Avoided crossings

Intuition

Change the bias $\epsilon$ slowly from large negative to large positive. Without coupling the two levels are straight lines $\pm\epsilon/2$ that cross at $\epsilon=0$ . With a coupling $\Delta$ they do not cross: the levels $\pm\tfrac12\sqrt{\epsilon^{2}+\Delta^{2}}$ bend away from each other and keep a gap of at least $\Delta$ , the smallest at $\epsilon=0$ . The eigenstates swap character across the gap: the lower level is $\ket{a}$ on one side and $\ket{b}$ on the other. For a system with one knob, coupled levels of the same symmetry avoid crossing each other; a true crossing needs the coupling to vanish, which symmetry can force or a second knob can tune.

Two roads approaching each other on a map, joined by a short bridge, do not cross but swap lanes: a car keeping to the lower road ends up on what began as the other road. Coupled levels meet the same way at an avoided crossing.

The two levels of $\tfrac12(\epsilon\sigma_{z}+\Delta\sigma_{x})$ against the bias , with . Without coupling they would be the dashed lines, crossing at ; with it they avoid each other, and the gap is never less than .

As the bias $\epsilon$ is varied at fixed coupling $\Delta$ :

\begin{aligned}E_{+}-E_{-}&=\sqrt{\epsilon^{2}+\Delta^{2}}\ \ge\ \Delta\\ &\text{with equality at}\ \epsilon=0\end{aligned}

Properties

Far from the crossing, $|\epsilon|\gg\Delta$ , the levels are close to $\pm\epsilon/2$ , pushed apart by about $\Delta^{2}/4|\epsilon|$ each.

The gap never closes

\Delta\neq0\ \Longrightarrow\ E_{+}(\epsilon)-E_{-}(\epsilon)\ge\Delta\ \ \text{for every}\ \epsilon

The gap is the length of the field, $\sqrt{\epsilon^{2}+\Delta^{2}}$ , which is smallest when the bias vanishes. Far away the square root is close to , with a small correction, and the angle that fixes the eigenstates turns through as the bias goes from one side to the other, so the eigenstates exchange characters.

Proof steps

E_{+}-E_{-}=\sqrt{\epsilon^{2}+\Delta^{2}}

From the energies of a two-level Hamiltonian.

\sqrt{\epsilon^{2}+\Delta^{2}}\ge\sqrt{\Delta^{2}}=|\Delta|

$\epsilon^{2}\ge0$ , with equality only at $\epsilon=0$ .

\tfrac12\sqrt{\epsilon^{2}+\Delta^{2}}\approx\tfrac12|\epsilon|+\frac{\Delta^{2}}{4|\epsilon|}\quad(|\epsilon|\gg\Delta)

Expand the square root.

\theta:\ \pi\to0\ \Longrightarrow\ \ket{E_{-}}:\ -\ket{a}\to\ket{b}

$\tan\theta=\Delta/\epsilon$ runs from $0^{-}$ to $0^{+}$ through , so runs from to .

Applications

Practice

The Smallest Gap

Coupled levels keep apart. Their gap is $\sqrt{\epsilon^{2}+\Delta^{2}}$ , smallest at , where it equals the coupling.

\min_{\epsilon}(E_{+}-E_{-})=\Delta

Try it

Two coupled levels have $\Delta=0.3$ in some unit of energy. What is their smallest gap as the bias is varied?