_{ij}=\mathrm{cov}(X_i,\,X_j)$, then the diagonal elements are just the familiar variances, $C_{ii} = \mathrm{cov}(X_i,\,X_i) = \mathrm{var}(X_i)$. It turns out that all the off-diagonal elements are zero if the stochastic variables are uncorrelated. This is easy to show, keeping in mind the linearity of the expectation value. Consider the stochastic variables $X_i$ and $X_j$, ($i\neq j$):

 
$$\begin{align} \mathrm{cov}(X_i,\,X_j) &= \langle(x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle \tag{17}\\ &=\langle x_i x_j - x_i\langle x_j\rangle - \langle x_i\rangle x_j + \langle x_i\rangle\langle x_j\rangle\rangle \tag{18}\\ &=\langle x_i x_j\rangle - \langle x_i\langle x_j\rangle\rangle - \langle \langle x_i\rangle x_j\rangle + \langle \langle x_i\rangle\langle x_j\rangle\rangle \tag{19}\\ &=\langle x_i x_j\rangle - \langle x_i\rangle\langle x_j\rangle - \langle x_i\rangle\langle x_j\rangle + \langle x_i\rangle\langle x_j\rangle \tag{20}\\ &=\langle x_i x_j\rangle - \langle x_i\rangle\langle x_j\rangle \tag{21} \end{align}$$

If $X_i$ and $X_j$ are independent, we get $\langle x_i x_j\rangle =\langle x_i\rangle\langle x_j\rangle$, resulting in $\mathrm{cov}(X_i, X_j) = 0\ \ (i\neq j)$.

Also useful for us is the covariance of linear combinations of stochastic variables. Let $\{X_i\}$ and $\{Y_i\}$ be two sets of stochastic variables. Let also $\{a_i\}$ and $\{b_i\}$ be two sets of scalars. Consider the linear combination:

 
$$U = \sum_i a_i X_i \qquad V = \sum_j b_j Y_j$$

 
By the linearity of the expectation value

 
$$\mathrm{cov}(U, V) = \sum_{i,j}a_i b_j \mathrm{cov}(X_i, Y_j)$$

Now, since the variance is just $\mathrm{var}(X_i) = \mathrm{cov}(X_i, X_i)$, we get the variance of the linear combination $U = \sum_i a_i X_i$:

 
$$\begin{equation} \mathrm{var}(U) = \sum_{i,j}a_i a_j \mathrm{cov}(X_i, X_j) \tag{22} \end{equation}$$

 
And in the special case when the stochastic variables are uncorrelated, the off-diagonal elements of the covariance are as we know zero, resulting in:

 
$$\mathrm{var}(U) = \sum_i a_i^2 \mathrm{cov}(X_i, X_i) = \sum_i a_i^2 \mathrm{var}(X_i)$$

 
$$\mathrm{var}(\sum_i a_i X_i) = \sum_i a_i^2 \mathrm{var}(X_i)$$

 
which will become very useful in our study of the error in the mean value of a set of measurements.

A stochastic process is a process that produces sequentially a chain of values:

 
$$\{x_1, x_2,\dots\,x_k,\dots\}.$$

 
We will call these values our measurements and the entire set as our measured sample. The action of measuring all the elements of a sample we will call a stochastic experiment since, operationally, they are often associated with results of empirical observation of some physical or mathematical phenomena; precisely an experiment. We assume that these values are distributed according to some PDF $p_X^{\phantom X}(x)$, where $X$ is just the formal symbol for the stochastic variable whose PDF is $p_X^{\phantom X}(x)$. Instead of trying to determine the full distribution $p$ we are often only interested in finding the few lowest moments, like the mean $\mu_X^{\phantom X}$ and the variance $\sigma_X^{\phantom X}$.

In practical situations a sample is always of finite size. Let that size be $n$. The expectation value of a sample, the sample mean, is then defined as follows:

 
$$\bar{x}_n \equiv \frac{1}{n}\sum_{k=1}^n x_k$$

 
The sample variance is:

 
$$\mathrm{var}(x) \equiv \frac{1}{n}\sum_{k=1}^n (x_k - \bar{x}_n)^2$$

 
its square root being the standard deviation of the sample. The sample covariance is:

 
$$\mathrm{cov}(x)\equiv\frac{1}{n}\sum_{kl}(x_k - \bar{x}_n)(x_l - \bar{x}_n)$$

Note that the sample variance is the sample covariance without the cross terms. In a similar manner as the covariance in Eq. (16) is a measure of the correlation between two stochastic variables, the above defined sample covariance is a measure of the sequential correlation between succeeding measurements of a sample.

These quantities, being known experimental values, differ significantly from and must not be confused with the similarly named quantities for stochastic variables, mean $\mu_X$, variance $\mathrm{var}(X)$ and covariance $\mathrm{cov}(X,Y)$.

The law of large numbers states that as the size of our sample grows to infinity, the sample mean approaches the true mean $\mu_X^{\phantom X}$ of the chosen PDF:

 
$$\lim_{n\to\infty}\bar{x}_n = \mu_X^{\phantom X}$$

 
The sample mean $\bar{x}_n$ works therefore as an estimate of the true mean $\mu_X^{\phantom X}$.

What we need to find out is how good an approximation $\bar{x}_n$ is to $\mu_X^{\phantom X}$. In any stochastic measurement, an estimated mean is of no use to us without a measure of its error. A quantity that tells us how well we can reproduce it in another experiment. We are therefore interested in the PDF of the sample mean itself. Its standard deviation will be a measure of the spread of sample means, and we will simply call it the error of the sample mean, or just sample error, and denote it by $\mathrm{err}_X^{\phantom X}$. In practice, we will only be able to produce an estimate of the sample error since the exact value would require the knowledge of the true PDFs behind, which we usually do not have.

The straight forward brute force way of estimating the sample error is simply by producing a number of samples, and treating the mean of each as a measurement. The standard deviation of these means will then be an estimate of the original sample error. If we are unable to produce more than one sample, we can split it up sequentially into smaller ones, treating each in the same way as above. This procedure is known as blocking and will be given more attention shortly. At this point it is worth while exploring more indirect methods of estimation that will help us understand some important underlying principles of correlational effects.

Let us first take a look at what happens to the sample error as the size of the sample grows. In a sample, each of the measurements $x_i$ can be associated with its own stochastic variable $X_i$. The stochastic variable $\overline X_n$ for the sample mean $\bar{x}_n$ is then just a linear combination, already familiar to us:

 
$$\overline X_n = \frac{1}{n}\sum_{i=1}^n X_i$$

 
All the coefficients are just equal $1/n$. The PDF of $\overline X_n$, denoted by $p_{\overline X_n}(x)$ is the desired PDF of the sample means.

The probability density of obtaining a sample mean $\bar x_n$ is the product of probabilities of obtaining arbitrary values $x_1, x_2,\dots,x_n$ with the constraint that the mean of the set $\{x_i\}$ is $\bar x_n$:

 
$$p_{\overline X_n}(x) = \int p_X^{\phantom X}(x_1)\cdots \int p_X^{\phantom X}(x_n)\ \delta\!\left(x - \frac{x_1+x_2+\dots+x_n}{n}\right)dx_n \cdots dx_1$$

 
And in particular we are interested in its variance $\mathrm{var}(\overline X_n)$.

It is generally not possible to express $p_{\overline X_n}(x)$ in a closed form given an arbitrary PDF $p_X^{\phantom X}$ and a number $n$. But for the limit $n\to\infty$ it is possible to make an approximation. The very important result is called the central limit theorem. It tells us that as $n$ goes to infinity, $p_{\overline X_n}(x)$ approaches a Gaussian distribution whose mean and variance equal the true mean and variance, $\mu_{X}^{\phantom X}$ and $\sigma_{X}^{2}$, respectively:

 
$$\begin{equation} \lim_{n\to\infty} p_{\overline X_n}(x) = \left(\frac{n}{2\pi\mathrm{var}(X)}\right)^{1/2} e^{-\frac{n(x-\bar x_n)^2}{2\mathrm{var}(X)}} \tag{23} \end{equation}$$

The desired variance $\mathrm{var}(\overline X_n)$, i.e. the sample error squared $\mathrm{err}_X^2$, is given by:

 
$$\begin{equation} \mathrm{err}_X^2 = \mathrm{var}(\overline X_n) = \frac{1}{n^2} \sum_{ij} \mathrm{cov}(X_i, X_j) \tag{24} \end{equation}$$

 
We see now that in order to calculate the exact error of the sample with the above expression, we would need the true means $\mu_{X_i}^{\phantom X}$ of the stochastic variables $X_i$. To calculate these requires that we know the true multivariate PDF of all the $X_i$. But this PDF is unknown to us, we have only got the measurements of one sample. The best we can do is to let the sample itself be an estimate of the PDF of each of the $X_i$, estimating all properties of $X_i$ through the measurements of the sample.

Our estimate of $\mu_{X_i}^{\phantom X}$ is then the sample mean $\bar x$ itself, in accordance with the the central limit theorem:

 
$$\mu_{X_i}^{\phantom X} = \langle x_i\rangle \approx \frac{1}{n}\sum_{k=1}^n x_k = \bar x$$

 
Using $\bar x$ in place of $\mu_{X_i}^{\phantom X}$ we can give an estimate of the covariance in Eq. (24)

 
$$\mathrm{cov}(X_i, X_j) = \langle (x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle \approx\langle (x_i - \bar x)(x_j - \bar{x})\rangle,$$

 
resulting in

 
$$\frac{1}{n} \sum_{l}^n \left(\frac{1}{n}\sum_{k}^n (x_k -\bar x_n)(x_l - \bar x_n)\right)=\frac{1}{n}\frac{1}{n} \sum_{kl} (x_k -\bar x_n)(x_l - \bar x_n)=\frac{1}{n}\mathrm{cov}(x)$$

By the same procedure we can use the sample variance as an estimate of the variance of any of the stochastic variables $X_i$

 
$$\mathrm{var}(X_i)=\langle x_i - \langle x_i\rangle\rangle \approx \langle x_i - \bar x_n\rangle\nonumber,$$

 
which is approximated as

 
$$\begin{equation} \mathrm{var}(X_i)\approx \frac{1}{n}\sum_{k=1}^n (x_k - \bar x_n)=\mathrm{var}(x) \tag{25} \end{equation}$$

Now we can calculate an estimate of the error $\mathrm{err}_X^{\phantom X}$ of the sample mean $\bar x_n$:

 
$$\begin{align} \mathrm{err}_X^2 &=\frac{1}{n^2}\sum_{ij} \mathrm{cov}(X_i, X_j) \nonumber \\ &\approx&\frac{1}{n^2}\sum_{ij}\frac{1}{n}\mathrm{cov}(x) =\frac{1}{n^2}n^2\frac{1}{n}\mathrm{cov}(x)\nonumber\\ &=\frac{1}{n}\mathrm{cov}(x) \tag{26} \end{align}$$

 
which is nothing but the sample covariance divided by the number of measurements in the sample.

In the special case that the measurements of the sample are uncorrelated (equivalently the stochastic variables $X_i$ are uncorrelated) we have that the off-diagonal elements of the covariance are zero. This gives the following estimate of the sample error:

 
$$\mathrm{err}_X^2=\frac{1}{n^2}\sum_{ij} \mathrm{cov}(X_i, X_j) = \frac{1}{n^2} \sum_i \mathrm{var}(X_i),$$

 
resulting in

 
$$\begin{equation} \mathrm{err}_X^2\approx \frac{1}{n^2} \sum_i \mathrm{var}(x)= \frac{1}{n}\mathrm{var}(x) \tag{27} \end{equation}$$

 
where in the second step we have used Eq. (25). The error of the sample is then just its standard deviation divided by the square root of the number of measurements the sample contains. This is a very useful formula which is easy to compute. It acts as a first approximation to the error, but in numerical experiments, we cannot overlook the always present correlations.

For computational purposes one usually splits up the estimate of $\mathrm{err}_X^2$, given by Eq. (26), into two parts

 
$$\mathrm{err}_X^2 = \frac{1}{n}\mathrm{var}(x) + \frac{1}{n}(\mathrm{cov}(x)-\mathrm{var}(x)),$$

 
which equals

 
$$\begin{equation} \frac{1}{n^2}\sum_{k=1}^n (x_k - \bar x_n)^2 +\frac{2}{n^2}\sum_{k < l} (x_k - \bar x_n)(x_l - \bar x_n) \tag{28} \end{equation}$$

 
The first term is the same as the error in the uncorrelated case, Eq. (27). This means that the second term accounts for the error correction due to correlation between the measurements. For uncorrelated measurements this second term is zero.

Computationally the uncorrelated first term is much easier to treat efficiently than the second.

 
$$\mathrm{var}(x) = \frac{1}{n}\sum_{k=1}^n (x_k - \bar x_n)^2 = \left(\frac{1}{n}\sum_{k=1}^n x_k^2\right) - \bar x_n^2$$

 
We just accumulate separately the values $x^2$ and $x$ for every measurement $x$ we receive. The correlation term, though, has to be calculated at the end of the experiment since we need all the measurements to calculate the cross terms. Therefore, all measurements have to be stored throughout the experiment.

Let us analyze the problem by splitting up the correlation term into partial sums of the form:

 
$$f_d = \frac{1}{n-d}\sum_{k=1}^{n-d}(x_k - \bar x_n)(x_{k+d} - \bar x_n)$$

 
The correlation term of the error can now be rewritten in terms of $f_d$

 
$$\frac{2}{n}\sum_{k < l} (x_k - \bar x_n)(x_l - \bar x_n) = 2\sum_{d=1}^{n-1} f_d$$

 
The value of $f_d$ reflects the correlation between measurements separated by the distance $d$ in the sample samples. Notice that for $d=0$, $f$ is just the sample variance, $\mathrm{var}(x)$. If we divide $f_d$ by $\mathrm{var}(x)$, we arrive at the so called autocorrelation function

 
$$\kappa_d = \frac{f_d}{\mathrm{var}(x)}$$

 
which gives us a useful measure of the correlation pair correlation starting always at $1$ for $d=0$.

The sample error (see eq. (28)) can now be written in terms of the autocorrelation function:

 
$$\begin{align} \mathrm{err}_X^2 &= \frac{1}{n}\mathrm{var}(x)+\frac{2}{n}\cdot\mathrm{var}(x)\sum_{d=1}^{n-1} \frac{f_d}{\mathrm{var}(x)}\nonumber\\ &=& \left(1+2\sum_{d=1}^{n-1}\kappa_d\right)\frac{1}{n}\mathrm{var}(x)\nonumber\\ &=\frac{\tau}{n}\cdot\mathrm{var}(x) \tag{29} \end{align}$$

 
and we see that $\mathrm{err}_X$ can be expressed in terms the uncorrelated sample variance times a correction factor $\tau$ which accounts for the correlation between measurements. We call this correction factor the autocorrelation time:

 
$$\begin{equation} \tau = 1+2\sum_{d=1}^{n-1}\kappa_d \tag{30} \end{equation}$$

For a correlation free experiment, $\tau$ equals 1. From the point of view of eq. (29) we can interpret a sequential correlation as an effective reduction of the number of measurements by a factor $\tau$. The effective number of measurements becomes:

 
$$n_\mathrm{eff} = \frac{n}{\tau}$$

 
To neglect the autocorrelation time $\tau$ will always cause our simple uncorrelated estimate of $\mathrm{err}_X^2\approx \mathrm{var}(x)/n$ to be less than the true sample error. The estimate of the error will be too good. On the other hand, the calculation of the full autocorrelation time poses an efficiency problem if the set of measurements is very large.

The so-called time-displacement autocorrelation $\phi(t)$ for a quantity $\mathbf{M}$ is given by

 
$$\phi(t) = \int dt' \left[\mathbf{M}(t')-\langle \mathbf{M} \rangle\right]\left[\mathbf{M}(t'+t)-\langle \mathbf{M} \rangle\right],$$

 
which can be rewritten as

 
$$\phi(t) = \int dt' \left[\mathbf{M}(t')\mathbf{M}(t'+t)-\langle \mathbf{M} \rangle^2\right],$$

 
where $\langle \mathbf{M} \rangle$ is the average value and $\mathbf{M}(t)$ its instantaneous value. We can discretize this function as follows, where we used our set of computed values $\mathbf{M}(t)$ for a set of discretized times (our Monte Carlo cycles corresponding to moving all electrons?)

 
$$\phi(t) = \frac{1}{t_{\mathrm{max}}-t}\sum_{t'=0}^{t_{\mathrm{max}}-t}\mathbf{M}(t')\mathbf{M}(t'+t) -\frac{1}{t_{\mathrm{max}}-t}\sum_{t'=0}^{t_{\mathrm{max}}-t}\mathbf{M}(t')\times \frac{1}{t_{\mathrm{max}}-t}\sum_{t'=0}^{t_{\mathrm{max}}-t}\mathbf{M}(t'+t). \tag{31}$$

One should be careful with times close to $t_{\mathrm{max}}$, the upper limit of the sums becomes small and we end up integrating over a rather small time interval. This means that the statistical error in $\phi(t)$ due to the random nature of the fluctuations in $\mathbf{M}(t)$ can become large.

One should therefore choose $t \ll t_{\mathrm{max}}$.

Note that the variable $\mathbf{M}$ can be any expectation values of interest.

The time-correlation function gives a measure of the correlation between the various values of the variable at a time $t'$ and a time $t'+t$. If we multiply the values of $\mathbf{M}$ at these two different times, we will get a positive contribution if they are fluctuating in the same direction, or a negative value if they fluctuate in the opposite direction. If we then integrate over time, or use the discretized version of, the time correlation function $\phi(t)$ should take a non-zero value if the fluctuations are correlated, else it should gradually go to zero. For times a long way apart the different values of $\mathbf{M}$ are most likely uncorrelated and $\phi(t)$ should be zero.

We can derive the correlation time by observing that our Metropolis algorithm is based on a random walk in the space of all possible spin configurations. Our probability distribution function $\mathbf{\hat{w}}(t)$ after a given number of time steps $t$ could be written as

 
$$\mathbf{\hat{w}}(t) = \mathbf{\hat{W}^t\hat{w}}(0),$$

 
with $\mathbf{\hat{w}}(0)$ the distribution at $t=0$ and $\mathbf{\hat{W}}$ representing the transition probability matrix. We can always expand $\mathbf{\hat{w}}(0)$ in terms of the right eigenvectors of $\mathbf{\hat{v}}$ of $\mathbf{\hat{W}}$ as

 
$$\mathbf{\hat{w}}(0) = \sum_i\alpha_i\mathbf{\hat{v}}_i,$$

 
resulting in

 
$$\mathbf{\hat{w}}(t) = \mathbf{\hat{W}}^t\mathbf{\hat{w}}(0)=\mathbf{\hat{W}}^t\sum_i\alpha_i\mathbf{\hat{v}}_i= \sum_i\lambda_i^t\alpha_i\mathbf{\hat{v}}_i,$$

 
with $\lambda_i$ the $i^{\mathrm{th}}$ eigenvalue corresponding to the eigenvector $\mathbf{\hat{v}}_i$.

If we assume that $\lambda_0$ is the largest eigenvector we see that in the limit $t\rightarrow \infty$, $\mathbf{\hat{w}}(t)$ becomes proportional to the corresponding eigenvector $\mathbf{\hat{v}}_0$. This is our steady state or final distribution.

We can relate this property to an observable like the mean energy. With the probabilty $\mathbf{\hat{w}}(t)$ (which in our case is the squared trial wave function) we can write the expectation values as

 
$$\langle \mathbf{M}(t) \rangle = \sum_{\mu} \mathbf{\hat{w}}(t)_{\mu}\mathbf{M}_{\mu},$$

 
or as the scalar of a vector product

 
$$\langle \mathbf{M}(t) \rangle = \mathbf{\hat{w}}(t)\mathbf{m},$$

 
with $\mathbf{m}$ being the vector whose elements are the values of $\mathbf{M}_{\mu}$ in its various microstates $\mu$.

We rewrite this relation as

 
$$\langle \mathbf{M}(t) \rangle = \mathbf{\hat{w}}(t)\mathbf{m}=\sum_i\lambda_i^t\alpha_i\mathbf{\hat{v}}_i\mathbf{m}_i.$$

 
If we define $m_i=\mathbf{\hat{v}}_i\mathbf{m}_i$ as the expectation value of $\mathbf{M}$ in the $i^{\mathrm{th}}$ eigenstate we can rewrite the last equation as

 
$$\langle \mathbf{M}(t) \rangle = \sum_i\lambda_i^t\alpha_im_i.$$

 
Since we have that in the limit $t\rightarrow \infty$ the mean value is dominated by the the largest eigenvalue $\lambda_0$, we can rewrite the last equation as

 
$$\langle \mathbf{M}(t) \rangle = \langle \mathbf{M}(\infty) \rangle+\sum_{i\ne 0}\lambda_i^t\alpha_im_i.$$

 
We define the quantity

 
$$\tau_i=-\frac{1}{log\lambda_i},$$

 
and rewrite the last expectation value as

 
$$\langle \mathbf{M}(t) \rangle = \langle \mathbf{M}(\infty) \rangle+\sum_{i\ne 0}\alpha_im_ie^{-t/\tau_i}. \tag{32}$$

The quantities $\tau_i$ are the correlation times for the system. They control also the auto-correlation function discussed above. The longest correlation time is obviously given by the second largest eigenvalue $\tau_1$, which normally defines the correlation time discussed above. For large times, this is the only correlation time that survives. If higher eigenvalues of the transition matrix are well separated from $\lambda_1$ and we simulate long enough, $\tau_1$ may well define the correlation time. In other cases we may not be able to extract a reliable result for $\tau_1$. Coming back to the time correlation function $\phi(t)$ we can present a more general definition in terms of the mean magnetizations $ \langle \mathbf{M}(t) \rangle$. Recalling that the mean value is equal to $ \langle \mathbf{M}(\infty) \rangle$ we arrive at the expectation values

 
$$\phi(t) =\langle \mathbf{M}(0)-\mathbf{M}(\infty)\rangle \langle \mathbf{M}(t)-\mathbf{M}(\infty)\rangle,$$

 
resulting in

 
$$\phi(t) =\sum_{i,j\ne 0}m_i\alpha_im_j\alpha_je^{-t/\tau_i},$$

 
which is appropriate for all times.

If the correlation function decays exponentially

 
$$\phi (t) \sim \exp{(-t/\tau)}$$

 
then the exponential correlation time can be computed as the average

 
$$\tau_{\mathrm{exp}} = -\langle \frac{t}{log|\frac{\phi(t)}{\phi(0)}|} \rangle.$$

 
If the decay is exponential, then

 
$$\int_0^{\infty} dt \phi(t) = \int_0^{\infty} dt \phi(0)\exp{(-t/\tau)} = \tau \phi(0),$$

 
which suggests another measure of correlation

 
$$\tau_{\mathrm{int}} = \sum_k \frac{\phi(k)}{\phi(0)},$$

 
called the integrated correlation time.

• Say that we have a set of samples from a Monte Carlo experiment
• Assuming (wrongly) that our samples are uncorrelated our best estimate of the standard deviation of the mean $\langle \mathbf{M}\rangle$ is given by

 
$$\sigma=\sqrt{\frac{1}{n}\left(\langle \mathbf{M}^2\rangle-\langle \mathbf{M}\rangle^2\right)}$$

• If the samples are correlated we can rewrite our results to show that

 
$$\sigma=\sqrt{\frac{1+2\tau/\Delta t}{n}\left(\langle \mathbf{M}^2\rangle-\langle \mathbf{M}\rangle^2\right)}$$

 
where $\tau$ is the correlation time (the time between a sample and the next uncorrelated sample) and $\Delta t$ is time between each sample

• If $\Delta t\gg\tau$ our first estimate of $\sigma$ still holds
• Much more common that $\Delta t < \tau$
• In the method of data blocking we divide the sequence of samples into blocks
• We then take the mean $\langle \mathbf{M}_i\rangle$ of block $i=1\ldots n_{blocks}$ to calculate the total mean and variance
• The size of each block must be so large that sample $j$ of block $i$ is not correlated with sample $j$ of block $i+1$
• The correlation time $\tau$ would be a good choice

• Problem: We don't know $\tau$ or it is too expensive to compute
• Solution: Make a plot of std. dev. as a function of blocksize
• The estimate of std. dev. of correlated data is too low $\to$ the error will increase with increasing block size until the blocks are uncorrelated, where we reach a plateau
• When the std. dev. stops increasing the blocks are uncorrelated

• Do a Monte Carlo simulation, storing all samples to file
• Do the statistical analysis on this file, independently of your Monte Carlo program
• Read the file into an array
• Loop over various block sizes
• For each block size $n_b$, loop over the array in steps of $n_b$ taking the mean of elements $i n_b,\ldots,(i+1) n_b$
• Take the mean and variance of the resulting array
• Write the results for each block size to file for later analysis

The potentially most time-consuming part is the evaluation of the gradient and the Laplacian of an $N$-particle Slater determinant.

We have to differentiate the determinant with respect to all spatial coordinates of all particles. A brute force differentiation would involve $N\cdot d$ evaluations of the entire determinant which would even worsen the already undesirable time scaling, making it $Nd\cdot O(N^3)\sim O(d\cdot N^4)$.

This poses serious hindrances to the overall efficiency of our code.

The efficiency can be improved however if we move only one electron at the time. The Slater determinant matrix $\hat{D}$ is defined by the matrix elements

 
$$d_{ij}=\phi_j(x_i)$$

 
where $\phi_j(\mathbf{r}_i)$ is a single particle wave function. The columns correspond to the position of a given particle while the rows stand for the various quantum numbers.

What we need to realize is that when differentiating a Slater determinant with respect to some given coordinate, only one row of the corresponding Slater matrix is changed.

Therefore, by recalculating the whole determinant we risk producing redundant information. The solution turns out to be an algorithm that requires to keep track of the inverse of the Slater matrix.

Let the current position in phase space be represented by the $(N\cdot d)$-element vector $\mathbf{r}^{\mathrm{old}}$ and the new suggested position by the vector $\mathbf{r}^{\mathrm{new}}$.

The inverse of $\hat{D}$ can be expressed in terms of its cofactors $C_{ij}$ and its determinant (this our notation for a determinant) $\vert\hat{D}\vert$:

 
$$\begin{equation} d_{ij}^{-1} = \frac{C_{ji}}{\vert\hat{D}\vert} \tag{33} \end{equation}$$

 
Notice that the interchanged indices indicate that the matrix of cofactors is to be transposed.

If $\hat{D}$ is invertible, then we must obviously have $\hat{D}^{-1}\hat{D}= \mathbf{1}$, or explicitly in terms of the individual elements of $\hat{D}$ and $\hat{D}^{-1}$:

 
$$\begin{equation} \sum_{k=1}^N d_{ik}^{\phantom X}d_{kj}^{-1} = \delta_{ij}^{\phantom X} \tag{34} \end{equation}$$

Consider the ratio, which we shall call $R$, between $\vert\hat{D}(\mathbf{r}^{\mathrm{new}})\vert$ and $\vert\hat{D}(\mathbf{r}^{\mathrm{old}})\vert$. By definition, each of these determinants can individually be expressed in terms of the i-th row of its cofactor matrix

 
$$\begin{equation} R\equiv\frac{\vert\hat{D}(\mathbf{r}^{\mathrm{new}})\vert} {\vert\hat{D}(\mathbf{r}^{\mathrm{old}})\vert} = \frac{\sum_{j=1}^N d_{ij}(\mathbf{r}^{\mathrm{new}})\, C_{ij}(\mathbf{r}^{\mathrm{new}})} {\sum_{j=1}^N d_{ij}(\mathbf{r}^{\mathrm{old}})\, C_{ij}(\mathbf{r}^{\mathrm{old}})} \tag{35} \end{equation}$$

Suppose now that we move only one particle at a time, meaning that $\mathbf{r}^{\mathrm{new}}$ differs from $\mathbf{r}^{\mathrm{old}}$ by the position of only one, say the i-th, particle . This means that $\hat{D}(\mathbf{r}^{\mathrm{new}})$ and $\hat{D}(\mathbf{r}^{\mathrm{old}})$ differ only by the entries of the i-th row. Recall also that the i-th row of a cofactor matrix $\hat{C}$ is independent of the entries of the i-th row of its corresponding matrix $\hat{D}$. In this particular case we therefore get that the i-th row of $\hat{C}(\mathbf{r}^{\mathrm{new}})$ and $\hat{C}(\mathbf{r}^{\mathrm{old}})$ must be equal. Explicitly, we have:

 
$$\begin{equation} C_{ij}(\mathbf{r}^{\mathrm{new}}) = C_{ij}(\mathbf{r}^{\mathrm{old}})\quad \forall\ j\in\{1,\dots,N\} \tag{36} \end{equation}$$

Inserting this into the numerator of eq. (35) and using eq. (33) to substitute the cofactors with the elements of the inverse matrix, we get:

 
$$\begin{equation} R =\frac{\sum_{j=1}^N d_{ij}(\mathbf{r}^{\mathrm{new}})\, C_{ij}(\mathbf{r}^{\mathrm{old}})} {\sum_{j=1}^N d_{ij}(\mathbf{r}^{\mathrm{old}})\, C_{ij}(\mathbf{r}^{\mathrm{old}})} = \frac{\sum_{j=1}^N d_{ij}(\mathbf{r}^{\mathrm{new}})\, d_{ji}^{-1}(\mathbf{r}^{\mathrm{old}})} {\sum_{j=1}^N d_{ij}(\mathbf{r}^{\mathrm{old}})\, d_{ji}^{-1}(\mathbf{r}^{\mathrm{old}})} \tag{37} \end{equation}$$

Now by eq. (34) the denominator of the rightmost expression must be unity, so that we finally arrive at:

 
$$\begin{equation} R = \sum_{j=1}^N d_{ij}(\mathbf{r}^{\mathrm{new}})\, d_{ji}^{-1}(\mathbf{r}^{\mathrm{old}}) = \sum_{j=1}^N \phi_j(\mathbf{r}_i^{\mathrm{new}})\, d_{ji}^{-1}(\mathbf{r}^{\mathrm{old}}) \tag{38} \end{equation}$$

 
What this means is that in order to get the ratio when only the i-th particle has been moved, we only need to calculate the dot product of the vector $\left(\phi_1(\mathbf{r}_i^\mathrm{new}),\,\dots,\,\phi_N(\mathbf{r}_i^\mathrm{new})\right)$ of single particle wave functions evaluated at this new position with the i-th column of the inverse matrix $\hat{D}^{-1}$ evaluated at the original position. Such an operation has a time scaling of $O(N)$. The only extra thing we need to do is to maintain the inverse matrix $\hat{D}^{-1}(\mathbf{x}^{\mathrm{old}})$.

If the new position $\mathbf{r}^{\mathrm{new}}$ is accepted, then the inverse matrix can by suitably updated by an algorithm having a time scaling of $O(N^2)$. This algorithm goes as follows. First we update all but the i-th column of $\hat{D}^{-1}$. For each column $j\neq i$, we first calculate the quantity:

 
$$\begin{equation} S_j = (\hat{D}(\mathbf{r}^{\mathrm{new}})\times \hat{D}^{-1}(\mathbf{r}^{\mathrm{old}}))_{ij} = \sum_{l=1}^N d_{il}(\mathbf{r}^{\mathrm{new}})\, d^{-1}_{lj}(\mathbf{r}^{\mathrm{old}}) \tag{39} \end{equation}$$

The new elements of the j-th column of $\hat{D}^{-1}$ are then given by:

 
$$\begin{equation} d_{kj}^{-1}(\mathbf{r}^{\mathrm{new}}) = d_{kj}^{-1}(\mathbf{r}^{\mathrm{old}}) - \frac{S_j}{R}\,d_{ki}^{-1}(\mathbf{r}^{\mathrm{old}})\quad \begin{array}{ll} \forall\ \ k\in\{1,\dots,N\}\\j\neq i \end{array} \tag{40} \end{equation}$$

Finally the elements of the i-th column of $\hat{D}^{-1}$ are updated simply as follows:

 
$$\begin{equation} d_{ki}^{-1}(\mathbf{r}^{\mathrm{new}}) = \frac{1}{R}\,d_{ki}^{-1}(\mathbf{r}^{\mathrm{old}})\quad \forall\ \ k\in\{1,\dots,N\} \tag{41} \end{equation}$$

 
We see from these formulas that the time scaling of an update of $\hat{D}^{-1}$ after changing one row of $\hat{D}$ is $O(N^2)$.

The scheme is also applicable for the calculation of the ratios involving derivatives. It turns out that differentiating the Slater determinant with respect to the coordinates of a single particle $\mathbf{r}_i$ changes only the i-th row of the corresponding Slater matrix.

The gradient and the Laplacian can therefore be calculated as follows:

 
$$\frac{\vec\nabla_i\vert\hat{D}(\mathbf{r})\vert}{\vert\hat{D}(\mathbf{r})\vert} = \sum_{j=1}^N \vec\nabla_i d_{ij}(\mathbf{r})d_{ji}^{-1}(\mathbf{r}) = \sum_{j=1}^N \vec\nabla_i \phi_j(\mathbf{r}_i)d_{ji}^{-1}(\mathbf{r})$$

 
and

 
$$\frac{\nabla^2_i\vert\hat{D}(\mathbf{r})\vert}{\vert\hat{D}(\mathbf{r})\vert} = \sum_{j=1}^N \nabla^2_i d_{ij}(\mathbf{r})d_{ji}^{-1}(\mathbf{r}) = \sum_{j=1}^N \nabla^2_i \phi_j(\mathbf{r}_i)\,d_{ji}^{-1}(\mathbf{r})$$

Thus, to calculate all the derivatives of the Slater determinant, we only need the derivatives of the single particle wave functions ($\vec\nabla_i \phi_j(\mathbf{r}_i)$ and $\nabla^2_i \phi_j(\mathbf{r}_i)$) and the elements of the corresponding inverse Slater matrix ($\hat{D}^{-1}(\mathbf{r}_i)$). A calculation of a single derivative is by the above result an $O(N)$ operation. Since there are $d\cdot N$ derivatives, the time scaling of the total evaluation becomes $O(d\cdot N^2)$. With an $O(N^2)$ updating algorithm for the inverse matrix, the total scaling is no worse, which is far better than the brute force approach yielding $O(d\cdot N^4)$.

Important note: In most cases you end with closed form expressions for the single-particle wave functions. It is then useful to calculate the various derivatives and make separate functions for them.

The Slater determinant takes the form

 
$$\Phi(\mathbf{r}_1,\mathbf{r}_2,,\mathbf{r}_3,\mathbf{r}_4, \alpha,\beta,\gamma,\delta)=\frac{1}{\sqrt{4!}} \left| \begin{array}{cccc} \psi_{100\uparrow}(\mathbf{r}_1)& \psi_{100\uparrow}(\mathbf{r}_2)& \psi_{100\uparrow}(\mathbf{r}_3)&\psi_{100\uparrow}(\mathbf{r}_4) \\ \psi_{100\downarrow}(\mathbf{r}_1)& \psi_{100\downarrow}(\mathbf{r}_2)& \psi_{100\downarrow}(\mathbf{r}_3)&\psi_{100\downarrow}(\mathbf{r}_4) \\ \psi_{200\uparrow}(\mathbf{r}_1)& \psi_{200\uparrow}(\mathbf{r}_2)& \psi_{200\uparrow}(\mathbf{r}_3)&\psi_{200\uparrow}(\mathbf{r}_4) \\ \psi_{200\downarrow}(\mathbf{r}_1)& \psi_{200\downarrow}(\mathbf{r}_2)& \psi_{200\downarrow}(\mathbf{r}_3)&\psi_{200\downarrow}(\mathbf{r}_4) \end{array} \right|.$$

 
The Slater determinant as written is zero since the spatial wave functions for the spin up and spin down states are equal. But we can rewrite it as the product of two Slater determinants, one for spin up and one for spin down.

We can rewrite it as

 
$$\Phi(\mathbf{r}_1,\mathbf{r}_2,,\mathbf{r}_3,\mathbf{r}_4, \alpha,\beta,\gamma,\delta)=\det\uparrow(1,2)\det\downarrow(3,4)-\det\uparrow(1,3)\det\downarrow(2,4)$$

 
$$-\det\uparrow(1,4)\det\downarrow(3,2)+\det\uparrow(2,3)\det\downarrow(1,4)-\det\uparrow(2,4)\det\downarrow(1,3)$$

 
$$+\det\uparrow(3,4)\det\downarrow(1,2),$$

 
where we have defined

 
$$\det\uparrow(1,2)=\frac{1}{\sqrt{2}}\left| \begin{array}{cc} \psi_{100\uparrow}(\mathbf{r}_1)& \psi_{100\uparrow}(\mathbf{r}_2)\\ \psi_{200\uparrow}(\mathbf{r}_1)& \psi_{200\uparrow}(\mathbf{r}_2) \end{array} \right|,$$

 
and

 
$$\det\downarrow(3,4)=\frac{1}{\sqrt{2}}\left| \begin{array}{cc} \psi_{100\downarrow}(\mathbf{r}_3)& \psi_{100\downarrow}(\mathbf{r}_4)\\ \psi_{200\downarrow}(\mathbf{r}_3)& \psi_{200\downarrow}(\mathbf{r}_4) \end{array} \right|.$$

 
The total determinant is still zero!

We want to avoid to sum over spin variables, in particular when the interaction does not depend on spin.

It can be shown, see for example Moskowitz and Kalos, Int. J. Quantum Chem. 20 1107 (1981), that for the variational energy we can approximate the Slater determinant as

 
$$\Phi(\mathbf{r}_1,\mathbf{r}_2,,\mathbf{r}_3,\mathbf{r}_4, \alpha,\beta,\gamma,\delta) \propto \det\uparrow(1,2)\det\downarrow(3,4),$$

 
or more generally as

 
$$\Phi(\mathbf{r}_1,\mathbf{r}_2,\dots \mathbf{r}_N) \propto \det\uparrow \det\downarrow,$$

 
where we have the Slater determinant as the product of a spin up part involving the number of electrons with spin up only (2 for beryllium and 5 for neon) and a spin down part involving the electrons with spin down.

This ansatz is not antisymmetric under the exchange of electrons with opposite spins but it can be shown (show this) that it gives the same expectation value for the energy as the full Slater determinant.

As long as the Hamiltonian is spin independent, the above is correct. It is rather straightforward to see this if you go back to the equations for the energy discussed earlier this semester.

We will thus factorize the full determinant $\vert\hat{D}\vert$ into two smaller ones, where each can be identified with $\uparrow$ and $\downarrow$ respectively:

 
$$\vert\hat{D}\vert = \vert\hat{D}\vert_\uparrow\cdot \vert\hat{D}\vert_\downarrow$$

The combined dimensionality of the two smaller determinants equals the dimensionality of the full determinant. Such a factorization is advantageous in that it makes it possible to perform the calculation of the ratio $R$ and the updating of the inverse matrix separately for $\vert\hat{D}\vert_\uparrow$ and $\vert\hat{D}\vert_\downarrow$:

 
$$\frac{\vert\hat{D}\vert^\mathrm{new}}{\vert\hat{D}\vert^\mathrm{old}} = \frac{\vert\hat{D}\vert^\mathrm{new}_\uparrow} {\vert\hat{D}\vert^\mathrm{old}_\uparrow}\cdot \frac{\vert\hat{D}\vert^\mathrm{new}_\downarrow }{\vert\hat{D}\vert^\mathrm{old}_\downarrow}$$

This reduces the calculation time by a constant factor. The maximal time reduction happens in a system of equal numbers of $\uparrow$ and $\downarrow$ particles, so that the two factorized determinants are half the size of the original one.

Consider the case of moving only one particle at a time which originally had the following time scaling for one transition:

 
$$O_R(N)+O_\mathrm{inverse}(N^2)$$

 
For the factorized determinants one of the two determinants is obviously unaffected by the change so that it cancels from the ratio $R$.

Therefore, only one determinant of size $N/2$ is involved in each calculation of $R$ and update of the inverse matrix. The scaling of each transition then becomes:

 
$$O_R(N/2)+O_\mathrm{inverse}(N^2/4)$$

 
and the time scaling when the transitions for all $N$ particles are put together:

 
$$O_R(N^2/2)+O_\mathrm{inverse}(N^3/4)$$

 
which gives the same reduction as in the case of moving all particles at once.

Computing the ratios discussed above requires that we maintain the inverse of the Slater matrix evaluated at the current position. Each time a trial position is accepted, the row number $i$ of the Slater matrix changes and updating its inverse has to be carried out. Getting the inverse of an $N \times N$ matrix by Gaussian elimination has a complexity of order of $\mathcal{O}(N^3)$ operations, a luxury that we cannot afford for each time a particle move is accepted. We will use the expression $$\begin{equation*} \tag{42} d^{-1}_{kj}(\mathbf{x^{new}}) = \left\{\begin{array}{l l} d^{-1}_{kj}(\mathbf{x^{old}}) - \frac{d^{-1}_{ki}(\mathbf{x^{old}})}{R} \sum_{l=1}^{N} d_{il}(\mathbf{x^{new}}) d^{-1}_{lj}(\mathbf{x^{old}}) & \mbox{if $j \neq i$}\nonumber \\ \\ \frac{d^{-1}_{ki}(\mathbf{x^{old}})}{R} \sum_{l=1}^{N} d_{il}(\mathbf{x^{old}}) d^{-1}_{lj}(\mathbf{x^{old}}) & \mbox{if $j=i$} \end{array} \right. \end{equation*}$$

This equation scales as $O(N^2)$. The evaluation of the determinant of an $N \times N$ matrix by standard Gaussian elimination requires $\mathbf{O}(N^3)$ calculations. As there are $Nd$ independent coordinates we need to evaluate $Nd$ Slater determinants for the gradient (quantum force) and $Nd$ for the Laplacian (kinetic energy). With the updating algorithm we need only to invert the Slater determinant matrix once. This can be done by standard LU decomposition methods.

Determining a determinant of an $N \times N$ matrix by standard Gaussian elimination is of the order of $\mathbf{O}(N^3)$ calculations. As there are $N\cdot d$ independent coordinates we need to evaluate $Nd$ Slater determinants for the gradient (quantum force) and $N\cdot d$ for the Laplacian (kinetic energy)

With the updating algorithm we need only to invert the Slater determinant matrix once. This is done by calling standard LU decomposition methods.

If you choose to implement the above recipe for the computation of the Slater determinant, you need to LU decompose the Slater matrix. This is described in chapter 6 of the lecture notes from FYS3150.

You need to call the function ludcmp in lib.cpp. You need to transfer the Slater matrix and its dimension. You get back an LU decomposed matrix.

The LU decomposition method means that we can rewrite this matrix as the product of two matrices $\hat{B}$ and $\hat{C}$ where

 
$$\left(\begin{array}{cccc} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{array} \right) = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ b_{21} & 1 & 0 & 0 \\ b_{31} & b_{32} & 1 & 0 \\ b_{41} & b_{42} & b_{43} & 1 \end{array} \right) \left( \begin{array}{cccc} c_{11} & c_{12} & c_{13} & c_{14} \\ 0 & c_{22} & c_{23} & c_{24} \\ 0 & 0 & c_{33} & c_{34} \\ 0 & 0 & 0 & c_{44} \end{array} \right).$$

The matrix $\hat{A}\in \mathbb{R}^{n\times n}$ has an LU factorization if the determinant is different from zero. If the LU factorization exists and $\hat{A}$ is non-singular, then the LU factorization is unique and the determinant is given by

 
$$\vert\hat{A}\vert = c_{11}c_{22}\dots c_{nn}.$$

The expectation value of the kinetic energy expressed in atomic units for electron $i$ is

 
$$\langle \hat{K}_i \rangle = -\frac{1}{2}\frac{\langle\Psi|\nabla_{i}^2|\Psi \rangle}{\langle\Psi|\Psi \rangle},$$

 
$$\begin{equation} \tag{43} K_i = -\frac{1}{2}\frac{\nabla_{i}^{2} \Psi}{\Psi}. \end{equation}$$

 
$$\begin{align} \frac{\nabla^2 \Psi}{\Psi} & = \frac{\nabla^2 ({\Psi_{D} \, \Psi_C})}{\Psi_{D} \, \Psi_C} = \frac{\nabla \cdot [\nabla {(\Psi_{D} \, \Psi_C)}]}{\Psi_{D} \, \Psi_C} = \frac{\nabla \cdot [ \Psi_C \nabla \Psi_{D} + \Psi_{D} \nabla \Psi_C]}{\Psi_{D} \, \Psi_C}\nonumber\\ & = \frac{\nabla \Psi_C \cdot \nabla \Psi_{D} + \Psi_C \nabla^2 \Psi_{D} + \nabla \Psi_{D} \cdot \nabla \Psi_C + \Psi_{D} \nabla^2 \Psi_C}{\Psi_{D} \, \Psi_C}\nonumber\\ \tag{44} \end{align}$$

 
$$\begin{align} \frac{\nabla^2 \Psi}{\Psi} & = \frac{\nabla^2 \Psi_{D}}{\Psi_{D}} + \frac{\nabla^2 \Psi_C}{ \Psi_C} + 2 \frac{\nabla \Psi_{D}}{\Psi_{D}}\cdot\frac{\nabla \Psi_C}{ \Psi_C} \tag{45} \end{align}$$

The second derivative of the Jastrow factor divided by the Jastrow factor (the way it enters the kinetic energy) is

 
$$\left[\frac{\nabla^2 \Psi_C}{\Psi_C}\right]_x =\ 2\sum_{k=1}^{N} \sum_{i=1}^{k-1}\frac{\partial^2 g_{ik}}{\partial x_k^2}\ +\ \sum_{k=1}^N \left( \sum_{i=1}^{k-1}\frac{\partial g_{ik}}{\partial x_k} - \sum_{i=k+1}^{N}\frac{\partial g_{ki}}{\partial x_i} \right)^2$$

But we have a simple form for the function, namely

 
$$\Psi_{C}=\prod_{i < j}\exp{f(r_{ij})}= \exp{\left\{\sum_{i < j}\frac{ar_{ij}}{1+\beta r_{ij}}\right\}},$$

 
and it is easy to see that for particle $k$ we have

 
$$\frac{\nabla^2_k \Psi_C}{\Psi_C }= \sum_{ij\ne k}\frac{(\mathbf{r}_k-\mathbf{r}_i)(\mathbf{r}_k-\mathbf{r}_j)}{r_{ki}r_{kj}}f'(r_{ki})f'(r_{kj})+ \sum_{j\ne k}\left( f''(r_{kj})+\frac{2}{r_{kj}}f'(r_{kj})\right)$$

Using

 
$$f(r_{ij})= \frac{ar_{ij}}{1+\beta r_{ij}},$$

 
and $g'(r_{kj})=dg(r_{kj})/dr_{kj}$ and $g''(r_{kj})=d^2g(r_{kj})/dr_{kj}^2$ we find that for particle $k$ we have

 
$$\frac{\nabla^2_k \Psi_C}{\Psi_C }= \sum_{ij\ne k}\frac{(\mathbf{r}_k-\mathbf{r}_i)(\mathbf{r}_k-\mathbf{r}_j)}{r_{ki}r_{kj}}\frac{a}{(1+\beta r_{ki})^2} \frac{a}{(1+\beta r_{kj})^2}+ \sum_{j\ne k}\left(\frac{2a}{r_{kj}(1+\beta r_{kj})^2}-\frac{2a\beta}{(1+\beta r_{kj})^3}\right)$$

The gradient and Laplacian can be calculated as follows:

 
$$\frac{\mathbf{\nabla}_i\vert\hat{D}(\mathbf{r})\vert} {\vert\hat{D}(\mathbf{r})\vert} = \sum_{j=1}^N \vec\nabla_i d_{ij}(\mathbf{r})\, d_{ji}^{-1}(\mathbf{r}) = \sum_{j=1}^N \vec\nabla_i \phi_j(\mathbf{r}_i)\, d_{ji}^{-1}(\mathbf{r})$$

 
and

 
$$\frac{\nabla^2_i\vert\hat{D}(\mathbf{r})\vert} {\vert\hat{D}(\mathbf{r})\vert} = \sum_{j=1}^N \nabla^2_i d_{ij}(\mathbf{r})\, d_{ji}^{-1}(\mathbf{r}) = \sum_{j=1}^N \nabla^2_i \phi_j(\mathbf{r}_i)\, d_{ji}^{-1}(\mathbf{r})$$

The gradient for the determinant is

 
$$\frac{\mathbf{\nabla}_i\vert\hat{D}(\mathbf{r})\vert} {\vert\hat{D}(\mathbf{r})\vert} = \sum_{j=1}^N \mathbf{\nabla}_i d_{ij}(\mathbf{r})\, d_{ji}^{-1}(\mathbf{r}) = \sum_{j=1}^N \mathbf{\nabla}_i \phi_j(\mathbf{r}_i)\, d_{ji}^{-1}(\mathbf{r}).$$

We have

 
$$\Psi_C=\prod_{i < j}g(r_{ij})= \exp{\left\{\sum_{i < j}\frac{ar_{ij}}{1+\beta r_{ij}}\right\}},$$

 
the gradient needed for the quantum force and local energy is easy to compute. We get for particle $k$

 
$$\frac{ \nabla_k \Psi_C}{ \Psi_C }= \sum_{j\ne k}\frac{\mathbf{r}_{kj}}{r_{kj}}\frac{a}{(1+\beta r_{kj})^2},$$

 
which is rather easy to code. Remember to sum over all particles when you compute the local energy.

We need to compute the ratio between wave functions, in particular for the Slater determinants.

 
$$R =\sum_{j=1}^N d_{ij}(\mathbf{r}^{\mathrm{new}})\, d_{ji}^{-1}(\mathbf{r}^{\mathrm{old}}) = \sum_{j=1}^N \phi_j(\mathbf{r}_i^{\mathrm{new}})\, d_{ji}^{-1}(\mathbf{r}^{\mathrm{old}})$$

 
What this means is that in order to get the ratio when only the i-th particle has been moved, we only need to calculate the dot product of the vector $\left(\phi_1(\mathbf{r}_i^\mathrm{new}),\,\dots,\, \phi_N(\mathbf{r}_i^\mathrm{new})\right)$ of single particle wave functions evaluated at this new position with the i-th column of the inverse matrix $\hat{D}^{-1}$ evaluated at the original position. Such an operation has a time scaling of $O(N)$. The only extra thing we need to do is to maintain the inverse matrix $\hat{D}^{-1}(\mathbf{x}^{\mathrm{old}})$.