From a258c1c5642d0c3b8bf0aa7bc61bcb1b70fa3979 Mon Sep 17 00:00:00 2001 From: Anthony Scemama Date: Thu, 4 Feb 2021 15:19:56 +0100 Subject: [PATCH] Show solutions --- QMC.org | 109 ++++++++++++++++++++++++++++++++++++++++++++++++++++++-- 1 file changed, 106 insertions(+), 3 deletions(-) diff --git a/QMC.org b/QMC.org index 8acaef2..fe84601 100644 --- a/QMC.org +++ b/QMC.org @@ -681,7 +681,7 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen ./energy_hydrogen #+end_src -**** Solution :solution: +**** Solution :solution2: *Python* #+BEGIN_SRC python :results none :exports both #!/usr/bin/env python3 @@ -816,7 +816,7 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen $$\langle \left( E - \langle E \rangle_{\Psi^2} \right)^2\rangle_{\Psi^2} = \langle E^2 \rangle_{\Psi^2} - \langle E \rangle_{\Psi^2}^2 $$ #+end_exercise -**** Solution :solution: +**** DONE Solution :solution2: $\bar{E} = \langle E \rangle$ is a constant, so $\langle \bar{E} \rangle = \bar{E}$ . @@ -893,7 +893,7 @@ gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen #+end_src **** Solution :solution: - *Python* + *Python* #+BEGIN_SRC python :results none :exports both #!/usr/bin/env python3 @@ -2768,6 +2768,109 @@ gfortran hydrogen.f90 qmc_stats.f90 pdmc.f90 -o pdmc And compute the ground state energy. +* Exam :noexport: + + +** Question 1 + + Consider the hydrogen atom. You are using Monte Carlo sampling to + compute the energy associated with a wave function $\Psi(\mathbf{r})$. + If you use a Gaussian with mean 0 and variance 1 (centered on the + nucleus) to generate the random samples, the correct weight + $w(\mathbf{r})$ involved in the expectation value of the energy + $\frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r_i}) \times w(\mathbf{r_i})$ is: + + A - $w(\mathbf{r})= \left|\Psi(\mathbf{r})\right|^2$ + + B - $w(\mathbf{r})= \left( 2 \pi \right)^{3/2} \exp \left( \frac{\mathbf{r}^2}{2} \right) \left|\Psi(\mathbf{r})\right|^2$ + + C - $w(\mathbf{r})= \frac{1}{\left( 2 \pi \right)^{3/2}} \exp \left( -\frac{\mathbf{r}^2}{2} \right) \left|\Psi(\mathbf{r})\right|^2$ + + D - $w(\mathbf{r})= \frac{1}{\left( 2 \pi \right)^{3/2}} \exp \left( -\frac{\mathbf{r}^2}{2} \right)$ + + + + +** Question 2 + + In the exercises, you only considered "bosonic" wave functions. + Let's assume that you now deal with a system where the wave + function has nodes ($\Psi(\mathbf{r})=0$). + + i ) Does the local energy diverge at the nodes? + ii) The drift $\nabla \Psi / \Psi$ diverge at the nodes. Does it + push the electrons towards the nodes or away from the nodes? + + A - i) $E_L$ diverges, ii) $\nabla \Psi / \Psi$ pushes in the direction of the nodes + + B - i) $E_L$ diverges, ii) $\nabla \Psi / \Psi$ pushes away from the nodes + + C - i) $E_L$ is finite, ii) $\nabla \Psi / \Psi$ pushes in the direction of the nodes + + D - i) $E_L$ is finite, ii) $\nabla \Psi / \Psi$ pushes away from the nodes + + *Hint*: You can also think in one dimension if this helps you. + +** Question 3 + + Consider the helium atom in its singlet ground state with the wave function + \[ + \Psi(\mathbf{r}_1, \mathbf{r}_2) = \exp \left( - ( \mathbf{r}_1 + + \mathbf{r}_2 ) \right) + \]. + + When an electron approaches + i ) the nucleus or, + ii) the other electron, + + the local energy diverges to: + + A - i) $+\infty$ and ii) $-\infty$ + B - i) $+\infty$ and ii) $+\infty$ + C - i) $-\infty$ and ii) $-\infty$ + D - i) $-\infty$ and ii) $+\infty$ + + *Hint 1* : Recall the expression of the Laplacian for the single + electron case (hydrogen). + + *Hint 2* : Helium has $Z=2$. + +** Question 4 + + Consider a 2-electron system. + We propose a move of the 2-electron configuration according to a uniform + distribution $[-\delta L/2, \delta L/2]$ in all directions. + + What is the expression of the transition probability for + the i) forward and ii) reverse move? + + A - i ) Forward : $T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \frac{1}{(\delta L)^3}$ + + ii) Reverse : $T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) = -\frac{1}{(\delta L)^3}$ + + B - i ) Forward : $T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \frac{1}{(\delta L)^6}$ + + ii) Reverse : $T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) = (\delta L)^6$ + + C - i ) Forward : $T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \frac{1}{(\delta L)^6}$ + + ii) Reverse : $T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) = \frac{1}{(\delta L)^6}$ + + C - i ) Forward : $T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = -\frac{1}{(\delta L)^3}$ + + ii) Reverse : $T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) = \frac{1}{(\delta L)^3}$ + +** Question 5 + + If you run a single DMC calculation on the Li$^+$ ion in the singlet + ground state, which approximations impact the final energy: + + - None + - The fixed-node approximation + - The time-step approximation + - The fixed-node approximation and the time-step approximation + + * Schedule :noexport: |------------------------------+---------|