Ehrenfest's theorem & the classical limit
Statement
Expectation values obey equations of the same form as the classical equations of motion — but not, in general, the classical equations themselves.
Why it matters
It is the sharpest available statement of how Newton's laws sit inside quantum mechanics. Read carelessly it looks like a proof that classical mechanics falls out of the quantum theory for free; read carefully it shows exactly why that is not so, and pins the correspondence to a condition on the potential and the width of the state. This is the bridge every later discussion of the classical limit and decoherence must cross.
Assumptions
Derivation
Result
Reading. Read carelessly, this says quantum mechanics contains Newton's laws. Read carefully, it does not. The classical statement would be d⟨p⟩/dt = −V′(⟨x⟩) — the force at the mean position. What the theorem gives is ⟨V′(x)⟩, the mean of the force. These agree only when V′ is linear over the width of the packet: when V is at most quadratic, or the packet is narrow compared with the scale on which the force varies.
Units check. d⟨x⟩/dt is [length]/[time] = velocity; ⟨p⟩/m is [momentum]/[mass] = velocity. ✓ d⟨p⟩/dt is [momentum]/[time] = force; ⟨dV/dx⟩ is [energy]/[length] = force. ✓
Limiting cases
- Free particle (V = 0): the two statements coincide exactly — the centroid drifts at constant ⟨p⟩/m.
- Harmonic oscillator (V quadratic, so V′ linear): ⟨V′(x)⟩ = V′(⟨x⟩) exactly, and the centroid follows the classical orbit. These two cases are exactly why the theorem is so often misquoted.
- Narrow packet in a slowly-varying potential: ⟨V′(x)⟩ ≈ V′(⟨x⟩) to leading order — the approximate classical regime.
Where the distinction bites
For a particle approaching a barrier, or near an unstable equilibrium, ⟨V′(x)⟩ ≠ V′(⟨x⟩) and the centroid of the packet does not follow a classical trajectory at all. A wave packet split by a barrier has a mean position sitting between the two halves — a place the particle is never actually found.
Breaks when
- The potential varies appreciably across the packet width, so V′ is not linear over the state.
- The state is a superposition of macroscopically distinct outcomes — the expectation value is then physically meaningless even though the equation still holds.
Failure modes
- Quoting the theorem as "quantum mechanics reduces to classical mechanics". It establishes a formal correspondence between averages; the true classical limit needs decoherence as well.
- Sign errors in [V, p̂] — the single most common algebra slip in an introductory quantum course.
Worked number
Take a cubic anharmonic potential V = ½ k x2 + λ x3, so V′ = k x + 3λ x2. Then ⟨V′⟩ = k⟨x⟩ + 3λ⟨x2⟩ = k⟨x⟩ + 3λ( ⟨x⟩2 + σx2 ), whereas the classical force at the mean is V′(⟨x⟩) = k⟨x⟩ + 3λ⟨x⟩2. The two differ by 3λ σx2 — the discrepancy is set directly by the variance of the packet, and vanishes only as σx → 0.
Discussion
Ehrenfest's theorem is best read as a statement about the algebra of averages, not about trajectories. The two equations d⟨x⟩/dt = ⟨p⟩/m and d⟨p⟩/dt = −⟨V′(x)⟩ reproduce the form of Hamilton's equations, but with every dynamical quantity replaced by its expectation value and, decisively, with the force averaged as a whole. Nothing in the derivation ever assumes a particle localised at a point; it assumes only a normalisable state, Hermiticity of Ĥ, and vanishing boundary terms. The classical-looking output is therefore a property of the first moments of an extended object, which is a much weaker thing than a Newtonian orbit.
The way the result splits apart is dictated entirely by the two commutators of step 2 and step 3. The centroid ⟨x⟩ is driven only by the kinetic energy, because the potential commutes with position, [V(x̂), x̂] = 0; the momentum ⟨p⟩ is driven only by the potential, because the kinetic term commutes with itself, [p̂2, p̂] = 0. Each expectation value evolves through exactly the piece of the Hamiltonian that fails to commute with it. This is why the structure of the pair mirrors the classical phase-space flow so cleanly: position is transported by momentum, momentum is transported by the gradient of the potential.
Underneath both lines sits the master equation d⟨Â⟩/dt = (i/ℏ)⟨[Ĥ,Â]⟩ + ⟨∂Â/∂t⟩, whose commutator term is the operator image of the classical Poisson bracket: (i/ℏ)[Ĥ, ·] ↔ {·, H}. The theorem is thus a special case of the Heisenberg equation of motion, and the "correspondence" it advertises is the correspondence between commutators and brackets, not between quantum and classical worlds as such. The honest content is the identity ⟨V′(x)⟩ = V′(⟨x⟩) + ½V⁈(⟨x⟩) Var(x) + …, obtained by Taylor-expanding the force about the mean position: the term linear in the fluctuation averages to zero, so the leading correction is set by the curvature of the force V⁈ weighted by the variance of the packet. The mean force equals the force at the mean position only when V⁈ = 0 — i.e. V at most quadratic — or when Var(x) is negligible on the scale over which the force curves. This is also why the classical limit needs more than a narrow packet: a superposition of two well-separated lumps can have a tiny variance-per-lump yet a huge total spread, and only decoherence, which destroys the cross terms, restores a genuine trajectory.
Common misconceptions. First, that the theorem proves quantum mechanics "contains" Newtonian mechanics — it establishes a correspondence between averages, and reduces to Newton's second law for the centroid only in the linear-force regime. Second, that a narrow initial packet stays narrow: under a non-quadratic potential the variance grows, so the correction term switches on with time even if it starts negligible. Third, treating ⟨x⟩ as "where the particle is" for a split packet — the mean can sit in a classically forbidden gap where the probability density vanishes. Fourth, the perennial sign slip in [V, p̂] = iℏ V′, which flips the force and turns restoring into anti-restoring.
Worked examples
Example 1 — A trapped ion: the centroid is exactly classical. A single 40Ca+ ion (m = 6.64×10−26 kg) sits in a harmonic trap of frequency ω = 2π×1.0 MHz. Its wave packet is prepared with ⟨x⟩0 = 100 nm and ⟨p⟩0 = 0. Because V = ½mω2x2 is quadratic, V⁈ = 0 and Ehrenfest is exact. Find ⟨x⟩ and ⟨p⟩ after t = 0.10 μs.
Answer. The centroid traces a perfect classical ellipse in phase space: ⟨x⟩ = 80.9 nm and ⟨p⟩ = −2.45×10−26 kg m/s after 0.10 μs. No quantum correction appears — the harmonic oscillator is precisely the case where Ehrenfest coincides with Newton for every state, which is exactly why it is a misleading example.
Example 2 — An anharmonic well: the "quantum force" correction. An electron (m = 9.11×10−31 kg) moves in V(x) = ½mω2x2 + b x3 with ω = 1.0×1015 s−1 and b = 3.0×107 J/m3. Its Gaussian packet has centroid ⟨x⟩ = 1.0 nm and width (standard deviation) σ = 0.50 nm. By how much does the mean force differ from the force at the mean position?
Answer. The packet's centroid feels an anomalous force 2.25×10−11 N — about 2.2% of the force a classical electron at 1.0 nm would feel — giving an extra acceleration ΔF/m ≈ 2.5×1019 m/s2. Because ΔF ∝ σ2, the deviation grows as the packet spreads, and the "classical" trajectory of ⟨x⟩ steadily fails.
Problems
- (Free particle.) An electron travels in a region of zero potential with ⟨p⟩ = 2.0×10−24 kg m/s and ⟨x⟩0 = 0. Using Ehrenfest, show ⟨p⟩ is conserved and find ⟨x⟩ after 1.0 ns.
Solution
With V = 0, d⟨p⟩/dt = −⟨V′⟩ = 0, so ⟨p⟩ is constant. Then d⟨x⟩/dt = ⟨p⟩/m is constant, giving ⟨x⟩(t) = ⟨p⟩t/m. Numerically ⟨x⟩ = (2.0×10−24)(1.0×10−9)/(9.11×10−31) = 2.2×10−3 m = 2.2 mm. The centroid moves at ⟨p⟩/m = 2.2×106 m/s. (The packet also spreads, but its mean is exactly ballistic.)
- (Uniform field — exactly classical.) A neutron (m = 1.675×10−27 kg) is released from rest in gravity, V = mgx. Show that ⟨p⟩ obeys the classical free-fall law regardless of the packet shape, and find ⟨p⟩ and ⟨v⟩ after 0.50 s.
Solution
V′ = mg is constant, so ⟨V′⟩ = mg = V′(⟨x⟩) exactly (all derivatives above the first vanish; no width correction). Hence d⟨p⟩/dt = −mg, so ⟨p⟩(t) = −mgt and ⟨v⟩ = ⟨p⟩/m = −gt. After 0.50 s: ⟨v⟩ = −(9.81)(0.50) = −4.9 m/s, ⟨p⟩ = m⟨v⟩ = −(1.675×10−27)(4.905) = −8.2×10−27 kg m/s. A linear potential is the boundary case: quadratic and below, so the centroid is always exactly classical.
- (Nanomechanical oscillator.) A resonator of mass m = 1.0×10−18 kg and frequency ω = 2π×2.0 MHz is prepared with ⟨x⟩0 = 5.0 nm, ⟨p⟩0 = 0. Find ⟨x⟩ and ⟨p⟩ at t = 0.20 μs.
Solution
Harmonic, so ⟨x⟩(t) = ⟨x⟩0 cos ωt and ⟨p⟩(t) = −mω⟨x⟩0 sin ωt, exactly. Here ω = 1.2566×107 s−1, so ωt = 2.513 rad (cos = −0.809, sin = 0.588). Thus ⟨x⟩ = (5.0 nm)(−0.809) = −4.0 nm. And mω⟨x⟩0 = (1.0×10−18)(1.2566×107)(5.0×10−9) = 6.28×10−20, so ⟨p⟩ = −(6.28×10−20)(0.588) = −3.7×10−20 kg m/s. The packet is just past its first quarter-period, heading back through the origin on the negative side.
- (Onset of anharmonic failure.) A particle sits in V = ½kx2 + bx3 with k = 0.50 N/m and b = 1.0×107 J/m3, packet centred at ⟨x⟩ = 2.0 nm. (a) For width σ = 1.0 nm, find the fractional Ehrenfest deviation ΔF/V′(⟨x⟩). (b) At what width does the correction reach 10% of the linear restoring force k⟨x⟩?
Solution
(a) ΔF = ½V⁈σ2 = 3bσ2 = 3(1.0×107)(1.0×10−9)2 = 3.0×10−11 N. The classical force V′(⟨x⟩) = k⟨x⟩ + 3b⟨x⟩2 = (0.50)(2.0×10−9) + 3(1.0×107)(2.0×10−9)2 = 1.0×10−9 + 1.2×10−10 = 1.12×10−9 N. Fractional deviation = 3.0×10−11/1.12×10−9 = 2.7%. (b) Set 3bσ2 = 0.10 k⟨x⟩: σ2 = 0.10(0.50)(2.0×10−9)/(3×1.0×107) = 3.33×10−18 m2, so σ = 1.8 nm. Beyond a width comparable to the displacement itself, the centroid ceases to track any classical orbit.
- (Derivation and the exact harmonic case.) (a) Taylor-expand V′(x) about ⟨x⟩ and take the expectation value to derive ⟨V′(x)⟩ = V′(⟨x⟩) + ½V⁈(⟨x⟩)Var(x) + … , explaining why the first-order term drops. (b) Hence prove that for a harmonic oscillator d2⟨x⟩/dt2 = −ω2⟨x⟩ holds for any state, not just a Gaussian. (c) State in one line why this makes the SHO a poor advertisement for the classical limit.
Solution
(a) Write x = ⟨x⟩ + δx with ⟨δx⟩ = 0. Then V′(x) = V′(⟨x⟩) + V⁈(⟨x⟩)δx + ½V⁈′(⟨x⟩)δx2 + … Taking ⟨·⟩: the constant term stays, the linear term gives V⁈(⟨x⟩)⟨δx⟩ = 0, and the quadratic term gives ½V⁈′(⟨x⟩)⟨δx2⟩ = ½V⁈′(⟨x⟩)Var(x). (Here V⁈ denotes the third derivative of V, i.e. the curvature of the force.) So ⟨V′⟩ = V′(⟨x⟩) + ½V⁈(⟨x⟩)Var(x) + … (b) For the SHO, V′ = mω2x is exactly linear, so V⁈ and all higher derivatives vanish and the expansion truncates: ⟨V′⟩ = mω2⟨x⟩ identically, whatever the state. Chaining Ehrenfest, d2⟨x⟩/dt2 = (1/m)d⟨p⟩/dt = −(1/m)⟨V′⟩ = −ω2⟨x⟩. No assumption about the packet was used. (c) Precisely because the centroid is classical for every state — even a wildly non-classical two-lump superposition — the harmonic oscillator hides the very feature (the width-dependent correction ½V⁈Var) that distinguishes averages from trajectories.