diff --git a/index.html b/index.html index 1343275..e538905 100644 --- a/index.html +++ b/index.html @@ -3,7 +3,7 @@ "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
- +This website contains the QMC tutorial of the 2021 LTTC winter school @@ -514,8 +514,8 @@ coordinates, etc).
For a given system with Hamiltonian \(\hat{H}\) and wave function \(\Psi\), we define the local energy as @@ -598,8 +598,8 @@ energy computed over these configurations:
In this section, we consider the hydrogen atom with the following @@ -628,8 +628,8 @@ To do that, we will compute the local energy and check whether it is constant.
You will now program all quantities needed to compute the local energy of the H atom for the given wave function. @@ -656,8 +656,8 @@ to catch the error.
@@ -702,8 +702,8 @@ and returns the potential.
Python @@ -744,8 +744,8 @@ and returns the potential.
@@ -780,8 +780,8 @@ input arguments, and returns a scalar.
Python @@ -808,8 +808,8 @@ input arguments, and returns a scalar.
@@ -890,8 +890,8 @@ Therefore, the local kinetic energy is
Python @@ -932,8 +932,8 @@ Therefore, the local kinetic energy is
@@ -992,8 +992,8 @@ are calling is yours.
Python @@ -1024,8 +1024,8 @@ are calling is yours.
@@ -1035,8 +1035,8 @@ Find the theoretical value of \(a\) for which \(\Psi\) is an eigenfunction of \(
The program you will write in this section will be written in @@ -1088,8 +1088,8 @@ In Fortran, you will need to compile all the source files together:
@@ -1183,8 +1183,8 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \
Python @@ -1261,8 +1261,8 @@ plt.savefig("plot_py.png")
If the space is discretized in small volume elements \(\mathbf{r}_i\) @@ -1292,8 +1292,8 @@ The energy is biased because:
@@ -1364,8 +1364,8 @@ To compile the Fortran and run it:
Python @@ -1482,8 +1482,8 @@ a = 2.0000000000000000 E = -8.0869806678448772E-002
The variance of the local energy is a functional of \(\Psi\) @@ -1510,8 +1510,8 @@ energy can be used as a measure of the quality of a wave function.
@@ -1522,8 +1522,8 @@ Prove that :
\(\bar{E} = \langle E \rangle\) is a constant, so \(\langle \bar{E} @@ -1542,8 +1542,8 @@ Prove that :
@@ -1619,8 +1619,8 @@ To compile and run:
Python @@ -1759,8 +1759,8 @@ a = 2.0000000000000000 E = -8.0869806678448772E-002 s2 = 1.8068814
Numerical integration with deterministic methods is very efficient @@ -1776,8 +1776,8 @@ interval.
To compute the statistical error, you need to perform \(M\) @@ -1817,8 +1817,8 @@ And the confidence interval is given by
@@ -1858,8 +1858,8 @@ input array.
Python @@ -1920,8 +1920,8 @@ input array.
We will now perform our first Monte Carlo calculation to compute the @@ -1982,8 +1982,8 @@ compute the statistical error.
@@ -2085,8 +2085,8 @@ well as the index of the current step.
Python @@ -2192,8 +2192,8 @@ E = -0.48084122147238995 +/- 2.4983775878329355E-003
We will now use the square of the wave function to sample random @@ -2312,8 +2312,8 @@ All samples should be kept, from both accepted and rejected moves.
If the box is infinitely small, the ratio will be very close @@ -2348,8 +2348,8 @@ the same variable later on to store a time step.
@@ -2458,8 +2458,8 @@ Can you observe a reduction in the statistical error?
Python @@ -2606,8 +2606,8 @@ A = 0.50762633333333318 +/- 3.4601756760043725E-004
One can use more efficient numerical schemes to move the electrons by choosing a smarter expression for the transition probability. @@ -2728,8 +2728,8 @@ The algorithm of the previous exercise is only slighlty modified as:
To obtain Gaussian-distributed random numbers, you can apply the
@@ -2793,8 +2793,8 @@ In Python, you can use the
-
@@ -2836,8 +2836,8 @@ Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\P
Python
@@ -2870,8 +2870,8 @@ Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\P
@@ -2967,8 +2967,8 @@ Modify the previous program to introduce the drift-diffusion scheme.
Python
@@ -3156,8 +3156,8 @@ A = 0.62037333333333333 +/- 4.8970160591451110E-004
As we have seen, Variational Monte Carlo is a powerful method to
@@ -3174,8 +3174,8 @@ finding a near-exact numerical solution to the Schrödinger equation.
Consider the time-dependent Schrödinger equation:
@@ -3243,8 +3243,8 @@ system.
The diffusion equation of particles is given by
@@ -3324,8 +3324,8 @@ Therefore, in both cases, you are dealing with a "Bosonic" ground state.
In a molecular system, the potential is far from being constant
@@ -3423,8 +3423,8 @@ energies computed with the trial wave function.
\[
@@ -3485,8 +3485,8 @@ Defining \(\Pi(\mathbf{r},t) = \psi(\mathbf{r},\tau)
Instead of having a variable number of particles to simulate the
@@ -3575,13 +3575,13 @@ the DMC algorithm. However, its use reduces significantly the time-step error.
@@ -3683,8 +3683,8 @@ time \(\tau_{\text{max}}\) =100 a.u.
Python
@@ -3904,8 +3904,8 @@ A = 0.98963533333333342 +/- 6.3052128284666221E-005
-
Change your PDMC code for one of the following:
@@ -3923,8 +3923,8 @@ And compute the ground state energy.
-3.4.2 Exercise 1
+3.4.2 Exercise 1
3.4.2.1 Solution solution
+3.4.2.1 Solution solution
3.4.3 Exercise 2
+3.4.3 Exercise 2
3.4.3.1 Solution solution
+3.4.3.1 Solution solution
4 Diffusion Monte Carlo
+4 Diffusion Monte Carlo
4.1 Schrödinger equation in imaginary time
+4.1 Schrödinger equation in imaginary time
4.2 Relation to diffusion
+4.2 Relation to diffusion
4.3 Importance sampling
+4.3 Importance sampling
4.3.1 Appendix : Details of the Derivation
+4.3.1 Appendix : Details of the Derivation
4.4 Pure Diffusion Monte Carlo
+4.4 Pure Diffusion Monte Carlo
4.5 Hydrogen atom
+4.5 Hydrogen atom
4.5.1 Exercise
+4.5.1 Exercise
4.5.1.1 Solution solution
+4.5.1.1 Solution solution
5 Project
+5 Project
6 Acknowledgments
+6 Acknowledgments