#+TITLE: Quantum Monte Carlo
#+AUTHOR: Anthony Scemama, Claudia Filippi
#+SETUPFILE: https://fniessen.github.io/org-html-themes/org/theme-readtheorg.setup

#+STARTUP: latexpreview
#+STARTUP: indent


* 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.
** Python

** Fortran

- 1.d0
- external
- r(:) = 0.d0
- a = (/ 0.1, 0.2 /)
- size(x)


* 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.

  
** Local energy
:PROPERTIES:
:header-args:python: :tangle hydrogen.py
:header-args:f90:    :tangle hydrogen.f90
:END:
*** Write a function which computes the potential at $\mathbf{r}$
   The function accepts q 3-dimensional vector =r= as input arguments
   and returns the potential.

   $\mathbf{r}=\sqrt{x^2 + y^2 + z^2})$, so
   $$
   V(x,y,z) = -\frac{1}{\sqrt{x^2 + y^2 + z^2})$
   $$

   #+BEGIN_SRC python 
import numpy as np

def potential(r):
    return -1. / np.sqrt(np.dot(r,r))
   #+END_SRC

   #+BEGIN_SRC f90 
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
   #+END_SRC

*** 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.

   #+BEGIN_SRC python 
def psi(a, r):
    return np.exp(-a*np.sqrt(np.dot(r,r)))
   #+END_SRC

   #+BEGIN_SRC f90 
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
   #+END_SRC
     
*** 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 $$-\frac{1}{2}\frac{\Delta \Psi}{\Psi}$$.
     
   $$
   \Psi(x,y,z) = \exp(-a\,\sqrt{x^2 + y^2 + z^2}).
   $$

   We differentiate $\Psi$ with respect to $x$:
     
   $$
   \frac{\partial \Psi}{\partial x}
   = \frac{\partial \Psi}{\partial r} \frac{\partial r}{\partial x}   
   = - \frac{a\,x}{|\mathbf{r}|} \Psi(x,y,z) 
   $$

   and we differentiate a second time:

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

   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:

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

   So the local kinetic energy is
   $$
   -\frac{1}{2} \frac{\Delta \Psi}{\Psi} (x,y,z) = -\frac{1}{2}\left(a^2 - \frac{2a}{\mathbf{|r|}} \right) 
   $$
     
   #+BEGIN_SRC python 
def kinetic(a,r):
    return -0.5 * (a**2 - (2.*a)/np.sqrt(np.dot(r,r)))
   #+END_SRC

   #+BEGIN_SRC f90 
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
   #+END_SRC

*** 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.
   
   $$
   E_L(x,y,z) = -\frac{1}{2} \frac{\Delta \Psi}{\Psi} (x,y,z) + V(x,y,z)
   $$

   #+BEGIN_SRC python
def e_loc(a,r):
    return kinetic(a,r) + potential(r)
   #+END_SRC

   #+BEGIN_SRC f90
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
   #+END_SRC
   
** Plot the local energy along the x axis
:PROPERTIES:
:header-args:python: :tangle plot_hydrogen.py
:header-args:f90:    :tangle plot_hydrogen.f90
:END:
:LOGBOOK:
CLOCK: [2021-01-03 Sun 17:48]
:END:

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

#+begin_src 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")
#+end_src

[[./plot_py.png]]


#+begin_src f90 
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
#+end_src

To compile and run:

#+begin_src sh :exports both
gfortran hydrogen.f90 plot_hydrogen.f90 -o plot_hydrogen
./plot_hydrogen > data
#+end_src

#+RESULTS:

To plot the data using gnuplot"

#+begin_src gnuplot :file plot.png :exports both
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'
#+end_src

#+RESULTS:
[[file:plot.png]]

** Compute numerically the average energy
:PROPERTIES:
:header-args:python: :tangle energy_hydrogen.py
:header-args:f90:    :tangle energy_hydrogen.f90
:END:

   We want to compute

   \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}}
   \end{eqnarray}

   If the space is discretized in small volume elements
   $\delta x\, \delta y\, \delta z$, this last equation corresponds
   to a weighted average of the local energy, where the weights are
   the values of the square of the wave function at $(x,y,z)$
   multiplied by the volume element:
     
   $$
   E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
   w_i = \left[\Psi(\mathbf{r}_i)\right]^2 \delta x\, \delta y\, \delta z
   $$
     
   We now compute an numerical estimate of the energy in a grid of
   $50\times50\times50$ points in the range    $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.

   Note: the energy is biased because:
   - The energy is evaluated only inside the box
   - The volume elements are not infinitely small
     
   #+BEGIN_SRC python :results output :exports both
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}")                
                
   #+end_src
   
   #+RESULTS:
   : 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

#+begin_src f90 
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
#+end_src

To compile and run:

#+begin_src sh :results output :exports both
gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
./energy_hydrogen 
#+end_src

#+RESULTS:
:  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
:PROPERTIES:
:header-args:python: :tangle variance_hydrogen.py
:header-args:f90:    :tangle variance_hydrogen.f90
:END:

   The variance of the local energy measures the intensity of the
   fluctuations of the local energy around the average. If the local
   energy is constant (i.e. $\Psi$ is an eigenfunction of $\hat{H}$)
   the variance is zero.

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

   Compute an numerical estimate of the variance of the local energy
   in a grid of $50\times50\times50$ points in the range    $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.
     
   #+BEGIN_SRC python :results output :exports both
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}")
                
   #+end_src

   #+RESULTS:
   : 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

#+begin_src f90 
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
#+end_src

To compile and run:

#+begin_src sh :results output :exports both
gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
./variance_hydrogen 
#+end_src

#+RESULTS:
:  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

  Instead of computing the average energy as a numerical integration
  on a grid, we will do a Monte Carlo sampling, which is an extremely
  efficient method to compute integrals in large dimensions.

  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.

#+BEGIN_SRC python :results output
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))
#+END_SRC

#+RESULTS:
   
** Uniform sampling in the box

   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.

   Then, 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 $a=0.9$.
   
   #+BEGIN_SRC python 
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
   #+END_SRC

   #+BEGIN_SRC python 
a = 0.9
nmax = 100000
X = [MonteCarlo(a,nmax) for i in range(30)]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")
   #+END_SRC

   #+RESULTS:
   : E = -0.4952626284319677 +/- 0.0006877988969872546
   
** 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. 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

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

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

   #+RESULTS:

   #+BEGIN_SRC python
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
   #+END_SRC

   #+RESULTS:

   #+BEGIN_SRC python :results output
a = 0.9
nmax = 100000
X = [MonteCarlo(a,nmax) for i in range(30)]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")
   #+END_SRC

   #+RESULTS:
   : E = -0.4952488228427792 +/- 0.00011913174676540714
   
** Sampling with $\Psi^2$

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

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

   Now, 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 can use a drifted
   diffusion scheme:

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

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

   First, write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$.
   
   #+BEGIN_SRC python
def drift(a,r):
  ar_inv = -a/np.sqrt(np.dot(r,r))
  return r * ar_inv
   #+END_SRC

   #+RESULTS:

   #+BEGIN_SRC 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
   #+END_SRC

   #+RESULTS:

   #+BEGIN_SRC python :results output
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}")
   #+END_SRC

   #+RESULTS:
   : E = -0.4951783346213532 +/- 0.00022067316984271938
   
  
* 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.