The first-order energy shift is the simplest formula in the subject: the average of the perturbation in the unperturbed state. It needs nothing but the state you already have. A particle in a box whose floor is raised on one half has its every level lifted by half the step, because every density in a box spends half its time in each half. An oscillator with a small quartic term has its ground level raised by the quartic’s average.

Raise the floor of half a room by a centimetre and the average height of someone walking about the whole room rises by half a centimetre. To first order, that is all a small perturbation does.

A box whose floor is raised by $V_{0}$ on its left half, and the density of its ground state drawn above it. The density is symmetric, so exactly half of it lies over the raised half: to first order every level rises by $V_{0}/2$.

Take the first-order equation and project it on the unperturbed state. The unknown state correction drops out, leaving the shift.

• A constant perturbation $V_{0}$ shifts every level by $V_{0}$, exactly.
• A box with its floor raised by $V_{0}$ on one half: every level shifts by to first order.

Take the inner product of the first-order equation with the unperturbed state. Because $\hat{H}_{0}$ is Hermitian, its left side vanishes, and what remains is the first-order energy against the average of the perturbation.

The first-order equation.

Take the inner product with $\bra{n^{(0)}}$, which is normalised.

$\hat{H}_{0}$ is Hermitian and $\ket{n^{(0)}}$ is its eigenstate.

The left side vanishes.

To first order, a level shifts by the average of the perturbation in the unperturbed state.

In the oscillator ground state the average of $x$ to the fourth is three quarters of $x_{0}$ to the fourth, so a small $\epsilon x^{4}$ raises the ground level by three quarters of $$.

In a problem symmetric about the origin, a perturbation odd in $x$ has zero average in every state of definite parity, so its first-order shift is zero.

The first-order energy uses only the diagonal matrix element of the perturbation in the state itself. Couplings to other states matter from second order on.

Great work! You now know how to:

• derive the first-order energy shift
• compute it for boxes, oscillators and constant or odd perturbations
• use symmetry to see when it vanishes