University Physics IV · The Schrödinger Equation · 6.8
Expectation Values & Ehrenfest's Theorem
Every prediction quantum mechanics makes about a measurable quantity is an average over many identical runs. Learn to compute those averages, then differentiate one and watch Newton's second law fall out — carrying a correction term that says exactly when the classical picture is safe.
Build the model
Connect the measurement to the mechanism.
Quantum mechanics does not hand you the value of an observable; it hands you a distribution, and the first number you can pull out of it is the mean, ⟨A⟩ = ∫ψ*Âψ dx. The operator sits between the two copies of ψ because it has to act before the integral is taken — for position that collapses harmlessly to ∫x|ψ|²dx, but for momentum it does not, since p is not a function of x. Once you can compute a mean you can differentiate it.
Feeding the time-dependent Schrödinger equation into d⟨A⟩/dt and leaning on the hermiticity of Ĥ gives one line that governs every observable at once: d⟨A⟩/dt = (i/ħ)⟨[Ĥ,Â]⟩ + ⟨∂Â/∂t⟩. Two commutators then do the rest. [Ĥ, x̂] = −iħp̂/m gives d⟨x⟩/dt = ⟨p⟩/m, and [Ĥ, p̂] = iħ dV/dx gives d⟨p⟩/dt = ⟨−dV/dx⟩, so m d²⟨x⟩/dt² = ⟨F(x)⟩. That is Ehrenfest's theorem, and it is exact for any normalised state in any potential.
What it is not is Newton's law: the right-hand side is the force averaged over the packet, ⟨F(x)⟩, not the force at the mean, F(⟨x⟩). The two differ by ½F″(⟨x⟩)σₓ² plus higher terms, so classical motion is recovered only where the force is near-linear across the width of the state — and narrowing the packet to buy that costs a momentum spread that widens it again.
- Simple definition
- The expectation value of an observable is the mean of its measured values over many identically prepared systems, computed as ⟨A⟩ = ∫ψ*Âψ dx with the operator acting on ψ before the integral is done.
- Example
- For the n = 2 state of an infinite well of width L = 0.50 nm, ⟨x⟩ = 0.25 nm — the exact midpoint, which is where ψ₂ has a node and |ψ|² is zero. The mean is a value the particle is never found at.
The ensemble mean of A over many identical preparations — never the result of any single run.
ψ normalised; Â acts on ψ first. ⟨A⟩ carries A's unit and is real for Hermitian Â.
Says whether the mean is a prediction or a shrug: σA = 0 exactly when ψ is an eigenfunction of Â.
Units of A². Apply the operator twice inside the sandwich; ⟨²⟩ ≠ ⟨A⟩² in general.
Gives the rate of change of any mean without solving for ψ(t); if [Ĥ,Â] = 0 that mean is conserved.
[Ĥ,Â] = ĤÂ − ÂĤ; ħ = 1.055 × 10⁻³⁴ J s. The last term is zero for x̂, p̂ and Ĥ.
Combined they give m d²⟨x⟩/dt² = ⟨F(x)⟩ — the centroid of the packet obeys a force law.
From [Ĥ, x̂] = −iħp̂/m and [Ĥ, p̂] = iħ dV/dx. Both hold exactly for any normalised ψ.
Exact classical motion for free, uniform-field and harmonic potentials; every other potential pays this term.
σₓ² is the position variance of the state. The correction vanishes identically when F″ = 0.
Splits a known total energy into kinetic and potential parts with no integral evaluated at all.
Set d⟨xp⟩/dt = 0, true in any bound stationary state. Coulomb has n = −1, so ⟨T⟩ = −E.
The operator sits inside the sandwich
⟨A⟩ = ∫ψ*Âψ dx, with the operator wedged between the two copies of ψ, because  has to act on ψ before the integral is taken. For position that detail is invisible: x̂ψ is just x times ψ, the x slides out, and ⟨x⟩ = ∫x|ψ|²dx — the familiar weighted mean of the density. Momentum refuses. p̂ = −iħ ∂/∂x is not a function of x, so there is no number p(x) available to weight by |ψ|²; writing ⟨p⟩ = ∫p|ψ|²dx is not a rough approximation but a sentence with no meaning. You differentiate ψ, then multiply by ψ*, then integrate, in that order. The reward is that the answer comes out real: hermiticity says ∫ψ*(Âψ)dx = ∫(Âψ)*ψ dx, so ⟨A⟩* = ⟨A⟩ for every Hermitian observable — the least you can ask of a predicted measurement.
A mean is an ensemble statement, so quote its spread
No single run returns ⟨A⟩. Prepare N systems in the same ψ, measure A on each, average: that is what the integral predicts, and it says nothing whatever about run number seven. The mean can even be a value the system never takes. In the n = 2 state of an infinite well, |ψ|² is symmetric about the midpoint, so ⟨x⟩ = L/2 — precisely where ψ₂ has a node and the density is zero. So always quote the spread alongside: σA² = ⟨²⟩ − ⟨A⟩², where ⟨²⟩ means applying the operator twice inside the sandwich. For the ground state of a well of width L = 0.50 nm, ⟨x⟩ = 0.25 nm and ⟨x²⟩ = L²(1/3 − 1/2π²) = 0.2827L², giving σₓ = 0.181L = 0.090 nm. Momentum is starker still: ⟨p⟩ = 0 while ⟨p²⟩ = (πħ/L)², so σₚ = 6.63 × 10⁻²⁵ kg m s⁻¹. A vanishing mean sitting on a large spread is the ordinary case, not a paradox.
Differentiate the mean and the Schrödinger equation answers
Now differentiate: d⟨A⟩/dt = ∫[(∂ψ*/∂t)Âψ + ψ*(∂Â/∂t)ψ + ψ*Â(∂ψ/∂t)]dx. The Schrödinger equation supplies both time derivatives — ∂ψ/∂t = Ĥψ/(iħ) and its conjugate ∂ψ*/∂t = −(Ĥψ)*/(iħ), the minus sign coming from conjugating the i. The first and third terms become (i/ħ)[∫(Ĥψ)*Âψ dx − ∫ψ*ÂĤψ dx], and hermiticity of Ĥ moves it off the starred factor to give ∫ψ*ĤÂψ dx. What survives is d⟨A⟩/dt = (i/ħ)⟨[Ĥ,Â]⟩ + ⟨∂Â/∂t⟩. Two consequences arrive free. An observable with no explicit time dependence that commutes with Ĥ has a constant mean — energy first of all, since [Ĥ,Ĥ] = 0. And in a stationary state ψ = φ(x)e(−iEt/ħ), the two phase factors cancel inside every sandwich, so every mean is frozen. That is the whole content of the word stationary.
Two commutators hand back Newton's two laws
For position only the kinetic term fails to commute: [p̂², x̂] = p̂[p̂, x̂] + [p̂, x̂]p̂ = −2iħp̂, so [Ĥ, x̂] = −iħp̂/m and d⟨x⟩/dt = (i/ħ)(−iħ⟨p⟩/m) = ⟨p⟩/m. Notice what that is — the claim that ⟨p⟩ equals m times the velocity of the centroid is a result of the theorem, not a definition smuggled in beforehand. For momentum only the potential matters, and the cleanest route is to act on a test function: (V̂p̂ − p̂V̂)f = −iħVf′ + iħ(Vf)′ = iħV′f, so [Ĥ, p̂] = iħ dV/dx and d⟨p⟩/dt = (i/ħ)(iħ⟨V′⟩) = −⟨dV/dx⟩ = ⟨F⟩. Differentiate the first relation once more and substitute the second: m d²⟨x⟩/dt² = ⟨F(x)⟩. Newton's second law, with angle brackets on both sides, exact for any normalised state in any potential.
Why ⟨F(x)⟩ is not F(⟨x⟩)
Everything now turns on whether ⟨F(x)⟩ equals F(⟨x⟩). Expand F about the mean: F(x) = F(⟨x⟩) + F′(⟨x⟩)(x − ⟨x⟩) + ½F″(⟨x⟩)(x − ⟨x⟩)² + …. Averaging kills the linear term by the definition of the mean, leaving ⟨F⟩ = F(⟨x⟩) + ½F″(⟨x⟩)σₓ² + …. If F″ = 0 everywhere the series stops at the first term and ⟨x⟩(t) traces an exact classical trajectory whatever the packet looks like — which covers the free particle, a uniform gravitational or electric field, and the harmonic oscillator, and is why those three are always the worked examples. Anywhere else you pay. In V = ¼λx⁴ the force is F = −λx³ with F″ = −6λx, and for a Gaussian the correction is exact rather than approximate: ⟨F⟩ = −λ(⟨x⟩³ + 3⟨x⟩σₓ²). At ⟨x⟩ = 0.50 nm and σₓ = 0.10 nm that is 12% more force than the classical reading gives. The criterion is the packet width against the scale over which F bends — not the smallness of ħ.
The virial bonus, and the limit Ehrenfest cannot cross
The same machinery pays out elsewhere. Put  = x̂p̂ into the general relation in a bound stationary state, where every mean is constant: d⟨xp⟩/dt = 2⟨T⟩ − ⟨x dV/dx⟩ = 0 delivers the virial theorem 2⟨T⟩ = ⟨x dV/dx⟩, which for V ∝ xⁿ is 2⟨T⟩ = n⟨V⟩. Hydrogen's ground state has E = −13.6 eV and n = −1, so ⟨T⟩ = +13.6 eV and ⟨V⟩ = −27.2 eV, split apart without evaluating a single integral. What the theorem does not deliver is classical mechanics. ⟨x⟩ is a centroid, and a centroid can live where nothing does: fire a packet at a barrier that transmits half of it and reflects the other half, and ⟨x⟩ drifts slowly through the barrier region while every actual detection lands far to one side or the other. Ehrenfest tells you how averages move. Turning averages into the single definite outcomes a classical world is made of takes decoherence and measurement, which this theorem does not contain.
Change one variable at a time
Make the relationship visible.
Hold the centre fixed and widen σx: the packet does not move, but the open dot slides down the curve away from the filled one. That gap is −3λ⟨x⟩σx². Drag the centre to zero and it vanishes — the correction needs both an off-centre packet and curvature in F.
F(⟨x⟩) CLASSICAL-0.043 λ nm³
⟨F(x)⟩ EHRENFEST-0.070 λ nm³
CORRECTION −3λ⟨x⟩σx²-0.027 λ nm³
CORRECTION / |F(⟨x⟩)|63 %
Live interpretationF(⟨x⟩) CLASSICAL: −0.043 λ nm³. ⟨F(x)⟩ EHRENFEST: −0.070 λ nm³. CORRECTION −3λ⟨x⟩σx²: −0.027 λ nm³. CORRECTION / |F(⟨x⟩)|: 63 %
Catch the common trap
Explain before calculating.
A wave packet moves in V(x) = ¼λx⁴, so the force is F(x) = −λx³. At one instant the packet is Gaussian with ⟨x⟩ = 0.50 nm and position spread σₓ = 0.10 nm. What is d⟨p⟩/dt at that instant?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron sits in the ground state of an infinite square well of width L = 0.50 nm, ψ₁ = √(2/L) sin(πx/L). Given ⟨x²⟩ = L²(1/3 − 1/2π²) and ⟨p²⟩ = (πħ/L)², find ⟨x⟩, σₓ, ⟨p⟩ and σₚ, then use Ehrenfest's theorem to say how ⟨x⟩ moves.
- |ψ₁|² is symmetric about the midpoint of the well, so ⟨x⟩ = L/2 = 0.25 nm.
- ⟨x²⟩ = L²(1/3 − 1/2π²) = L²(0.33333 − 0.05066) = 0.28267 L². Then σₓ² = ⟨x²⟩ − ⟨x⟩² = (0.28267 − 0.25000)L² = 0.03267 L², so σₓ = 0.1808 L = 0.090 nm.
- ψ₁ is real, so ⟨p⟩ = ∫ψ₁(−iħ dψ₁/dx)dx = −iħ[ψ₁²/2] evaluated at the walls = 0. This is not ∫p|ψ|²dx, which has no meaning: p̂ must act on ψ before the integral is taken.
- σₚ² = ⟨p²⟩ − ⟨p⟩² = (πħ/L)², so σₚ = πħ/L = π(1.055 × 10⁻³⁴ J s)/(0.50 × 10⁻⁹ m) = 6.63 × 10⁻²⁵ kg m s⁻¹.
- Ehrenfest gives d⟨x⟩/dt = ⟨p⟩/m = 0: the mean position is pinned at the centre of the well for all time. A stationary state has frozen means even though σₚ is large — and σₓσₚ = 5.99 × 10⁻³⁵ J s, a factor 1.14 above the ħ/2 floor.
Answer⟨x⟩ = 0.25 nm, σₓ = 0.090 nm, ⟨p⟩ = 0, σₚ = 6.63 × 10⁻²⁵ kg m s⁻¹; d⟨x⟩/dt = ⟨p⟩/m = 0, so the centroid never moves.
MediumA proton (m = 1.67 × 10⁻²⁷ kg) is bound in a harmonic potential V = ½mω²x² with ω = 2.0 × 10¹⁴ rad s⁻¹. A packet of unspecified shape is prepared with ⟨x⟩ = 2.0 × 10⁻¹¹ m and ⟨p⟩ = 0. Find ⟨x⟩(t), and evaluate ⟨p⟩ a quarter of a period later.
- Ehrenfest gives d⟨x⟩/dt = ⟨p⟩/m and d⟨p⟩/dt = ⟨−dV/dx⟩ = ⟨−mω²x⟩ = −mω²⟨x⟩. The force is linear, so F″ = 0 and the average of the force equals the force at the average — exactly, for any packet shape or width.
- Combining the two: d²⟨x⟩/dt² = −ω²⟨x⟩. With ⟨x⟩(0) = x₀ and ⟨p⟩(0) = 0 the solution is ⟨x⟩(t) = x₀ cos ωt, and ⟨p⟩(t) = m d⟨x⟩/dt = −mωx₀ sin ωt.
- Period T = 2π/ω = 2π/(2.0 × 10¹⁴ s⁻¹) = 3.14 × 10⁻¹⁴ s, so a quarter period is 7.85 × 10⁻¹⁵ s = 7.9 fs.
- At t = T/4 the phase is ωt = π/2, so ⟨x⟩ = 0 and ⟨p⟩ = −mωx₀ = −(1.67 × 10⁻²⁷)(2.0 × 10¹⁴)(2.0 × 10⁻¹¹) = −6.7 × 10⁻²⁴ kg m s⁻¹.
- The width of the packet never entered the calculation. That is special to the harmonic oscillator: in any potential whose force has curvature, the answer would carry a ½F″(⟨x⟩)σₓ² correction.
Answer⟨x⟩(t) = (2.0 × 10⁻¹¹ m) cos ωt, exactly classical for any packet; at t = T/4 = 7.9 fs, ⟨x⟩ = 0 and ⟨p⟩ = −6.7 × 10⁻²⁴ kg m s⁻¹.
HardA Gaussian packet moves in the quartic well V(x) = ¼λx⁴, so F(x) = −λx³. At one instant ⟨x⟩ = 0.50 nm and σₓ = 0.10 nm. (a) Find ⟨F⟩ exactly and its fractional excess over F(⟨x⟩). (b) How narrow must σₓ be for that excess to drop below 1%? (c) What does the uncertainty bound say about the packet you would then need?
- Write x = ⟨x⟩ + u with ⟨u⟩ = 0 and ⟨u²⟩ = σₓ². Then x³ = ⟨x⟩³ + 3⟨x⟩²u + 3⟨x⟩u² + u³, and for a Gaussian ⟨u³⟩ = 0, so ⟨x³⟩ = ⟨x⟩³ + 3⟨x⟩σₓ² exactly — no truncation of the series is involved.
- (a) ⟨F⟩ = −λ(⟨x⟩³ + 3⟨x⟩σₓ²) = −λ(0.125 + 3 × 0.50 × 0.010) nm³ = −λ(0.125 + 0.015) nm³ = −0.140 λ nm³, against F(⟨x⟩) = −0.125 λ nm³.
- Fractional excess = 3σₓ²/⟨x⟩² = 3(0.010)/(0.250) = 0.12, i.e. 12%. The Taylor form agrees: ½F″(⟨x⟩)σₓ² = ½(−6λ⟨x⟩)σₓ² = −3λ(0.50)(0.010) = −0.015 λ nm³.
- (b) Require 3σₓ²/⟨x⟩² < 0.01, so σₓ < ⟨x⟩/√300 = 0.50 nm ÷ 17.32 = 0.029 nm.
- (c) That packet has σₚ ≥ ħ/(2σₓ) = (1.055 × 10⁻³⁴ J s)/(2 × 2.89 × 10⁻¹¹ m) = 1.8 × 10⁻²⁴ kg m s⁻¹, against 5.3 × 10⁻²⁵ kg m s⁻¹ at σₓ = 0.10 nm.
- Squeezing x until the classical reading is accurate to 1% costs a momentum spread about 3.5 times larger, and the packet then re-widens that much faster. Classical behaviour here is bought on credit, not free.
Answer(a) ⟨F⟩ = −0.140 λ nm³ against F(⟨x⟩) = −0.125 λ nm³, a 12% excess. (b) σₓ < ⟨x⟩/√300 = 0.029 nm. (c) The momentum-spread floor rises from 5.3 × 10⁻²⁵ to 1.8 × 10⁻²⁴ kg m s⁻¹.