qmc-lttc/QMC.org

Quantum Monte Carlo

• Introduction
• Numerical evaluation of the energy
  • Local energy
    • Write a function which computes the potential at $\mathbf{r}$
    • Write a function which computes the wave function at $\mathbf{r}$
    • Write a function which computes the local kinetic energy at $\mathbf{r}$
    • Write a function which computes the local energy at $\mathbf{r}$
  • Plot of the local energy along the $x$ axis
  • Compute numerically the expectation value of the energy
  • Compute the variance of the local energy
• Variational Monte Carlo
  • Computation of the statistical error
  • Uniform sampling in the box
  • Gaussian sampling
  • Sampling with $\Psi^2$
• Diffusion Monte Carlo

Introduction

We propose different exercises to understand quantum Monte Carlo (QMC) methods. In the first section, we propose to compute the energy of a hydrogen atom using numerical integration. The goal of this section is to introduce the 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. Finally, we introduce the diffusion Monte Carlo (DMC) method which gives the exact energy of the $H_2$ molecule.

Code examples will be given in Python and Fortran. Whatever language can be chosen.

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.

Note

In Fortran, when you use a double precision constant, don't forget to put d0 as a suffix (for example 2.0d0), or it will be interpreted as a single precision value

Numerical evaluation of the energy

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 $\Psi$ is an eigenfunction of the Hamiltonian

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

when $a=1$, by checking that $\hat{H}\Psi(\mathbf{r}) = E\Psi(\mathbf{r})$ for all $\mathbf{r}$. We will check that the local energy, defined as

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

is constant. We will also see that when $a \ne 1$ the local energy is not constant, so $\hat{H} \Psi \ne E \Psi$.

The probabilistic expected value of an arbitrary function $f(x)$ with respect to a probability density function $p(x)$ is given by

$$ \langle f \rangle_p = \int_{-\infty}^\infty p(x)\, f(x)\,dx $$.

Recall that a probability density function $p(x)$ is non-negative and integrates to one:

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

The electronic energy of a system is the expectation value of the local energy $E(\mathbf{r})$ with respect to the 3N-dimensional electron density given by the square of the wave function:

\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}} = \langle E_L \rangle_{\Psi^2} \end{eqnarray*}

Local energy

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):
return -1. / np.sqrt(np.dot(r,r))

Fortran

double precision function potential(r)
implicit none
double precision, intent(in) :: r(3)
potential = -1.d0 / dsqrt( r(1)*r(1) + r(2)*r(2) + r(3)*r(3) )
end function potential

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):
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

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:

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

So 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):
return -0.5 * (a**2 - (2.*a)/np.sqrt(np.dot(r,r)))

Fortran

double precision function kinetic(a,r)
implicit none
double precision, intent(in) :: a, r(3)
kinetic = -0.5d0 * (a*a - (2.d0*a) / &
  dsqrt( r(1)*r(1) + r(2)*r(2) + r(3)*r(3) ) ) 
end function kinetic

Write a function which computes the local energy at $\mathbf{r}$

The function accepts x,y,z as input arguments and returns the local energy.

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

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

Plot of the local energy along the $x$ axis

For multiple values of $a$ (0.1, 0.2, 0.5, 1., 1.5, 2.), plot the local energy along the $x$ axis.

Python

import numpy as np
import matplotlib.pyplot as plt

from hydrogen import e_loc

x=np.linspace(-5,5)

def make_array(a):
y=np.array([ e_loc(a, np.array([t,0.,0.]) ) for t in x])
return y

plt.figure(figsize=(10,5))
for a in [0.1, 0.2, 0.5, 1., 1.5, 2.]:
y = make_array(a)
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

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'

Compute numerically the expectation value 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 probability density at $\mathbf{r}_i$ multiplied by the volume element:

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

In this section, we will compute a numerical estimate of the energy in a grid of $50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.

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)

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

To compile the Fortran and run it:

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

Compute the 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 the average:

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

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.

Compute a numerical estimate of the variance of the local energy in 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.]:
 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
             El = e_loc(a, r)
             E  += w * El
             norm += w 
 E = E / norm
 s2 = 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
             El = e_loc(a, r)
             s2 += w * (El - E)**2
 s2 = s2 / norm
 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, external :: e_loc, psi
double precision :: x(50), w, delta, energy, dx, r(3), a(6), norm, s2
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

  s2 = 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
           s2 = s2 + w * ( e_loc(a(j), r) - energy )**2
           norm   = norm   + w 
        end do
     end do
  end do
  s2 = s2 / norm
  print *, 'a = ', a(j), ' E = ', energy, ' s2 = ', s2
end do

end program variance_hydrogen

To compile and run:

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

 a =   0.10000000000000001       E =  -0.24518438948809140       s2 =    2.6965218719733813E-002
 a =   0.20000000000000001       E =  -0.26966057967803236       s2 =    3.7197072370217653E-002
 a =   0.50000000000000000       E =  -0.38563576125173815       s2 =    5.3185967578488862E-002
 a =    1.0000000000000000       E =  -0.50000000000000000       s2 =    0.0000000000000000     
 a =    1.5000000000000000       E =  -0.39242967082602065       s2 =   0.31449670909180444     
 a =    2.0000000000000000       E =   -8.0869806678448772E-002  s2 =    1.8068814270851303     

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 do a Monte Carlo sampling.

Moreover, a Monte Carlo sampling will alow 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 by 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

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

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

Python

from math import sqrt
def ave_error(arr):
 M = len(arr)
 assert (M>1)
 average = sum(arr)/M
 variance = 1./(M-1) * sum( [ (x - average)**2 for x in arr ] )
 return (average, sqrt(variance/M))

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
  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

In this section we write a function to perform a Monte Carlo calculation of the average energy. At every Monte Carlo step:

• Draw 3 uniform random numbers in the interval $(-5,-5,-5) \le (x,y,z) \le (5,5,5)$
• Compute $\Psi^2 \times E_L$ at this point and accumulate the result in E
• Compute $\Psi^2$ at this point and accumulate the result in N Once all the steps have been computed, return the average energy computed on the Monte Carlo calculation. In the main program, write a loop to perform 30 Monte Carlo runs, and compute the average energy and the associated statistical error. Compute the energy of the wave function with $a=0.9$.

Python

from hydrogen  import *
from qmc_stats import *

def MonteCarlo(a, nmax):
  E = 0.
  N = 0.
  for istep in range(nmax):
       r = np.random.uniform(-5., 5., (3))
       w = psi(a,r)
       w = w*w
       N += w
       E += w * e_loc(a,r)
return E/N

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

subroutine uniform_montecarlo(a,nmax,energy)
implicit none
double precision, intent(in)  :: a
integer         , 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
  norm = norm + w
  energy = energy + w * e_loc(a,r)
end do
energy = energy / norm
end subroutine uniform_montecarlo

program qmc
implicit none
double precision, parameter :: a = 0.9
integer         , 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

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

 E =  -0.49588321986667677      +/-   7.1758863546737969E-004

Gaussian sampling

We will now improve the sampling and allow to sample in the whole 3D space, correcting the bias related to the sampling in the box.

Instead of drawing uniform random numbers, we will draw Gaussian random numbers centered on 0 and with a variance of 1.

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*}

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

Now the equation for the energy is changed into

\[ E = \frac{\int P(\mathbf{r}) \frac{\left[\Psi(\mathbf{r})\right]^2}{P(\mathbf{r})}\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int P(\mathbf{r}) \frac{\left[\Psi(\mathbf{r}) \right]^2}{P(\mathbf{r})} d\mathbf{r}} \] with \[ P(\mathbf{r}) = \frac{1}{(2 \pi)^{3/2}}\exp\left( -\frac{\mathbf{r}^2}{2} \right) \]

As the coordinates are drawn with probability $P(\mathbf{r})$, the average energy can be computed as

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

Python

from hydrogen  import *
from qmc_stats import *

norm_gauss = 1./(2.*np.pi)**(1.5)
def gaussian(r):
 return norm_gauss * np.exp(-np.dot(r,r)*0.5)

def MonteCarlo(a,nmax):
 E = 0.
 N = 0.
 for istep in range(nmax):
     r = np.random.normal(loc=0., scale=1.0, size=(3))
     w = psi(a,r)
     w = w*w / gaussian(r)
     N += w
     E += w * e_loc(a,r)
 return E/N

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 = -0.49507506093129827 +/- 0.00014164037765553668

Fortran

double precision function gaussian(r)
implicit none
double precision, intent(in) :: r(3)
double precision, parameter :: norm_gauss = 1.d0/(2.d0*dacos(-1.d0))**(1.5d0)
gaussian = norm_gauss * dexp( -0.5d0 * dsqrt(r(1)*r(1) + r(2)*r(2) + r(3)*r(3) ))
end function gaussian


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

integer*8 :: istep

double precision :: norm, r(3), w

double precision, external :: e_loc, psi, gaussian

energy = 0.d0
norm   = 0.d0
do istep = 1,nmax
  call random_gauss(r,3)
  w = psi(a,r) 
  w = w*w / gaussian(r)
  norm = norm + w
  energy = energy + w * e_loc(a,r)
end do
energy = energy / norm
end subroutine gaussian_montecarlo

program qmc
implicit none
double precision, parameter :: a = 0.9
integer         , parameter :: nmax = 100000
integer         , parameter :: nruns = 30

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

do irun=1,nruns
  call gaussian_montecarlo(a,nmax,X(irun))
enddo
call ave_error(X,nruns,ave,err)
print *, 'E = ', ave, '+/-', err
end program qmc

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

 E =  -0.49606057056767766      +/-   1.3918807547836872E-004

Sampling with $\Psi^2$

We will now use the square of the wave function to make the sampling:

\[ P(\mathbf{r}) = \left[\Psi(\mathbf{r})\right]^2 \]

The expression for the energy will be simplified to the average of the local energies, each with a weight of 1.

$$ E \approx \frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r}_i) $$

To generate the probability density $\Psi^2$, we consider a diffusion process characterized by a time-dependent density $[\Psi(\mathbf{r},t)]^2$, which obeys the Fokker-Planck equation:

\[ \frac{\partial \Psi^2}{\partial t} = \sum_i D \frac{\partial}{\partial \mathbf{r}_i} \left( \frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right) [\Psi(\mathbf{r},t)]^2. \]

$D$ is the diffusion constant and $F_i$ is the i-th component of a drift velocity caused by an external potential. For a stationary density, \( \frac{\partial \Psi^2}{\partial t} = 0 \), so

\begin{eqnarray*} 0 & = & \sum_i D \frac{\partial}{\partial \mathbf{r}_i} \left( \frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right) [\Psi(\mathbf{r})]^2 \\ 0 & = & \sum_i D \frac{\partial}{\partial \mathbf{r}_i} \left( \frac{\partial [\Psi(\mathbf{r})]^2}{\partial \mathbf{r}_i} - F_i(\mathbf{r})\,[\Psi(\mathbf{r})]^2 \right) \\ 0 & = & \frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} - \frac{\partial F_i }{\partial \mathbf{r}_i}[\Psi(\mathbf{r})]^2 - \frac{\partial \Psi^2}{\partial \mathbf{r}_i} F_i(\mathbf{r}) \end{eqnarray*}

we search for a drift function which satisfies

\[ \frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} = [\Psi(\mathbf{r})]^2 \frac{\partial F_i }{\partial \mathbf{r}_i} + \frac{\partial \Psi^2}{\partial \mathbf{r}_i} F_i(\mathbf{r}) \]

to obtain a second derivative on the left, we need the drift to be of the form \[ F_i(\mathbf{r}) = g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i} \]

\[ \frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} = [\Psi(\mathbf{r})]^2 \frac{\partial g(\mathbf{r})}{\partial \mathbf{r}_i}\frac{\partial \Psi^2}{\partial \mathbf{r}_i} + [\Psi(\mathbf{r})]^2 g(\mathbf{r}) \frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} + \frac{\partial \Psi^2}{\partial \mathbf{r}_i} g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i} \]

$g = 1 / \Psi^2$ satisfies this equation, so

\[ F(\mathbf{r}) = \frac{\nabla [\Psi(\mathbf{r})]^2}{[\Psi(\mathbf{r})]^2} = 2 \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} = 2 \nabla \left( \log \Psi(\mathbf{r}) \right) \]

following drifted diffusion scheme:

\[ \mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \eta \sqrt{\tau} \]

where $\eta$ is a normally-distributed Gaussian random number.

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

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

Now we can write the Monte Carlo sampling:

Python

def MonteCarlo(a,tau,nmax):
 E = 0.
 N = 0.
 sq_tau = sqrt(tau)
 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):
     eta = np.random.normal(loc=0., scale=1.0, size=(3))
     r_new = r_old + tau * d_old + sq_tau * eta
     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)*tau + prod
     q = psi_new / psi_old
     q = np.exp(-argexpo) * q*q
     if np.random.uniform() < q:
         r_old = r_new
         d_old = d_new
         d2_old = d2_new
         psi_old = psi_new
         N += 1.
         E += e_loc(a,r_old)
 return E/N


nmax = 100000
tau = 0.1
X = [MonteCarlo(a,tau,nmax) for i in range(30)]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")

E = -0.4951783346213532 +/- 0.00022067316984271938

Fortran

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

integer*8 :: istep

double precision :: norm, r(3), w

double precision, external :: e_loc, psi, gaussian

energy = 0.d0
norm   = 0.d0
do istep = 1,nmax
  call random_gauss(r,3)
  w = psi(a,r) 
  w = w*w / gaussian(r)
  norm = norm + w
  energy = energy + w * e_loc(a,r)
end do
energy = energy / norm
end subroutine variational_montecarlo

program qmc
implicit none
double precision, parameter :: a = 0.9
integer         , parameter :: nmax = 100000
integer         , parameter :: nruns = 30

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

do irun=1,nruns
  call gaussian_montecarlo(a,nmax,X(irun))
enddo
call ave_error(X,nruns,ave,err)
print *, 'E = ', ave, '+/-', err
end program qmc

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

Diffusion Monte Carlo

We will now consider the H_2 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.