qmc-lttc/QMC.org

#+TITLE: Quantum Monte Carlo
#+AUTHOR: Anthony Scemama, Claudia Filippi
#+LANGUAGE:  en
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  #+RESULTS:
  : /home/scemama/TREX/qmc-lttc/QMC.pdf

* Introduction

  This website contains the QMC tutorial of the 2021 LTTC winter school
  [[https://www.irsamc.ups-tlse.fr/lttc/Luchon][Tutorials in Theoretical Chemistry]].

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

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

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

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

** Energy and local energy

  For a given system with Hamiltonian $\hat{H}$ and wave function $\Psi$, we define the local energy as
  
  $$
  E_L(\mathbf{r}) = \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})},
  $$

  where $\mathbf{r}$ denotes the 3N-dimensional electronic coordinates.
  
  The electronic energy of a system, $E$, can be rewritten in terms of the 
  local energy $E_L(\mathbf{r})$ as

  \begin{eqnarray*}
  E & = & \frac{\langle \Psi| \hat{H} | \Psi\rangle}{\langle \Psi |\Psi \rangle} 
      =   \frac{\int \Psi(\mathbf{r})\, \hat{H} \Psi(\mathbf{r})\, d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}} \\
    & = & \frac{\int |\Psi(\mathbf{r})|^2\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}} 
      =   \frac{\int |\Psi(\mathbf{r})|^2\, E_L(\mathbf{r})\,d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}}  
  \end{eqnarray*}
   
  For few dimensions, one can easily compute $E$ by evaluating the
  integrals on a grid but, for a high number of dimensions, one can
  resort to Monte Carlo techniques to compute $E$.
  
  To this aim, recall that the probabilistic /expected value/ of an
  arbitrary function $f(x)$ with respect to a probability density
  function $P(x)$ is given by

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

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

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

  Similarly, we can view the the energy of a system, $E$, as the expected value of the local energy with respect to 
  a probability density $P(\mathbf{r})$ defined in 3$N$ dimensions:
  
  $$ E =  \int E_L(\mathbf{r}) P(\mathbf{r})\,d\mathbf{r} \equiv  \langle E_L \rangle_{P}\,, $$
  
  where the probability density is given by the square of the wave function:
  
  $$ P(\mathbf{r}) = \frac{|\Psi(\mathbf{r})|^2}{\int |\Psi(\mathbf{r})|^2 d\mathbf{r}}\,. $$
  
  If we can sample $N_{\rm MC}$ configurations $\{\mathbf{r}\}$
  distributed as $P$, we can estimate $E$ as the average of the local
  energy computed over these configurations:
  
  $$ E \approx \frac{1}{N_{\rm MC}} \sum_{i=1}^{N_{\rm MC}} E_L(\mathbf{r}_i) \,. $$
  
* Numerical evaluation of the energy of the hydrogen atom

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

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

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

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

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

** Local energy
   :PROPERTIES:
   :header-args:python: :tangle hydrogen.py
   :header-args:f90: :tangle hydrogen.f90
   :END:

   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.
   
   #+begin_note
   - 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.
   #+end_note
   
*** Exercise 1

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

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

 *Python*
     #+BEGIN_SRC python :results none :tangle none
#!/usr/bin/env python3
import numpy as np

def potential(r):
    # TODO
     #+END_SRC

 *Fortran*
     #+BEGIN_SRC f90 :tangle none
double precision function potential(r)
  implicit none
  double precision, intent(in) :: r(3)

  ! TODO

end function potential
     #+END_SRC

**** Solution                                                      :solution2:
 *Python*
     #+BEGIN_SRC python :results none
#!/usr/bin/env python3
import numpy as np

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

 *Fortran*
     #+BEGIN_SRC f90 
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
     #+END_SRC

*** Exercise 2
    #+begin_exercise
    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.
    #+end_exercise
    
 *Python*
     #+BEGIN_SRC python :results none  :tangle none
def psi(a, r):
    # TODO
     #+END_SRC

 *Fortran*
     #+BEGIN_SRC f90  :tangle none
double precision function psi(a, r)
  implicit none
  double precision, intent(in) :: a, r(3)

  ! TODO

end function psi
     #+END_SRC
     
**** Solution                                                      :solution2:
 *Python*
     #+BEGIN_SRC python :results none
def psi(a, r):
    return np.exp(-a*np.sqrt(np.dot(r,r)))
     #+END_SRC

 *Fortran*
     #+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
     
*** Exercise 3
    #+begin_exercise
    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.
    #+end_exercise

    The local kinetic energy is defined as $$T_L(\mathbf{r}) = -\frac{1}{2}\frac{\Delta \Psi(\mathbf{r})}{\Psi(\mathbf{r})}.$$
     
    We differentiate $\Psi$ with respect to $x$:
     
    \[ \Psi(\mathbf{r})  =  \exp(-a\,|\mathbf{r}|) \]
    \[\frac{\partial \Psi}{\partial x}
      = \frac{\partial \Psi}{\partial |\mathbf{r}|} \frac{\partial |\mathbf{r}|}{\partial x}   
      =  - \frac{a\,x}{|\mathbf{r}|} \Psi(\mathbf{r}) \]

    and we differentiate a second time:

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

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

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

    Therefore, the local kinetic energy is
    $$
    T_L (\mathbf{r}) = -\frac{1}{2}\left(a^2 - \frac{2a}{\mathbf{|r|}} \right) 
    $$
     
 *Python*
     #+BEGIN_SRC python :results none :tangle none
def kinetic(a,r):
    # TODO
     #+END_SRC

 *Fortran*
     #+BEGIN_SRC f90  :tangle none
double precision function kinetic(a,r)
  implicit none
  double precision, intent(in) :: a, r(3)

  ! TODO

end function kinetic
     #+END_SRC

**** Solution                                                      :solution2:
 *Python*
     #+BEGIN_SRC python :results none
def kinetic(a,r):
    distance = np.sqrt(np.dot(r,r))
    assert (distance > 0.)

    return a * (1./distance - 0.5 * a)
     #+END_SRC

 *Fortran*
     #+BEGIN_SRC f90 
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
     #+END_SRC

*** Exercise 4
    #+begin_exercise
    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.
    #+end_exercise
   
    $$
    E_L(\mathbf{r}) = -\frac{1}{2} \frac{\Delta \Psi}{\Psi} (\mathbf{r}) + V(\mathbf{r})
    $$

    
 *Python*
     #+BEGIN_SRC python :results none :tangle none
def e_loc(a,r):
    #TODO
     #+END_SRC

 *Fortran*

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

    #+BEGIN_SRC f90 :tangle none
double precision function e_loc(a,r)
  implicit none
  double precision, intent(in) :: a, r(3)

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

  ! TODO

end function e_loc
    #+END_SRC
   
**** Solution                                                      :solution2:
 *Python*
     #+BEGIN_SRC python :results none
def e_loc(a,r):
    return kinetic(a,r) + potential(r)
     #+END_SRC

 *Fortran*
     #+BEGIN_SRC f90
double precision function e_loc(a,r)
  implicit none
  double precision, intent(in) :: a, r(3)

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

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

end function e_loc
     #+END_SRC
   
*** Exercise 5

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

**** Solution                                                      :solution2:

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

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

** Plot of the local energy along the $x$ axis
   :PROPERTIES:
   :header-args:python: :tangle plot_hydrogen.py
   :header-args:f90: :tangle plot_hydrogen.f90
   :END:
   
   The program you will write in this section will be written in
   another file (~plot_hydrogen.py~ or ~plot_hydrogen.f90~ for
   example).
   It will use the functions previously defined.

   In Python, you should put at the beginning of the file
   #+BEGIN_SRC python :results none :tangle none
#!/usr/bin/env python3

from hydrogen import e_loc
   #+END_SRC
   to be able to use the ~e_loc~ function of the ~hydrogen.py~ file.
   
   In Fortran, you will need to compile all the source files together:
   #+begin_src sh :exports both
gfortran hydrogen.f90 plot_hydrogen.f90 -o plot_hydrogen
   #+end_src

*** Exercise

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

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

 *Python*
     #+BEGIN_SRC python :results none :tangle none
#!/usr/bin/env python3

import numpy as np
import matplotlib.pyplot as plt

from hydrogen import e_loc

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

# TODO

plt.tight_layout()
plt.legend()
plt.savefig("plot_py.png")
     #+end_src

 *Fortran*
     #+begin_src f90  :tangle none
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
     #+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

     To plot the data using Gnuplot:

     #+begin_src gnuplot :file plot.png :exports code
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

**** Solution                                                      :solution2:
 *Python*
     #+BEGIN_SRC python :results none
#!/usr/bin/env python3

import numpy as np
import matplotlib.pyplot as plt

from hydrogen import e_loc

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

for a in [0.1, 0.2, 0.5, 1., 1.5, 2.]:
  y=np.array([ e_loc(a, np.array([t,0.,0.]) ) for t in x])
  plt.plot(x,y,label=f"a={a}")
  
plt.tight_layout()
plt.legend()
plt.savefig("plot_py.png")
     #+end_src

     #+RESULTS:

     [[./plot_py.png]]

 *Fortran*
     #+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

     #+begin_src sh :exports none
gfortran hydrogen.f90 plot_hydrogen.f90 -o plot_hydrogen
./plot_hydrogen > data
     #+end_src

     #+begin_src gnuplot :file plot.png :exports results
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]]

** Numerical estimation of the energy
   :PROPERTIES:
   :header-args:python: :tangle energy_hydrogen.py
   :header-args:f90: :tangle energy_hydrogen.f90
   :END:

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

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

 *Python*
     #+BEGIN_SRC python :results none :tangle none
#!/usr/bin/env python3

import numpy as np
from hydrogen import e_loc, psi

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

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

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

     #+end_src

 *Fortran*
     #+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

  do j=1,size(a)

     ! TODO

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

end program energy_hydrogen
     #+end_src

     To compile the Fortran and run it:

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

**** Solution                                                      :solution:
 *Python*
     #+BEGIN_SRC python :results none :exports both
#!/usr/bin/env python3

import numpy as np
from hydrogen import e_loc, psi

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

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

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

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

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

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

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

     #+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

 *Fortran*
     #+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

     #+begin_src sh :results output :exports results
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

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

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

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

*** Exercise (optional)
   #+begin_exercise
   Prove that :
   $$\langle \left( E - \langle E \rangle_{\Psi^2} \right)^2\rangle_{\Psi^2}  = \langle E^2 \rangle_{\Psi^2} - \langle E \rangle_{\Psi^2}^2 $$
   #+end_exercise
   
**** Solution                                                      :solution:

   $\bar{E} = \langle E \rangle$ is a constant, so $\langle \bar{E}
   \rangle = \bar{E}$ .
   
   \begin{eqnarray*}
   \langle (E - \bar{E})^2 \rangle & = & 
   \langle E^2 - 2 E \bar{E} + \bar{E}^2 \rangle \\
   &=& \langle E^2 \rangle - 2 \langle E \bar{E} \rangle + \langle \bar{E}^2 \rangle \\
   &=& \langle E^2 \rangle - 2 \langle E \rangle \bar{E}  + \bar{E}^2 \\
   &=& \langle E^2 \rangle - 2 \bar{E}^2  + \bar{E}^2 \\
   &=& \langle E^2 \rangle - \bar{E}^2 \\
   &=& \langle E^2 \rangle - \langle E \rangle^2 \\
   \end{eqnarray*}
*** Exercise
   #+begin_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 $a$.
   #+end_exercise
     
 *Python*
     #+begin_src python :results none :tangle none
#!/usr/bin/env python3

import numpy as np from hydrogen import e_loc, psi

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

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

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

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

    # TODO

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

 *Fortran*
     #+begin_src f90 :tangle none
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
     #+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
    
**** Solution                                                      :solution:
 *Python*
     #+BEGIN_SRC python :results none :exports both
#!/usr/bin/env python3

import numpy as np
from hydrogen import e_loc, psi

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

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

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

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

    for x in interval:
        r[0] = x

        for y in interval:
            r[1] = y

            for z in interval:
                r[2] = z

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

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

    E  = E  / norm
    E2 = E2 / norm

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

     #+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

 *Fortran*

    #+begin_src f90 
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
    #+end_src

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

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

* Variational Monte Carlo

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

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

** Computation of the statistical error
   :PROPERTIES:
   :header-args:python: :tangle qmc_stats.py
   :header-args:f90: :tangle qmc_stats.f90
   :END:

   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 [[https://en.wikipedia.org/wiki/Central_limit_theorem][Central Limit Theorem]].

   The estimate of the energy is

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

   The variance of the average energies can be computed as

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

   And the confidence interval is given by

   $$
   E \pm \delta E, \text{ where } \delta E = \frac{\sigma}{\sqrt{M}}
   $$
   
*** Exercise
   #+begin_exercise
   Write a function returning the average and statistical error of an
   input array.
   #+end_exercise

 *Python*
     #+BEGIN_SRC python :results none :tangle none
#!/usr/bin/env python3

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

 *Fortran*
    #+BEGIN_SRC f90 :tangle none
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
    #+END_SRC
   
**** Solution                                                      :solution:
 *Python*
     #+BEGIN_SRC python :results none :exports code
#!/usr/bin/env python3

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

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

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

    return (average, error)
     #+END_SRC

 *Fortran*
        #+BEGIN_SRC f90 :exports both
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
        #+END_SRC
   
** Uniform sampling in the box
   :PROPERTIES:
   :header-args:python: :tangle qmc_uniform.py
   :header-args:f90: :tangle qmc_uniform.f90
   :END:

   We will now perform our first Monte Carlo calculation to compute the
   energy of the hydrogen atom. 
   
   Consider again the expression of the energy
   
   \begin{eqnarray*}
   E & = & \frac{\int E_L(\mathbf{r})|\Psi(\mathbf{r})|^2\,d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}}\,. 
   \end{eqnarray*}
   
   Clearly, the square of the wave function is a good choice of probability density to sample but we will start with something simpler and rewrite the energy as 
   
   \begin{eqnarray*}
   E & = & \frac{\int E_L(\mathbf{r})\frac{|\Psi(\mathbf{r})|^2}{P(\mathbf{r})}P(\mathbf{r})\, \,d\mathbf{r}}{\int \frac{|\Psi(\mathbf{r})|^2 }{P(\mathbf{r})}P(\mathbf{r})d\mathbf{r}}\,. 
   \end{eqnarray*}
   
   Here, we will sample a uniform probability $P(\mathbf{r})$ in a cube of volume $L^3$ centered at the origin:
   
   $$ P(\mathbf{r}) = \frac{1}{L^3}\,, $$
   
   and zero outside the cube.
   
   One Monte Carlo run will consist of $N_{\rm MC}$ Monte Carlo iterations. At every Monte Carlo iteration:

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

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

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

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

 *Python*
     #+begin_note
     To draw a uniform random number in Python, you can use
     the [[https://numpy.org/doc/stable/reference/random/generated/numpy.random.uniform.html][~random.uniform~]] function of Numpy.
     #+end_note

     #+BEGIN_SRC python :tangle none :exports code
#!/usr/bin/env python3

from hydrogen  import *
from qmc_stats import *

def MonteCarlo(a, nmax):
     # TODO

a    = 1.2
nmax = 100000

#TODO

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

 *Fortran*
     #+begin_note
     To draw a uniform random number in Fortran, you can use
     the [[https://gcc.gnu.org/onlinedocs/gfortran/RANDOM_005fNUMBER.html][~RANDOM_NUMBER~]] subroutine.
     #+end_note

     #+begin_note
     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.
     #+end_note

     #+BEGIN_SRC f90 :tangle none
subroutine uniform_montecarlo(a,nmax,energy)
  implicit none
  double precision, intent(in)  :: a
  integer*8       , intent(in)  :: nmax 
  double precision, intent(out) :: energy

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

  double precision, external :: e_loc, psi

  ! TODO
end subroutine uniform_montecarlo

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

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

  !TODO

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

end program qmc
     #+END_SRC

     #+begin_src sh :results output :exports code
gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
./qmc_uniform
     #+end_src

**** Solution                                                      :solution:
 *Python*
     #+BEGIN_SRC python :results output :exports both
#!/usr/bin/env python3

from hydrogen  import *
from qmc_stats import *

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

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

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

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

     return energy / normalization

a    = 1.2
nmax = 100000

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

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

     #+RESULTS:
     : E = -0.48363807880008725 +/- 0.002330876047368999

 *Fortran*
     #+BEGIN_SRC f90 :exports code
subroutine uniform_montecarlo(a,nmax,energy)
  implicit none
  double precision, intent(in)  :: a
  integer*8       , intent(in)  :: nmax 
  double precision, intent(out) :: energy

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

  double precision, external :: e_loc, psi

  energy = 0.d0
  norm   = 0.d0

  do istep = 1,nmax

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

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

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

  end do

  energy = energy / norm

end subroutine uniform_montecarlo

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

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

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

  call ave_error(X, nruns, ave, err)

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

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

     #+RESULTS:
     :  E =  -0.48084122147238995      +/-   2.4983775878329355E-003

** Metropolis sampling with $\Psi^2$
   :PROPERTIES:
   :header-args:python: :tangle qmc_metropolis.py
   :header-args:f90: :tangle qmc_metropolis.f90
   :END:

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

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

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

   To sample a chosen probability density, an efficient method is the 
   [[https://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm][Metropolis-Hastings sampling algorithm]]. Starting from a random
   initial position $\mathbf{r}_0$, we will realize a random walk:
   
   $$ \mathbf{r}_0 \rightarrow \mathbf{r}_1 \rightarrow \mathbf{r}_2 \ldots \rightarrow \mathbf{r}_{N_{\rm MC}}\,, $$
   
   according to the following algorithm.
   
   At every step, we propose a new move according to a transition probability $T(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1})$ of our choice.
   
   For simplicity, we will move the electron in a 3-dimensional box of side $2\delta L$ centered at the current position
   of the electron:

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

   where $\delta L$ is a fixed constant, and
   $\mathbf{u}$ is a uniform random number in a 3-dimensional box
   $(-1,-1,-1) \le \mathbf{u} \le (1,1,1)$. 
   
   After having moved the electron, we add the
   accept/reject step that guarantees that the distribution of the
   $\mathbf{r}_n$ is $\Psi^2$. This amounts to accepting the move with
   probability
   
   $$
   A(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1}) = \min\left(1,\frac{T(\mathbf{r}_{n+1}\rightarrow\mathbf{r}_{n}) P(\mathbf{r}_{n+1})}{T(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1})P(\mathbf{r}_{n})}\right)\,,
   $$
   
   which, for our choice of transition probability, becomes
   
   $$
   A(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1}) = \min\left(1,\frac{P(\mathbf{r}_{n+1})}{P(\mathbf{r}_{n})}\right)= \min\left(1,\frac{|\Psi(\mathbf{r}_{n+1})|^2}{|\Psi(\mathbf{r}_{n})|^2}\right)\,.
   $$
   
   #+begin_exercise
   Explain why the transition probability cancels out in the
   expression of $A$.
   #+end_exercise
   Also note that we do not need to compute the norm of the wave function!
   
   The algorithm is summarized as follows:
   
   1) Evaluate the local energy at $\mathbf{r}_n$ and accumulate it
   2) Compute a new position $\mathbf{r'} = \mathbf{r}_n + \delta L\, \mathbf{u}$
   3) Evaluate $\Psi(\mathbf{r}')$ at the new position
   4) Compute the ratio $A = \frac{\left|\Psi(\mathbf{r'})\right|^2}{\left|\Psi(\mathbf{r}_{n})\right|^2}$
   5) Draw a uniform random number $v \in [0,1]$
   6) if $v \le A$, accept the move : set $\mathbf{r}_{n+1} = \mathbf{r'}$
   7) else, reject the move : set $\mathbf{r}_{n+1} = \mathbf{r}_n$
   
   #+begin_note
    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.
   #+end_note
   
   
*** Optimal step size
    
    If the box is infinitely small, the ratio will be very close
    to one and all the steps will be accepted. However, the moves will be 
    very correlated and you will visit the configurational space very slowly.

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

    The size of the move should be adjusted so that it is as large as
    possible, keeping the number of accepted steps not too small. To
    achieve that, we define the acceptance rate as the number of
    accepted steps over the total number of steps. Adjusting the time
    step such that the acceptance rate is close to 0.5 is a good 
    compromise for the current problem.
   
   #+begin_note
    Below, we use the symbol $\delta t$ to denote $\delta L$ since we will use
    the same variable later on to store a time step.
   #+end_note
   
   
*** Exercise
    
    #+begin_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?
    #+end_exercise

 *Python*
     #+BEGIN_SRC python :results output :tangle none
#!/usr/bin/env python3

from hydrogen  import *
from qmc_stats import *

def MonteCarlo(a,nmax,dt):

    # TODO

    return energy/nmax, N_accep/nmax


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

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

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

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

     #+RESULTS:

 *Fortran*
     #+BEGIN_SRC f90 :tangle none
subroutine metropolis_montecarlo(a,nmax,dt,energy,accep)
  implicit none
  double precision, intent(in)  :: a
  integer*8       , intent(in)  :: nmax 
  double precision, intent(in)  :: dt 
  double precision, intent(out) :: energy
  double precision, intent(out) :: accep

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

  double precision, external :: e_loc, psi, gaussian

  ! TODO

end subroutine metropolis_montecarlo

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

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

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

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

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

end program qmc
     #+END_SRC

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

**** Solution                                                      :solution:
 *Python*
     #+BEGIN_SRC python :results output :exports both
#!/usr/bin/env python3

from hydrogen  import *
from qmc_stats import *

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

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

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

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

        ratio = (psi_new / psi_old)**2

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

            r_old   = r_new
            psi_old = psi_new

    return energy/nmax, N_accep/nmax

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

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

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

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

     #+RESULTS:
     : E = -0.4802595860693983 +/- 0.0005124420418289207
     : A = 0.5074913333333334 +/- 0.000350889422714878

 *Fortran*
     #+BEGIN_SRC f90 :exports code
subroutine metropolis_montecarlo(a,nmax,dt,energy,accep)
  implicit none
  double precision, intent(in)  :: a
  integer*8       , intent(in)  :: nmax 
  double precision, intent(in)  :: dt
  double precision, intent(out) :: energy
  double precision, intent(out) :: accep

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

  double precision, external :: e_loc, psi, gaussian

  energy  = 0.d0
  n_accep = 0_8

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

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

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

     psi_new = psi(a,r_new)

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

     if (v <= ratio) then

        n_accep = n_accep + 1_8

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

     endif

  end do

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

end subroutine metropolis_montecarlo

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

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

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

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

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

end program qmc
     #+END_SRC

     #+begin_src sh :results output :exports results
gfortran hydrogen.f90 qmc_stats.f90 qmc_metropolis.f90 -o qmc_metropolis
./qmc_metropolis
     #+end_src
     #+RESULTS:
     :  E =  -0.47948142754168033      +/-   4.8410177863919307E-004
     :  A =   0.50762633333333318      +/-   3.4601756760043725E-004

** Generalized Metropolis algorithm
   :PROPERTIES:
   :header-args:python: :tangle vmc_metropolis.py
   :header-args:f90: :tangle vmc_metropolis.f90
   :END:

   One can use more efficient numerical schemes to move the electrons by choosing a smarter expression for the transition probability.
   
   The Metropolis acceptance step has to be adapted accordingly to ensure that the detailed balance condition is satisfied. This means that
   the acceptance probability $A$ is chosen so that it is consistent with the
   probability of leaving $\mathbf{r}_n$ and the probability of
   entering $\mathbf{r}_{n+1}$:

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

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

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

   so the expression of $A$ was simplified to the ratios of the squared
   wave functions.
    
   Now, if instead of drawing uniform random numbers, we
   choose to draw Gaussian random numbers with zero mean and variance
   $\delta t$, the transition probability becomes:
    
   \[
   T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1})  = 
   \frac{1}{(2\pi\,\delta t)^{3/2}} \exp \left[ - \frac{\left(
   \mathbf{r}_{n+1} - \mathbf{r}_{n} \right)^2}{2\delta t} \right]\,.
   \]


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

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

   \[
   \frac{\nabla [ \Psi^2 ]}{\Psi^2} = 2 \frac{\nabla \Psi}{\Psi}\,,
   \]
    
   and add the so-called drift vector, $\frac{\nabla \Psi}{\Psi}$, so that the numerical scheme becomes a 
   drifted diffusion with transition probability:
   
    \[
   T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1})  = 
   \frac{1}{(2\pi\,\delta t)^{3/2}} \exp \left[ - \frac{\left(
   \mathbf{r}_{n+1} - \mathbf{r}_{n} - \delta t\frac{\nabla
   \Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\delta t} \right]\,.
   \]

   The corresponding move is proposed as
   
   \[
   \mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta t\, \frac{\nabla
   \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi \,,
   \]

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

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

*** Gaussian random number generator
   
    To obtain Gaussian-distributed random numbers, you can apply the
    [[https://en.wikipedia.org/wiki/Box%E2%80%93Muller_transform][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*
    #+BEGIN_SRC f90 :tangle qmc_stats.f90
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
    #+END_SRC

    In Python, you can use the [[https://numpy.org/doc/stable/reference/random/generated/numpy.random.normal.html][~random.normal~]] function of Numpy.
   

*** Exercise 1
    
     #+begin_exercise
     If you use Fortran, copy/paste the ~random_gauss~ function in
     a Fortran file.
     #+end_exercise
     
     #+begin_exercise
     Write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$.
     #+end_exercise
   
 *Python*
     #+BEGIN_SRC python :tangle hydrogen.py :tangle none
def drift(a,r):
   # TODO
     #+END_SRC

 *Fortran*
     #+BEGIN_SRC f90 :tangle hydrogen.f90 :tangle none
subroutine drift(a,r,b)
  implicit none
  double precision, intent(in)  :: a, r(3)
  double precision, intent(out) :: b(3)

  ! TODO

end subroutine drift
     #+END_SRC

**** Solution                                                      :solution:
 *Python*
     #+BEGIN_SRC python :tangle hydrogen.py
def drift(a,r):
   ar_inv = -a/np.sqrt(np.dot(r,r))
   return r * ar_inv
     #+END_SRC

 *Fortran*
     #+BEGIN_SRC f90 :tangle hydrogen.f90
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
     #+END_SRC

*** Exercise 2

    #+begin_exercise
    Modify the previous program to introduce the drift-diffusion scheme.
    (This is a necessary step for the next section on diffusion Monte Carlo).
    #+end_exercise
   
 *Python*
     #+BEGIN_SRC python :results output :tangle none
#!/usr/bin/env python3

from hydrogen  import *
from qmc_stats import *

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

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

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

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

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

 *Fortran*
     #+BEGIN_SRC f90 :tangle none
subroutine variational_montecarlo(a,dt,nmax,energy,accep)
  implicit none
  double precision, intent(in)  :: a, dt
  integer*8       , intent(in)  :: nmax 
  double precision, intent(out) :: energy, accep

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

  double precision, external :: e_loc, psi

  ! TODO

end subroutine variational_montecarlo

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

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

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

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

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

end program qmc
     #+END_SRC

     #+begin_src sh :results output :exports code
gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
./vmc_metropolis
     #+end_src

**** Solution                                                      :solution:
 *Python*
     #+BEGIN_SRC python :results output :exports both
#!/usr/bin/env python3

from hydrogen  import *
from qmc_stats import *

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

    energy  = 0.
    N_accep = 0

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

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

        energy += e_loc(a,r_old)

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

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

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

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

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

    return energy/nmax, N_accep/nmax


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

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

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

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

     #+RESULTS:
     : E = -0.48034171558629885 +/- 0.0005286038561061781
     : A = 0.6210380000000001 +/- 0.0005457375900937905
   
 *Fortran*
     #+BEGIN_SRC f90
subroutine variational_montecarlo(a,dt,nmax,energy,accep)
  implicit none
  double precision, intent(in)  :: a, dt
  integer*8       , intent(in)  :: nmax 
  double precision, intent(out) :: energy, accep

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

  double precision, external :: e_loc, psi

  sq_dt = dsqrt(dt)

  ! Initialization
  energy  = 0.d0
  n_accep = 0_8

  call random_gauss(r_old,3)

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

  psi_old = psi(a,r_old)

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

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

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

     psi_new = psi(a,r_new)

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

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

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

     call random_number(u)

     if (u <= q) then

        n_accep = n_accep + 1_8

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

     end if

  end do

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

end subroutine variational_montecarlo

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

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

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

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

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

end program qmc
     #+END_SRC

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

     #+RESULTS:
     :  E =  -0.47940635575542706      +/-   5.5613594433433764E-004
     :  A =   0.62037333333333333      +/-   4.8970160591451110E-004
     
* Diffusion Monte Carlo

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

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

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

    where we introduced a shift in the energy, $E_{\rm ref}$, for reasons which will become apparent below.
    
    We can expand a given starting wave function, $\Psi(\mathbf{r},0)$, in the basis of the eigenstates
    of the time-independent Hamiltonian, $\Phi_k$, with energies $E_k$:

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

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

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

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

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

    where $\psi(\mathbf{r},\tau) = \Psi(\mathbf{r},-i\,\tau)$
    and
    
    \begin{eqnarray*}
    \psi(\mathbf{r},\tau) &=& \sum_k a_k \exp( -(E_k-E_{\rm ref})\, \tau) \Phi_k(\mathbf{r})\\
                          &=& \exp(-(E_0-E_{\rm ref})\, \tau)\sum_k a_k \exp( -(E_k-E_0)\, \tau) \Phi_k(\mathbf{r})\,.
    \end{eqnarray*}
    
    For large positive values of $\tau$, $\psi$ is dominated by the
    $k=0$ term, namely, the lowest eigenstate. If we adjust $E_{\rm ref}$ to the running estimate of $E_0$,
    we can expect that simulating the differetial equation in
    imaginary time will converge to the exact ground state of the
    system.

** Relation to diffusion

    The [[https://en.wikipedia.org/wiki/Diffusion_equation][diffusion equation]] of particles is given by

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

    where $D$ is the diffusion coefficient. When the imaginary-time
    Schrödinger equation is written in terms of the kinetic energy and
    potential,
    
    \[
    \frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} =
    \left(\frac{1}{2}\Delta - [V(\mathbf{r}) -E_{\rm ref}]\right) \psi(\mathbf{r}, \tau)\,, 
    \]
    
    it can be identified as the combination of:
    - a diffusion equation (Laplacian)
    - an equation whose solution is an exponential (potential)

    The diffusion equation can be simulated by a Brownian motion:
    
    \[ \mathbf{r}_{n+1} = \mathbf{r}_{n} + \sqrt{\delta t}\, \chi \]
    
    where $\chi$ is a Gaussian random variable, and the potential term 
    can be simulated by creating or destroying particles over time (a
    so-called branching process) or by simply considering it as a
    cumulative multiplicative weight along the diffusion trajectory
    (pure Diffusion Monte Carlo):

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

    
    We note that the ground-state wave function of a Fermionic system is
    antisymmetric and changes sign. Therefore, its interpretation as a probability 
    distribution is somewhat problematic. In fact, mathematically, since
    the Bosonic ground state is lower in energy than the Fermionic one, for
    large $\tau$, the system will evolve towards the Bosonic solution.
    
    For the systems you will study, this is not an issue:
    
    - Hydrogen atom: You only have one electron! 
    - Two-electron system ($H_2$ or He): The ground-wave function is
      antisymmetric in the spin variables but symmetric in the space ones.
    
    Therefore, in both cases, you are dealing with a "Bosonic" ground state.
    
** Importance sampling
   
    In a molecular system, the potential is far from being constant
    and, in fact, diverges at the inter-particle coalescence points. Hence,
    it results in very large fluctuations of the erm weight associated with
    the potental, making the calculations impossible in practice.
    Fortunately, if we multiply the Schrödinger equation by a chosen
 /trial wave function/ $\Psi_T(\mathbf{r})$ (Hartree-Fock, Kohn-Sham
    determinant, CI wave function, /etc/), one obtains

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

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

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

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

  This equation generates the /N/-electron density $\Pi$, which is the
  product of the ground state solution with the trial wave
  function. You may then ask: how can we compute the total energy of
  the system?
  
  To this aim, we use the /mixed estimator/ of the energy:
  
  \begin{eqnarray*}
   E(\tau)  &=&  \frac{\langle \psi(\tau) | \hat{H} | \Psi_T \rangle}{\langle \psi(\tau) | \Psi_T \rangle}\\
            &=& \frac{\int \psi(\mathbf{r},\tau) \hat{H} \Psi_T(\mathbf{r}) d\mathbf{r}}
                {\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}} \\
            &=& \frac{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) E_L(\mathbf{r}) d\mathbf{r}}
                {\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}} \,.
   \end{eqnarray*}
  
   For large $\tau$, we have that 
   
   \[ 
   \Pi(\mathbf{r},\tau) =\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \rightarrow \Phi_0(\mathbf{r}) \Psi_T(\mathbf{r})\,,
   \]
   
   and, using that $\hat{H}$ is Hermitian and that $\Phi_0$ is an
   eigenstate of the Hamiltonian, we obtain for large $\tau$

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

   Therefore, we can compute the energy within DMC by generating the
   density $\Pi$ with random walks, and simply averaging the local
   energies computed with the trial wave function.
  
*** Appendix : Details of the Derivation
    
  \[
    -\frac{\partial \psi(\mathbf{r},\tau)}{\partial \tau} \Psi_T(\mathbf{r}) =
    \left[ -\frac{1}{2} \Delta \psi(\mathbf{r},\tau) + V(\mathbf{r}) \psi(\mathbf{r},\tau) \right] \Psi_T(\mathbf{r}) 
  \]

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

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

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

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

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

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

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

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

   The PDMC algorithm is less stable than the DMC algorithm: it 
   requires to have a value of $E_\text{ref}$ which is close to the
   fixed-node energy, and a good trial wave function. Moreover, we
   can't let $\tau$ become too large because the weight whether
   explode or vanish: we need to have a fixed value of $\tau$
   (projection time).
   The big advantage of PDMC is that it is rather simple to implement
   starting from a VMC code:
   
   0) Start with $W(\mathbf{r}_0)=1, \tau_0 = 0$
   1) Evaluate the local energy at $\mathbf{r}_{n}$
   2) Compute the contribution to the weight $w(\mathbf{r}_n) =
      \exp(-\delta t(E_L(\mathbf{r}_n)-E_\text{ref}))$
   3) Update $W(\mathbf{r}_{n}) = W(\mathbf{r}_{n-1}) \times w(\mathbf{r}_n)$
   4) Accumulate the weighted energy $W(\mathbf{r}_n) \times
      E_L(\mathbf{r}_n)$,
      and the weight $W(\mathbf{r}_n)$ for the normalization
   5) Update $\tau_n = \tau_{n-1} + \delta t$
   6) If $\tau_{n} > \tau_\text{max}$, the long projection time has
      been reached and we can start an new trajectory from the current
      position: reset $W(r_n) = 1$ and $\tau_n
      = 0$
   7) Compute a new position $\mathbf{r'} = \mathbf{r}_n +
      \delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi$
   8) Evaluate $\Psi(\mathbf{r}')$ and $\frac{\nabla \Psi(\mathbf{r'})}{\Psi(\mathbf{r'})}$ at the new position
   9) Compute the ratio $A = \frac{T(\mathbf{r}' \rightarrow \mathbf{r}_{n}) P(\mathbf{r}')}{T(\mathbf{r}_{n} \rightarrow \mathbf{r}') P(\mathbf{r}_{n})}$
  10) Draw a uniform random number $v \in [0,1]$
  11) if $v \le A$, accept the move : set $\mathbf{r}_{n+1} = \mathbf{r'}$
  12) else, reject the move : set $\mathbf{r}_{n+1} = \mathbf{r}_n$

   
   Some comments are needed:
   
   - You estimate the energy as 
   
     \begin{eqnarray*}
     E = \frac{\sum_{k=1}^{N_{\rm MC}} E_L(\mathbf{r}_k) W(\mathbf{r}_k, k\delta t)}{\sum_{k=1}^{N_{\rm MC}} W(\mathbf{r}_k, k\delta t)}
     \end{eqnarray*}
   
   - The result will be affected by a time-step error
     (the finite size of $\delta t$) due to the discretization of the
     integral, and one has in principle to extrapolate to the limit
     $\delta t \rightarrow 0$. This amounts to fitting the energy
     computed for multiple values of $\delta t$.
   
     Here, you will be using a small enough time-step and you should not worry about the extrapolation.
   - The accept/reject step (steps 9-12 in the algorithm) is in principle not needed for the correctness of 
     the DMC algorithm. However, its use reduces significantly the time-step error.
   
   
** Hydrogen atom
  :PROPERTIES:
  :header-args:python: :tangle pdmc.py
  :header-args:f90: :tangle pdmc.f90
  :END:
   
*** Exercise 

     #+begin_exercise
     Modify the Metropolis VMC program into a PDMC program.
     In the limit $\delta t \rightarrow 0$, you should recover the exact
     energy of H for any value of $a$, as long as the simulation is stable. 
     We choose here a time step of 0.05 a.u. and a fixed projection
     time $\tau$ =100 a.u. 
     #+end_exercise
   
 *Python*
         #+BEGIN_SRC python :results output
from hydrogen  import *
from qmc_stats import *

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

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

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

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

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

         #+RESULTS:

 *Fortran*
     #+BEGIN_SRC f90 :tangle none
subroutine pdmc(a, dt, nmax, energy, accep, tau, E_ref)
  implicit none
  double precision, intent(in)  :: a, dt, tau
  integer*8       , intent(in)  :: nmax 
  double precision, intent(out) :: energy, accep
  double precision, intent(in)  :: E_ref

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

  double precision, external :: e_loc, psi

  ! TODO

end subroutine pdmc

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

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

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

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

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

end program qmc
     #+END_SRC

     #+begin_src sh :results output :exports code
gfortran hydrogen.f90 qmc_stats.f90 pdmc.f90 -o pdmc
./pdmc
     #+end_src

**** Solution                                                      :solution:

 *Python*
     #+BEGIN_SRC python :results output
#!/usr/bin/env python3

from hydrogen  import *
from qmc_stats import *

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

    energy        = 0.
    N_accep       = 0
    normalization = 0.

    w           = 1.
    tau_current = 0.

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

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

        normalization += w
        energy        += w * el

        tau_current += dt

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

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

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

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

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

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

    return energy/normalization, N_accep/nmax


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

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

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

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

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

 *Fortran* 
     #+BEGIN_SRC f90
subroutine pdmc(a, dt, nmax, energy, accep, tau, E_ref)
  implicit none
  double precision, intent(in)  :: a, dt, tau
  integer*8       , intent(in)  :: nmax 
  double precision, intent(out) :: energy, accep
  double precision, intent(in)  :: E_ref

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

  double precision, external :: e_loc, psi

  sq_dt = dsqrt(dt)

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

  w           = 1.d0
  tau_current = 0.d0

  call random_gauss(r_old,3)

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

  psi_old = psi(a,r_old)

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

     norm   = norm + w
     energy = energy + w*e
     
     tau_current = tau_current + dt

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

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

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

     psi_new = psi(a,r_new)

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

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

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

     call random_number(u)

     if (u <= q) then

        n_accep = n_accep + 1_8

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

     end if

  end do

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

end subroutine pdmc

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

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

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

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

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

end program qmc
     #+END_SRC
   
     #+begin_src sh :results output :exports results
gfortran hydrogen.f90 qmc_stats.f90 pdmc.f90 -o pdmc
./pdmc
     #+end_src

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


   
* Project

  Change your PDMC code for one of the following:
  - the Helium atom, or
  - the H_2 molecule at $R$ =1.401 bohr.
    
  And compute the ground state energy.
  
      
* Schedule                                                         :noexport:

 |------------------------------+---------|
 | <2021-02-04 Thu 09:00-10:30> | Lecture |
 |------------------------------+---------|
 | <2021-02-04 Thu 11:00-11:20> |     2.1 |
 | <2021-02-04 Thu 11:20-11:40> |     2.2 |
 | <2021-02-04 Thu 11:40-12:15> |     2.3 |
 | <2021-02-04 Thu 12:15-12:30> |     2.4 |
 |------------------------------+---------|
 | <2021-02-04 Thu 14:00-14:10> |     3.1 |
 | <2021-02-04 Thu 14:10-14:30> |     3.2 |
 | <2021-02-04 Thu 14:30-15:30> |     3.3 |
 | <2021-02-04 Thu 15:30-16:30> |     3.4 |
 | <2021-02-04 Thu 16:30-18:30> |     4.5 |
 |------------------------------+---------|

* Acknowledgments

[[https://trex-coe.eu/sites/default/files/inline-images/euflag.jpg]]

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