Landau–Zener transitions · Quantum Mechanics

Intuition

Now sweep the bias through an avoided crossing at a steady rate. A slow sweep follows the adiabatic level and ends in the other diabatic state; a fast one does not notice the coupling and stays in the diabatic state it started in, jumping the gap. Landau and Zener found the exact probability of the jump: $e^{-\pi\Delta^{2}/2\hbar v}$ , where $\Delta$ is the smallest gap and $v$ the rate at which the bias changes. It depends on one number, the square of the gap over the sweep rate. It is exact, and it comes from summing the whole Dyson series of the problem.

A car taking a bend in the road: slowly, it follows the bend; fast, it goes straight on. How fast is too fast depends on how sharp the bend is — for an avoided crossing, on the square of the smallest gap.

The probabilities of jumping the gap, $e^{-u}$ , and of following the adiabatic level, $1-e^{-u}$ , against $u = π Δ^{2} / 2 ℏ v$ . Fast sweeps, small , jump; slow sweeps, large , follow.

The Landau–Zener formula

For $\hat{H}(t)=\tfrac12(vt\,\sigma_{z}+\Delta\,\sigma_{x})$ , swept from to with the system starting in a diabatic state:

P_{\text{jump}}=\exp\Bigl(-\frac{\pi\Delta^{2}}{2\hbar v}\Bigr)

Properties

$P_{\text{jump}}$ is the probability of ending in the same diabatic state, crossing the gap; $1-P_{\text{jump}}$ is the probability of following the adiabatic level.

The Landau–Zener formula through the Dyson series

\begin{aligned}&\hat{H}=\begin{pmatrix}\alpha t&\Gamma\\ \Gamma&-\alpha t\end{pmatrix}\\ &\Longrightarrow\ W=\bigl|a(+\infty)\bigr|^{2}=e^{-\pi\Gamma^{2}/\hbar\alpha}\end{aligned}

Remove the diagonal by a change of phase, which leaves a Hamiltonian with only off-diagonal entries that turn in time. Its odd powers are off-diagonal and its even powers diagonal, so in the Dyson series only even orders keep the system in its first state. The $2n$ -th order is $(-\gamma)^{n}$ times a nested integral equal to $(\pi/2)^{n}/n!$ , shown here for and stated beyond it, and the series sums to an exponential.

Proof steps

\hat{H}_{\chi}=\begin{pmatrix}0&\Gamma e^{i\alpha t^{2}/\hbar}\\ \Gamma e^{-i\alpha t^{2}/\hbar}&0\end{pmatrix}

The change of phase $e^{\pm i\alpha t^{2}/2\hbar}$ on the two components removes the diagonal.

\frac{a(+\infty)}{a(-\infty)}=1-\gamma\,T_{2}+\gamma^{2}\,T_{4}-\dots,\qquad \gamma=\frac{\Gamma^{2}}{\hbar\alpha}

Odd powers of an off-diagonal matrix are off-diagonal: only even orders of the Dyson series keep $a$ .

T_{2n}=\frac{1}{n!}\Bigl(\frac{\pi}{2}\Bigr)^{n}

For $n=1$ , with times in units of $\sqrt{\hbar/\alpha}$ , the ordered integral is half of $∣$ , the two orderings being equal by symmetry; the general case is stated here.

\frac{a(+\infty)}{a(-\infty)}=\sum_{n}\frac{(-\gamma\pi/2)^{n}}{n!}=e^{-\pi\gamma/2}

The exponential series.

W=e^{-\pi\gamma}=\exp\Bigl(-\frac{\pi\Gamma^{2}}{\hbar\alpha}\Bigr)

Square the modulus.

Applications

Practice

The Formula

A linear sweep through an avoided crossing jumps the gap with probability $e^{-u}$ , where $u=\pi\Delta^{2}/2\hbar v$ .

P_{\text{jump}}=e^{-\pi\Delta^{2}/2\hbar v}

Try it

A sweep has $\pi\Delta^{2}/2\hbar v=1.5$ . What is the probability of jumping the gap? Give three decimal places.