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@ 33,18 +33,20 @@




* Introduction




This web site is the QMC tutorial of the LTTC winter school


This website contains the QMC tutorial of the 2021 LTTC winter school


[[https://www.irsamc.upstlse.fr/lttc/Luchon][Tutorials in Theoretical Chemistry]].




We propose different exercises to understand quantum Monte Carlo (QMC)


methods. In the first section, we propose to compute the energy of a


methods. In the first section, we start with the computation of the energy of a


hydrogen atom using numerical integration. The goal of this section is


to introduce the /local energy/.


Then we introduce the variational Monte Carlo (VMC) method which


to familarize yourself with the concept of /local energy/.


Then, we introduce the variational Monte Carlo (VMC) method which


computes a statistical estimate of the expectation value of the energy


associated with a given wave function.


Finally, we introduce the diffusion Monte Carlo (DMC) method which


gives the exact energy of the hydrogen atom and of the H_2 molecule.


associated with a given wave function, and apply this approach to the


hydrogen atom.


Finally, we present the diffusion Monte Carlo (DMC) method which


we use here to estimate the exact energy of the hydrogen atom and of the H_2 molecule,


starting from an approximate wave function.




Code examples will be given in Python and Fortran. You can use


whatever language you prefer to write the program.


@ 53,58 +55,73 @@


the wave functions considered here are real: for an $N$ electron


system where the electrons move in the 3dimensional space,


$\Psi : \mathbb{R}^{3N} \rightarrow \mathbb{R}$. In addition, $\Psi$


is defined everywhere, continuous and infinitely differentiable.


is defined everywhere, continuous, and infinitely differentiable.




All the quantities are expressed in /atomic units/ (energies,


coordinates, etc).




* Numerical evaluation of the energy




In this section we consider the Hydrogen atom with the following


** Energy and local energy




For a given system with Hamiltonian $\hat{H}$ and wave function $\Psi$, we define the local energy as




$$


E_L(\mathbf{r}) = \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})},


$$




where $\mathbf{r}$ denotes the 3Ndimensional electronic coordinates.




The electronic energy of a system, $E$, can be rewritten in terms of the


local energy $E_L(\mathbf{r})$ as




\begin{eqnarray*}


E & = & \frac{\langle \Psi \hat{H}  \Psi\rangle}{\langle \Psi \Psi \rangle}


= \frac{\int \Psi(\mathbf{r})\, \hat{H} \Psi(\mathbf{r})\, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}} \\


& = & \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}


= \frac{\int \left[\Psi(\mathbf{r})\right]^2\, E_L(\mathbf{r})\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}


\end{eqnarray*}




For few dimensions, one can easily compute $E$ by evaluating the integrals on a grid but, for a high number of dimensions, one can resort to Monte Carlo techniques to compute $E$.




To this aim, recall that the probabilistic /expected value/ of an arbitrary function $f(x)$


with respect to a probability density function $P(x)$ is given by




$$ \langle f \rangle_p = \int_{\infty}^\infty P(x)\, f(x)\,dx, $$




where a probability density function $p(x)$ is nonnegative


and integrates to one:




$$ \int_{\infty}^\infty P(x)\,dx = 1. $$




Similarly, we can view the the energy of a system, $E$, as the expected value of the local energy with respect to


a probability density $P(\mathbf{r}}$ defined in 3$N$ dimensions:




$$ E = \int E_L(\mathbf{r}) P(\mathbf{r})\,d\mathbf{r}} \equiv \langle E_L \rangle_{\Psi^2}\,, $$




where the probability density is given by the square of the wave function:




$$ P(\mathbf{r}) = \frac{Psi(\mathbf{r}^2){\int \left \Psi(\mathbf{r})^2 d\mathbf{r}}\,. $$




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:




$$ E \approx \frac{1}{N_{\rm MC}} \sum_{i=1}^{N_{\rm MC}} E_L(\mathbf{r}_i} \,.




* Numerical evaluation of the energy of the hydrogen atom




In this section, we consider the hydrogen atom with the following


wave function:




$$


\Psi(\mathbf{r}) = \exp(a \mathbf{r})


$$




We will first verify that, for a given value of $a$, $\Psi$ is an


We will first verify that, for a particular value of $a$, $\Psi$ is an


eigenfunction of the Hamiltonian




$$


\hat{H} = \hat{T} + \hat{V} =  \frac{1}{2} \Delta  \frac{1}{\mathbf{r}}


$$




To do that, we will check if the local energy, defined as




$$


E_L(\mathbf{r}) = \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})},


$$




is constant.






The probabilistic /expected value/ of an arbitrary function $f(x)$


with respect to a probability density function $p(x)$ is given by




$$ \langle f \rangle_p = \int_{\infty}^\infty p(x)\, f(x)\,dx. $$




Recall that a probability density function $p(x)$ is nonnegative


and integrates to one:




$$ \int_{\infty}^\infty p(x)\,dx = 1. $$






The electronic energy of a system is the expectation value of the


local energy $E(\mathbf{r})$ with respect to the 3Ndimensional


electron density given by the square of the wave function:




\begin{eqnarray*}


E & = & \frac{\langle \Psi \hat{H}  \Psi\rangle}{\langle \Psi \Psi \rangle}


= \frac{\int \Psi(\mathbf{r})\, \hat{H} \Psi(\mathbf{r})\, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}} \\


& = & \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}


= \frac{\int \left[\Psi(\mathbf{r})\right]^2\, E_L(\mathbf{r})\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}


= \langle E_L \rangle_{\Psi^2}


\end{eqnarray*}


To do that, we will compute the local energy and check whether it is constant.




** Local energy


:PROPERTIES:


@ 112,6 +129,8 @@


:headerargs:f90: :tangle hydrogen.f90


:END:




You will now program all quantities needed to compute the local energy of the H atom for the given wave function.




Write all the functions of this section in a single file :


~hydrogen.py~ if you use Python, or ~hydrogen.f90~ is you use


Fortran.


@ 187,7 +206,7 @@ double precision function potential(r)


end function potential


#+END_SRC




*** Exercise 2


*** Exercise 2


#+begin_exercise


Write a function which computes the wave function at $\mathbf{r}$.


The function accepts a scalar =a= and a 3dimensional vector =r= as


@ 257,10 +276,10 @@ end function psi


applied to the wave function gives:




$$


\Delta \Psi (\mathbf{r}) = \left(a^2  \frac{2a}{\mathbf{r}} \right) \Psi(\mathbf{r})


\Delta \Psi (\mathbf{r}) = \left(a^2  \frac{2a}{\mathbf{r}} \right) \Psi(\mathbf{r})\,.


$$




So the local kinetic energy is


Therefore, the local kinetic energy is


$$


\frac{1}{2} \frac{\Delta \Psi}{\Psi} (\mathbf{r}) = \frac{1}{2}\left(a^2  \frac{2a}{\mathbf{r}} \right)


$$


@ 544,7 +563,7 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \


If the space is discretized in small volume elements $\mathbf{r}_i$


of size $\delta \mathbf{r}$, the expression of $\langle E_L \rangle_{\Psi^2}$


becomes a weighted average of the local energy, where the weights


are the values of the probability density at $\mathbf{r}_i$


are the values of the wave function square at $\mathbf{r}_i$


multiplied by the volume element:




$$


@ 561,7 +580,7 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \




*** Exercise


#+begin_exercise


Compute a numerical estimate of the energy in a grid of


Compute a numerical estimate of the energy using a grid of


$50\times50\times50$ points in the range $(5,5,5) \le


\mathbf{r} \le (5,5,5)$.


#+end_exercise


@ 764,7 +783,7 @@ gfortran hydrogen.f90 energy_hydrogen.f90 o energy_hydrogen


*** Exercise


#+begin_exercise


Add the calculation of the variance to the previous code, and


compute a numerical estimate of the variance of the local energy in


compute a numerical estimate of the variance of the local energy using


a grid of $50\times50\times50$ points in the range $(5,5,5) \le


\mathbf{r} \le (5,5,5)$ for different values of $a$.


#+end_exercise


@ 952,9 +971,9 @@ gfortran hydrogen.f90 variance_hydrogen.f90 o variance_hydrogen


Numerical integration with deterministic methods is very efficient


in low dimensions. When the number of dimensions becomes large,


instead of computing the average energy as a numerical integration


on a grid, it is usually more efficient to do a Monte Carlo sampling.


on a grid, it is usually more efficient to use Monte Carlo sampling.




Moreover, a Monte Carlo sampling will alow us to remove the bias due


Moreover, Monte Carlo sampling will alow us to remove the bias due


to the discretization of space, and compute a statistical confidence


interval.




@ 967,7 +986,7 @@ gfortran hydrogen.f90 variance_hydrogen.f90 o variance_hydrogen


To compute the statistical error, you need to perform $M$


independent Monte Carlo calculations. You will obtain $M$ different


estimates of the energy, which are expected to have a Gaussian


distribution according to the [[https://en.wikipedia.org/wiki/Central_limit_theorem][Central Limit Theorem]].


distribution for large $M$, according to the [[https://en.wikipedia.org/wiki/Central_limit_theorem][Central Limit Theorem]].




The estimate of the energy is




@ 1068,10 +1087,28 @@ end subroutine ave_error


:headerargs:f90: :tangle qmc_uniform.f90


:END:




We will now do our first Monte Carlo calculation to compute the


energy of the hydrogen atom.


We will now perform our first Monte Carlo calculation to compute the


energy of the hydrogen atom.




At every Monte Carlo iteration:


Consider again the expression of the energy




\begin{eqnarray*}


E & = & \frac{\int E_L(\mathbf{r})\left[\Psi(\mathbf{r})\right]^2\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}\,.


\end{eqnarray*}




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}}\,.


\end{eqnarray*}




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




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




and zero outside the cube.




One Monte Carlo run will consist of $N_{\rm MC}$ Monte Carlo iterations. At every Monte Carlo iteration:




 Draw a random point $\mathbf{r}_i$ in the box $(5,5,5) \le


(x,y,z) \le (5,5,5)$


@ 1080,9 +1117,8 @@ end subroutine ave_error


 Compute $[\Psi(\mathbf{r}_i)]^2 \times E_L(\mathbf{r}_i)$, and accumulate the


result in a variable =energy=




One Monte Carlo run will consist of $N$ Monte Carlo iterations. Once all the


iterations have been computed, the run returns the average energy


$\bar{E}_k$ over the $N$ iterations of the run.


Once all the iterations have been computed, the run returns the average energy


$\bar{E}_k$ over the $N_{\rm MC}$ iterations of the run.




To compute the statistical error, perform $M$ independent runs. The


final estimate of the energy will be the average over the


@ 1279,37 +1315,62 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 o qmc_uniform


We will now use the square of the wave function to sample random


points distributed with the probability density


\[


P(\mathbf{r}) = \left[\Psi(\mathbf{r})\right]^2


P(\mathbf{r}) = \frac{Psi(\mathbf{r}^2){\int \left \Psi(\mathbf{r})^2 d\mathbf{r}}


\]




The expression of the average energy is now simplified as the average of


the local energies, since the weights are taken care of by the


sampling :


sampling:




$$


E \approx \frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r}_i)


E \approx \frac{1}{N_{\rm MC}}\sum_{i=1}^{N_{\rm MC} E_L(\mathbf{r}_i)


$$






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:




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




following the following algorithm.




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




For simplicity, we will 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, we 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}\rightarrow\mathbf{r}_{n+1}) = \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}\rightarrow\mathbf{r}_{n+1}) = \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 $\mathbf{r}_{n+1}$




@ 1320,20 +1381,24 @@ 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.




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


@ 1343,7 +1408,7 @@ 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


@ 1613,16 +1678,17 @@ 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 $\mathbf{r}_{n+1}$:




@ 1635,57 +1701,74 @@ end subroutine random_gauss


probability of transition from $\mathbf{r}_n$ to


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




In the previous example, we were using uniform random


numbers. Hence, the transition probability was


In the previous example, we were using uniform sampling in a box centered


at the current position. Hence, the transition probability was symmetric




\[


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


\text{constant}


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


\text{constant}\,,


\]




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


wave functions.




Now, if instead of drawing uniform random numbers we


Now, if instead of drawing uniform random numbers, we


choose to draw Gaussian random numbers with zero mean and variance


$\delta t$, the transition probability becomes:




\[


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]\,.


\]






To sample even better the density, we can "push" the electrons




Furthermore, to sample the density even better, we can "push" the electrons


into in the regions of high probability, and "pull" them away from


the lowprobability regions. This will mechanically increase the


the lowprobability regions. This will ncrease 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 with transition probability:




\[


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]\,.


\]




and the corrsponding move is proposed as




\[


\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


variance $\delta t$.


The transition probability becomes:




\[


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]


\]










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




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




@ 1737,8 +1820,8 @@ end subroutine drift


*** Exercise 2




#+begin_exercise


Modify the previous program to introduce the drifted diffusion scheme.


(This is a necessary step for the next section).


Modify the previous program to introduce the driftdiffusion scheme.


(This is a necessary step for the next section on diffusion Monte Carlo).


#+end_exercise




*Python*


@ 2000,11 +2083,13 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 o vmc_metropolis


Consider the timedependent Schrödinger equation:




\[


i\frac{\partial \Psi(\mathbf{r},t)}{\partial t} = \hat{H} \Psi(\mathbf{r},t)


i\frac{\partial \Psi(\mathbf{r},t)}{\partial t} = (\hat{H} E_T) \Psi(\mathbf{r},t)\,.


\]




We can expand $\Psi(\mathbf{r},0)$, in the basis of the eigenstates


of the timeindependent Hamiltonian:


where we introduced a shift in the energy, $E_T$, which will come useful below.




We can expand a given starting wave function, $\Psi(\mathbf{r},0)$, in the basis of the eigenstates


of the timeindependent Hamiltonian, $\Phi_k$, with energies $E_k$:




\[


\Psi(\mathbf{r},0) = \sum_k a_k\, \Phi_k(\mathbf{r}).


@ 2013,36 +2098,49 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 o vmc_metropolis


The solution of the Schrödinger equation at time $t$ is




\[


\Psi(\mathbf{r},t) = \sum_k a_k \exp \left( i\, E_k\, t \right) \Phi_k(\mathbf{r}).


\Psi(\mathbf{r},t) = \sum_k a_k \exp \left( i\, (E_kE_T)\, t \right) \Phi_k(\mathbf{r}).


\]




Now, let's replace the time variable $t$ by an imaginary time variable


Now, if we replace the time variable $t$ by an imaginary time variable


$\tau=i\,t$, we obtain




\[


\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = \hat{H} \psi(\mathbf{r}, \tau)


\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = (\hat{H} E_T) \psi(\mathbf{r}, \tau)


\]




where $\psi(\mathbf{r},\tau) = \Psi(\mathbf{r},i\tau) = \Psi(\mathbf{r},t)$


where $\psi(\mathbf{r},\tau) = \Psi(\mathbf{r},i\,)$


and


\[


\psi(\mathbf{r},\tau) = \sum_k a_k \exp( E_k\, \tau) \phi_k(\mathbf{r}).


\]




\begin{eqnarray*}


\psi(\mathbf{r},\tau) &=& \sum_k a_k \exp( E_k\, \tau) \phi_k(\mathbf{r})\\


&=& \exp((E_0E_T)\, \tau)\sum_k a_k \exp( (E_kE_0)\, \tau) \phi_k(\mathbf{r})\,.


\end{eqnarray*}




For large positive values of $\tau$, $\psi$ is dominated by the


$k=0$ term, namely the lowest eigenstate.


So we can expect that simulating the differetial equation in


$k=0$ term, namely, the lowest eigenstate. If we adjust $E_T$ to the running estimate of $E_0$,


we can expect that simulating the differetial equation in


imaginary time will converge to the exact ground state of the


system.




** Diffusion and branching




The [[https://en.wikipedia.org/wiki/Diffusion_equation][diffusion equation]] of particles is given by


The imaginarytime Schrödinger equation can be explicitly written in terms of the kinetic and


potential energies as




\[


\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = \left(\frac{1}{2}\Delta  [V(\mathbf{r}) E_T]\right) \psi(\mathbf{r}, \tau)\,.


\]




We can simulate this differential equation as a diffusionbranching process.






To see this, recall that the [[https://en.wikipedia.org/wiki/Diffusion_equation][diffusion equation]] of particles is given by




\[


\frac{\partial \phi(\mathbf{r},t)}{\partial t} = D\, \Delta \phi(\mathbf{r},t).


\]




The [[https://en.wikipedia.org/wiki/Reaction_rate][rate of reaction]] $v$ is the speed at which a chemical reaction


Furthermore, the [[https://en.wikipedia.org/wiki/Reaction_rate][rate of reaction]] $v$ is the speed at which a chemical reaction


takes place. In a solution, the rate is given as a function of the


concentration $[A]$ by




@ 2059,7 +2157,9 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 o vmc_metropolis


 a rate equation for the potential.




The diffusion equation can be simulated by a Brownian motion:




\[ \mathbf{r}_{n+1} = \mathbf{r}_{n} + \sqrt{\delta t}\, \chi \]




where $\chi$ is a Gaussian random variable, and the rate equation


can be simulated by creating or destroying particles over time (a


socalled branching process).


@ 2068,9 +2168,23 @@ 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 that the groundstate 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.




For the systems you will study this is not an issue:




 Hydrogen atom: You only have one electron!


 Twoelectron system ($H_2$ or He): The groundwave 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




In a molecular system, the potential is far from being constant,


In a molecular system, the potential is far from being constant


and diverges at interparticle 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


@ 2090,28 +2204,55 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 o vmc_metropolis


\frac{\partial \Pi(\mathbf{r},\tau)}{\partial \tau}


= \frac{1}{2} \Delta \Pi(\mathbf{r},\tau) +


\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)


\right] + (E_L(\mathbf{r})E_T)\Pi(\mathbf{r},\tau)


\]




The new "kinetic energy" can be simulated by the drifted diffusion


The new "kinetic energy" can be simulated by the driftdiffusion


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


changing the number of particles according to $\exp\left[ \delta t\,


\left(E_L(\mathbf{r})  E_\text{ref}\right)\right]$


where $E_{\text{ref}}$ is a constant introduced so that the average


of this term is close to one, keeping the number of particles rather


constant.


\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 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




\[


@ 2157,52 +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)


\]






** Fixednode 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 fixednode 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:




\[



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