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@ 1397,6 +1397,9 @@ 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 for dL for reasons which will


become clear later.






*** Exercise




@ 2080,11 +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)\,.


\]




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:


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




\[


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


@ 2093,36 +2098,48 @@ 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, 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 $\psi(\mathbf{r},\tau) = \Psi(\mathbf{r},i\,)$


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})\,.


\begin{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 this 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




@ 2139,7 +2156,9 @@ 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).


@ 2150,7 +2169,7 @@ 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,


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



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