From 72e18f780b44037a677f1242bd589e922f2a4b66 Mon Sep 17 00:00:00 2001 From: filippi-claudia <44236509+filippi-claudia@users.noreply.github.com> Date: Mon, 1 Feb 2021 09:07:13 +0100 Subject: [PATCH] Update QMC.org --- QMC.org | 106 +++++++++++++++++++++++++++----------------------------- 1 file changed, 51 insertions(+), 55 deletions(-) diff --git a/QMC.org b/QMC.org index 32df533..c938fe0 100644 --- a/QMC.org +++ b/QMC.org @@ -1397,8 +1397,8 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform step such that the acceptance rate is close to 0.5 is a good compromise for the current problem. - NOTE: below, we use the symbol dt for dL for reasons which will - become clear later. + NOTE: below, we use the symbol dt to denote dL since we will use + the same variable later on to store a time step. *** Exercise @@ -2168,10 +2168,19 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis system by simulating the Schrödinger equation in imaginary time, by the combination of a diffusion process and a branching process. - We note here that the ground-state wave function of a Fermionic system is - antisymmetric and changes sign. + We note that the ground-state wave function of a Fermionic system is + antisymmetric and changes sign. Therefore, it is interpretation as a probability + distribution is somewhat problematic. In fact, mathematically, since + the Bosonic ground state is lower in energy than the Fermionic one, for + large $\tau$, the system will evolve towards the Bosonic solution. - I AM HERE + For the systems you will study this is not an issue: + + - Hydrogen atom: You only have one electron! + - Two-electron system ($H_2$ or He): The ground-wave function is antisymmetric + in the spin variables but symmetric in the space ones. + + Therefore, in both cases, you are dealing with a "Bosonic" ground state. ** Importance sampling @@ -2205,18 +2214,45 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis changing the number of particles according to $\exp\left[ -\delta t\, \left(E_L(\mathbf{r}) - E_T\right)\right]$ where $E_T$ is the constant we had introduced above, which is adjusted to - the running average energy and is introduced to keep the number of particles + the running average energy to keep the number of particles reasonably constant. This equation generates the /N/-electron density $\Pi$, which is the - product of the ground state with the trial wave function. It - introduces the constraint that $\Pi(\mathbf{r},\tau)=0$ where - $\Psi_T(\mathbf{r})=0$. In the few cases where the wave function has no nodes, - such as in the hydrogen atom or the H_2 molecule, this - constraint is harmless and we can obtain the exact energy. But for - systems where the wave function has nodes, this scheme introduces an - error known as the /fixed node error/. + product of the ground state with the trial wave function. You may then ask: how + can we compute the total energy of the system? + + To this aim, we use the mixed estimator of the energy: + + \begin{eqnarray*} + E(\tau) &=& \frac{\langle \psi(tau) | \hat{H} | \Psi_T \rangle}{\frac{\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}} \\ + &=& \int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) E_L(\mathbf{r}) d\mathbf{r}} + {\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}} + \end{eqnarray*} + + Since, for large $\tau$, we have that + + \[ + \Pi(\mathbf{r},\tau) =\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \rightarrow \Phi_0(\mathbf{r}) \Psi_T(\mathbf{r})\,, + \] + + and, using that $\hat{H}$ is Hermitian and that $\Phi_0$ is an eigenstate of the Hamiltonian, we obtain + \[ + E(\tau) = \frac{\langle \psi_\tau | \hat{H} | \Psi_T \rangle} + {\langle \psi_\tau | \Psi_T \rangle} + = \frac{\langle \Psi_T | \hat{H} | \psi_\tau \rangle} + {\langle \Psi_T | \psi_\tau \rangle} + \rightarrow E_0 \frac{\langle \Psi_T | \psi_\tau \rangle} + {\langle \Psi_T | \psi_\tau \rangle} + = E_0 + \] + + Therefore, we can compute the energy within DMC by generating the + density $\Pi$ with random walks, and simply averaging the local + energies computed with the trial wave function. + *** Appendix : Details of the Derivation \[ @@ -2262,52 +2298,12 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis \nabla \left[ \Pi(\mathbf{r},\tau) \frac{\nabla \Psi_T(\mathbf{r})}{\Psi_T(\mathbf{r})} \right] + E_L(\mathbf{r}) \Pi(\mathbf{r},\tau) \] - - -** Fixed-node DMC energy - - Now that we have a process to sample $\Pi(\mathbf{r},\tau) = - \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})$, we can compute the exact - energy of the system, within the fixed-node constraint, as: - - \[ - E = \lim_{\tau \to \infty} \frac{\int \Pi(\mathbf{r},\tau) E_L(\mathbf{r}) d\mathbf{r}} - {\int \Pi(\mathbf{r},\tau) d\mathbf{r}} = \lim_{\tau \to - \infty} E(\tau). - \] - - - \[ - E(\tau) = \frac{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) E_L(\mathbf{r}) d\mathbf{r}} - {\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}} - = \frac{\int \psi(\mathbf{r},\tau) \hat{H} \Psi_T(\mathbf{r}) d\mathbf{r}} - {\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}} - = \frac{\langle \psi_\tau | \hat{H} | \Psi_T \rangle} - {\langle \psi_\tau | \Psi_T \rangle} - \] - - As $\hat{H}$ is Hermitian, - - \[ - E(\tau) = \frac{\langle \psi_\tau | \hat{H} | \Psi_T \rangle} - {\langle \psi_\tau | \Psi_T \rangle} - = \frac{\langle \Psi_T | \hat{H} | \psi_\tau \rangle} - {\langle \Psi_T | \psi_\tau \rangle} - = E[\psi_\tau] \frac{\langle \Psi_T | \psi_\tau \rangle} - {\langle \Psi_T | \psi_\tau \rangle} - = E[\psi_\tau] - \] - - So computing the energy within DMC consists in generating the - density $\Pi$ with random walks, and simply averaging the local - energies computed with the trial wave function. ** Pure Diffusion Monte Carlo (PDMC) Instead of having a variable number of particles to simulate the - branching process, one can choose to sample $[\Psi_T(\mathbf{r})]^2$ instead of - $\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})$, and consider the term - $\exp \left( -\delta t\,( E_L(\mathbf{r}) - E_{\text{ref}} \right)$ as a + branching process, one can consider the term + $\exp \left( -\delta t\,( E_L(\mathbf{r}) - E_T} \right)$ as a cumulative product of weights: \[