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@ -1397,6 +1397,9 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
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step such that the acceptance rate is close to 0.5 is a good
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compromise for the current problem.
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NOTE: below, we use the symbol dt for dL for reasons which will
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become clear later.
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*** Exercise
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@ -2080,11 +2083,13 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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Consider the time-dependent Schrödinger equation:
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\[
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i\frac{\partial \Psi(\mathbf{r},t)}{\partial t} = \hat{H} \Psi(\mathbf{r},t)\,.
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i\frac{\partial \Psi(\mathbf{r},t)}{\partial t} = (\hat{H} -E_T) \Psi(\mathbf{r},t)\,.
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\]
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where we introduced a shift in the energy, $E_T$, which will come useful below.
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We can expand a given starting wave function, $\Psi(\mathbf{r},0)$, in the basis of the eigenstates
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of the time-independent Hamiltonian:
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of the time-independent Hamiltonian, $\Phi_k$, with energies $E_k$:
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\[
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\Psi(\mathbf{r},0) = \sum_k a_k\, \Phi_k(\mathbf{r}).
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@ -2093,36 +2098,48 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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The solution of the Schrödinger equation at time $t$ is
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\[
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\Psi(\mathbf{r},t) = \sum_k a_k \exp \left( -i\, E_k\, t \right) \Phi_k(\mathbf{r}).
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\Psi(\mathbf{r},t) = \sum_k a_k \exp \left( -i\, (E_k-E_T)\, t \right) \Phi_k(\mathbf{r}).
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\]
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Now, if we replace the time variable $t$ by an imaginary time variable
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$\tau=i\,t$, we obtain
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\[
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-\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = \hat{H} \psi(\mathbf{r}, \tau)
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-\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = (\hat{H} -E_T) \psi(\mathbf{r}, \tau)
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\]
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where $\psi(\mathbf{r},\tau) = \Psi(\mathbf{r},-i\,)$
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and
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\[
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\psi(\mathbf{r},\tau) = \sum_k a_k \exp( -E_k\, \tau) \phi_k(\mathbf{r}).
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\]
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\begin{eqnarray*}
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\psi(\mathbf{r},\tau) &=& \sum_k a_k \exp( -E_k\, \tau) \phi_k(\mathbf{r})\\
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&=& \exp(-(E_0-E_T)\, \tau)\sum_k a_k \exp( -(E_k-E_0)\, \tau) \phi_k(\mathbf{r})\,.
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\begin{eqnarray*}
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For large positive values of $\tau$, $\psi$ is dominated by the
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$k=0$ term, namely the lowest eigenstate.
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So we can expect that simulating the differetial equation in
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$k=0$ term, namely, the lowest eigenstate. If we adjust $E_T$ to the running estimate of $E_0$,
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we can expect that simulating the differetial equation in
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imaginary time will converge to the exact ground state of the
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system.
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** Diffusion and branching
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The [[https://en.wikipedia.org/wiki/Diffusion_equation][diffusion equation]] of particles is given by
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The imaginary-time Schrödinger equation can be explicitly written in terms of the kinetic and
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potential energies as
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\[
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\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = \left(\frac{1}{2}\Delta - [V(\mathbf{r}) -E_T]\right) \psi(\mathbf{r}, \tau)\,.
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\]
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We can simulate this differential equation as a diffusion-branching process.
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To this this, recall that the [[https://en.wikipedia.org/wiki/Diffusion_equation][diffusion equation]] of particles is given by
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\[
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\frac{\partial \phi(\mathbf{r},t)}{\partial t} = D\, \Delta \phi(\mathbf{r},t).
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\]
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The [[https://en.wikipedia.org/wiki/Reaction_rate][rate of reaction]] $v$ is the speed at which a chemical reaction
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Furthermore, the [[https://en.wikipedia.org/wiki/Reaction_rate][rate of reaction]] $v$ is the speed at which a chemical reaction
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takes place. In a solution, the rate is given as a function of the
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concentration $[A]$ by
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@ -2139,7 +2156,9 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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- a rate equation for the potential.
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The diffusion equation can be simulated by a Brownian motion:
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\[ \mathbf{r}_{n+1} = \mathbf{r}_{n} + \sqrt{\delta t}\, \chi \]
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where $\chi$ is a Gaussian random variable, and the rate equation
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can be simulated by creating or destroying particles over time (a
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so-called branching process).
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@ -2150,7 +2169,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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** Importance sampling
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In a molecular system, the potential is far from being constant,
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In a molecular system, the potential is far from being constant
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and diverges at inter-particle coalescence points. Hence, when the
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rate equation is simulated, it results in very large fluctuations
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in the numbers of particles, making the calculations impossible in
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