Checkpoint: Perturb a System

Bring it together

This checkpoint combines everything from the module on a single concrete system. You will compute the ground-state energy of a two-level Hamiltonian to second order in perturbation theory and compare against the exact answer.

The system

The unperturbed Hamiltonian is diagonal,

\hat{H}^{(0)} = \begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix},

with ground state $|0\rangle$ at $E_0^{(0)} = 2$ and excited state $|1\rangle$ at $E_1^{(0)} = 5$ . The perturbation, in the same basis, is

\hat{H}' = \begin{pmatrix} 0.4 & 0.3 \\ 0.3 & -0.2 \end{pmatrix}.

The recipe, step by step

First order — the diagonal expectation value:

E_0^{(1)} = \langle 0|\hat{H}'|0\rangle = 0.4.

Second order — sum over the other state (here just $|1\rangle$ ):

E_0^{(2)} = \frac{|\langle 1|\hat{H}'|0\rangle|^2}{E_0^{(0)} - E_1^{(0)}} = \frac{(0.3)^2}{2 - 5} = \frac{0.09}{-3} = -0.03.

Adding everything, the perturbative ground-state energy is

E_0 \approx E_0^{(0)} + E_0^{(1)} + E_0^{(2)} = 2 + 0.4 - 0.03 = 2.37.

Checking against the exact answer

This $2\times 2$ problem can be diagonalized exactly. The full ground Hamiltonian is $\hat{H} = \begin{pmatrix} 2.4 & 0.3 \\ 0.3 & 4.8 \end{pmatrix}$ , whose lower eigenvalue is

E_0^{\text{exact}} = \frac{2.4 + 4.8}{2} - \sqrt{\left(\frac{4.8 - 2.4}{2}\right)^2 + 0.3^2} = 3.6 - \sqrt{1.2^2 + 0.09} \approx 2.3631.

Our second-order estimate $2.37$ sits within about $0.007$ of the exact $2.3631$ — the residual is the third-order-and-beyond contribution. The agreement confirms the perturbation is genuinely small here (coupling $0.3$ versus gap $\approx 2.6$ ), exactly the validity condition from earlier in the module.

Try it

For the system above, compute the ground-state energy to second order, $E_0 = E_0^{(0)} + E_0^{(1)} + E_0^{(2)}$ , and return it as a number.

Run your code to see the quantum state.