First-Order State Corrections

How the state shifts

In the previous lesson we projected the first-order equation onto $\langle n^{(0)}|$ to extract the energy. To find the state correction $|n^{(1)}\rangle$ we project onto a different unperturbed state $\langle m^{(0)}|$ with $m \neq n$ . Starting from

\hat{H}^{(0)}|n^{(1)}\rangle + \hat{H}'|n^{(0)}\rangle = E_n^{(0)}|n^{(1)}\rangle + E_n^{(1)}|n^{(0)}\rangle,

and using $\langle m^{(0)}|\hat{H}^{(0)} = E_m^{(0)}\langle m^{(0)}|$ together with $\langle m^{(0)}|n^{(0)}\rangle = 0$ , the $E_n^{(1)}$ term drops out and we get

\big(E_m^{(0)} - E_n^{(0)}\big)\langle m^{(0)}|n^{(1)}\rangle + \langle m^{(0)}|\hat{H}'|n^{(0)}\rangle = 0.

Solving for the overlap of $|n^{(1)}\rangle$ with each $|m^{(0)}\rangle$ gives the coefficients. We choose the normalization $\langle n^{(0)}|n^{(1)}\rangle = 0$ (the correction has no component along the original state — that would only rescale it). Expanding $|n^{(1)}\rangle$ in the unperturbed basis:

\boxed{\,|n^{(1)}\rangle = \sum_{m \neq n} \frac{\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle}{E_n^{(0)} - E_m^{(0)}}\,|m^{(0)}\rangle\,.}

Reading the formula

Each unperturbed state $|m^{(0)}\rangle$ mixes into $|n\rangle$ by an amount set by two factors:

the coupling $\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle$ — how strongly the perturbation connects the two states, and
the inverse energy gap $1/(E_n^{(0)} - E_m^{(0)})$ — nearby levels mix much more than distant ones.

This is why a vanishing gap (degeneracy) is dangerous: the coefficient diverges, and ordinary perturbation theory fails. It also explains the convergence condition from the first lesson — the ratio of coupling to gap must be small for the correction to be a genuine small admixture.

A two-level example

Take $\hat{H}^{(0)} = \mathrm{diag}(E_0^{(0)}, E_1^{(0)})$ with $E_0^{(0)} = 1$ , $E_1^{(0)} = 4$ , and a perturbation with off-diagonal coupling $\langle 1^{(0)}|\hat{H}'|0^{(0)}\rangle = 0.6$ . The only state that can mix into $|0\rangle$ is $|1\rangle$ , with coefficient

c_1 = \frac{\langle 1^{(0)}|\hat{H}'|0^{(0)}\rangle}{E_0^{(0)} - E_1^{(0)}} = \frac{0.6}{1 - 4} = \frac{0.6}{-3} = -0.2,

so $|0^{(1)}\rangle = -0.2\,|1\rangle$ and the corrected state is approximately $|0\rangle - 0.2\,|1\rangle$ (before renormalization).

Try it

Using the example above ( $E_0^{(0)} = 1$ , $E_1^{(0)} = 4$ , coupling $\langle 1^{(0)}|\hat{H}'|0^{(0)}\rangle = 0.6$ ), compute the mixing coefficient $c_1$ of $|1\rangle$ in the first-order correction to the ground state, and return it as a number.

Run your code to see the quantum state.