From afe3c3e4de14d2d2929bfd34ef08fc5041f95636 Mon Sep 17 00:00:00 2001 From: scemama Date: Tue, 2 Feb 2021 23:06:01 +0000 Subject: [PATCH] deploy: 707338aa8af92563b60b74e8532f330575a281bc --- index.html | 345 +++++++++++++++++++++++++---------------------------- 1 file changed, 162 insertions(+), 183 deletions(-) diff --git a/index.html b/index.html index db0ab47..f720613 100644 --- a/index.html +++ b/index.html @@ -3,7 +3,7 @@ "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> - + Quantum Monte Carlo @@ -329,153 +329,152 @@ for the JavaScript code in this tag.

Table of Contents

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1 Introduction

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1 Introduction

This website contains the QMC tutorial of the 2021 LTTC winter school @@ -515,8 +514,8 @@ coordinates, etc).

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1.1 Energy and local energy

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1.1 Energy and local energy

For a given system with Hamiltonian \(\hat{H}\) and wave function \(\Psi\), we define the local energy as @@ -599,8 +598,8 @@ energy computed over these configurations:

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2 Numerical evaluation of the energy of the hydrogen atom

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2 Numerical evaluation of the energy of the hydrogen atom

In this section, we consider the hydrogen atom with the following @@ -629,8 +628,8 @@ To do that, we will compute the local energy and check whether it is constant.

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2.1 Local energy

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2.1 Local energy

You will now program all quantities needed to compute the local energy of the H atom for the given wave function. @@ -657,8 +656,8 @@ to catch the error.

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2.1.1 Exercise 1

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2.1.1 Exercise 1

@@ -703,8 +702,8 @@ and returns the potential.

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2.1.1.1 Solution   solution
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2.1.1.1 Solution   solution

Python @@ -745,8 +744,8 @@ and returns the potential.

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2.1.2 Exercise 2

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2.1.2 Exercise 2

@@ -781,8 +780,8 @@ input arguments, and returns a scalar.

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2.1.2.1 Solution   solution
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2.1.2.1 Solution   solution

Python @@ -809,8 +808,8 @@ input arguments, and returns a scalar.

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2.1.3 Exercise 3

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2.1.3 Exercise 3

@@ -891,8 +890,8 @@ Therefore, the local kinetic energy is

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2.1.3.1 Solution   solution
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2.1.3.1 Solution   solution

Python @@ -933,8 +932,8 @@ Therefore, the local kinetic energy is

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2.1.4 Exercise 4

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2.1.4 Exercise 4

@@ -993,8 +992,8 @@ are calling is yours.

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2.1.4.1 Solution   solution
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2.1.4.1 Solution   solution

Python @@ -1025,8 +1024,8 @@ are calling is yours.

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2.1.5 Exercise 5

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2.1.5 Exercise 5

@@ -1036,8 +1035,8 @@ Find the theoretical value of \(a\) for which \(\Psi\) is an eigenfunction of \(

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2.1.5.1 Solution   solution
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2.1.5.1 Solution   solution
\begin{eqnarray*} E &=& \frac{\hat{H} \Psi}{\Psi} = - \frac{1}{2} \frac{\Delta \Psi}{\Psi} - @@ -1057,8 +1056,8 @@ equal to -0.5 atomic units.
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2.2 Plot of the local energy along the \(x\) axis

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2.2 Plot of the local energy along the \(x\) axis

The program you will write in this section will be written in @@ -1089,8 +1088,8 @@ In Fortran, you will need to compile all the source files together:

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2.2.1 Exercise

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2.2.1 Exercise

@@ -1184,8 +1183,8 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \

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2.2.1.1 Solution   solution
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2.2.1.1 Solution   solution

Python @@ -1262,8 +1261,8 @@ plt.savefig("plot_py.png")

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2.3 Numerical estimation of the energy

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2.3 Numerical estimation of the energy

If the space is discretized in small volume elements \(\mathbf{r}_i\) @@ -1293,8 +1292,8 @@ The energy is biased because:

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2.3.1 Exercise

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2.3.1 Exercise

@@ -1365,8 +1364,8 @@ To compile the Fortran and run it:

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2.3.1.1 Solution   solution
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2.3.1.1 Solution   solution

Python @@ -1483,8 +1482,8 @@ a = 2.0000000000000000 E = -8.0869806678448772E-002

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2.4 Variance of the local energy

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2.4 Variance of the local energy

The variance of the local energy is a functional of \(\Psi\) @@ -1511,8 +1510,8 @@ energy can be used as a measure of the quality of a wave function.

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2.4.1 Exercise (optional)

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2.4.1 Exercise (optional)

@@ -1523,8 +1522,8 @@ Prove that :

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2.4.1.1 Solution   solution
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2.4.1.1 Solution   solution

\(\bar{E} = \langle E \rangle\) is a constant, so \(\langle \bar{E} @@ -1543,8 +1542,8 @@ Prove that :

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2.4.2 Exercise

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2.4.2 Exercise

@@ -1620,8 +1619,8 @@ To compile and run:

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2.4.2.1 Solution   solution
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2.4.2.1 Solution   solution

Python @@ -1760,8 +1759,8 @@ a = 2.0000000000000000 E = -8.0869806678448772E-002 s2 = 1.8068814

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3 Variational Monte Carlo

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3 Variational Monte Carlo

Numerical integration with deterministic methods is very efficient @@ -1777,8 +1776,8 @@ interval.

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3.1 Computation of the statistical error

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3.1 Computation of the statistical error

To compute the statistical error, you need to perform \(M\) @@ -1818,8 +1817,8 @@ And the confidence interval is given by

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3.1.1 Exercise

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3.1.1 Exercise

@@ -1859,8 +1858,8 @@ input array.

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3.1.1.1 Solution   solution
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3.1.1.1 Solution   solution

Python @@ -1921,8 +1920,8 @@ input array.

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3.2 Uniform sampling in the box

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3.2 Uniform sampling in the box

We will now perform our first Monte Carlo calculation to compute the @@ -1983,8 +1982,8 @@ compute the statistical error.

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3.2.1 Exercise

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3.2.1 Exercise

@@ -2086,8 +2085,8 @@ well as the index of the current step.

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3.2.1.1 Solution   solution
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3.2.1.1 Solution   solution

Python @@ -2193,8 +2192,8 @@ E = -0.48084122147238995 +/- 2.4983775878329355E-003

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3.3 Metropolis sampling with \(\Psi^2\)

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3.3 Metropolis sampling with \(\Psi^2\)

We will now use the square of the wave function to sample random @@ -2313,8 +2312,8 @@ All samples should be kept, from both accepted and rejected moves.

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3.3.1 Optimal step size

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3.3.1 Optimal step size

If the box is infinitely small, the ratio will be very close @@ -2349,8 +2348,8 @@ the same variable later on to store a time step.

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3.3.2 Exercise

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3.3.2 Exercise

@@ -2459,8 +2458,8 @@ Can you observe a reduction in the statistical error?

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3.3.2.1 Solution   solution
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3.3.2.1 Solution   solution

Python @@ -2607,8 +2606,8 @@ A = 0.50762633333333318 +/- 3.4601756760043725E-004

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3.4 Generalized Metropolis algorithm

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3.4 Generalized Metropolis algorithm

One can use more efficient numerical schemes to move the electrons by choosing a smarter expression for the transition probability. @@ -2729,8 +2728,8 @@ The algorithm of the previous exercise is only slighlty modified as:

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3.4.1 Gaussian random number generator

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3.4.1 Gaussian random number generator

To obtain Gaussian-distributed random numbers, you can apply the @@ -2794,8 +2793,8 @@ In Python, you can use the -

3.4.2 Exercise 1

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3.4.2 Exercise 1

@@ -2837,8 +2836,8 @@ Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\P

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3.4.2.1 Solution   solution
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3.4.2.1 Solution   solution

Python @@ -2871,8 +2870,8 @@ Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\P

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3.4.3 Exercise 2

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3.4.3 Exercise 2

@@ -2968,8 +2967,8 @@ Modify the previous program to introduce the drift-diffusion scheme.

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3.4.3.1 Solution   solution
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3.4.3.1 Solution   solution

Python @@ -3157,8 +3156,8 @@ A = 0.62037333333333333 +/- 4.8970160591451110E-004

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4 Diffusion Monte Carlo   solution

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4 Diffusion Monte Carlo   solution

As we have seen, Variational Monte Carlo is a powerful method to @@ -3175,8 +3174,8 @@ finding a near-exact numerical solution to the Schrödinger equation.

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4.1 Schrödinger equation in imaginary time

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4.1 Schrödinger equation in imaginary time

Consider the time-dependent Schrödinger equation: @@ -3244,8 +3243,8 @@ system.

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4.2 Relation to diffusion

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4.2 Relation to diffusion

The diffusion equation of particles is given by @@ -3325,8 +3324,8 @@ Therefore, in both cases, you are dealing with a "Bosonic" ground state.

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4.3 Importance sampling

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4.3 Importance sampling

In a molecular system, the potential is far from being constant @@ -3424,8 +3423,8 @@ energies computed with the trial wave function.

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4.3.1 Appendix : Details of the Derivation

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4.3.1 Appendix : Details of the Derivation

\[ @@ -3486,8 +3485,8 @@ Defining \(\Pi(\mathbf{r},t) = \psi(\mathbf{r},\tau)

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4.4 Pure Diffusion Monte Carlo

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4.4 Pure Diffusion Monte Carlo

Instead of having a variable number of particles to simulate the @@ -3578,13 +3577,13 @@ the DMC algorithm. However, its use reduces significantly the time-step error. -

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4.5 Hydrogen atom

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4.5 Hydrogen atom

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4.5.1 Exercise

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4.5.1 Exercise

@@ -3685,8 +3684,8 @@ We choose here a fixed projection time \(\tau=10\) a.u.

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4.5.1.1 Solution   solution
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4.5.1.1 Solution   solution

Python @@ -3902,32 +3901,12 @@ A = 0.98963533333333342 +/- 6.3052128284666221E-005

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4.6 H2

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-We will now consider the H2 molecule in a minimal basis composed of the -\(1s\) orbitals of the hydrogen atoms: -

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-\[ - \Psi(\mathbf{r}_1, \mathbf{r}_2) = - \exp(-(\mathbf{r}_1 - \mathbf{R}_A)) + - \] -where \(\mathbf{r}_1\) and \(\mathbf{r}_2\) denote the electron -coordinates and \(\mathbf{R}_A\) and \(\mathbf{R}_B\) the coordinates of -the nuclei. -

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5 Project

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5 Project

Change your PDMC code for one of the following: @@ -3944,8 +3923,8 @@ And compute the ground state energy.

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6 Schedule

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6 Schedule

@@ -4014,7 +3993,7 @@ And compute the ground state energy.

Author: Anthony Scemama, Claudia Filippi

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Created: 2021-02-02 Tue 23:04

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Created: 2021-02-02 Tue 23:06

Validate