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@ 1099,12 +1099,12 @@ end subroutine ave_error


Clearly, the square of the wave function is a good choice of probability density to sample but we will start with something simpler and rewrite the energy as




\begin{eqnarray*}


E & = & \frac{\int E_L(\mathbf{r})\frac{\Psi(\mathbf{r})^2}{p(\mathbf{r})}p(\mathbf{r})\, \,d\mathbf{r}}{\int \frac{\Psi(\mathbf{r})^2 }{p(\mathbf{r})}p(\mathbf{r})d\mathbf{r}}\,.


E & = & \frac{\int E_L(\mathbf{r})\frac{\Psi(\mathbf{r})^2}{P(\mathbf{r})}P(\mathbf{r})\, \,d\mathbf{r}}{\int \frac{\Psi(\mathbf{r})^2 }{P(\mathbf{r})}P(\mathbf{r})d\mathbf{r}}\,.


\end{eqnarray*}




Here, we will sample a uniform probability $p(\mathbf{r})$ in a cube of volume $L^3$ centered at the origin:


Here, we will sample a uniform probability $P(\mathbf{r})$ in a cube of volume $L^3$ centered at the origin:




$$ p(\mathbf{r}) = \frac{1}{L^3}\,, $$


$$ P(\mathbf{r}) = \frac{1}{L^3}\,, $$




and zero outside the cube.




@ 1328,23 +1328,49 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 o qmc_uniform




To sample a chosen probability density, an efficient method is the


[[https://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm][MetropolisHastings sampling algorithm]]. Starting from a random


initial position $\mathbf{r}_0$, we will realize a random walk as follows:


initial position $\mathbf{r}_0$, we will realize a random walk:




$$ \mathbf{r}_0 \rightarrow \mathbf{r}_1 \rightarrow \mathbf{r}_2 \ldots \mathbf{r}_{N_{\rm MC}}\,, $$




according to the following algorithm.




At every step, we propose a new move according to a transition probability $T(\mathbf{r}_{n+1},\mathbf{r}_n)$ of our choice.




For simplicity, let us move the electron in a 3dimensional box of side $2\delta L$ centered at the current position


of the electron:




$$


\mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta t\, \mathbf{u}


\mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta L \, \mathbf{u}


$$




where $\delta t$ is a fixed constant (the socalled /timestep/), and


where $\delta L$ is a fixed constant, and


$\mathbf{u}$ is a uniform random number in a 3dimensional box


$(1,1,1) \le \mathbf{u} \le (1,1,1)$. We will then add the


$(1,1,1) \le \mathbf{u} \le (1,1,1)$.




After having moved the electron, add the


accept/reject step that guarantees that the distribution of the


$\mathbf{r}_n$ is $\Psi^2$:




$\mathbf{r}_n$ is $\Psi^2$. This amounts to accepting the move with


probability




$$


A{\mathbf{r}_{n+1},\mathbf{r}_n) = \min\left(1,\frac{T(\mathbf{r}_{n},\mathbf{r}_{n+1}) P(\mathbf{r}_{n+1})}{T(\mathbf{r}_{n+1},\mathbf{r}_n)P(\mathbf{r}_{n})}\right)\,,


$$




which, for our choice of transition probability, becomes




$$


A{\mathbf{r}_{n+1},\mathbf{r}_n) = \min\left(1,\frac{P(\mathbf{r}_{n+1})}{P(\mathbf{r}_{n})}\right)= \min\left(1,\frac{\Psi(\mathbf{r}_{n+1})^2}{\Psi(\mathbf{r}_{n})^2}


$$




Explain why the transition probability cancels out in the expression of $A$. Also note that we do not need to compute the norm of the wave function!




The algorithm is summarized as follows:




1) Compute $\Psi$ at a new position $\mathbf{r'} = \mathbf{r}_n +


\delta t\, \mathbf{u}$


2) Compute the ratio $R = \frac{\left[\Psi(\mathbf{r'})\right]^2}{\left[\Psi(\mathbf{r}_{n})\right]^2}$


\delta L\, \mathbf{u}$


2) Compute the ratio $A = \frac{\left[\Psi(\mathbf{r'})\right]^2}{\left[\Psi(\mathbf{r}_{n})\right]^2}$


3) Draw a uniform random number $v \in [0,1]$


4) if $v \le R$, accept the move : set $\mathbf{r}_{n+1} = \mathbf{r'}$


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 $\mathbf{r}_{n+1}$




@ 1355,20 +1381,20 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 o qmc_uniform


All samples should be kept, from both accepted and rejected moves.


#+end_note




If the time step is infinitely small, the ratio will be very close


to one and all the steps will be accepted. But the trajectory will


be infinitely too short to have statistical significance.


If the box is infinitely small, the ratio will be very close


to one and all the steps will be accepted. However, the moves will be


very correlated and you will visit the configurational space very slowly.




On the other hand, as the time step increases, the number of


On the other hand, if you propose too large moves, the number of


accepted steps will decrease because the ratios might become


small. If the number of accepted steps is close to zero, then the


space is not well sampled either.




The time step should be adjusted so that it is as large as


The size of the move should be adjusted so that it is as large as


possible, keeping the number of accepted steps not too small. To


achieve that, we define the acceptance rate as the number of


accepted steps over the total number of steps. Adjusting the time


step such that the acceptance rate is close to 0.5 is a good compromise.


step such that the acceptance rate is close to 0.5 is a good compromise for the current problem.






*** Exercise


@ 1378,7 +1404,7 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 o qmc_uniform


sampled with $\Psi^2$.




Compute also the acceptance rate, so that you can adapt the time


step in order to have an acceptance rate close to 0.5 .


step in order to have an acceptance rate close to 0.5.




Can you observe a reduction in the statistical error?


#+end_exercise


@ 1648,16 +1674,17 @@ end subroutine random_gauss


#+END_SRC




In Python, you can use the [[https://numpy.org/doc/stable/reference/random/generated/numpy.random.normal.html][~random.normal~]] function of Numpy.




** Generalized Metropolis algorithm


:PROPERTIES:


:headerargs:python: :tangle vmc_metropolis.py


:headerargs:f90: :tangle vmc_metropolis.f90


:END:




One can use more efficient numerical schemes to move the electrons,


but the Metropolis accepation step has to be adapted accordingly:


the acceptance


probability $A$ is chosen so that it is consistent with the


One can use more efficient numerical schemes to move the electrons by choosing a smarter expression for the transition probability.




The Metropolis acceptance step has to be adapted accordingly to ensure that the detailed balance condition is satisfied. This means that


the acceptance probability $A$ is chosen so that it is consistent with the


probability of leaving $\mathbf{r}_n$ and the probability of


entering $\mathbf{r}_{n+1}$:




@ 1675,7 +1702,7 @@ end subroutine random_gauss




\[


T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) =


\text{constant}


\text{constant}\,,


\]




so the expression of $A$ was simplified to the ratios of the squared


@ 1688,7 +1715,7 @@ end subroutine random_gauss


\[


T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) =


\frac{1}{(2\pi\,\delta t)^{3/2}} \exp \left[  \frac{\left(


\mathbf{r}_{n+1}  \mathbf{r}_{n} \right)^2}{2\delta t} \right]


\mathbf{r}_{n+1}  \mathbf{r}_{n} \right)^2}{2\delta t} \right]\,.


\]






@ 1697,17 +1724,17 @@ end subroutine random_gauss


the lowprobability regions. This will mechanically increase the


acceptance ratios and improve the sampling.




To do this, we can add the drift vector


To do this, we can use the gradient of the probability density




\[


\frac{\nabla [ \Psi^2 ]}{\Psi^2} = 2 \frac{\nabla \Psi}{\Psi}.


\frac{\nabla [ \Psi^2 ]}{\Psi^2} = 2 \frac{\nabla \Psi}{\Psi}\,,


\]




The numerical scheme becomes a drifted diffusion:


and add the socalled drift vector, so that the numerical scheme becomes a drifted diffusion:




\[


\mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta t\, \frac{\nabla


\Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi


\Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi \,,


\]




where $\chi$ is a Gaussian random variable with zero mean and


@ 1718,9 +1745,23 @@ end subroutine random_gauss


T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) =


\frac{1}{(2\pi\,\delta t)^{3/2}} \exp \left[  \frac{\left(


\mathbf{r}_{n+1}  \mathbf{r}_{n}  \frac{\nabla


\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\delta t} \right]


\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\delta t} \right]\,.


\]






The algorithm of the previous exercise is only slighlty modified summarized:




0) For the starting position compute $\Psi$ and the drifvector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$


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 $\mathbf{r}_{n+1}$






*** Exercise 1





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