--- title: "How much probability changes channel in one passage" description: "Define gap slopes and channel labels, calculate the Landau–Zener parameter, and test fast and slow limits with a worked example." canonical_url: "https://sandboxphysics.com/stories/nonadiabatic-dynamics/landau-zener-passage/" language: "en" page_type: "story" alternate_languages: zh-CN: "https://sandboxphysics.com/zh/stories/nonadiabatic-dynamics/landau-zener-passage/" last_modified: "2026-10-28" --- [Nonadiabatic molecular dynamics](https://sandboxphysics.com/stories/nonadiabatic-dynamics/) L2 · Chapter 3 # How much probability changes channel in one passage Sandbox Physics · 2026-10-28 Slow motion through a finite avoided crossing often lets the electrons follow an instantaneous eigenstate. Fast motion more readily preserves the original diabatic character. The Landau–Zener parameter compares the time available in the mixing region with the electronic response time. ## Before you begin Read chapter two and distinguish adiabatic and diabatic states. You need exponential functions and basic integration. Nuclear motion is prescribed as uniform passage here; reflection and wave-packet splitting are outside this local model. [Review the preceding chapter](https://sandboxphysics.com/stories/nonadiabatic-dynamics/moving-electronic-basis/) - Relate diabatic survival to an adiabatic transition. - Calculate a passage probability with an explicit slope convention. - Use the linear crossing as a limited check of a finite model. ## 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. $$ H(t)=\begin{pmatrix}\alpha t/2&\Delta\\\Delta&-\alpha t/2\end{pmatrix} $$ The slope is the time derivative of the full diabatic energy difference, twice the magnitude of either diagonal slope. 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. $$ \hbar^2\ddot a+\left(\Delta^2+\frac{\alpha^2t^2}{4}+\frac{i\hbar\alpha}{2}\right)a=0 $$ No further approximation enters this elimination; prescribed linear sweep and constant coupling were specified at the start. 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, $\alpha=|vF|$. $$ P_{\mathrm{diab}}=e^{-2\pi\Gamma},\qquad \Gamma=\frac{\Delta^2}{\hbar|vF|} $$ For lower-adiabatic incidence, this is the upper-adiabatic outgoing probability. The complementary probability exits on the lower surface. ## 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. $$ \exp\!\left[-\frac{2\pi(0.005)^2}{0.01\times0.032}\right]\simeq0.612 $$ These numbers illustrate the local linear model. Finite Tully packets require propagation with the full potential and their momentum spread. ## 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. ## Try the prediction ### [Compare slow and fast passage](https://sandboxphysics.com/labs/nonadiabatic-avoided-crossing/) 1. Hold Packet width and Electronic mixing fixed. Predict the upper-adiabatic outgoing trend as incoming momentum increases. 2. Change Incoming momentum and compare quantum, Ehrenfest and surface-hopping outcomes. 3. Decrease coupling and first interpret the trend through the decoupled limit, then inspect reflection and exit thresholds. ## Exercises and explanations 1. Why must the slope of either symmetric diabatic curve be doubled before entering this formula? Read the explanation The defined slope belongs to the energy difference. Opposite slopes add when one curve is subtracted from the other. 2. At fixed speed, how does doubling the minimum gap change the exponent? Read the explanation The gap is twice the coupling. Doubling it doubles coupling and quadruples the exponent’s magnitude, reducing diabatic survival. [Previous: What changes when the electronic basis moves](https://sandboxphysics.com/stories/nonadiabatic-dynamics/moving-electronic-basis/) [Next: Three dynamics methods for the same preparation](https://sandboxphysics.com/stories/nonadiabatic-dynamics/three-dynamics/) ## Sources and calculation scope 1. [John C. Tully, Molecular dynamics with electronic transitions, J. Chem. Phys. 93, 1061–1071 (1990)](https://doi.org/10.1063/1.459170) Published primary method paper. Sections II–III define fewest switches and its one-dimensional test models. Our three-method comparison uses the simple avoided-crossing Hamiltonian; the nuclear packet, finite windows and numerical settings are stated with each saved run. 2. [C. Zener, Non-adiabatic crossing of energy levels, Proc. R. Soc. Lond. A 137, 696–702 (1932)](https://doi.org/10.1098/rspa.1932.0165) Published original result for a linear-in-time diabatic gap and constant coupling. The course defines the slope as the slope of the full diabatic energy difference, avoiding a factor-of-two ambiguity.