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QMC.org
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@ -99,7 +99,7 @@
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where the probability density is given by the square of the wave function:
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$$ P(\mathbf{r}) = \frac{|Psi(\mathbf{r}|^2)}{\int \left |\Psi(\mathbf{r})|^2 d\mathbf{r}}\,. $$
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$$ P(\mathbf{r}) = \frac{|Psi(\mathbf{r}|^2)}{\int |\Psi(\mathbf{r})|^2 d\mathbf{r}}\,. $$
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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:
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@ -763,7 +763,7 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
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*** Exercise (optional)
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#+begin_exercise
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Prove that :
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$$\left( \langle E - \langle E \rangle_{\Psi^2} \rangle_{\Psi^2} \right)^2 = \langle E^2 \rangle_{\Psi^2} - \langle E \rangle_{\Psi^2}^2 $$
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$$\langle \left( E - \langle E \rangle_{\Psi^2} \right)^2\rangle_{\Psi^2} = \langle E^2 \rangle_{\Psi^2} - \langle E \rangle_{\Psi^2}^2 $$
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#+end_exercise
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**** Solution :solution:
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@ -772,7 +772,7 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
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\rangle = \bar{E}$ .
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\begin{eqnarray*}
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\langle E - \bar{E} \rangle^2 & = &
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\langle (E - \bar{E})^2 \rangle & = &
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\langle E^2 - 2 E \bar{E} + \bar{E}^2 \rangle \\
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&=& \langle E^2 \rangle - 2 \langle E \bar{E} \rangle + \langle \bar{E}^2 \rangle \\
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&=& \langle E^2 \rangle - 2 \langle E \rangle \bar{E} + \bar{E}^2 \\
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@ -991,13 +991,13 @@ gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
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The estimate of the energy is
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$$
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E = \frac{1}{M} \sum_{i=1}^M E_M
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E = \frac{1}{M} \sum_{i=1}^M E_i
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$$
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The variance of the average energies can be computed as
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$$
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\sigma^2 = \frac{1}{M-1} \sum_{i=1}^{M} (E_M - E)^2
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\sigma^2 = \frac{1}{M-1} \sum_{i=1}^{M} (E_i - E)^2
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$$
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And the confidence interval is given by
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@ -1315,7 +1315,7 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
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We will now use the square of the wave function to sample random
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points distributed with the probability density
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\[
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P(\mathbf{r}) = \frac{|Psi(\mathbf{r})|^2)}{\int |\Psi(\mathbf{r})|^2 d\mathbf{r}}\,.
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P(\mathbf{r}) = \frac{|\Psi(\mathbf{r})|^2}{\int |\Psi(\mathbf{r})|^2 d\mathbf{r}}\,.
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\]
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The expression of the average energy is now simplified as the average of
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@ -1353,7 +1353,7 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
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probability
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$$
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A(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1}) = \min\left(1,\frac{T(\mathbf{r}_{n},\mathbf{r}_{n+1}) P(\mathbf{r}_{n+1})}{T(\mathbf{r}_{n+1},\mathbf{r}_n)P(\mathbf{r}_{n})}\right)\,,
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A(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1}) = \min\left(1,\frac{T(\mathbf{r}_{n+1}\rightarrow\mathbf{r}_{n}) P(\mathbf{r}_{n+1})}{T(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1})P(\mathbf{r}_{n})}\right)\,,
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$$
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which, for our choice of transition probability, becomes
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@ -2083,10 +2083,10 @@ 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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\[
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i\frac{\partial \Psi(\mathbf{r},t)}{\partial t} = (\hat{H} -E_T) \Psi(\mathbf{r},t)\,.
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i\frac{\partial \Psi(\mathbf{r},t)}{\partial t} = (\hat{H} -E_{\rm ref}) \Psi(\mathbf{r},t)\,.
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\]
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where we introduced a shift in the energy, $E_T$, which will come useful below.
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where we introduced a shift in the energy, $E_{\rm ref}$, 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, $\Phi_k$, with energies $E_k$:
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@ -2098,26 +2098,26 @@ 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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\[
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\Psi(\mathbf{r},t) = \sum_k a_k \exp \left( -i\, (E_k-E_T)\, t \right) \Phi_k(\mathbf{r}).
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\Psi(\mathbf{r},t) = \sum_k a_k \exp \left( -i\, (E_k-E_{\rm ref})\, t \right) \Phi_k(\mathbf{r}).
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\]
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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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\[
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-\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = (\hat{H} -E_T) \psi(\mathbf{r}, \tau)
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-\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = (\hat{H} -E_{\rm ref}) \psi(\mathbf{r}, \tau)
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\]
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where $\psi(\mathbf{r},\tau) = \Psi(\mathbf{r},-i\,)$
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where $\psi(\mathbf{r},\tau) = \Psi(\mathbf{r},-i\,t)$
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and
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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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\psi(\mathbf{r},\tau) &=& \sum_k a_k \exp( -(E_k-E_{\rm ref})\, \tau) \phi_k(\mathbf{r})\\
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&=& \exp(-(E_0-E_{\rm ref})\, \tau)\sum_k a_k \exp( -(E_k-E_0)\, \tau) \phi_k(\mathbf{r})\,.
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\end{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. If we adjust $E_T$ to the running estimate of $E_0$,
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$k=0$ term, namely, the lowest eigenstate. If we adjust $E_{\rm ref}$ 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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@ -2128,7 +2128,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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potential energies as
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\[
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\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = \left(\frac{1}{2}\Delta - [V(\mathbf{r}) -E_T]\right) \psi(\mathbf{r}, \tau)\,.
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\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = \left(\frac{1}{2}\Delta - [V(\mathbf{r}) -E_{\rm ref}]\right) \psi(\mathbf{r}, \tau)\,.
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\]
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We can simulate this differential equation as a diffusion-branching process.
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@ -2137,7 +2137,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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To see this, recall that the [[https://en.wikipedia.org/wiki/Diffusion_equation][diffusion equation]] of particles is given by
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\[
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\frac{\partial \phi(\mathbf{r},t)}{\partial t} = D\, \Delta \phi(\mathbf{r},t).
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\frac{\partial \psi(\mathbf{r},t)}{\partial t} = D\, \Delta \psi(\mathbf{r},t).
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\]
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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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@ -2203,7 +2203,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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-\frac{\partial \Pi(\mathbf{r},\tau)}{\partial \tau}
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= -\frac{1}{2} \Delta \Pi(\mathbf{r},\tau) +
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\nabla \left[ \Pi(\mathbf{r},\tau) \frac{\nabla \Psi_T(\mathbf{r})}{\Psi_T(\mathbf{r})}
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\right] + (E_L(\mathbf{r})-E_T)\Pi(\mathbf{r},\tau)
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\right] + (E_L(\mathbf{r})-E_{\rm ref})\Pi(\mathbf{r},\tau)
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\]
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The new "kinetic energy" can be simulated by the drift-diffusion
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@ -2211,8 +2211,8 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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The new "potential" is the local energy, which has smaller fluctuations
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when $\Psi_T$ gets closer to the exact wave function. It can be simulated by
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changing the number of particles according to $\exp\left[ -\delta t\,
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\left(E_L(\mathbf{r}) - E_T\right)\right]$
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where $E_T$ is the constant we had introduced above, which is adjusted to
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\left(E_L(\mathbf{r}) - E_{\rm ref}\right)\right]$
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where $E_{\rm ref}$ is the constant we had introduced above, which is adjusted to
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the running average energy to keep the number of particles
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reasonably constant.
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@ -2223,28 +2223,28 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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To this aim, we use the mixed estimator of the energy:
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\begin{eqnarray*}
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E(\tau) &=& \frac{\langle \psi(tau) | \hat{H} | \Psi_T \rangle}{\frac{\langle \psi(tau) | \Psi_T \rangle}\\
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E(\tau) &=& \frac{\langle \psi(tau) | \hat{H} | \Psi_T \rangle}{\langle \psi(tau) | \Psi_T \rangle}\\
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&=& \frac{\int \psi(\mathbf{r},\tau) \hat{H} \Psi_T(\mathbf{r}) d\mathbf{r}}
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{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}} \\
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&=& \int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) E_L(\mathbf{r}) d\mathbf{r}}
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{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}}
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&=& \frac{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) E_L(\mathbf{r}) d\mathbf{r}}
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{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}} \,.
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\end{eqnarray*}
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Since, for large $\tau$, we have that
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For large $\tau$, we have that
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\[
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\Pi(\mathbf{r},\tau) =\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \rightarrow \Phi_0(\mathbf{r}) \Psi_T(\mathbf{r})\,,
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\]
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and, using that $\hat{H}$ is Hermitian and that $\Phi_0$ is an eigenstate of the Hamiltonian, we obtain
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and, using that $\hat{H}$ is Hermitian and that $\Phi_0$ is an eigenstate of the Hamiltonian, we obtain for large $\tau$
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\[
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E(\tau) = \frac{\langle \psi_\tau | \hat{H} | \Psi_T \rangle}
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{\langle \psi_\tau | \Psi_T \rangle}
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= \frac{\langle \Psi_T | \hat{H} | \psi_\tau \rangle}
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{\langle \Psi_T | \psi_\tau \rangle}
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\rightarrow E_0 \frac{\langle \Psi_T | \psi_\tau \rangle}
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{\langle \Psi_T | \psi_\tau \rangle}
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\rightarrow E_0 \frac{\langle \Psi_T | \Phi_0 \rangle}
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{\langle \Psi_T | \Phi_0 \rangle}
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= E_0
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\]
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@ -2302,7 +2302,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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Instead of having a variable number of particles to simulate the
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branching process, one can consider the term
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$\exp \left( -\delta t\,( E_L(\mathbf{r}) - E_T} \right)$ as a
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$\exp \left( -\delta t\,( E_L(\mathbf{r}) - E_{\rm ref}) \right)$ as a
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cumulative product of weights:
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\[
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