Stationary states, and superpositions that are not
Eigenstates freeze the statistics; motion lives in their beats
Every motion a quantum particle makes is built out of states that do not move. This lesson pins down exactly what freezes in an energy eigenstate, what starts to slosh the moment a second level is added, how fast it sloshes, and which assumption lets you write any wavefunction as such a sum.
Build the model
Energy eigenstates hold still; motion lives in the beat between two of them.
1The big idea
Separating variables in a static potential gave solutions Ψₙ = ψₙ(x)e(−iEₙt/ħ), whose entire time dependence is one phase turning at Eₙ/ħ. A modulus squared cannot see a phase, so |Ψₙ|² = |ψₙ|² stands still, and so does the expectation value of every operator that does not itself depend on t: ⟨x⟩ is fixed, ⟨p⟩ = 0 and the energy is sharp. That is all stationary means. The particle is not at rest; its statistics are frozen. Because the Schrödinger equation is linear, sums of stationary states are solutions too, and they are not stationary. In |c₁Ψ₁ + c₂Ψ₂|² each absolute phase cancels against its own conjugate, but the cross term keeps e(−i(E₂ − E₁)t/ħ), so the density sloshes at ω₂₁ = (E₂ − E₁)/ħ, a rate set by an energy difference and therefore blind to where the zero of V was put. Orthogonality makes that cross term integrate to zero, so the norm and ⟨H⟩ stay fixed while the density moves, and at large quantum numbers the beat between neighbouring levels approaches the classical orbital frequency. Writing an arbitrary Ψ(x,0) as Σcₙψₙ assumes the ψₙ are complete, which for the infinite well is Fourier's theorem and in general is taken on trust until Unit 3. The picture needs a real E and a potential fixed in time: let V depend on t and there are no stationary states to expand in.
2In one sentence
A stationary state is a solution Ψ(x, t) = ψ(x)e(−iEt/ħ) of definite energy E: the wavefunction keeps turning in phase, but its probability density and the expectation value of every time-independent quantity never change.
Example. An electron in the ground state of a 1.0 nm infinite well has E₁ = 0.376 eV, so its phase turns once every h/E₁ = 11.0 fs, yet its density (2/L)sin²(πx/L) peaks at 2.0 nm⁻¹ at the centre at every instant.
3The phase turns; the density does not
Separation of variables in a static potential produced Ψₙ(x, t) = ψₙ(x)e(−iEₙt/ħ). Form the density: Ψₙ*Ψₙ = ψₙ* e(+iEₙt/ħ) ψₙ e(−iEₙt/ħ) = |ψₙ|², and t has gone. The wavefunction itself has not stopped. With ψₙ real, Re Ψₙ = ψₙ cos(Eₙt/ħ) and Im Ψₙ = −ψₙ sin(Eₙt/ħ): the real part swings through zero twice a turn while the sum of the two squares holds still. For an electron in the ground state of a 1.0 nm infinite well, E₁ = 0.376 eV and the phase turns once every h/E₁ = 11.0 fs. That rate is not observable. Raise V everywhere by a constant V₀ and every Eₙ rises by V₀, every state picks up the same extra factor e(−iV₀t/ħ), and nothing computed from |Ψ|² changes. The cancellation does need E real: an energy E₀ − iΓ/2 would leave |Ψ|² ∝ e(−Γt/ħ), and the norm would drain away.
4Frozen statistics, not a particle at rest
The same cancellation runs through any expectation value. For an operator Q̂ with no explicit t, ⟨Q⟩ = ∫ψₙ* e(+iEₙt/ħ) Q̂ ψₙ e(−iEₙt/ħ) dx; Q̂ acts on x alone, so the phases pass through it and cancel. ⟨x⟩ is therefore constant, and ⟨p⟩ = m d⟨x⟩/dt = 0. The energy is sharp as well: ĤΨₙ = EₙΨₙ gives ⟨H⟩ = Eₙ and ⟨H²⟩ = Eₙ², a spread of exactly zero. None of this says the particle is still. In the 1.0 nm ground state ⟨p²⟩ = 2mE₁ = (πħ/L)², an rms momentum of 3.31 × 10⁻²⁵ kg m s⁻¹, which for an electron corresponds to 3.6 × 10⁵ m s⁻¹. ⟨p⟩ vanishes because motion to the left and to the right carry equal weight, not because nothing moves. What a stationary state freezes is the distribution of results.
5Add a second level and the density sloshes
The Schrödinger equation is linear, so Ψ = c₁ψ₁e(−iE₁t/ħ) + c₂ψ₂e(−iE₂t/ħ) is also a solution, but it is neither separable nor stationary. Multiply it by its conjugate with ψ₁ and ψ₂ real: the squares give the frozen terms |c₁|²ψ₁² and |c₂|²ψ₂², and the two cross products combine into 2|c₁c₂|ψ₁ψ₂cos(ω₂₁t − φ), with ω₂₁ = (E₂ − E₁)/ħ and φ the relative phase of c₂ and c₁. Each absolute frequency Eₙ/ħ has cancelled against its own conjugate and only the difference survives, so a shift V₀ drops out. In the 1.0 nm well E₂ − E₁ = 3E₁ = 1.128 eV, so ω₂₁ = 1.71 × 10¹⁵ rad s⁻¹ and the density repeats every 3.67 fs. Because ψ₁ψ₂ is positive on the left half of the well and negative on the right, equal real coefficients start the lump on the left, and half a period later it sits on the right; φ only sets where in that cycle it starts. Let c₂ → 0 and the cross term vanishes, returning the frozen ground state.
6What the cross term moves, and what it cannot
Integrate the two-level density over the well. The frozen terms give |c₁|² + |c₂|², and the cross term gives 2|c₁c₂|cos(ω₂₁t − φ)∫ψ₁ψ₂ dx = 0 by orthogonality, at every instant. The norm is constant, and |c₁|² + |c₂|² = 1 is the normalisation to check first. The energy behaves the same way: the cross terms of ∫Ψ*ĤΨ dx carry ∫ψ₁ψ₂ dx again, leaving ⟨H⟩ = |c₁|²E₁ + |c₂|²E₂, which is 0.940 eV at all times for equal weights in the 1.0 nm well. Position is different, because ∫xψ₁ψ₂ dx = −16L/(9π²) is not zero: for equal real weights ⟨x⟩ = L/2 − 0.180L cos(ω₂₁t), swinging between 0.320L and 0.680L. Mix levels 1 and 3 instead and both are symmetric about L/2, so ∫xψ₁ψ₃ dx = 0 and ⟨x⟩ never leaves the centre while the density breathes in and out of it. A constant ⟨x⟩ does not certify a stationary state; only a constant density does.
7Completeness is the assumption behind every expansion
Linearity lets you add any finite set of stationary states. The stronger claim, that every admissible Ψ(x,0) can be written as Σcₙψₙ(x), is completeness: a property of the set of ψₙ that the Schrödinger equation does not hand you. For the infinite well the ψₙ are √(2/L)sin(nπx/L), and completeness is Fourier's theorem for sine series on the interval from 0 to L. Granted it, orthonormality ∫ψₘ*ψₙ dx = δₘₙ picks out each coefficient, cₙ = ∫ψₙ*Ψ(x,0) dx, normalisation becomes Σ|cₙ|² = 1, and time evolution costs nothing: attach e(−iEₙt/ħ) to each term. Solving the time-independent equation once solves every initial-value problem in that potential. The recipe stops in three places. A potential that changes in time has no separable solutions to expand in. A free particle has a continuous spectrum, so the sum becomes an integral over states that cannot themselves be normalised (Unit 5). And for a general potential, completeness does not follow from Hermiticity alone: it is proved case by case, as Fourier's theorem proves it here, or else assumed, and Unit 3 examines it.
8Classical motion lives in the beats
A classical electron in the well bounces from wall to wall; no single stationary state shows that, because |ψₙ|² never moves. At large n, |ψₙ|² = (2/L)sin²(nπx/L) oscillates so rapidly that its average over any stretch holding many of its oscillations is 1/L, the uniform density of the classical particle averaged over its motion. The motion itself appears only in superpositions of neighbouring levels. Levels n and n + 1 beat at (2n + 1)π²ħ/(2mL²), while a classical particle with energy Eₙ = n²π²ħ²/(2mL²) has speed nπħ/(mL) and a round-trip angular frequency of nπ²ħ/(mL²). Their ratio is (2n + 1)/(2n): 1.5 at n = 1, 1.05 at n = 10 and 1.005 at n = 100, approaching the classical value as correspondence demands. For the harmonic oscillator of Unit 9 the levels are evenly spaced by ħω₀, so every neighbouring pair beats at exactly the classical frequency ω₀, at every n.
Key equations
What each relation says, and the units it is written in.
Stationary state
Ψₙ(x, t) = ψₙ(x) e(−iEₙt/ħ)
Solve the time-independent equation once and you hold the whole future of that state: a single phase that turns once every h/Eₙ.
ψₙ in m(−1/2) in one dimension; Eₙ in J; ħ = 1.055 × 10⁻³⁴ J s; the exponent Eₙt/ħ is a pure number
The density stands still
|Ψₙ(x, t)|² = |ψₙ(x)|²
Probabilities computed at t = 0 hold for all time. An imaginary part in Eₙ would make |Ψₙ|² grow or decay, so real energies are what keep the density static.
units m⁻¹; e(+iEₙt/ħ) from Ψₙ* cancels e(−iEₙt/ħ) from Ψₙ, which requires Eₙ to be real
Static observables stay constant
⟨Q⟩ = ∫Ψₙ* Q̂ Ψₙ dx = ∫ψₙ* Q̂ ψₙ dx
So ⟨x⟩ is fixed, ⟨p⟩ = m d⟨x⟩/dt = 0, and ĤΨₙ = EₙΨₙ makes ⟨H²⟩ − ⟨H⟩² = 0: the energy is sharp.
Q̂ any operator with no explicit t, such as x, p̂ = −iħ ∂/∂x or Ĥ; ⟨Q⟩ carries the unit of Q
Density of a two-level superposition
|Ψ|² = |c₁|²ψ₁² + |c₂|²ψ₂² + 2|c₁c₂| ψ₁ψ₂ cos(ω₂₁t − φ)
The first two terms are frozen. Only the cross term moves, and because ∫ψ₁ψ₂ dx = 0 it moves probability around without changing the total.
real ψ₁, ψ₂ in m(−1/2); cₙ = |cₙ|e(iφₙ) dimensionless, φ = φ₂ − φ₁; normalised when |c₁|² + |c₂|² = 1
Beat angular frequency
ω₂₁ = (E₂ − E₁)/ħ
Only the spacing appears, so raising V by a constant changes no observable. For an electron in a 1.0 nm well, levels 1 and 2 beat every 3.67 fs.
ω₂₁ in rad s⁻¹, energies in J; the density repeats every h/(E₂ − E₁), which is 4.14 fs for a 1.00 eV spacing
Expansion on stationary states
Ψ(x, t) = Σₙ cₙ ψₙ(x) e(−iEₙt/ħ)
Valid for every admissible Ψ(x,0) only if the ψₙ are complete. Granted that, any initial-value problem reduces to the time-independent equation plus one overlap integral per level.
cₙ = ∫ψₙ* Ψ(x,0) dx, dimensionless; normalisation requires Σ|cₙ|² = 1
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Set w to 0 or 1 and drag t: Re Ψ keeps turning, yet |Ψ|² never moves. At w = 0.5, k = 2 the lump crosses the well and back every 3.67 fs; at k = 3 it only breathes about the centre and ⟨x⟩ holds at 0.500 L. No setting makes ⟨H⟩ depend on t.
Live results
- BEAT PERIOD h/(Eₖ − E₁)
- 3.67 fs
- BEAT ω = (Eₖ − E₁)/ħ
- 1.71 10¹⁵ rad s⁻¹
- ⟨x⟩ / L
- 0.320
- ⟨H⟩, FIXED IN TIME
- 0.940 eV
Catch the common trap
The mistake this lesson is built to prevent.
The tempting shortcut is to add densities: each piece of c₁Ψ₁ + c₂Ψ₂ is stationary, so |Ψ|² must be |c₁|²ψ₁² + |c₂|²ψ₂², which is frozen. That drops the interference term. The modulus squared of a sum is not the sum of the moduli squared: |a + b|² = |a|² + |b|² + 2Re(a*b), and 2Re(a*b) is exactly where e(−i(E₂ − E₁)t/ħ) survives. The error hides well, because the dropped term integrates to zero: a normalisation check passes and ⟨H⟩ comes out right. It fails where it matters. For equal weights of levels 1 and 2 in a 1.0 nm well, the shortcut gives a steady 1.50 nm⁻¹ at x = L/4 and puts ⟨x⟩ at L/2 for ever; the true density there swings between 2.91 nm⁻¹ and 0.086 nm⁻¹ every 3.67 fs, and ⟨x⟩ swings with it.
Test it: what changes?
- ⟨H⟩ rises toward Eₖ
- ⟨H⟩ stays constant in time
- Beat period h/(Eₖ − E₁) unchanged
- Beat period h/(Eₖ − E₁) shortens
- ⟨H⟩ rises
- At k = 3, ⟨x⟩ stays at L/2
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron is in the n = 2 stationary state of an infinite square well of width L = 0.60 nm, with V = 0 inside and Ψ(x,0) = ψ₂(x) real. Find E₂, the time its phase factor takes to turn once, and the first time at which Re Ψ is zero everywhere. Then find the probability of finding the electron in 0 ≤ x ≤ L/4 at t = 0 and at that time.
View solution →
- Ground level: E₁ = π²ħ²/(2mL²) = π²(1.0546 × 10⁻³⁴ J s)² / [2(9.109 × 10⁻³¹ kg)(0.60 × 10⁻⁹ m)²] = 1.674 × 10⁻¹⁹ J = 1.045 eV. Levels go as n², so E₂ = 4E₁ = 4.178 eV.
- The factor e(−iE₂t/ħ) completes a turn when E₂t/ħ = 2π, at t = h/E₂ = (4.136 × 10⁻¹⁵ eV s)/(4.178 eV) = 0.990 fs.
- With ψ₂ real, Re Ψ = ψ₂(x)cos(E₂t/ħ), which is zero at every x once E₂t/ħ = π/2, a quarter turn: t = 0.990 fs/4 = 0.247 fs. At that instant Im Ψ = −ψ₂(x), so the wavefunction is purely imaginary but no smaller.
- The density is ψ₂² = (2/L)sin²(2πx/L) at both times, so P = (2/L)∫₀(L/4) sin²(2πx/L) dx = (2/L)[x/2 − L sin(4πx/L)/(8π)] from 0 to L/4 = (2/L)(L/8) = 0.250.
- Checks: E₂t/ħ is (J)(s)/(J s), a pure number as a phase must be; sin²(2πx/L) is symmetric about its peak at L/4 and repeats every L/2, so each quarter of the well holds the same share, 1/4. Neither time is observable: raise V by a constant and both change, while P does not.
Answer. E₂ = 4.18 eV; the phase turns once every 0.990 fs; Re Ψ vanishes everywhere at t = 0.247 fs; P(0 ≤ x ≤ L/4) = 0.250 at both times, because the density never changed.
MediumExample 2
An electron in an infinite well of width L = 0.50 nm is prepared at t = 0 in Ψ(x,0) = √0.8 ψ₁(x) + √0.2 ψ₂(x). Find the period with which its density repeats, the value of ⟨H⟩, the range over which ⟨x⟩ swings, and the largest and smallest values the density takes at x = L/4.
View solution →
- Normalisation first: the ψₙ are orthonormal and 0.8 + 0.2 = 1, so Ψ is normalised at t = 0, and the cross term integrates to zero at every later t. For L = 0.50 nm, E₁ = π²ħ²/(2mL²) = 1.504 eV and E₂ = 4E₁ = 6.016 eV.
- The density repeats every T = h/(E₂ − E₁) = (4.136 × 10⁻¹⁵ eV s)/(4.512 eV) = 0.917 fs, from ω₂₁ = (E₂ − E₁)/ħ = 6.86 × 10¹⁵ rad s⁻¹.
- Energy: ⟨H⟩ = 0.8E₁ + 0.2E₂ = 0.8(1.504 eV) + 0.2(6.016 eV) = 1.203 eV + 1.203 eV = 2.41 eV, the same at every t.
- Position: ⟨x⟩ = (0.8 + 0.2)(L/2) + 2√(0.8 × 0.2) cos(ω₂₁t) ∫xψ₁ψ₂ dx. Writing sin a sin b = ½[cos(a − b) − cos(a + b)] and integrating by parts gives ∫xψ₁ψ₂ dx = −16L/(9π²) = −0.1801L, so ⟨x⟩ = L/2 − 0.1441L cos(ω₂₁t), between 0.356L = 0.178 nm and 0.644L = 0.322 nm.
- At x = L/4: ψ₁² = (2/L)(½), ψ₂² = (2/L)(1) and ψ₁ψ₂ = (2/L)(1/√2), so |Ψ|² = (2/L)[0.8(0.5) + 0.2(1) + 0.8(0.7071)cos(ω₂₁t)] = (4.00 nm⁻¹)[0.600 + 0.566 cos(ω₂₁t)], from 4.66 nm⁻¹ down to 0.137 nm⁻¹. Dropping the cross term would have predicted a frozen 2.40 nm⁻¹.
Answer. The density repeats every 0.917 fs; ⟨H⟩ = 2.41 eV at all times; ⟨x⟩ swings between 0.178 nm and 0.322 nm; the density at L/4 swings between 4.66 nm⁻¹ and 0.137 nm⁻¹, a factor of 34.
HardExample 3
An electron in an infinite well of width L = 1.2 nm starts in Ψ(x,0) = [ψ₁(x) + ψ₂(x) + ψ₃(x)]/√3. Find ⟨H⟩; list the angular frequencies at which the density oscillates and say which of them move ⟨x⟩; find ⟨x⟩ at t = 0 and at t = h/(2E₁); and find the shortest time after which the whole density repeats.
View solution →
- Normalisation: three orthonormal terms with |cₙ|² = 1/3 each sum to 1. For L = 1.2 nm, E₁ = π²ħ²/(2mL²) = 0.2611 eV, with E₂ = 4E₁ and E₃ = 9E₁, so ⟨H⟩ = (1 + 4 + 9)E₁/3 = 14E₁/3 = 1.219 eV, fixed in time.
- Three pairs give three cross terms, each beating at its own spacing: (E₂ − E₁)/ħ = 3E₁/ħ, (E₃ − E₂)/ħ = 5E₁/ħ and (E₃ − E₁)/ħ = 8E₁/ħ, with periods h/(3E₁) = 5.28 fs, h/(5E₁) = 3.17 fs and h/(8E₁) = 1.98 fs.
- Parity decides which pairs move ⟨x⟩. Write x = (x − L/2) + L/2. The L/2 part integrates to zero against any ψₘψₙ with m ≠ n by orthogonality. About the centre, ψ₁ and ψ₃ are even and ψ₂ is odd, so (x − L/2)ψ₁ψ₃ is odd and ∫xψ₁ψ₃ dx = 0: the 8E₁/ħ term reshapes the density symmetrically without shifting ⟨x⟩. The opposite-parity pairs give ∫xψ₁ψ₂ dx = −16L/(9π²) = −0.1801L and ∫xψ₂ψ₃ dx = −48L/(25π²) = −0.1945L.
- So ⟨x⟩ = L/2 − (2/3)[0.1801L cos(3E₁t/ħ) + 0.1945L cos(5E₁t/ħ)]. At t = 0 both cosines equal 1: ⟨x⟩ = L/2 − 0.2498L = 0.2502L = 0.300 nm.
- At t = h/(2E₁) = 7.92 fs, E₁t/ħ = π, so the phases 3π and 5π make both cosines −1 and ⟨x⟩ = L/2 + 0.2498L = 0.7498L = 0.900 nm, the mirror image of the start. The (1,3) term's phase 8π is back to +1, so the whole density is the t = 0 density reflected about the centre.
- The density repeats only when 3E₁t/ħ, 5E₁t/ħ and 8E₁t/ħ are all whole multiples of 2π. Since 3, 5 and 8 share no common factor, that first happens at E₁t/ħ = 2π, t = h/E₁ = 15.8 fs, when every phase e(−in²E₁t/ħ) is back to 1.
Answer. ⟨H⟩ = 1.22 eV at all times; beats at 3E₁/ħ, 5E₁/ħ and 8E₁/ħ (periods 5.28, 3.17 and 1.98 fs), only the first two moving ⟨x⟩; ⟨x⟩ = 0.300 nm at t = 0 and 0.900 nm at t = 7.92 fs; the density first repeats at 15.8 fs.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Only energy differences set how fast a density moves.
One stationary state freezes |Ψ|² and every expectation value of an operator with no explicit t. Add a second level and the cross term oscillates at (E₂ − E₁)/ħ, while orthogonality keeps the norm and ⟨H⟩ fixed. Expanding any Ψ(x,0) as Σcₙψₙ assumes completeness and a static V.