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@ -75,9 +75,9 @@
\begin{eqnarray*} \begin{eqnarray*}
E & = & \frac{\langle \Psi| \hat{H} | \Psi\rangle}{\langle \Psi |\Psi \rangle} E & = & \frac{\langle \Psi| \hat{H} | \Psi\rangle}{\langle \Psi |\Psi \rangle}
= \frac{\int \Psi(\mathbf{r})\, \hat{H} \Psi(\mathbf{r})\, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}} \\ = \frac{\int \Psi(\mathbf{r})\, \hat{H} \Psi(\mathbf{r})\, d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}} \\
& = & \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}} & = & \frac{\int |\Psi(\mathbf{r})|^2\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}}
= \frac{\int \left[\Psi(\mathbf{r})\right]^2\, E_L(\mathbf{r})\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}} = \frac{\int |\Psi(\mathbf{r})|^2\, E_L(\mathbf{r})\,d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}}
\end{eqnarray*} \end{eqnarray*}
For few dimensions, one can easily compute $E$ by evaluating the integrals on a grid but, for a high number of dimensions, one can resort to Monte Carlo techniques to compute $E$. For few dimensions, one can easily compute $E$ by evaluating the integrals on a grid but, for a high number of dimensions, one can resort to Monte Carlo techniques to compute $E$.
@ -93,17 +93,17 @@
$$ \int_{-\infty}^\infty P(x)\,dx = 1. $$ $$ \int_{-\infty}^\infty P(x)\,dx = 1. $$
Similarly, we can view the the energy of a system, $E$, as the expected value of the local energy with respect to Similarly, we can view the the energy of a system, $E$, as the expected value of the local energy with respect to
a probability density $P(\mathbf{r}}$ defined in 3$N$ dimensions: a probability density $P(\mathbf{r})$ defined in 3$N$ dimensions:
$$ E = \int E_L(\mathbf{r}) P(\mathbf{r})\,d\mathbf{r}} \equiv \langle E_L \rangle_{\Psi^2}\,, $$ $$ E = \int E_L(\mathbf{r}) P(\mathbf{r})\,d\mathbf{r} \equiv \langle E_L \rangle_{\Psi^2}\,, $$
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:
$$ P(\mathbf{r}) = \frac{|Psi(\mathbf{r}|^2){\int \left |\Psi(\mathbf{r})|^2 d\mathbf{r}}\,. $$ $$ P(\mathbf{r}) = \frac{|Psi(\mathbf{r}|^2)}{\int \left |\Psi(\mathbf{r})|^2 d\mathbf{r}}\,. $$
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:
$$ E \approx \frac{1}{N_{\rm MC}} \sum_{i=1}^{N_{\rm MC}} E_L(\mathbf{r}_i} \,. $$ E \approx \frac{1}{N_{\rm MC}} \sum_{i=1}^{N_{\rm MC}} E_L(\mathbf{r}_i) \,. $$
* Numerical evaluation of the energy of the hydrogen atom * Numerical evaluation of the energy of the hydrogen atom
@ -399,7 +399,7 @@ end function e_loc
-\frac{1}{2} a^2 + \frac{a-1}{\mathbf{|r|}} -\frac{1}{2} a^2 + \frac{a-1}{\mathbf{|r|}}
\end{eqnarray*} \end{eqnarray*}
$a=1$ cancels the $1/|r|$ term, and makes the energy constant, $a=1$ cancels the $1/|r|$ term, and makes the energy constant and
equal to -0.5 atomic units. equal to -0.5 atomic units.
** Plot of the local energy along the $x$ axis ** Plot of the local energy along the $x$ axis
@ -749,8 +749,8 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
energy associated with $\Psi$ around its average: energy associated with $\Psi$ around its average:
$$ $$
\sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[ \sigma^2(E_L) = \frac{\int |\Psi(\mathbf{r})|^2\, \left[
E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}} E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}}
$$ $$
which can be simplified as which can be simplified as
@ -1093,7 +1093,7 @@ end subroutine ave_error
Consider again the expression of the energy Consider again the expression of the energy
\begin{eqnarray*} \begin{eqnarray*}
E & = & \frac{\int E_L(\mathbf{r})\left[\Psi(\mathbf{r})\right]^2\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}\,. E & = & \frac{\int E_L(\mathbf{r})|\Psi(\mathbf{r})|^2\,d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}}\,.
\end{eqnarray*} \end{eqnarray*}
Clearly, the square of the wave function is a good choice of probability density to sample but we will start with something simpler and rewrite the energy as Clearly, the square of the wave function is a good choice of probability density to sample but we will start with something simpler and rewrite the energy as
@ -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
\[ \[
P(\mathbf{r}) = \frac{|Psi(\mathbf{r}|^2){\int \left |\Psi(\mathbf{r})|^2 d\mathbf{r}} P(\mathbf{r}) = \frac{|Psi(\mathbf{r})|^2)}{\int |\Psi(\mathbf{r})|^2 d\mathbf{r}}\,.
\] \]
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
@ -1323,16 +1323,16 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
sampling: sampling:
$$ $$
E \approx \frac{1}{N_{\rm MC}}\sum_{i=1}^{N_{\rm MC} E_L(\mathbf{r}_i) E \approx \frac{1}{N_{\rm MC}}\sum_{i=1}^{N_{\rm MC}} E_L(\mathbf{r}_i)\,.
$$ $$
To sample a chosen probability density, an efficient method is the To sample a chosen probability density, an efficient method is the
[[https://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm][Metropolis-Hastings sampling algorithm]]. Starting from a random [[https://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm][Metropolis-Hastings sampling algorithm]]. Starting from a random
initial position $\mathbf{r}_0$, we will realize a random walk: initial position $\mathbf{r}_0$, we will realize a random walk:
$$ \mathbf{r}_0 \rightarrow \mathbf{r}_1 \rightarrow \mathbf{r}_2 \ldots \mathbf{r}_{N_{\rm MC}}\,, $$ $$ \mathbf{r}_0 \rightarrow \mathbf{r}_1 \rightarrow \mathbf{r}_2 \ldots \rightarrow \mathbf{r}_{N_{\rm MC}}\,, $$
following the following algorithm. according to the following algorithm.
At every step, we propose a new move according to a transition probability $T(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1})$ of our choice. At every step, we propose a new move according to a transition probability $T(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1})$ of our choice.
@ -1353,13 +1353,13 @@ 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},\mathbf{r}_{n+1}) P(\mathbf{r}_{n+1})}{T(\mathbf{r}_{n+1},\mathbf{r}_n)P(\mathbf{r}_{n})}\right)\,,
$$ $$
which, for our choice of transition probability, becomes which, for our choice of transition probability, becomes
$$ $$
A{\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1}) = \min\left(1,\frac{P(\mathbf{r}_{n+1})}{P(\mathbf{r}_{n})}\right)= \min\left(1,\frac{\Psi(\mathbf{r}_{n+1})^2}{\Psi(\mathbf{r}_{n})^2} A(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1}) = \min\left(1,\frac{P(\mathbf{r}_{n+1})}{P(\mathbf{r}_{n})}\right)= \min\left(1,\frac{\Psi(\mathbf{r}_{n+1})^2}{\Psi(\mathbf{r}_{n})^2}\right)\,.
$$ $$
Explain why the transition probability cancels out in the expression of $A$. Also note that we do not need to compute the norm of the wave function! Explain why the transition probability cancels out in the expression of $A$. Also note that we do not need to compute the norm of the wave function!
@ -1734,13 +1734,13 @@ end subroutine random_gauss
\frac{\nabla [ \Psi^2 ]}{\Psi^2} = 2 \frac{\nabla \Psi}{\Psi}\,, \frac{\nabla [ \Psi^2 ]}{\Psi^2} = 2 \frac{\nabla \Psi}{\Psi}\,,
\] \]
and add the so-called drift vector, so that the numerical scheme becomes a and add the so-called drift vector, $\frac{\nabla \Psi}{\Psi}$, so that the numerical scheme becomes a
drifted diffusion with transition probability: drifted diffusion with transition probability:
\[ \[
T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) =
\frac{1}{(2\pi\,\delta t)^{3/2}} \exp \left[ - \frac{\left( \frac{1}{(2\pi\,\delta t)^{3/2}} \exp \left[ - \frac{\left(
\mathbf{r}_{n+1} - \mathbf{r}_{n} - \frac{\nabla \mathbf{r}_{n+1} - \mathbf{r}_{n} - \delta t\frac{\nabla
\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\delta t} \right]\,. \Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\delta t} \right]\,.
\] \]