The first-order state correction

Intuition

The perturbation also changes the states: to first order, each state mixes in a little of every other state the perturbation connects it to. Each admixture is the connecting matrix element divided by the energy gap, so near levels mix strongly and far ones weakly. A uniform force on an oscillator shifts its ground-state Gaussian sideways; to first order that shift is a little of the first excited state added in.

A pendulum hanging in a gentle breeze leans slightly downwind. To first order the lean is the breeze’s push divided by the stiffness: the same shape as an admixture, a coupling divided by a gap.

The ground state of an oscillator with an added force, $\hat{V}=\lambda\hat{x}$ with $\lambda=0.5$ in units of $\hbar\omega$ and $x_{0}$ : exactly the Gaussian shifted by $-\lambda$ . First-order perturbation theory adds $-\frac{\lambda}{\sqrt2}\ket{1}$ to $\ket{0}$ , which reproduces the shift for small $\lambda$ .

Expand the first-order correction in the unperturbed states. Projecting the first-order equation on each other state gives its coefficient.

\ket{n^{(1)}}=\sum_{m\neq n}\frac{\bra{m^{(0)}}\hat{V}\ket{n^{(0)}}}{E_{n}^{(0)}-E_{m}^{(0)}}\ket{m^{(0)}}

Properties

Each admixture is a matrix element over an energy gap: nearby levels mix most.
It is small only if $|\bra{m^{(0)}}\hat{V}\ket{n^{(0)}}|\ll|E_{n}^{(0)}-E_{m}^{(0)}|$ for every .

The first-order admixtures

\braket{m^{(0)}|n^{(1)}}=\frac{\bra{m^{(0)}}\hat{V}\ket{n^{(0)}}}{E_{n}^{(0)}-E_{m}^{(0)}},\qquad m\neq n

Project the first-order equation on another unperturbed state. On the left, $\hat{H}_{0}$ acts on the bra and gives its energy; on the right, the term with the first-order energy vanishes by orthogonality. What remains is the coefficient times the energy gap, set equal to the matrix element.

Proof steps

(\hat{H}_{0}-E_{n}^{(0)})\ket{n^{(1)}}=(E_{n}^{(1)}-\hat{V})\ket{n^{(0)}}

The first-order equation.

\begin{aligned}&\bra{m^{(0)}}(\hat{H}_{0}-E_{n}^{(0)})\ket{n^{(1)}}\\ &=(E_{m}^{(0)}-E_{n}^{(0)})\braket{m^{(0)}|n^{(1)}}\end{aligned}

$\hat{H}_{0}$ acts to the left on its eigenbra.

\bra{m^{(0)}}(E_{n}^{(1)}-\hat{V})\ket{n^{(0)}}=-\bra{m^{(0)}}\hat{V}\ket{n^{(0)}}

$\braket{m^{(0)}|n^{(0)}}=0$ for $m\neq n$ .

\braket{m^{(0)}|n^{(1)}}=\frac{\bra{m^{(0)}}\hat{V}\ket{n^{(0)}}}{E_{n}^{(0)}-E_{m}^{(0)}}

Divide by the gap, which is not zero for a non-degenerate level.

Applications

Practice

Coupling Over Gap

To first order, a state picks up each other state with a coefficient equal to the connecting matrix element divided by the energy difference.

c_{m}=\frac{\bra{m^{(0)}}\hat{V}\ket{n^{(0)}}}{E_{n}^{(0)}-E_{m}^{(0)}}

Try it

A perturbation connects the ground state, $E=0$ , to a state at $E=4$ with matrix element $0.2$ . What is the admixture coefficient of that state in the ground state?