LESSON NOTES · 01

Variational Monte Carlo and local energy

Prerequisites: Variational principle; Metropolis–Hastings

Prerequisite lessons: Detailed balance and Metropolis–Hastings · Variational principle and finite bases

1. The bound applies to the exact expectation

Let \(H\) be self-adjoint and bounded below, with nonzero admissible trial \(\Psi_T\) in its domain. A spectral expansion gives a weighted mean of energies and a lower bound \(E_0\). In a symmetry sector use its lowest allowed energy. Electronic fermions require antisymmetry. A finite noisy sample mean can fall below the bound by chance without contradicting the theorem.

\[ E_V=\frac{\langle\Psi_T|H|\Psi_T\rangle}{\langle\Psi_T|\Psi_T\rangle}=\frac{\sum_n|c_n|^2E_n}{\sum_n|c_n|^2}\geq E_0. \]

2. Sample squared amplitude, average local energy

Divide the integrand by \(\Psi_T\) where it is nonzero and use its squared modulus as normalized sampling density. Sample by a suitable chain and use correlated errors. Ordinary nodes have zero probability measure, but nearby divergences can cause heavy tails; the ratio is not defined on a node.

\[ p_T(R)=\frac{|\Psi_T(R)|^2}{\int|\Psi_T|^2dR},\quad E_L(R)=\frac{H\Psi_T(R)}{\Psi_T(R)},\quad E_V=\int p_T E_L\,dR. \]

3. The zero-variance property has conditions

For an exact eigenstate satisfying the full operator domain and boundaries, local energy is constant almost everywhere and its variance is zero. Piecewise solutions with artificial nodal boundaries are not automatically full-Hamiltonian eigenstates. Zero variance alone does not establish the ground state.

Variational Monte Carlo and local energy — schematic under the stated model assumptions

Teaching schematic drawn from the equations or algorithm steps in this lesson; not measured data.

Worked Gaussian benchmark

In one dimension take \(\hbar=m=\omega=1\) and \(\Psi_a=e^{-ax^2/2}\), \(a>0\). Differentiate twice, divide by the trial function and average \(x^2\) under \(e^{-ax^2}\). The resulting energy is minimized at \(a=1\). At \(a=2\) it is \(5/8\), above \(1/2\) because the state is too narrow. These are analytic implementation benchmarks, not executed VMC results.

\[ \begin{aligned}H&=-\tfrac12\frac{d^2}{dx^2}+\tfrac12x^2,\quad\Psi_a^{\prime\prime}=(a^2x^2-a)\Psi_a,\\E_L(x)&=\tfrac a2+\tfrac{1-a^2}{2}x^2,\quad\langle x^2\rangle=\frac1{2a},\\E_V(a)&=\tfrac14(a+a^{-1})\geq\tfrac12,\\\operatorname{Var}(E_L)&=\frac{(1-a^2)^2}{8a^2}.\end{aligned} \]

Check yourself

Does low energy variance prove a ground state?

Answer and reasoning

No, an exact excited eigenstate also has zero variance.

Is \(E_L\) defined exactly on a node?

Answer and reasoning

No.

Can a sampled energy be below the exact ground energy?

Answer and reasoning

Statistical fluctuations can do this; report uncertainty and distinguish the sample mean from the exact variational expectation.

References and further reading

Derivations and numerical examples here are original teaching constructions, not copied passages or reported research data.

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