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OK up to Gaussian RNG

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Anthony Scemama 2021-01-26 10:02:38 +01:00
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QMC.org
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@ -2,11 +2,14 @@
#+AUTHOR: Anthony Scemama, Claudia Filippi
# SETUPFILE: https://fniessen.github.io/org-html-themes/org/theme-readtheorg.setup
# SETUPFILE: https://fniessen.github.io/org-html-themes/org/theme-bigblow.setup
#+LANGUAGE: en
#+INFOJS_OPT: toc:t mouse:underline path:http://orgmode.org/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: H:3 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: Python solution
# EXCLUDE_TAGS: Fortran solution
@ -123,8 +126,6 @@
to catch the error.
#+end_note
#+end_note
*** Exercise 1
#+begin_exercise
@ -964,7 +965,7 @@ subroutine ave_error(x,n,ave,err)
end subroutine ave_error
#+END_SRC
** TODO Uniform sampling in the box
** Uniform sampling in the box
:PROPERTIES:
:header-args:python: :tangle qmc_uniform.py
:header-args:f90: :tangle qmc_uniform.f90
@ -973,7 +974,7 @@ end subroutine ave_error
We will now do our first Monte Carlo calculation to compute the
energy of the hydrogen atom.
At every Monte Carlo step:
At every Monte Carlo iteration:
- Draw a random point $\mathbf{r}_i$ in the box $(-5,-5,-5) \le
(x,y,z) \le (5,5,5)$
@ -982,14 +983,14 @@ end subroutine ave_error
- Compute $[\Psi(\mathbf{r}_i)]^2 \times E_L(\mathbf{r}_i)$, and accumulate the
result in a variable =energy=
One Monte Carlo run will consist of $N$ Monte Carlo steps. Once all the
steps have been computed, the run returns the average energy
$\bar{E}_k$ over the $N$ steps of the run.
One Monte Carlo run will consist of $N$ Monte Carlo iterations. Once all the
iterations have been computed, the run returns the average energy
$\bar{E}_k$ over the $N$ iterations of the run.
To compute the statistical error, perform $M$ 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.
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
@ -1000,41 +1001,113 @@ end subroutine ave_error
compute the average energy and the associated error bar.
#+end_exercise
*Python*
#+BEGIN_SRC python :results output
**** 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 :results output :tangle none
from hydrogen import *
from qmc_stats import *
def MonteCarlo(a, nmax):
E = 0.
N = 0.
# TODO
a = 0.9
nmax = 100000
#TODO
print(f"E = {E} +/- {deltaE}")
#+END_SRC
#+RESULTS:
: E = -0.4956255109300764 +/- 0.0007082875482711226
**** Python :solution:
#+BEGIN_SRC python :results output
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
N += w
E += w * e_loc(a,r)
return E/N
normalization += w
energy += w * e_loc(a,r)
return energy/normalization
a = 0.9
nmax = 100000
X = [MonteCarlo(a,nmax) for i in range(30)]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")
#+END_SRC
#+END_SRC
#+RESULTS:
: E = -0.4956255109300764 +/- 0.0007082875482711226
#+RESULTS:
: E = -0.4956255109300764 +/- 0.0007082875482711226
*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
**** 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_SRC f90
#+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 = 0.9
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 both
gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
./qmc_uniform
#+end_src
**** Fortran :solution:
#+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
subroutine uniform_montecarlo(a,nmax,energy)
implicit none
double precision, intent(in) :: a
@ -1076,17 +1149,17 @@ program qmc
call ave_error(X,nruns,ave,err)
print *, 'E = ', ave, '+/-', err
end program qmc
#+END_SRC
#+END_SRC
#+begin_src sh :results output :exports both
#+begin_src sh :results output :exports both
gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
./qmc_uniform
#+end_src
#+end_src
#+RESULTS:
: E = -0.49588321986667677 +/- 7.1758863546737969E-004
#+RESULTS:
: E = -0.49588321986667677 +/- 7.1758863546737969E-004
** TODO Metropolis sampling with $\Psi^2$
** Metropolis sampling with $\Psi^2$
:PROPERTIES:
:header-args:python: :tangle qmc_metropolis.py
:header-args:f90: :tangle qmc_metropolis.f90
@ -1098,7 +1171,7 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
P(\mathbf{r}) = \left[\Psi(\mathbf{r})\right]^2
\]
The expression of the average energy is now simplified to the average of
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 :
@ -1118,15 +1191,16 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
where $\tau$ is a fixed constant (the so-called /time-step/), and
$\mathbf{u}$ is a uniform random number in a 3-dimensional box
$(-1,-1,-1) \le \mathbf{u} \le (1,1,1)$. We will then add the
accept/reject step that will guarantee that the distribution of the
accept/reject step that guarantees that the distribution of the
$\mathbf{r}_n$ is $\Psi^2$:
- Compute a new position $\mathbf{r}_{n+1}$
- Draw a uniform random number $v \in [0,1]$
- Compute the ratio $R = \frac{\left[\Psi(\mathbf{r}_{n+1})\right]^2}{\left[\Psi(\mathbf{r}_{n})\right]^2}$
- if $v \le R$, accept the move (do nothing)
- else, reject the move (set $\mathbf{r}_{n+1} = \mathbf{r}_n$)
- evaluate the local energy at $\mathbf{r}_{n+1}$
1) Compute $\Psi$ at a new position $\mathbf{r'} = \mathbf{r}_n +
\tau \mathbf{u}$
2) Compute the ratio $R = \frac{\left[\Psi(\mathbf{r'})\right]^2}{\left[\Psi(\mathbf{r}_{n})\right]^2}$
3) Draw a uniform random number $v \in [0,1]$
4) if $v \le R$, 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
@ -1146,7 +1220,7 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
The time step 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
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.
@ -1154,55 +1228,90 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
*** Exercise
#+begin_exercise
Modify the program of the previous section to compute the energy, sampling with
$Psi^2$.
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
**** Python
#+BEGIN_SRC python :results output :tangle none
from hydrogen import *
from qmc_stats import *
def MonteCarlo(a,nmax,tau):
E = 0.
N = 0.
N_accep = 0.
# TODO
return energy/nmax, N_accep/nmax
# Run simulation
a = 0.9
nmax = 100000
tau = 1.3
X0 = [ MonteCarlo(a,nmax,tau) 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 = -0.4950720838131573 +/- 0.00019089638602238043
: A = 0.5172960000000001 +/- 0.0003443446549306529
**** Python :solution:
#+BEGIN_SRC python :results output
from hydrogen import *
from qmc_stats import *
def MonteCarlo(a,nmax,tau):
energy = 0.
N_accep = 0
r_old = np.random.uniform(-tau, tau, (3))
psi_old = psi(a,r_old)
for istep in range(nmax):
r_new = r_old + np.random.uniform(-tau,tau,(3))
psi_new = psi(a,r_new)
ratio = (psi_new / psi_old)**2
v = np.random.uniform(0,1,(1))
if v < ratio:
N_accep += 1.
v = np.random.uniform()
if v <= ratio:
N_accep += 1
r_old = r_new
psi_old = psi_new
N += 1.
E += e_loc(a,r_old)
return E/N, N_accep/N
energy += e_loc(a,r_old)
return energy/nmax, N_accep/nmax
# Run simulation
a = 0.9
nmax = 100000
tau = 1.3
X0 = [ MonteCarlo(a,nmax,tau) for i in range(30)]
X = [ x for x, _ in X0 ]
A = [ x for _, x in X0 ]
# Energy
X = [ x for (x, _) in X0 ]
E, deltaE = ave_error(X)
A, deltaA = ave_error(A)
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
#+END_SRC
#+RESULTS:
: E = -0.4950720838131573 +/- 0.00019089638602238043
: A = 0.5172960000000001 +/- 0.0003443446549306529
#+RESULTS:
: E = -0.4950720838131573 +/- 0.00019089638602238043
: A = 0.5172960000000001 +/- 0.0003443446549306529
*Fortran*
#+BEGIN_SRC f90
**** Fortran
#+BEGIN_SRC f90 :tangle none
subroutine metropolis_montecarlo(a,nmax,tau,energy,accep)
implicit none
double precision, intent(in) :: a
@ -1213,32 +1322,12 @@ subroutine metropolis_montecarlo(a,nmax,tau,energy,accep)
integer*8 :: istep
double precision :: norm, r_old(3), r_new(3), psi_old, psi_new
double precision :: v, ratio, n_accep
double precision :: r_old(3), r_new(3), psi_old, psi_new
double precision :: v, ratio
integer*8 :: n_accep
double precision, external :: e_loc, psi, gaussian
energy = 0.d0
norm = 0.d0
n_accep = 0.d0
call random_number(r_old)
r_old(:) = tau * (2.d0*r_old(:) - 1.d0)
psi_old = psi(a,r_old)
do istep = 1,nmax
call random_number(r_new)
r_new(:) = r_old(:) + tau * (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
r_old(:) = r_new(:)
psi_old = psi_new
n_accep = n_accep + 1.d0
endif
norm = norm + 1.d0
energy = energy + e_loc(a,r_old)
end do
energy = energy / norm
accep = n_accep / norm
! TODO
end subroutine metropolis_montecarlo
program qmc
@ -1255,20 +1344,89 @@ program qmc
do irun=1,nruns
call metropolis_montecarlo(a,nmax,tau,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
#+END_SRC
#+begin_src sh :results output :exports both
#+begin_src sh :results output :exports both
gfortran hydrogen.f90 qmc_stats.f90 qmc_metropolis.f90 -o qmc_metropolis
./qmc_metropolis
#+end_src
#+RESULTS:
: E = -0.49478505004797046 +/- 2.0493795299184956E-004
: A = 0.51737800000000000 +/- 4.1827406733181444E-004
#+end_src
**** Fortran :solution:
#+BEGIN_SRC f90
subroutine metropolis_montecarlo(a,nmax,tau,energy,accep)
implicit none
double precision, intent(in) :: a
integer*8 , intent(in) :: nmax
double precision, intent(in) :: tau
double precision, intent(out) :: energy
double precision, intent(out) :: accep
integer*8 :: istep
double precision :: r_old(3), r_new(3), psi_old, psi_new
double precision :: v, ratio
integer*8 :: n_accep
double precision, external :: e_loc, psi, gaussian
energy = 0.d0
n_accep = 0_8
call random_number(r_old)
r_old(:) = tau * (2.d0*r_old(:) - 1.d0)
psi_old = psi(a,r_old)
do istep = 1,nmax
call random_number(r_new)
r_new(:) = r_old(:) + tau * (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
r_old(:) = r_new(:)
psi_old = psi_new
n_accep = n_accep + 1_8
endif
energy = energy + e_loc(a,r_old)
end do
energy = energy / dble(nmax)
accep = dble(n_accep) / dble(nmax)
end subroutine metropolis_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9d0
double precision, parameter :: tau = 1.3d0
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,tau,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
#+RESULTS:
: E = -0.49515370205041676 +/- 1.7660819245720729E-004
: A = 0.51713866666666664 +/- 3.7072551835783688E-004
** TODO Gaussian random number generator

2
docs/worg.css
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@ -17,8 +17,8 @@
line-height: 22pt;
color: black;
margin-top: 0;
}
body #content {
padding-top: 2em;
margin: auto;

19
qmc_metropolis.f90
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@ -8,13 +8,13 @@ subroutine metropolis_montecarlo(a,nmax,tau,energy,accep)
integer*8 :: istep
double precision :: norm, r_old(3), r_new(3), psi_old, psi_new
double precision :: v, ratio, n_accep
double precision :: r_old(3), r_new(3), psi_old, psi_new
double precision :: v, ratio
integer*8 :: n_accep
double precision, external :: e_loc, psi, gaussian
energy = 0.d0
norm = 0.d0
n_accep = 0.d0
n_accep = 0_8
call random_number(r_old)
r_old(:) = tau * (2.d0*r_old(:) - 1.d0)
psi_old = psi(a,r_old)
@ -24,16 +24,15 @@ subroutine metropolis_montecarlo(a,nmax,tau,energy,accep)
psi_new = psi(a,r_new)
ratio = (psi_new / psi_old)**2
call random_number(v)
if (v < ratio) then
if (v <= ratio) then
r_old(:) = r_new(:)
psi_old = psi_new
n_accep = n_accep + 1.d0
n_accep = n_accep + 1_8
endif
norm = norm + 1.d0
energy = energy + e_loc(a,r_old)
end do
energy = energy / norm
accep = n_accep / norm
energy = energy / dble(nmax)
accep = dble(n_accep) / dble(nmax)
end subroutine metropolis_montecarlo
program qmc
@ -50,8 +49,10 @@ program qmc
do irun=1,nruns
call metropolis_montecarlo(a,nmax,tau,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

27
qmc_metropolis.py
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@ -2,31 +2,34 @@ from hydrogen import *
from qmc_stats import *
def MonteCarlo(a,nmax,tau):
E = 0.
N = 0.
N_accep = 0.
energy = 0.
N_accep = 0
r_old = np.random.uniform(-tau, tau, (3))
psi_old = psi(a,r_old)
for istep in range(nmax):
r_new = r_old + np.random.uniform(-tau,tau,(3))
psi_new = psi(a,r_new)
ratio = (psi_new / psi_old)**2
v = np.random.uniform(0,1,(1))
if v < ratio:
N_accep += 1.
v = np.random.uniform()
if v <= ratio:
N_accep += 1
r_old = r_new
psi_old = psi_new
N += 1.
E += e_loc(a,r_old)
return E/N, N_accep/N
energy += e_loc(a,r_old)
return energy/nmax, N_accep/nmax
# Run simulation
a = 0.9
nmax = 100000
tau = 1.3
X0 = [ MonteCarlo(a,nmax,tau) for i in range(30)]
X = [ x for x, _ in X0 ]
A = [ x for _, x in X0 ]
# Energy
X = [ x for (x, _) in X0 ]
E, deltaE = ave_error(X)
A, deltaA = ave_error(A)
print(f"E = {E} +/- {deltaE}")
# Acceptance rate
X = [ x for (_, x) in X0 ]
A, deltaA = ave_error(X)
print(f"A = {A} +/- {deltaA}")

10
qmc_uniform.py
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@ -2,15 +2,15 @@ from hydrogen import *
from qmc_stats import *
def MonteCarlo(a, nmax):
E = 0.
N = 0.
energy = 0.
normalization = 0.
for istep in range(nmax):
r = np.random.uniform(-5., 5., (3))
w = psi(a,r)
w = w*w
N += w
E += w * e_loc(a,r)
return E/N
normalization += w
energy += w * e_loc(a,r)
return energy/normalization
a = 0.9
nmax = 100000