Expectation values that move, and ones that do not
In a superposition, ⟨x⟩ can swing but ⟨H⟩ cannot.
This is the lesson where the infinite well finally moves. Superpose two stationary states and the mean position swings from side to side while the mean energy does not budge. The job is to predict, before any integral, which averages will move, how fast, and what the moving ones say, and do not say, about one particle.
Build the model
Only a cross term between two different energies can make an average change in time.
1The big idea
Put the evolved state Ψ = Σ cₙψₙ e(−iEₙt/ħ) into ⟨Q⟩ = ∫ Ψ* Q̂ Ψ dx and the answer is a double sum of terms cₘ* cₙ Qₘₙ e(iωₘₙt), with ωₘₙ = (Eₘ − Eₙ)/ħ. Every diagonal term has lost its phase, so it is frozen. Every off-diagonal term turns at its Bohr frequency ωₘₙ, but it contributes only if the matrix element Qₘₙ = ∫ ψₘ* Q̂ ψₙ dx is nonzero. That single rule sorts the averages. Ĥ has no off-diagonal elements in its own eigenbasis, so ⟨H⟩ = Σ |cₙ|² Eₙ, and each |cₙ|² is fixed for all time because its phase factor has modulus 1. Position does connect ψ₁ to ψ₂, with x₁₂ = −16L/(9π²), so in (ψ₁ + ψ₂)/√2 the mean swings as L/2 − 0.180L cos ωt with ω = (E₂ − E₁)/ħ: a period of 3.67 fs for an electron in a 1.00 nm well. Position does not connect ψ₁ to ψ₃, so in (ψ₁ + ψ₃)/√2 the density changes while ⟨x⟩ stays at L/2. The limiting cases hold: one term gives no motion, and for neighbouring levels at large n the beat frequency approaches the classical bounce frequency. The limit of the picture is interpretive. ⟨x⟩(t) is an average over many identically prepared systems, not the path of one particle, and the σₓσₚ ≥ ħ/2 floor that accompanies it is fixed by [x̂, p̂] = iħ acting on the state, not by any probe disturbing the particle.
2In one sentence
An expectation value in a superposition is the average of one quantity over many identically prepared systems, and it changes in time only through cross terms that join two occupied levels of different energy.
Example. For an electron in a 1.00 nm well in (ψ₁ + ψ₂)/√2, ⟨x⟩ swings between 0.320 nm and 0.680 nm with period 3.67 fs, while ⟨H⟩ stays at 0.940 eV, the mean of E₁ = 0.376 eV and E₂ = 1.504 eV.
3Diagonal terms lose their phase, cross terms keep it
Put Ψ(x, t) = Σ cₙψₙ e(−iEₙt/ħ) into ⟨Q⟩ = ∫ Ψ* Q̂ Ψ dx. The double sum Σₘ Σₙ cₘ* cₙ Qₘₙ e(iωₘₙt), with ωₘₙ = (Eₘ − Eₙ)/ħ, sorts itself at once. A term with m = n carries e(iEₙt/ħ) e(−iEₙt/ħ) = 1, so it is frozen at |cₙ|² Qₙₙ. A term with m ≠ n keeps a phase turning at the Bohr angular frequency ωₘₙ, and together with its partner n, m it makes the real quantity 2 Re[cₘ* cₙ Qₘₙ e(iωₘₙt)]. So every change in every average comes from cross terms, and a single stationary state, which has none, has constant expectation values: ψ₁ alone keeps ⟨x⟩ = L/2 and ⟨p⟩ = 0 forever. Only differences of energy appear, so adding a constant V₀ to the potential multiplies Ψ by the global phase e(−iV₀t/ħ), which shifts every energy by V₀ but leaves every probability density and every Bohr frequency unchanged. And (E/ħ)t is J ÷ (J s) × s, dimensionless, as an exponent must be.
4The mean position of ψ₁ + ψ₂
Take Ψ(x,0) = c₁ψ₁ + c₂ψ₂ with |c₁|² + |c₂|² = 1. The diagonal terms give |c₁|²(L/2) + |c₂|²(L/2) = L/2, because each sine density is symmetric about the centre. The cross term needs one integral over the well, x₁₂ = (2/L) ∫₀ᴸ x sin(πx/L) sin(2πx/L) dx = −16L/(9π²) = −0.180L, negative because ψ₁ψ₂ is positive on the left half and negative on the right. With φ = arg c₂ − arg c₁ this gives ⟨x⟩(t) = L/2 + 2|c₁c₂| x₁₂ cos(ωt − φ) and ω = (E₂ − E₁)/ħ = 3π²ħ/(2mL²). For equal real weights ⟨x⟩ = L/2 − 0.180L cos ωt: the mean starts at 0.320L, where the two sines reinforce, and reaches 0.680L half a period later. For an electron in a 1.00 nm well, E₁ = 0.376 eV, E₂ − E₁ = 1.128 eV, ω = 1.71 × 10¹⁵ rad s⁻¹ and the period is 3.67 fs. The checks pass: x₁₂ is a length, the swing stays well inside the walls, and c₂ → 0 removes the swing and returns the stationary L/2.
5Symmetry decides which averages can move
A cross term moves ⟨Q⟩ only if Qₘₙ ≠ 0, and in the well a symmetry settles that before any integral. About the centre x = L/2, ψₙ is even for odd n and odd for even n, while x − L/2 is odd. So xₘₙ with m ≠ n survives only when ψₘψₙ is odd, which means m + n odd, and then it equals −8Lmn/(π²(m² − n²)²). Prepare (ψ₁ + ψ₃)/√2 and the density still changes at 8E₁/ħ, but symmetrically: σₓ breathes between 0.131L and 0.305L while ⟨x⟩ never leaves L/2 and ⟨p⟩ never leaves zero. Energy obeys a stronger rule. Ĥψₙ = Eₙψₙ and orthogonality make every Hₘₙ with m ≠ n vanish, so ⟨H⟩ = Σ |cₙ|² Eₙ for any state. Inside the well V = 0, so ⟨p²⟩ = 2m⟨H⟩ is frozen with it. In (ψ₁ + ψ₂)/√2 that means ⟨p²⟩ is constant while ⟨p⟩ swings, so σₚ² = ⟨p²⟩ − ⟨p⟩² must dip each time the mean momentum peaks.
6Ehrenfest holds, and the beat becomes the bounce
Two checks tie the moving mean to classical mechanics. First Ehrenfest: multiply d⟨x⟩/dt by m to get 16mLω sin ωt/(9π²) = (8ħ/3L) sin ωt. Computing ⟨p⟩ directly instead needs p₁₂ = −iħ ∫₀ᴸ ψ₁ ψ₂′ dx = 8iħ/(3L), and 2 Re[c₁* c₂ p₁₂ e(−iωt)] returns the same (8ħ/3L) sin ωt, so m d⟨x⟩/dt = ⟨p⟩ holds term by term. Second, the frequency. A classical particle of energy Eₙ moves at nπħ/(mL) and covers the round trip 2L in 2mL²/(nπħ), an angular frequency nπ²ħ/(mL²). Superposing the neighbours n and n + 1 gives (Eₙ₊₁ − Eₙ)/ħ = (n + ½)π²ħ/(mL²), so quantum over classical is (n + ½)/n: 1.50 for the 1–2 pair, 1.05 by n = 10 and 1.005 by n = 100. The swing amplitude |xₙ,ₙ₊₁| climbs toward 2L/π² = 0.203L at the same time. At low n the beat is fixed by the discrete ladder; only at large n does it settle onto the classical bounce, as the correspondence principle demands.
7An average over an ensemble, not the path of one particle
⟨x⟩(t) is measured the only way an expectation value can be: prepare many systems identically, measure x once on each at the same time t, and average. At t = 0 in the 1.00 nm well that average is 0.320 nm, yet each single result may lie anywhere in the well, with spread σₓ = 0.139 nm, comparable to the 0.180 nm swing itself. Measuring the same particle again a little later does not trace the curve, because the first result has already replaced Ψ with a new state. The swing also needs both amplitudes at once. A collection of systems, half in ψ₁ and half in ψ₂, gives the same energy statistics, 50% E₁ and 50% E₂, but its mean position sits at L/2 for all time, since a mixture has no cross term. Measure the energy of every system in the superposition and the ensemble becomes exactly that mixture, and ⟨x⟩ stops moving. The oscillation is interference between ψ₁ and ψ₂, not the motion of a particle that is secretly in one level.
8The ħ/2 floor is a property of the state
In the ground state σₓ = 0.181L and σₚ = πħ/L, so σₓσₚ = 0.568ħ; in ψ₂ it is 1.670ħ, and in general (ħ/2)√(n²π²/3 − 2), independent of L and m because shrinking the well narrows σₓ exactly as it widens σₚ. Nothing in that calculation touches an apparatus. Both spreads are integrals over the same ψ, and the floor they cannot beat comes from the commutator: for any state σₓσₚ ≥ ½|⟨[x̂, p̂]⟩|, and [x̂, p̂] = iħ makes the right side ħ/2. The experiment that makes this concrete never measures one system twice. Measure x on half of an ensemble and p on the other half, and the two histograms have exactly these widths, so no gentler probe could narrow them, since no system was measured twice. What the floor forbids is a state with both distributions narrow at once. In a superposition the product itself moves, because σₓ and σₚ both depend on t, yet it stays above ħ/2 at every instant, as the figure's readout shows.
Key equations
What each relation says, and the units it is written in.
Any expectation value in a superposition
⟨Q⟩(t) = Σₘ Σₙ cₘ* cₙ Qₘₙ e(iωₘₙt)
Terms with m = n lose their phase and freeze. Anything that moves is an off-diagonal term, turning at a Bohr frequency.
Qₘₙ = ∫ ψₘ* Q̂ ψₙ dx, taken over 0 ≤ x ≤ L in the well, carries the units of Q. ωₘₙ = (Eₘ − Eₙ)/ħ in rad s⁻¹, with Eₙ in J and ħ in J s; cₙ is dimensionless.
Mean position of c₁ψ₁ + c₂ψ₂ in the well
⟨x⟩(t) = L/2 + 2|c₁c₂| x₁₂ cos(ωt − φ)
Equal weights give the largest swing, ±0.180L about the centre. The phase φ changes when the swing happens, not how far it goes.
x₁₂ = −16L/(9π²) = −0.180L, in m. φ = arg c₂ − arg c₁ in rad, and ω in rad s⁻¹.
Beat angular frequency in the well
ω = (E₂ − E₁)/ħ = 3π²ħ/(2mL²)
Only an energy difference enters, so shifting every level by a constant leaves the motion unchanged.
m in kg, L in m, ω in rad s⁻¹. The period 2π/ω = 4mL²/(3πħ) is 3.67 fs for an electron in a 1.00 nm well.
Position matrix elements of the well
xₘₙ = −8Lmn/(π²(m² − n²)²) for m + n odd, and xₘₙ = 0 for m + n even, with m ≠ n
A selection rule: in (ψ₁ + ψ₃)/√2 the density still changes, but symmetrically, and ⟨x⟩ stays at L/2.
L in m, so xₘₙ is in m. x₁₂ = −0.1801L, x₂₃ = −0.1945L, and x₁₃ = 0 because ψ₁ψ₃ is even about L/2.
Mean energy of any superposition
⟨H⟩ = Σ |cₙ|² Eₙ
Each |cₙ|² is frozen for the same reason, so energy statistics never change between preparation and measurement.
Eₙ in J, |cₙ|² dimensionless with Σ |cₙ|² = 1. No t appears, because Ĥψₙ = Eₙψₙ and orthogonality remove every cross term.
Uncertainty product in the nth stationary state
σₓσₚ = (ħ/2)√(n²π²/3 − 2)
The floor ħ/2 = ½|⟨[x̂, p̂]⟩| comes from [x̂, p̂] = iħ applied to the state, with no apparatus in the calculation.
σₓ in m, σₚ in kg m s⁻¹, product in J s. n = 1 gives 0.568ħ and n = 2 gives 1.670ħ, whatever L and m are.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. At w = 0.50 drag t through one period: the density sloshes and ⟨x⟩ swings from 0.32L to 0.68L while the bars and ⟨H⟩ stay put. Set φ to 90° and the whole swing slides a quarter period later; push w to 0 or 1 and every curve freezes.
Live results
- ⟨x⟩ / L
- 0.320
- ⟨p⟩
- 0.00 ħ/L
- ⟨H⟩
- 2.50 E₁
- σₓσₚ
- 0.689 ħ
Catch the common trap
The mistake this lesson is built to prevent.
The tempting reading is that the particle sloshes back and forth and ⟨x⟩(t) is its path, so that watching one particle would trace the curve. It would not. In (ψ₁ + ψ₂)/√2 the mean stays between 0.320L and 0.680L, yet single position measurements land anywhere in the well, scattered about the mean by a σₓ that runs from 0.139L to 0.227L during each beat. The first measurement also replaces Ψ with a new state, so a second look at the same particle samples a different distribution, not the same curve a moment later. And a half-and-half mixture of ψ₁ and ψ₂, whose members each sit in one definite level, has ⟨x⟩ = L/2 at every time. The swing exists only in averages over many fresh systems, each measured once at time t, and it comes from two amplitudes interfering, not from anything travelling along the curve.
Test it: what changes?
- ⟨x⟩ swing grows to ±0.180L about L/2
- ⟨H⟩ = (1 + 3w)E₁ rises
- Swing period 2π/ω unchanged
- x₁₃ = 0, so ⟨x⟩ stays at L/2
- ⟨p⟩ stays at zero
- Density still changes at 8E₁/ħ
- ⟨H⟩ stays at 5E₁
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron in an infinite well of width 1.50 nm is prepared in Ψ(x,0) = (2ψ₁ + iψ₂)/√5. Find ⟨H⟩ in eV, the period of ⟨x⟩(t) in femtoseconds, the range of positions that ⟨x⟩ sweeps through, and which way ⟨x⟩ moves first.
View solution →
- Normalisation first: |c₁|² + |c₂|² = 4/5 + 1/5 = 1, so the weights are probabilities as they stand. Ground energy: E₁ = π²ħ²/(2mL²) = π²(1.05457 × 10⁻³⁴ J s)² ÷ (2 × 9.1094 × 10⁻³¹ kg × (1.50 × 10⁻⁹ m)²) = 2.678 × 10⁻²⁰ J = 0.1671 eV, so E₂ = 4E₁ = 0.6685 eV.
- The mean energy has no cross term: ⟨H⟩ = (4/5)E₁ + (1/5)E₂ = 1.6E₁ = 0.267 eV at every t. No single energy measurement returns 0.267 eV; it is the mean of the outcome 0.1671 eV, found with probability 0.8, and 0.6685 eV, found with probability 0.2.
- The cross term turns at ω = (E₂ − E₁)/ħ whatever the weights and phase. With E₂ − E₁ = 3E₁ = 0.5014 eV the period is T = 2πħ/(E₂ − E₁) = h/(E₂ − E₁) = (4.136 × 10⁻¹⁵ eV s) ÷ (0.5014 eV) = 8.25 × 10⁻¹⁵ s.
- Here c₁ = 2/√5 and c₂ = i/√5, so 2|c₁c₂| = 4/5 = 0.800 and φ = arg c₂ − arg c₁ = π/2. Then ⟨x⟩ = L/2 + 0.800 x₁₂ cos(ωt − π/2) = L/2 − 0.800 × 16L/(9π²) × sin ωt. With 16L/(9π²) = 0.1801 × 1.50 nm = 0.2702 nm the amplitude is 0.800 × 0.2702 nm = 0.216 nm, so ⟨x⟩ runs from 0.750 − 0.216 = 0.534 nm to 0.750 + 0.216 = 0.966 nm.
- At t = 0 the sine is zero, so ⟨x⟩ starts at the centre, 0.750 nm, and then falls, reaching 0.534 nm at T/4 = 2.06 fs: the mean moves first toward x = 0. With c₂ real it would have started at 0.534 nm; the factor i delays the swing by a quarter period without changing its size. Checks: the swing is 0.800 of the equal-weight 0.270 nm and stays inside the well, and T scales as L², so the 1.00 nm well's 3.666 fs times (1.50)² = 2.25 gives 8.25 fs again.
Answer. ⟨H⟩ = 0.267 eV, constant in time; ⟨x⟩ oscillates with period 8.25 fs between 0.534 nm and 0.966 nm, starting at the centre and moving first toward x = 0.
MediumExample 2
An electron in a 1.20 nm infinite well is prepared in Ψ(x,0) = (ψ₁ + ψ₂ + ψ₃)/√3. Which angular frequencies appear in ⟨x⟩(t), and with what amplitudes? After what time does the whole curve ⟨x⟩(t) repeat, and what is ⟨H⟩?
View solution →
- Three levels give three Bohr frequencies: (E₂ − E₁)/ħ = 3E₁/ħ, (E₃ − E₂)/ħ = 5E₁/ħ and (E₃ − E₁)/ħ = 8E₁/ħ. Each appears in ⟨x⟩ only if the matching xₘₙ is nonzero.
- Parity decides: 1 + 2 and 2 + 3 are odd, so x₁₂ = −16L/(9π²) = −0.1801L and x₂₃ = −48L/(25π²) = −0.1945L, but 1 + 3 is even, so x₁₃ = 0. The 8E₁/ħ beat is present in |Ψ|² and absent from ⟨x⟩.
- Each surviving term is 2|cₘcₙ| xₘₙ cos(ωₘₙt) with |cₘcₙ| = 1/3. The amplitudes are (2/3)(0.1801)(1.20 nm) = 0.144 nm at 3E₁/ħ and (2/3)(0.1945)(1.20 nm) = 0.156 nm at 5E₁/ħ, and ⟨x⟩ starts at 0.600 − 0.144 − 0.156 = 0.300 nm.
- Both frequencies are whole multiples of E₁/ħ, and 3 and 5 share no factor, so the curve first repeats after 2πħ/E₁ = h/E₁. Here E₁ = 0.3760 eV ÷ (1.20)² = 0.2611 eV, so h/E₁ = (4.136 × 10⁻¹⁵ eV s) ÷ (0.2611 eV) = 1.58 × 10⁻¹⁴ s.
- The energy average has no cross terms: ⟨H⟩ = (E₁ + 4E₁ + 9E₁)/3 = 14E₁/3 = 1.22 eV, the same at every t while ⟨x⟩ wanders.
Answer. ⟨x⟩ = L/2 − (0.144 nm) cos(3E₁t/ħ) − (0.156 nm) cos(5E₁t/ħ), with no 8E₁/ħ term; the curve repeats every 15.8 fs, and ⟨H⟩ = 1.22 eV.
HardExample 3
For Ψ(x,0) = (ψ₁ + ψ₂)/√2 in a well of width L, find σₓσₚ at t = 0 and at t = T/4, where T = 2πħ/(E₂ − E₁). Confirm that ⟨p⟩ at T/4 equals m d⟨x⟩/dt, and that both products respect ħ/2.
View solution →
- Momentum squared is frozen: inside the well V = 0, so ⟨p²⟩ = 2m⟨H⟩ = 2m(E₁ + E₂)/2 = 5π²ħ²/(2L²) = 24.674ħ²/L² at every t.
- Position moments need matrix elements. (x²)ₙₙ = L²(1/3 − 1/(2n²π²)) gives (x²)₁₁ = 0.2827L² and (x²)₂₂ = 0.3207L², and the cross element is (x²)₁₂ = −16L²/(9π²) = −0.1801L². So ⟨x²⟩ = 0.30167L² − 0.18013L² cos ωt, alongside ⟨x⟩ = 0.5L − 0.18013L cos ωt.
- At t = 0: ⟨x⟩ = 0.31987L and ⟨x²⟩ = 0.12154L², so σₓ² = (0.12154 − 0.31987²)L² = 0.019223L² and σₓ = 0.13865L. ⟨p⟩ = 0, so σₚ = √24.674 ħ/L = 4.9673ħ/L, and σₓσₚ = 0.13865 × 4.9673 ħ = 0.689ħ.
- At t = T/4, ωt = π/2: ⟨x⟩ = 0.5L and ⟨x²⟩ = 0.30167L², so σₓ = √(0.30167 − 0.25) L = 0.22731L. Here m d⟨x⟩/dt = 16mLω sin ωt/(9π²) = 8ħ/(3L), matching the direct 2 Re[c₁* c₂ p₁₂ e(−iωt)] with p₁₂ = 8iħ/(3L).
- Then σₚ² = (24.674 − (8/3)²)ħ²/L² = (24.674 − 7.111)ħ²/L² = 17.563ħ²/L², so σₚ = 4.1908ħ/L and σₓσₚ = 0.22731 × 4.1908 ħ = 0.953ħ. Both products exceed 0.500ħ, and the product is back to 0.689ħ at T/2, so it oscillates twice per beat.
Answer. σₓσₚ = 0.689ħ at t = 0 and 0.953ħ at t = T/4, both above ħ/2; ⟨p⟩ at T/4 is 8ħ/(3L), equal to m d⟨x⟩/dt.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
An average moves only through cross terms with Qₘₙ ≠ 0.
Expand ⟨Q⟩(t) = Σ cₘ* cₙ Qₘₙ e(iωₘₙt): diagonal terms freeze, cross terms turn at ωₘₙ. ⟨H⟩ never moves, and in the well ⟨x⟩ moves only through pairs with m + n odd. Read ⟨x⟩(t) as an average over many fresh systems, each measured once, not as one particle's path.
Next: The finite square well: three regions and two energy sectors →