diff --git a/index.html b/index.html index d4476b5..e3c21b0 100644 --- a/index.html +++ b/index.html @@ -3,7 +3,7 @@ "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> - + Quantum Monte Carlo @@ -329,152 +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 @@ -514,8 +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 @@ -593,8 +593,8 @@ $$ E ≈ \frac{1}{M} ∑i=1M EL(\mathbf{r

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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 @@ -623,8 +623,8 @@ 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. @@ -651,8 +651,8 @@ to catch the error.

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2.1.1 Exercise 1

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2.1.1 Exercise 1

@@ -696,8 +696,8 @@ 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 @@ -737,8 +737,8 @@ and returns the potential.

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2.1.2 Exercise 2

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2.1.2 Exercise 2

@@ -773,8 +773,8 @@ 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 @@ -801,8 +801,8 @@ input arguments, and returns a scalar.

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2.1.3 Exercise 3

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2.1.3 Exercise 3

@@ -848,12 +848,12 @@ applied to the wave function gives:

\[ - \Delta \Psi (\mathbf{r}) = \left(a^2 - \frac{2a}{\mathbf{|r|}} \right) \Psi(\mathbf{r}) + \Delta \Psi (\mathbf{r}) = \left(a^2 - \frac{2a}{\mathbf{|r|}} \right) \Psi(\mathbf{r})\,. \]

-So the local kinetic energy is +Therefore, the local kinetic energy is \[ -\frac{1}{2} \frac{\Delta \Psi}{\Psi} (\mathbf{r}) = -\frac{1}{2}\left(a^2 - \frac{2a}{\mathbf{|r|}} \right) \] @@ -883,8 +883,8 @@ So 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 @@ -925,8 +925,8 @@ So the local kinetic energy is

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2.1.4 Exercise 4

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2.1.4 Exercise 4

@@ -969,8 +969,8 @@ local kinetic energy.

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2.1.4.1 Solution   solution
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2.1.4.1 Solution   solution

Python @@ -1000,8 +1000,8 @@ local kinetic energy.

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2.1.5 Exercise 5

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2.1.5 Exercise 5

@@ -1011,8 +1011,8 @@ 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} - @@ -1032,8 +1032,8 @@ 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

@@ -1044,8 +1044,8 @@ choose a grid which does not contain the origin.

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2.2.1 Exercise

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2.2.1 Exercise

@@ -1128,8 +1128,8 @@ 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 @@ -1204,14 +1204,14 @@ 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 \(\mathbf{r}_i\) of size \(\delta \mathbf{r}\), the expression of \(\langle E_L \rangle_{\Psi^2}\) becomes a weighted average of the local energy, where the weights -are the values of the probability density at \(\mathbf{r}_i\) +are the values of the wave function square at \(\mathbf{r}_i\) multiplied by the volume element:

@@ -1235,12 +1235,12 @@ The energy is biased because:
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2.3.1 Exercise

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2.3.1 Exercise

-Compute a numerical estimate of the energy in a grid of +Compute a numerical estimate of the energy using a grid of \(50\times50\times50\) points in the range \((-5,-5,-5) \le \mathbf{r} \le (5,5,5)\).

@@ -1305,8 +1305,8 @@ 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 @@ -1421,8 +1421,8 @@ 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 \(\Psi\) @@ -1449,8 +1449,8 @@ 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)

@@ -1461,8 +1461,8 @@ 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} @@ -1481,13 +1481,13 @@ Prove that :

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2.4.2 Exercise

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2.4.2 Exercise

Add the calculation of the variance to the previous code, and -compute a numerical estimate of the variance of the local energy in +compute a numerical estimate of the variance of the local energy using a grid of \(50\times50\times50\) points in the range \((-5,-5,-5) \le \mathbf{r} \le (5,5,5)\) for different values of \(a\).

@@ -1556,8 +1556,8 @@ 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 @@ -1694,8 +1694,8 @@ 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 @@ -1711,8 +1711,8 @@ 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 \(M\) @@ -1752,8 +1752,8 @@ And the confidence interval is given by

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3.1.1 Exercise

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3.1.1 Exercise

@@ -1791,8 +1791,8 @@ input array.

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3.1.1.1 Solution   solution
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3.1.1.1 Solution   solution

Python @@ -1851,8 +1851,8 @@ 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 do our first Monte Carlo calculation to compute the @@ -1886,8 +1886,8 @@ compute the statistical error.

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3.2.1 Exercise

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3.2.1 Exercise

@@ -1987,8 +1987,8 @@ 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 @@ -2102,8 +2102,8 @@ E = -0.49518773675598715 +/- 5.2391494923686175E-004

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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 @@ -2191,8 +2191,8 @@ step such that the acceptance rate is close to 0.5 is a good compromise.

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3.3.1 Exercise

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3.3.1 Exercise

@@ -2299,8 +2299,8 @@ Can you observe a reduction in the statistical error?

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3.3.1.1 Solution   solution
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3.3.1.1 Solution   solution

Python @@ -2445,8 +2445,8 @@ A = 0.51695266666666673 +/- 4.0445505648997396E-004

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3.4 Gaussian random number generator

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3.4 Gaussian random number generator

To obtain Gaussian-distributed random numbers, you can apply the @@ -2508,8 +2508,8 @@ In Python, you can use the -

3.5 Generalized Metropolis algorithm

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3.5 Generalized Metropolis algorithm

One can use more efficient numerical schemes to move the electrons, @@ -2608,8 +2608,8 @@ The transition probability becomes:

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3.5.1 Exercise 1

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3.5.1 Exercise 1

@@ -2643,8 +2643,8 @@ Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\P

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3.5.1.1 Solution   solution
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3.5.1.1 Solution   solution

Python @@ -2677,8 +2677,8 @@ Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\P

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3.5.2 Exercise 2

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3.5.2 Exercise 2

@@ -2772,8 +2772,8 @@ Modify the previous program to introduce the drifted diffusion scheme.

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3.5.2.1 Solution   solution
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3.5.2.1 Solution   solution

Python @@ -2959,12 +2959,12 @@ A = 0.78839866666666658 +/- 3.2503783452043152E-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: @@ -3023,8 +3023,8 @@ system.

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4.2 Diffusion and branching

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4.2 Diffusion and branching

The diffusion equation of particles is given by @@ -3078,8 +3078,8 @@ the combination of a diffusion process and a branching process.

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4.3 Importance sampling

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