From 18d529e80caadf01c5d9500724e6ba9705038d25 Mon Sep 17 00:00:00 2001 From: scemama Date: Tue, 2 Feb 2021 16:05:21 +0000 Subject: [PATCH] deploy: d3d1e84680f9d04e07ff54c557069bbce5a9fca3 --- index.html | 349 ++++++++++++++++++++++++++--------------------------- 1 file changed, 173 insertions(+), 176 deletions(-) diff --git a/index.html b/index.html index d17b387..76dabc5 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 +329,153 @@ 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 +515,8 @@ 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 +599,8 @@ 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 +629,8 @@ 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 +657,8 @@ to catch the error.

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2.1.1 Exercise 1

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2.1.1 Exercise 1

@@ -703,8 +703,8 @@ 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 +745,8 @@ and returns the potential.

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2.1.2 Exercise 2

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2.1.2 Exercise 2

@@ -781,8 +781,8 @@ 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 +809,8 @@ 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 +891,8 @@ 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 +933,8 @@ 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 +993,8 @@ 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 +1025,8 @@ are calling is yours.

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2.1.5 Exercise 5

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2.1.5 Exercise 5

@@ -1036,8 +1036,8 @@ 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 +1057,8 @@ 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 +1089,8 @@ 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 +1184,8 @@ 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 +1262,8 @@ 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 +1293,8 @@ The energy is biased because:

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2.3.1 Exercise

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2.3.1 Exercise

@@ -1365,8 +1365,8 @@ 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 +1483,8 @@ 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 +1511,8 @@ 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 +1523,8 @@ 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 +1543,8 @@ Prove that :

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2.4.2 Exercise

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2.4.2 Exercise

@@ -1620,8 +1620,8 @@ 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 +1760,8 @@ 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 +1777,8 @@ 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 +1818,8 @@ 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 +1859,8 @@ 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 +1921,8 @@ 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 +1983,8 @@ compute the statistical error.

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3.2.1 Exercise

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3.2.1 Exercise

@@ -2086,8 +2086,8 @@ 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 +2193,8 @@ 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 +2313,8 @@ 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 +2349,8 @@ 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 +2459,8 @@ 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 +2607,8 @@ 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 +2729,8 @@ 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 +2794,8 @@ In Python, you can use the -

3.4.2 Exercise 1

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3.4.2 Exercise 1

@@ -2837,8 +2837,8 @@ 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 +2871,8 @@ 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 +2968,8 @@ 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 @@ -3157,12 +3157,12 @@ A = 0.62037333333333333 +/- 4.8970160591451110E-004

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4 Diffusion Monte Carlo   solution

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4 Diffusion Monte Carlo   solution

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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: @@ -3230,8 +3230,8 @@ system.

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4.2 Diffusion and branching

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4.2 Diffusion and branching

The imaginary-time Schrödinger equation can be explicitly written in terms of the kinetic and @@ -3328,8 +3328,8 @@ 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 @@ -3425,8 +3425,8 @@ 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

\[ @@ -3487,8 +3487,8 @@ Defining \(\Pi(\mathbf{r},t) = \psi(\mathbf{r},\tau)

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4.4 Pure Diffusion Monte Carlo (PDMC)

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4.4 Pure Diffusion Monte Carlo (PDMC)

Instead of having a variable number of particles to simulate the @@ -3514,23 +3514,20 @@ The algorithm can be rather easily built on top of your VMC code:

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  1. -Compute a new position \(\mathbf{r'} = \mathbf{r}_n + - \delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi\) -

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    -Evaluate \(\Psi\) and \(\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\) at the new position -

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  2. Compute the ratio \(A = \frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})}{T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})}\)
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  3. Start with \(W=1\)
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  4. Evaluate the local energy at \(\mathbf{r}_{n}\) and accumulate it
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  5. Compute the weight \(w(\mathbf{r}_n)\)
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  6. Update \(W\)
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  7. Compute a new position \(\mathbf{r'} = \mathbf{r}_n + + \delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi\)
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  8. Evaluate \(\Psi(\mathbf{r}')\) and \(\frac{\nabla \Psi(\mathbf{r'})}{\Psi(\mathbf{r'})}\) at the new position
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  9. Compute the ratio \(A = \frac{T(\mathbf{r}' \rightarrow \mathbf{r}_{n}) P(\mathbf{r}')}{T(\mathbf{r}_{n} \rightarrow \mathbf{r}') P(\mathbf{r}_{n})}\)
  10. Draw a uniform random number \(v \in [0,1]\)
  11. if \(v \le A\), accept the move : set \(\mathbf{r}_{n+1} = \mathbf{r'}\)
  12. else, reject the move : set \(\mathbf{r}_{n+1} = \mathbf{r}_n\)
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  13. evaluate the local energy at \(\mathbf{r}_{n+1}\)
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  14. compute the weight \(w(\mathbf{r}_i)\)
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  15. update \(W\)
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Some comments are needed:

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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

@@ -3672,8 +3669,8 @@ energy of H for any value of \(a\).

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4.5.1.1 Solution   solution
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4.5.1.1 Solution   solution

Python @@ -3891,8 +3888,8 @@ A = 0.98788066666666663 +/- 7.2889356133441110E-005

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4.6 TODO H2

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4.6 TODO H2

We will now consider the H2 molecule in a minimal basis composed of the @@ -3913,8 +3910,8 @@ the nuclei.

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5 TODO [0/3] Last things to do

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5 TODO [0/3] Last things to do

  • [ ] Give some hints of how much time is required for each section
    • @@ -3928,8 +3925,8 @@ the H\(_2\) molecule at $R$=1.4010 bohr. Answer: 0.17406 a.u.
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6 Schedule

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6 Schedule

@@ -3993,7 +3990,7 @@ the H\(_2\) molecule at $R$=1.4010 bohr. Answer: 0.17406 a.u.

Author: Anthony Scemama, Claudia Filippi

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Created: 2021-02-02 Tue 16:01

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Created: 2021-02-02 Tue 16:05

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