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@ -99,7 +99,7 @@
where the probability density is given by the square of the wave function:
$$ P(\mathbf{r}) = \frac{|Psi(\mathbf{r}|^2)}{\int |\Psi(\mathbf{r})|^2 d\mathbf{r}}\,. $$
$$ P(\mathbf{r}) = \frac{|\Psi(\mathbf{r})|^2}{\int |\Psi(\mathbf{r})|^2 d\mathbf{r}}\,. $$
If we can sample $N_{\rm MC}$ configurations $\{\mathbf{r}\}$ distributed as $p$, we can estimate $E$ as the average of the local energy computed over these configurations:
@ -1397,7 +1397,7 @@ 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 to denote dL since we will use
NOTE: below, we use the symbol $\delta t$ to denote $\delta L$ since we will use
the same variable later on to store a time step.
@ -1762,8 +1762,7 @@ end subroutine random_gauss
\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 $A = \frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})}
{T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})}$
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 $\mathbf{r}_{n+1} = \mathbf{r}_n$
@ -2086,7 +2085,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
i\frac{\partial \Psi(\mathbf{r},t)}{\partial t} = (\hat{H} -E_{\rm ref}) \Psi(\mathbf{r},t)\,.
\]
where we introduced a shift in the energy, $E_{\rm ref}$, which will come useful below.
where we introduced a shift in the energy, $E_{\rm ref}$, for reasons which will become apparent below.
We can expand a given starting wave function, $\Psi(\mathbf{r},0)$, in the basis of the eigenstates
of the time-independent Hamiltonian, $\Phi_k$, with energies $E_k$:
@ -2108,12 +2107,12 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
-\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = (\hat{H} -E_{\rm ref}) \psi(\mathbf{r}, \tau)
\]
where $\psi(\mathbf{r},\tau) = \Psi(\mathbf{r},-i\,t)$
where $\psi(\mathbf{r},\tau) = \Psi(\mathbf{r},-i\,\tau)$
and
\begin{eqnarray*}
\psi(\mathbf{r},\tau) &=& \sum_k a_k \exp( -(E_k-E_{\rm ref})\, \tau) \phi_k(\mathbf{r})\\
&=& \exp(-(E_0-E_{\rm ref})\, \tau)\sum_k a_k \exp( -(E_k-E_0)\, \tau) \phi_k(\mathbf{r})\,.
\psi(\mathbf{r},\tau) &=& \sum_k a_k \exp( -(E_k-E_{\rm ref})\, \tau) \Phi_k(\mathbf{r})\\
&=& \exp(-(E_0-E_{\rm ref})\, \tau)\sum_k a_k \exp( -(E_k-E_0)\, \tau) \Phi_k(\mathbf{r})\,.
\end{eqnarray*}
For large positive values of $\tau$, $\psi$ is dominated by the
@ -2164,8 +2163,8 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
can be simulated by creating or destroying particles over time (a
so-called branching process).
/Diffusion Monte Carlo/ (DMC) consists in obtaining the ground state of a
system by simulating the Schrödinger equation in imaginary time, by
In /Diffusion Monte Carlo/ (DMC), one onbtains the ground state of a
system by simulating the Schrödinger equation in imaginary time via
the combination of a diffusion process and a branching process.
We note that the ground-state wave function of a Fermionic system is
@ -2184,7 +2183,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
** Importance sampling
In a molecular system, the potential is far from being constant
and diverges at inter-particle coalescence points. Hence, when the
and, in fact, diverges at the inter-particle coalescence points. Hence, when the
rate equation is simulated, it results in very large fluctuations
in the numbers of particles, making the calculations impossible in
practice.
@ -2209,7 +2208,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
The new "kinetic energy" can be simulated by the drift-diffusion
scheme presented in the previous section (VMC).
The new "potential" is the local energy, which has smaller fluctuations
when $\Psi_T$ gets closer to the exact wave function. It can be simulated by
when $\Psi_T$ gets closer to the exact wave function. This term can be simulated by
changing the number of particles according to $\exp\left[ -\delta t\,
\left(E_L(\mathbf{r}) - E_{\rm ref}\right)\right]$
where $E_{\rm ref}$ is the constant we had introduced above, which is adjusted to
@ -2321,8 +2320,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
\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 $A = \frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})}
{T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})}$
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 $\mathbf{r}_{n+1} = \mathbf{r}_n$
@ -2335,36 +2333,18 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
- 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}
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*}
- 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 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.
- 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 wave function becomes
\[
\psi(\mathbf{r},\tau) = \Psi_T(\mathbf{r}) W(\mathbf{r},\tau)
\]
and the expression of the fixed-node DMC energy is
\begin{eqnarray*}
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 \left[ \Psi_T(\mathbf{r}) \right]^2 W(\mathbf{r},\tau) E_L(\mathbf{r}) d\mathbf{r}}
{\int \left[ \Psi_T(\mathbf{r}) \right]^2 W(\mathbf{r},\tau) d\mathbf{r}} \\
\end{eqnarray*}
This algorithm is less stable than the branching algorithm: it
PDMC algorithm is less stable than the branching algorithm: it
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
advantage is that it is very easy to program starting from a VMC