Part 2 · full derivation
Deriving the Landau–Zener Formula
Part 2 quoted the hopping probability and used it to argue that a conical intersection sends the wavepacket straight through. This page derives it. The exponent — including the factor of 2π — comes out of one Gaussian integral; the exponentiation needs the exact solution, and this page is explicit about where that line is.
Reduced units with ℏ = 1 at the end, but ℏ is kept visible through the derivation so the dimensions can be checked.
\[ P_{\text{diabatic}} = \exp\!\left(-\frac{2\pi V^{2}}{\hbar\,\alpha}\right), \qquad \alpha \equiv \left|\frac{d(E_1-E_2)}{dt}\right| \]
Step 1 — The model, and what “diabatic” means
Keep the two electronic states that matter near the crossing and let their energies sweep linearly through one another at rate α, with a constant coupling V:
\[ H(t) = \begin{pmatrix} \tfrac{1}{2}\alpha t & V \\ V & -\tfrac{1}{2}\alpha t \end{pmatrix} \]
The basis states here are diabatic: they keep their electronic character as the nuclei move, and their energies cross at t = 0. The adiabatic states are the instantaneous eigenvalues from Part 2,
\[ E_\pm(t) = \pm\sqrt{\left(\tfrac{1}{2}\alpha t\right)^{2} + V^{2}} \]
which never touch: they are separated by 2V at closest approach. Staying diabatic therefore means hopping between adiabatic surfaces — the two descriptions label the same event with opposite words, which is the single most common source of confusion in this subject.
Step 2 — First-order perturbation theory
Start entirely in state 1 at t = −∞ and treat V as a perturbation. Writing the state as c1|1〉 + c2|2〉 and removing the diagonal phases, the amplitude to find the system in state 2 is
\[ c_2 = -\frac{i}{\hbar}\int_{-\infty}^{\infty} V\,e^{\,i\varphi(t)}\,dt, \qquad \varphi(t) = \frac{1}{\hbar}\int_0^{t}\big(E_1-E_2\big)dt' \]
The whole point of the linear sweep is that this phase is exactly quadratic. With E1 − E2 = αt:
\[ \varphi(t) = \frac{\alpha t^{2}}{2\hbar} \qquad\Longrightarrow\qquad c_2 = -\frac{iV}{\hbar}\int_{-\infty}^{\infty} e^{\,i\alpha t^{2}/2\hbar}\,dt \]
Step 3 — One Gaussian integral, and the 2π appears
That is a Fresnel integral, the oscillatory cousin of a Gaussian:
\[ \int_{-\infty}^{\infty} e^{\,iax^{2}}dx = \sqrt{\frac{\pi}{a}};e^{\,i\pi/4} \qquad (a>0) \]
Here a = α/2ℏ, so the integral is √(2πℏ/α) eiπ/4 and
\[ |c_2|^{2} = \frac{V^{2}}{\hbar^{2}}\cdot\frac{2\pi\hbar}{\alpha} = \frac{2\pi V^{2}}{\hbar\,\alpha} \]
There is the exponent, coefficient and all. Note it is dimensionless as required: V²/ℏα has units of (energy)² / (energy × time × energy/time) = 1.
The physical content is a competition between two times. The coupling needs roughly ℏ/V to move population across; the crossing region, where the gap is smaller than V, lasts about V/α. Their ratio is V²/ℏα. Sweep quickly and nothing has time to transfer — the system carries straight on in its diabatic state.
Step 4 — Where first order breaks, and how it tells you
To first order the system stays diabatic with probability 1 − 2πV²/ℏα. That is fine while the coupling is weak and absurd when it is not — the expression goes negative. A probability cannot. The theory is announcing its own failure, and the shape of the failure says what to do: a quantity that should decay but is only known to first order is almost always an exponential seen through its first two terms.
| V | 1 − 2πV²/α | exp(−2πV²/α) | numerical solution |
|---|---|---|---|
| 0.02 | 0.9975 | 0.9975 | 0.9974 |
| 0.05 | 0.9843 | 0.9844 | 0.9841 |
| 0.10 | 0.9372 | 0.9391 | 0.9379 |
| 0.20 | 0.7487 | 0.7778 | 0.7748 |
| 0.40 | −0.0053 | 0.3659 | 0.3718 |
α = 1, ℏ = 1. The last row is the tell: first order returns a negative probability, while the exponential tracks the numerical solution to a few per cent. Both agree to four decimal places when the coupling is weak, which is what confirms the exponent was derived correctly rather than fitted.
Step 5 — The step this course quotes
The exponentiation is not a guess, but proving it needs the exact solution. Substituting the linear sweep into the two-state Schrödinger equation turns it into Weber's equation, whose solutions are parabolic cylinder functions; matching their asymptotic forms as t → ±∞ gives
\[ P_{\text{diabatic}} = \exp\!\left(-\frac{2\pi V^{2}}{\hbar\,\alpha}\right) \]
C. Zener, “Non-adiabatic crossing of energy levels”, Proc. R. Soc. Lond. A 137, 696 (1932); L. D. Landau, Phys. Z. Sowjetunion 2, 46 (1932). The exact result holds for all coupling strengths, and reduces to Step 3 when the exponent is small — which is the consistency check the table performs.
Back to retinal
In a molecule the sweep rate is set by how fast the nuclei move through the crossing, so α = v|ΔF| with v the speed along the reaction coordinate and ΔF the difference in slopes of the diabatic curves. That is the form quoted in Part 2:
\[ P_{\text{hop}} = \exp\!\left(-\frac{2\pi V^{2}}{\hbar\,v\,|\Delta F|}\right) \]
Now the point of the whole exercise. At a conical intersection the coupling V does not merely become small — it vanishes identically, because that is the defining condition for the two surfaces to touch. The exponent goes to zero and P → 1. The wavepacket crosses on its first pass, with certainty.
This is why retinal reaches the ground state within a few hundred femtoseconds instead of rattling about on the excited surface long enough to fluoresce. The protein did not evolve a faster molecule; it evolved a funnel where the coupling vanishes.
What this leaves out
The sweep is taken as exactly linear and the coupling as constant, over infinite time; a real wavepacket crosses in finite time with a varying gap, which is why the numerical solution on the simulations page still shows ringing after the crossing. The model is also strictly two-state and one-dimensional, whereas a conical intersection is an intrinsically two-dimensional object: Landau–Zener describes a slice through it, not the funnel itself.