Quantum Monte Carlo

For a given system with Hamiltonian \(\hat{H}\) and wave function \(\Psi\), we define the local energy as

\[ E_L(\mathbf{r}) = \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}, \]

where \(\mathbf{r}\) denotes the 3N-dimensional electronic coordinates.

The electronic energy of a system, \(E\), can be rewritten in terms of the local energy \(E_L(\mathbf{r})\) as

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 \(E\).

To this aim, recall that the probabilistic expected value of an arbitrary function \(f(x)\) with respect to a probability density function \(P(x)\) is given by

\[ \langle f \rangle_P = \int_{-\infty}^\infty P(x)\, f(x)\,dx, \]

where a probability density function \(P(x)\) is non-negative and integrates to one:

\[ \int_{-\infty}^\infty P(x)\,dx = 1. \]

Similarly, we can view the the energy of a system, \(E\), as the expected value of the local energy with respect to a probability density \(P(\mathbf{r})\) defined in 3\(N\) dimensions:

\[ E = \int E_L(\mathbf{r}) P(\mathbf{r})\,d\mathbf{r} \equiv \langle E_L \rangle_{P}\,, \]

where the probability density is given by the square of the wave function:

\[ P(\mathbf{r}) = \frac{|\Psi(\mathbf{r})|^2}{\int |\Psi(\mathbf{r})|^2 d\mathbf{r}}\,. \]

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:

\[ E \approx \frac{1}{N_{\rm MC}} \sum_{i=1}^{N_{\rm MC}} E_L(\mathbf{r}_i) \,. \]

You will now program all quantities needed to compute the local energy of the H atom for the given wave function.

Write all the functions of this section in a single file : hydrogen.py if you use Python, or hydrogen.f90 is you use Fortran.

The program you will write in this section will be written in another file (plot_hydrogen.py or plot_hydrogen.f90 for example). It will use the functions previously defined.

In Python, you should put at the beginning of the file

#!/usr/bin/env python3

from hydrogen import e_loc

to be able to use the e_loc function of the hydrogen.py file.

In Fortran, you will need to compile all the source files together:

gfortran hydrogen.f90 plot_hydrogen.f90 -o plot_hydrogen

If the space is discretized in small volume elements \(\mathbf{r}_i\) of size \(\delta \mathbf{r}\), the expression of \(\langle E_L \rangle_{\Psi^2}\) becomes a weighted average of the local energy, where the weights are the values of the wave function square at \(\mathbf{r}_i\) multiplied by the volume element:

\[ \langle E \rangle_{\Psi^2} \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\; w_i = \left|\Psi(\mathbf{r}_i)\right|^2 \delta \mathbf{r} \]

The variance of the local energy is a functional of \(\Psi\) which measures the magnitude of the fluctuations of the local energy associated with \(\Psi\) around its average:

\[ \sigma^2(E_L) = \frac{\int |\Psi(\mathbf{r})|^2\, \left[ E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}} \] which can be simplified as

\[ \sigma^2(E_L) = \langle E_L^2 \rangle_{\Psi^2} - \langle E_L \rangle_{\Psi^2}^2.\]

If the local energy is constant (i.e. \(\Psi\) is an eigenfunction of \(\hat{H}\)) the variance is zero, so the variance of the local energy can be used as a measure of the quality of a wave function.

To compute the statistical error, you need to perform \(M\) independent Monte Carlo calculations. You will obtain \(M\) different estimates of the energy, which are expected to have a Gaussian distribution for large \(M\), according to the Central Limit Theorem.

The estimate of the energy is

\[ E = \frac{1}{M} \sum_{i=1}^M E_i \]

The variance of the average energies can be computed as

\[ \sigma^2 = \frac{1}{M-1} \sum_{i=1}^{M} (E_i - E)^2 \]

And the confidence interval is given by

\[ E \pm \delta E, \text{ where } \delta E = \frac{\sigma}{\sqrt{M}} \]

We will now perform our first Monte Carlo calculation to compute the energy of the hydrogen atom.

Consider again the expression of the energy

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

Here, we will sample a uniform probability \(P(\mathbf{r})\) in a cube of volume \(L^3\) centered at the origin:

\[ P(\mathbf{r}) = \frac{1}{L^3}\,, \]

and zero outside the cube.

One Monte Carlo run will consist of \(N_{\rm MC}\) Monte Carlo iterations. At every Monte Carlo iteration:

• Draw a random point \(\mathbf{r}_i\) in the box \((-5,-5,-5) \le (x,y,z) \le (5,5,5)\)
• Compute \(|\Psi(\mathbf{r}_i)|^2\) and accumulate the result in a variable normalization
• Compute \(|\Psi(\mathbf{r}_i)|^2 \times E_L(\mathbf{r}_i)\), and accumulate the result in a variable energy

Once all the iterations have been computed, the run returns the average energy \(\bar{E}_k\) over the \(N_{\rm MC}\) iterations of the run.

To compute the statistical error, perform \(M\) independent runs. The final estimate of the energy will be the average over the \(\bar{E}_k\), and the variance of the \(\bar{E}_k\) will be used to compute the statistical error.

We will now use the square of the wave function to sample random points distributed with the probability density \[ P(\mathbf{r}) = \frac{|\Psi(\mathbf{r})|^2}{\int |\Psi(\mathbf{r})|^2 d\mathbf{r}}\,. \]

The expression of the average energy is now simplified as the average of the local energies, since the weights are taken care of by the sampling:

\[ E \approx \frac{1}{N_{\rm MC}}\sum_{i=1}^{N_{\rm MC}} E_L(\mathbf{r}_i)\,. \]

To sample a chosen probability density, an efficient method is the Metropolis-Hastings sampling algorithm. Starting from a random initial position \(\mathbf{r}_0\), we will realize a random walk:

\[ \mathbf{r}_0 \rightarrow \mathbf{r}_1 \rightarrow \mathbf{r}_2 \ldots \rightarrow \mathbf{r}_{N_{\rm MC}}\,, \]

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.

For simplicity, we will move the electron in a 3-dimensional box of side \(2\delta L\) centered at the current position of the electron:

\[ \mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta L \, \mathbf{u} \]

where \(\delta L\) is a fixed constant, and \(\mathbf{u}\) is a uniform random number in a 3-dimensional box \((-1,-1,-1) \le \mathbf{u} \le (1,1,1)\).

After having moved the electron, we add the accept/reject step that guarantees that the distribution of the \(\mathbf{r}_n\) is \(\Psi^2\). This amounts to accepting the move with probability

\[ A(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1}) = \min\left(1,\frac{T(\mathbf{r}_{n+1}\rightarrow\mathbf{r}_{n}) P(\mathbf{r}_{n+1})}{T(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1})P(\mathbf{r}_{n})}\right)\,, \]

which, for our choice of transition probability, becomes

\[ A(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1}) = \min\left(1,\frac{P(\mathbf{r}_{n+1})}{P(\mathbf{r}_{n})}\right)= \min\left(1,\frac{|\Psi(\mathbf{r}_{n+1})|^2}{|\Psi(\mathbf{r}_{n})|^2}\right)\,. \]

Also note that we do not need to compute the norm of the wave function!

The algorithm is summarized as follows:

1. Evaluate the local energy at \(\mathbf{r}_n\) and accumulate it
2. Compute a new position \(\mathbf{r'} = \mathbf{r}_n + \delta L\, \mathbf{u}\)
3. Evaluate \(\Psi(\mathbf{r}')\) at the new position
4. Compute the ratio \(A = \frac{\left|\Psi(\mathbf{r'})\right|^2}{\left|\Psi(\mathbf{r}_{n})\right|^2}\)
5. Draw a uniform random number \(v \in [0,1]\)
6. if \(v \le A\), accept the move : set \(\mathbf{r}_{n+1} = \mathbf{r'}\)
7. else, reject the move : set \(\mathbf{r}_{n+1} = \mathbf{r}_n\)

One can use more efficient numerical schemes to move the electrons by choosing a smarter expression for the transition probability.

The Metropolis acceptance step has to be adapted accordingly to ensure that the detailed balance condition is satisfied. This means that the acceptance probability \(A\) is chosen so that it is consistent with the probability of leaving \(\mathbf{r}_n\) and the probability of entering \(\mathbf{r}_{n+1}\):

\[ A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \min \left( 1, \frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})} {T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})} \right) \] where \(T(\mathbf{r}_n \rightarrow \mathbf{r}_{n+1})\) is the probability of transition from \(\mathbf{r}_n\) to \(\mathbf{r}_{n+1}\).

In the previous example, we were using uniform sampling in a box centered at the current position. Hence, the transition probability was symmetric

\[ T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) = \text{constant}\,, \]

so the expression of \(A\) was simplified to the ratios of the squared wave functions.

Now, if instead of drawing uniform random numbers, we choose to draw Gaussian random numbers with zero mean and variance \(\delta t\), the transition probability becomes:

\[ T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \frac{1}{(2\pi\,\delta t)^{3/2}} \exp \left[ - \frac{\left( \mathbf{r}_{n+1} - \mathbf{r}_{n} \right)^2}{2\delta t} \right]\,. \]

Furthermore, to sample the density even better, we can "push" the electrons into in the regions of high probability, and "pull" them away from the low-probability regions. This will increase the acceptance ratios and improve the sampling.

To do this, we can use the gradient of the probability density

\[ \frac{\nabla [ \Psi^2 ]}{\Psi^2} = 2 \frac{\nabla \Psi}{\Psi}\,, \]

and add the so-called drift vector, \(\frac{\nabla \Psi}{\Psi}\), so that the numerical scheme becomes a drifted diffusion with transition probability:

\[ T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \frac{1}{(2\pi\,\delta t)^{3/2}} \exp \left[ - \frac{\left( \mathbf{r}_{n+1} - \mathbf{r}_{n} - \delta t\frac{\nabla \Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\delta t} \right]\,. \]

The corresponding move is proposed as

\[ \mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi \,, \]

where \(\chi\) is a Gaussian random variable with zero mean and variance \(\delta t\).

The algorithm of the previous exercise is only slighlty modified as:

1. Evaluate the local energy at \(\mathbf{r}_{n}\) and accumulate it
2. Compute a new position \(\mathbf{r'} = \mathbf{r}_n + \delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi\)
3. Evaluate \(\Psi(\mathbf{r}')\) and \(\frac{\nabla \Psi(\mathbf{r'})}{\Psi(\mathbf{r'})}\) at the new position
4. Compute the ratio \(A = \frac{T(\mathbf{r}' \rightarrow \mathbf{r}_{n}) P(\mathbf{r}')}{T(\mathbf{r}_{n} \rightarrow \mathbf{r}') P(\mathbf{r}_{n})}\)
5. Draw a uniform random number \(v \in [0,1]\)
6. if \(v \le A\), accept the move : set \(\mathbf{r}_{n+1} = \mathbf{r'}\)
7. else, reject the move : set \(\mathbf{r}_{n+1} = \mathbf{r}_n\)