Time evolution: each eigenstate carries its own phase
Each coefficient keeps its length and turns its own phase.
Expanding Ψ(x,0) on the sine basis looks like bookkeeping, but it is the whole solution. Once the coefficients are known, the Schrödinger equation hands each one its own clock and never lets two of them interact, so predicting the well at any later time costs one phase factor per term.
Build the model
Expand once at t = 0; after that, time only turns phases.
1The big idea
Write the state at any instant as Ψ(x, t) = Σ cₙ(t)ψₙ(x), which completeness of the sine basis always allows, and substitute it into iħ∂Ψ/∂t = ĤΨ. Each ψₙ is an eigenfunction of Ĥ, so the right side is Σ Eₙcₙψₙ; multiply by ψₘ* and integrate, and orthonormality strips both sums to one term, iħ dcₘ/dt = Eₘcₘ. A partial differential equation in x and t has become one first-order equation per level, none of which mentions any other, and each is solved by cₘ(t) = cₘ(0)e(−iEₘt/ħ). Everything else follows. Every |cₙ| is frozen, so the energy probabilities and the norm never change. Every phase turns at its own rate Eₙ/ħ, so a single factor can be pulled out front only when one term is present; with two or more, |Ψ|² moves at the differences (Eₙ − Eₘ)/ħ. In this well the Eₙ = n²E₁ are whole multiples of E₁, so all the phases realign after h/E₁ and any state returns exactly. The derivation leaned on two things that can fail: eigenfunctions that stay put, which needs V independent of time, and an evolution left undisturbed, which a measurement interrupts by replacing the whole set of cₙ. Between such events the rule is exact; across one it must be restarted from the new state.
2In one sentence
In a time-independent potential, a state written as Σ cₙψₙ evolves by multiplying each coefficient by its own phase factor e(−iEₙt/ħ), so every |cₙ| stays fixed and only the relative phases between the terms change.
Example. An electron in a 1.00 nm well has E₁ = 0.376 eV and E₂ = 1.504 eV, so c₁ turns once every h/E₁ = 11.0 fs and c₂ once every 2.75 fs; their relative phase, the only part |Ψ|² can see, cycles every h/(E₂ − E₁) = 3.67 fs.
3Project the equation onto one eigenfunction at a time
Completeness lets the state at every instant be written Ψ(x, t) = Σ cₙ(t)ψₙ(x), with all the time dependence in the coefficients; the ψₙ themselves are fixed because V is. Substitute into iħ∂Ψ/∂t = ĤΨ. The left side is iħ Σ (dcₙ/dt)ψₙ. On the right Ĥ meets each ψₙ separately and returns Eₙψₙ, giving Σ Eₙcₙψₙ. Multiply both sides by ψₘ* and integrate from 0 to L: since ∫₀ᴸ ψₘ*ψₙ dx is 1 for n = m and 0 otherwise, every term but one disappears and iħ dcₘ/dt = Eₘcₘ. Now expand instead on some other orthonormal set φₖ that fits the walls, polynomials for instance. Ĥφₖ is no longer a multiple of φₖ, and the same projection gives iħ dbₖ/dt = Σⱼ Hₖⱼbⱼ with Hₖⱼ = ∫₀ᴸ φₖ*Ĥφⱼ dx: every coefficient is driven by all the others. The energy eigenfunctions are exactly the set for which Hₖⱼ vanishes off the diagonal, and that, not habit, is why states are expanded on them.
4Each coefficient keeps its length and turns its own phase
iħ dcₙ/dt = Eₙcₙ is exponential change with a purely imaginary rate, so cₙ(t) = cₙ(0)e(−iEₙt/ħ). Check the dimensions first: Eₙt/ħ is J × s ÷ J s, a pure number, as any exponent must be. Then picture each coefficient as the hand of its own dial in the complex plane: its length |cₙ| never changes, and it turns clockwise at Eₙ/ħ radians per second. For an electron in a 1.00 nm well E₁ = 0.376 eV, so c₁ turns at 5.71 × 10¹⁴ rad s⁻¹, once every 11.0 fs; c₂ turns four times as fast and c₃ nine times. Because the lengths are frozen, P(Eₙ) = |cₙ|² is the same at every instant, and so is Σ|cₙ|² = 1: normalisation checked once at t = 0 holds for good. The rates themselves are not observable. Raise the floor of the well by a constant V₀ and every Eₙ grows by V₀, multiplying every term by the same e(−iV₀t/ħ); a factor shared by every term cancels from |Ψ|² and from every probability.
5One phase comes out front only when there is one term
Pull e(−iE₁t/ħ) out of the sum and what is left is Σ cₙψₙe(−iωₙ₁t), with ωₙ₁ = (Eₙ − E₁)/ħ: the common factor is harmless, but the relative phases stay inside, and they are what |Ψ|² can see. With one term no relative phase is left, Ψ is ψₙ times a phase, and the density is frozen: that is a stationary state. With two, |Ψ|² gains a cross term 2|c₁c₂|ψ₁ψ₂cos(ω₂₁t − φ), φ being the starting relative phase, and it moves. Take (ψ₁ + ψ₂)/√2 for the electron in a 1.00 nm well. The two sines add on the left and largely cancel on the right, putting 92.4% of the probability in the left half. At t = h/(2(E₂ − E₁)) = 1.83 fs the relative phase is π, the state is proportional to ψ₁ − ψ₂, and 92.4% sits in the right half, while P(E₁) = P(E₂) = 1/2 throughout. A relative phase set by hand and one gained in time are interchangeable: replacing c₂ by ic₂ prepares, up to an overall phase, the state the original held a quarter of a beat earlier.
6The n² ladder brings every phase back together
Because Eₙ = n²E₁ and every n² is a whole number, the time Tᵣₑᵥ = h/E₁ = 4mL²/(πħ) turns every phase through a whole number of cycles, and Ψ(x, Tᵣₑᵥ) = Ψ(x, 0) for any starting state whatever. For an electron that is 11.0 fs in a 1.00 nm well and 1.10 ps in a 10.0 nm one. Halfway there each phase is e(−iπn²) = (−1)ⁿ, and since ψₙ(L − x) = (−1)ⁿ⁺¹ψₙ(x), the sum becomes −Ψ(L − x, 0): the starting state reflected about the centre, its density exactly mirrored. A particular density can repeat sooner. A state built from ψ₁ and ψ₂ alone repeats its density every h/(E₂ − E₁) = Tᵣₑᵥ/3, because it has only one relative phase to realign. The revival also marks the classical limit: Tᵣₑᵥ grows as mL², so for a 1 μg grain in a 1 cm box it is 1.2 × 10²¹ s, about 2800 times the age of the universe, and nothing macroscopic is ever seen to revive.
7Neighbouring beats become the classical bounce
The density oscillates at the gaps (Eₙ − Eₘ)/ħ, so for a state built from high levels the motion is set by neighbouring gaps, ω = (2n + 1)E₁/ħ. A classical particle of energy Eₙ moves at v = √(2Eₙ/m) = nπħ/(mL) and returns after 2L/v, an angular frequency πv/L = 2nE₁/ħ. The ratio of quantum beat to classical bounce is (2n + 1)/(2n): 1.5 for n = 1, where the correspondence is poor, 1.05 for n = 10 and 1.005 for n = 100. The same algebra shows where the correspondence stops. Write n = n₀ + j, so n² = n₀² + 2n₀j + j². The n₀² part is a common phase; the 2n₀j part is an evenly spaced ladder that repeats every classical bounce; the leftover j² part makes a packet of several levels drift out of step and spread, and then, at Tᵣₑᵥ, puts it back together exactly. A classical cloud with the same spread of speeds would smear out and never reassemble.
8Two conditions: a V fixed in time, and no measurement
The derivation needed Ĥψₙ = Eₙψₙ with ψₙ that never change, so V must not depend on time. One exception is instructive: a uniform V₀(t), the same at every x, shifts every Eₙ equally and adds only the common phase e(−i∫₀ᵗV₀(t′)dt′/ħ); raising the whole floor by 0.50 eV for 2.00 fs turns every term by the same 1.52 rad and changes nothing measurable. A V that changes shape, a wall that moves or a field that tilts the floor, has eigenfunctions that change with it; the projection no longer separates the levels and the |cₙ| begin to change, which is how any transition happens. A measurement breaks the rule differently. If energy is measured and E₂ is found, the set of coefficients is replaced by a single one of modulus 1 and the state is stationary from then on; a position measurement leaves a narrow spike that needs many new cₙ. Either way the rule restarts from the new state and holds exactly until the next interruption.
Key equations
What each relation says, and the units it is written in.
One equation per coefficient
iħ dcₙ/dt = Eₙ cₙ
What the Schrödinger equation becomes once it is projected onto ψₙ: in the energy eigenbasis one partial differential equation splits into uncoupled first-order ones.
cₙ dimensionless; Eₙ in J; ħ = 1.055 × 10⁻³⁴ J s; t in s. No other coefficient appears on the right.
Each term carries its own phase
Ψ(x, t) = Σₙ cₙ ψₙ(x) e(−iEₙt/ħ)
The complete time dependence for a static V: expand once at t = 0, then attach e(−iEₙt/ħ) to the term it belongs to and to no other.
cₙ = ∫₀ᴸ ψₙ* Ψ(x,0) dx, the t = 0 coefficient; ψₙ in m(−1/2); Eₙt/ħ in radians
Moduli are frozen
|cₙ(t)|² = |cₙ(0)|²
The energy statistics and the norm are constants of this evolution; only a change in V or a measurement can alter them.
dimensionless; P(Eₙ) = |cₙ|², so Σₙ|cₙ(t)|² = 1 at every t
Relative phase of level n against the ground state
(Eₙ − E₁)t/ħ = (n² − 1)E₁t/ħ
Only differences of energy reach |Ψ|²: shifting V by a constant adds the same phase to every term and changes nothing measurable.
radians; E₁ = π²ħ²/(2mL²). Electron, L = 1.00 nm: (E₂ − E₁)/ħ = 1.71 × 10¹⁵ rad s⁻¹
Revival time of the infinite well
Tᵣₑᵥ = h/E₁ = 4mL²/(πħ)
Every Eₙ = n²E₁ is a whole multiple of E₁, so after Tᵣₑᵥ every phase has made whole turns and Ψ(x, Tᵣₑᵥ) = Ψ(x, 0) for any starting state.
s; m in kg, L in m. Electron: 11.0 fs for L = 1.00 nm, 1.10 ps for L = 10.0 nm
Mirror image at half the revival time
Ψ(x, Tᵣₑᵥ/2) = −Ψ(L − x, 0)
Halfway to its revival any state is reflected about x = L/2, its density exactly mirrored: a prediction of the phase rule alone that a classical particle does not share.
from e(−iπn²) = (−1)ⁿ and ψₙ(L − x) = (−1)ⁿ⁺¹ψₙ(x); x and L in m
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Set |cₙ|² to 0 or 1 and drag the time: a dial turns, the density never moves. At 0.50 with n = 2, turning 0° to 60° moves the density from 92% in the left half to 92% in the right; 180° mirrors any state and 360° restores it. With n = 3 the density breathes about the centre; each half keeps 0.500.
Live results
- RELATIVE PHASE (Eₙ − E₁)t/ħ
- 60.0 °
- PROBABILITY IN LEFT HALF
- 0.712
- t FOR AN ELECTRON, L = 1.00 nm
- 0.61 fs
- BEAT h/(Eₙ − E₁), SAME WELL
- 3.67 fs
Catch the common trap
The mistake this lesson is built to prevent.
The tempting shortcut is Ψ(x, t) = Ψ(x,0)e(−iEt/ħ) with one E for the whole state, usually E₁ or ⟨H⟩. It does not solve the equation it is meant to solve. For Ψ(x,0) = (ψ₁ + ψ₂)/√2, ĤΨ = (E₁ψ₁ + 4E₁ψ₂)/√2, which is not any single number times Ψ, so no single exponential satisfies iħ∂Ψ/∂t = ĤΨ; ⟨H⟩ = 2.5E₁ is not even an energy that either term has. It also predicts the wrong physics. A factor shared by the whole state cancels from |Ψ|², so every state would be stationary and nothing in the well could ever move, yet for an electron in a 1.00 nm well this one goes from 92.4% of its probability in the left half to 92.4% in the right in 1.83 fs. The phase factor belongs to the eigenfunction it came with and is attached term by term; it comes out front only when one energy is present.
Test it: what changes?
- P(E₁) = P(E₂) = ½ throughout
- Relative phase grows at (E₂ − E₁)/ħ
- Left-half probability swings between 0.924 and 0.076
- Density repeats every h/(E₂ − E₁) = Tᵣₑᵥ/3
- Relative phase turns at 8E₁/ħ
- Density breathes symmetrically about the centre
- Each half keeps probability 0.500
- Density repeats every h/(8E₁)
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron in an infinite well of width L = 0.500 nm is prepared in Ψ(x,0) = 0.6ψ₁ + 0.8iψ₂. Write Ψ(x, t), find the phase each term has turned through at t = 1.00 fs and the relative phase between them, give P(E₁) and P(E₂) at that time, and find |Ψ|² at x = L/4 at t = 0 and at t = 1.00 fs.
View solution →
- Normalisation first: |0.6|² + |0.8i|² = 0.36 + 0.64 = 1, so the coefficients can be read as probability amplitudes as they stand.
- Energies: E₁ = π²ħ²/(2mL²) = π² × 0.0381 eV nm² ÷ (0.500 nm)² = 1.504 eV and E₂ = 4E₁ = 6.016 eV. Attaching each phase to its own term, Ψ(x, t) = 0.6ψ₁e(−iE₁t/ħ) + 0.8iψ₂e(−iE₂t/ħ).
- At t = 1.00 fs, with ħ = 6.582 × 10⁻¹⁶ eV s: E₁t/ħ = 2.285 rad (130.9°) and E₂t/ħ = 9.141 rad, which is 2.857 rad (163.7°) once a whole turn is removed. Dimension check: eV × s ÷ eV s leaves a pure number.
- Relative phase: (E₂ − E₁)t/ħ = 6.855 rad, one full turn plus 0.572 rad (32.8°), so the beat period h/(E₂ − E₁) = 0.917 fs has passed just over once. The moduli have not moved, so P(E₁) = 0.36 and P(E₂) = 0.64 still.
- At x = L/4, ψ₁ = √(1/L) and ψ₂ = √(2/L). Writing ω₂₁ = (E₂ − E₁)/ħ, the cross term is 2Re[(0.6)(0.8i)e(−iω₂₁t)]ψ₁ψ₂ = 0.96 sin(ω₂₁t) × √2/L, so |Ψ(L/4)|² = [1.64 + 1.358 sin(ω₂₁t)]/L: 1.64/L = 3.28 nm⁻¹ at t = 0, and with sin(6.855 rad) = 0.542, 2.375/L = 4.75 nm⁻¹ at 1.00 fs.
Answer. P(E₁) = 0.36 and P(E₂) = 0.64, unchanged; the phases have turned by 2.285 rad and 9.141 rad and the relative phase by 6.855 rad, so |Ψ(L/4)|² has risen from 3.28 nm⁻¹ to 4.75 nm⁻¹.
MediumExample 2
An electron in an infinite well of width L = 1.50 nm starts in Ψ(x,0) = (2ψ₁ + ψ₂ + iψ₄)/√6, with 78.3% of its probability in the left half. Find the revival time, the earliest time at which |Ψ|² repeats, and the earliest time at which it appears mirrored about x = L/2, and give the probabilities of E₁, E₂ and E₄ at those times.
View solution →
- Normalisation: (4 + 1 + 1)/6 = 1. The probabilities are 4/6, 1/6 and 1/6 for E₁, E₂ and E₄, and since every |cₙ| is frozen they keep those values at every time asked for.
- E₁ = π² × 0.0381 eV nm² ÷ (1.50 nm)² = 0.1671 eV, and Tᵣₑᵥ = h/E₁ = 4.136 × 10⁻¹⁵ eV s ÷ 0.1671 eV = 24.7 fs. At Tᵣₑᵥ each phase e(−2πin²) equals 1, so Ψ itself returns exactly.
- The density sees only relative phases: 3E₁t/ħ, 15E₁t/ħ and 12E₁t/ħ for the pairs (1, 2), (1, 4) and (2, 4). Since 15 and 12 are multiples of 3, all three are whole multiples of 2π first when 3E₁t/ħ = 2π: t = Tᵣₑᵥ/3 = 8.25 fs. There Ψ = e(−2πi/3)Ψ(x,0), a common phase only.
- Mirroring: ψ₁ is even about the centre while ψ₂ and ψ₄ are odd, so the mirror image needs both odd terms reversed against ψ₁ and unchanged against each other: 3E₁t/ħ and 15E₁t/ħ odd multiples of π, 12E₁t/ħ a multiple of 2π. All hold first at E₁t/ħ = π/3, t = Tᵣₑᵥ/6 = 4.12 fs, where Ψ ∝ 2ψ₁ − ψ₂ − iψ₄.
- Check with the left half: at t = 0 the ψ₁ψ₂ cross term adds 2 × (2/6) × 4/(3π) = 0.283 to 0.500, giving 0.783, and the iψ₄ terms add nothing. At 4.12 fs that cross term has flipped sign, giving 0.217 = 1 − 0.783, as a mirror image must. The mirror recurs at Tᵣₑᵥ/2 = 12.4 fs, as Ψ(x, Tᵣₑᵥ/2) = −Ψ(L − x, 0) requires.
Answer. Tᵣₑᵥ = 24.7 fs; |Ψ|² first repeats at 8.25 fs and first appears mirrored at 4.12 fs (and again at 12.4 fs), with P(E₁) = 2/3 and P(E₂) = P(E₄) = 1/6 at all of these times.
HardExample 3
An electron in an infinite well of width 4.00 nm is prepared as a packet of levels n = 12 to 18, centred on n = 15, near x = L/4 and moving right. Compare the classical bounce period at E₁₅ with the beat periods of the pairs (14, 15) and (15, 16); find the revival time in bounces; find when the packet's edges slip half a cycle out of step; and say where the packet is at Tᵣₑᵥ/2, compared with a classical electron.
View solution →
- E₁ = π² × 0.0381 eV nm² ÷ (4.00 nm)² = 0.02350 eV, so E₁₅ = 225E₁ = 5.288 eV = 8.47 × 10⁻¹⁹ J. Classically v = √(2E/m) = √(2 × 8.47 × 10⁻¹⁹ J ÷ 9.109 × 10⁻³¹ kg) = 1.364 × 10⁶ m s⁻¹, and one round trip takes 2L/v = 8.00 × 10⁻⁹ m ÷ 1.364 × 10⁶ m s⁻¹ = 5.87 fs.
- Beats: h/(E₁₅ − E₁₄) = h/(29E₁) = 6.07 fs and h/(E₁₆ − E₁₅) = h/(31E₁) = 5.68 fs. The classical period, which is exactly h/(30E₁), sits between them, about 3% from each: this is the correspondence at n = 15.
- Tᵣₑᵥ = h/E₁ = 4.136 × 10⁻¹⁵ eV s ÷ 0.02350 eV = 176 fs, which is 30 bounces, 2n₀ for n₀ = 15.
- Write n = 15 + j, so n² = 225 + 30j + j². The 225 is common to every term, a global phase; the 30j part is an evenly spaced ladder that would repeat the packet every h/(30E₁), the classical bounce. The j² part is left over: for the edges, j = ±3, it reaches π when 9E₁t/ħ = π, at t = h/(18E₁) = 9.78 fs, about 1.7 bounces, after which the packet no longer holds together.
- At Tᵣₑᵥ/2 = 88.0 fs, exactly 15 bounces, the phase rule gives Ψ(x, Tᵣₑᵥ/2) = −Ψ(L − x, 0): a packet at L − L/4 = 3.00 nm, moving left, because reflecting e(ik₀x) about the centre reverses k₀. A classical electron after 15 round trips is back at 1.00 nm moving right, so the packet is half a bounce out of step with it.
Answer. Bounce period 5.87 fs, between the beats of 6.07 fs and 5.68 fs; Tᵣₑᵥ = 176 fs = 30 bounces; the edges dephase after 9.78 fs; at 88.0 fs the packet is at 3.00 nm moving left, where a classical electron would be at 1.00 nm moving right.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Give each term its own phase e(−iEₙt/ħ).
Expand Ψ(x,0) once, then multiply each cₙ by e(−iEₙt/ħ). Every |cₙ|², and so every energy probability, stays fixed; relative phases move |Ψ|² at (Eₙ − Eₘ)/ħ. Valid only while V is fixed in time and nothing measures the system; in this well all states revive at h/E₁.