The second-order energy shift · Quantum Mechanics

Intuition

When the first-order shift vanishes, or for more accuracy, go to second order. The second-order shift is a sum over every other level: the squared coupling divided by the gap. Levels above push a state down, levels below push it up — levels repel. The ground state has only levels above it, so its second-order shift is always downward. This is how an electric field lowers an atom’s energy in proportion to the field squared, its polarisability.

Two magnets with like poles facing push each other apart. Coupled quantum levels do the same: the closer they are and the stronger the coupling, the harder they push.

The second-order shift

Projecting the second-order equation on the unperturbed state and using the first-order correction gives a sum over all other levels.

E_{n}^{(2)}=\sum_{m\neq n}\frac{\bigl|\bra{m^{(0)}}\hat{V}\ket{n^{(0)}}\bigr|^{2}}{E_{n}^{(0)}-E_{m}^{(0)}}

Properties

For the ground state every denominator is negative, so $E_{0}^{(2)}\le0$ : second order always lowers the ground state.
Each pair of coupled levels pushes apart: the one above is raised, the one below lowered, by the same amount $|V_{mn}|^{2}/|E_{m}-E_{n}|$ .

The second-order energy

\begin{aligned}&E_{n}^{(2)}=\sum_{m\neq n}\frac{|V_{mn}|^{2}}{E_{n}^{(0)}-E_{m}^{(0)}},\\ &V_{mn}=\bra{m^{(0)}}\hat{V}\ket{n^{(0)}}\end{aligned}

At second order the same projection trick removes the second-order state correction and leaves the average of the perturbation between the zeroth- and first-order states. Inserting the first-order admixtures gives each squared coupling over its gap.

Proof steps

(\hat{H}_{0}-E_{n}^{(0)})\ket{n^{(2)}}=(E_{n}^{(1)}-\hat{V})\ket{n^{(1)}}+E_{n}^{(2)}\ket{n^{(0)}}

The terms with $\lambda^{2}$ .

0=E_{n}^{(1)}\braket{n^{(0)}|n^{(1)}}-\bra{n^{(0)}}\hat{V}\ket{n^{(1)}}+E_{n}^{(2)}

Project on $\bra{n^{(0)}}$ : the left side vanishes as before.

E_{n}^{(2)}=\bra{n^{(0)}}\hat{V}\ket{n^{(1)}}

With $\braket{n^{(0)}|n^{(1)}}=0$ .

E_{n}^{(2)}=\sum_{m\neq n}\bra{n^{(0)}}\hat{V}\ket{m^{(0)}}\frac{\bra{m^{(0)}}\hat{V}\ket{n^{(0)}}}{E_{n}^{(0)}-E_{m}^{(0)}}

Insert the first-order admixtures.

E_{n}^{(2)}=\sum_{m\neq n}\frac{|V_{mn}|^{2}}{E_{n}^{(0)}-E_{m}^{(0)}}

$V_{nm}V_{mn}=|V_{mn}|^{2}$ for a Hermitian perturbation.

Applications

Practice

Squared Coupling Over Gap

The second-order shift adds up, over every other level, the squared matrix element divided by the energy difference.

E_{n}^{(2)}=\sum_{m\neq n}\frac{|V_{mn}|^{2}}{E_{n}^{(0)}-E_{m}^{(0)}}

Try it

With $\hbar\omega=1$ and $x_{0}=1$ , the perturbation $\lambda\hat{x}$ has $\bra{1}\hat{x}\ket{0}=1/\sqrt2$ . What is the second-order shift of the ground state, in units of $\lambda^{2}$ ?