Variational Methods

For any trial wavefunction $\psi_{\text{trial}}$ (normalised), the expectation value of the Hamiltonian is an upper bound on the true ground state energy:

$E_{\text{trial} = \langle\psi_{\text{trial}|\hat{H}|\psi_{\text{trial}\rangle \geq E_0}}}$

The equality holds if and only if $\psi_{\text{trial} = \psi_0}$.

The simplest molecule: one electron in the field of two protons separated by distance $R$. The Hamiltonian:

$\hat{H} = -\frac{\hbar^2}{2m_e}\nabla^2 - \frac{e^2}{4\pi\varepsilon_0 r_A} - \frac{e^2}{4\pi\varepsilon_0 r_B} + \frac{e^2}{4\pi\varepsilon_0 R}$

LCAO trial function: $\psi_\pm = N_\pm[\psi_{1s}(\mathbf{r}_A) \pm \psi_{1s}(\mathbf{r}_B)]$

The energies:

$E_\pm(R) = E_{1s} + \frac{e^2}{4\pi\varepsilon_0 R} + \frac{J \pm K}{1 \pm S}$

Where $S = \langle\psi_A|\psi_B\rangle$ is the overlap integral, $J$ is the Coulomb integral, and $K$ is the exchange integral.

• $E_-$ (bonding): has a minimum at $R \approx 2.5\,a_0$Giving a binding energy of $\sim 1.8$ eV (experiment: 2.8 eV).
• $E_+$ (antibonding): monotonically decreases, no bound state.

With two electrons, the full Hamiltonian includes the electron-electron repulsion. Using the variational method with properly (anti)symmetrised spatial-spin wavefunctions:

Bonding (singlet): $E_{\text{singlet} = 2E_{1s} + \frac{e^2}{R} + \frac{2J + 2K}{1 + S^2}}$

Antibonding (triplet): $E_{\text{triplet} = 2E_{1s} + \frac{e^2}{R} + \frac{2J - 2K}{1 - S^2}}$

The equilibrium bond length is $R_e \approx 1.4\,a_0$ with binding energy $\sim 3.5$ eV (experiment: 4.75 eV).

• Using a non-normalised trial function without adjusting. The variational principle requires $\langle\psi|\psi\rangle = 1$. If the trial function is not normalised, use $E = \langle\psi|\hat{H}|\psi\rangle / \langle\psi|\psi\rangle$ instead.
• Forgetting the variational principle only gives ground state bounds. For excited states, the trial function must be orthogonal to lower-energy states, which is hard to enforce.
• Choosing a trial function with the wrong symmetry. If the true ground state has a different parity or angular momentum than the trial function, the variational estimate can be very poor.
• Assuming more variational parameters always improves accuracy. Additional parameters can lead to overfitting, numerical instability, and may only marginally improve the energy while obscuring the physics.

• Quantum chemistry: The Hartree-Fock method is a variational approach where the trial function is a Slater determinant. Post-Hartree-Fock methods (CI, MP2, CC) systematically improve the variational bound.
• Condensed matter physics: The Hubbard model is studied variationally using Gutzwiller wavefunctions and density matrix renormalisation group (DMRG) methods.
• Nuclear physics: The nuclear shell model uses variational calculations with configurational mixing to predict nuclear spectra and binding energies.
• Machine learning: Variational autoencoders (VAEs) use the variational principle to approximate intractable posterior distributions by minimising the evidence lower bound (ELBO).

Problem. Estimate the ground state energy of a particle in a 1D finite square well $V(x) = -V_0$ for $|x| < a$, $V(x) = 0$ for $|x| \geq a$, using the Gaussian trial function $\psi(x) = (\beta/\pi)^{1/4} e^{-\beta x^2/2}$.

Solution. The Hamiltonian is $\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)$. The kinetic energy: $\langle T\rangle = \frac{\hbar^2\beta}{4m}$. The potential energy: $\langle V\rangle = -\frac{V_0}{\sqrt{\pi}}\int_{-\sqrt{\beta}a}^{\sqrt{\beta}a} e^{-u^2}\,du = -V_0\,\text{erf}(\sqrt{\beta}a)$.

Thus $E(\beta) = \frac{\hbar^2\beta}{4m} - V_0\,\text{erf}(\sqrt{\beta}a)$. Minimising numerically for typical parameters ($V_0 = 10$ eV, $a = 2$ Å, $m = m_e$) gives $\beta_{\text{opt}} \approx 0.8$ Å$^{-2}$ and $E \approx -6.2$ eV, compared to the exact value $-7.0$ eV. The Gaussian trial function cannot capture the exponential decay outside the well, leading to a 12% error. $\blacksquare$