diff --git a/index.html b/index.html index 7694664..8fd3089 100644 --- a/index.html +++ b/index.html @@ -3,7 +3,7 @@ "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> - + Quantum Monte Carlo @@ -329,151 +329,151 @@ 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 @@ -513,8 +513,8 @@ 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 @@ -592,8 +592,8 @@ If we can sample \(N_{\rm MC}\) configurations \(\{\mathbf{r}\}\) distributed as

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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 @@ -622,8 +622,8 @@ 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. @@ -650,8 +650,8 @@ to catch the error.

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2.1.1 Exercise 1

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2.1.1 Exercise 1

@@ -695,8 +695,8 @@ 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 @@ -736,8 +736,8 @@ and returns the potential.

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2.1.2 Exercise 2

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2.1.2 Exercise 2

@@ -772,8 +772,8 @@ 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 @@ -800,8 +800,8 @@ input arguments, and returns a scalar.

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2.1.3 Exercise 3

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2.1.3 Exercise 3

@@ -882,8 +882,8 @@ 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 @@ -924,8 +924,8 @@ Therefore, the local kinetic energy is

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2.1.4 Exercise 4

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2.1.4 Exercise 4

@@ -968,8 +968,8 @@ local kinetic energy.

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2.1.4.1 Solution   solution
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2.1.4.1 Solution   solution

Python @@ -999,8 +999,8 @@ local kinetic energy.

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2.1.5 Exercise 5

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2.1.5 Exercise 5

@@ -1010,8 +1010,8 @@ 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} - @@ -1031,8 +1031,8 @@ 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

@@ -1043,8 +1043,8 @@ 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

@@ -1127,8 +1127,8 @@ 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 @@ -1203,8 +1203,8 @@ 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 \(\mathbf{r}_i\) @@ -1234,8 +1234,8 @@ The energy is biased because:

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2.3.1 Exercise

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2.3.1 Exercise

@@ -1304,8 +1304,8 @@ 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 @@ -1420,8 +1420,8 @@ 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 \(\Psi\) @@ -1448,8 +1448,8 @@ 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)

@@ -1460,8 +1460,8 @@ 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} @@ -1480,8 +1480,8 @@ Prove that :

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2.4.2 Exercise

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2.4.2 Exercise

@@ -1555,8 +1555,8 @@ 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 @@ -1693,8 +1693,8 @@ 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 @@ -1710,8 +1710,8 @@ 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 \(M\) @@ -1751,8 +1751,8 @@ And the confidence interval is given by

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3.1.1 Exercise

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3.1.1 Exercise

@@ -1790,8 +1790,8 @@ input array.

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3.1.1.1 Solution   solution
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3.1.1.1 Solution   solution

Python @@ -1850,8 +1850,8 @@ 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 @@ -1912,8 +1912,8 @@ compute the statistical error.

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3.2.1 Exercise

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3.2.1 Exercise

@@ -2013,8 +2013,8 @@ 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 @@ -2128,8 +2128,8 @@ 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 @@ -2268,8 +2268,8 @@ the same variable later on to store a time step.

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3.3.1 Exercise

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3.3.1 Exercise

@@ -2376,8 +2376,8 @@ 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 @@ -2522,8 +2522,8 @@ 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 @@ -2586,8 +2586,8 @@ 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 by choosing a smarter expression for the transition probability. @@ -2714,8 +2714,8 @@ Evaluate \(\Psi\) and \(\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\) at th

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3.5.1 Exercise 1

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3.5.1 Exercise 1

@@ -2749,8 +2749,8 @@ 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 @@ -2783,8 +2783,8 @@ 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

@@ -2878,8 +2878,8 @@ Modify the previous program to introduce the drift-diffusion scheme.

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3.5.2.1 Solution   solution
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3.5.2.1 Solution   solution

Python @@ -3065,12 +3065,12 @@ 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: @@ -3138,8 +3138,8 @@ 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 @@ -3236,8 +3236,8 @@ 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 @@ -3293,7 +3293,7 @@ To this aim, we use the mixed estimator of the energy:

\begin{eqnarray*} - E(\tau) &=& \frac{\langle \psi(tau) | \hat{H} | \Psi_T \rangle}{\langle \psi(tau) | \Psi_T \rangle}\\ + E(\tau) &=& \frac{\langle \psi(\tau) | \hat{H} | \Psi_T \rangle}{\langle \psi(\tau) | \Psi_T \rangle}\\ &=& \frac{\int \psi(\mathbf{r},\tau) \hat{H} \Psi_T(\mathbf{r}) d\mathbf{r}} {\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}} \\ &=& \frac{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) E_L(\mathbf{r}) d\mathbf{r}} @@ -3333,8 +3333,8 @@ 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

\[ @@ -3395,8 +3395,8 @@ 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 @@ -3446,33 +3446,29 @@ Some comments are needed:

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  • You estimate the energy as
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+

  • +You estimate the energy as +

    \begin{eqnarray*} -E = \frac{\sum_{k=1}{N_{\rm MC}} E_L(\mathbf{r}_k) W(\mathbf{r}_k, k\delta t)}{\sum_{k=1}{N_{\rm MC}} W(\mathbf{r}_k, k\delta t)} -\end{eqnarray*} +E = \frac{\sum_{k=1}^{N_{\rm MC}} E_L(\mathbf{r}_k) W(\mathbf{r}_k, k\delta t)}{\sum_{k=1}^{N_{\rm MC}} W(\mathbf{r}_k, k\delta t)} +\end{eqnarray*}
    • -
      -
    • The result will be affected by a time-step error (the finite size of \(\delta t\)) and one
      • -
    -

    +

  • +The result will be affected by a time-step error (the finite size of \(\delta t\)) and one has in principle to extrapolate to the limit \(\delta t \rightarrow 0\). This amounts to fitting the energy computed for multiple values of \(\delta t\).

    Here, you will be using a small enough time-step and you should not worry about the extrapolation. -

    -
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    • The accept/reject step (steps 2-5 in the algorithm) is in principle not needed for the correctness of
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    • The accept/reject step (steps 2-5 in the algorithm) is in principle not needed for the correctness of +the DMC algorithm. However, its use reduces significantly the time-step error.
    -

    -the DMC algorithm. However, its use reduces significantly the time-step error. -

    -PDMC algorithm is less stable than the branching algorithm: it +The PDMC algorithm is less stable than the branching algorithm: it requires to have a value of \(E_\text{ref}\) which is close to the fixed-node energy, and a good trial wave function. Its big advantage is that it is very easy to program starting from a VMC @@ -3481,13 +3477,13 @@ code, so this is what we will do in the next section.

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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

    @@ -3586,8 +3582,8 @@ 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 @@ -3803,8 +3799,8 @@ 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 @@ -3825,8 +3821,8 @@ 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
      • @@ -3842,7 +3838,7 @@ the H\(_2\) molecule at $R$=1.4010 bohr. Answer: 0.17406 a.u.

    Author: Anthony Scemama, Claudia Filippi

    -

    Created: 2021-02-01 Mon 20:57

    +

    Created: 2021-02-01 Mon 21:10

    Validate