University Physics IV · Quantum Potentials · 7.3
Superpositions & Time Evolution in a Box
The box's levels are an alphabet. This is where you learn to spell an arbitrary state in it, read the measurement statistics straight off the spelling, and then let it run — one phase per letter, each turning at its own rate, with every visible motion hiding in the cross terms.
Build the model
Connect the measurement to the mechanism.
The infinite well hands you a complete orthonormal alphabet: ψₙ(x) = √(2/L) sin(nπx/L), with Eₙ = n²E₁. Because the walls quantise k, that alphabet is countable, so any state the box can hold is a sum — a Fourier sine series with coefficients cₙ = ∫₀L ψₙΨ(x,0) dx — and not the integral a free particle would need. Two rules then finish the model.
Born's rule reads |cₙ|² as the probability that an energy measurement returns Eₙ, so the coefficients fix the statistics once and for all. And the Schrodinger equation, linear with a time-independent H, evolves each term by nothing more than a phase: Ψ(x, t) = Σ cₙψₙ(x) e(−iEₙt/ħ). The moduli never move, so P(Eₙ) and ⟨H⟩ are constants of the motion; what moves is the relative phase between terms, and it moves only because the levels differ in energy.
Expand |Ψ|² and every time-dependent piece is a cross term beating at ωₘₙ = (Eₘ − Eₙ)/ħ. That is the cost as much as the content: the model is closed and unitary, so nothing inside it can make an excited amplitude decay, give a level a width, or single out the ground state. Decay needs a coupling this box does not have.
- Simple definition
- A superposition in a box is a normalised sum Ψ = Σ cₙψₙ over the well's energy eigenfunctions, in which |cₙ|² is the probability that an energy measurement returns Eₙ and each term evolves by its own phase factor e(−iEₙt/ħ).
- Example
- For an electron in a 1.00 nm box, Ψ = (1/√5)ψ₁ + (2/√5)ψ₂ gives P(E₁) = 0.20 and P(E₂) = 0.80 at every instant, while ⟨x⟩ swings between 0.356 nm and 0.644 nm with a period of 3.67 fs.
Orthonormality is the tool that pulls a single coefficient out of an infinite sum.
n = 1, 2, 3, …; E₁ = π²ħ²/2mL² in J; ∫₀L ψₘψₙ dx = δₘₙ
One number per level: a discrete sine series on (0, L), never a Fourier integral.
Ψ and ψₙ both carry m⁻¹⁄², so cₙ is a pure number; the sum rule is just normalisation.
A single eigenstate factors out a global phase and freezes; a sum of two cannot.
ωₙ = Eₙ/ħ in rad s⁻¹. |cₙ| is untouched, so P(Eₙ) = |cₙ|² is fixed for all t.
Set at t = 0 and conserved, because the box's H carries no time dependence.
⟨H⟩ in J or eV. It is a mean over many identically prepared copies, not a member of the spectrum.
Delete the last term and you have the statistical mixture, which never moves.
ω₂₁ = (E₂ − E₁)/ħ = 3E₁/ħ; in general ωₘₙ = (m² − n²)E₁/ħ, in rad s⁻¹.
Parity switches half the pairs off; Eₙ ∝ n² brings every phase home together.
⟨1|x|2⟩ = −16L/9π² = −0.180L; Tᵣₑᵥ = 4mL²/πħ = 11.0 fs for an electron in 1.00 nm.
Why a discrete sum, not a Fourier integral
The walls force ψ(0) = ψ(L) = 0, which quantises k = nπ/L. The resulting set √(2/L) sin(nπx/L) is complete on the interval (0, L): every square-integrable function that vanishes at both ends can be written as Σ cₙψₙ, which is exactly the Fourier sine series you met in mathematics. A free particle has a continuum of k and needs Ψ(x,0) = ∫ φ(k)e(ikx) dk/√(2π), an integral over an amplitude density. The box replaces that integral with a sum, and replaces φ(k) with a list of numbers c₁, c₂, c₃, and so on. The difference is not cosmetic. It is why the box has levels you can label and count, why an energy measurement has a discrete menu of possible answers rather than a probability density over a continuum, and why the whole of this lesson is bookkeeping over an index n rather than calculus over a variable k.
Orthonormality extracts one coefficient at a time
To find cₘ, multiply Ψ(x,0) = Σₙ cₙψₙ by ψₘ and integrate across the well. Orthonormality, ∫₀L ψₘψₙ dx = δₘₙ, kills every term but one and leaves cₘ = ∫₀L ψₘ(x)Ψ(x,0) dx — the overlap of the state with that level. (In general it is ψₘ* under the integral; the box's real sines make the conjugate invisible here, but it will not stay invisible in later units.) Two shortcuts save most of the labour. Parity: about the midpoint x = L/2, ψₙ is even for odd n and odd for even n, so a state symmetric about the centre has cₙ = 0 for every even n before you integrate anything at all. And Σ|cₙ|² = 1 is not an extra condition to impose; it is the statement that Ψ was normalised, so use it as an arithmetic check on the coefficients you just computed.
The coefficients are the measurement statistics
Born's rule for a discrete spectrum reads P(Eₙ) = |cₙ|². Take Ψ = (ψ₁ + 2ψ₂ + 3ψ₃)/√14: the probabilities are 1/14 = 0.071, 4/14 = 0.286 and 9/14 = 0.643, and they sum to 1. The mean energy is ⟨H⟩ = Σ|cₙ|²Eₙ = E₁(1 + 16 + 81)/14 = 7E₁, which for an electron in a 1.00 nm box is 7 × 0.376 eV = 2.63 eV. Now notice what 7E₁ is not. The spectrum is E₁, 4E₁, 9E₁, 16E₁, …, and 7 is not a perfect square, so no measurement ever returns 2.63 eV. ⟨H⟩ is the average over many identically prepared copies, and the spread about it, √(⟨H²⟩ − ⟨H⟩²), is a genuine physical width of the prepared state — not an instrumental uncertainty, and not something a better apparatus can reduce.
One phase per level, not one phase overall
Because H is time-independent and the ψₙ are its eigenfunctions, the full solution is Ψ(x, t) = Σ cₙψₙ(x) e(−iEₙt/ħ). Each coefficient becomes cₙ(t) = cₙ e(−iEₙt/ħ), and since |e(−iEₙt/ħ)| = 1 the modulus is untouched: the energy statistics and ⟨H⟩ are frozen for ever. What is not frozen is the set of phases, which turn at different rates ωₙ = Eₙ/ħ. Only their differences are observable. If the state is a single eigenfunction, the one phase multiplies the whole wavefunction, cancels in |Ψ|², and nothing whatever happens — that is precisely what the word stationary means. Put two levels in and the relative phase ω₂₁t = (E₂ − E₁)t/ħ advances at 1.71 × 10¹⁵ rad s⁻¹ for the 1.00 nm electron. That relative phase is the clock the state runs on.
The cross terms are where the motion lives
Square the two-term state: |Ψ(x, t)|² = c₁²ψ₁² + c₂²ψ₂² + 2c₁c₂ψ₁ψ₂ cos(ω₂₁t). The first two pieces are exactly the statistical mixture — the density you would get from an ensemble that is 20% in ψ₁ and 80% in ψ₂ — and they never move. Everything that moves is the third piece. Feed the whole thing into ⟨x⟩: the diagonal terms each contribute L/2, and only the off-diagonal matrix element survives, giving ⟨x⟩(t) = L/2 + 2c₁c₂⟨1|x|2⟩cos(ω₂₁t) with ⟨1|x|2⟩ = −16L/9π² = −0.1801L. For c₁ = 1/√5 and c₂ = 2/√5 that is ⟨x⟩ = L(0.500 − 0.144 cos ω₂₁t): the packet sloshes between 0.356L and 0.644L every 3.67 fs. Change the pair and parity can switch the motion off entirely, since ⟨n|x|m⟩ = 0 whenever n + m is even — a ψ₁ + ψ₃ state breathes but never slides.
Revivals, and what a closed box will not do
Eₙ = n²E₁ with n² an integer, so at Tᵣₑᵥ = 2πħ/E₁ = 4mL²/πħ every phase has turned through a whole number of cycles and the state returns exactly: 11.0 fs for an electron in a 1.00 nm box. Nothing dephases permanently, because the box has no dispersion to lose coherence to. Halfway there, at Tᵣₑᵥ/2, the phases are (−1)ⁿ and the density is the exact mirror image about x = L/2. This is a peculiarity of the hard-wall spectrum; in a finite well, or an atom, the levels are not proportional to n² and only partial revivals survive. The same closedness forbids decay. Unitary evolution conserves every |cₙ|², so no amplitude drains from ψ₂ into ψ₁, no line acquires a width, and no level is preferred. Spontaneous emission, linewidths and thermalisation all arrive later, through a coupling this Hamiltonian does not contain.
Change one variable at a time
Make the relationship visible.
Set n = 2 and drag t from 0 to 0.5: the solid curve sloshes from the left wall to the right while the dashed mixture never budges. Now set n = 3 — the density still breathes, but the dot stays pinned at the centre, because ⟨1|x|3⟩ vanishes by parity.
P(E₁) = |c₁|²0.250
P(Eₙ) = |cₙ|²0.750
MEAN ENERGY ⟨H⟩3.25 E₁
BEAT PERIOD, 1 nm e⁻ box3.67 fs
Live interpretationP(E₁) = |c₁|²: 0.250. P(Eₙ) = |cₙ|²: 0.750. MEAN ENERGY ⟨H⟩: 3.25 E₁. BEAT PERIOD, 1 nm e⁻ box: 3.67 fs
Catch the common trap
Explain before calculating.
An electron in a rigid box of width 1.00 nm is prepared in Ψ(x,0) = (1/√5)ψ₁ + (2/√5)ψ₂ and then left completely alone. Take E₁ = 0.376 eV. What is true 5.0 fs later, a time that is not a whole number of beat periods?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron sits in a rigid one-dimensional box of width L = 1.00 nm, where E₁ = 0.376 eV. At t = 0 it is prepared in the unnormalised state Ψ(x,0) ∝ ψ₁ + 2ψ₂ + 3ψ₃. Normalise it, list the possible outcomes of an energy measurement with their probabilities, and find ⟨H⟩.
- Normalise. Write Ψ = A(ψ₁ + 2ψ₂ + 3ψ₃). Orthonormality collapses the cross terms, so ∫|Ψ|² dx = A²(1² + 2² + 3²) = 14A² = 1, giving A = 1/√14 and c₁ = 1/√14, c₂ = 2/√14, c₃ = 3/√14.
- Read the statistics straight off the coefficients: P(Eₙ) = |cₙ|², so P(E₁) = 1/14 = 0.071, P(E₂) = 4/14 = 0.286 and P(E₃) = 9/14 = 0.643. They sum to 1, which is the normalisation check.
- Use the box spectrum Eₙ = n²E₁: E₁ = 0.376 eV, E₂ = 4E₁ = 1.504 eV, E₃ = 9E₁ = 3.384 eV. Those three numbers are the only readings the meter can ever give.
- Average them: ⟨H⟩ = Σ|cₙ|²Eₙ = E₁(1×1 + 4×4 + 9×9)/14 = 98E₁/14 = 7E₁ = 2.63 eV. Note that 7E₁ is not a level, since n² = 7 has no integer solution, so no single measurement returns it.
- Every one of these numbers is time-independent. Evolution multiplies each cₙ by e(−iEₙt/ħ), which changes no modulus, so the probabilities and ⟨H⟩ are the same at t = 0 and at t = 1 s.
AnswerP(E₁) = 0.071 at 0.376 eV, P(E₂) = 0.286 at 1.504 eV, P(E₃) = 0.643 at 3.384 eV. ⟨H⟩ = 7E₁ = 2.63 eV — a value the spectrum never contains, and a number that never changes.
MediumThe same electron (L = 1.00 nm, E₁ = 0.376 eV) is prepared in Ψ(x,0) = (1/√5)ψ₁ + (2/√5)ψ₂. Find ⟨H⟩, the angular frequency and period of the beat in the probability density, and the range over which ⟨x⟩ oscillates. Use ⟨1|x|2⟩ = −16L/9π² and ħ = 1.055 × 10⁻³⁴ J s.
- Statistics first: |c₁|² = 1/5 = 0.20 and |c₂|² = 4/5 = 0.80, so ⟨H⟩ = 0.20E₁ + 0.80(4E₁) = 3.4E₁ = 3.4 × 0.376 = 1.28 eV. Both the probabilities and ⟨H⟩ are fixed for all time.
- The density is |Ψ(x, t)|² = c₁²ψ₁² + c₂²ψ₂² + 2c₁c₂ψ₁ψ₂ cos(ω₂₁t), with ω₂₁ = (E₂ − E₁)/ħ = 3E₁/ħ. Only the cross term carries any t at all.
- Numbers: 3E₁ = 1.128 eV = 1.807 × 10⁻¹⁹ J, so ω₂₁ = 1.807 × 10⁻¹⁹ / 1.055 × 10⁻³⁴ = 1.71 × 10¹⁵ rad s⁻¹, and the beat period is T = 2π/ω₂₁ = 3.67 × 10⁻¹⁵ s = 3.67 fs.
- For ⟨x⟩ the two diagonal terms each give L/2 and only the off-diagonal element survives: ⟨x⟩(t) = L/2 + 2c₁c₂⟨1|x|2⟩cos(ω₂₁t), where 2c₁c₂ = 2(1/√5)(2/√5) = 0.800 and ⟨1|x|2⟩ = −16L/9π² = −0.1801L.
- So ⟨x⟩(t) = L(0.500 − 0.144 cos ω₂₁t), swinging between 0.356L and 0.644L — that is 0.356 nm to 0.644 nm, an excursion of 0.288 nm, completed once every 3.67 fs.
Answer⟨H⟩ = 3.4E₁ = 1.28 eV, constant. ω₂₁ = 1.71 × 10¹⁵ rad s⁻¹ and T = 3.67 fs. ⟨x⟩ = L(0.500 − 0.144 cos ω₂₁t), oscillating between 0.356 nm and 0.644 nm while P(E₁) and P(E₂) stay at 0.20 and 0.80.
HardA particle in a box of width L is released from the symmetric triangular state Ψ(x,0) = Ax for 0 ≤ x ≤ L/2 and Ψ(x,0) = A(L − x) for L/2 ≤ x ≤ L. Normalise it, find the cₙ, give the three largest energy probabilities, compute ⟨H⟩ in units of E₁ (and in eV for an electron with L = 1.00 nm, E₁ = 0.376 eV), and say how the density and ⟨x⟩ behave in time.
- Normalise: ∫₀L |Ψ|² dx = 2A²∫₀(L/2) x² dx = 2A²(L³/24) = A²L³/12 = 1, so A = 2√3/L³⁄².
- Symmetry does half the work. Ψ is even about x = L/2 while ψₙ is odd about the midpoint for every even n, so cₙ = 0 for n = 2, 4, 6, … before any integration is done.
- For odd n the overlap integral gives cₙ = ∫₀L ψₙΨ dx = (4√6/n²π²) sin(nπ/2), so c₁ = 4√6/π² = 0.9927, c₃ = −4√6/9π² = −0.1103 and c₅ = 4√6/25π² = 0.0397.
- Probabilities: |c₁|² = 96/π⁴ = 0.9855, |c₃|² = 0.0122, |c₅|² = 0.0016. They already sum to 0.9993, so a tent is almost entirely ground state — the sign of c₃ never appears in a probability.
- Mean energy: ⟨H⟩ = Σ n²|cₙ|²E₁ = (96/π⁴)E₁ × Σ over odd n of 1/n² = (96/π⁴)(π²/8)E₁ = 12E₁/π² = 1.216E₁. For the 1.00 nm electron box that is 1.216 × 0.376 = 0.457 eV. The same number comes out of ⟨H⟩ = (ħ²/2m)∫₀L |dΨ/dx|² dx = 6ħ²/mL².
- In time each term picks up e(−iEₙt/ħ), and with only odd n present every relative phase is (n² − 1)E₁t/ħ, always a multiple of 8E₁t/ħ. So |Ψ|² repeats exactly every Tᵣₑᵥ/8 = 1.37 fs. Meanwhile ⟨n|x|m⟩ = 0 whenever n + m is even, and here it always is: ⟨x⟩ = L/2 for all t. The state breathes; it never slides.
AnswerA = 2√3/L³⁄² and cₙ = (4√6/n²π²)sin(nπ/2), zero for even n. P(E₁) = 0.9855, P(E₃) = 0.0122, P(E₅) = 0.0016. ⟨H⟩ = 12E₁/π² = 1.216E₁ = 0.457 eV. The density breathes with period 1.37 fs while ⟨x⟩ stays pinned at L/2.