Second-Order Energy Corrections

Going to second order

The first-order energy used only the unperturbed states. The second-order correction is the first place the state correction feeds back into the energy. Return to the order-by-order expansion and collect the $\lambda^2$ terms of the eigenvalue equation:

Project onto $\langle n^{(0)}|$. As before the $\hat{H}^{(0)}$ term cancels, and with the normalization choice $\langle n^{(0)}|n^{(1)}\rangle = 0$ the $E_n^{(1)}$ term drops too. What survives is

Now substitute the explicit first-order state from the previous lesson, $|n^{(1)}\rangle = \sum_{m\neq n} \frac{\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle}{E_n^{(0)}-E_m^{(0)}}|m^{(0)}\rangle$. Using $\langle n^{(0)}|\hat{H}'|m^{(0)}\rangle = \langle m^{(0)}|\hat{H}'|n^{(0)}\rangle^*$ (Hermiticity), the numerator becomes a squared modulus:

What the formula tells you

Two features are worth memorizing:

1. The numerator is always non-negative because it is $|\cdot|^2$. The sign of each term is therefore set entirely by the energy denominator.
2. The ground state always shifts down. For the lowest level, every other state has higher energy, so $E_0^{(0)} - E_m^{(0)} < 0$ for all $m$, making every term negative. Second-order perturbation theory lowers the ground-state energy — a manifestation of level repulsion: nearby states "push apart," and the lower one is pushed further down.

The single-term case

For a two-level system there is only one other state, so the sum collapses to one term. With $\hat{H}^{(0)} = \mathrm{diag}(E_0^{(0)}, E_1^{(0)})$ and real coupling $V = \langle 1^{(0)}|\hat{H}'|0^{(0)}\rangle$, the ground state's second-order shift is

If $E_0^{(0)} = 1$, $E_1^{(0)} = 4$, and $V = 0.6$, then $E_0^{(2)} = 0.36 / (1 - 4) = -0.12$ — negative, as expected for a ground state.

Try it

Compute the second-order ground-state shift $E_0^{(2)}$ for $E_0^{(0)} = 1$, $E_1^{(0)} = 4$, and real coupling $V = 0.6$, and return it as a number.

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