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:

\[ E = \int E_L(\mathbf{r}) P(\mathbf{r})\,d\mathbf{r} \equiv \langle E_L \rangle_{\Psi^2}\,, \]

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.

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

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

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!

The algorithm is summarized as follows:

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

If the box is infinitely small, the ratio will be very close to one and all the steps will be accepted. However, the moves will be very correlated and you will visit the configurational space very slowly.

On the other hand, if you propose too large moves, the number of accepted steps will decrease because the ratios might become small. If the number of accepted steps is close to zero, then the space is not well sampled either.

The size of the move should be adjusted so that it is as large as possible, keeping the number of accepted steps not too small. To achieve that, we define the acceptance rate as the number of accepted steps over the total number of steps. Adjusting the time step such that the acceptance rate is close to 0.5 is a good compromise for the current problem.

NOTE: below, we use the symbol \(\delta t\) to denote \(\delta L\) since we will use the same variable later on to store a time step.

To obtain Gaussian-distributed random numbers, you can apply the Box Muller transform to uniform random numbers:

Below is a Fortran implementation returning a Gaussian-distributed n-dimensional vector \(\mathbf{z}\). This will be useful for the following sections.

Fortran

subroutine random_gauss(z,n)
  implicit none
  integer, intent(in) :: n
  double precision, intent(out) :: z(n)
  double precision :: u(n+1)
  double precision, parameter :: two_pi = 2.d0*dacos(-1.d0)
  integer :: i

  call random_number(u)

  if (iand(n,1) == 0) then
     ! n is even
     do i=1,n,2
        z(i)   = dsqrt(-2.d0*dlog(u(i))) 
        z(i+1) = z(i) * dsin( two_pi*u(i+1) )
        z(i)   = z(i) * dcos( two_pi*u(i+1) )
     end do

  else
     ! n is odd
     do i=1,n-1,2
        z(i)   = dsqrt(-2.d0*dlog(u(i))) 
        z(i+1) = z(i) * dsin( two_pi*u(i+1) )
        z(i)   = z(i) * dcos( two_pi*u(i+1) )
     end do

     z(n)   = dsqrt(-2.d0*dlog(u(n))) 
     z(n)   = z(n) * dcos( two_pi*u(n+1) )

  end if

end subroutine random_gauss

In Python, you can use the random.normal function of Numpy.

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 ncrease 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 corrsponding 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. Compute a new position \(\mathbf{r'} = \mathbf{r}_n + \delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi\)

   Evaluate \(\Psi\) and \(\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\) at the new position

2. Compute the ratio \(A = \frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})}{T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})}\)
3. Draw a uniform random number \(v \in [0,1]\)
4. if \(v \le A\), accept the move : set \(\mathbf{r}_{n+1} = \mathbf{r'}\)
5. else, reject the move : set \(\mathbf{r}_{n+1} = \mathbf{r}_n\)
6. evaluate the local energy at \(\mathbf{r}_{n+1}\)

Consider the time-dependent Schrödinger equation:

\[ i\frac{\partial \Psi(\mathbf{r},t)}{\partial t} = (\hat{H} -E_{\rm ref}) \Psi(\mathbf{r},t)\,. \]

where we introduced a shift in the energy, \(E_{\rm ref}\), for reasons which will become apparent below.

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 \(E_k\):

\[ \Psi(\mathbf{r},0) = \sum_k a_k\, \Phi_k(\mathbf{r}). \]

The solution of the Schrödinger equation at time \(t\) is

\[ \Psi(\mathbf{r},t) = \sum_k a_k \exp \left( -i\, (E_k-E_{\rm ref})\, t \right) \Phi_k(\mathbf{r}). \]

Now, if we replace the time variable \(t\) by an imaginary time variable \(\tau=i\,t\), we obtain

\[ -\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = (\hat{H} -E_{\rm ref}) \psi(\mathbf{r}, \tau) \]

where \(\psi(\mathbf{r},\tau) = \Psi(\mathbf{r},-i\,\tau)\) and

For large positive values of \(\tau\), \(\psi\) is dominated by the \(k=0\) 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 imaginary time will converge to the exact ground state of the system.

The imaginary-time Schrödinger equation can be explicitly written in terms of the kinetic and potential energies as

\[ \frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = \left(\frac{1}{2}\Delta - [V(\mathbf{r}) -E_{\rm ref}]\right) \psi(\mathbf{r}, \tau)\,. \]

We can simulate this differential equation as a diffusion-branching process.

To see this, recall that the diffusion equation of particles is given by

\[ \frac{\partial \psi(\mathbf{r},t)}{\partial t} = D\, \Delta \psi(\mathbf{r},t). \]

Furthermore, the rate of reaction \(v\) is the speed at which a chemical reaction takes place. In a solution, the rate is given as a function of the concentration \([A]\) by

\[ v = \frac{d[A]}{dt}, \]

where the concentration \([A]\) is proportional to the number of particles.

These two equations allow us to interpret the Schrödinger equation in imaginary time as the combination of:

• a diffusion equation with a diffusion coefficient \(D=1/2\) for the kinetic energy, and
• a rate equation for the potential.

The diffusion equation can be simulated by a Brownian motion:

\[ \mathbf{r}_{n+1} = \mathbf{r}_{n} + \sqrt{\delta t}\, \chi \]

where \(\chi\) is a Gaussian random variable, and the rate equation can be simulated by creating or destroying particles over time (a so-called branching process).

In Diffusion Monte Carlo (DMC), one onbtains the ground state of a system by simulating the Schrödinger equation in imaginary time via the combination of a diffusion process and a branching process.

We note that the ground-state wave function of a Fermionic system is antisymmetric and changes sign. Therefore, its interpretation as a probability distribution is somewhat problematic. In fact, mathematically, since the Bosonic ground state is lower in energy than the Fermionic one, for large \(\tau\), the system will evolve towards the Bosonic solution.

For the systems you will study, this is not an issue:

• Hydrogen atom: You only have one electron!
• Two-electron system (\(H_2\) or He): The ground-wave function is antisymmetric in the spin variables but symmetric in the space ones.

Therefore, in both cases, you are dealing with a "Bosonic" ground state.

In a molecular system, the potential is far from being constant and, in fact, diverges at the inter-particle coalescence points. Hence, when the rate equation is simulated, it results in very large fluctuations in the numbers of particles, making the calculations impossible in practice. Fortunately, if we multiply the Schrödinger equation by a chosen trial wave function \(\Psi_T(\mathbf{r})\) (Hartree-Fock, Kohn-Sham determinant, CI wave function, etc), one obtains

\[ -\frac{\partial \psi(\mathbf{r},\tau)}{\partial \tau} \Psi_T(\mathbf{r}) = \left[ -\frac{1}{2} \Delta \psi(\mathbf{r},\tau) + V(\mathbf{r}) \psi(\mathbf{r},\tau) \right] \Psi_T(\mathbf{r}) \]

Defining \(\Pi(\mathbf{r},\tau) = \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})\), (see appendix for details)

\[ -\frac{\partial \Pi(\mathbf{r},\tau)}{\partial \tau} = -\frac{1}{2} \Delta \Pi(\mathbf{r},\tau) + \nabla \left[ \Pi(\mathbf{r},\tau) \frac{\nabla \Psi_T(\mathbf{r})}{\Psi_T(\mathbf{r})} \right] + (E_L(\mathbf{r})-E_{\rm ref})\Pi(\mathbf{r},\tau) \]

The new "kinetic energy" can be simulated by the drift-diffusion scheme presented in the previous section (VMC). The new "potential" is the local energy, which has smaller fluctuations when \(\Psi_T\) gets closer to the exact wave function. This term can be simulated by changing the number of particles according to \(\exp\left[ -\delta t\, \left(E_L(\mathbf{r}) - E_{\rm ref}\right)\right]\) 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 reasonably constant.

This equation generates the N-electron density \(\Pi\), which is the product of the ground state with the trial wave function. You may then ask: how can we compute the total energy of the system?

To this aim, we use the mixed estimator of the energy:

For large \(\tau\), we have that

\[ \Pi(\mathbf{r},\tau) =\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \rightarrow \Phi_0(\mathbf{r}) \Psi_T(\mathbf{r})\,, \]

and, using that \(\hat{H}\) is Hermitian and that \(\Phi_0\) is an eigenstate of the Hamiltonian, we obtain for large \(\tau\)

\[ E(\tau) = \frac{\langle \psi_\tau | \hat{H} | \Psi_T \rangle} {\langle \psi_\tau | \Psi_T \rangle} = \frac{\langle \Psi_T | \hat{H} | \psi_\tau \rangle} {\langle \Psi_T | \psi_\tau \rangle} \rightarrow E_0 \frac{\langle \Psi_T | \Phi_0 \rangle} {\langle \Psi_T | \Phi_0 \rangle} = E_0 \]

Therefore, we can compute the energy within DMC by generating the density \(\Pi\) with random walks, and simply averaging the local energies computed with the trial wave function.

Instead of having a variable number of particles to simulate the branching process, one can consider the term \(\exp \left( -\delta t\,( E_L(\mathbf{r}) - E_{\rm ref}) \right)\) as a cumulative product of weights:

\[ W(\mathbf{r}_n, \tau) = \exp \left( \int_0^\tau - (E_L(\mathbf{r}_t) - E_{\text{ref}}) dt \right) \approx \prod_{i=1}^{n} \exp \left( -\delta t\, (E_L(\mathbf{r}_i) - E_{\text{ref}}) \right) = \prod_{i=1}^{n} w(\mathbf{r}_i) \]

where \(\mathbf{r}_i\) are the coordinates along the trajectory and we introduced a time-step \(\delta t\).

The algorithm can be rather easily built on top of your VMC code:

1. Compute a new position \(\mathbf{r'} = \mathbf{r}_n + \delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi\)

   Evaluate \(\Psi\) and \(\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\) at the new position

2. Compute the ratio \(A = \frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})}{T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})}\)
3. Draw a uniform random number \(v \in [0,1]\)
4. if \(v \le A\), accept the move : set \(\mathbf{r}_{n+1} = \mathbf{r'}\)
5. else, reject the move : set \(\mathbf{r}_{n+1} = \mathbf{r}_n\)
6. evaluate the local energy at \(\mathbf{r}_{n+1}\)
7. compute the weight \(w(\mathbf{r}_i)\)
8. update \(W\)

Some comments are needed:

• You estimate the energy as

  \begin{eqnarray*} E = \frac{\sum_{k=1}^{N_{\rm MC}} E_L(\mathbf{r}_k) W(\mathbf{r}_k, k\delta t)}{\sum_{k=1}^{N_{\rm MC}} W(\mathbf{r}_k, k\delta t)} \end{eqnarray*}

• The result will be affected by a time-step error (the finite size of \(\delta t\)) and one has in principle to extrapolate to the limit \(\delta t \rightarrow 0\). This amounts to fitting the energy computed for multiple values of \(\delta t\).

  Here, you will be using a small enough time-step and you should not worry about the extrapolation.

• The accept/reject step (steps 2-5 in the algorithm) is in principle not needed for the correctness of the DMC algorithm. However, its use reduces significantly the time-step error.

The PDMC algorithm is less stable than the branching algorithm: it requires to have a value of \(E_\text{ref}\) which is close to the fixed-node energy, and a good trial wave function. Its big advantage is that it is very easy to program starting from a VMC code, so this is what we will do in the next section.

We will now consider the H₂ molecule in a minimal basis composed of the \(1s\) orbitals of the hydrogen atoms:

\[ \Psi(\mathbf{r}_1, \mathbf{r}_2) = \exp(-(\mathbf{r}_1 - \mathbf{R}_A)) + \] where \(\mathbf{r}_1\) and \(\mathbf{r}_2\) denote the electron coordinates and \(\mathbf{R}_A\) and \(\mathbf{R}_B\) the coordinates of the nuclei.