Hartree–Fock Initial States

Every variational algorithm needs a starting state, and for molecules the natural choice is the Hartree–Fock (HF) reference. It is the best single-configuration description of the molecule and costs only a layer of $X$ gates to prepare — making it the universal initialization for VQE.

What Hartree–Fock is

Hartree–Fock approximates the many-electron wavefunction by a single Slater determinant: each electron occupies one spin-orbital, and the orbitals are optimized (self-consistently) to minimize the energy under the constraint that the wavefunction stays a single determinant. The HF procedure yields a set of ordered spin-orbital energies; the determinant fills the lowest $N$ of them with the $N$ electrons. This is the "Aufbau" filling familiar from introductory chemistry, applied to the self-consistent molecular orbitals.

The HF state as an occupation string

A single Slater determinant is exactly one occupation-number state. For $\text{H}_2$ in the STO-3G basis there are 4 spin-orbitals and 2 electrons, so the HF determinant fills orbitals 0 and 1:

|\Psi_{\text{HF}}\rangle = a_1^\dagger a_0^\dagger |\text{vac}\rangle = |1\,1\,0\,0\rangle.

Under Jordan–Wigner this maps to the qubit state with qubits 0 and 1 in $|1\rangle$ and qubits 2, 3 in $|0\rangle$ . Preparing it requires only an $X$ gate on each occupied orbital — no entanglement, no parameters.

Why HF is the right starting point

Hartree–Fock already recovers the bulk of the molecular energy (typically over 99%). The remaining piece is the correlation energy, which the off-diagonal terms of the Hamiltonian generate by mixing in other configurations such as the doubly-excited $|0\,0\,1\,1\rangle$ . A good ansatz therefore starts at $|\Psi_{\text{HF}}\rangle$ and applies parametrized rotations that add a small amount of those excited configurations — searching the neighbourhood of HF rather than all of Hilbert space.

In the two-qubit encoding

After tapering $\text{H}_2$ to two qubits (previous lesson), the HF determinant becomes the basis state $|01\rangle$ . The single-parameter ansatz you will use for VQE rotates this toward $\cos\tfrac{\theta}{2}|01\rangle - \sin\tfrac{\theta}{2}|10\rangle$ , sweeping between the HF configuration and its double excitation — exactly the two configurations the off-diagonal $X_0X_1$ / $Y_0Y_1$ terms couple.

Try it

Prepare the 4-qubit Hartree–Fock reference state for $\text{H}_2$ : 2 electrons in the 2 lowest spin-orbitals, i.e. qubits 0 and 1 in $|1\rangle$ and qubits 2, 3 in $|0\rangle$ . The grader checks the full statevector.

Run your code to see the quantum state.

The statevector should have a single non-zero amplitude at the occupation string with qubits 0 and 1 high — the Hartree–Fock determinant for molecular hydrogen.