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Variational Methods | Physics - Wyatt's Notes

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

Etrial=ψtrialH^ψtrialE0E_{\text{trial} = \langle\psi_{\text{trial}|\hat{H}|\psi_{\text{trial}\rangle \geq E_0}}}

The equality holds if and only if ψtrial=ψ0\psi_{\text{trial} = \psi_0}.

10.2 The Hydrogen Molecule Ion H2+H_2^+

Section titled “10.2 The Hydrogen Molecule Ion H2+H_2^+H2+​”

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

H^=22me2e24πε0rAe24πε0rB+e24πε0R\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: ψ±=N±[ψ1s(rA)±ψ1s(rB)]\psi_\pm = N_\pm[\psi_{1s}(\mathbf{r}_A) \pm \psi_{1s}(\mathbf{r}_B)]

The energies:

E±(R)=E1s+e24πε0R+J±K1±SE_\pm(R) = E_{1s} + \frac{e^2}{4\pi\varepsilon_0 R} + \frac{J \pm K}{1 \pm S}

Where S=ψAψBS = \langle\psi_A|\psi_B\rangle is the overlap integral, JJ is the Coulomb integral, and KK is the exchange integral.

  • EE_- (bonding): has a minimum at R2.5a0R \approx 2.5\,a_0Giving a binding energy of 1.8\sim 1.8 eV (experiment: 2.8 eV).
  • E+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): Esinglet=2E1s+e2R+2J+2K1+S2E_{\text{singlet} = 2E_{1s} + \frac{e^2}{R} + \frac{2J + 2K}{1 + S^2}}

Antibonding (triplet): Etriplet=2E1s+e2R+2J2K1S2E_{\text{triplet} = 2E_{1s} + \frac{e^2}{R} + \frac{2J - 2K}{1 - S^2}}

The equilibrium bond length is Re1.4a0R_e \approx 1.4\,a_0 with binding energy 3.5\sim 3.5 eV (experiment: 4.75 eV).

MethodTypeGuaranteeAccuracy
VariationalGround stateUpper bound EtrialE0E_{\text{trial}} \geq E_0Depends on trial function
Perturbation theoryAny stateNo bound (asymptotic series)Depends on smallness
WKBSemiclassicalLeading order in \hbarGood for large nn
Hartree-FockMany-bodyVariational (single determinant)99% of energy
  • Using a non-normalised trial function without adjusting. The variational principle requires ψψ=1\langle\psi|\psi\rangle = 1. If the trial function is not normalised, use E=ψH^ψ/ψψE = \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.
SystemTrial functionVariational parameterEnergy (theory)Energy (exact)
He atomeZeffr1/a0eZeffr2/a0e^{-Z_{\text{eff}} r_1/a_0} e^{-Z_{\text{eff}} r_2/a_0}ZeffZ_{\text{eff}}77.5-77.5 eV79.0-79.0 eV
H2+H_2^+LCAO ψ1s(rA)±ψ1s(rB)\psi_{1s}(r_A) \pm \psi_{1s}(r_B)RR1.8-1.8 eV binding2.8-2.8 eV
H2H_2Singlet Heitler-LondonRR3.5-3.5 eV binding4.75-4.75 eV
  • 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).
Worked Example 10.1: Variational Estimate for Helium Ground State

Use the trial function ψtrial=(Zeff3/πa03)exp(Zeffr1/a0)exp(Zeffr2/a0)\psi_{\text{trial} = (Z_{\text{eff}^3/\pi a_0^3)\exp(-Z_{\text{eff}r_1/a_0)\exp(-Z_{\text{eff}r_2/a_0)}}}} where ZeffZ_{\text{eff}} is a variational parameter.

The energy expectation value (treating the electron-electron repulsion as a perturbation):

E(Z_{\text{eff}) = 2\times\frac{Z_{\text{eff}^2}}{2}\text{Ry} - 2\times\frac{Z_{\text{eff} Z}{1}\text{Ry} + \frac{5}{8}Z_{\text{eff}\text{Ry}}}}}

= \left(Z_{\text{eff}^2 - 4Z_{\text{eff} + \frac{5}{4}Z_{\text{eff}\right)\text{Ry} = \left(Z_{\text{eff}^2 - \frac{11}{4}Z_{\text{eff}\right)\text{Ry}}}}}}

Minimising: E/Zeff=(2Zeff11/4)=0    Zeff=11/8=1.375\partial E/\partial Z_{\text{eff} = (2Z_{\text{eff} - 11/4) = 0 \implies Z_{\text{eff} = 11/8 = 1.375}}}.

E=(1216412132)Ry=12164Ry=2.848Ry=77.5 eVE = \left(\frac{121}{64} - \frac{121}{32}\right)\text{Ry} = -\frac{121}{64}\text{Ry} = -2.848\text{Ry} = -77.5\ \text{eV}

The exact (non-relativistic) ground state energy is 79.0-79.0 eV, so the variational result is within 2%.

The effective charge Zeff=1.375<2Z_{\text{eff} = 1.375 < 2} reflects the screening of the nuclear charge by the other electron: each electron partially shields the nucleus from the other, reducing the effective charge from Z=2Z = 2 to Zeff1.375Z_{\text{eff} \approx 1.375}.

10.8 Worked Example: Variational Estimate for Particle in a Finite Well

Section titled “10.8 Worked Example: Variational Estimate for Particle in a Finite Well”

Problem. Estimate the ground state energy of a particle in a 1D finite square well V(x)=V0V(x) = -V_0 for x<a|x| < a, V(x)=0V(x) = 0 for xa|x| \geq a, using the Gaussian trial function ψ(x)=(β/π)1/4eβx2/2\psi(x) = (\beta/\pi)^{1/4} e^{-\beta x^2/2}.

Solution. The Hamiltonian is H^=22md2dx2+V(x)\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x). The kinetic energy: T=2β4m\langle T\rangle = \frac{\hbar^2\beta}{4m}. The potential energy: V=V0πβaβaeu2du=V0erf(βa)\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(β)=2β4mV0erf(βa)E(\beta) = \frac{\hbar^2\beta}{4m} - V_0\,\text{erf}(\sqrt{\beta}a). Minimising numerically for typical parameters (V0=10V_0 = 10 eV, a=2a = 2 Å, m=mem = m_e) gives βopt0.8\beta_{\text{opt}} \approx 0.8 Å2^{-2} and E6.2E \approx -6.2 eV, compared to the exact value 7.0-7.0 eV. The Gaussian trial function cannot capture the exponential decay outside the well, leading to a 12% error. \blacksquare

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The variational principle states that any trial wavefunction gives an energy estimate that is always greater than or equal to the true ground state energy. This transforms finding the ground state into an optimization problem: minimize the energy over a family of trial functions. The better the trial function, the closer the estimate. This method is exact in the limit of a complete basis set, but even simple trial functions give remarkably accurate results. It is the foundation of quantum chemistry calculations, where clever choices of trial wavefunctions and basis sets balance accuracy against computational cost.

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

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