Two levels, exactly and approximately · Quantum Mechanics

Intuition

Two coupled levels can be solved exactly, which makes them the best place to see what perturbation theory gets right and where it breaks. When the coupling is small against the gap, the exact energies agree with the second-order ones; when the gap is zero the exact answer is the degenerate one, split by twice the coupling. In between, the exact formula interpolates smoothly, and the perturbation series — whose radius of convergence is set by the ratio of coupling to half the gap — gives out.

Two tuning forks of nearly the same pitch, joined by a spring, sound two new notes. With a weak spring and different pitches the notes barely move; with identical pitches any spring at all splits them evenly. Two coupled levels behave the same way.

Two levels at 0 and 1 coupled by $w$ : the exact energies $\tfrac12\pm\sqrt{\tfrac14+w^{2}}$ (solid) and the second-order estimates $1+w^{2}$ and $-w^{2}$ (dashed). They agree while $w$ is small against the gap and part company once $w$ approaches half of it.

The exactly solvable two-level problem

For two levels $E_{1},E_{2}$ coupled by a real $w$ , the eigenvalues follow from a quadratic. Expanding them for small or large coupling recovers the two kinds of perturbation theory.

\begin{aligned}\hat{H}&=\begin{pmatrix}E_{1}&w\\ w&E_{2}\end{pmatrix}\\ E_{\pm}&=\frac{E_{1}+E_{2}}{2}\pm\sqrt{\Bigl(\frac{E_{1}-E_{2}}{2}\Bigr)^{2}+w^{2}}\end{aligned}

The two limits

$|w|\ll|E_{1}-E_{2}|$ : , the second-order shifts — the levels repel.

Expanding the exact answer

\begin{aligned}E_{\pm}&=\bar E\pm\sqrt{\Delta^{2}+w^{2}}\\ &\approx\bar E\pm\Bigl(\Delta+\frac{w^{2}}{2\Delta}\Bigr)\quad(|w|\ll\Delta)\end{aligned}

Write $\bar E$ for the average and $\Delta$ for half the gap. Expanding the square root for small $w$ gives the second-order shift; setting $\Delta=0$ gives the degenerate splitting at once.

Proof steps

\bar E=\frac{E_{1}+E_{2}}{2},\qquad \Delta=\frac{E_{1}-E_{2}}{2}>0

Average and half-gap.

E_{\pm}=\bar E\pm\sqrt{\Delta^{2}+w^{2}}

The eigenvalues of the $2\times2$ matrix.

\sqrt{\Delta^{2}+w^{2}}=\Delta\sqrt{1+\frac{w^{2}}{\Delta^{2}}}\approx\Delta+\frac{w^{2}}{2\Delta}

For $|w|\ll\Delta$ .

E_{+}\approx E_{1}+\frac{w^{2}}{E_{1}-E_{2}},\qquad E_{-}\approx E_{2}-\frac{w^{2}}{E_{1}-E_{2}}

Exactly the second-order shifts, since $2\Delta=E_{1}-E_{2}$ .

\Delta=0:\ \ E_{\pm}=\bar E\pm|w|

The degenerate answer: a splitting linear in the coupling.

Applications

Practice

The Exact Energies

Two levels coupled by $w$ have exact energies at their average plus or minus $\sqrt{\Delta^{2}+w^{2}}$ , with half the gap.

E_{\pm}=\bar E\pm\sqrt{\Delta^{2}+w^{2}}

Try it

Levels at 0 and 1 are coupled by $w=0.5$ . What is the exact lower energy? Give three decimal places.