University Quantum Mechanics II · The Adiabatic Approximation and Berry's Phase · 14.4
Landau-Zener transitions at a swept avoided crossing
Any experiment that drives one level through another does it at a finite speed, and the question is always whether the state kept up. This lesson hands you the exponent that answers it, shows that the adiabatic condition and the formula measure the same number, and has you confirm it with a few lines of numerical integration.
Build the model
Connect the measurement to the mechanism.
Take Quantum Mechanics I's avoided crossing and let the detuning run linearly in time: diabatic states |1⟩ and |2⟩ whose energies pass each other at a constant relative rate α, joined by a constant coupling V, so the adiabatic curves never come closer than 2|V|. A system that starts in |1⟩ has two ways out. Swept slowly, it obeys the adiabatic theorem and stays on its adiabatic curve, and because that curve changes character at the crossing, the whole population is carried into |2⟩.
Swept fast, it is left as it was, the sudden approximation's verdict: still |1⟩, now on the upper curve, having jumped the gap. Landau, Zener, Stückelberg and Majorana found in 1932 that the jump has probability PD = exp(−2π|V|²/(ħα)), exactly, for a sweep from t = −∞ to +∞. This lesson quotes that result without solving the parabolic-cylinder equation behind it, derives its fast limit from first-order perturbation theory, shows that the unit's adiabatic condition at the crossing, ħα ≪ 8|V|², is the exponent in another form, and confirms the formula numerically.
No power of the fine-structure constant enters. The formula is exact inside the model and fails with it: when V varies across the crossing, when the sweep is not linear over a few times the longer of 2|V|/α and √(ħ/α) around it, when a third level comes near, when the sweep stops too close to the gap, or when the crossing is passed twice and the two routes interfere.
- Simple definition
- A Landau-Zener transition is the jump of a two-level system across an avoided crossing whose diabatic energies are swept past each other at a constant rate α; for coupling V it happens with probability exp(−2π|V|²/(ħα)), and otherwise the system follows the adiabatic curve into the other state.
- Example
- With |V| = 0.50 μeV and α = 2.0 μeV ns⁻¹, ħα = 0.658 μeV ns × 2.0 μeV ns⁻¹ = 1.32 μeV², so the exponent is 2π × 0.25/1.32 = 1.19 and the system jumps with probability 0.30, ending in the other diabatic state with 0.70.
The model, set up rather than derived: Quantum Mechanics I's crossing with its detuning linear in t. Its eigenvalues ±√((αt/2)² + |V|²) are the adiabatic curves, 2|V| apart at t = 0.
α in J s⁻¹ is the rate at which ε₁ − ε₂ changes; V in J is constant; t in s, zero at the crossing
Exact for the model. Integrated numerically from c₁ = 1 long before the crossing, they are the independent check on the quoted formula.
c₁, c₂ are dimensionless amplitudes on |1⟩ and |2⟩ with |c₁|² + |c₂|² = 1; every term is in J
Quoted (Landau, Zener, Stückelberg, Majorana, 1932), exact for a sweep from t = −∞ to +∞. The rest, 1 − PD, follows the adiabatic curve into |2⟩.
|V|² in J² and ħα in J s × J s⁻¹ = J², so the exponent is a pure number; PD is the probability of ending in |1⟩
Derived from this unit's condition. The exponent is (π/4)(8|V|²/ħα), so condition and formula read one number: at ħα = 8|V|² the jump still has probability 0.46.
at t = 0, ħ|⟨−|dH/dt|+⟩| = ħα/2 and (E₊ − E₋)² = 4|V|²; both sides in J²
Derived with Unit 12's first-order amplitude. It is the first term of 1 − exp(−2π|V|²/ħα), and it passes 1 for slow sweeps, where only the exact result survives.
the Fresnel integral over all t has modulus √(2πħ/α), in s; valid only while 2π|V|²/(ħα) ≪ 1
Derived from the tilt, about |V|/(αT), of the adiabatic states from |1⟩ and |2⟩. Stopping at αT = 40|V| leaves wobbles up to 0.025; read populations in the adiabatic basis instead.
αT in J is the diabatic separation when the sweep stops at t = T; the ± term oscillates at the gap frequency
Two ways to leave a swept crossing
Quantum Mechanics I's two-state Hamiltonian, with diagonal energies ε₁ and ε₂ and coupling V, has eigenvalues split by √((ε₁ − ε₂)² + 4|V|²). Now make the detuning linear in time, ε₁ − ε₂ = αt. The diabatic lines ±αt/2 cross at t = 0; the adiabatic curves ±√((αt/2)² + |V|²) do not, and pass 2|V| apart. Far from the crossing each adiabatic state is one diabatic state, but not the same one on both sides: the lower curve is |1⟩ for t ≪ −2|V|/α and |2⟩ for t ≫ 2|V|/α. So a system prepared in |1⟩ has exactly two ways out. It can stay on the lower adiabatic curve and arrive in |2⟩, a complete transfer of population, or stay in |1⟩ and arrive on the upper curve, having jumped the gap. With |V| = 1.0 μeV and α = 2.0 μeV ns⁻¹ the curves are within a gap of each other for about 2|V|/α = 1.0 ns, and the whole question is what the state does in that nanosecond.
The adiabatic condition, evaluated along the sweep
This unit's condition, ħ|⟨m|dH/dt|n⟩| ≪ (Eₙ − Eₘ)², can be checked at every instant. Here dH/dt = (α/2)(|1⟩⟨1| − |2⟩⟨2|) never changes, and with the gap written Δ(t) = √((αt)² + 4|V|²), its matrix element between the two adiabatic states is α|V|/Δ. The condition becomes ħα|V| ≪ Δ³. It is hardest to meet at t = 0, where Δ = 2|V| and it reads ħα ≪ 8|V|². Away from the crossing Δ³ grows as |t|³: by |t| = 2|V|/α the margin has improved by 2³⁄² = 2.8, and by |t| = 10|V|/α by about 130. For a sweep slow enough that this margin starts near or above 1, everything non-adiabatic therefore happens within a few times 2|V|/α of the crossing. A fast sweep starts far below 1 and takes longer to recover: its window is set by √(ħ/α), the time over which the diabatic phase αt²/2ħ stays near its stationary value. Either way the adiabatic populations settle within a few times the longer of 2|V|/α and √(ħ/α), and outside that window they are frozen however long the sweep goes on. That is why a formula for one crossing can exist at all, and why it can depend only on α and V there.
One exponent, both limits
The exact answer, found in 1932 by Landau, Zener, Stückelberg and Majorana, is quoted here: solving the amplitude equations with parabolic-cylinder functions gives PD = exp(−2π|V|²/(ħα)) for the jump and 1 − PD for transfer into |2⟩. Check its dimensions: |V|² is J², and ħα is J s × J s⁻¹ = J². Check its limits: α → ∞ gives PD → 1, the sudden approximation's verdict that the state is left as it was, and α → 0 gives PD → 0, the adiabatic theorem's complete following. The exponent is (π/4) × 8|V|²/(ħα), the adiabatic ratio itself. At the edge, ħα = 8|V|², the jump still happens with e(−π/4) = 0.46; at ħα = 0.8|V|² it falls to 3.9 × 10⁻⁴. For 99% transfer the exponent must reach ln 100 = 4.6, which needs 8|V|²/(ħα) ≥ 5.9: that is what ≪ means here. No power of the fine-structure constant appears; the result holds for any two-level system, and the physics of a particular one enters only through V and α.
The fast limit from first-order perturbation theory
When the sweep is fast, |1⟩ barely changes, and Unit 12's first-order amplitude applies with the diabatic energies in the phase: c₂ ≈ −(i/ħ) ∫ V* e(iφ) dt, with the phase φ(t) = ∫(ε₂ − ε₁) dt′/ħ = −αt²/(2ħ). The Fresnel integral ∫ e(−iαt²/2ħ) dt over all t has modulus √(2πħ/α), so |c₂|² = (|V|/ħ)² × 2πħ/α = 2π|V|²/(ħα). That is exactly the first term of 1 − exp(−2π|V|²/(ħα)), so this limit is derived, not quoted. It also shows when the transfer is made: the phase stays within π of its stationary value only for |t| ≲ √(2πħ/α), which is 1.4 ns for α = 2.0 μeV ns⁻¹. First order fails in the other direction. At 2π|V|²/(ħα) = 1 it predicts complete transfer where the exact result still leaves e⁻¹ = 0.37 in |1⟩, and beyond that it gives probabilities above 1.
Confirming the formula numerically
Measure time in √(ħ/α) and energy in √(ħα). With τ = t√(α/ħ) and v = |V|/√(ħα) the equations become i dc₁/dτ = (τ/2)c₁ + vc₂ and i dc₂/dτ = vc₁ − (τ/2)c₂, and the prediction is PD = exp(−2πv²), with v² the only parameter. Start in the lower adiabatic state at τ = −40, integrate with fourth-order Runge-Kutta at step 0.002, and project onto the upper adiabatic state at τ = +40. For v² = 0.05, 0.10, 0.25 and 0.50 this gives 0.7304, 0.5335, 0.2079 and 0.0432, equal to exp(−2πv²) within 2 × 10⁻⁵, with the norm held to 10⁻⁶. Read |c₁|² in the diabatic basis instead and you get 0.7254, 0.5265, 0.2089 and 0.0499, off by up to 0.007, because at τ = 40 each adiabatic state still carries an amplitude of about v/40 of the other diabatic state and the populations beat at the gap frequency, as the finite-sweep card says. The step matters at the ends, not the crossing: at τ = ±80 each amplitude turns 40 rad per unit of τ, and a step of 0.01 returns 0.776 instead of 0.882 for v² = 0.02.
Where the formula stops, and where it was measured
Each assumption can fail. If V changes across the crossing window, a few times the longer of 2|V|/α and √(ħ/α), or the sweep is curved there, the exponent changes; if a third level lies within a few |V| of the crossing, two levels are not enough; if the sweep stops too near the gap, the finite-sweep wobble remains. Sweep through and back again, and the two routes to each outcome pick up different dynamic phases, so the final populations oscillate with the time between passages. That is Stückelberg interference, which no single PD describes. An environment that dephases |1⟩ and |2⟩ faster than the crossing takes spoils it too. Where the conditions hold, the formula has been tested: Rubbmark, Kash, Littman and Kleppner (1981) swept an electric field through avoided crossings of alkali Rydberg levels and found the diabatic fraction followed it as the slew rate changed, and Wernsdorfer and Sessoli (1999) turned it round in Fe₈ molecular magnets, reading tunnel splittings near 10⁻⁷ K off measured jump probabilities.
Change one variable at a time
Make the relationship visible.
At |V| = 0.5 μeV slide α down from 8 to 0.5: the population moves from the upper circle to the lower one, and the dashed first-order line leaves the exact curve as 2πδ grows. Then compare the default, α = 2 and |V| = 0.5, with α = 4 and |V| = 0.7: the picture changes, but PD only moves from 0.30 to 0.31, because only |V|²/α enters.
δ = |V|²/ħα0.190
JUMP PROBABILITY PD0.303
ħα / 8|V|²0.66
CROSSING TIME 2|V|/α0.50 ns
Live interpretationδ = |V|²/ħα: 0.190. JUMP PROBABILITY PD: 0.303. ħα / 8|V|²: 0.66. CROSSING TIME 2|V|/α: 0.50 ns
Catch the common trap
Explain before calculating.
A two-level system swept through an avoided crossing jumps the gap with probability 0.30. The coupling |V| is then doubled, which doubles the minimum gap, and the sweep rate α is doubled as well. What is the new jump probability?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTwo diabatic levels are swept through each other at a relative rate α = 2.0 μeV ns⁻¹ and coupled by |V| = 1.0 μeV. Find the probability that the system jumps the gap, check the adiabatic condition at the crossing, and say what doubling the sweep rate does.
- Work in μeV and ns, where ħ = 0.6582 μeV ns. Then ħα = 0.6582 μeV ns × 2.0 μeV ns⁻¹ = 1.316 μeV², the same unit as |V|² = 1.0 μeV², so the exponent is a pure number.
- Exponent: 2π|V|²/(ħα) = 2π × 1.0/1.316 = 4.77. Jump probability PD = e(−4.77) = 0.0085, so the system follows the lower curve into the other diabatic state with probability 0.9915.
- Adiabatic condition at the crossing: ħα/(8|V|²) = 1.316/8.0 = 0.16, well below 1 and consistent with a jump of under 1%. The curves stay within a gap of each other for about 2|V|/α = 1.0 ns.
- Doubling α to 4.0 μeV ns⁻¹ halves the exponent to 2.39, so PD = e(−2.39) = 0.092, eleven times larger, because PD is exponential in 1/α. The ratio ħα/(8|V|²) is now 0.33, no longer small.
AnswerPD = 0.0085, so 99.2% of the population is transferred; doubling the sweep rate raises the jump probability to 0.092.
MediumAn electron spin, γₑ = gₑμB/ħ = 1.761 × 10¹¹ s⁻¹ T⁻¹ with the measured gₑ = 2.0023, sits in a steady transverse field Bₓ = 5.0 μT while Bz is swept linearly from −1.0 mT to +1.0 mT. Find the shortest sweep that reverses the spin along z, following the field, with probability at least 0.99, and check that the window is wide enough for the formula.
- With H = (ħγₑ/2)σ⋅B, the diabatic states are |↑⟩ and |↓⟩ along z with energies ±(ħγₑ/2)Bz. They separate at the rate α = ħγₑ dBz/dt, and the transverse field couples them with V = ħγₑ Bₓ/2.
- The exponent is 2π|V|²/(ħα) = (π/2)γₑ Bₓ²/(dBz/dt). Planck's constant cancels, as it should: the expectation value of a spin obeys the classical precession equation exactly, so this probability is also a classical result for a magnetic moment.
- Following the field is the adiabatic outcome, so we need PD ≤ 0.01, an exponent of at least ln 100 = 4.605: dBz/dt ≤ (π/2)(1.761 × 10¹¹ s⁻¹ T⁻¹)(5.0 × 10⁻⁶ T)²/4.605 = 1.50 T s⁻¹.
- Sweeping 2.0 mT at 1.50 T s⁻¹ takes 2.0 × 10⁻³ T/(1.50 T s⁻¹) = 1.33 ms. The crossing itself lasts only 2|V|/α = Bₓ/(dBz/dt) = 3.3 μs, and ħα/(8|V|²) = (π/4)/4.605 = 0.17.
- Window: at the ends the diabatic separation is ħγₑ × 1.0 mT, 400 times |V| = ħγₑ × 2.5 μT, so the finite-sweep wobble is at most 2√(0.01 × 0.99)/400 = 5 × 10⁻⁴, far below the 0.01 being designed for.
AnswerdBz/dt ≤ 1.50 T s⁻¹, so the sweep must last at least 1.33 ms; the ends sit 400 couplings from the crossing, and the formula holds to better than 10⁻³.
HardA sodium atom meets an iodine atom head-on. The covalent curve Na + I is flat; the ionic curve Na⁺ + I⁻ lies IE − EA = 5.139 − 3.059 = 2.080 eV above it at large R and falls as −e²/(4πε₀R), with e²/(4πε₀) = 1.440 eV nm. Take the coupling where they cross as |V| = 0.050 eV and the radial speed as constant through it. Find the crossing distance, the collision energy at which one passage is a coin toss, and the ion-pair yield there.
- The diabatic curves cross where e²/(4πε₀Rc) = 2.080 eV, so Rc = 1.440 eV nm/2.080 eV = 0.692 nm. Their separation changes there at e²/(4πε₀Rc²) = (2.080 eV)²/(1.440 eV nm) = 3.004 eV nm⁻¹, so an atom at radial speed v sweeps the crossing at α = (3.004 eV nm⁻¹)v.
- A coin toss means PD = 1/2, an exponent of ln 2: ħα = 2π(0.050 eV)²/0.693 = 0.0227 eV², so α = 0.0227 eV²/(6.582 × 10⁻¹⁶ eV s) = 3.44 × 10¹³ eV s⁻¹.
- Then v = α/(3.004 × 10⁹ eV m⁻¹) = 1.146 × 10⁴ m s⁻¹. With the reduced mass μ = (22.99 × 126.90/149.89) u = 19.46 u = 3.23 × 10⁻²⁶ kg, the collision energy is ½μv² = 2.12 × 10⁻¹⁸ J = 13.2 eV.
- The crossing is passed going in and coming out, at the same speed because the two diabatic curves have equal energy there. Ions leave only if exactly one passage is diabatic, so Pᵢₒₙ = 2PD(1 − PD) = 0.50, the most any speed can give. The interference of the two routes is dropped, as it averages out over a beam's spread of energies and impact parameters.
- Limits: at 2.5 eV, just above the 2.080 eV threshold, v = 4.98 × 10³ m s⁻¹, the exponent is 1.60, PD = 0.20 and Pᵢₒₙ = 0.32. At 0.05 eV, PD = 1.3 × 10⁻⁵ and the atoms follow the adiabatic curves in and out, while at very high energy PD → 1 and the collision never leaves the covalent curve.
AnswerRc = 0.692 nm; one passage is a coin toss at v = 1.15 × 10⁴ m s⁻¹, a collision energy of about 13 eV, where the ion-pair yield reaches its maximum of 0.50.