Eigenvalue equations: solving Âψ = aψ
Shape test from the equation, eigenvalues from the boundaries.
Before this course solves a single system, it needs one test done fluently: apply the operator and ask whether the same function comes back times a constant. Here you drill that test on plane waves, box sines and delta spikes, and see that the boundary conditions, not the differential equation, decide which constants are allowed.
Build the model
Act on the function, then ask whether it came back only rescaled.
1The big idea
Most operators scramble a function: p̂ turns a Gaussian into x times a Gaussian, and Ĥ turns a sum of two sines into a differently weighted sum. The eigenvalue equation Âψ = aψ picks out the rare functions that come back with their shape intact, rescaled by one constant a that carries the operator's units. For a differential operator the equation is an ordinary differential equation, and on its own it quantises nothing: −iħψ′ = pψ has a solution for every complex p, and −(ħ²/2m)ψ″ = Eψ has one for every E. The eigenvalues are chosen by what else ψ must do. Staying bounded on the whole line leaves every real p, a continuum; returning to itself round a ring of circumference L leaves p = nh/L; vanishing at two walls a apart leaves k = nπ/a and Eₙ = n²π²ħ²/(2ma²). A spectrum therefore belongs to an operator together with its domain, never to the formula alone. The picture has two edges. On the whole line, x̂ and p̂ have no normalisable eigenfunction at all, only a delta spike and an endless plane wave, so their eigenvalues label a continuum that physical states can approach as normalisable packets but never reach. And being an eigenfunction is fragile: add two eigenfunctions with different eigenvalues and the sum is no longer an eigenfunction of that operator, which is the condition of a typical state.
2In one sentence
An eigenfunction of an operator  is a nonzero function that  returns multiplied by a single constant a, the eigenvalue, while also meeting the boundary and normalisability conditions the problem imposes.
Example. In a box of width a = 0.50 nm, Ĥ = −(ħ²/2m) d²/dx² acting on sin(3πx/a) returns (9π²ħ²/2ma²) sin(3πx/a), which is 13.5 eV × sin(3πx/a) for an electron: the same function times one constant, so it is an eigenfunction with eigenvalue 13.5 eV.
3The test: act, then compare shapes
The eigenvalue equation Âψ = aψ asks whether  hands ψ back unchanged in shape, multiplied by one constant. Two exclusions make it a real question. ψ must not be identically zero, because Â0 = a0 for every a. And a must not depend on x, because any Âψ can be written as f(x)ψ(x) with f = Âψ/ψ. So the working test is: act, form the ratio Âψ/ψ wherever ψ ≠ 0, and see whether it is constant. On the Gaussian ψ = e(−x²/2b²), p̂ gives (iħx/b²)ψ; the ratio iħx/b² changes with x, so the Gaussian is not a momentum eigenfunction. d²/dx² on the same Gaussian gives the ratio x²/b⁴ − 1/b², again x-dependent. Linearity then hands over two facts for free: 5ψ has the same eigenvalue as ψ, so normalisation never sets an eigenvalue; and the eigenvalue carries the units of the operator, joules for Ĥ and kg m s⁻¹ for p̂, whatever units ψ has.
4Solving for p̂: the equation accepts every number
Turn the question round and solve −iħ dψ/dx = pψ. It is first order, so ψ = A e(ipx/ħ) for any constant p whatever, real or complex. Real p = ħk gives Re ψ = A cos(kx), Im ψ = A sin(kx) for real A, and |ψ|² = |A|² at every x; for k = 1.0 × 10¹⁰ m⁻¹ the eigenvalue is ħk = 1.05 × 10⁻²⁴ kg m s⁻¹ and the wavelength 2π/k = 0.628 nm, while e(−ikx) returns −ħk, the same magnitude aimed the other way. Complex p = pᵣ + ipᵢ gives e(ipᵣ x/ħ) e(−pᵢ x/ħ), which grows without limit towards one end of the line. Requiring ψ to stay bounded removes every complex p and keeps every real one, a continuum. The −iħ is what turns the ik that d/dx returns into a real ħk; why only real p survive when the boundary condition makes p̂ Hermitian, while a condition such as ψ(a) = 2ψ(0) on an interval admits complex p, is the business of the next topic.
5Boundary conditions choose the eigenvalues: the ring
Put the same operator on a ring of circumference L, with x the arc length. Going once round must return the same ψ, so ψ(x + L) = ψ(x), which for A e(ipx/ħ) means e(ipL/ħ) = 1 and pL/ħ = 2πn: pₙ = 2πnħ/L = nh/L with n = 0, ±1, ±2, …. The operator has not changed; only its domain has, and the continuum has become a ladder. For L = 2.0 nm the rungs are 3.31 × 10⁻²⁵ kg m s⁻¹ apart. Every eigenfunction is now normalisable, since ∫₀L |A|² dx = |A|²L = 1 gives A = 1/√L, and n = 0, the constant function, is a perfectly good eigenfunction with p = 0. As L grows the spacing h/L shrinks and the whole-line continuum returns, the limiting-case check. The lesson generalises: an eigenvalue problem is an operator plus the conditions its functions must meet, and changing the conditions changes the spectrum.
6Testing sin(nπx/a) against Ĥ in a box
Inside a box 0 < x < a with V = 0, Ĥ = −(ħ²/2m) d²/dx², and two derivatives of sin(kx) return −k² sin(kx), so Ĥ sin(kx) = (ħ²k²/2m) sin(kx) for every k. The shape test passes; the walls decide. With ψ = 0 required outside and ψ continuous, ψ(0) = 0 is met by the sine (cos kx would fail it), and ψ(a) = 0 demands sin(ka) = 0, so k = nπ/a. n = 0 gives the zero function and is excluded; negative n return −sin(|n|πx/a), the same eigenfunction times −1. So Eₙ = n²π²ħ²/(2ma²), n = 1, 2, 3, …, which for an electron in a 0.50 nm box gives 1.50 eV, 6.02 eV and 13.5 eV. Checks: ħ²/(ma²) has units J² s²/(kg m²) = J; widening the box drives every level towards zero; and the relative gap (2n + 1)/n² between neighbours vanishes at large n, where the ladder looks as continuous as a classical particle's energy.
7Sums of eigenfunctions: same eigenvalue or not
Linearity decides what happens to combinations. If Âψ₁ = aψ₁ and Âψ₂ = aψ₂ with the same a, then Â(c₁ψ₁ + c₂ψ₂) = a(c₁ψ₁ + c₂ψ₂): the combination is an eigenfunction too, and the eigenvalue is called degenerate. The free Ĥ gives cos(kx), sin(kx), e(ikx) and e(−ikx) all the eigenvalue ħ²k²/2m, while p̂ accepts only the two exponentials, so a degenerate eigenvalue of one operator can be split by another, a thread picked up with commutators. With different eigenvalues the combination fails. In the 0.50 nm box take ψ = ψ₁ + ψ₂, the n = 1 and n = 2 sines: Ĥψ = (1.50 eV)ψ₁ + (6.02 eV)ψ₂, and Ĥψ/ψ is 1.50 eV at x = a/2, where ψ₂ has a node, 4.15 eV at x = a/4 and 16.9 eV at x = 3a/4. No single constant, so no eigenfunction. Measuring energy on an eigenfunction returns its eigenvalue with certainty; what it returns on a sum like this is Unit 4's question.
8On the whole line, x̂ and p̂ have no normalisable eigenfunction
For x̂ the eigenvalue equation reads xψ(x) = x₀ψ(x), that is (x − x₀)ψ(x) = 0 at every x. Wherever x ≠ x₀ this forces ψ = 0, so an eigenfunction could live only at the single point x₀, and no ordinary function qualifies. Normalisable functions can come close: a rectangle of width w and height 1/√w centred on x₀ has ∫|ψ|² dx = 1 and ∫|(x − x₀)ψ|² dx = w²/12, so it misses being an eigenfunction by only w/√12, which is 2.9 pm for w = 10 pm. The miss vanishes only as w → 0, and that limit is no function at all: the height 1/√w diverges while the area √w shrinks to zero. Rescaled to unit area, height 1/w, the rectangle tends instead to the Dirac delta δ(x − x₀), which does satisfy (x − x₀)δ(x − x₀) = 0 but has ∫δ² dx divergent, so it cannot be normalised either. On the whole line p̂ fails the same way from the other side: |e(ipx/ħ)|² is constant, so ∫|ψ|² dx over the line is infinite. Both operators have a continuous spectrum, every real x₀ and every real p, and no normalisable eigenfunction. Physical states are normalisable packets built from these eigenfunctions, which is how they serve as a basis later.
Key equations
What each relation says, and the units it is written in.
Eigenvalue equation
Âψ = aψ, with ψ not identically zero and a independent of x
The test: act with  and form Âψ/ψ wherever ψ ≠ 0. A ratio that varies with x means ψ is not an eigenfunction.
a carries the units of Â: J for Ĥ, kg m s⁻¹ for p̂. Rescaling ψ never changes a.
Momentum eigenfunctions on the whole line
−iħ dψ/dx = pψ gives ψₚ(x) = A e(ipx/ħ)
Boundedness removes every complex p and keeps every real one: a continuous spectrum with no member that can be normalised.
p in kg m s⁻¹, ħ = 1.055 × 10⁻³⁴ J s, so p/ħ = k in m⁻¹. |ψₚ|² = |A|² at every x.
Momentum on a ring of circumference L
ψ(x + L) = ψ(x) gives pₙ = 2πnħ/L, n = 0, ±1, ±2, …
Same operator, new domain: the continuum becomes a ladder of spacing h/L, which closes up as L → ∞.
L and the arc length x in m; normalised ψₙ = e(2πinx/L)/√L, in m⁻¹⁄².
Free Hamiltonian acting on a sine
−(ħ²/2m) d²/dx² sin(kx) = (ħ²k²/2m) sin(kx)
The constant ratio is what lets sin(kx) pass the shape test inside a box; whether it is allowed is the walls' decision.
m in kg, k in m⁻¹, eigenvalue in J. True for every k, so the equation alone selects nothing.
Wall condition in a box of width a
sin(ka) = 0 gives kₙ = nπ/a and Eₙ = n²π²ħ²/(2ma²), n = 1, 2, 3, …
The boundary condition turns the continuum of k that pass the shape test into a discrete set of energies.
a in m. Electron with a = 0.50 nm: E₁ = 1.50 eV, E₂ = 6.02 eV. n = 0 is the zero function, excluded.
Position eigenfunctions
(x − x₀)ψ(x) = 0 for every x gives ψ ∝ δ(x − x₀)
Every real x₀ is an eigenvalue and none has a normalisable eigenfunction: a normalised rectangle of width w misses by w/√12, never by zero.
x₀ in m, any real value. δ carries m⁻¹, and ∫δ² dx diverges.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Set c to 0 and sweep k: the dashed curve sits on ψ at every k, yet the wall dot leaves zero and the arrow misses the rungs except at k = π/a, 2π/a and 3π/a. Then park k at 2π/a and raise c: the wall still reads zero while the dashed curve peels away.
Live results
- TRIAL k
- 8.17 nm⁻¹
- ħ²k²/2m
- 2.54 eV
- WALL VALUE ψ(a)
- -0.62
- LARGEST SHAPE GAP
- 0.60
Catch the common trap
The mistake this lesson is built to prevent.
Because p̂² sin(kx) = ħ²k² sin(kx), students take the square root and declare sin(kx), or a box state sin(nπx/a), an eigenfunction of p̂ with eigenvalue ħk, or ±ħk. Act with p̂ once and the claim fails: −iħ d/dx sin(kx) = −iħk cos(kx), a different shape, and no constant turns a sine into a cosine. Being an eigenfunction of ² is a weaker condition than being one of Â. sin(kx) = (e(ikx) − e(−ikx))/2i is an equal mix of two p̂ eigenfunctions with eigenvalues +ħk and −ħk, and p̂² cannot tell them apart because it squares both to ħ²k². Only the forward direction is safe: if Âψ = aψ, then ²ψ = a²ψ. So a box state has a definite energy and a definite p², but it is not a momentum eigenfunction; what a momentum measurement on it returns is a question for Unit 4.
Test it: what changes?
- Ĥψ ∝ ψ at every k: shape test passes
- ħ²k²/2m grows as k²
- ψ(a) = 0 only at k = π/a, 2π/a, 3π/a
- Allowed energies only Eₙ = n²π²ħ²/(2ma²)
- Wall value ψ(a) stays zero
- Ĥψ peels away from ψ: no eigenfunction
- Largest shape gap grows as 3|c|
- ħ²k²/2m stays at E₂
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
On the whole line, with b = 0.25 nm, test three trial functions against p̂ = −iħ d/dx: ψ₁ = 3e(−ix/b), ψ₂ = cos(x/b) and ψ₃ = e(−x/b). For each, decide whether it solves the eigenvalue equation and whether it survives the requirement that ψ stay bounded; for any survivor give the eigenvalue and the de Broglie wavelength.
View solution →
- ψ₁: dψ₁/dx = (−i/b)ψ₁, so p̂ψ₁ = −iħ(−i/b)ψ₁ = −(ħ/b)ψ₁. The ratio p̂ψ₁/ψ₁ is the constant −ħ/b, so ψ₁ is an eigenfunction; the factor 3 only rescales it and leaves the eigenvalue alone.
- Eigenvalue: −ħ/b = −(1.055 × 10⁻³⁴ J s)/(0.25 × 10⁻⁹ m) = −4.2 × 10⁻²⁵ kg m s⁻¹, negative because the phase winds towards −x. |ψ₁|² = 9 at every x: bounded, though not normalisable, as every p̂ eigenfunction on the line must be. Wavelength λ = h/|p| = 2πb = 1.6 nm.
- ψ₂: p̂ cos(x/b) = (iħ/b) sin(x/b). A cosine has come back as a sine, and the ratio (iħ/b) tan(x/b) varies with x, so ψ₂ is not an eigenfunction. It is the average of e(ix/b) and e(−ix/b), whose eigenvalues +ħ/b and −ħ/b differ.
- ψ₃: p̂ e(−x/b) = (iħ/b) e(−x/b), a constant ratio, so ψ₃ solves the equation with the imaginary eigenvalue iħ/b. But e(−x/b) grows without bound as x → −∞; at x = −5.0 nm it is already e²⁰ ≈ 4.9 × 10⁸. It fails the boundary condition and is rejected.
- Checks: ħ/b has units J s m⁻¹ = kg m s⁻¹, a momentum. The root that boundedness throws out is the complex one, which the next topic shows is no accident.
Answer. Only ψ₁ is an allowed eigenfunction: p = −ħ/b = −4.2 × 10⁻²⁵ kg m s⁻¹, λ = 1.6 nm. ψ₂ is not an eigenfunction; ψ₃ solves the equation with eigenvalue iħ/b but is unbounded, so it is rejected.
MediumExample 2
An electron moves freely (V = 0) on a thin conducting ring of circumference L = 2.4 nm, so ψ(x + L) = ψ(x) with x the arc length. Find the eigenvalues and normalised eigenfunctions of p̂, the three lowest distinct eigenvalues of Ĥ = p̂²/2m with their degeneracies, and say whether cos(2πx/L) is an eigenfunction of either operator.
View solution →
- The eigenvalue equation −iħψ′ = pψ gives ψ = A e(ipx/ħ). Periodicity needs e(ipL/ħ) = 1, so pL/ħ = 2πn and pₙ = nh/L, n = 0, ±1, ±2, …, with h/L = (6.626 × 10⁻³⁴ J s)/(2.4 × 10⁻⁹ m) = 2.76 × 10⁻²⁵ kg m s⁻¹.
- Normalise over one circuit: ∫₀L |A|² dx = |A|²L = 1, so ψₙ = e(2πinx/L)/√L, with 1/√L = 0.645 nm⁻¹⁄². Every one is normalisable, n = 0 included: the constant function is periodic and has p = 0.
- Applying p̂ twice returns pₙ²ψₙ, so each ψₙ is also an eigenfunction of Ĥ, with Eₙ = n²h²/(2mL²) = n² × (6.626 × 10⁻³⁴)²/(2 × 9.109 × 10⁻³¹ × (2.4 × 10⁻⁹)²) J = n² × 4.18 × 10⁻²⁰ J = n² × 0.261 eV.
- Lowest three distinct values: 0 from n = 0 (one state), 0.261 eV from n = ±1 (two states), 1.04 eV from n = ±2 (two states). +n and −n share an energy because p̂² cannot see the sign of p.
- cos(2πx/L) is proportional to ψ₁ + ψ₋₁, two Ĥ eigenfunctions with the same eigenvalue, so it is an eigenfunction of Ĥ with 0.261 eV. Under p̂ it returns (2πiħ/L) sin(2πx/L), a new shape, so it is not an eigenfunction of p̂.
- Checks: h²/(mL²) has units J² s²/(kg m²) = J. As L → ∞ the spacing h/L → 0 and the whole-line continuum returns. Classically, p₁ times the ring radius L/2π is (h/L)(L/2π) = ħ: angular momentum comes in units of ħ, as Unit 12 derives.
Answer. pₙ = n × 2.76 × 10⁻²⁵ kg m s⁻¹ with ψₙ = e(2πinx/L)/√L. Ĥ: 0 (single), 0.261 eV (double), 1.04 eV (double). cos(2πx/L) is an eigenfunction of Ĥ with 0.261 eV but not of p̂.
HardExample 3
The trial function ψ(x) = e(−x²/(2β)), with β a real constant in m², is tested against Ĥ = −(ħ²/2m) d²/dx² + ½mω²x² for an electron with ħω = 0.20 eV. Find every β that makes ψ an eigenfunction, the eigenvalue each gives, and which survive normalisability. Then show that a Gaussian 20% wider than the survivor fails the test, by evaluating Ĥψ/ψ at two points.
View solution →
- Differentiate: ψ′ = −(x/β)ψ and ψ″ = (x²/β² − 1/β)ψ. So Ĥψ/ψ = ħ²/(2mβ) + x²(½mω² − ħ²/(2mβ²)).
- The ratio is independent of x only if the x² coefficient vanishes: β² = ħ²/(m²ω²), so β = +ħ/(mω) or β = −ħ/(mω). The eigenvalues are ħ²/(2mβ) = +½ħω and −½ħω respectively.
- Normalisability: β = −ħ/(mω) makes ψ = e(+mωx²/2ħ), which grows without bound, so it is rejected together with its eigenvalue −0.10 eV. β = ħ/(mω) gives a Gaussian e(−x²/2b²) of width b = √(ħ/(mω)) and eigenvalue ½ħω = 0.10 eV.
- Numbers: ω = (0.20 × 1.602 × 10⁻¹⁹ J)/(1.055 × 10⁻³⁴ J s) = 3.04 × 10¹⁴ rad s⁻¹, so b = √[(1.055 × 10⁻³⁴)/(9.109 × 10⁻³¹ × 3.04 × 10¹⁴)] m = 0.617 nm. Normalising, ∫e(−x²/b²) dx = b√π gives A = (πb²)(−1/4) = 0.956 nm⁻¹⁄², a rescaling that leaves 0.10 eV untouched.
- Too wide: with width 1.2b, β = 1.44b² and Ĥψ/ψ = ½ħω[1/1.44 + (x/b)²(1 − 1/1.44²)]. At x = 0 this is 0.10 eV/1.44 = 0.069 eV; at x = 1.2b = 0.741 nm it is ½ħω × 1.44 = 0.144 eV. The ratio more than doubles across the packet, so this Gaussian is no eigenfunction.
- Checks: ħ/(mω) has units J s/(kg s⁻¹) = m². As ω → 0 the width grows without limit and the eigenvalue falls to 0, the flat free-particle limit. Classically an oscillator with energy ½ħω turns round where ½mω²x² = ½ħω, at x = b: the eigenfunction's width is the classical amplitude at its own energy.
Answer. β = ħ/(mω): the normalisable Gaussian of width b = 0.617 nm, eigenvalue ½ħω = 0.10 eV. β = −ħ/(mω) gives −0.10 eV but an unbounded ψ, rejected. At width 1.2b, Ĥψ/ψ runs from 0.069 eV at x = 0 to 0.144 eV at x = 0.741 nm: not an eigenfunction.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
A constant Âψ/ψ plus the boundary conditions make an eigenfunction.
Act with Â, form Âψ/ψ wherever ψ ≠ 0, and require one constant. Then impose boundedness, periodicity or the walls: they select the eigenvalues, such as kₙ = nπ/a in a box. Exclude ψ = 0, and remember an eigenfunction of ² need not be one of Â.
Next: Hermitian operators, and why their eigenvalues are real →