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