University Physics IV · Foundations of Quantum Mechanics · 5.9

Energy–Time Uncertainty & 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.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

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 σ_E · τ_A ≥ ħ/2, where τ_A = σ_A / |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 turn a superposition into an outcome. That takes an environment.

Simple definition: The energy–time uncertainty relation says a state whose energy is spread by σ_E cannot change appreciably in less than a time of order ħ/σ_E — 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.

Rate of change of an expectation valued⟨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 boundσ_E · τ_A ≥ ħ/2, with τ_A ≡ σ_A / |d⟨A⟩/dt|

Defines Δt instead of assuming it: the time ⟨A⟩ needs to shift by one standard deviation of A.

σ_E in J or eV, τ_A in s; ħ/2 = 3.29×10⁻¹⁶ eV s. Every observable A supplies its own τ_A.

Natural linewidth of a decaying levelΓ = 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 orthogonalityt⊥ ≥ πħ/(2σ_E) 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 correctionS/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.

No time operator, so no Robertson relation

Robertson's bound σ_A σ_B ≥ 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.

Mandelstam 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): σ_E σ_A ≥ ½|⟨[H, A]⟩|. Ehrenfest supplies d⟨A⟩/dt = (i/ħ)⟨[H, A]⟩, so |⟨[H, A]⟩| = ħ|d⟨A⟩/dt|. Substitute and the bound becomes σ_E σ_A ≥ (ħ/2)|d⟨A⟩/dt|, that is σ_E τ_A ≥ ħ/2 with τ_A = σ_A / |d⟨A⟩/dt|. Read τ_A 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 τ_A is infinite and the bound holds trivially, which is exactly right, since σ_E = 0 there and nothing about the state ever changes. Different observables give different τ_A for the same state, and the relation holds separately for each.

Lifetime 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. Numbers set the scale. Hydrogen 2p, τ = 1.60 ns, gives Γ = 0.41 μeV and Δν = 99.5 MHz. Sodium's D2 line, τ = 16.2 ns, gives 40.6 neV and 9.8 MHz. The Z boson runs the other way: its measured Γ = 2.50 GeV implies τ = ħ/Γ = 2.6×10⁻²⁵ s, a lifetime no clock could ever time, read instead off the width of a resonance curve. Γ counts every channel that empties the level, so collisions and quenching broaden a line as surely as spontaneous emission does.

Read 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/(2σ_E). Margolus and Levitin found a second, independent floor with σ_E 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: σ_E = 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.

The 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.

Small ħ does not make outcomes definite — decoherence 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. Estimate the rate for a 1.0 μm dust grain in air at 300 K and 1 atm: number density n = P/kT = 2.4×10²⁵ m⁻³, geometric cross-section πa² = 7.9×10⁻¹³ m², mean molecular speed 476 m s⁻¹, so collisions arrive at 9.1×10¹⁵ s⁻¹. A nitrogen molecule's thermal de Broglie wavelength is 19 pm, far below a 1 μm separation, so a single collision already resolves the branches and coherence dies in about 1.1×10⁻¹⁶ s. ħ was never the variable here. Pump the chamber to 10⁻¹⁰ mbar and the same grain stays coherent for roughly a millisecond, which is why matter interferometry lives in ultra-high vacuum.

Change one variable at a time

Make the relationship visible.

Interactive model

Radiative lifetime τ_rad1.6 ns

Collisional quench rate0.00 1/ns

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 counts every channel that empties it, not only the radiative one.

TOTAL LIFETIME τ1.60 ns

NATURAL WIDTH Γ0.411 μeV

LINE FWHM Δν99.5 MHz

RADIATIVE BRANCH1.00

Live interpretationTOTAL LIFETIME τ: 1.60 ns. NATURAL WIDTH Γ: 0.411 μeV. LINE FWHM Δν: 99.5 MHz. RADIATIVE BRANCH: 1.00

Catch the common trap

Explain before calculating.

Misconception check

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.

Hydrogen's 2p level decays to 1s with mean lifetime τ = 1.60 ns, and the Lyman-α line it emits has a natural width of 99.5 MHz. Which account of that width is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasySodium'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.

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.

AnswerΓ = 4.06×10⁻⁸ eV (40.6 neV), Δν = 9.8 MHz, and Δν/ν₀ = 1.93×10⁻⁸ — a line sharp to eight decimal places, and still 173 times narrower than Doppler broadening at 500 K.

MediumThe 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 σ_E, 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/(2σ_E).

Energy statistics of an equal two-level superposition: ⟨H⟩ = (E₁ + E₂)/2 and ⟨H²⟩ = (E₁² + E₂²)/2, so σ_E² = ⟨H²⟩ − ⟨H⟩² = (E₂ − E₁)²/4 and σ_E = ΔE/2.
Numbers: ΔE = (4.136×10⁻¹⁵ eV s)(2.387×10¹⁰ s⁻¹) = 9.87×10⁻⁵ eV, so σ_E = 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/(2σ_E) = 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.

Answerσ_E = 4.94×10⁻⁵ eV and t⊥ = πħ/ΔE = 21 ps, exactly the Mandelstam–Tamm floor πħ/(2σ_E). An equal two-level superposition changes as fast as any state with that energy spread possibly can.

HardA 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.

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.

AnswerS ≈ 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.