How much probability changes channel in one passage

Specify which state survives

Let $t=0$ mark the crossing. The diabatic energy difference is $\alpha t$ with $\alpha>0$ in energy per time. Constant coupling $\Delta>0$ gives minimum adiabatic gap $2\Delta$. Remove the common energy to obtain the two-state Hamiltonian below.

An initially lower adiabatic state changes its diabatic character under perfect adiabatic following. Preserving the original diabatic character instead leads to the upper outgoing adiabatic state. The survival probability below refers to the latter process.

Reading: C. Zener · John C. Tully

Reduce the two amplitudes to one second-order equation

Call the diabatic electronic amplitudes $a(t)$ and $b(t)$. The first row gives $b=(i\hbar\dot a-\alpha t a/2)/\Delta$. Differentiate that row, use the second to eliminate $\dot b$, then substitute the expression for $b$. This produces the equation below.

After rescaling time, it is a parabolic-cylinder equation. Its asymptotic connection coefficients give the exact exponential in the next section: prepare one incoming diabatic state sufficiently far before the crossing and read both outgoing probabilities sufficiently far afterward. We derive the reduction here and take the special-function connection from Zener’s original result, with fast and slow limits checking the channel convention.

Reading: C. Zener · John C. Tully

A dimensionless ratio controls the passage

Strong mixing occurs over an interval of order $\Delta/\alpha$. The electronic response time is of order $\hbar/\Delta$. Their ratio gives $\Gamma=\Delta^2/(\hbar\alpha)$.

Solving the linear-sweep equation from sufficiently early to sufficiently late times gives the Landau–Zener result. It assumes constant coupling and one passage. For nuclear speed $v$ and spatial slope $F$ of the diabatic energy difference, $\alpha=|vF|$.

Worked example with twice the speed

In atomic units, take $\hbar=1$, $\Delta=0.005$, $F=0.032$ and $v=0.01$. The product of speed and slope is $0.00032$, giving $\Gamma=0.078125$. The exponent is about negative 0.491 and diabatic survival is approximately $0.612$.

Doubling the speed halves the exponent’s magnitude and gives approximately $0.782$. Alternatively, doubling coupling at the original speed quadruples that magnitude and gives approximately $0.140$. Predicting the direction before calculating helps detect interchanged channel labels.

A wave packet contains a momentum distribution

For a minimum-uncertainty Gaussian with position standard deviation $\sigma_R$, momentum standard deviation is $\sigma_p=\hbar/(2\sigma_R)$. Different momentum components pass through at different speeds.

Substituting the central speed is a useful check when momentum spread is narrow. Averaging the formula over a classical momentum distribution still omits reflection, the finite coupling region and coherent return passages. These distinctions matter near an outgoing-channel threshold.

Check known limits before interpreting a discrepancy

At zero coupling the fixed diabatic channels decouple. At increasingly slow sweep with a finite nonzero gap, adiabatic following improves. A solver should recover each limit under its own assumptions.

For Tully’s finite crossing, identify the local slope and whether the chosen speed is incident or local. Nuclear motion changes speed on a surface. A difference from the linear formula can reflect model assumptions or numerical error; the limits help separate them.