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OK up to DMC
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@ -1428,7 +1428,7 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_metropolis.f90 -o qmc_metropolis
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: E = -0.49515370205041676 +/- 1.7660819245720729E-004
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: A = 0.51713866666666664 +/- 3.7072551835783688E-004
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** TODO Gaussian random number generator
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** Gaussian random number generator
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To obtain Gaussian-distributed random numbers, you can apply the
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[[https://en.wikipedia.org/wiki/Box%E2%80%93Muller_transform][Box Muller transform]] to uniform random numbers:
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@ -1473,15 +1473,16 @@ subroutine random_gauss(z,n)
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end subroutine random_gauss
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#+END_SRC
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** TODO Generalized Metropolis algorithm
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In Python, you can use the [[https://numpy.org/doc/stable/reference/random/generated/numpy.random.normal.html][~random.normal~]] function of Numpy.
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** Generalized Metropolis algorithm
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:PROPERTIES:
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:header-args:python: :tangle vmc_metropolis.py
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:header-args:f90: :tangle vmc_metropolis.f90
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:END:
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One can use more efficient numerical schemes to move the electrons.
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But in that case, the Metropolis accepation step has to be adapted
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accordingly: the acceptance
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One can use more efficient numerical schemes to move the electrons,
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but the Metropolis accepation step has to be adapted accordingly:
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the acceptance
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probability $A$ is chosen so that it is consistent with the
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probability of leaving $\mathbf{r}_n$ and the probability of
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entering $\mathbf{r}_{n+1}$:
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@ -1499,19 +1500,19 @@ end subroutine random_gauss
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numbers. Hence, the transition probability was
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\[
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T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = &
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T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) =
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\text{constant}
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\]
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So the expression of $A$ was simplified to the ratios of the squared
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so the expression of $A$ was simplified to the ratios of the squared
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wave functions.
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Now, if instead of drawing uniform random numbers
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choose to draw Gaussian random numbers with mean 0 and variance
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Now, if instead of drawing uniform random numbers we
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choose to draw Gaussian random numbers with zero mean and variance
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$\tau$, the transition probability becomes:
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\[
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T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = &
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T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) =
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\frac{1}{(2\pi\,\tau)^{3/2}} \exp \left[ - \frac{\left(
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\mathbf{r}_{n+1} - \mathbf{r}_{n} \right)^2}{2\tau} \right]
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\]
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@ -1525,8 +1526,8 @@ end subroutine random_gauss
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To do this, we can add the drift vector
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\[
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\frac{\nabla [ \Psi^2 ]}{\Psi^2} = 2 \frac{\nabla \Psi}{\Psi}
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\].
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\frac{\nabla [ \Psi^2 ]}{\Psi^2} = 2 \frac{\nabla \Psi}{\Psi}.
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\]
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The numerical scheme becomes a drifted diffusion:
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@ -1540,7 +1541,7 @@ end subroutine random_gauss
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The transition probability becomes:
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\[
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T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = &
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T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) =
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\frac{1}{(2\pi\,\tau)^{3/2}} \exp \left[ - \frac{\left(
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\mathbf{r}_{n+1} - \mathbf{r}_{n} - \frac{\nabla
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\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\tau} \right]
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@ -1553,14 +1554,30 @@ end subroutine random_gauss
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Write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$.
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#+end_exercise
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*Python*
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**** Python
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#+BEGIN_SRC python :tangle hydrogen.py :tangle none
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def drift(a,r):
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# TODO
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#+END_SRC
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**** Python :solution:
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#+BEGIN_SRC python :tangle hydrogen.py
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def drift(a,r):
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ar_inv = -a/np.sqrt(np.dot(r,r))
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return r * ar_inv
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#+END_SRC
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*Fortran*
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**** Fortran
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#+BEGIN_SRC f90 :tangle hydrogen.f90 :tangle none
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subroutine drift(a,r,b)
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implicit none
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double precision, intent(in) :: a, r(3)
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double precision, intent(out) :: b(3)
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! TODO
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end subroutine drift
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#+END_SRC
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**** Fortran :solution:
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#+BEGIN_SRC f90 :tangle hydrogen.f90
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subroutine drift(a,r,b)
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implicit none
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@ -1579,14 +1596,38 @@ end subroutine drift
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(This is a necessary step for the next section).
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#+end_exercise
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*Python*
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**** Python
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#+BEGIN_SRC python :results output :tangle none
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from hydrogen import *
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from qmc_stats import *
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def MonteCarlo(a,nmax,tau):
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# TODO
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# Run simulation
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a = 0.9
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nmax = 100000
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tau = 1.3
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X0 = [ MonteCarlo(a,nmax,tau) for i in range(30)]
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# Energy
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X = [ x for (x, _) in X0 ]
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E, deltaE = ave_error(X)
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print(f"E = {E} +/- {deltaE}")
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# Acceptance rate
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X = [ x for (_, x) in X0 ]
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A, deltaA = ave_error(X)
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print(f"A = {A} +/- {deltaA}")
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#+END_SRC
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**** Python :solution:
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#+BEGIN_SRC python :results output
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from hydrogen import *
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from qmc_stats import *
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def MonteCarlo(a,tau,nmax):
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E = 0.
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N = 0.
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def MonteCarlo(a,nmax,tau):
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energy = 0.
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accep_rate = 0.
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sq_tau = np.sqrt(tau)
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r_old = np.random.normal(loc=0., scale=1.0, size=(3))
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@ -1610,25 +1651,77 @@ def MonteCarlo(a,tau,nmax):
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d_old = d_new
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d2_old = d2_new
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psi_old = psi_new
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N += 1.
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E += e_loc(a,r_old)
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return E/N, accep_rate/N
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energy += e_loc(a,r_old)
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return energy/nmax, accep_rate/nmax
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# Run simulation
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a = 0.9
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nmax = 100000
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tau = 1.0
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X = [MonteCarlo(a,tau,nmax) for i in range(30)]
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E, deltaE = ave_error([x[0] for x in X])
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A, deltaA = ave_error([x[1] for x in X])
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print(f"E = {E} +/- {deltaE}\nA = {A} +/- {deltaA}")
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tau = 1.3
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X0 = [ MonteCarlo(a,nmax,tau) for i in range(30)]
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# Energy
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X = [ x for (x, _) in X0 ]
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E, deltaE = ave_error(X)
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print(f"E = {E} +/- {deltaE}")
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# Acceptance rate
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X = [ x for (_, x) in X0 ]
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A, deltaA = ave_error(X)
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print(f"A = {A} +/- {deltaA}")
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#+END_SRC
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#+RESULTS:
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: E = -0.4949730317138491 +/- 0.00012478601801760644
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: A = 0.7887163333333334 +/- 0.00026834549840347617
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: E = -0.4951317910667116 +/- 0.00014045774335059988
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: A = 0.7200673333333333 +/- 0.00045942791345632793
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*Fortran*
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**** Fortran
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#+BEGIN_SRC f90 :tangle none
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subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate)
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implicit none
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double precision, intent(in) :: a, tau
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integer*8 , intent(in) :: nmax
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double precision, intent(out) :: energy, accep_rate
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integer*8 :: istep
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double precision :: sq_tau, chi(3)
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double precision :: psi_old, psi_new
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double precision :: r_old(3), r_new(3)
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double precision :: d_old(3), d_new(3)
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double precision, external :: e_loc, psi
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! TODO
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end subroutine variational_montecarlo
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program qmc
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implicit none
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double precision, parameter :: a = 0.9
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double precision, parameter :: tau = 1.0
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integer*8 , parameter :: nmax = 100000
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integer , parameter :: nruns = 30
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integer :: irun
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double precision :: X(nruns), accep(nruns)
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double precision :: ave, err
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do irun=1,nruns
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call variational_montecarlo(a,tau,nmax,X(irun),accep(irun))
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enddo
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call ave_error(X,nruns,ave,err)
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print *, 'E = ', ave, '+/-', err
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call ave_error(accep,nruns,ave,err)
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print *, 'A = ', ave, '+/-', err
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end program qmc
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#+END_SRC
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#+begin_src sh :results output :exports both
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gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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./vmc_metropolis
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#+end_src
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**** Fortran :solution:
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#+BEGIN_SRC f90
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subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate)
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implicit none
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@ -1637,7 +1730,7 @@ subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate)
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double precision, intent(out) :: energy, accep_rate
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integer*8 :: istep
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double precision :: norm, sq_tau, chi(3), d2_old, prod, u
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double precision :: sq_tau, chi(3), d2_old, prod, u
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double precision :: psi_old, psi_new, d2_new, argexpo, q
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double precision :: r_old(3), r_new(3)
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double precision :: d_old(3), d_new(3)
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@ -1647,7 +1740,6 @@ subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate)
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! Initialization
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energy = 0.d0
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norm = 0.d0
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accep_rate = 0.d0
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call random_gauss(r_old,3)
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call drift(a,r_old,d_old)
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@ -1675,11 +1767,10 @@ subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate)
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d2_old = d2_new
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psi_old = psi_new
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end if
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norm = norm + 1.d0
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energy = energy + e_loc(a,r_old)
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end do
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energy = energy / norm
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accep_rate = accep_rate / norm
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energy = energy / dble(nmax)
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accep_rate = dble(accep_rate) / dble(nmax)
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end subroutine variational_montecarlo
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program qmc
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@ -1709,8 +1800,8 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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#+end_src
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#+RESULTS:
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: E = -0.49499990423528023 +/- 1.5958250761863871E-004
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: A = 0.78861366666666655 +/- 3.5096729498002445E-004
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: E = -0.49495906384751226 +/- 1.5257646086088266E-004
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: A = 0.78861366666666666 +/- 3.7855335138754813E-004
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* TODO Diffusion Monte Carlo
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:PROPERTIES:
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@ -1733,10 +1824,10 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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from hydrogen import *
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from qmc_stats import *
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def MonteCarlo(a,tau,nmax,Eref):
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E = 0.
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N = 0.
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accep_rate = 0.
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def MonteCarlo(a,nmax,tau,Eref):
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energy = 0.
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normalization = 0.
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accep_rate = 0
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sq_tau = np.sqrt(tau)
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r_old = np.random.normal(loc=0., scale=1.0, size=(3))
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d_old = drift(a,r_old)
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@ -1747,8 +1838,8 @@ def MonteCarlo(a,tau,nmax,Eref):
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chi = np.random.normal(loc=0., scale=1.0, size=(3))
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el = e_loc(a,r_old)
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w *= np.exp(-tau*(el - Eref))
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N += w
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E += w * el
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normalization += w
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energy += w * el
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r_new = r_old + tau * d_old + sq_tau * chi
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d_new = drift(a,r_new)
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@ -1761,18 +1852,19 @@ def MonteCarlo(a,tau,nmax,Eref):
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q = np.exp(-argexpo) * q*q
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# PDMC weight
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if np.random.uniform() < q:
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accep_rate += w
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accep_rate += 1
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r_old = r_new
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d_old = d_new
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d2_old = d2_new
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psi_old = psi_new
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return E/N, accep_rate/N
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return energy/normalization, accep_rate/nmax
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a = 0.9
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nmax = 10000
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tau = .1
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X = [MonteCarlo(a,tau,nmax,-0.5) for i in range(30)]
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E_ref = -0.5
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X = [MonteCarlo(a,nmax,tau,E_ref) for i in range(30)]
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E, deltaE = ave_error([x[0] for x in X])
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A, deltaA = ave_error([x[1] for x in X])
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print(f"E = {E} +/- {deltaE}\nA = {A} +/- {deltaA}")
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@ -1782,7 +1874,7 @@ print(f"E = {E} +/- {deltaE}\nA = {A} +/- {deltaA}")
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: E = -0.49654807434947584 +/- 0.0006868522447409156
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: A = 0.9876193891840709 +/- 0.00041857361650995804
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*Fortran*
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**** Fortran
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#+BEGIN_SRC f90
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subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate)
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implicit none
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@ -1924,7 +2016,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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the statistical error?
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#+end_exercise
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*Python*
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**** Python
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#+BEGIN_SRC python :results output
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from hydrogen import *
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from qmc_stats import *
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@ -1954,7 +2046,7 @@ print(f"E = {E} +/- {deltaE}")
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#+RESULTS:
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: E = -0.49511014287471955 +/- 0.00012402022172236656
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*Fortran*
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**** Fortran
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#+BEGIN_SRC f90
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double precision function gaussian(r)
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implicit none
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@ -5,7 +5,7 @@ subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate)
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double precision, intent(out) :: energy, accep_rate
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integer*8 :: istep
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double precision :: norm, sq_tau, chi(3), d2_old, prod, u
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double precision :: sq_tau, chi(3), d2_old, prod, u
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double precision :: psi_old, psi_new, d2_new, argexpo, q
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double precision :: r_old(3), r_new(3)
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double precision :: d_old(3), d_new(3)
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@ -15,7 +15,6 @@ subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate)
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! Initialization
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energy = 0.d0
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norm = 0.d0
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accep_rate = 0.d0
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call random_gauss(r_old,3)
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call drift(a,r_old,d_old)
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@ -43,11 +42,10 @@ subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate)
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d2_old = d2_new
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psi_old = psi_new
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end if
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norm = norm + 1.d0
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energy = energy + e_loc(a,r_old)
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end do
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energy = energy / norm
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accep_rate = accep_rate / norm
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energy = energy / dble(nmax)
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accep_rate = dble(accep_rate) / dble(nmax)
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end subroutine variational_montecarlo
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program qmc
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@ -1,9 +1,8 @@
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from hydrogen import *
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from qmc_stats import *
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def MonteCarlo(a,tau,nmax):
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def MonteCarlo(a,nmax,tau):
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E = 0.
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N = 0.
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accep_rate = 0.
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sq_tau = np.sqrt(tau)
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r_old = np.random.normal(loc=0., scale=1.0, size=(3))
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@ -27,15 +26,22 @@ def MonteCarlo(a,tau,nmax):
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d_old = d_new
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d2_old = d2_new
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psi_old = psi_new
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N += 1.
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E += e_loc(a,r_old)
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return E/N, accep_rate/N
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return E/nmax, accep_rate/nmax
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# Run simulation
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a = 0.9
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nmax = 100000
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tau = 1.0
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X = [MonteCarlo(a,tau,nmax) for i in range(30)]
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E, deltaE = ave_error([x[0] for x in X])
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A, deltaA = ave_error([x[1] for x in X])
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print(f"E = {E} +/- {deltaE}\nA = {A} +/- {deltaA}")
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tau = 1.3
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X0 = [ MonteCarlo(a,nmax,tau) for i in range(30)]
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# Energy
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X = [ x for (x, _) in X0 ]
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E, deltaE = ave_error(X)
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print(f"E = {E} +/- {deltaE}")
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# Acceptance rate
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X = [ x for (_, x) in X0 ]
|
||||
A, deltaA = ave_error(X)
|
||||
print(f"A = {A} +/- {deltaA}")
|
||||
|
||||
Loading…
Reference in New Issue
Block a user