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Changing a=0.9 to a=1.2 to avoid population explosion in DMC

This commit is contained in:
Anthony Scemama 2021-02-02 13:57:53 +01:00
parent e697ff2da4
commit 9c937da308
17 changed files with 99 additions and 66 deletions
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QMC.org
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@ -1185,7 +1185,7 @@ end subroutine ave_error
*** Exercise
#+begin_exercise
Parameterize the wave function with $a=0.9$. Perform 30
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.
@ -1206,7 +1206,7 @@ from qmc_stats import *
def MonteCarlo(a, nmax):
# TODO
a = 0.9
a = 1.2
nmax = 100000
#TODO
@ -1244,7 +1244,7 @@ end subroutine uniform_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9
double precision, parameter :: a = 1.2d0
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30
@ -1287,7 +1287,7 @@ def MonteCarlo(a, nmax):
return energy / normalization
a = 0.9
a = 1.2
nmax = 100000
X = [MonteCarlo(a,nmax) for i in range(30)]
@ -1297,7 +1297,7 @@ print(f"E = {E} +/- {deltaE}")
#+END_SRC
#+RESULTS:
: E = -0.4956255109300764 +/- 0.0007082875482711226
: E = -0.4773221805255284 +/- 0.0022489426510479975
*Fortran*
#+begin_note
@ -1341,7 +1341,7 @@ end subroutine uniform_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9
double precision, parameter :: a = 1.2d0
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30
@ -1365,7 +1365,7 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
#+end_src
#+RESULTS:
: E = -0.49518773675598715 +/- 5.2391494923686175E-004
: E = -0.48084122147238995 +/- 2.4983775878329355E-003
** Metropolis sampling with $\Psi^2$
:PROPERTIES:
@ -1420,10 +1420,14 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
which, for our choice of transition probability, becomes
$$
A(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1}) = \min\left(1,\frac{P(\mathbf{r}_{n+1})}{P(\mathbf{r}_{n})}\right)= \min\left(1,\frac{\Psi(\mathbf{r}_{n+1})^2}{\Psi(\mathbf{r}_{n})^2}\right)\,.
A(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1}) = \min\left(1,\frac{P(\mathbf{r}_{n+1})}{P(\mathbf{r}_{n})}\right)= \min\left(1,\frac{|\Psi(\mathbf{r}_{n+1})|^2}{|\Psi(\mathbf{r}_{n})|^2}\right)\,.
$$
Explain why the transition probability cancels out in the expression of $A$. Also note that we do not need to compute the norm of the wave function!
#+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:
@ -1439,27 +1443,32 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
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.
All samples should be kept, from both accepted /and/ rejected moves.
#+end_note
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.
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.
*** 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
@ -1489,7 +1498,7 @@ def MonteCarlo(a,nmax,dt):
# Run simulation
a = 0.9
a = 1.2
nmax = 100000
dt = #TODO
@ -1506,6 +1515,8 @@ 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)
@ -1529,7 +1540,7 @@ end subroutine metropolis_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9d0
double precision, parameter :: a = 1.2d0
double precision, parameter :: dt = ! TODO
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30
@ -1588,9 +1599,9 @@ def MonteCarlo(a,nmax,dt):
return energy/nmax, N_accep/nmax
# Run simulation
a = 0.9
a = 1.2
nmax = 100000
dt = 1.3
dt = 1.0
X0 = [ MonteCarlo(a,nmax,dt) for i in range(30)]
@ -1606,8 +1617,8 @@ print(f"A = {A} +/- {deltaA}")
#+END_SRC
#+RESULTS:
: E = -0.4950720838131573 +/- 0.00019089638602238043
: A = 0.5172960000000001 +/- 0.0003443446549306529
: E = -0.4802595860693983 +/- 0.0005124420418289207
: A = 0.5074913333333334 +/- 0.000350889422714878
*Fortran*
#+BEGIN_SRC f90 :exports code
@ -1662,8 +1673,8 @@ end subroutine metropolis_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9d0
double precision, parameter :: dt = 1.3d0
double precision, parameter :: a = 1.2d0
double precision, parameter :: dt = 1.0d0
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30
@ -1689,8 +1700,8 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_metropolis.f90 -o qmc_metropolis
./qmc_metropolis
#+end_src
#+RESULTS:
: E = -0.49503130891988767 +/- 1.7160104275040037E-004
: A = 0.51695266666666673 +/- 4.0445505648997396E-004
: E = -0.47948142754168033 +/- 4.8410177863919307E-004
: A = 0.50762633333333318 +/- 3.4601756760043725E-004
** Gaussian random number generator
@ -1899,7 +1910,7 @@ def MonteCarlo(a,nmax,dt):
# TODO
# Run simulation
a = 0.9
a = 1.2
nmax = 100000
dt = # TODO
@ -1939,7 +1950,7 @@ end subroutine variational_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9
double precision, parameter :: a = 1.2d0
double precision, parameter :: dt = ! TODO
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30
@ -2014,7 +2025,7 @@ def MonteCarlo(a,nmax,dt):
# Run simulation
a = 0.9
a = 1.2
nmax = 100000
dt = 1.3
@ -2112,8 +2123,8 @@ end subroutine variational_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9
double precision, parameter :: dt = 1.0
double precision, parameter :: a = 1.2d0
double precision, parameter :: dt = 1.0d0
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30
@ -2442,7 +2453,7 @@ def MonteCarlo(a, nmax, dt, Eref):
# TODO
# Run simulation
a = 0.9
a = 1.2
nmax = 100000
dt = 0.01
E_ref = -0.5
@ -2484,7 +2495,7 @@ end subroutine pdmc
program qmc
implicit none
double precision, parameter :: a = 0.9
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
@ -2576,7 +2587,7 @@ def MonteCarlo(a, nmax, dt, tau, Eref):
# Run simulation
a = 0.9
a = 1.2
nmax = 100000
dt = 0.1
tau = 10.
@ -2694,7 +2705,7 @@ end subroutine pdmc
program qmc
implicit none
double precision, parameter :: a = 0.9
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
@ -2814,7 +2825,7 @@ def MonteCarlo(a,nmax):
E += w * e_loc(a,r)
return E/N
a = 0.9
a = 1.2
nmax = 100000
X = [MonteCarlo(a,nmax) for i in range(30)]
E, deltaE = ave_error(X)
@ -2860,7 +2871,7 @@ end subroutine gaussian_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9
double precision, parameter :: a = 1.2d0
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30
@ -3003,7 +3014,7 @@ def MonteCarlo(a,dt,nmax):
return E/N
a = 0.9
a = 1.2
nmax = 100000
dt = 0.2
X = [MonteCarlo(a,dt,nmax) for i in range(30)]
@ -3046,8 +3057,8 @@ end subroutine variational_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9
double precision, parameter :: dt = 0.2
double precision, parameter :: a = 1.2d0
double precision, parameter :: dt = 0.2d0
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30
@ -3093,4 +3104,6 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc.f90 -o vmc
| <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 |
|------------------------------+---------|

2
energy_hydrogen.py
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@ -1,3 +1,5 @@
#!/usr/bin/env python3
import numpy as np
from hydrogen import e_loc, psi

3
hydrogen.f90
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@ -43,7 +43,8 @@ double precision function e_loc(a,r)
implicit none
double precision, intent(in) :: a, r(3)
double precision, external :: kinetic, potential
double precision, external :: kinetic
double precision, external :: potential
e_loc = kinetic(a,r) + potential(r)

1
hydrogen.py
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@ -1,3 +1,4 @@
#!/usr/bin/env python3
import numpy as np
def potential(r):

2
plot_hydrogen.py
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@ -1,3 +1,5 @@
#!/usr/bin/env python3
import numpy as np
import matplotlib.pyplot as plt

2
qmc_gaussian.f90
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@ -32,7 +32,7 @@ end subroutine gaussian_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9
double precision, parameter :: a = 1.2d0
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30

4
qmc_gaussian.py
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@ -1,3 +1,5 @@
#!/usr/bin/env python3
from hydrogen import *
from qmc_stats import *
@ -16,7 +18,7 @@ def MonteCarlo(a,nmax):
E += w * e_loc(a,r)
return E/N
a = 0.9
a = 1.2
nmax = 100000
X = [MonteCarlo(a,nmax) for i in range(30)]
E, deltaE = ave_error(X)

4
qmc_metropolis.f90
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@ -49,8 +49,8 @@ end subroutine metropolis_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9d0
double precision, parameter :: dt = 1.3d0
double precision, parameter :: a = 1.2d0
double precision, parameter :: dt = 1.0d0
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30

6
qmc_metropolis.py
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@ -1,3 +1,5 @@
#!/usr/bin/env python3
from hydrogen import *
from qmc_stats import *
@ -25,9 +27,9 @@ def MonteCarlo(a,nmax,dt):
return energy/nmax, N_accep/nmax
# Run simulation
a = 0.9
a = 1.2
nmax = 100000
dt = 1.3
dt = 1.0
X0 = [ MonteCarlo(a,nmax,dt) for i in range(30)]

2
qmc_stats.py
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@ -1,3 +1,5 @@
#!/usr/bin/env python3
from math import sqrt
def ave_error(arr):
M = len(arr)

2
qmc_uniform.f90
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@ -31,7 +31,7 @@ end subroutine uniform_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9
double precision, parameter :: a = 1.2d0
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30

4
qmc_uniform.py
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@ -1,3 +1,5 @@
#!/usr/bin/env python3
from hydrogen import *
from qmc_stats import *
@ -16,7 +18,7 @@ def MonteCarlo(a, nmax):
return energy / normalization
a = 0.9
a = 1.2
nmax = 100000
X = [MonteCarlo(a,nmax) for i in range(30)]

2
variance_hydrogen.py
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@ -1,3 +1,5 @@
#!/usr/bin/env python3
import numpy as np
from hydrogen import e_loc, psi

4
vmc.f90
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@ -28,8 +28,8 @@ end subroutine variational_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9
double precision, parameter :: dt = 0.2
double precision, parameter :: a = 1.2d0
double precision, parameter :: dt = 0.2d0
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30

4
vmc.py
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@ -1,3 +1,5 @@
#!/usr/bin/env python3
from hydrogen import *
from qmc_stats import *
@ -19,7 +21,7 @@ def MonteCarlo(a,dt,nmax):
return E/N
a = 0.9
a = 1.2
nmax = 100000
dt = 0.2
X = [MonteCarlo(a,dt,nmax) for i in range(30)]

4
vmc_metropolis.f90
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@ -73,8 +73,8 @@ end subroutine variational_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9
double precision, parameter :: dt = 1.0
double precision, parameter :: a = 1.2d0
double precision, parameter :: dt = 1.0d0
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30

4
vmc_metropolis.py
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@ -1,3 +1,5 @@
#!/usr/bin/env python3
from hydrogen import *
from qmc_stats import *
@ -41,7 +43,7 @@ def MonteCarlo(a,nmax,dt):
# Run simulation
a = 0.9
a = 1.2
nmax = 100000
dt = 1.3