Exact variational solution in a two-level system ($E(\theta )=2{\sin }^2\theta$)

In a two-level system, let the Hamiltonian operator have matrix $H=\mathrm{diag}(0,2)$ in the orthonormal basis $|0⟩,|1⟩$. The normalized real trial quantum states and their Rayleigh quotient areThe excited-state probability amplitude is $\sin \theta$, so its probability is ${\sin }^2\theta$. Thus $E(\theta )\ge 0=E_0$, with equality if and only if the excited-state probability amplitude vanishes: $\theta \in \pi \mathbb{Z}$. The remaining quantum state is $|0⟩$ 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.

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 $c_0|0⟩+c_1|1⟩$ 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.

The eigenvalues of $\left(\begin{array}{cc}{E}_1& h\\ h& E_2\end{array}\right)$ are their mean plus or minus $\sqrt{(E_2-E_1{)}^2/4+h^2}$, so nonzero coupling increases their separation.