Two-Level Perturbation

A qubit is a two-level system

Every two-level quantum system — a spin-1/2, a two-state molecule, a superconducting transmon in its lowest two levels — is mathematically a qubit. Its most general Hermitian Hamiltonian is a combination of the Pauli matrices,

\hat{H} = a\,\sigma_z + b\,\sigma_x + c\,\sigma_y + d\,\mathbb{1}.

The identity term $d\,\mathbb{1}$ only shifts all levels equally, so we drop it. Take the simplest nontrivial case, a "longitudinal" splitting plus a "transverse" coupling:

\hat{H} = a\,\sigma_z + b\,\sigma_x = \begin{pmatrix} a & b \\ b & -a \end{pmatrix}.

The $a\,\sigma_z$ part is our solvable $\hat{H}^{(0)}$ (eigenstates $|0\rangle, |1\rangle$ with energies $\pm a$ ); the $b\,\sigma_x$ part is the perturbation $\hat{H}'$ that mixes them.

Perturbative vs exact

Because $\sigma_x$ is purely off-diagonal, $\langle 0|\hat{H}'|0\rangle = 0$ , so there is no first-order energy shift. The leading correction is second order:

E_0^{(2)} = \frac{|\langle 1|b\sigma_x|0\rangle|^2}{E_0^{(0)} - E_1^{(0)}} = \frac{b^2}{a - (-a)} = \frac{b^2}{2a}.

Wait — for the ground state we want the lower level. With $a > 0$ the lower unperturbed level is $|1\rangle$ at $-a$ , and its second-order shift is $b^2/(-a - a) = -b^2/(2a)$ , pushing it down. This two-level model is special because we can also solve it exactly: the eigenvalues are

E_\pm = \pm\sqrt{a^2 + b^2}.

Expanding the lower one, $-\sqrt{a^2 + b^2} = -a\sqrt{1 + b^2/a^2} \approx -a - \tfrac{b^2}{2a}$ , exactly reproduces the perturbative result. The two-level system is the cleanest place to see perturbation theory agree with the truth, term by term.

The exact eigenstate as a rotation

The exact ground state is a rotation of $|0\rangle$ about the $y$ -axis of the Bloch sphere. Writing it as $\cos(\theta/2)|0\rangle + \sin(\theta/2)|1\rangle$ , the angle is fixed by the eigenvector of the lower eigenvalue. For $a = 1$ , $b = 0.5$ one finds $\theta \approx -2.678$ rad. In the small-perturbation limit $b \ll a$ , $\theta \to \pi$ (the ground state $\approx |1\rangle$ ), and the small admixture of $|0\rangle$ is exactly the first-order state correction.

Try it

Build a 1-qubit circuit that prepares the exact ground state of $\hat{H} = \sigma_z + 0.5\,\sigma_x$ . Use a single $RY(\theta)$ rotation with the angle determined by the lower-eigenvalue eigenvector ( $\theta \approx -2.678$ rad). The simulator will compare your prepared state to the reference.

Run your code to see the quantum state.