Quantum Monte Carlo

Table of Contents

1 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 H2 molecule, starting from an approximate wave function.

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

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).

1.1 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 \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}} \\ & = & \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}} = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, E_L(\mathbf{r})\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^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:

\[ E = \int E_L(\mathbf{r}) P(\mathbf{r})\,d\mathbf{r}} \equiv \langle E_L \rangle_{\Psi^2}\,, \]

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

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

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 ≈ \frac{1}{N\rm MC} ∑i=1N\rm MC EL(\mathbf{r}i} \,.

2 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.

2.1 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.

2.1.1 Exercise 1

Write a function which computes the potential at \(\mathbf{r}\). The function accepts a 3-dimensional vector r as input arguments 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

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
2.1.1.1 Solution   solution

Python

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

2.1.2 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
2.1.2.1 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

2.1.3 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 \[-\frac{1}{2}\frac{\Delta \Psi}{\Psi}.\]

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 \[ -\frac{1}{2} \frac{\Delta \Psi}{\Psi} (\mathbf{r}) = -\frac{1}{2}\left(a^2 - \frac{2a}{\mathbf{|r|}} \right) \]

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
2.1.3.1 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

2.1.4 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

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

  ! TODO

end function e_loc
2.1.4.1 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, potential

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

end function e_loc

2.1.5 Exercise 5

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

2.1.5.1 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, equal to -0.5 atomic units.

2.2 Plot of the local energy along the \(x\) axis

The potential and the kinetic energy both diverge at \(r=0\), so we choose a grid which does not contain the origin.

2.2.1 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.

Python

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'
2.2.1.1 Solution   solution

Python

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")

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

plot.png

2.3 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)

2.3.1 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

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 
2.3.1.1 Solution   solution

Python

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

2.4 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:

\[ \sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[ E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}} \] 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.

2.4.1 Exercise (optional)

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

2.4.1.1 Solution   solution

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

\begin{eqnarray*} \langle E - \bar{E} \rangle^2 & = & \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*}

2.4.2 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

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
2.4.2.1 Solution   solution

Python

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     

3 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 alow us to remove the bias due to the discretization of space, and compute a statistical confidence interval.

3.1 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

\[ E = \frac{1}{M} \sum_{i=1}^M E_M \]

The variance of the average energies can be computed as

\[ \sigma^2 = \frac{1}{M-1} \sum_{i=1}^{M} (E_M - E)^2 \]

And the confidence interval is given by

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

3.1.1 Exercise

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

Python

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
3.1.1.1 Solution   solution

Python

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

3.2 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})\left[\Psi(\mathbf{r})\right]^2\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^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.

3.2.1 Exercise

Parameterize the wave function with \(a=0.9\). 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.

from hydrogen  import *
from qmc_stats import *

def MonteCarlo(a, nmax):
     # TODO

a    = 0.9
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 = 0.9
  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
3.2.1.1 Solution   solution

Python

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    = 0.9
nmax = 100000

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

print(f"E = {E} +/- {deltaE}")
E = -0.4956255109300764 +/- 0.0007082875482711226

Fortran

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

  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     = 0.9
  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 =  -0.49518773675598715      +/-   5.2391494923686175E-004

3.3 Metropolis sampling with \(\Psi^2\)

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

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:

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

according to the following algorithm.

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

For simplicity, let us 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, 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

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

which, for our choice of transition probability, becomes

\[ A{\mathbf{r}_{n+1},\mathbf{r}_n) = \min\left(1,\frac{P(\mathbf{r}_{n+1})}{P(\mathbf{r}_{n})}\right)= \min\left(1,\frac{\Psi(\mathbf{r}_{n+1})^2}{\Psi(\mathbf{r}_{n})^2} \]

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

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.

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.

3.3.1 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

from hydrogen  import *
from qmc_stats import *

def MonteCarlo(a,nmax,dt):

    # TODO

    return energy/nmax, N_accep/nmax


# Run simulation
a    = 0.9
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     = 0.9d0
  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
3.3.1.1 Solution   solution

Python

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    = 0.9
nmax = 100000
dt   = 1.3

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 = -0.4950720838131573 +/- 0.00019089638602238043
A = 0.5172960000000001 +/- 0.0003443446549306529

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 = 0.9d0
  double precision, parameter :: dt = 1.3d0
  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 =  -0.49503130891988767      +/-   1.7160104275040037E-004
A =   0.51695266666666673      +/-   4.0445505648997396E-004

3.4 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.

3.5 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 random numbers. Hence, the transition probability was

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

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:

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

To sample even better the density, we can "push" the electrons into in the regions of high probability, and "pull" them away from the low-probability regions. This will mechanically 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, so that the numerical scheme becomes a drifted diffusion:

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

where \(\chi\) is a Gaussian random variable with zero mean and variance \(\delta t\). The transition probability becomes:

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

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

  1. For the starting position compute \(\Psi\) and the drif-vector \(\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\)
  2. Compute a new position \(\mathbf{r'} = \mathbf{r}_n + \delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi\)

    Evaluate \(\Psi\) and \(\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\) at the new position

  3. Compute the ratio $A = \frac{T(\mathbf{r}n+1 → \mathbf{r}n) P(\mathbf{r}n+1)}

{T(\mathbf{r}n → \mathbf{r}n+1) 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 \(\mathbf{r}_{n+1} = \mathbf{r}_n\)
  4. evaluate the local energy at \(\mathbf{r}_{n+1}\)

3.5.1 Exercise 1

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
3.5.1.1 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

3.5.2 Exercise 2

Modify the previous program to introduce the drifted diffusion scheme. (This is a necessary step for the next section).

Python

from hydrogen  import *
from qmc_stats import *

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

# Run simulation
a    = 0.9
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     = 0.9
  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
3.5.2.1 Solution   solution

Python

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, accep_rate/nmax


# Run simulation
a    = 0.9
nmax = 100000
dt   = 1.3

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 = -0.4951317910667116 +/- 0.00014045774335059988
A = 0.7200673333333333 +/- 0.00045942791345632793

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     = 0.9
  double precision, parameter :: dt    = 1.0
  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 =  -0.49497258331144794      +/-   1.0973395750688713E-004
A =   0.78839866666666658      +/-   3.2503783452043152E-004

4 Diffusion Monte Carlo   solution

4.1 Schrödinger equation in imaginary time

Consider the time-dependent Schrödinger equation:

\[ i\frac{\partial \Psi(\mathbf{r},t)}{\partial t} = \hat{H} \Psi(\mathbf{r},t) \]

We can expand \(\Psi(\mathbf{r},0)\), in the basis of the eigenstates of the time-independent Hamiltonian:

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

The solution of the Schrödinger equation at time \(t\) is

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

Now, let's replace the time variable \(t\) by an imaginary time variable \(\tau=i\,t\), we obtain

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

where \(\psi(\mathbf{r},\tau) = \Psi(\mathbf{r},-i\tau) = \Psi(\mathbf{r},t)\) and \[ \psi(\mathbf{r},\tau) = \sum_k a_k \exp( -E_k\, \tau) \phi_k(\mathbf{r}). \] For large positive values of \(\tau\), \(\psi\) is dominated by the \(k=0\) term, namely the lowest eigenstate. So we can expect that simulating the differetial equation in imaginary time will converge to the exact ground state of the system.

4.2 Diffusion and branching

The diffusion equation of particles is given by

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

The rate of reaction \(v\) is the speed at which a chemical reaction takes place. In a solution, the rate is given as a function of the concentration \([A]\) by

\[ v = \frac{d[A]}{dt}, \]

where the concentration \([A]\) is proportional to the number of particles.

These two equations allow us to interpret the Schrödinger equation in imaginary time as the combination of:

  • a diffusion equation with a diffusion coefficient \(D=1/2\) for the kinetic energy, and
  • a rate equation for the 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 rate equation can be simulated by creating or destroying particles over time (a so-called branching process).

Diffusion Monte Carlo (DMC) consists in obtaining the ground state of a system by simulating the Schrödinger equation in imaginary time, by the combination of a diffusion process and a branching process.

4.3 Importance sampling

In a molecular system, the potential is far from being constant, and diverges at inter-particle coalescence points. Hence, when the rate equation is simulated, it results in very large fluctuations in the numbers of particles, 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)

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

The new "kinetic energy" can be simulated by the drifted 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. It can be simulated by changing the number of particles according to \(\exp\left[ -\delta t\, \left(E_L(\mathbf{r}) - E_\text{ref}\right)\right]\) where \(E_{\text{ref}}\) is a constant introduced so that the average of this term is close to one, keeping the number of particles rather constant.

This equation generates the N-electron density \(\Pi\), which is the product of the ground state with the trial wave function. It introduces the constraint that \(\Pi(\mathbf{r},\tau)=0\) where \(\Psi_T(\mathbf{r})=0\). In the few cases where the wave function has no nodes, such as in the hydrogen atom or the H2 molecule, this constraint is harmless and we can obtain the exact energy. But for systems where the wave function has nodes, this scheme introduces an error known as the fixed node error.

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

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

4.4 Fixed-node DMC energy

Now that we have a process to sample \(\Pi(\mathbf{r},\tau) = \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})\), we can compute the exact energy of the system, within the fixed-node constraint, as:

\[ E = \lim_{\tau \to \infty} \frac{\int \Pi(\mathbf{r},\tau) E_L(\mathbf{r}) d\mathbf{r}} {\int \Pi(\mathbf{r},\tau) d\mathbf{r}} = \lim_{\tau \to \infty} E(\tau). \]

\[ E(\tau) = \frac{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) E_L(\mathbf{r}) d\mathbf{r}} {\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}} = \frac{\int \psi(\mathbf{r},\tau) \hat{H} \Psi_T(\mathbf{r}) d\mathbf{r}} {\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}} = \frac{\langle \psi_\tau | \hat{H} | \Psi_T \rangle} {\langle \psi_\tau | \Psi_T \rangle} \]

As \(\hat{H}\) is Hermitian,

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

So computing the energy within DMC consists in generating the density \(\Pi\) with random walks, and simply averaging the local energies computed with the trial wave function.

4.5 Pure Diffusion Monte Carlo (PDMC)

Instead of having a variable number of particles to simulate the branching process, one can choose to sample \([\Psi_T(\mathbf{r})]^2\) instead of \(\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})\), and consider the term \(\exp \left( -\delta t\,( E_L(\mathbf{r}) - E_{\text{ref}} \right)\) as a cumulative product of weights:

\[ W(\mathbf{r}_n, \tau) = \exp \left( \int_0^\tau - (E_L(\mathbf{r}_t) - E_{\text{ref}}) dt \right) \approx \prod_{i=1}^{n} \exp \left( -\delta t\, (E_L(\mathbf{r}_i) - E_{\text{ref}}) \right) = \prod_{i=1}^{n} w(\mathbf{r}_i) \]

where \(\mathbf{r}_i\) are the coordinates along the trajectory. The wave function becomes

\[ \psi(\mathbf{r},\tau) = \Psi_T(\mathbf{r}) W(\mathbf{r},\tau) \]

and the expression of the fixed-node DMC energy is

\begin{eqnarray*} E(\tau) & = & \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}} \\ & = & \frac{\int \left[ \Psi_T(\mathbf{r}) \right]^2 W(\mathbf{r},\tau) E_L(\mathbf{r}) d\mathbf{r}} {\int \left[ \Psi_T(\mathbf{r}) \right]^2 W(\mathbf{r},\tau) d\mathbf{r}} \\ \end{eqnarray*}

This algorithm is less stable than the branching 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. Its big advantage is that it is very easy to program starting from a VMC code, so this is what we will do in the next section.

4.6 Hydrogen atom

4.6.1 Exercise

Modify the Metropolis VMC program to introduce the PDMC weight. In the limit \(\delta t \rightarrow 0\), you should recover the exact energy of H for any value of \(a\).

Python

from hydrogen  import *
from qmc_stats import *

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

# Run simulation
a     = 0.9
nmax  = 100000
dt    = 0.01
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     = 0.9
  double precision, parameter :: dt    = 0.1d0
  double precision, parameter :: E_ref = -0.5d0
  double precision, parameter :: tau   = 10.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
4.6.1.1 Solution   solution

Python

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     = 0.9
nmax  = 100000
dt    = 0.1
tau   = 10.
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}")

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))

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

     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     = 0.9
  double precision, parameter :: dt    = 0.1d0
  double precision, parameter :: E_ref = -0.5d0
  double precision, parameter :: tau   = 10.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 =  -0.50067519934141380      +/-   7.9390940184720371E-004
A =   0.98788066666666663      +/-   7.2889356133441110E-005

4.7 TODO H2

We will now consider the H2 molecule in a minimal basis composed of the \(1s\) orbitals of the hydrogen atoms:

\[ \Psi(\mathbf{r}_1, \mathbf{r}_2) = \exp(-(\mathbf{r}_1 - \mathbf{R}_A)) + \] where \(\mathbf{r}_1\) and \(\mathbf{r}_2\) denote the electron coordinates and \(\mathbf{R}_A\) and \(\mathbf{R}_B\) the coordinates of the nuclei.

5 TODO [0/3] Last things to do

  • [ ] Give some hints of how much time is required for each section
  • [ ] Prepare 4 questions for the exam: multiple-choice questions with 4 possible answers. Questions should be independent because they will be asked in a random order.
  • [ ] Propose a project for the students to continue the programs. Idea: Modify the program to compute the exact energy of the H\(_2\) molecule at $R$=1.4010 bohr. Answer: 0.17406 a.u.

Author: Anthony Scemama, Claudia Filippi

Created: 2021-01-31 Sun 09:26

Validate