
Quantum core · institution-dependent · Interactive lesson
Energy–time uncertainty and the classical limit
Every other uncertainty relation pairs two operators. This one cannot, because a clock is not an observable of the system. Find out what Δt does mean — a lifetime, or the time an expectation value needs to move — and then where the classical world comes from once the bound is settled.
UPIV-05.09 · Lesson 9 of 9 in this unit. Position is not a completion score.Quantum core · institution-dependent. Departments differ in how much quantum mechanics they include in Physics IV or teach in a dedicated course. Follow your university’s required mathematical depth and assessment; these numbered units are not automatically optional.
Build the model
There is no time operator — so this bound has to mean something else.
The key idea
The energy–time uncertainty relation says a state whose energy is spread by σE cannot change appreciably in less than a time of order ħ/σE — Δt is a property of the state's own evolution, never the reading of a clock operator.
Example. Hydrogen's 2p level empties in τ = 1.60 ns, so its energy is not sharp: Γ = ħ/τ = 6.582×10⁻¹⁶ eV s ÷ 1.60×10⁻⁹ s = 0.41 μeV, a Lyman-α line 99.5 MHz wide on a centre frequency of 2.47×10¹⁵ Hz.
The complete physical model and assumptions
Every relation in the previous topic came from a commutator: two Hermitian operators, two variances, and Robertson's bound between them. ΔE Δt ≥ ħ/2 looks the same and is not, because there is no self-adjoint time operator to put on the other side. Pauli's argument is short — if T obeyed [T, H] = iħ then e(−iεT/ħ) would translate the spectrum of H by ε for every real ε, forcing the energy of every system to run from −∞ to +∞, and hydrogen has a ground state. So Δt has to be built out of something the system already has. Two constructions do it. Mandelstam and Tamm feed Ehrenfest's theorem into Robertson's bound and get σE · τA ≥ ħ/2, where τA = σA / |d⟨A⟩/dt| is the time a chosen observable needs to shift by one standard deviation: a speed limit on change, not an error bar on a clock. Separately, a state that empties exponentially with lifetime τ carries a Lorentzian energy profile of full width Γ = ħ/τ — an exact Fourier equality, not an inequality. Neither reading licenses borrowing energy from anywhere. And the classical limit is a separate question: large actions make the quantum corrections small, but they never remove the interference between branches of a superposition. That takes an environment.
1No time operator, so no Robertson relation
Robertson's bound σA σB ≥ ½|⟨[A, B]⟩| needs two Hermitian operators acting on the state. Energy supplies one: H. Time supplies nothing. Pauli's argument shows the gap is not an oversight. Suppose a self-adjoint T existed with [T, H] = iħ. Then the unitary family U(ε) = e(−iεT/ħ) obeys U(ε) H U(ε)† = H + ε for every real ε, so if E lies in the spectrum of H then so does E + ε for all real ε — the spectrum is the entire real line. Every Hamiltonian in this course is bounded below: hydrogen bottoms out at −13.6 eV, the oscillator at ½ħω. So T does not exist, and ΔE Δt ≥ ħ/2 cannot be read off the commutator machinery. What travels under that name is a family of theorems sharing a shape, not one theorem. Any symbol you write as Δt must be defined before the inequality means anything at all.
2Mandelstam and Tamm build Δt out of the state itself
Take an observable A with no explicit time dependence. Robertson still applies to the pair (H, A): σE σA ≥ ½|⟨[H, A]⟩|. Ehrenfest supplies d⟨A⟩/dt = (i/ħ)⟨[H, A]⟩, so |⟨[H, A]⟩| = ħ|d⟨A⟩/dt|. Substitute and the bound becomes σE σA ≥ (ħ/2)|d⟨A⟩/dt|, that is σE τA ≥ ħ/2 with τA = σA / |d⟨A⟩/dt|. Read τA as the time ⟨A⟩ needs to move by one standard deviation of A — the time after which the change is statistically visible above the state's own spread. Notice what has not appeared: no clock, no apparatus, no duration of an interaction. A stationary state has d⟨A⟩/dt = 0 for every A, so τA is infinite and the bound holds trivially, which is exactly right, since σE = 0 there and nothing about the state ever changes. Different observables give different τA for the same state, and the relation holds separately for each.
3Lifetime and linewidth: an exact Fourier equality
An excited level coupled to a continuum has no sharp energy, because it is not an eigenstate of the full atom-plus-field Hamiltonian. Model its amplitude as ψ(t) = e(−iE₀t/ħ) e(−t/2τ) for t > 0, so that |ψ|² = e(−t/τ) reproduces the observed exponential decay. Its Fourier transform gives an energy distribution |ψ̃(E)|² ∝ 1/[(E − E₀)² + (ħ/2τ)²] — a Lorentzian of full width at half maximum Γ = ħ/τ. Note the equals sign: the profile is fixed and Γτ = ħ exactly, with Δν = 1/(2πτ) in frequency. Nothing here is saturated and nothing is claimed about a clock; the width is a property the level carries before it decays and after it has gone. The relation also runs backwards, which is how it earns its keep at the short end: the Z boson's measured width Γ = 2.50 GeV implies τ = ħ/Γ = 2.6×10⁻²⁵ s, a lifetime no apparatus could ever time directly, read instead off the shape of a resonance curve. One warning about Γ. It is built from the total rate at which the level empties, so every open channel adds — collisions, quenching and photoionisation broaden a line exactly as spontaneous emission does, and a measured width is only an upper bound on the radiative part.
4Read as a speed limit on how fast a state can change
Push Mandelstam–Tamm to its extreme case by letting A be the projector onto the initial state. The bound becomes a floor on t⊥, the time for a state to become orthogonal to where it started: t⊥ ≥ πħ/(2σE). Margolus and Levitin found a second, independent floor with σE replaced by ⟨E⟩ − E₀, and whichever is larger is the real limit. An equal superposition of two levels split by ΔE saturates the first exactly: σE = ΔE/2, the overlap is |cos(ΔE t/2ħ)|, and its first zero sits at t⊥ = πħ/ΔE. This is the sense in which energy spread is a resource. A state with no energy spread is a stationary state and does nothing at all; the more spread you are willing to carry, the faster the state can travel to a distinguishable one. The same inequality bounds gate times in a quantum computer — and being a bound, it says nothing about whether any particular Hamiltonian gets anywhere near it.
5The classical limit is a correspondence check
Ehrenfest's theorem gives d⟨x⟩/dt = ⟨p⟩/m and d⟨p⟩/dt = −⟨V′(x)⟩. That is Newton's second law for the means only when ⟨V′(x)⟩ = V′(⟨x⟩), which is exact for a free particle, a uniform field and a harmonic oscillator, and otherwise carries a leading correction −½σₓ²V‴(⟨x⟩) that stays negligible while the packet is narrow on the scale over which the force curves. The other half of the check is counting quanta. A 10 g bob swinging at 1.0 Hz with 5.0 cm amplitude holds E = ½mω²A² = 4.9×10⁻⁴ J, which is n = E/ħω = 7.4×10²⁹ quanta at a level spacing of ħω = 6.6×10⁻³⁴ J, or 4.1×10⁻¹⁵ eV — a discreteness no instrument will resolve. Both statements are correspondence checks: the quantum description reproduces the classical one where the classical one already worked. Neither derives classical mechanics, and neither explains why the bob has a definite position in the first place.
6Small ħ does not remove interference — the environment does
Suppose the bob were prepared in a superposition of two positions 1 mm apart. Ehrenfest returns the mean, which sits between them, where the bob is not; the interference terms remain in the state, and no smallness of ħ removes them. What removes them is the environment. Every air molecule and thermal photon that scatters off the bob carries away a record of which position it took, and once that record exists elsewhere the two branches can no longer interfere locally: the off-diagonal elements of the object's reduced density matrix decay, and they decay faster for a wider separation and for a shorter scattering wavelength. When that wavelength is short compared with the separation, one scattering event already carries a complete which-path record, so the coherence time is simply the mean time between collisions — and in air at room pressure that is shorter than any dynamical time the object has. Lower the pressure and the same superposition survives proportionally longer, which is why matter interferometry lives in ultra-high vacuum. Two cautions. This is ordinary unitary evolution of object plus environment, so it accounts for why interference becomes unobservable and which basis survives, not for why one outcome rather than another is the one you see; that last question stays open and interpretation-dependent. And ħ never moved.
Key equations
Read the units, assumptions and limits before substituting.
Rate of change of an expectation value
d⟨A⟩/dt = (i/ħ)⟨[H, A]⟩
The step that turns Robertson's static bound into a statement about how fast a state can move.
For A with no explicit time dependence; H in J, A in its own unit, ħ = 1.055×10⁻³⁴ J s.Mandelstam–Tamm bound
σE · τA ≥ ħ/2, with τA ≡ σA / |d⟨A⟩/dt|
Defines Δt instead of assuming it: the time ⟨A⟩ needs to shift by one standard deviation of A.
σE in J or eV, τA in s; ħ/2 = 3.29×10⁻¹⁶ eV s. Every observable A supplies its own τA.Natural linewidth of a decaying level
Γ = ħ/τ, Δν = Γ/h = 1/(2πτ)
An equality, not a bound — Fourier's answer for an amplitude damped as e(−t/2τ).
Γ is the Lorentzian FWHM in energy, τ the mean lifetime in s; ħ = 6.582×10⁻¹⁶ eV s.Quantum speed limit to orthogonality
t⊥ ≥ πħ/(2σE) and t⊥ ≥ πħ/[2(⟨E⟩ − E₀)]
Two independent floors, Mandelstam–Tamm and Margolus–Levitin; whichever is larger is the real limit.
t⊥ in s is the time to reach a perfectly distinguishable state; E₀ is the ground-state energy.Pauli's obstruction to a time operator
[T, H] = iħ ⇒ e(−iεT/ħ) H e(iεT/ħ) = H + ε
Why this is not a Robertson relation at all: there is no second variance available to bound.
That would place every real ε in the spectrum of H, so no such H could be bounded below.Classical-limit criterion and its correction
S/ħ ≫ 1d⟨p⟩/dt = −V′(⟨x⟩) − ½σₓ²V‴(⟨x⟩) − …
Newton's law for the means, plus the first term telling you when a packet stops obeying it.
S is the action in J s; the correction needs σₓ small on the scale over which V″ varies.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Leave the quench rate at zero and set the radiative lifetime to 1.6 ns: the width reads 99.5 MHz, hydrogen's Lyman-α natural linewidth. Now raise the quench rate — the level empties faster and the line broadens, because Γ answers to the total decay rate, not to the radiative one alone.
Live results
- TOTAL LIFETIME τ
- 1.60 ns
- NATURAL WIDTH Γ
- 0.411 μeV
- LINE FWHM Δν
- 99.5 MHz
- RADIATIVE BRANCH
- 1.00
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Radiative lifetime τrad: 1.6 ns; Collisional quench rate: 0.00 1/ns
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The error is reading ΔE Δt ≥ ħ/2 as permission to break energy conservation — a system "borrows" ΔE for a time ħ/2ΔE and pays it back, with virtual particles living on the loan. Nothing in quantum mechanics says this. For a closed system with a time-independent H, every moment of the energy distribution is frozen: d⟨Hⁿ⟩/dt = (i/ħ)⟨[H, Hⁿ]⟩ = 0 for all n. The distribution you would measure at t = 0 is the distribution you would measure an hour later, so there is nothing to borrow and nothing to return. What the relation actually constrains is either the energy spread of a level that genuinely is not stationary — a decaying state carries a real Lorentzian width, there before the decay and after it — or the speed at which an expectation value can move. Both are properties of one prepared state, and neither creates energy for anybody.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Sodium's D2 line at 589.0 nm is emitted by a level of mean lifetime τ = 16.2 ns. Find its natural linewidth Γ in energy and Δν in frequency, then the fractional width Δν/ν₀. Use ħ = 6.582×10⁻¹⁶ eV s and c = 2.998×10⁸ m s⁻¹.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- The level empties exponentially, so its amplitude is damped as e(−t/2τ) and its energy profile is a Lorentzian of full width Γ = ħ/τ. This is an equality fixed by the Fourier transform, not a bound to be saturated.
- Γ = (6.582×10⁻¹⁶ eV s)/(16.2×10⁻⁹ s) = 4.06×10⁻⁸ eV = 40.6 neV.
- In frequency, Δν = Γ/h = 1/(2πτ) = 1/(2π × 16.2×10⁻⁹ s) = 9.82×10⁶ Hz = 9.8 MHz.
- The line centre is ν₀ = c/λ = (2.998×10⁸ m s⁻¹)/(589.0×10⁻⁹ m) = 5.09×10¹⁴ Hz, so Δν/ν₀ = 9.82×10⁶/5.09×10¹⁴ = 1.93×10⁻⁸.
- For contrast, Doppler broadening in a 500 K sodium vapour has FWHM ν₀√(8 ln2 · kT/Mc²) = 1.70×10⁹ Hz, some 173 times wider. The natural width is the floor, and it is usually not what a spectrometer sees.
Answer. Γ = 4.06×10⁻⁸ eV (40.6 neV), Δν = 9.8 MHz, and Δν/ν₀ = 1.93×10⁻⁸ — two parts in a hundred million, and still 173 times narrower than Doppler broadening at 500 K.
MediumExample 2
The ammonia inversion doublet splits the molecular ground state by ΔE = h × 23.87 GHz. A molecule is prepared in the equal superposition (|1⟩ + |2⟩)/√2 of the two levels. Find σE, find the time t⊥ at which the state first becomes orthogonal to its initial state, and check the result against the Mandelstam–Tamm floor t⊥ ≥ πħ/(2σE).
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Energy statistics of an equal two-level superposition: ⟨H⟩ = (E₁ + E₂)/2 and ⟨H²⟩ = (E₁² + E₂²)/2, so σE² = ⟨H²⟩ − ⟨H⟩² = (E₂ − E₁)²/4 and σE = ΔE/2.
- Numbers: ΔE = (4.136×10⁻¹⁵ eV s)(2.387×10¹⁰ s⁻¹) = 9.87×10⁻⁵ eV, so σE = 4.94×10⁻⁵ eV.
- Evolve: |ψ(t)⟩ = (e(−iE₁t/ħ)|1⟩ + e(−iE₂t/ħ)|2⟩)/√2, so ⟨ψ(0)|ψ(t)⟩ = ½(e(−iE₁t/ħ) + e(−iE₂t/ħ)) and taking the modulus leaves |cos(ΔE t/2ħ)|.
- The first zero is at ΔE t/2ħ = π/2, so t⊥ = πħ/ΔE = h/(2ΔE) = 1/(2 × 23.87×10⁹ s⁻¹) = 2.09×10⁻¹¹ s = 21 ps.
- The floor: πħ/(2σE) = π(6.582×10⁻¹⁶ eV s)/(2 × 4.94×10⁻⁵ eV) = 2.09×10⁻¹¹ s. Identical, so the equal two-level superposition saturates Mandelstam–Tamm exactly.
- Note what 21 ps is not. Nothing decays, so it is not a lifetime; no instrument is involved, so it is not a measurement time. It is how long this state needs to become perfectly distinguishable from where it started.
Answer. σE = 4.94×10⁻⁵ eV and t⊥ = πħ/ΔE = 21 ps, exactly the Mandelstam–Tamm floor πħ/(2σE). An equal two-level superposition changes as fast as any state with that energy spread possibly can.
HardExample 3
A spherical dust grain of diameter 1.0 μm and density 2.0×10³ kg m⁻³ drifts in air at 300 K and 1 atm. (a) Compare its action scale with ħ for a coherent separation of 1.0 μm. (b) Estimate how long that superposition survives collisions with air molecules. (c) Give the answer to (b) at 10⁻¹⁰ mbar, and say what the comparison shows.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Mass: m = ρ(4/3)πa³ with a = 0.50 μm gives m = 2.0×10³ × (4/3)π(5.0×10⁻⁷ m)³ = 1.05×10⁻¹⁵ kg.
- Action scale: vᵣₘₛ = √(3kT/m) = √(3 × 1.381×10⁻²³ × 300 / 1.05×10⁻¹⁵) = 3.4×10⁻³ m s⁻¹, so p = 3.6×10⁻¹⁸ kg m s⁻¹ and S ≈ p⋅d = 3.6×10⁻²⁴ J s = 3.4×10¹⁰ ħ. The de Broglie wavelength h/p = 1.8×10⁻¹⁶ m puts the fringes 1.8×10⁻¹⁰ rad apart — unresolvable, but not absent from the state.
- Collision rate: n = P/kT = 1.013×10⁵/(1.381×10⁻²³ × 300) = 2.45×10²⁵ m⁻³; cross-section πa² = 7.9×10⁻¹³ m²; mean N₂ speed v̄ = √(8kT/πM) = 476 m s⁻¹. Rate = nσv̄ = 9.1×10¹⁵ s⁻¹.
- Does one collision resolve the paths? A nitrogen molecule's thermal de Broglie wavelength is h/√(2πMkT) = 1.9×10⁻¹¹ m, far shorter than the 1.0 μm separation, so a single scattering carries a full which-path record and the decoherence time is one collision time, 1.1×10⁻¹⁶ s.
- At 10⁻¹⁰ mbar the density falls by 1.0×10⁻¹⁰/1013 ≈ 1×10⁻¹³, so the rate drops to about 9.0×10² s⁻¹ and the coherence time rises to about 1.1×10⁻³ s.
- ħ took the same value in every line above. What moved the answer by thirteen orders of magnitude was the pressure — so the classical look of the world is an environmental fact, not a consequence of ħ being small.
Answer. S ≈ 3.4×10¹⁰ ħ, deep in the classical regime; yet a 1.0 μm superposition decoheres in 1.1×10⁻¹⁶ s at 1 atm and survives about 1.1 ms at 10⁻¹⁰ mbar. The environment sets classicality, not the size of ħ.
Quick check
Select an answer and explain it before checking.
Review & continue
The energy–time uncertainty relation says a state whose energy is spread by σE cannot change appreciably in less than a time of order ħ/σE — Δt is a property of the state's own evolution, never the reading of a clock operator.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.