qmc-lttc/QMC.org

Quantum Monte Carlo

• Introduction
  • Energy and local energy
• Numerical evaluation of the energy of the hydrogen atom
  • Local energy
    • Exercise 1
      • Solution
    • Exercise 2
      • Solution
    • Exercise 3
      • Solution
    • Exercise 4
      • Solution
    • Exercise 5
      • Solution
  • Plot of the local energy along the $x$ axis
    • Exercise
      • Solution
  • Numerical estimation of the energy
    • Exercise
      • Solution
  • Variance of the local energy
    • Exercise (optional)
      • Solution
    • Exercise
      • Solution
• Variational Monte Carlo
  • Computation of the statistical error
    • Exercise
      • Solution
  • Uniform sampling in the box
    • Exercise
      • Solution
  • Metropolis sampling with $\Psi^2$
    • Optimal step size
    • Exercise
      • Solution
  • Generalized Metropolis algorithm
    • Gaussian random number generator
    • Exercise 1
      • Solution
    • Exercise 2
      • Solution
• Diffusion Monte Carlo
  • Schrödinger equation in imaginary time
  • Relation to diffusion
  • Importance sampling
    • Appendix : Details of the Derivation
  • Pure Diffusion Monte Carlo
  • Hydrogen atom
    • Exercise
      • Solution
• Project
• Acknowledgments

Introduction

This website contains the QMC tutorial of the 2021 LTTC winter school Tutorials in Theoretical Chemistry.

We propose different exercises to understand quantum Monte Carlo (QMC) methods. In the first section, we start with the computation of the energy of a hydrogen atom using numerical integration. The goal of this section is to familarize yourself with the concept of local energy. Then, we introduce the variational Monte Carlo (VMC) method which computes a statistical estimate of the expectation value of the energy associated with a given wave function, and apply this approach to the hydrogen atom. Finally, we present the diffusion Monte Carlo (DMC) method which we use here to estimate the exact energy of the hydrogen atom and of the H_2 molecule, starting from an approximate wave function.

Code examples will be given in Python3 and Fortran. You can use whatever language you prefer to write the programs.

We consider the stationary solution of the Schrödinger equation, so the wave functions considered here are real: for an $N$ electron system where the electrons move in the 3-dimensional space, $\Psi : \mathbb{R}^{3N} \rightarrow \mathbb{R}$. In addition, $\Psi$ is defined everywhere, continuous, and infinitely differentiable.

All the quantities are expressed in atomic units (energies, coordinates, etc).

Energy and local energy

For a given system with Hamiltonian $\hat{H}$ and wave function $\Psi$, we define the local energy as

$$ E_L(\mathbf{r}) = \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}, $$

where $\mathbf{r}$ denotes the 3N-dimensional electronic coordinates.

The electronic energy of a system, $E$, can be rewritten in terms of the local energy $E_L(\mathbf{r})$ as

\begin{eqnarray*} E & = & \frac{\langle \Psi| \hat{H} | \Psi\rangle}{\langle \Psi |\Psi \rangle} = \frac{\int \Psi(\mathbf{r})\, \hat{H} \Psi(\mathbf{r})\, d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}} \\ & = & \frac{\int |\Psi(\mathbf{r})|^2\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}} = \frac{\int |\Psi(\mathbf{r})|^2\, E_L(\mathbf{r})\,d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}} \end{eqnarray*}

For few dimensions, one can easily compute $E$ by evaluating the integrals on a grid but, for a high number of dimensions, one can resort to Monte Carlo techniques to compute $E$.

To this aim, recall that the probabilistic expected value of an arbitrary function $f(x)$ with respect to a probability density function $P(x)$ is given by

$$ \langle f \rangle_P = \int_{-\infty}^\infty P(x)\, f(x)\,dx, $$

where a probability density function $P(x)$ is non-negative and integrates to one:

$$ \int_{-\infty}^\infty P(x)\,dx = 1. $$

Similarly, we can view the the energy of a system, $E$, as the expected value of the local energy with respect to a probability density $P(\mathbf{r})$ defined in 3$N$ dimensions:

$$ E = \int E_L(\mathbf{r}) P(\mathbf{r})\,d\mathbf{r} \equiv \langle E_L \rangle_{P}\,, $$

where the probability density is given by the square of the wave function:

$$ P(\mathbf{r}) = \frac{|\Psi(\mathbf{r})|^2}{\int |\Psi(\mathbf{r})|^2 d\mathbf{r}}\,. $$

If we can sample $N_{\rm MC}$ configurations $\{\mathbf{r}\}$ distributed as $P$, we can estimate $E$ as the average of the local energy computed over these configurations:

$$ E \approx \frac{1}{N_{\rm MC}} \sum_{i=1}^{N_{\rm MC}} E_L(\mathbf{r}_i) \,. $$

Numerical evaluation of the energy of the hydrogen atom

In this section, we consider the hydrogen atom with the following wave function:

$$ \Psi(\mathbf{r}) = \exp(-a |\mathbf{r}|) $$

We will first verify that, for a particular value of $a$, $\Psi$ is an eigenfunction of the Hamiltonian

$$ \hat{H} = \hat{T} + \hat{V} = - \frac{1}{2} \Delta - \frac{1}{|\mathbf{r}|} $$

To do that, we will compute the local energy and check whether it is constant.

Local energy

You will now program all quantities needed to compute the local energy of the H atom for the given wave function.

Write all the functions of this section in a single file : hydrogen.py if you use Python, or hydrogen.f90 is you use Fortran.

• When computing a square root in $\mathbb{R}$, always make sure that the argument of the square root is non-negative.
• When you divide, always make sure that you will not divide by zero

If a floating-point exception can occur, you should make a test to catch the error.

Exercise 1

Write a function which computes the potential at $\mathbf{r}$. The function accepts a 3-dimensional vector r as input argument and returns the potential.

$\mathbf{r}=\left( \begin{array}{c} x \\ y\\ z\end{array} \right)$, so $$ V(\mathbf{r}) = -\frac{1}{\sqrt{x^2 + y^2 + z^2}} $$

Python

#!/usr/bin/env python3
import numpy as np

def potential(r):
# TODO

Fortran

double precision function potential(r)
implicit none
double precision, intent(in) :: r(3)

! TODO

end function potential

Solution   solution

Python

#!/usr/bin/env python3
import numpy as np

def potential(r):
distance = np.sqrt(np.dot(r,r))
assert (distance > 0)
return -1. / distance

Fortran

double precision function potential(r)
implicit none
double precision, intent(in) :: r(3)

double precision             :: distance

distance = dsqrt( r(1)*r(1) + r(2)*r(2) + r(3)*r(3) )

if (distance > 0.d0) then
potential = -1.d0 / distance
else
stop 'potential at r=0.d0 diverges'
end if

end function potential

Exercise 2

Write a function which computes the wave function at $\mathbf{r}$. The function accepts a scalar a and a 3-dimensional vector r as input arguments, and returns a scalar.

Python

def psi(a, r):
# TODO

Fortran

double precision function psi(a, r)
implicit none
double precision, intent(in) :: a, r(3)

! TODO

end function psi

Solution   solution

Python

def psi(a, r):
return np.exp(-a*np.sqrt(np.dot(r,r)))

Fortran

double precision function psi(a, r)
implicit none
double precision, intent(in) :: a, r(3)

psi = dexp(-a * dsqrt( r(1)*r(1) + r(2)*r(2) + r(3)*r(3) ))
end function psi

Exercise 3

Write a function which computes the local kinetic energy at $\mathbf{r}$. The function accepts a and r as input arguments and returns the local kinetic energy.

The local kinetic energy is defined as $$T_L(\mathbf{r}) = -\frac{1}{2}\frac{\Delta \Psi(\mathbf{r})}{\Psi(\mathbf{r})}.$$

We differentiate $\Psi$ with respect to $x$:

\[ \Psi(\mathbf{r}) = \exp(-a\,|\mathbf{r}|) \] \[\frac{\partial \Psi}{\partial x} = \frac{\partial \Psi}{\partial |\mathbf{r}|} \frac{\partial |\mathbf{r}|}{\partial x} = - \frac{a\,x}{|\mathbf{r}|} \Psi(\mathbf{r}) \]

and we differentiate a second time:

$$ \frac{\partial^2 \Psi}{\partial x^2} = \left( \frac{a^2\,x^2}{|\mathbf{r}|^2} - \frac{a(y^2+z^2)}{|\mathbf{r}|^{3}} \right) \Psi(\mathbf{r}). $$

The Laplacian operator $\Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$ applied to the wave function gives:

$$ \Delta \Psi (\mathbf{r}) = \left(a^2 - \frac{2a}{\mathbf{|r|}} \right) \Psi(\mathbf{r})\,. $$

Therefore, the local kinetic energy is $$ T_L (\mathbf{r}) = -\frac{1}{2}\left(a^2 - \frac{2a}{\mathbf{|r|}} \right) $$

Python

def kinetic(a,r):
# TODO

Fortran

double precision function kinetic(a,r)
implicit none
double precision, intent(in) :: a, r(3)

! TODO

end function kinetic

Solution   solution

Python

def kinetic(a,r):
distance = np.sqrt(np.dot(r,r))
assert (distance > 0.)

return a * (1./distance - 0.5 * a)

Fortran

double precision function kinetic(a,r)
implicit none
double precision, intent(in) :: a, r(3)

double precision             :: distance

distance = dsqrt( r(1)*r(1) + r(2)*r(2) + r(3)*r(3) ) 

if (distance > 0.d0) then

kinetic =  a * (1.d0 / distance - 0.5d0 * a)

else
stop 'kinetic energy diverges at r=0'
end if

end function kinetic

Exercise 4

Write a function which computes the local energy at $\mathbf{r}$, using the previously defined functions. The function accepts a and r as input arguments and returns the local kinetic energy.

$$ E_L(\mathbf{r}) = -\frac{1}{2} \frac{\Delta \Psi}{\Psi} (\mathbf{r}) + V(\mathbf{r}) $$

Python

def e_loc(a,r):
#TODO

Fortran

When you call a function in Fortran, you need to declare its return type. You might by accident choose a function name which is the same as an internal function of Fortran. So it is recommended to always use the keyword external to make sure the function you are calling is yours.

double precision function e_loc(a,r)
implicit none
double precision, intent(in) :: a, r(3)

double precision, external :: kinetic
double precision, external :: potential

! TODO

end function e_loc

Solution   solution

Python

def e_loc(a,r):
return kinetic(a,r) + potential(r)

Fortran

double precision function e_loc(a,r)
implicit none
double precision, intent(in) :: a, r(3)

double precision, external :: kinetic
double precision, external :: potential

e_loc = kinetic(a,r) + potential(r)

end function e_loc

Exercise 5

Find the theoretical value of $a$ for which $\Psi$ is an eigenfunction of $\hat{H}$.

Solution   solution

\begin{eqnarray*} E &=& \frac{\hat{H} \Psi}{\Psi} = - \frac{1}{2} \frac{\Delta \Psi}{\Psi} - \frac{1}{|\mathbf{r}|} \\ &=& -\frac{1}{2}\left(a^2 - \frac{2a}{\mathbf{|r|}} \right) - \frac{1}{|\mathbf{r}|} \\ &=& -\frac{1}{2} a^2 + \frac{a-1}{\mathbf{|r|}} \end{eqnarray*}

$a=1$ cancels the $1/|r|$ term, and makes the energy constant and equal to -0.5 atomic units.

Plot of the local energy along the $x$ axis

The program you will write in this section will be written in another file (plot_hydrogen.py or plot_hydrogen.f90 for example). It will use the functions previously defined.

In Python, you should put at the beginning of the file

#!/usr/bin/env python3

from hydrogen import e_loc

to be able to use the e_loc function of the hydrogen.py file.

In Fortran, you will need to compile all the source files together:

gfortran hydrogen.f90 plot_hydrogen.f90 -o plot_hydrogen

Exercise

For multiple values of $a$ (0.1, 0.2, 0.5, 1., 1.5, 2.), plot the local energy along the $x$ axis. In Python, you can use matplotlib for example. In Fortran, it is convenient to write in a text file the values of $x$ and $E_L(\mathbf{r})$ for each point, and use Gnuplot to plot the files. With Gnuplot, you will need 2 blank lines to separate the data corresponding to different values of $a$.

The potential and the kinetic energy both diverge at $r=0$, so we choose a grid which does not contain the origin to avoid numerical issues.

Python

#!/usr/bin/env python3

import numpy as np
import matplotlib.pyplot as plt

from hydrogen import e_loc

x=np.linspace(-5,5)
plt.figure(figsize=(10,5))

# TODO

plt.tight_layout()
plt.legend()
plt.savefig("plot_py.png")

Fortran

program plot
implicit none
double precision, external :: e_loc

double precision :: x(50), dx
integer :: i, j

dx = 10.d0/(size(x)-1)
do i=1,size(x)
x(i) = -5.d0 + (i-1)*dx
end do

! TODO

end program plot

To compile and run:

gfortran hydrogen.f90 plot_hydrogen.f90 -o plot_hydrogen
./plot_hydrogen > data

To plot the data using Gnuplot:

set grid
set xrange [-5:5]
set yrange [-2:1]
plot './data' index 0 using 1:2 with lines title 'a=0.1', \
'./data' index 1 using 1:2 with lines title 'a=0.2', \
'./data' index 2 using 1:2 with lines title 'a=0.5', \
'./data' index 3 using 1:2 with lines title 'a=1.0', \
'./data' index 4 using 1:2 with lines title 'a=1.5', \
'./data' index 5 using 1:2 with lines title 'a=2.0'

Solution   solution

Python

#!/usr/bin/env python3

import numpy as np
import matplotlib.pyplot as plt

from hydrogen import e_loc

x=np.linspace(-5,5)
plt.figure(figsize=(10,5))

for a in [0.1, 0.2, 0.5, 1., 1.5, 2.]:
y=np.array([ e_loc(a, np.array([t,0.,0.]) ) for t in x])
plt.plot(x,y,label=f"a={a}")

plt.tight_layout()
plt.legend()
plt.savefig("plot_py.png")

Fortran

program plot
implicit none
double precision, external :: e_loc

double precision :: x(50), energy, dx, r(3), a(6)
integer :: i, j

a = (/ 0.1d0, 0.2d0, 0.5d0, 1.d0, 1.5d0, 2.d0 /)

dx = 10.d0/(size(x)-1)
do i=1,size(x)
x(i) = -5.d0 + (i-1)*dx
end do

r(:) = 0.d0

do j=1,size(a)
print *, '# a=', a(j)
do i=1,size(x)
   r(1) = x(i)
   energy = e_loc( a(j), r )
   print *, x(i), energy
end do
print *, ''
print *, ''
end do

end program plot

Numerical estimation of the energy

If the space is discretized in small volume elements $\mathbf{r}_i$ of size $\delta \mathbf{r}$, the expression of $\langle E_L \rangle_{\Psi^2}$ becomes a weighted average of the local energy, where the weights are the values of the wave function square at $\mathbf{r}_i$ multiplied by the volume element:

$$ \langle E \rangle_{\Psi^2} \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\; w_i = \left|\Psi(\mathbf{r}_i)\right|^2 \delta \mathbf{r} $$

The energy is biased because:

• The volume elements are not infinitely small (discretization error)
• The energy is evaluated only inside the box (incompleteness of the space)

Exercise

Compute a numerical estimate of the energy using a grid of $50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.

Python

#!/usr/bin/env python3

import numpy as np
from hydrogen import e_loc, psi

interval = np.linspace(-5,5,num=50)
delta = (interval[1]-interval[0])**3

r = np.array([0.,0.,0.])

for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
# TODO
print(f"a = {a} \t E = {E}")

Fortran

program energy_hydrogen
implicit none
double precision, external :: e_loc, psi
double precision :: x(50), w, delta, energy, dx, r(3), a(6), norm
integer :: i, k, l, j

a = (/ 0.1d0, 0.2d0, 0.5d0, 1.d0, 1.5d0, 2.d0 /)

dx = 10.d0/(size(x)-1)
do i=1,size(x)
x(i) = -5.d0 + (i-1)*dx
end do

do j=1,size(a)

! TODO

print *, 'a = ', a(j), '    E = ', energy
end do

end program energy_hydrogen

To compile the Fortran and run it:

gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
./energy_hydrogen

Solution   solution

Python

#!/usr/bin/env python3

import numpy as np
from hydrogen import e_loc, psi

interval = np.linspace(-5,5,num=50)
delta = (interval[1]-interval[0])**3

r = np.array([0.,0.,0.])

for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
E    = 0.
norm = 0.

for x in interval:
   r[0] = x
   for y in interval:
       r[1] = y
       for z in interval:
           r[2] = z

           w = psi(a,r)
           w = w * w * delta

           E    += w * e_loc(a,r)
           norm += w 

E = E / norm
print(f"a = {a} \t E = {E}")

a = 0.1 	 E = -0.24518438948809218
a = 0.2 	 E = -0.26966057967803525
a = 0.5 	 E = -0.3856357612517407
a = 0.9 	 E = -0.49435709786716214
a = 1.0 	 E = -0.5
a = 1.5 	 E = -0.39242967082602226
a = 2.0 	 E = -0.08086980667844901

Fortran

program energy_hydrogen
implicit none
double precision, external :: e_loc, psi
double precision :: x(50), w, delta, energy, dx, r(3), a(6), norm
integer          :: i, k, l, j

a = (/ 0.1d0, 0.2d0, 0.5d0, 1.d0, 1.5d0, 2.d0 /)

dx = 10.d0/(size(x)-1)
do i=1,size(x)
x(i) = -5.d0 + (i-1)*dx
end do

delta = dx**3

r(:) = 0.d0

do j=1,size(a)
energy = 0.d0
norm   = 0.d0

do i=1,size(x)
   r(1) = x(i)

   do k=1,size(x)
      r(2) = x(k)

      do l=1,size(x)
         r(3) = x(l)

         w = psi(a(j),r)
         w = w * w * delta

         energy = energy + w * e_loc(a(j), r)
         norm   = norm   + w 
      end do

   end do

end do

energy = energy / norm
print *, 'a = ', a(j), '    E = ', energy
end do

end program energy_hydrogen

 a =   0.10000000000000001          E =  -0.24518438948809140     
 a =   0.20000000000000001          E =  -0.26966057967803236     
 a =   0.50000000000000000          E =  -0.38563576125173815     
 a =    1.0000000000000000          E =  -0.50000000000000000     
 a =    1.5000000000000000          E =  -0.39242967082602065     
 a =    2.0000000000000000          E =   -8.0869806678448772E-002

Variance of the local energy

The variance of the local energy is a functional of $\Psi$ which measures the magnitude of the fluctuations of the local energy associated with $\Psi$ around its average:

$$ \sigma^2(E_L) = \frac{\int |\Psi(\mathbf{r})|^2\, \left[ E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}} $$ which can be simplified as

$$ \sigma^2(E_L) = \langle E_L^2 \rangle_{\Psi^2} - \langle E_L \rangle_{\Psi^2}^2.$$

If the local energy is constant (i.e. $\Psi$ is an eigenfunction of $\hat{H}$) the variance is zero, so the variance of the local energy can be used as a measure of the quality of a wave function.

Exercise (optional)

Prove that : $$\langle \left( E - \langle E \rangle_{\Psi^2} \right)^2\rangle_{\Psi^2} = \langle E^2 \rangle_{\Psi^2} - \langle E \rangle_{\Psi^2}^2 $$

Solution   solution

$\bar{E} = \langle E \rangle$ is a constant, so $\langle \bar{E} \rangle = \bar{E}$ .

\begin{eqnarray*} \langle (E - \bar{E})^2 \rangle & = & \langle E^2 - 2 E \bar{E} + \bar{E}^2 \rangle \\ &=& \langle E^2 \rangle - 2 \langle E \bar{E} \rangle + \langle \bar{E}^2 \rangle \\ &=& \langle E^2 \rangle - 2 \langle E \rangle \bar{E} + \bar{E}^2 \\ &=& \langle E^2 \rangle - 2 \bar{E}^2 + \bar{E}^2 \\ &=& \langle E^2 \rangle - \bar{E}^2 \\ &=& \langle E^2 \rangle - \langle E \rangle^2 \\ \end{eqnarray*}

Exercise

Add the calculation of the variance to the previous code, and compute a numerical estimate of the variance of the local energy using a grid of $50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$ for different values of $a$.

Python

#!/usr/bin/env python3

import numpy as np from hydrogen import e_loc, psi

interval = np.linspace(-5,5,num=50)

delta = (interval[1]-interval[0])**3

r = np.array([0.,0.,0.])

for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:

# TODO

print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}")

Fortran

program variance_hydrogen
implicit none

double precision :: x(50), w, delta, energy, energy2
double precision :: dx, r(3), a(6), norm, e_tmp, s2
integer          :: i, k, l, j

double precision, external :: e_loc, psi

a = (/ 0.1d0, 0.2d0, 0.5d0, 1.d0, 1.5d0, 2.d0 /)

dx = 10.d0/(size(x)-1)
do i=1,size(x)
x(i) = -5.d0 + (i-1)*dx
end do

do j=1,size(a)

! TODO

print *, 'a = ', a(j), '    E = ', energy
end do

end program variance_hydrogen

To compile and run:

gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
./variance_hydrogen

Solution   solution

Python

#!/usr/bin/env python3

import numpy as np
from hydrogen import e_loc, psi

interval = np.linspace(-5,5,num=50)

delta = (interval[1]-interval[0])**3

r = np.array([0.,0.,0.])

for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
E    = 0.
E2   = 0.
norm = 0.

for x in interval:
   r[0] = x

   for y in interval:
       r[1] = y

       for z in interval:
           r[2] = z

           w = psi(a,r)
           w = w * w * delta

           e_tmp = e_loc(a,r)
           E    += w * e_tmp
           E2   += w * e_tmp * e_tmp
           norm += w 

E  = E  / norm
E2 = E2 / norm

s2 = E2 - E**2
print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}")

a = 0.1 	 E = -0.24518439 	 \sigma^2 = 0.02696522
a = 0.2 	 E = -0.26966058 	 \sigma^2 = 0.03719707
a = 0.5 	 E = -0.38563576 	 \sigma^2 = 0.05318597
a = 0.9 	 E = -0.49435710 	 \sigma^2 = 0.00577812
a = 1.0 	 E = -0.50000000 	 \sigma^2 = 0.00000000
a = 1.5 	 E = -0.39242967 	 \sigma^2 = 0.31449671
a = 2.0 	 E = -0.08086981 	 \sigma^2 = 1.80688143

Fortran

program variance_hydrogen
implicit none

double precision :: x(50), w, delta, energy, energy2
double precision :: dx, r(3), a(6), norm, e_tmp, s2
integer          :: i, k, l, j

double precision, external :: e_loc, psi

a = (/ 0.1d0, 0.2d0, 0.5d0, 1.d0, 1.5d0, 2.d0 /)

dx = 10.d0/(size(x)-1)
do i=1,size(x)
 x(i) = -5.d0 + (i-1)*dx
end do

delta = dx**3

r(:) = 0.d0

do j=1,size(a)
 energy  = 0.d0
 energy2 = 0.d0
 norm    = 0.d0

 do i=1,size(x)
    r(1) = x(i)

    do k=1,size(x)
       r(2) = x(k)

       do l=1,size(x)
          r(3) = x(l)

          w = psi(a(j),r)
          w = w * w * delta

          e_tmp = e_loc(a(j), r)

          energy  = energy  + w * e_tmp
          energy2 = energy2 + w * e_tmp * e_tmp
          norm   = norm     + w 
       end do

    end do

 end do

 energy  = energy  / norm
 energy2 = energy2 / norm

 s2 = energy2 - energy*energy

 print *, 'a = ', a(j), ' E = ', energy, ' s2 = ', s2
end do

end program variance_hydrogen

 a =   0.10000000000000001      E =  -0.24518438948809140       s2 =    2.6965218719722767E-002
 a =   0.20000000000000001      E =  -0.26966057967803236       s2 =    3.7197072370201284E-002
 a =   0.50000000000000000      E =  -0.38563576125173815       s2 =    5.3185967578480653E-002
 a =    1.0000000000000000      E =  -0.50000000000000000       s2 =    0.0000000000000000     
 a =    1.5000000000000000      E =  -0.39242967082602065       s2 =   0.31449670909172917     
 a =    2.0000000000000000      E =   -8.0869806678448772E-002  s2 =    1.8068814270846534     

Variational Monte Carlo

Numerical integration with deterministic methods is very efficient in low dimensions. When the number of dimensions becomes large, instead of computing the average energy as a numerical integration on a grid, it is usually more efficient to use Monte Carlo sampling.

Moreover, Monte Carlo sampling will allow us to remove the bias due to the discretization of space, and compute a statistical confidence interval.

Computation of the statistical error

To compute the statistical error, you need to perform $M$ independent Monte Carlo calculations. You will obtain $M$ different estimates of the energy, which are expected to have a Gaussian distribution for large $M$, according to the Central Limit Theorem.

The estimate of the energy is

$$ E = \frac{1}{M} \sum_{i=1}^M E_i $$

The variance of the average energies can be computed as

$$ \sigma^2 = \frac{1}{M-1} \sum_{i=1}^{M} (E_i - E)^2 $$

And the confidence interval is given by

$$ E \pm \delta E, \text{ where } \delta E = \frac{\sigma}{\sqrt{M}} $$

Exercise

Write a function returning the average and statistical error of an input array.

Python

#!/usr/bin/env python3

from math import sqrt
def ave_error(arr):
#TODO
return (average, error)

Fortran

subroutine ave_error(x,n,ave,err)
implicit none
integer, intent(in)           :: n 
double precision, intent(in)  :: x(n) 
double precision, intent(out) :: ave, err

! TODO

end subroutine ave_error

Solution   solution

Python

#!/usr/bin/env python3

from math import sqrt
def ave_error(arr):
M = len(arr)
assert(M>0)

if M == 1:
   average = arr[0]
   error   = 0.

else:
   average = sum(arr)/M
   variance = 1./(M-1) * sum( [ (x - average)**2 for x in arr ] )
   error = sqrt(variance/M)

return (average, error)

Fortran

subroutine ave_error(x,n,ave,err)
implicit none

integer, intent(in)           :: n 
double precision, intent(in)  :: x(n) 
double precision, intent(out) :: ave, err

double precision              :: variance

if (n < 1) then
stop 'n<1 in ave_error'

else if (n == 1) then
ave = x(1)
err = 0.d0

else
ave      = sum(x(:)) / dble(n)

variance = sum((x(:) - ave)**2) / dble(n-1)
err      = dsqrt(variance/dble(n))

endif
end subroutine ave_error

Uniform sampling in the box

We will now perform our first Monte Carlo calculation to compute the energy of the hydrogen atom.

Consider again the expression of the energy

\begin{eqnarray*} E & = & \frac{\int E_L(\mathbf{r})|\Psi(\mathbf{r})|^2\,d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}}\,. \end{eqnarray*}

Clearly, the square of the wave function is a good choice of probability density to sample but we will start with something simpler and rewrite the energy as

\begin{eqnarray*} E & = & \frac{\int E_L(\mathbf{r})\frac{|\Psi(\mathbf{r})|^2}{P(\mathbf{r})}P(\mathbf{r})\, \,d\mathbf{r}}{\int \frac{|\Psi(\mathbf{r})|^2 }{P(\mathbf{r})}P(\mathbf{r})d\mathbf{r}}\,. \end{eqnarray*}

Here, we will sample a uniform probability $P(\mathbf{r})$ in a cube of volume $L^3$ centered at the origin:

$$ P(\mathbf{r}) = \frac{1}{L^3}\,, $$

and zero outside the cube.

One Monte Carlo run will consist of $N_{\rm MC}$ Monte Carlo iterations. At every Monte Carlo iteration:

• Draw a random point $\mathbf{r}_i$ in the box $(-5,-5,-5) \le (x,y,z) \le (5,5,5)$
• Compute $|\Psi(\mathbf{r}_i)|^2$ and accumulate the result in a variable normalization
• Compute $|\Psi(\mathbf{r}_i)|^2 \times E_L(\mathbf{r}_i)$, and accumulate the result in a variable energy

Once all the iterations have been computed, the run returns the average energy $\bar{E}_k$ over the $N_{\rm MC}$ iterations of the run.

To compute the statistical error, perform $M$ independent runs. The final estimate of the energy will be the average over the $\bar{E}_k$, and the variance of the $\bar{E}_k$ will be used to compute the statistical error.

Exercise

Parameterize the wave function with $a=1.2$. Perform 30 independent Monte Carlo runs, each with 100 000 Monte Carlo steps. Store the final energies of each run and use this array to compute the average energy and the associated error bar.

Python

To draw a uniform random number in Python, you can use the random.uniform function of Numpy.

#!/usr/bin/env python3

from hydrogen  import *
from qmc_stats import *

def MonteCarlo(a, nmax):
# TODO

a    = 1.2
nmax = 100000

#TODO

print(f"E = {E} +/- {deltaE}")

Fortran

To draw a uniform random number in Fortran, you can use the RANDOM_NUMBER subroutine.

When running Monte Carlo calculations, the number of steps is usually very large. We expect nmax to be possibly larger than 2 billion, so we use 8-byte integers (integer*8) to represent it, as well as the index of the current step.

subroutine uniform_montecarlo(a,nmax,energy)
implicit none
double precision, intent(in)  :: a
integer*8       , intent(in)  :: nmax 
double precision, intent(out) :: energy

integer*8        :: istep
double precision :: norm, r(3), w

double precision, external :: e_loc, psi

! TODO
end subroutine uniform_montecarlo

program qmc
implicit none
double precision, parameter :: a = 1.2d0
integer*8       , parameter :: nmax = 100000
integer         , parameter :: nruns = 30

integer          :: irun
double precision :: X(nruns)
double precision :: ave, err

!TODO

print *, 'E = ', ave, '+/-', err

end program qmc

gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
./qmc_uniform

Solution   solution

Python

#!/usr/bin/env python3

from hydrogen  import *
from qmc_stats import *

def MonteCarlo(a, nmax):
energy = 0.
normalization = 0.

for istep in range(nmax):
     r = np.random.uniform(-5., 5., (3))

     w = psi(a,r)
     w = w*w

     energy        += w * e_loc(a,r)
     normalization += w

return energy / normalization

a    = 1.2
nmax = 100000

X = [MonteCarlo(a,nmax) for i in range(30)]
E, deltaE = ave_error(X)

print(f"E = {E} +/- {deltaE}")

E = -0.48363807880008725 +/- 0.002330876047368999

Fortran

subroutine uniform_montecarlo(a,nmax,energy)
implicit none
double precision, intent(in)  :: a
integer*8       , intent(in)  :: nmax 
double precision, intent(out) :: energy

integer*8        :: istep
double precision :: norm, r(3), w

double precision, external :: e_loc, psi

energy = 0.d0
norm   = 0.d0

do istep = 1,nmax

call random_number(r)
r(:) = -5.d0 + 10.d0*r(:)

w = psi(a,r)
w = w*w

energy = energy + w * e_loc(a,r)
norm   = norm   + w

end do

energy = energy / norm

end subroutine uniform_montecarlo

program qmc
implicit none
double precision, parameter :: a     = 1.2d0
integer*8       , parameter :: nmax  = 100000
integer         , parameter :: nruns = 30

integer          :: irun
double precision :: X(nruns)
double precision :: ave, err

do irun=1,nruns
call uniform_montecarlo(a, nmax, X(irun))
enddo

call ave_error(X, nruns, ave, err)

print *, 'E = ', ave, '+/-', err
end program qmc

 E =  -0.48084122147238995      +/-   2.4983775878329355E-003

Metropolis sampling with $\Psi^2$

We will now use the square of the wave function to sample random points distributed with the probability density \[ P(\mathbf{r}) = \frac{|\Psi(\mathbf{r})|^2}{\int |\Psi(\mathbf{r})|^2 d\mathbf{r}}\,. \]

The expression of the average energy is now simplified as the average of the local energies, since the weights are taken care of by the sampling:

$$ E \approx \frac{1}{N_{\rm MC}}\sum_{i=1}^{N_{\rm MC}} E_L(\mathbf{r}_i)\,. $$

To sample a chosen probability density, an efficient method is the Metropolis-Hastings sampling algorithm. Starting from a random initial position $\mathbf{r}_0$, we will realize a random walk:

$$ \mathbf{r}_0 \rightarrow \mathbf{r}_1 \rightarrow \mathbf{r}_2 \ldots \rightarrow \mathbf{r}_{N_{\rm MC}}\,, $$

according to the following algorithm.

At every step, we propose a new move according to a transition probability $T(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1})$ of our choice.

For simplicity, we will move the electron in a 3-dimensional box of side $2\delta L$ centered at the current position of the electron:

$$ \mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta L \, \mathbf{u} $$

where $\delta L$ is a fixed constant, and $\mathbf{u}$ is a uniform random number in a 3-dimensional box $(-1,-1,-1) \le \mathbf{u} \le (1,1,1)$.

After having moved the electron, we add the accept/reject step that guarantees that the distribution of the $\mathbf{r}_n$ is $\Psi^2$. This amounts to accepting the move with probability

$$ A(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1}) = \min\left(1,\frac{T(\mathbf{r}_{n+1}\rightarrow\mathbf{r}_{n}) P(\mathbf{r}_{n+1})}{T(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1})P(\mathbf{r}_{n})}\right)\,, $$

which, for our choice of transition probability, becomes

$$ A(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1}) = \min\left(1,\frac{P(\mathbf{r}_{n+1})}{P(\mathbf{r}_{n})}\right)= \min\left(1,\frac{|\Psi(\mathbf{r}_{n+1})|^2}{|\Psi(\mathbf{r}_{n})|^2}\right)\,. $$

Explain why the transition probability cancels out in the expression of $A$.

Also note that we do not need to compute the norm of the wave function!

The algorithm is summarized as follows:

1. Evaluate the local energy at $\mathbf{r}_n$ and accumulate it
2. Compute a new position $\mathbf{r'} = \mathbf{r}_n + \delta L\, \mathbf{u}$
3. Evaluate $\Psi(\mathbf{r}')$ at the new position
4. Compute the ratio $A = \frac{\left|\Psi(\mathbf{r'})\right|^2}{\left|\Psi(\mathbf{r}_{n})\right|^2}$
5. Draw a uniform random number $v \in [0,1]$
6. if $v \le A$, accept the move : set $\mathbf{r}_{n+1} = \mathbf{r'}$
7. else, reject the move : set $\mathbf{r}_{n+1} = \mathbf{r}_n$

A common error is to remove the rejected samples from the calculation of the average. Don't do it!

All samples should be kept, from both accepted and rejected moves.

Optimal step size

If the box is infinitely small, the ratio will be very close to one and all the steps will be accepted. However, the moves will be very correlated and you will visit the configurational space very slowly.

On the other hand, if you propose too large moves, the number of accepted steps will decrease because the ratios might become small. If the number of accepted steps is close to zero, then the space is not well sampled either.

The size of the move should be adjusted so that it is as large as possible, keeping the number of accepted steps not too small. To achieve that, we define the acceptance rate as the number of accepted steps over the total number of steps. Adjusting the time step such that the acceptance rate is close to 0.5 is a good compromise for the current problem.

Below, we use the symbol $\delta t$ to denote $\delta L$ since we will use the same variable later on to store a time step.

Exercise

Modify the program of the previous section to compute the energy, sampled with $\Psi^2$.

Compute also the acceptance rate, so that you can adapt the time step in order to have an acceptance rate close to 0.5.

Can you observe a reduction in the statistical error?

Python

#!/usr/bin/env python3

from hydrogen  import *
from qmc_stats import *

def MonteCarlo(a,nmax,dt):

# TODO

return energy/nmax, N_accep/nmax


# Run simulation
a    = 1.2
nmax = 100000
dt   = #TODO

X0 = [ MonteCarlo(a,nmax,dt) for i in range(30)]

# Energy
X = [ x for (x, _) in X0 ]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")

# Acceptance rate
X = [ x for (_, x) in X0 ]
A, deltaA = ave_error(X)
print(f"A = {A} +/- {deltaA}")

Fortran

subroutine metropolis_montecarlo(a,nmax,dt,energy,accep)
implicit none
double precision, intent(in)  :: a
integer*8       , intent(in)  :: nmax 
double precision, intent(in)  :: dt 
double precision, intent(out) :: energy
double precision, intent(out) :: accep

integer*8        :: istep
integer*8        :: n_accep
double precision :: r_old(3), r_new(3), psi_old, psi_new
double precision :: v, ratio

double precision, external :: e_loc, psi, gaussian

! TODO

end subroutine metropolis_montecarlo

program qmc
implicit none
double precision, parameter :: a     = 1.2d0
double precision, parameter :: dt    = ! TODO
integer*8       , parameter :: nmax  = 100000
integer         , parameter :: nruns = 30

integer          :: irun
double precision :: X(nruns), Y(nruns)
double precision :: ave, err

do irun=1,nruns
call metropolis_montecarlo(a,nmax,dt,X(irun),Y(irun))
enddo

call ave_error(X,nruns,ave,err)
print *, 'E = ', ave, '+/-', err

call ave_error(Y,nruns,ave,err)
print *, 'A = ', ave, '+/-', err

end program qmc

gfortran hydrogen.f90 qmc_stats.f90 qmc_metropolis.f90 -o qmc_metropolis
./qmc_metropolis

Solution   solution

Python

#!/usr/bin/env python3

from hydrogen  import *
from qmc_stats import *

def MonteCarlo(a,nmax,dt):
energy  = 0.
N_accep = 0

r_old = np.random.uniform(-dt, dt, (3))
psi_old = psi(a,r_old)

for istep in range(nmax):
   energy += e_loc(a,r_old)

   r_new = r_old + np.random.uniform(-dt,dt,(3))
   psi_new = psi(a,r_new)

   ratio = (psi_new / psi_old)**2

   if np.random.uniform() <= ratio:
       N_accep += 1

       r_old   = r_new
       psi_old = psi_new

return energy/nmax, N_accep/nmax

# Run simulation
a    = 1.2
nmax = 100000
dt   = 1.0

X0 = [ MonteCarlo(a,nmax,dt) for i in range(30)]

# Energy
X = [ x for (x, _) in X0 ]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")

# Acceptance rate
X = [ x for (_, x) in X0 ]
A, deltaA = ave_error(X)
print(f"A = {A} +/- {deltaA}")

E = -0.4802595860693983 +/- 0.0005124420418289207
A = 0.5074913333333334 +/- 0.000350889422714878

Fortran

subroutine metropolis_montecarlo(a,nmax,dt,energy,accep)
implicit none
double precision, intent(in)  :: a
integer*8       , intent(in)  :: nmax 
double precision, intent(in)  :: dt
double precision, intent(out) :: energy
double precision, intent(out) :: accep

double precision :: r_old(3), r_new(3), psi_old, psi_new
double precision :: v, ratio
integer*8        :: n_accep
integer*8        :: istep

double precision, external :: e_loc, psi, gaussian

energy  = 0.d0
n_accep = 0_8

call random_number(r_old)
r_old(:) = dt * (2.d0*r_old(:) - 1.d0)
psi_old = psi(a,r_old)

do istep = 1,nmax
energy = energy + e_loc(a,r_old)

call random_number(r_new)
r_new(:) = r_old(:) + dt*(2.d0*r_new(:) - 1.d0)

psi_new = psi(a,r_new)

ratio = (psi_new / psi_old)**2
call random_number(v)

if (v <= ratio) then

   n_accep = n_accep + 1_8

   r_old(:) = r_new(:)
   psi_old = psi_new

endif

end do

energy = energy / dble(nmax)
accep  = dble(n_accep) / dble(nmax)

end subroutine metropolis_montecarlo

program qmc
implicit none
double precision, parameter :: a = 1.2d0
double precision, parameter :: dt = 1.0d0
integer*8       , parameter :: nmax = 100000
integer         , parameter :: nruns = 30

integer          :: irun
double precision :: X(nruns), Y(nruns)
double precision :: ave, err

do irun=1,nruns
call metropolis_montecarlo(a,nmax,dt,X(irun),Y(irun))
enddo

call ave_error(X,nruns,ave,err)
print *, 'E = ', ave, '+/-', err

call ave_error(Y,nruns,ave,err)
print *, 'A = ', ave, '+/-', err

end program qmc

 E =  -0.47948142754168033      +/-   4.8410177863919307E-004
 A =   0.50762633333333318      +/-   3.4601756760043725E-004

Generalized Metropolis algorithm

One can use more efficient numerical schemes to move the electrons by choosing a smarter expression for the transition probability.

The Metropolis acceptance step has to be adapted accordingly to ensure that the detailed balance condition is satisfied. This means that the acceptance probability $A$ is chosen so that it is consistent with the probability of leaving $\mathbf{r}_n$ and the probability of entering $\mathbf{r}_{n+1}$:

\[ A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \min \left( 1, \frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})} {T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})} \right) \] where $T(\mathbf{r}_n \rightarrow \mathbf{r}_{n+1})$ is the probability of transition from $\mathbf{r}_n$ to $\mathbf{r}_{n+1}$.

In the previous example, we were using uniform sampling in a box centered at the current position. Hence, the transition probability was symmetric

\[ T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) = \text{constant}\,, \]

so the expression of $A$ was simplified to the ratios of the squared wave functions.

Now, if instead of drawing uniform random numbers, we choose to draw Gaussian random numbers with zero mean and variance $\delta t$, the transition probability becomes:

\[ T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \frac{1}{(2\pi\,\delta t)^{3/2}} \exp \left[ - \frac{\left( \mathbf{r}_{n+1} - \mathbf{r}_{n} \right)^2}{2\delta t} \right]\,. \]

Furthermore, to sample the density even better, we can "push" the electrons into in the regions of high probability, and "pull" them away from the low-probability regions. This will increase the acceptance ratios and improve the sampling.

To do this, we can use the gradient of the probability density

\[ \frac{\nabla [ \Psi^2 ]}{\Psi^2} = 2 \frac{\nabla \Psi}{\Psi}\,, \]

and add the so-called drift vector, $\frac{\nabla \Psi}{\Psi}$, so that the numerical scheme becomes a drifted diffusion with transition probability:

\[ T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \frac{1}{(2\pi\,\delta t)^{3/2}} \exp \left[ - \frac{\left( \mathbf{r}_{n+1} - \mathbf{r}_{n} - \delta t\frac{\nabla \Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\delta t} \right]\,. \]

The corresponding move is proposed as

\[ \mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi \,, \]

where $\chi$ is a Gaussian random variable with zero mean and variance $\delta t$.

The algorithm of the previous exercise is only slighlty modified as:

1. Evaluate the local energy at $\mathbf{r}_{n}$ and accumulate it
2. Compute a new position $\mathbf{r'} = \mathbf{r}_n + \delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi$
3. Evaluate $\Psi(\mathbf{r}')$ and $\frac{\nabla \Psi(\mathbf{r'})}{\Psi(\mathbf{r'})}$ at the new position
4. Compute the ratio $A = \frac{T(\mathbf{r}' \rightarrow \mathbf{r}_{n}) P(\mathbf{r}')}{T(\mathbf{r}_{n} \rightarrow \mathbf{r}') P(\mathbf{r}_{n})}$
5. Draw a uniform random number $v \in [0,1]$
6. if $v \le A$, accept the move : set $\mathbf{r}_{n+1} = \mathbf{r'}$
7. else, reject the move : set $\mathbf{r}_{n+1} = \mathbf{r}_n$

Gaussian random number generator

To obtain Gaussian-distributed random numbers, you can apply the Box Muller transform to uniform random numbers:

\begin{eqnarray*} z_1 &=& \sqrt{-2 \ln u_1} \cos(2 \pi u_2) \\ z_2 &=& \sqrt{-2 \ln u_1} \sin(2 \pi u_2) \end{eqnarray*}

Below is a Fortran implementation returning a Gaussian-distributed n-dimensional vector $\mathbf{z}$. This will be useful for the following sections.

Fortran

subroutine random_gauss(z,n)
implicit none
integer, intent(in) :: n
double precision, intent(out) :: z(n)
double precision :: u(n+1)
double precision, parameter :: two_pi = 2.d0*dacos(-1.d0)
integer :: i

call random_number(u)

if (iand(n,1) == 0) then
 ! n is even
 do i=1,n,2
    z(i)   = dsqrt(-2.d0*dlog(u(i))) 
    z(i+1) = z(i) * dsin( two_pi*u(i+1) )
    z(i)   = z(i) * dcos( two_pi*u(i+1) )
 end do

else
 ! n is odd
 do i=1,n-1,2
    z(i)   = dsqrt(-2.d0*dlog(u(i))) 
    z(i+1) = z(i) * dsin( two_pi*u(i+1) )
    z(i)   = z(i) * dcos( two_pi*u(i+1) )
 end do

 z(n)   = dsqrt(-2.d0*dlog(u(n))) 
 z(n)   = z(n) * dcos( two_pi*u(n+1) )

end if

end subroutine random_gauss

In Python, you can use the random.normal function of Numpy.

Exercise 1

If you use Fortran, copy/paste the random_gauss function in a Fortran file.

Write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$.

Python

def drift(a,r):
# TODO

Fortran

subroutine drift(a,r,b)
implicit none
double precision, intent(in)  :: a, r(3)
double precision, intent(out) :: b(3)

! TODO

end subroutine drift

Solution   solution

Python

def drift(a,r):
ar_inv = -a/np.sqrt(np.dot(r,r))
return r * ar_inv

Fortran

subroutine drift(a,r,b)
implicit none
double precision, intent(in)  :: a, r(3)
double precision, intent(out) :: b(3)

double precision :: ar_inv

ar_inv = -a / dsqrt(r(1)*r(1) + r(2)*r(2) + r(3)*r(3))
b(:)   = r(:) * ar_inv

end subroutine drift

Exercise 2

Modify the previous program to introduce the drift-diffusion scheme. (This is a necessary step for the next section on diffusion Monte Carlo).

Python

#!/usr/bin/env python3

from hydrogen  import *
from qmc_stats import *

def MonteCarlo(a,nmax,dt):
# TODO

# Run simulation
a    = 1.2
nmax = 100000
dt   = # TODO

X0 = [ MonteCarlo(a,nmax,dt) for i in range(30)]

# Energy
X = [ x for (x, _) in X0 ]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")

# Acceptance rate
X = [ x for (_, x) in X0 ]
A, deltaA = ave_error(X)
print(f"A = {A} +/- {deltaA}")

Fortran

subroutine variational_montecarlo(a,dt,nmax,energy,accep)
implicit none
double precision, intent(in)  :: a, dt
integer*8       , intent(in)  :: nmax 
double precision, intent(out) :: energy, accep

integer*8        :: istep
integer*8        :: n_accep
double precision :: sq_dt, chi(3)
double precision :: psi_old, psi_new
double precision :: r_old(3), r_new(3)
double precision :: d_old(3), d_new(3)

double precision, external :: e_loc, psi

! TODO

end subroutine variational_montecarlo

program qmc
implicit none
double precision, parameter :: a     = 1.2d0
double precision, parameter :: dt    = ! TODO
integer*8       , parameter :: nmax  = 100000
integer         , parameter :: nruns = 30

integer          :: irun
double precision :: X(nruns), accep(nruns)
double precision :: ave, err

do irun=1,nruns
call variational_montecarlo(a,dt,nmax,X(irun),accep(irun))
enddo

call ave_error(X,nruns,ave,err)
print *, 'E = ', ave, '+/-', err

call ave_error(accep,nruns,ave,err)
print *, 'A = ', ave, '+/-', err

end program qmc

gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
./vmc_metropolis

Solution   solution

Python

#!/usr/bin/env python3

from hydrogen  import *
from qmc_stats import *

def MonteCarlo(a,nmax,dt):
sq_dt = np.sqrt(dt)

energy  = 0.
N_accep = 0

r_old   = np.random.normal(loc=0., scale=1.0, size=(3))
d_old   = drift(a,r_old)
d2_old  = np.dot(d_old,d_old)
psi_old = psi(a,r_old)

for istep in range(nmax):
   chi = np.random.normal(loc=0., scale=1.0, size=(3))

   energy += e_loc(a,r_old)

   r_new   = r_old + dt * d_old + sq_dt * chi
   d_new   = drift(a,r_new)
   d2_new  = np.dot(d_new,d_new)
   psi_new = psi(a,r_new)

   # Metropolis
   prod    = np.dot((d_new + d_old), (r_new - r_old))
   argexpo = 0.5 * (d2_new - d2_old)*dt + prod

   q = psi_new / psi_old
   q = np.exp(-argexpo) * q*q

   if np.random.uniform() <= q:
       N_accep += 1

       r_old   = r_new
       d_old   = d_new
       d2_old  = d2_new
       psi_old = psi_new

return energy/nmax, N_accep/nmax


# Run simulation
a    = 1.2
nmax = 100000
dt   = 1.0

X0 = [ MonteCarlo(a,nmax,dt) for i in range(30)]

# Energy
X = [ x for (x, _) in X0 ]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")

# Acceptance rate
X = [ x for (_, x) in X0 ]
A, deltaA = ave_error(X)
print(f"A = {A} +/- {deltaA}")

E = -0.48034171558629885 +/- 0.0005286038561061781
A = 0.6210380000000001 +/- 0.0005457375900937905

Fortran

subroutine variational_montecarlo(a,dt,nmax,energy,accep)
implicit none
double precision, intent(in)  :: a, dt
integer*8       , intent(in)  :: nmax 
double precision, intent(out) :: energy, accep

integer*8        :: istep
integer*8        :: n_accep
double precision :: sq_dt, chi(3), d2_old, prod, u
double precision :: psi_old, psi_new, d2_new, argexpo, q
double precision :: r_old(3), r_new(3)
double precision :: d_old(3), d_new(3)

double precision, external :: e_loc, psi

sq_dt = dsqrt(dt)

! Initialization
energy  = 0.d0
n_accep = 0_8

call random_gauss(r_old,3)

call drift(a,r_old,d_old)
d2_old  = d_old(1)*d_old(1) + &
       d_old(2)*d_old(2) + &
       d_old(3)*d_old(3)

psi_old = psi(a,r_old)

do istep = 1,nmax
energy = energy + e_loc(a,r_old)

call random_gauss(chi,3)
r_new(:) = r_old(:) + dt*d_old(:) + chi(:)*sq_dt

call drift(a,r_new,d_new)
d2_new = d_new(1)*d_new(1) + &
         d_new(2)*d_new(2) + &
         d_new(3)*d_new(3)

psi_new = psi(a,r_new)

! Metropolis
prod = (d_new(1) + d_old(1))*(r_new(1) - r_old(1)) + &
       (d_new(2) + d_old(2))*(r_new(2) - r_old(2)) + &
       (d_new(3) + d_old(3))*(r_new(3) - r_old(3))

argexpo = 0.5d0 * (d2_new - d2_old)*dt + prod

q = psi_new / psi_old
q = dexp(-argexpo) * q*q

call random_number(u)

if (u <= q) then

   n_accep = n_accep + 1_8

   r_old(:) = r_new(:)
   d_old(:) = d_new(:)
   d2_old   = d2_new
   psi_old  = psi_new

end if

end do

energy = energy / dble(nmax)
accep  = dble(n_accep) / dble(nmax)

end subroutine variational_montecarlo

program qmc
implicit none
double precision, parameter :: a     = 1.2d0
double precision, parameter :: dt    = 1.0d0
integer*8       , parameter :: nmax  = 100000
integer         , parameter :: nruns = 30

integer          :: irun
double precision :: X(nruns), accep(nruns)
double precision :: ave, err

do irun=1,nruns
call variational_montecarlo(a,dt,nmax,X(irun),accep(irun))
enddo

call ave_error(X,nruns,ave,err)
print *, 'E = ', ave, '+/-', err

call ave_error(accep,nruns,ave,err)
print *, 'A = ', ave, '+/-', err

end program qmc

 E =  -0.47940635575542706      +/-   5.5613594433433764E-004
 A =   0.62037333333333333      +/-   4.8970160591451110E-004

Diffusion Monte Carlo

As we have seen, Variational Monte Carlo is a powerful method to compute integrals in large dimensions. It is often used in cases where the expression of the wave function is such that the integrals can't be evaluated (multi-centered Slater-type orbitals, correlation factors, etc).

Diffusion Monte Carlo is different. It goes beyond the computation of the integrals associated with an input wave function, and aims at finding a near-exact numerical solution to the Schrödinger equation.

Schrödinger equation in imaginary time

Consider the time-dependent Schrödinger equation:

\[ i\frac{\partial \Psi(\mathbf{r},t)}{\partial t} = (\hat{H} -E_{\rm ref}) \Psi(\mathbf{r},t)\,. \]

where we introduced a shift in the energy, $E_{\rm ref}$, for reasons which will become apparent below.

We can expand a given starting wave function, $\Psi(\mathbf{r},0)$, in the basis of the eigenstates of the time-independent Hamiltonian, $\Phi_k$, with energies $E_k$:

\[ \Psi(\mathbf{r},0) = \sum_k a_k\, \Phi_k(\mathbf{r}). \]

The solution of the Schrödinger equation at time $t$ is

\[ \Psi(\mathbf{r},t) = \sum_k a_k \exp \left( -i\, (E_k-E_{\rm ref})\, t \right) \Phi_k(\mathbf{r}). \]

Now, if we replace the time variable $t$ by an imaginary time variable $\tau=i\,t$, we obtain

\[ -\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = (\hat{H} -E_{\rm ref}) \psi(\mathbf{r}, \tau) \]

where $\psi(\mathbf{r},\tau) = \Psi(\mathbf{r},-i\,\tau)$ and

\begin{eqnarray*} \psi(\mathbf{r},\tau) &=& \sum_k a_k \exp( -(E_k-E_{\rm ref})\, \tau) \Phi_k(\mathbf{r})\\ &=& \exp(-(E_0-E_{\rm ref})\, \tau)\sum_k a_k \exp( -(E_k-E_0)\, \tau) \Phi_k(\mathbf{r})\,. \end{eqnarray*}

For large positive values of $\tau$, $\psi$ is dominated by the $k=0$ term, namely, the lowest eigenstate. If we adjust $E_{\rm ref}$ to the running estimate of $E_0$, we can expect that simulating the differetial equation in imaginary time will converge to the exact ground state of the system.

Relation to diffusion

The diffusion equation of particles is given by

\[ \frac{\partial \psi(\mathbf{r},t)}{\partial t} = D\, \Delta \psi(\mathbf{r},t) \]

where $D$ is the diffusion coefficient. When the imaginary-time Schrödinger equation is written in terms of the kinetic energy and potential,

\[ \frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = \left(\frac{1}{2}\Delta - [V(\mathbf{r}) -E_{\rm ref}]\right) \psi(\mathbf{r}, \tau)\,, \]

it can be identified as the combination of:

• a diffusion equation (Laplacian)
• an equation whose solution is an exponential (potential)

The diffusion equation can be simulated by a Brownian motion:

\[ \mathbf{r}_{n+1} = \mathbf{r}_{n} + \sqrt{\delta t}\, \chi \]

where $\chi$ is a Gaussian random variable, and the potential term can be simulated by creating or destroying particles over time (a so-called branching process) or by simply considering it as a cumulative multiplicative weight along the diffusion trajectory (pure Diffusion Monte Carlo):

\[ \prod_i \exp \left( - (V(\mathbf{r}_i) - E_{\text{ref}}) \delta t \right). \]

We note that the ground-state wave function of a Fermionic system is antisymmetric and changes sign. Therefore, its interpretation as a probability distribution is somewhat problematic. In fact, mathematically, since the Bosonic ground state is lower in energy than the Fermionic one, for large $\tau$, the system will evolve towards the Bosonic solution.

For the systems you will study, this is not an issue:

• Hydrogen atom: You only have one electron!
• Two-electron system ($H_2$ or He): The ground-wave function is antisymmetric in the spin variables but symmetric in the space ones.

Therefore, in both cases, you are dealing with a "Bosonic" ground state.

Importance sampling

In a molecular system, the potential is far from being constant and, in fact, diverges at the inter-particle coalescence points. Hence, it results in very large fluctuations of the erm weight associated with the potental, making the calculations impossible in practice. Fortunately, if we multiply the Schrödinger equation by a chosen trial wave function $\Psi_T(\mathbf{r})$ (Hartree-Fock, Kohn-Sham determinant, CI wave function, etc), one obtains

\[ -\frac{\partial \psi(\mathbf{r},\tau)}{\partial \tau} \Psi_T(\mathbf{r}) = \left[ -\frac{1}{2} \Delta \psi(\mathbf{r},\tau) + V(\mathbf{r}) \psi(\mathbf{r},\tau) \right] \Psi_T(\mathbf{r}) \]

Defining $\Pi(\mathbf{r},\tau) = \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})$, (see appendix for details)

\[ -\frac{\partial \Pi(\mathbf{r},\tau)}{\partial \tau} = -\frac{1}{2} \Delta \Pi(\mathbf{r},\tau) + \nabla \left[ \Pi(\mathbf{r},\tau) \frac{\nabla \Psi_T(\mathbf{r})}{\Psi_T(\mathbf{r})} \right] + (E_L(\mathbf{r})-E_{\rm ref})\Pi(\mathbf{r},\tau) \]

The new "kinetic energy" can be simulated by the drift-diffusion scheme presented in the previous section (VMC). The new "potential" is the local energy, which has smaller fluctuations when $\Psi_T$ gets closer to the exact wave function. This term can be simulated by \[ \prod_i \exp \left( - (E_L(\mathbf{r}_i) - E_{\text{ref}}) \delta t \right). \] where $E_{\rm ref}$ is the constant we had introduced above, which is adjusted to an estimate of the average energy to keep the weights close to one.

This equation generates the N-electron density $\Pi$, which is the product of the ground state solution with the trial wave function. You may then ask: how can we compute the total energy of the system?

To this aim, we use the mixed estimator of the energy:

\begin{eqnarray*} E(\tau) &=& \frac{\langle \psi(\tau) | \hat{H} | \Psi_T \rangle}{\langle \psi(\tau) | \Psi_T \rangle}\\ &=& \frac{\int \psi(\mathbf{r},\tau) \hat{H} \Psi_T(\mathbf{r}) d\mathbf{r}} {\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}} \\ &=& \frac{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) E_L(\mathbf{r}) d\mathbf{r}} {\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}} \,. \end{eqnarray*}

For large $\tau$, we have that

\[ \Pi(\mathbf{r},\tau) =\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \rightarrow \Phi_0(\mathbf{r}) \Psi_T(\mathbf{r})\,, \]

and, using that $\hat{H}$ is Hermitian and that $\Phi_0$ is an eigenstate of the Hamiltonian, we obtain for large $\tau$

\[ E(\tau) = \frac{\langle \psi_\tau | \hat{H} | \Psi_T \rangle} {\langle \psi_\tau | \Psi_T \rangle} = \frac{\langle \Psi_T | \hat{H} | \psi_\tau \rangle} {\langle \Psi_T | \psi_\tau \rangle} \rightarrow E_0 \frac{\langle \Psi_T | \Phi_0 \rangle} {\langle \Psi_T | \Phi_0 \rangle} = E_0 \]

Therefore, we can compute the energy within DMC by generating the density $\Pi$ with random walks, and simply averaging the local energies computed with the trial wave function.

Appendix : Details of the Derivation

\[ -\frac{\partial \psi(\mathbf{r},\tau)}{\partial \tau} \Psi_T(\mathbf{r}) = \left[ -\frac{1}{2} \Delta \psi(\mathbf{r},\tau) + V(\mathbf{r}) \psi(\mathbf{r},\tau) \right] \Psi_T(\mathbf{r}) \]

\[ -\frac{\partial \big[ \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \big]}{\partial \tau} = -\frac{1}{2} \Big( \Delta \big[ \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \big] - \psi(\mathbf{r},\tau) \Delta \Psi_T(\mathbf{r}) - 2 \nabla \psi(\mathbf{r},\tau) \nabla \Psi_T(\mathbf{r}) \Big) + V(\mathbf{r}) \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \]

\[ -\frac{\partial \big[ \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \big]}{\partial \tau} = -\frac{1}{2} \Delta \big[\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \big] + \frac{1}{2} \psi(\mathbf{r},\tau) \Delta \Psi_T(\mathbf{r}) + \Psi_T(\mathbf{r})\nabla \psi(\mathbf{r},\tau) \frac{\nabla \Psi_T(\mathbf{r})}{\Psi_T(\mathbf{r})} + V(\mathbf{r}) \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \]

\[ -\frac{\partial \big[ \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \big]}{\partial \tau} = -\frac{1}{2} \Delta \big[\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \big] + \psi(\mathbf{r},\tau) \Delta \Psi_T(\mathbf{r}) + \Psi_T(\mathbf{r})\nabla \psi(\mathbf{r},\tau) \frac{\nabla \Psi_T(\mathbf{r})}{\Psi_T(\mathbf{r})} + E_L(\mathbf{r}) \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \] \[ -\frac{\partial \big[ \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \big]}{\partial \tau} = -\frac{1}{2} \Delta \big[\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \big] + \nabla \left[ \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \frac{\nabla \Psi_T(\mathbf{r})}{\Psi_T(\mathbf{r})} \right] + E_L(\mathbf{r}) \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \]

Defining $\Pi(\mathbf{r},t) = \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})$,

\[ -\frac{\partial \Pi(\mathbf{r},\tau)}{\partial \tau} = -\frac{1}{2} \Delta \Pi(\mathbf{r},\tau) + \nabla \left[ \Pi(\mathbf{r},\tau) \frac{\nabla \Psi_T(\mathbf{r})}{\Psi_T(\mathbf{r})} \right] + E_L(\mathbf{r}) \Pi(\mathbf{r},\tau) \]

Pure Diffusion Monte Carlo

Instead of having a variable number of particles to simulate the branching process as in the Diffusion Monte Carlo (DMC) algorithm, we use variant called pure Diffusion Monte Carlo (PDMC) where the potential term is considered as a cumulative product of weights:

\begin{eqnarray*} W(\mathbf{r}_n, \tau) = \prod_{i=1}^{n} \exp \left( -\delta t\, (E_L(\mathbf{r}_i) - E_{\text{ref}}) \right) = \prod_{i=1}^{n} w(\mathbf{r}_i) \end{eqnarray*}

where $\mathbf{r}_i$ are the coordinates along the trajectory and we introduced a time-step variable $\delta t$ to discretize the integral.

The PDMC algorithm is less stable than the DMC algorithm: it requires to have a value of $E_\text{ref}$ which is close to the fixed-node energy, and a good trial wave function. Moreover, we can't let $\tau$ become too large because the weight whether explode or vanish: we need to have a fixed value of $\tau$ (projection time). The big advantage of PDMC is that it is rather simple to implement starting from a VMC code:

1. Start with $W(\mathbf{r}_0)=1, \tau_0 = 0$
2. Evaluate the local energy at $\mathbf{r}_{n}$
3. Compute the contribution to the weight $w(\mathbf{r}_n) = \exp(-\delta t(E_L(\mathbf{r}_n)-E_\text{ref}))$
4. Update $W(\mathbf{r}_{n}) = W(\mathbf{r}_{n-1}) \times w(\mathbf{r}_n)$
5. Accumulate the weighted energy $W(\mathbf{r}_n) \times E_L(\mathbf{r}_n)$, and the weight $W(\mathbf{r}_n)$ for the normalization
6. Update $\tau_n = \tau_{n-1} + \delta t$
7. If $\tau_{n} > \tau_\text{max}$ ($\tau_\text{max}$ is an input parameter), the long projection time has been reached and we can start an new trajectory from the current position: reset $W(r_n) = 1$ and $\tau_n = 0$
8. Compute a new position $\mathbf{r'} = \mathbf{r}_n + \delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi$
9. Evaluate $\Psi(\mathbf{r}')$ and $\frac{\nabla \Psi(\mathbf{r'})}{\Psi(\mathbf{r'})}$ at the new position
10. Compute the ratio $A = \frac{T(\mathbf{r}' \rightarrow \mathbf{r}_{n}) P(\mathbf{r}')}{T(\mathbf{r}_{n} \rightarrow \mathbf{r}') P(\mathbf{r}_{n})}$

1. Draw a uniform random number $v \in [0,1]$
2. if $v \le A$, accept the move : set $\mathbf{r}_{n+1} = \mathbf{r'}$
3. else, reject the move : set $\mathbf{r}_{n+1} = \mathbf{r}_n$

Some comments are needed:

• You estimate the energy as

  \begin{eqnarray*} E = \frac{\sum_{k=1}^{N_{\rm MC}} E_L(\mathbf{r}_k) W(\mathbf{r}_k, k\delta t)}{\sum_{k=1}^{N_{\rm MC}} W(\mathbf{r}_k, k\delta t)} \end{eqnarray*}
• The result will be affected by a time-step error (the finite size of $\delta t$) due to the discretization of the integral, and one has in principle to extrapolate to the limit $\delta t \rightarrow 0$. This amounts to fitting the energy computed for multiple values of $\delta t$. Here, you will be using a small enough time-step and you should not worry about the extrapolation.
• The accept/reject step (steps 9-12 in the algorithm) is in principle not needed for the correctness of the DMC algorithm. However, its use reduces significantly the time-step error.

Hydrogen atom

Exercise

Modify the Metropolis VMC program into a PDMC program. In the limit $\delta t \rightarrow 0$, you should recover the exact energy of H for any value of $a$, as long as the simulation is stable. We choose here a time step of 0.05 a.u. and a fixed projection time $\tau_{\text{max}}$ =100 a.u.

Python

from hydrogen  import *
from qmc_stats import *

def MonteCarlo(a, nmax, dt, Eref):
# TODO

# Run simulation
a     = 1.2
nmax  = 100000
dt    = 0.05
tau   = 100.
E_ref = -0.5

X0 = [ MonteCarlo(a, nmax, dt, E_ref) for i in range(30)]

# Energy
X = [ x for (x, _) in X0 ]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")

# Acceptance rate
X = [ x for (_, x) in X0 ]
A, deltaA = ave_error(X)
print(f"A = {A} +/- {deltaA}")

Fortran

subroutine pdmc(a, dt, nmax, energy, accep, tau, E_ref)
implicit none
double precision, intent(in)  :: a, dt, tau
integer*8       , intent(in)  :: nmax 
double precision, intent(out) :: energy, accep
double precision, intent(in)  :: E_ref

integer*8        :: istep
integer*8        :: n_accep
double precision :: sq_dt, chi(3)
double precision :: psi_old, psi_new
double precision :: r_old(3), r_new(3)
double precision :: d_old(3), d_new(3)

double precision, external :: e_loc, psi

! TODO

end subroutine pdmc

program qmc
implicit none
double precision, parameter :: a     = 1.2d0
double precision, parameter :: dt    = 0.05d0
double precision, parameter :: E_ref = -0.5d0
double precision, parameter :: tau   = 100.d0
integer*8       , parameter :: nmax  = 100000
integer         , parameter :: nruns = 30

integer          :: irun
double precision :: X(nruns), accep(nruns)
double precision :: ave, err

do irun=1,nruns
call pdmc(a, dt, nmax, X(irun), accep(irun), tau, E_ref)
enddo

call ave_error(X,nruns,ave,err)
print *, 'E = ', ave, '+/-', err

call ave_error(accep,nruns,ave,err)
print *, 'A = ', ave, '+/-', err

end program qmc

gfortran hydrogen.f90 qmc_stats.f90 pdmc.f90 -o pdmc
./pdmc

Solution   solution

Python

#!/usr/bin/env python3

from hydrogen  import *
from qmc_stats import *

def MonteCarlo(a, nmax, dt, tau, Eref):
sq_dt = np.sqrt(dt)

energy        = 0.
N_accep       = 0
normalization = 0.

w           = 1.
tau_current = 0.

r_old   = np.random.normal(loc=0., scale=1.0, size=(3))
d_old   = drift(a,r_old)
d2_old  = np.dot(d_old,d_old)
psi_old = psi(a,r_old)

for istep in range(nmax):
   el = e_loc(a,r_old)
   w *= np.exp(-dt*(el - Eref))

   normalization += w
   energy        += w * el

   tau_current += dt

   # Reset when tau is reached
   if tau_current > tau:
       w           = 1.
       tau_current = 0.

   chi = np.random.normal(loc=0., scale=1.0, size=(3))

   r_new = r_old + dt * d_old + sq_dt * chi
   d_new = drift(a,r_new)
   d2_new = np.dot(d_new,d_new)
   psi_new = psi(a,r_new)

   # Metropolis
   prod = np.dot((d_new + d_old), (r_new - r_old))
   argexpo = 0.5 * (d2_new - d2_old)*dt + prod

   q = psi_new / psi_old
   q = np.exp(-argexpo) * q*q

   if np.random.uniform() <= q:
       N_accep += 1
       r_old   = r_new
       d_old   = d_new
       d2_old  = d2_new
       psi_old = psi_new

return energy/normalization, N_accep/nmax


# Run simulation
a     = 1.2
nmax  = 100000
dt    = 0.05
tau   = 100.
E_ref = -0.5

X0 = [ MonteCarlo(a, nmax, dt, tau, E_ref) for i in range(30)]

# Energy
X = [ x for (x, _) in X0 ]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")

# Acceptance rate
X = [ x for (_, x) in X0 ]
A, deltaA = ave_error(X)
print(f"A = {A} +/- {deltaA}")

E = -0.500188803288012 +/- 0.0010615739297642462
A = 0.9896496666666668 +/- 7.688845979106312e-05

Fortran

subroutine pdmc(a, dt, nmax, energy, accep, tau, E_ref)
implicit none
double precision, intent(in)  :: a, dt, tau
integer*8       , intent(in)  :: nmax 
double precision, intent(out) :: energy, accep
double precision, intent(in)  :: E_ref

integer*8        :: istep
integer*8        :: n_accep
double precision :: sq_dt, chi(3), d2_old, prod, u
double precision :: psi_old, psi_new, d2_new, argexpo, q
double precision :: r_old(3), r_new(3)
double precision :: d_old(3), d_new(3)
double precision :: e, w, norm, tau_current

double precision, external :: e_loc, psi

sq_dt = dsqrt(dt)

! Initialization
energy  = 0.d0
n_accep = 0_8
norm    = 0.d0

w           = 1.d0
tau_current = 0.d0

call random_gauss(r_old,3)

call drift(a,r_old,d_old)
d2_old  = d_old(1)*d_old(1) + &
       d_old(2)*d_old(2) + &
       d_old(3)*d_old(3)

psi_old = psi(a,r_old)

do istep = 1,nmax
e = e_loc(a,r_old)
w = w * dexp(-dt*(e - E_ref))

norm   = norm + w
energy = energy + w*e

tau_current = tau_current + dt

! Reset when tau is reached
if (tau_current > tau) then
   w           = 1.d0
   tau_current = 0.d0
endif

call random_gauss(chi,3)
r_new(:) = r_old(:) + dt*d_old(:) + chi(:)*sq_dt

call drift(a,r_new,d_new)
d2_new = d_new(1)*d_new(1) + &
         d_new(2)*d_new(2) + &
         d_new(3)*d_new(3)

psi_new = psi(a,r_new)

! Metropolis
prod = (d_new(1) + d_old(1))*(r_new(1) - r_old(1)) + &
       (d_new(2) + d_old(2))*(r_new(2) - r_old(2)) + &
       (d_new(3) + d_old(3))*(r_new(3) - r_old(3))

argexpo = 0.5d0 * (d2_new - d2_old)*dt + prod

q = psi_new / psi_old
q = dexp(-argexpo) * q*q

call random_number(u)

if (u <= q) then

   n_accep = n_accep + 1_8

   r_old(:) = r_new(:)
   d_old(:) = d_new(:)
   d2_old   = d2_new
   psi_old  = psi_new

end if

end do

energy = energy / norm
accep  = dble(n_accep) / dble(nmax)

end subroutine pdmc

program qmc
implicit none
double precision, parameter :: a     = 1.2d0
double precision, parameter :: dt    = 0.05d0
double precision, parameter :: E_ref = -0.5d0
double precision, parameter :: tau   = 100.d0
integer*8       , parameter :: nmax  = 100000
integer         , parameter :: nruns = 30

integer          :: irun
double precision :: X(nruns), accep(nruns)
double precision :: ave, err

do irun=1,nruns
call pdmc(a, dt, nmax, X(irun), accep(irun), tau, E_ref)
enddo

call ave_error(X,nruns,ave,err)
print *, 'E = ', ave, '+/-', err

call ave_error(accep,nruns,ave,err)
print *, 'A = ', ave, '+/-', err

end program qmc

 E =  -0.49963953547336709      +/-   6.8755513992017491E-004
 A =   0.98963533333333342      +/-   6.3052128284666221E-005

Project

Change your PDMC code for one of the following:

• the Helium atom, or
• the H_2 molecule at $R$ =1.401 bohr.

And compute the ground state energy.

Acknowledgments

TREX : Targeting Real Chemical Accuracy at the Exascale project has received funding from the European Union’s Horizon 2020 - Research and Innovation program - under grant agreement no. 952165. The content of this document does not represent the opinion of the European Union, and the European Union is not responsible for any use that might be made of such content.