diff --git a/QMC.org b/QMC.org index 5e6faf7..ee86c95 100644 --- a/QMC.org +++ b/QMC.org @@ -988,125 +988,140 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_metropolis.f90 -o qmc_metropolis : E = -0.49478505004797046 +/- 2.0493795299184956E-004 : A = 0.51737800000000000 +/- 4.1827406733181444E-004 - -** Sampling with $\Psi^2$ - - We will now use the square of the wave function to make the sampling: - - \[ - P(\mathbf{r}) = \left[\Psi(\mathbf{r})\right]^2 - \] - - The expression for the energy will be simplified to the average of - the local energies, each with a weight of 1. - - $$ - E \approx \frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r}_i) - $$ +** Gaussian random number generator - -*** Importance sampling + 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 + +** Generalized Metropolis algorithm :PROPERTIES: - :header-args:python: :tangle vmc.py - :header-args:f90: :tangle vmc.f90 + :header-args:python: :tangle vmc_metropolis.py + :header-args:f90: :tangle vmc_metropolis.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: + One can use more efficient numerical schemes to move the electrons. + But in that case, the Metropolis accepation step has to be adapted + accordingly: the acceptance + probability $A$ is chosen so that it is consistent with the + probability of leaving $\mathbf{r}_n$ and the probability of + entering $\mathbf{r}_{n+1}$: - \[ - \frac{\partial \Psi^2}{\partial t} = \sum_i D - \frac{\partial}{\partial \mathbf{r}_i} \left( - \frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right) - [\Psi(\mathbf{r},t)]^2. - \] + \[ A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \min \left( 1, + \frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})} + {T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})} + \right) + \] + 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 random + numbers. Hence, the transition probability was + + \[ + T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = & + \text{constant} + \] + + So the expression of $A$ was simplified to the ratios of the squared + wave functions. + + Now, if instead of drawing uniform random numbers + choose to draw Gaussian random numbers with mean 0 and variance + $\tau$, the transition probability becomes: + + \[ + T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = & + \frac{1}{(2\pi\,\tau)^{3/2}} \exp \left[ - \frac{\left( + \mathbf{r}_{n+1} - \mathbf{r}_{n} \right)^2}{2\tau} \right] + \] + + + To sample even better the density, we can "push" the electrons + into in the regions of high probability, and "pull" them away from + the low-probability regions. This will mechanically increase the + acceptance ratios and improve the sampling. + + To do this, we can add the drift vector + + \[ + \frac{\nabla [ \Psi^2 ]}{\Psi^2} = 2 \frac{\nabla \Psi}{\Psi} + \]. + + The numerical scheme becomes a drifted diffusion: + + \[ + \mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \frac{\nabla + \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi + \] + + where $\chi$ is a Gaussian random variable with zero mean and + variance $\tau$. + The transition probability becomes: + + \[ + T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = & + \frac{1}{(2\pi\,\tau)^{3/2}} \exp \left[ - \frac{\left( + \mathbf{r}_{n+1} - \mathbf{r}_{n} - \frac{\nabla + \Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\tau} \right] + \] + + +*** Exercise 1 + + #+begin_exercise + Write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$. + #+end_exercise - $D$ 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: - - \[ - \mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau\, 2D \frac{\nabla - \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi - \] - where $\chi$ is a Gaussian random variable with zero mean and - variance $\tau\,2D$. - -**** 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 + *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 + #+END_SRC - *Fortran* - #+BEGIN_SRC f90 :tangle hydrogen.f90 + *Fortran* + #+BEGIN_SRC f90 :tangle hydrogen.f90 subroutine drift(a,r,b) implicit none double precision, intent(in) :: a, r(3) @@ -1115,167 +1130,17 @@ subroutine drift(a,r,b) 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 + #+END_SRC -**** TODO Exercise 2 +*** 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 + #+begin_exercise + Modify the previous program to introduce the drifted diffusion scheme. + (This is a necessary step for the next section). + #+end_exercise - *Python* - #+BEGIN_SRC python :results output -from hydrogen import * -from qmc_stats import * - -def MonteCarlo(a,tau,nmax): - sq_tau = np.sqrt(tau) - - # 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 + tau * d_old + chi*sq_tau - N += 1. - E += e_loc(a,r_new) - r_old = r_new - return E/N - - -a = 0.9 -nmax = 100000 -tau = 0.2 -X = [MonteCarlo(a,tau,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,tau,nmax,energy) - implicit none - double precision, intent(in) :: a, tau - 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_tau, chi(3) - double precision, external :: e_loc - - sq_tau = dsqrt(tau) - - ! 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(:) + tau * d_old(:) + chi(:)*sq_tau - 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 = 0.9 - double precision, parameter :: tau = 0.2 - 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,tau,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 - -*** Generalized Metropolis algorithm - :PROPERTIES: - :header-args:python: :tangle vmc_metropolis.py - :header-args:f90: :tangle vmc_metropolis.f90 - :END: - - Discretizing the differential equation to generate the desired - probability density will suffer from a discretization error - leading to biases in the averages. The [[https://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm][Metropolis-Hastings - sampling algorithm]] removes exactly the discretization errors, so - large time steps can be employed. - - After the new position $\mathbf{r}_{n+1}$ has been computed, an - additional accept/reject step is performed. 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 our Hydrogen atom example, $P$ is $\Psi^2$ and $g$ is a - solution of the discretized Fokker-Planck equation: - - \begin{eqnarray*} - P(r_{n}) &=& \Psi^2(\mathbf{r}_n) \\ - T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = & - \frac{1}{(4\pi\,D\,\tau)^{3/2}} \exp \left[ - \frac{\left( - \mathbf{r}_{n+1} - \mathbf{r}_{n} - 2D \frac{\nabla - \Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{4D\,\tau} \right] - \end{eqnarray*} - - The accept/reject step is the following: - - Compute $A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1})$. - - Draw a uniform random number $u$ - - if $u \le A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1})$, accept - the move - - if $u>A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1})$, reject - the move: set $\mathbf{r}_{n+1} = \mathbf{r}_{n}$, but *don't remove the sample from the average!* - - The /acceptance rate/ is the ratio of the number of accepted step - over the total number of steps. The time step should be adapted so - that the acceptance rate is around 0.5 for a good efficiency of - the simulation. - -**** Exercise - - #+begin_exercise - Modify the previous program to introduce the accept/reject step. - You should recover the unbiased result. - Adjust the time-step so that the acceptance rate is 0.5. - #+end_exercise - - *Python* - #+BEGIN_SRC python :results output + *Python* + #+BEGIN_SRC python :results output from hydrogen import * from qmc_stats import * @@ -1305,8 +1170,8 @@ def MonteCarlo(a,tau,nmax): d_old = d_new d2_old = d2_new psi_old = psi_new - N += 1. - E += e_loc(a,r_old) + N += 1. + E += e_loc(a,r_old) return E/N, accep_rate/N @@ -1317,14 +1182,14 @@ X = [MonteCarlo(a,tau,nmax) for i in range(30)] E, deltaE = ave_error([x[0] for x in X]) A, deltaA = ave_error([x[1] for x in X]) print(f"E = {E} +/- {deltaE}\nA = {A} +/- {deltaA}") - #+END_SRC + #+END_SRC - #+RESULTS: - : E = -0.4949730317138491 +/- 0.00012478601801760644 - : A = 0.7887163333333334 +/- 0.00026834549840347617 + #+RESULTS: + : E = -0.4949730317138491 +/- 0.00012478601801760644 + : A = 0.7887163333333334 +/- 0.00026834549840347617 - *Fortran* - #+BEGIN_SRC f90 + *Fortran* + #+BEGIN_SRC f90 subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate) implicit none double precision, intent(in) :: a, tau @@ -1339,7 +1204,7 @@ subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate) double precision, external :: e_loc, psi sq_tau = dsqrt(tau) - + ! Initialization energy = 0.d0 norm = 0.d0 @@ -1357,8 +1222,8 @@ subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate) 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)) + (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)*tau + prod q = psi_new / psi_old q = dexp(-argexpo) * q*q @@ -1396,18 +1261,17 @@ program qmc call ave_error(accep,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 vmc_metropolis.f90 -o vmc_metropolis ./vmc_metropolis - #+end_src + #+end_src - #+RESULTS: - : E = -0.49499990423528023 +/- 1.5958250761863871E-004 - : A = 0.78861366666666655 +/- 3.5096729498002445E-004 + #+RESULTS: + : E = -0.49499990423528023 +/- 1.5958250761863871E-004 + : A = 0.78861366666666655 +/- 3.5096729498002445E-004 - * TODO Diffusion Monte Carlo :PROPERTIES: :header-args:python: :tangle dmc.py @@ -1576,12 +1440,8 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis coordinates and $\mathbf{R}_A$ and $\mathbf{R}_B$ the coordinates of the nuclei. - - - - -* Appendix +* Appendix :noexport: ** Gaussian sampling :noexport: :PROPERTIES: @@ -1594,49 +1454,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis Instead of drawing uniform random numbers, we will draw Gaussian random numbers centered on 0 and with a variance of 1. - - 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*} - - Here is a Fortran implementation returning a Gaussian-distributed - n-dimensional vector $\mathbf{z}$; - - *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 - + Now the sampling probability can be inserted into the equation of the energy: \[ @@ -1755,3 +1573,192 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_gaussian.f90 -o qmc_gaussian #+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: + + \[ + \mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau\, 2D \frac{\nabla + \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi + \] + where $\chi$ is a Gaussian random variable with zero mean and + variance $\tau\,2D$. + +**** 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 +from hydrogen import * +from qmc_stats import * + +def MonteCarlo(a,tau,nmax): + sq_tau = np.sqrt(tau) + + # 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 + tau * d_old + chi*sq_tau + N += 1. + E += e_loc(a,r_new) + r_old = r_new + return E/N + + +a = 0.9 +nmax = 100000 +tau = 0.2 +X = [MonteCarlo(a,tau,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,tau,nmax,energy) + implicit none + double precision, intent(in) :: a, tau + 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_tau, chi(3) + double precision, external :: e_loc + + sq_tau = dsqrt(tau) + + ! 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(:) + tau * d_old(:) + chi(:)*sq_tau + 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 = 0.9 + double precision, parameter :: tau = 0.2 + 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,tau,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 +