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@ -1397,8 +1397,8 @@ 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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NOTE: below, we use the symbol dt to denote dL since we will use
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the same variable later on to store a time step.
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*** Exercise
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@ -2168,10 +2168,19 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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system by simulating the Schrödinger equation in imaginary time, by
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the combination of a diffusion process and a branching process.
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We note here that the ground-state wave function of a Fermionic system is
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antisymmetric and changes sign.
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We note that the ground-state wave function of a Fermionic system is
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antisymmetric and changes sign. Therefore, it is interpretation as a probability
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distribution is somewhat problematic. In fact, mathematically, since
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the Bosonic ground state is lower in energy than the Fermionic one, for
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large $\tau$, the system will evolve towards the Bosonic solution.
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I AM HERE
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For the systems you will study this is not an issue:
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- Hydrogen atom: You only have one electron!
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- Two-electron system ($H_2$ or He): The ground-wave function is antisymmetric
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in the spin variables but symmetric in the space ones.
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Therefore, in both cases, you are dealing with a "Bosonic" ground state.
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** Importance sampling
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@ -2205,17 +2214,44 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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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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the running average energy and is introduced to keep the number of particles
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the running average energy to keep the number of particles
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reasonably constant.
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This equation generates the /N/-electron density $\Pi$, which is the
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product of the ground state with the trial wave function. It
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introduces the constraint that $\Pi(\mathbf{r},\tau)=0$ where
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$\Psi_T(\mathbf{r})=0$. In the few cases where the wave function has no nodes,
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such as in the hydrogen atom or the H_2 molecule, this
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constraint is harmless and we can obtain the exact energy. But for
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systems where the wave function has nodes, this scheme introduces an
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error known as the /fixed node error/.
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product of the ground state with the trial wave function. You may then ask: how
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can we compute the total energy of the system?
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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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&=& \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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\end{eqnarray*}
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Since, 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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\[
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E(\tau) = \frac{\langle \psi_\tau | \hat{H} | \Psi_T \rangle}
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{\langle \psi_\tau | \Psi_T \rangle}
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= \frac{\langle \Psi_T | \hat{H} | \psi_\tau \rangle}
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{\langle \Psi_T | \psi_\tau \rangle}
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\rightarrow E_0 \frac{\langle \Psi_T | \psi_\tau \rangle}
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{\langle \Psi_T | \psi_\tau \rangle}
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= E_0
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\]
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Therefore, we can compute the energy within DMC by generating the
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density $\Pi$ with random walks, and simply averaging the local
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energies computed with the trial wave function.
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*** Appendix : Details of the Derivation
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@ -2263,51 +2299,11 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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\right] + E_L(\mathbf{r}) \Pi(\mathbf{r},\tau)
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\]
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** Fixed-node DMC energy
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Now that we have a process to sample $\Pi(\mathbf{r},\tau) =
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\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})$, we can compute the exact
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energy of the system, within the fixed-node constraint, as:
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\[
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E = \lim_{\tau \to \infty} \frac{\int \Pi(\mathbf{r},\tau) E_L(\mathbf{r}) d\mathbf{r}}
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{\int \Pi(\mathbf{r},\tau) d\mathbf{r}} = \lim_{\tau \to
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\infty} E(\tau).
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\]
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\[
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E(\tau) = \frac{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) E_L(\mathbf{r}) d\mathbf{r}}
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{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}}
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= \frac{\int \psi(\mathbf{r},\tau) \hat{H} \Psi_T(\mathbf{r}) d\mathbf{r}}
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{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}}
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= \frac{\langle \psi_\tau | \hat{H} | \Psi_T \rangle}
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{\langle \psi_\tau | \Psi_T \rangle}
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\]
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As $\hat{H}$ is Hermitian,
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\[
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E(\tau) = \frac{\langle \psi_\tau | \hat{H} | \Psi_T \rangle}
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{\langle \psi_\tau | \Psi_T \rangle}
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= \frac{\langle \Psi_T | \hat{H} | \psi_\tau \rangle}
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{\langle \Psi_T | \psi_\tau \rangle}
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= E[\psi_\tau] \frac{\langle \Psi_T | \psi_\tau \rangle}
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{\langle \Psi_T | \psi_\tau \rangle}
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= E[\psi_\tau]
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\]
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So computing the energy within DMC consists in generating the
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density $\Pi$ with random walks, and simply averaging the local
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energies computed with the trial wave function.
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** Pure Diffusion Monte Carlo (PDMC)
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Instead of having a variable number of particles to simulate the
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branching process, one can choose to sample $[\Psi_T(\mathbf{r})]^2$ instead of
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$\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})$, and consider the term
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$\exp \left( -\delta t\,( E_L(\mathbf{r}) - E_{\text{ref}} \right)$ as a
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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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cumulative product of weights:
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\[
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