VQE for H₂ (Toy Model)

The Variational Quantum Eigensolver (VQE) finds the lowest eigenvalue of a Hamiltonian $H$ by minimizing the energy expectation value $E(\theta) = \langle \psi(\theta) | H | \psi(\theta) \rangle$ over the parameters $\theta$ of a trial circuit (the ansatz). In this lesson you apply the full idea to a two-qubit toy model of the hydrogen molecule.

The toy H₂ Hamiltonian

A real H₂ calculation maps molecular orbital integrals onto qubits via the Jordan-Wigner transformation. Even at the minimal STO-3G basis level the full mapped Hamiltonian has five Pauli terms. For this introductory lesson we strip everything down to the dominant spin-spin coupling term:

$H = -Z_0 \otimes Z_1.$

This operator is diagonal in the computational basis:

| State | $Z_0 Z_1$ eigenvalue | $H$ eigenvalue | |-------|---------------------|----------------| | $|00\rangle$ | $+1$ | $-1$ | | $|01\rangle$ | $-1$ | $+1$ | | $|10\rangle$ | $-1$ | $+1$ | | $|11\rangle$ | $+1$ | $-1$ |

The two lowest-energy states are $|00\rangle$ and $|11\rangle$ , both with $E_0 = -1$ .

Verifying the energy of $|\Phi^+\rangle$

The expectation value of $H = -Z_0 Z_1$ in the Bell state $|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ is

\langle \Phi^+ | H | \Phi^+ \rangle = -\langle \Phi^+ | Z_0 Z_1 | \Phi^+ \rangle.

\langle \Phi^+ | Z_0 Z_1 | \Phi^+ \rangle = 1 \quad\Longrightarrow\quad \langle \Phi^+ | H | \Phi^+ \rangle = -1 = E_0.

The variational principle guarantees $\langle H \rangle \geq E_0 = -1$ , and $|\Phi^+\rangle$ achieves this bound exactly.

The ansatz circuit

The general two-parameter ansatz for states in the $\{|00\rangle, |11\rangle\}$ subspace is

$|\psi(\theta, \phi)\rangle = \cos\!\tfrac{\theta}{2}|00\rangle + e^{i\phi}\sin\!\tfrac{\theta}{2}|11\rangle.$

Setting $\theta = \pi/2$ and $\phi = 0$ gives $|\Phi^+\rangle$ . The circuit that prepares this state from $|00\rangle$ requires exactly two gates:

Hadamard on qubit 0: puts qubit 0 into $\tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ , leaving the full register in $\tfrac{1}{\sqrt{2}}(|00\rangle + |10\rangle)$ .
CNOT with qubit 0 as control and qubit 1 as target: flips qubit 1 only when qubit 0 is $|1\rangle$ , entangling the pair to give $\tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ .

In a real VQE run, $\theta$ would be a free parameter tuned by the optimizer. Here the optimizer has already converged: the optimal angle is $\theta = \pi/2$ , which is exactly $H|0\rangle$ composed with a CNOT.

The hybrid VQE loop (conceptual)

A production VQE run on H₂ would iterate:

Quantum step: prepare $|\psi(\theta)\rangle$ with the ansatz; measure $\langle H \rangle$ by estimating each Pauli term's expectation value from shot statistics.
Classical step: feed $E(\theta)$ to a gradient-free optimizer (e.g., COBYLA or SPSA), which proposes a new $\theta$ .
Convergence: stop when $E(\theta)$ stops decreasing. The final value is the VQE estimate of the ground-state energy.

In this toy model, if one uses only a plain $R_Y(\theta)$ on qubit 0 without the CNOT (a product-state ansatz), the landscape is $E(\theta) = -\cos\theta$ , which has a single minimum at $\theta = 0$ ( $E = -1$ , state $|00\rangle$ ), so the optimizer finds it in a few steps.

Try it

Prepare the Bell state $|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ , which is the ground state of the toy $H = -Z_0 Z_1$ Hamiltonian. The grader checks the full statevector, so the two amplitudes must each equal $\tfrac{1}{\sqrt{2}} \approx 0.707$ .

Run your code to see the quantum state.

After running, inspect the amplitudes: only $|00\rangle$ and $|11\rangle$ are nonzero, each with coefficient $\tfrac{1}{\sqrt{2}}$ . The circuit has discovered the ground state using just a Hadamard and a CNOT — the simplest possible VQE ansatz for this Hamiltonian.