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@ -1744,7 +1744,7 @@ end subroutine random_gauss
\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\delta t} \right]\,.
\]
and the corrsponding move is proposed as
The corrsponding move is proposed as
\[
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta t\, \frac{\nabla
@ -2108,7 +2108,6 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
-\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = (\hat{H} -E_T) \psi(\mathbf{r}, \tau)
\]
# TODO Should we put 't' after i ?
where $\psi(\mathbf{r},\tau) = \Psi(\mathbf{r},-i\,)$
and
@ -2170,16 +2169,15 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
the combination of a diffusion process and a branching process.
We note that the ground-state wave function of a Fermionic system is
antisymmetric and changes sign. Therefore, it is interpretation as a probability
antisymmetric and changes sign. Therefore, its 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.
For the systems you will study this is not an issue:
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.
- 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.
@ -2315,7 +2313,42 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
\prod_{i=1}^{n} w(\mathbf{r}_i)
\]
where $\mathbf{r}_i$ are the coordinates along the trajectory.
where $\mathbf{r}_i$ are the coordinates along the trajectory and we introduced a time-step $\delta t$.
The algorithm can be rather easily built on top of your VMC code:
1) Compute a new position $\mathbf{r'} = \mathbf{r}_n +
\delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi$
Evaluate $\Psi$ and $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$ at the new position
2) Compute the ratio $A = \frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})}
{T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})}$
3) Draw a uniform random number $v \in [0,1]$
4) if $v \le A$, accept the move : set $\mathbf{r}_{n+1} = \mathbf{r'}$
5) else, reject the move : set $\mathbf{r}_{n+1} = \mathbf{r}_n$
6) evaluate the local energy at $\mathbf{r}_{n+1}$
7) compute the weight $w(\mathbf{r}_i)$
8) update $W$
Some comments are needed:
- You estimate the energy as
\begin{eqnarray*}
E = \frac{\sum_{i=1}{N_{\rm MC}} E_L(\mathbf{r}_i) W(\mathbf{r}_i, i\delta t)}{\sum_{i=1}{N_{\rm MC}} W(\mathbf{r}_i, i\delta t)}
\end{eqnarray}
- 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 $\delta t$.
- The accept/reject step (steps 2-5 in the algorithm) is not in principle needed for the correctness of
the DMC algorithm. However, its use reduces si
The wave function becomes
\[
@ -2356,7 +2389,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
from hydrogen import *
from qmc_stats import *
def MonteCarlo(a, nmax, dt, tau, Eref):
def MonteCarlo(a, nmax, dt, Eref):
# TODO
# Run simulation
@ -2365,7 +2398,7 @@ nmax = 100000
dt = 0.01
E_ref = -0.5
X0 = [ MonteCarlo(a, nmax, dt, tau, E_ref) for i in range(30)]
X0 = [ MonteCarlo(a, nmax, dt, E_ref) for i in range(30)]
# Energy
X = [ x for (x, _) in X0 ]