From 0fff33303731239b87677c469008fe463a22d49c Mon Sep 17 00:00:00 2001 From: scemama Date: Thu, 4 Feb 2021 16:27:33 +0000 Subject: [PATCH] deploy: 5baa6aaf74fa33d1f6dbb285af04d73683ad8183 --- index.html | 320 ++++++++++++++++++++++++++--------------------------- 1 file changed, 160 insertions(+), 160 deletions(-) diff --git a/index.html b/index.html index edeac2e..0ce0827 100644 --- a/index.html +++ b/index.html @@ -3,7 +3,7 @@ "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> - + Quantum Monte Carlo @@ -329,148 +329,148 @@ 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 @@ -510,8 +510,8 @@ 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 @@ -594,8 +594,8 @@ 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 @@ -624,8 +624,8 @@ 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. @@ -652,8 +652,8 @@ to catch the error.

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2.1.1 Exercise 1

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2.1.1 Exercise 1

@@ -698,8 +698,8 @@ and returns the potential.

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2.1.1.1 Solution   solution2
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2.1.1.1 Solution   solution2

Python @@ -740,8 +740,8 @@ and returns the potential.

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2.1.2 Exercise 2

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2.1.2 Exercise 2

@@ -776,8 +776,8 @@ input arguments, and returns a scalar.

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2.1.2.1 Solution   solution2
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2.1.2.1 Solution   solution2

Python @@ -804,8 +804,8 @@ input arguments, and returns a scalar.

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2.1.3 Exercise 3

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2.1.3 Exercise 3

@@ -886,8 +886,8 @@ Therefore, the local kinetic energy is

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2.1.3.1 Solution   solution2
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2.1.3.1 Solution   solution2

Python @@ -928,8 +928,8 @@ Therefore, the local kinetic energy is

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2.1.4 Exercise 4

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2.1.4 Exercise 4

@@ -988,8 +988,8 @@ are calling is yours.

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2.1.4.1 Solution   solution2
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2.1.4.1 Solution   solution2

Python @@ -1020,8 +1020,8 @@ are calling is yours.

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2.1.5 Exercise 5

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2.1.5 Exercise 5

@@ -1031,8 +1031,8 @@ Find the theoretical value of \(a\) for which \(\Psi\) is an eigenfunction of \(

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2.1.5.1 Solution   solution2
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2.1.5.1 Solution   solution2
\begin{eqnarray*} E &=& \frac{\hat{H} \Psi}{\Psi} = - \frac{1}{2} \frac{\Delta \Psi}{\Psi} - @@ -1052,8 +1052,8 @@ 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 @@ -1084,8 +1084,8 @@ 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

@@ -1179,8 +1179,8 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \

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2.2.1.1 Solution   solution2
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2.2.1.1 Solution   solution2

Python @@ -1257,8 +1257,8 @@ 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 \(\mathbf{r}_i\) @@ -1288,8 +1288,8 @@ The energy is biased because:

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2.3.1 Exercise

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2.3.1 Exercise

@@ -1360,8 +1360,8 @@ To compile the Fortran and run it:

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2.3.1.1 Solution   solution2
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2.3.1.1 Solution   solution2

Python @@ -1478,8 +1478,8 @@ 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 \(\Psi\) @@ -1506,8 +1506,8 @@ 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)

@@ -1518,8 +1518,8 @@ Prove that :

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2.4.1.1 DONE Solution   solution2
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2.4.1.1 DONE Solution   solution2

\(\bar{E} = \langle E \rangle\) is a constant, so \(\langle \bar{E} @@ -1538,8 +1538,8 @@ Prove that :

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2.4.2 Exercise

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2.4.2 Exercise

@@ -1615,8 +1615,8 @@ To compile and run:

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2.4.2.1 Solution   solution2
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2.4.2.1 Solution   solution2

Python @@ -1755,8 +1755,8 @@ 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 @@ -1772,8 +1772,8 @@ 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 \(M\) @@ -1813,8 +1813,8 @@ And the confidence interval is given by

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3.1.1 Exercise

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3.1.1 Exercise

@@ -1854,8 +1854,8 @@ input array.

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3.1.1.1 Solution   solution2
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3.1.1.1 Solution   solution2

Python @@ -1916,8 +1916,8 @@ 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 @@ -1978,8 +1978,8 @@ compute the statistical error.

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3.2.1 Exercise

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3.2.1 Exercise

@@ -2081,8 +2081,8 @@ well as the index of the current step.

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3.2.1.1 Solution   solution2
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3.2.1.1 Solution   solution2

Python @@ -2188,8 +2188,8 @@ 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 @@ -2308,8 +2308,8 @@ 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 @@ -2344,8 +2344,8 @@ 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

@@ -2454,8 +2454,8 @@ Can you observe a reduction in the statistical error?

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3.3.2.1 Solution   solution2
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3.3.2.1 Solution   solution2

Python @@ -2602,8 +2602,8 @@ 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. @@ -2724,8 +2724,8 @@ 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 @@ -2789,8 +2789,8 @@ In Python, you can use the -

3.4.2 Exercise 1

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3.4.2 Exercise 1

@@ -2832,8 +2832,8 @@ Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\P

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3.4.2.1 Solution   solution2
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3.4.2.1 Solution   solution2

Python @@ -2866,8 +2866,8 @@ 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

@@ -2963,8 +2963,8 @@ Modify the previous program to introduce the drift-diffusion scheme.

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3.4.3.1 Solution   solution2
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3.4.3.1 Solution   solution2

Python @@ -3152,8 +3152,8 @@ A = 0.62037333333333333 +/- 4.8970160591451110E-004

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4 Diffusion Monte Carlo

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4 Diffusion Monte Carlo

As we have seen, Variational Monte Carlo is a powerful method to @@ -3170,8 +3170,8 @@ 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: @@ -3239,8 +3239,8 @@ 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 @@ -3320,8 +3320,8 @@ 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 @@ -3419,8 +3419,8 @@ 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

\[ @@ -3481,8 +3481,8 @@ 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 @@ -3524,7 +3524,7 @@ starting from a VMC code: E_L(\mathbf{r}_n)\), and the weight \(W(\mathbf{r}_n)\) for the normalization

  • Update \(\tau_n = \tau_{n-1} + \delta t\)
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  • If \(\tau_{n} > \tau_\text{max}\), the long projection time has +
  • If \(\tau_{n} > \tau_\text{max}\) (\(\tau_\text{max}\) is an input parameter), the long projection time has been reached and we can start an new trajectory from the current position: reset \(W(r_n) = 1\) and \(\tau_n = 0\)
    • @@ -3571,13 +3571,13 @@ the DMC algorithm. However, its use reduces significantly the time-step error. -