Gaussian variational energy for the harmonic oscillator ($E(\sigma )=1/(8{\sigma }^2)+{\sigma }^2/2$)

Use the quantum harmonic oscillator in units with reduced Planck constant $\hslash =1$, mass $m=1$, and angular frequency $\omega =1$. Its Hamiltonian operator and trial Gaussian wave packet areThis is the same Gaussian wave packet used for the Heisenberg uncertainty relation: $|{\psi }_{\sigma }{|}^2$ is a normal distribution with mean zero and variance ${\sigma }^2$. The Gaussian integral gives wavefunction normalization and $⟨x^2⟩={\sigma }^2$. These smooth, rapidly decaying trials lie in the one-dimensional harmonic oscillator form domain and even in the operator domain.

Since ${\psi }_{\sigma }^{\prime }=-x{\psi }_{\sigma }/(2{\sigma }^2)$, integration by parts with vanishing boundary terms gives the kinetic energy and potential energy:Narrow trials cost kinetic energy, while broad trials cost potential energy. The exact differenceshows that the unique optimal width is ${\sigma }^2=1/2$, with $E=1/2$. At $\sigma =1/2$ and $\sigma =1$, the trial energies are both $5/8=0.625$; the kinetic energy and potential energy contributions exchange places.

Minimizing within this trial family alone establishes $E_0\le 1/2$. To establish exactness, use the Heisenberg uncertainty relation to obtain $E_{\psi }\ge 1/2$ for every admissible quantum state, as in its existing oscillator example. Together the bounds give $E_0=1/2$. Direct substitution also yields $H{\psi }_{1/\sqrt{2}}={\psi }_{1/\sqrt{2}}/2$, agreeing with the Gaussian eigenstate for a quadratic potential. This is a useful distinction: the variational method supplies the upper bound, while the Heisenberg uncertainty relation supplies the complementary lower bound.