diff --git a/index.html b/index.html index 97316a3..f746ae4 100644 --- a/index.html +++ b/index.html @@ -3,7 +3,7 @@ "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
- +We propose different exercises to understand quantum Monte Carlo (QMC) @@ -365,8 +366,8 @@ interpreted as a single precision value
In this section we consider the Hydrogen atom with the following @@ -440,13 +441,13 @@ E & = & \frac{\langle \Psi| \hat{H} | \Psi\rangle}{\langle \Psi |\Psi \rangle} \end{eqnarray*}
@@ -490,8 +491,8 @@ and returns the potential.
@@ -526,8 +527,8 @@ input arguments, and returns a scalar.
@@ -608,8 +609,8 @@ So the local kinetic energy is
@@ -652,14 +653,14 @@ local energy.
@@ -776,8 +777,8 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \
If the space is discretized in small volume elements \(\mathbf{r}_i\) @@ -807,8 +808,8 @@ The energy is biased because:
@@ -918,8 +919,8 @@ a = 2.0000000000000000 E = -8.0869806678448772E-002
The variance of the local energy is a functional of \(\Psi\) @@ -936,7 +937,7 @@ which can be simplified as
-\[ \sigma^2(E_L) = \langle E^2 \rangle - \langle E \rangle^2 \] +\[ \sigma^2(E_L) = \langle E_L^2 \rangle - \langle E_L \rangle^2 \]
@@ -946,8 +947,8 @@ energy can be used as a measure of the quality of a wave function.
@@ -959,8 +960,8 @@ Prove that :
@@ -1095,8 +1096,8 @@ a = 2.0000000000000000 E = -8.0869806678448772E-002 s2 = 1.806881
Numerical integration with deterministic methods is very efficient @@ -1112,8 +1113,8 @@ interval.
To compute the statistical error, you need to perform \(M\) @@ -1153,8 +1154,8 @@ And the confidence interval is given by
@@ -1203,8 +1204,8 @@ input array.
We will now do our first Monte Carlo calculation to compute the @@ -1238,8 +1239,8 @@ statistical error.
@@ -1349,19 +1350,228 @@ E = -0.49588321986667677 +/- 7.1758863546737969E-004
-We will now improve the sampling and allow to sample in the whole -3D space, correcting the bias related to the sampling in the box. +We will now use the square of the wave function to sample random +points distributed with the probability density +\[ + P(\mathbf{r}) = \left[\Psi(\mathbf{r})\right]^2 + \]
-Instead of drawing uniform random numbers, we will draw Gaussian -random numbers centered on 0 and with a variance of 1. +The expression of the average energy is now simplified to the average of +the local energies, since the weights are taken care of by the +sampling :
++\[ + E \approx \frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r}_i) + \] +
+ + ++To sample a chosen probability density, an efficient method is the +Metropolis-Hastings sampling algorithm. Starting from a random +initial position \(\mathbf{r}_0\), we will realize a random walk as follows: +
+ ++\[ + \mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \mathbf{u} + \] +
+ ++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 +\(\mathbf{r}_n\) is \(\Psi^2\): +
+ ++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. +
+ ++If the time step is infinitely small, the ratio will be very close +to one and all the steps will be accepted. But the trajectory will +be infinitely too short to have statistical significance. +
+ ++On the other hand, as the time step increases, 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 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 +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. +
++Modify the program of the previous section to compute the energy, sampling 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? +
+ ++Python +
+from hydrogen import * +from qmc_stats import * + +def MonteCarlo(a,nmax,tau): + E = 0. + N = 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. + r_old = r_new + psi_old = psi_new + N += 1. + E += e_loc(a,r_old) + return E/N, N_accep/N + +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 ] +E, deltaE = ave_error(X) +A, deltaA = ave_error(A) +print(f"E = {E} +/- {deltaE}") +print(f"A = {A} +/- {deltaA}") ++
+Fortran +
+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 :: norm, r_old(3), r_new(3), psi_old, psi_new + double precision :: v, ratio, 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 +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 ++
gfortran hydrogen.f90 qmc_stats.f90 qmc_metropolis.f90 -o qmc_metropolis +./qmc_metropolis ++
+E = -0.49478505004797046 +/- 2.0493795299184956E-004 +A = 0.51737800000000000 +/- 4.1827406733181444E-004 + ++
To obtain Gaussian-distributed random numbers, you can apply the Box Muller transform to uniform random numbers: @@ -1373,8 +1583,9 @@ 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 \(\mathbf{z}\); +Below is a Fortran implementation returning a Gaussian-distributed +n-dimensional vector \(\mathbf{z}\). This will be useful for the +following sections.
@@ -1410,296 +1621,112 @@ n-dimensional vector \(\mathbf{z}\); end subroutine random_gauss
+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 \(\mathbf{r}_{n+1}\): +
-Now the sampling probability can be inserted into the equation of the energy: +\[ A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \min \left( 1, + \frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})} + {T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})} + \right) + \] +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
\[ - E = \frac{\int P(\mathbf{r}) \frac{\left[\Psi(\mathbf{r})\right]^2}{P(\mathbf{r})}\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int P(\mathbf{r}) \frac{\left[\Psi(\mathbf{r}) \right]^2}{P(\mathbf{r})} d\mathbf{r}} + T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = & + \text{constant} \]
-with the Gaussian probability +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:
\[ - P(\mathbf{r}) = \frac{1}{(2 \pi)^{3/2}}\exp\left( -\frac{\mathbf{r}^2}{2} \right). + T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = & + \frac{1}{(2\pi\,\tau)^{3/2}} \exp \left[ - \frac{\left( + \mathbf{r}_{n+1} - \mathbf{r}_{n} \right)^2}{2\tau} \right] + \] +
+ + ++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: +
+ ++\[ + \mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \frac{\nabla + \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi \]
-As the coordinates are drawn with probability \(P(\mathbf{r})\), the -average energy can be computed as -
- --\[ - E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\; - w_i = \frac{\left[\Psi(\mathbf{r}_i)\right]^2}{P(\mathbf{r}_i)} \delta \mathbf{r} - \] -
--Modify the exercise of the previous section to sample with -Gaussian-distributed random numbers. Can you see an reduction in -the statistical error? -
- --Python -
-from hydrogen import * -from qmc_stats import * - -norm_gauss = 1./(2.*np.pi)**(1.5) -def gaussian(r): - return norm_gauss * np.exp(-np.dot(r,r)*0.5) - -def MonteCarlo(a,nmax): - E = 0. - N = 0. - for istep in range(nmax): - r = np.random.normal(loc=0., scale=1.0, size=(3)) - w = psi(a,r) - w = w*w / gaussian(r) - N += w - E += w * e_loc(a,r) - return E/N - -a = 0.9 -nmax = 100000 -X = [MonteCarlo(a,nmax) for i in range(30)] -E, deltaE = ave_error(X) -print(f"E = {E} +/- {deltaE}") --
-Fortran -
-double precision function gaussian(r) - implicit none - double precision, intent(in) :: r(3) - double precision, parameter :: norm_gauss = 1.d0/(2.d0*dacos(-1.d0))**(1.5d0) - gaussian = norm_gauss * dexp( -0.5d0 * (r(1)*r(1) + r(2)*r(2) + r(3)*r(3) )) -end function gaussian - - -subroutine gaussian_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, gaussian - - energy = 0.d0 - norm = 0.d0 - do istep = 1,nmax - call random_gauss(r,3) - w = psi(a,r) - w = w*w / gaussian(r) - norm = norm + w - energy = energy + w * e_loc(a,r) - end do - energy = energy / norm -end subroutine gaussian_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 - - do irun=1,nruns - call gaussian_montecarlo(a,nmax,X(irun)) - enddo - call ave_error(X,nruns,ave,err) - print *, 'E = ', ave, '+/-', err -end program qmc --
gfortran hydrogen.f90 qmc_stats.f90 qmc_gaussian.f90 -o qmc_gaussian -./qmc_gaussian --
-E = -0.49517104619091717 +/- 1.0685523607878961E-004 - --
-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. -
- --\[ - E \approx \frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r}_i) - \] -
--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 \(\tau\,2D\). +variance \(\tau\). +The transition probability becomes: +
+ ++\[ + T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = & + \frac{1}{(2\pi\,\tau)^{3/2}} \exp \left[ - \frac{\left( + \mathbf{r}_{n+1} - \mathbf{r}_{n} - \frac{\nabla + \Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\tau} \right] + \]
Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\). @@ -1732,189 +1759,15 @@ Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\P
-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. -
- --Python -
-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}") --
-Fortran -
-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 --
gfortran hydrogen.f90 qmc_stats.f90 vmc.f90 -o vmc -./vmc --
-E = -0.48584030499187431 +/- 1.0411743995438257E-004 - --
-Discretizing the differential equation to generate the desired -probability density will suffer from a discretization error -leading to biases in the averages. The 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 \(\mathbf{r}_{n+1}\): -
- --\[ A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \min \left( 1, - \frac{g(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})} - {g(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})} - \right) - \] -
- --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) \\ -g(\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: -
--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. -
--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. +Modify the previous program to introduce the drifted diffusion scheme. +(This is a necessary step for the next section).