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#+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 = -0.4773221805255284 +/- 0.0022489426510479975
*Fortran*
#+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 :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) Compute $\Psi$ at a new position $\mathbf{r'} = \mathbf{r}_n +
\delta L\, \mathbf{u}$
2) Compute the ratio $A = \frac{\left|\Psi(\mathbf{r'})\right|^2}{\left|\Psi(\mathbf{r}_{n})\right|^2}$
3) Draw a uniform random number $v \in [0,1]$
4) if $v \le A$, accept the move : set $\mathbf{r}_{n+1} = \mathbf{r'}$
5) else, reject the move : set $\mathbf{r}_{n+1} = \mathbf{r}_n$
6) evaluate the local energy at $\mathbf{r}_{n+1}$
#+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
** 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.
** 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 ncrease 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 corrsponding 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) 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 $\mathbf{r}_{n+1}$
*** Exercise 1
#+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, accep_rate/nmax
# Run simulation
a = 1.2
nmax = 100000
dt = 1.3
X0 = [ MonteCarlo(a,nmax,dt) for i in range(30)]
# Energy
X = [ x for (x, _) in X0 ]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")
# Acceptance rate
X = [ x for (_, x) in X0 ]
A, deltaA = ave_error(X)
print(f"A = {A} +/- {deltaA}")
#+END_SRC
#+RESULTS:
: E = -0.4951317910667116 +/- 0.00014045774335059988
: A = 0.7200673333333333 +/- 0.00045942791345632793 *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 = -0.49497258331144794 +/- 1.0973395750688713E-004
: A = 0.78839866666666658 +/- 3.2503783452043152E-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:
\[
W(\mathbf{r}_n, \tau)
= \exp \left( \int_0^\tau - (E_L(\mathbf{r}_t) - E_{\text{ref}}) dt \right)
\approx \prod_{i=1}^{n} \exp \left( -\delta t\,
(E_L(\mathbf{r}_i) - E_{\text{ref}}) \right) =
\prod_{i=1}^{n} w(\mathbf{r}_i)
\]
where $\mathbf{r}_i$ are the coordinates along the trajectory 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 |
|------------------------------+---------|
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