= Exact variational solution in a two-level system
{title2=$E(\theta)=2\sin^2\theta$}
In a <two-level system>, let the <Hamiltonian operator> have <matrix> $H=\operatorname{diag}(0,2)$ in the <orthonormal basis> $|0\rangle,|1\rangle$. The normalized real trial <quantum states> and their <Rayleigh quotient> are
$$
|\psi_\theta\rangle=\cos\theta\,|0\rangle+\sin\theta\,|1\rangle,\qquad E(\theta)=\langle\psi_\theta|H|\psi_\theta\rangle=2\sin^2\theta.
$$
The excited-state <probability amplitude> is $\sin\theta$, so its <probability> is $\sin^2\theta$. Thus $E(\theta)\geq0=E_0$, with equality if and only if the excited-state <probability amplitude> vanishes: $\theta\in\pi\mathbb Z$. The remaining <quantum state> is $|0\rangle$ up to a <global phase>. At $\theta=\pi/4$ the trial <energy> is $1$; at $\theta=\pi/2$ it is $2$. These are upper bounds on the <ground-state energy>, even though neither is a minimizing trial.
\Image[/two-level-variational-energy.png]
{title=Two-level variational energy is twice the excited-level probability and reaches zero only when the excited amplitude vanishes}
The real trial family contains the <ground state>, so it attains the exact minimum despite not covering every complex <quantum state>. More generally, a family covering all normalized <vectors> in a two-dimensional <Hilbert space> gives the exact lowest <eigenvalue> of any <Hermitian matrix> by minimizing its <Rayleigh quotient>. For a complex trial $c_0|0\rangle+c_1|1\rangle$ with $|c_0|^2+|c_1|^2=1$, the same calculation gives $E=2|c_1|^2$: a relative <quantum phase> cannot change this diagonal <Hamiltonian> energy. This finite example is the <weighted mean> proof of the <variational method> with just two <energy eigenstates>.
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