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Compute E_L before moving, and fix bug in Python code

This commit is contained in:
Anthony Scemama 2021-02-02 14:23:54 +01:00
parent dbf3f108a0
commit 4df30f51c0
2 changed files with 27 additions and 24 deletions

49
QMC.org
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@ -1424,13 +1424,13 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
The algorithm is summarized as follows: The algorithm is summarized as follows:
1) Compute $\Psi$ at a new position $\mathbf{r'} = \mathbf{r}_n + 1) Evaluate the local energy at $\mathbf{r}_n$ and accumulate it
\delta L\, \mathbf{u}$ 2) Compute 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) Evaluate $\Psi(\mathbf{r}')$ at the new position
3) Draw a uniform random number $v \in [0,1]$ 4) Compute the ratio $A = \frac{\left|\Psi(\mathbf{r'})\right|^2}{\left|\Psi(\mathbf{r}_{n})\right|^2}$
4) if $v \le A$, accept the move : set $\mathbf{r}_{n+1} = \mathbf{r'}$ 5) Draw a uniform random number $v \in [0,1]$
5) else, reject the move : set $\mathbf{r}_{n+1} = \mathbf{r}_n$ 6) if $v \le A$, accept the move : set $\mathbf{r}_{n+1} = \mathbf{r'}$
6) evaluate the local energy at $\mathbf{r}_{n+1}$ 7) else, reject the move : set $\mathbf{r}_{n+1} = \mathbf{r}_n$
#+begin_note #+begin_note
A common error is to remove the rejected samples from the A common error is to remove the rejected samples from the
@ -1742,7 +1742,7 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_metropolis.f90 -o qmc_metropolis
Furthermore, to sample the density even better, we can "push" the electrons Furthermore, to sample the density even better, we can "push" the electrons
into in the regions of high probability, and "pull" them away from into in the regions of high probability, and "pull" them away from
the low-probability regions. This will ncrease the the low-probability regions. This will increase the
acceptance ratios and improve the sampling. acceptance ratios and improve the sampling.
To do this, we can use the gradient of the probability density To do this, we can use the gradient of the probability density
@ -1761,7 +1761,7 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_metropolis.f90 -o qmc_metropolis
\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\delta t} \right]\,. \Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\delta t} \right]\,.
\] \]
The corrsponding move is proposed as The corresponding move is proposed as
\[ \[
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta t\, \frac{\nabla \mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta t\, \frac{\nabla
@ -1775,15 +1775,14 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_metropolis.f90 -o qmc_metropolis
The algorithm of the previous exercise is only slighlty modified as: The algorithm of the previous exercise is only slighlty modified as:
1) Compute a new position $\mathbf{r'} = \mathbf{r}_n + 1) Evaluate the local energy at $\mathbf{r}_{n}$ and accumulate it
2) Compute a new position $\mathbf{r'} = \mathbf{r}_n +
\delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi$ \delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi$
3) Evaluate $\Psi(\mathbf{r}')$ and $\frac{\nabla \Psi(\mathbf{r'})}{\Psi(\mathbf{r'})}$ at the new position
Evaluate $\Psi$ and $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$ at the new position 4) Compute the ratio $A = \frac{T(\mathbf{r}' \rightarrow \mathbf{r}_{n}) P(\mathbf{r}')}{T(\mathbf{r}_{n} \rightarrow \mathbf{r}') P(\mathbf{r}_{n})}$
2) Compute the ratio $A = \frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})}{T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})}$ 5) Draw a uniform random number $v \in [0,1]$
3) Draw a uniform random number $v \in [0,1]$ 6) if $v \le A$, accept the move : set $\mathbf{r}_{n+1} = \mathbf{r'}$
4) if $v \le A$, accept the move : set $\mathbf{r}_{n+1} = \mathbf{r'}$ 7) else, reject the move : set $\mathbf{r}_{n+1} = \mathbf{r}_n$
5) else, reject the move : set $\mathbf{r}_{n+1} = \mathbf{r}_n$
6) evaluate the local energy at $\mathbf{r}_{n+1}$
*** Gaussian random number generator *** Gaussian random number generator
@ -1840,6 +1839,11 @@ end subroutine random_gauss
*** Exercise 1 *** Exercise 1
#+begin_exercise
If you use Fortran, copy/paste the ~random_gauss~ function in
a Fortran file.
#+end_exercise
#+begin_exercise #+begin_exercise
Write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$. Write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$.
#+end_exercise #+end_exercise
@ -2014,13 +2018,13 @@ def MonteCarlo(a,nmax,dt):
d2_old = d2_new d2_old = d2_new
psi_old = psi_new psi_old = psi_new
return energy/nmax, accep_rate/nmax return energy/nmax, N_accep/nmax
# Run simulation # Run simulation
a = 1.2 a = 1.2
nmax = 100000 nmax = 100000
dt = 1.3 dt = 1.0
X0 = [ MonteCarlo(a,nmax,dt) for i in range(30)] X0 = [ MonteCarlo(a,nmax,dt) for i in range(30)]
@ -2036,8 +2040,6 @@ print(f"A = {A} +/- {deltaA}")
#+END_SRC #+END_SRC
#+RESULTS: #+RESULTS:
: E = -0.4951317910667116 +/- 0.00014045774335059988
: A = 0.7200673333333333 +/- 0.00045942791345632793
*Fortran* *Fortran*
#+BEGIN_SRC f90 #+BEGIN_SRC f90
@ -2144,8 +2146,8 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
#+end_src #+end_src
#+RESULTS: #+RESULTS:
: E = -0.49497258331144794 +/- 1.0973395750688713E-004 : E = -0.47940635575542706 +/- 5.5613594433433764E-004
: A = 0.78839866666666658 +/- 3.2503783452043152E-004 : A = 0.62037333333333333 +/- 4.8970160591451110E-004
* Diffusion Monte Carlo :solution: * Diffusion Monte Carlo :solution:
@ -3099,4 +3101,5 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc.f90 -o vmc
| <2021-02-04 Thu 14:00-14:10> | 3.1 | | <2021-02-04 Thu 14:00-14:10> | 3.1 |
| <2021-02-04 Thu 14:10-14:30> | 3.2 | | <2021-02-04 Thu 14:10-14:30> | 3.2 |
| <2021-02-04 Thu 14:30-15:30> | 3.3 | | <2021-02-04 Thu 14:30-15:30> | 3.3 |
| <2021-02-04 Thu 15:30-16:30> | 3.4 |
|------------------------------+---------| |------------------------------+---------|

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@ -45,7 +45,7 @@ def MonteCarlo(a,nmax,dt):
# Run simulation # Run simulation
a = 1.2 a = 1.2
nmax = 100000 nmax = 100000
dt = 1.3 dt = 1.0
X0 = [ MonteCarlo(a,nmax,dt) for i in range(30)] X0 = [ MonteCarlo(a,nmax,dt) for i in range(30)]