mirror of
https://github.com/TREX-CoE/qmc-lttc.git
synced 2024-10-19 06:21:56 +02:00
3108 lines
88 KiB
Org Mode
3108 lines
88 KiB
Org Mode
#+TITLE: Quantum Monte Carlo
|
|
#+AUTHOR: Anthony Scemama, Claudia Filippi
|
|
#+LANGUAGE: en
|
|
#+INFOJS_OPT: toc:t mouse:underline path:org-info.js
|
|
#+STARTUP: latexpreview
|
|
#+LATEX_CLASS: report
|
|
#+LATEX_HEADER_EXTRA: \usepackage{minted}
|
|
#+HTML_HEAD: <link rel="stylesheet" title="Standard" href="worg.css" type="text/css" />
|
|
|
|
#+OPTIONS: H:4 num:t toc:t \n:nil @:t ::t |:t ^:t -:t f:t *:t <:t
|
|
#+OPTIONS: TeX:t LaTeX:t skip:nil d:nil todo:t pri:nil tags:not-in-toc
|
|
# EXCLUDE_TAGS: solution
|
|
|
|
#+BEGIN_SRC elisp :output none :exports none
|
|
(setq org-latex-listings 'minted
|
|
org-latex-packages-alist '(("" "minted"))
|
|
org-latex-pdf-process
|
|
'("pdflatex -shell-escape -interaction nonstopmode -output-directory %o %f"
|
|
"pdflatex -shell-escape -interaction nonstopmode -output-directory %o %f"
|
|
"pdflatex -shell-escape -interaction nonstopmode -output-directory %o %f"))
|
|
(setq org-latex-minted-options '(("breaklines" "true")
|
|
("breakanywhere" "true")))
|
|
(setq org-latex-minted-options
|
|
'(("frame" "lines")
|
|
("fontsize" "\\scriptsize")
|
|
("linenos" "")))
|
|
(org-beamer-export-to-pdf)
|
|
|
|
#+END_SRC
|
|
|
|
#+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 :solution: *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 :solution: *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 :solution: *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 :solution: *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 :solution:
|
|
|
|
\begin{eqnarray*}
|
|
E &=& \frac{\hat{H} \Psi}{\Psi} = - \frac{1}{2} \frac{\Delta \Psi}{\Psi} -
|
|
\frac{1}{|\mathbf{r}|} \\
|
|
&=& -\frac{1}{2}\left(a^2 - \frac{2a}{\mathbf{|r|}} \right) -
|
|
\frac{1}{|\mathbf{r}|} \\
|
|
&=&
|
|
-\frac{1}{2} a^2 + \frac{a-1}{\mathbf{|r|}}
|
|
\end{eqnarray*}
|
|
|
|
$a=1$ cancels the $1/|r|$ term, and makes the energy constant and
|
|
equal to -0.5 atomic units.
|
|
|
|
** Plot of the local energy along the $x$ axis
|
|
: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 :solution: *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 :solution:
|
|
|
|
|
|
** 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.
|
|
|
|
** Diffusion and branching
|
|
|
|
The imaginary-time Schrödinger equation can be explicitly written in terms of the kinetic and
|
|
potential energies as
|
|
|
|
\[
|
|
\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = \left(\frac{1}{2}\Delta - [V(\mathbf{r}) -E_{\rm ref}]\right) \psi(\mathbf{r}, \tau)\,.
|
|
\]
|
|
|
|
We can simulate this differential equation as a diffusion-branching process.
|
|
|
|
|
|
To see this, recall that the [[https://en.wikipedia.org/wiki/Diffusion_equation][diffusion equation]] of particles is given by
|
|
|
|
\[
|
|
\frac{\partial \psi(\mathbf{r},t)}{\partial t} = D\, \Delta \psi(\mathbf{r},t).
|
|
\]
|
|
|
|
Furthermore, the [[https://en.wikipedia.org/wiki/Reaction_rate][rate of reaction]] $v$ is the speed at which a chemical reaction
|
|
takes place. In a solution, the rate is given as a function of the
|
|
concentration $[A]$ by
|
|
|
|
\[
|
|
v = \frac{d[A]}{dt},
|
|
\]
|
|
|
|
where the concentration $[A]$ is proportional to the number of particles.
|
|
|
|
These two equations allow us to interpret the Schrödinger equation
|
|
in imaginary time as the combination of:
|
|
- a diffusion equation with a diffusion coefficient $D=1/2$ for the
|
|
kinetic energy, and
|
|
- a rate equation for the potential.
|
|
|
|
The diffusion equation can be simulated by a Brownian motion:
|
|
|
|
\[ \mathbf{r}_{n+1} = \mathbf{r}_{n} + \sqrt{\delta t}\, \chi \]
|
|
|
|
where $\chi$ is a Gaussian random variable, and the rate equation
|
|
can be simulated by creating or destroying particles over time (a
|
|
so-called branching process).
|
|
|
|
In /Diffusion Monte Carlo/ (DMC), one onbtains the ground state of a
|
|
system by simulating the Schrödinger equation in imaginary time via
|
|
the combination of a diffusion process and a branching process.
|
|
|
|
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, when the
|
|
rate equation is simulated, it results in very large fluctuations
|
|
in the numbers of particles, making the calculations impossible in
|
|
practice.
|
|
Fortunately, if we multiply the Schrödinger equation by a chosen
|
|
/trial wave function/ $\Psi_T(\mathbf{r})$ (Hartree-Fock, Kohn-Sham
|
|
determinant, CI wave function, /etc/), one obtains
|
|
|
|
\[
|
|
-\frac{\partial \psi(\mathbf{r},\tau)}{\partial \tau} \Psi_T(\mathbf{r}) =
|
|
\left[ -\frac{1}{2} \Delta \psi(\mathbf{r},\tau) + V(\mathbf{r}) \psi(\mathbf{r},\tau) \right] \Psi_T(\mathbf{r})
|
|
\]
|
|
|
|
Defining $\Pi(\mathbf{r},\tau) = \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})$, (see appendix for details)
|
|
|
|
\[
|
|
-\frac{\partial \Pi(\mathbf{r},\tau)}{\partial \tau}
|
|
= -\frac{1}{2} \Delta \Pi(\mathbf{r},\tau) +
|
|
\nabla \left[ \Pi(\mathbf{r},\tau) \frac{\nabla \Psi_T(\mathbf{r})}{\Psi_T(\mathbf{r})}
|
|
\right] + (E_L(\mathbf{r})-E_{\rm ref})\Pi(\mathbf{r},\tau)
|
|
\]
|
|
|
|
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
|
|
changing the number of particles according to $\exp\left[ -\delta t\,
|
|
\left(E_L(\mathbf{r}) - E_{\rm ref}\right)\right]$
|
|
where $E_{\rm ref}$ is the constant we had introduced above, which is adjusted to
|
|
the running average energy to keep the number of particles
|
|
reasonably constant.
|
|
|
|
This equation generates the /N/-electron density $\Pi$, which is the
|
|
product of the ground state 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 (PDMC)
|
|
|
|
Instead of having a variable number of particles to simulate the
|
|
branching process, one can consider the term
|
|
$\exp \left( -\delta t\,( E_L(\mathbf{r}) - E_{\rm ref}) \right)$ as a
|
|
cumulative product of weights:
|
|
|
|
\begin{eqnarray*}
|
|
W(\mathbf{r}_n, \tau)
|
|
& = & \exp \left( \int_0^\tau - (E_L(\mathbf{r}_t) - E_{\text{ref}}) dt \right) \\
|
|
& \approx & \prod_{i=1}^{n} \exp \left( -\delta t\,
|
|
(E_L(\mathbf{r}_i) - E_{\text{ref}}) \right) =
|
|
\prod_{i=1}^{n} w(\mathbf{r}_i)
|
|
\end{eqnarray*}
|
|
|
|
where $\mathbf{r}_i$ are the coordinates along the trajectory and we introduced a time-step $\delta t$.
|
|
|
|
The algorithm can be rather easily built on top of your VMC code:
|
|
|
|
1) Compute a new position $\mathbf{r'} = \mathbf{r}_n +
|
|
\delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi$
|
|
|
|
Evaluate $\Psi$ and $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$ at the new position
|
|
2) Compute the ratio $A = \frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})}{T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})}$
|
|
3) Draw a uniform random number $v \in [0,1]$
|
|
4) if $v \le A$, accept the move : set $\mathbf{r}_{n+1} = \mathbf{r'}$
|
|
5) else, reject the move : set $\mathbf{r}_{n+1} = \mathbf{r}_n$
|
|
6) evaluate the local energy at $\mathbf{r}_{n+1}$
|
|
7) compute the weight $w(\mathbf{r}_i)$
|
|
8) update $W$
|
|
|
|
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 $\delta t$) 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 2-5 in the algorithm) is in principle not needed for the correctness of
|
|
the DMC algorithm. However, its use reduces significantly the time-step error.
|
|
|
|
The PDMC algorithm is less stable than the branching algorithm: it
|
|
requires to have a value of $E_\text{ref}$ which is close to the
|
|
fixed-node energy, and a good trial wave function. Its big
|
|
advantage is that it is very easy to program starting from a VMC
|
|
code, so this is what we will do in the next section.
|
|
|
|
** Hydrogen atom
|
|
:PROPERTIES:
|
|
:header-args:python: :tangle pdmc.py
|
|
:header-args:f90: :tangle pdmc.f90
|
|
:END:
|
|
|
|
*** Exercise
|
|
|
|
#+begin_exercise
|
|
Modify the Metropolis VMC program to introduce the PDMC weight.
|
|
In the limit $\delta t \rightarrow 0$, you should recover the exact
|
|
energy of H for any value of $a$.
|
|
#+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 = 0.01
|
|
E_ref = -0.5
|
|
|
|
X0 = [ MonteCarlo(a, nmax, dt, E_ref) for i in range(30)]
|
|
|
|
# Energy
|
|
X = [ x for (x, _) in X0 ]
|
|
E, deltaE = ave_error(X)
|
|
print(f"E = {E} +/- {deltaE}")
|
|
|
|
# Acceptance rate
|
|
X = [ x for (_, x) in X0 ]
|
|
A, deltaA = ave_error(X)
|
|
print(f"A = {A} +/- {deltaA}")
|
|
#+END_SRC *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 = 0.1d0
|
|
double precision, parameter :: E_ref = -0.5d0
|
|
double precision, parameter :: tau = 10.d0
|
|
integer*8 , parameter :: nmax = 100000
|
|
integer , parameter :: nruns = 30
|
|
|
|
integer :: irun
|
|
double precision :: X(nruns), accep(nruns)
|
|
double precision :: ave, err
|
|
|
|
do irun=1,nruns
|
|
call pdmc(a, dt, nmax, X(irun), accep(irun), tau, E_ref)
|
|
enddo
|
|
|
|
call ave_error(X,nruns,ave,err)
|
|
print *, 'E = ', ave, '+/-', err
|
|
|
|
call ave_error(accep,nruns,ave,err)
|
|
print *, 'A = ', ave, '+/-', err
|
|
|
|
end program qmc
|
|
#+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 = 0.1
|
|
tau = 10.
|
|
E_ref = -0.5
|
|
|
|
X0 = [ MonteCarlo(a, nmax, dt, tau, E_ref) for i in range(30)]
|
|
|
|
# Energy
|
|
X = [ x for (x, _) in X0 ]
|
|
E, deltaE = ave_error(X)
|
|
print(f"E = {E} +/- {deltaE}")
|
|
|
|
# Acceptance rate
|
|
X = [ x for (_, x) in X0 ]
|
|
A, deltaA = ave_error(X)
|
|
print(f"A = {A} +/- {deltaA}")
|
|
#+END_SRC
|
|
|
|
#+RESULTS:
|
|
: E = -0.5006126340155459 +/- 0.00042555955652480285
|
|
: A = 0.9878509999999998 +/- 6.920791596475006e-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))
|
|
|
|
energy = energy + w*e
|
|
norm = norm + w
|
|
|
|
tau_current = tau_current + dt
|
|
|
|
! Reset when tau is reached
|
|
if (tau_current >= tau) then
|
|
w = 1.d0
|
|
tau_current = 0.d0
|
|
endif
|
|
|
|
call random_gauss(chi,3)
|
|
r_new(:) = r_old(:) + dt*d_old(:) + chi(:)*sq_dt
|
|
|
|
call drift(a,r_new,d_new)
|
|
d2_new = d_new(1)*d_new(1) + &
|
|
d_new(2)*d_new(2) + &
|
|
d_new(3)*d_new(3)
|
|
|
|
psi_new = psi(a,r_new)
|
|
|
|
! Metropolis
|
|
prod = (d_new(1) + d_old(1))*(r_new(1) - r_old(1)) + &
|
|
(d_new(2) + d_old(2))*(r_new(2) - r_old(2)) + &
|
|
(d_new(3) + d_old(3))*(r_new(3) - r_old(3))
|
|
|
|
argexpo = 0.5d0 * (d2_new - d2_old)*dt + prod
|
|
|
|
q = psi_new / psi_old
|
|
q = dexp(-argexpo) * q*q
|
|
|
|
call random_number(u)
|
|
|
|
if (u <= q) then
|
|
|
|
n_accep = n_accep + 1_8
|
|
|
|
r_old(:) = r_new(:)
|
|
d_old(:) = d_new(:)
|
|
d2_old = d2_new
|
|
psi_old = psi_new
|
|
|
|
end if
|
|
|
|
end do
|
|
|
|
energy = energy / norm
|
|
accep = dble(n_accep) / dble(nmax)
|
|
|
|
end subroutine pdmc
|
|
|
|
program qmc
|
|
implicit none
|
|
double precision, parameter :: a = 1.2d0
|
|
double precision, parameter :: dt = 0.1d0
|
|
double precision, parameter :: E_ref = -0.5d0
|
|
double precision, parameter :: tau = 10.d0
|
|
integer*8 , parameter :: nmax = 100000
|
|
integer , parameter :: nruns = 30
|
|
|
|
integer :: irun
|
|
double precision :: X(nruns), accep(nruns)
|
|
double precision :: ave, err
|
|
|
|
do irun=1,nruns
|
|
call pdmc(a, dt, nmax, X(irun), accep(irun), tau, E_ref)
|
|
enddo
|
|
|
|
call ave_error(X,nruns,ave,err)
|
|
print *, 'E = ', ave, '+/-', err
|
|
|
|
call ave_error(accep,nruns,ave,err)
|
|
print *, 'A = ', ave, '+/-', err
|
|
|
|
end program qmc
|
|
#+END_SRC
|
|
|
|
#+begin_src sh :results output :exports results
|
|
gfortran hydrogen.f90 qmc_stats.f90 pdmc.f90 -o pdmc
|
|
./pdmc
|
|
#+end_src
|
|
|
|
#+RESULTS:
|
|
: E = -0.50067519934141380 +/- 7.9390940184720371E-004
|
|
: A = 0.98788066666666663 +/- 7.2889356133441110E-005
|
|
|
|
|
|
** TODO H_2
|
|
|
|
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.
|
|
|
|
|
|
* Old sections to be removed :noexport:
|
|
|
|
:PROPERTIES:
|
|
:header-args:python: :tangle none
|
|
:header-args:f90: :tangle none
|
|
:END:
|
|
|
|
** Gaussian sampling :noexport:
|
|
:PROPERTIES:
|
|
:header-args:python: :tangle qmc_gaussian.py
|
|
:header-args:f90: :tangle qmc_gaussian.f90
|
|
:END:
|
|
|
|
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 sampling probability can be inserted into the equation of the energy:
|
|
|
|
\[
|
|
E = \frac{\int P(\mathbf{r})
|
|
\frac{\left|\Psi(\mathbf{r})\right|^2}{P(\mathbf{r})}\,
|
|
\frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int P(\mathbf{r}) \frac{\left|\Psi(\mathbf{r}) \right|^2}{P(\mathbf{r})} d\mathbf{r}}
|
|
\]
|
|
|
|
with the Gaussian probability
|
|
|
|
\[
|
|
P(\mathbf{r}) = \frac{1}{(2 \pi)^{3/2}}\exp\left( -\frac{\mathbf{r}^2}{2} \right).
|
|
\]
|
|
|
|
As the coordinates are drawn with probability $P(\mathbf{r})$, the
|
|
average energy can be computed as
|
|
|
|
$$
|
|
E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
|
|
w_i = \frac{\left|\Psi(\mathbf{r}_i)\right|^2}{P(\mathbf{r}_i)} \delta \mathbf{r}
|
|
$$
|
|
|
|
|
|
*** Exercise
|
|
|
|
#+begin_exercise
|
|
Modify the program of the previous section to sample with
|
|
Gaussian-distributed random numbers. Can you see an reduction in
|
|
the statistical error?
|
|
#+end_exercise
|
|
|
|
**** Python
|
|
#+BEGIN_SRC python :results output
|
|
#!/usr/bin/env python3
|
|
|
|
from hydrogen import *
|
|
from qmc_stats import *
|
|
|
|
norm_gauss = 1./(2.*np.pi)**(1.5)
|
|
def gaussian(r):
|
|
return norm_gauss * np.exp(-np.dot(r,r)*0.5)
|
|
|
|
def MonteCarlo(a,nmax):
|
|
E = 0.
|
|
N = 0.
|
|
for istep in range(nmax):
|
|
r = np.random.normal(loc=0., scale=1.0, size=(3))
|
|
w = psi(a,r)
|
|
w = w*w / gaussian(r)
|
|
N += w
|
|
E += w * e_loc(a,r)
|
|
return E/N
|
|
|
|
a = 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 = -0.49511014287471955 +/- 0.00012402022172236656
|
|
|
|
**** Fortran
|
|
#+BEGIN_SRC f90
|
|
double precision function gaussian(r)
|
|
implicit none
|
|
double precision, intent(in) :: r(3)
|
|
double precision, parameter :: norm_gauss = 1.d0/(2.d0*dacos(-1.d0))**(1.5d0)
|
|
gaussian = norm_gauss * dexp( -0.5d0 * (r(1)*r(1) + r(2)*r(2) + r(3)*r(3) ))
|
|
end function gaussian
|
|
|
|
|
|
subroutine gaussian_montecarlo(a,nmax,energy)
|
|
implicit none
|
|
double precision, intent(in) :: a
|
|
integer*8 , intent(in) :: nmax
|
|
double precision, intent(out) :: energy
|
|
|
|
integer*8 :: istep
|
|
|
|
double precision :: norm, r(3), w
|
|
|
|
double precision, external :: e_loc, psi, gaussian
|
|
|
|
energy = 0.d0
|
|
norm = 0.d0
|
|
do istep = 1,nmax
|
|
call random_gauss(r,3)
|
|
w = psi(a,r)
|
|
w = w*w / gaussian(r)
|
|
norm = norm + w
|
|
energy = energy + w * e_loc(a,r)
|
|
end do
|
|
energy = energy / norm
|
|
end subroutine gaussian_montecarlo
|
|
|
|
program qmc
|
|
implicit none
|
|
double precision, parameter :: a = 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 gaussian_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 both
|
|
gfortran hydrogen.f90 qmc_stats.f90 qmc_gaussian.f90 -o qmc_gaussian
|
|
./qmc_gaussian
|
|
#+end_src
|
|
|
|
#+RESULTS:
|
|
: E = -0.49517104619091717 +/- 1.0685523607878961E-004
|
|
|
|
** Improved sampling with $\Psi^2$ :noexport:
|
|
|
|
*** Importance sampling
|
|
:PROPERTIES:
|
|
:header-args:python: :tangle vmc.py
|
|
:header-args:f90: :tangle vmc.f90
|
|
:END:
|
|
|
|
To generate the probability density $\Psi^2$, we consider a
|
|
diffusion process characterized by a time-dependent density
|
|
$|\Psi(\mathbf{r},t)|^2$, which obeys the Fokker-Planck equation:
|
|
|
|
\[
|
|
\frac{\partial \Psi^2}{\partial t} = \sum_i D
|
|
\frac{\partial}{\partial \mathbf{r}_i} \left(
|
|
\frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right) |\Psi(\mathbf{r},t)|^2.
|
|
\]
|
|
|
|
$D$ is the diffusion constant and $F_i$ is the i-th component of a
|
|
drift velocity caused by an external potential. For a stationary
|
|
density, \( \frac{\partial \Psi^2}{\partial t} = 0 \), so
|
|
|
|
\begin{eqnarray*}
|
|
0 & = & \sum_i D
|
|
\frac{\partial}{\partial \mathbf{r}_i} \left(
|
|
\frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right) |\Psi(\mathbf{r})|^2 \\
|
|
0 & = & \sum_i D
|
|
\frac{\partial}{\partial \mathbf{r}_i} \left(
|
|
\frac{\partial |\Psi(\mathbf{r})|^2}{\partial \mathbf{r}_i} -
|
|
F_i(\mathbf{r})\,|\Psi(\mathbf{r})|^2 \right) \\
|
|
0 & = &
|
|
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} -
|
|
\frac{\partial F_i }{\partial \mathbf{r}_i}|\Psi(\mathbf{r})|^2 -
|
|
\frac{\partial \Psi^2}{\partial \mathbf{r}_i} F_i(\mathbf{r})
|
|
\end{eqnarray*}
|
|
|
|
we search for a drift function which satisfies
|
|
|
|
\[
|
|
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} = |\Psi(\mathbf{r})|^2 \frac{\partial F_i }{\partial \mathbf{r}_i} +
|
|
\frac{\partial \Psi^2}{\partial \mathbf{r}_i} F_i(\mathbf{r})
|
|
\]
|
|
|
|
to obtain a second derivative on the left, we need the drift to be
|
|
of the form
|
|
\[
|
|
F_i(\mathbf{r}) = g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i}
|
|
\]
|
|
|
|
\[
|
|
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2}
|
|
= |\Psi(\mathbf{r})|^2 \frac{\partial g(\mathbf{r})}{\partial
|
|
\mathbf{r}_i}\frac{\partial \Psi^2}{\partial
|
|
\mathbf{r}_i} + |\Psi(\mathbf{r})|^2 g(\mathbf{r})
|
|
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} + \frac{\partial
|
|
\Psi^2}{\partial \mathbf{r}_i} g(\mathbf{r}) \frac{\partial
|
|
\Psi^2}{\partial \mathbf{r}_i}
|
|
\]
|
|
|
|
$g = 1 / \Psi^2$ satisfies this equation, so
|
|
|
|
\[
|
|
F(\mathbf{r}) = \frac{\nabla |\Psi(\mathbf{r})|^2}{|\Psi(\mathbf{r})|^2} = 2 \frac{\nabla
|
|
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} = 2 \nabla \left( \log \Psi(\mathbf{r}) \right)
|
|
\]
|
|
|
|
In statistical mechanics, Fokker-Planck trajectories are generated
|
|
by a Langevin equation:
|
|
|
|
\[
|
|
\frac{\partial \mathbf{r}(t)}{\partial t} = 2D \frac{\nabla
|
|
\Psi(\mathbf{r}(t))}{\Psi} + \eta
|
|
\]
|
|
|
|
where $\eta$ is a normally-distributed fluctuating random force.
|
|
|
|
Discretizing this differential equation gives the following drifted
|
|
diffusion scheme:
|
|
|
|
\[
|
|
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta t\, 2D \frac{\nabla
|
|
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi
|
|
\]
|
|
where $\chi$ is a Gaussian random variable with zero mean and
|
|
variance $\delta t\,2D$.
|
|
|
|
**** Exercise 2
|
|
|
|
#+begin_exercise
|
|
Sample $\Psi^2$ approximately using the drifted diffusion scheme,
|
|
with a diffusion constant $D=1/2$. You can use a time step of
|
|
0.001 a.u.
|
|
#+end_exercise *Python*
|
|
#+BEGIN_SRC python :results output
|
|
#!/usr/bin/env python3
|
|
|
|
from hydrogen import *
|
|
from qmc_stats import *
|
|
|
|
def MonteCarlo(a,dt,nmax):
|
|
sq_dt = np.sqrt(dt)
|
|
|
|
# Initialization
|
|
E = 0.
|
|
N = 0.
|
|
r_old = np.random.normal(loc=0., scale=1.0, size=(3))
|
|
|
|
for istep in range(nmax):
|
|
d_old = drift(a,r_old)
|
|
chi = np.random.normal(loc=0., scale=1.0, size=(3))
|
|
r_new = r_old + dt * d_old + chi*sq_dt
|
|
N += 1.
|
|
E += e_loc(a,r_new)
|
|
r_old = r_new
|
|
return E/N
|
|
|
|
|
|
a = 1.2
|
|
nmax = 100000
|
|
dt = 0.2
|
|
X = [MonteCarlo(a,dt,nmax) for i in range(30)]
|
|
E, deltaE = ave_error(X)
|
|
print(f"E = {E} +/- {deltaE}")
|
|
#+END_SRC
|
|
|
|
#+RESULTS:
|
|
: E = -0.4858534479298907 +/- 0.00010203236131158794
|
|
|
|
*Fortran*
|
|
#+BEGIN_SRC f90
|
|
subroutine variational_montecarlo(a,dt,nmax,energy)
|
|
implicit none
|
|
double precision, intent(in) :: a, dt
|
|
integer*8 , intent(in) :: nmax
|
|
double precision, intent(out) :: energy
|
|
|
|
integer*8 :: istep
|
|
double precision :: norm, r_old(3), r_new(3), d_old(3), sq_dt, chi(3)
|
|
double precision, external :: e_loc
|
|
|
|
sq_dt = dsqrt(dt)
|
|
|
|
! Initialization
|
|
energy = 0.d0
|
|
norm = 0.d0
|
|
call random_gauss(r_old,3)
|
|
|
|
do istep = 1,nmax
|
|
call drift(a,r_old,d_old)
|
|
call random_gauss(chi,3)
|
|
r_new(:) = r_old(:) + dt * d_old(:) + chi(:)*sq_dt
|
|
norm = norm + 1.d0
|
|
energy = energy + e_loc(a,r_new)
|
|
r_old(:) = r_new(:)
|
|
end do
|
|
energy = energy / norm
|
|
end subroutine variational_montecarlo
|
|
|
|
program qmc
|
|
implicit none
|
|
double precision, parameter :: a = 1.2d0
|
|
double precision, parameter :: dt = 0.2d0
|
|
integer*8 , parameter :: nmax = 100000
|
|
integer , parameter :: nruns = 30
|
|
|
|
integer :: irun
|
|
double precision :: X(nruns)
|
|
double precision :: ave, err
|
|
|
|
do irun=1,nruns
|
|
call variational_montecarlo(a,dt,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 both
|
|
gfortran hydrogen.f90 qmc_stats.f90 vmc.f90 -o vmc
|
|
./vmc
|
|
#+end_src
|
|
|
|
#+RESULTS:
|
|
: E = -0.48584030499187431 +/- 1.0411743995438257E-004
|
|
|
|
|
|
* TODO [0/3] Last things to do
|
|
|
|
- [ ] Give some hints of how much time is required for each section
|
|
- [ ] Prepare 4 questions for the exam: multiple-choice questions
|
|
with 4 possible answers. Questions should be independent because
|
|
they will be asked in a random order.
|
|
- [ ] Propose a project for the students to continue the
|
|
programs. Idea: Modify the program to compute the exact energy of
|
|
the H$_2$ molecule at $R$=1.4010 bohr. Answer: 0.17406 a.u.
|
|
|
|
* Schedule
|
|
|
|
|------------------------------+---------|
|
|
| <2021-02-04 Thu 09:00-10:30> | Lecture |
|
|
|------------------------------+---------|
|
|
| <2021-02-04 Thu 10:45-11:10> | 2.1 |
|
|
| <2021-02-04 Thu 11:10-11:30> | 2.2 |
|
|
| <2021-02-04 Thu 11:30-12:15> | 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 |
|
|
|------------------------------+---------|
|