Checkpoint: Run a VQE

You have worked through every building block of the Variational Quantum Eigensolver: the variational principle, parameterized ansatz circuits, Pauli-string measurements, classical optimization, convergence, and initialization. This checkpoint asks you to put them together and prepare the ground state of a two-qubit Hamiltonian.

The Hamiltonian

The system is described by

$H = -X_0 \otimes X_1,$

where $X$ is the Pauli- $X$ operator. To find the ground state we need the lowest eigenvalue of this $4 \times 4$ matrix.

The operator $X_0 \otimes X_1$ swaps pairs of basis states: it maps $|00\rangle \leftrightarrow |11\rangle$ and $|01\rangle \leftrightarrow |10\rangle$ . All four Bell states are eigenstates of $X_0 \otimes X_1$ :

|{\Phi^+}\rangle = \tfrac{1}{\sqrt{2}}\bigl(|00\rangle + |11\rangle\bigr), \quad \text{eigenvalue } +1,

|{\Phi^-}\rangle = \tfrac{1}{\sqrt{2}}\bigl(|00\rangle - |11\rangle\bigr), \quad \text{eigenvalue } -1,

|{\Psi^+}\rangle = \tfrac{1}{\sqrt{2}}\bigl(|01\rangle + |10\rangle\bigr), \quad \text{eigenvalue } +1,

|{\Psi^-}\rangle = \tfrac{1}{\sqrt{2}}\bigl(|01\rangle - |10\rangle\bigr), \quad \text{eigenvalue } -1.

Multiplying by the minus sign in $H$ , eigenstates with $X_0 X_1$ eigenvalue $+1$ get $H$ eigenvalue $-1$ and those with eigenvalue $-1$ get $H$ eigenvalue $+1$ . The ground energy is $E_0 = -1$ , and it is doubly degenerate: both $|\Phi^+\rangle$ and $|\Psi^+\rangle$ are ground states.

$E_0 = -1, \quad \text{ground subspace} = \mathrm{span}\bigl\{|\Phi^+\rangle,\,|\Psi^+\rangle\bigr\}.$

You can confirm this satisfies the variational bound: any two-qubit state $|\psi\rangle$ obeys $\langle \psi | H | \psi \rangle \geq -1$ , with equality for any state in the ground subspace.

What VQE does (conceptually)

A real VQE run would:

Initialize a hardware-efficient ansatz, e.g.\ $R_Y(\theta_0)|0\rangle \otimes R_Y(\theta_1)|0\rangle$ followed by a CNOT.
Measure the energy $E(\theta_0, \theta_1) = \langle \psi(\boldsymbol{\theta}) | H | \psi(\boldsymbol{\theta}) \rangle$ by sampling in the $X$ basis (basis-change circuit: Hadamard before measurement).
Optimize $\boldsymbol{\theta}$ with a classical optimizer (COBYLA, gradient descent via the parameter-shift rule, etc.) until $E$ stops decreasing.
The final circuit at convergence prepares the ground state.

For this two-qubit problem, the optimizer converges to a circuit whose output is $|\Phi^+\rangle$ . That circuit is simply a Hadamard followed by a CNOT — the same Bell-state preparation you first built in module 4.

Why the Bell circuit reaches the target state

The specific ground state the exercise grader checks for is $|\Phi^+\rangle$ , which is maximally entangled: it cannot be written as a product $|\psi_0\rangle \otimes |\psi_1\rangle$ . Reaching $|\Phi^+\rangle$ in particular therefore requires an entangling gate.

It is worth noting that the ground energy $E_0 = -1$ is not unique to entangled states. For example, the product state $|{+}\rangle|{+}\rangle = H|0\rangle \otimes H|0\rangle$ satisfies $\langle {+}{+}|H|{+}{+}\rangle = -\langle X\rangle^2 = -1$ , so it also achieves energy $-1$ and lies in the ground subspace. The ground subspace is two-dimensional, spanned by $|\Phi^+\rangle$ and $|\Psi^+\rangle$ , and contains both entangled and product-state representatives.

A single CNOT layer is sufficient to reach $|\Phi^+\rangle$ exactly (the target of this checkpoint), which is why the Bell circuit $H_0 \cdot CNOT_{01}$ is the natural choice.

Energies of all basis states

To build intuition, compare the energy of the VQE result with the energy of a random guess. For any computational basis state $|xy\rangle$ the expectation value of $H = -XX$ is

$\langle xy | (-X_0 X_1) | xy \rangle = -\langle xy | x'y' \rangle = 0,$

where $x'y'$ is the bit-flipped state. Since $|xy\rangle \perp |x'y'\rangle$ in the computational basis, all basis states give zero energy — much worse than the ground state energy of $-1$ . Ground states achieve their energy precisely because of quantum superposition: the off-diagonal action of $X$ on a superposition produces a non-zero overlap with the original state, yielding a negative energy. The Bell state $|\Phi^+\rangle$ additionally relies on entanglement, but any state in the two-dimensional ground subspace will do.

Try it

The classical optimizer has converged and found that the optimal circuit is $H_0 \cdot CNOT_{01}$ . Build that circuit to prepare the ground state of $H = -X_0 X_1$ . The grader checks the full statevector, so a blank circuit or a computational basis state will not pass.

Run your code to see the quantum state.

After running, inspect the statevector: you should see amplitudes $\tfrac{1}{\sqrt{2}} \approx 0.707$ on $|00\rangle$ and $|11\rangle$ , and zero on $|01\rangle$ and $|10\rangle$ . This is $|\Phi^+\rangle$ , a ground state of the $XX$ Hamiltonian (ground energy $E_0 = -1$ ), achieved by the VQE at its convergence point.