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@ -2222,7 +2222,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
To this aim, we use the mixed estimator of the energy: To this aim, we use the mixed estimator of the energy:
\begin{eqnarray*} \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}} &=& \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}) d\mathbf{r}} \\
&=& \frac{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) E_L(\mathbf{r}) d\mathbf{r}} &=& \frac{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) E_L(\mathbf{r}) d\mathbf{r}}
@ -2332,19 +2332,19 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
- You estimate the energy as - You estimate the energy as
\begin{eqnarray*} \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)} 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*} \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 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 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. Here, you will be using a small enough time-step and you should not worry about the extrapolation.
- The accept/reject step (steps 2-5 in the algorithm) is in principle not needed for the correctness of - 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 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 fixed-node energy, and a good trial wave function. Its big
advantage is that it is very easy to program starting from a VMC advantage is that it is very easy to program starting from a VMC