Eigenvalue equations and determinate states
Zero spread means an eigenfunction, and nothing else.
Before a wavefunction can give odds, you need to know when there are no odds at all. This lesson finds the preparations a perfect instrument answers the same way every time, shows that they are exactly the eigenfunctions of the observable, and asks which quantities a real particle can ever be prepared sharp in.
Build the model
One reading, every run: the eigenvalue equation is the condition for a sharp value.
1The big idea
Some preparations answer a question the same way every time. Measure Q once on each of a thousand identical copies and, if every ideal reading is the same q, the state is determinate for Q: its histogram is a single bar and its spread σQ is zero. Quantum mechanics says exactly which states these are. Substituting Q̂ψ = qψ into σQ² = ⟨(Q − q)²⟩ gives zero, and the identity σQ² = ∫ |(Q̂ − ⟨Q⟩)ψ|² dx of UQMI-03.08 runs the argument backwards: nothing but an eigenfunction has zero spread, and the value it always returns is its eigenvalue. So the eigenvalue equation is a forecast about an experiment, and the ratio Q̂ψ/ψ, flat for an eigenfunction, measures by its scatter how far any state is from sharp. Determinacy belongs to a state and an observable together: the ground state of a box is sharp in energy and spread in momentum, and a standing wave on a ring is sharp in p² but not in p. The model stops where the eigenfunction stops being a state. It must be normalisable and obey the boundary conditions, so on the open line no preparation is sharp in x or in p, or in the energy of a free particle; packets can approach those values as closely as you like, at a price paid in another spread, and never reach them.
2In one sentence
A determinate state of an observable Q is a normalisable state in which every ideal measurement of Q returns the same value q; such states are exactly the allowed eigenfunctions of Q̂, and q is the eigenvalue.
Example. On a ring of circumference 2.00 nm, ψ = e(2πix/L)/√L gives p = h/L = 3.31 × 10⁻²⁵ kg m s⁻¹ in every momentum measurement, so σₚ = 0. The plane wave e(ikx) on the open line has the same shape, but ∫ |ψ|² dx diverges, so it is not a state.
3Determinate means one bar in every histogram of Q
Prepare a thousand systems in the same state, measure one observable Q once on each with an ideal instrument, and plot the readings. For most states the bars spread over several eigenvalues. Sometimes every reading is the same number q: the state is then determinate for Q, with ⟨Q⟩ = q and σQ = 0. Classically that is the normal case, since a point in phase space fixes x, p and every function of them at once, and any scatter in a record belongs to the apparatus. In quantum mechanics it is special, and it belongs to a pair, a state and an observable. An electron in the ground state of a 0.40 nm box returns E₁ = 2.35 eV in all thousand energy runs, while momentum runs on the same preparation scatter with σₚ = πħ/a = 8.28 × 10⁻²⁵ kg m s⁻¹ (UQMI-03.08). Determinacy is fixed by the preparation before any run: a detector that blurs energies by 0.05 eV would widen the recorded bar without making the state any less sharp, the distinction UQMI-04.09 develops.
4Zero spread and the eigenvalue equation are one statement
Forwards, substitute. If Q̂ψ = qψ with ψ normalised, then ⟨Q⟩ = ∫ ψ*qψ dx = q and (Q̂ − q)ψ = 0, so σQ² = ∫ ψ*(Q̂ − q)²ψ dx = 0. The same substitution gives ⟨Qᵏ⟩ = qᵏ for every power k, the moments of a distribution sitting entirely at q: not a narrow bar but a single one. Backwards, quote UQMI-03.08, where it is proved: for Hermitian Q̂, σQ = 0 only if Q̂ψ = ⟨Q⟩ψ, so zero spread forces the eigenvalue equation, with the mean as the eigenvalue. Three things follow for an experiment. The only way to prepare a sharp value is to prepare an eigenfunction. The sharp value must be an eigenvalue: a box electron can be prepared so that every energy run reads E₁, or so that every run reads 4E₁, but no preparation makes every run read 2.5E₁. And a state that is not an eigenfunction, however close, has a nonzero σQ, so some runs return a different number.
5The ratio test measures how far a state is from sharp
UQMI-03.04 tests a candidate by forming Q̂ψ/ψ and asking whether it depends on x. That ratio, the local value of Q, carries more than a yes or no. Because (Q̂ − ⟨Q⟩)ψ = (Q̂ψ/ψ − ⟨Q⟩)ψ, the identity above becomes σQ² = ∫ |Q̂ψ/ψ − ⟨Q⟩|² |ψ|² dx: the spread is the |ψ|²-weighted scatter of the local value about its mean. Take the Gaussian ψ ∝ e(−x²/(2s²)) in the well V = ½mω²x², whose local energy UQMI-03.04 found: ħ²/(2ms²) + (½mω² − ħ²/(2ms⁴))x², flat only at s = b = √(ħ/mω), where it equals ħω/2. For an experiment, that width is a preparation in which every energy run reads ħω/2, found without solving the oscillator, which Unit 9 does. Any other width leaves the x² term, and its scatter is σE = |½mω²s² − ħ²/(2ms²)|/√2. A narrower trial, s = b/2, has a local energy falling as x² away from the centre, with σE = 1.33ħω about ⟨E⟩ = 1.06ħω. The local value is bookkeeping, not a reading: it sweeps through a continuum of values, some of them negative here, while every run returns an eigenvalue.
6Sharp in one observable, spread in another
A state determinate for Q is determinate for every function of Q, since Q̂ψ = qψ gives Q̂²ψ = q²ψ and so on through any power series (UQMI-03.04). On a free ring, e(2πinx/L)/√L is sharp in p = nh/L and therefore in E = p²/2m. The converse fails, and degeneracy is the reason. The energy n²h²/(2mL²) belongs to both e(±2πinx/L)/√L, so every combination of the two is still determinate in energy, including the standing wave √(2/L) cos(2πnx/L). That state is sharp in p² as well, yet it is not an eigenfunction of p̂: momentum runs return +nh/L and −nh/L half the time each, and σₚ = nh/L. Which observables can be sharp together, and how to tell without testing states one by one, is the subject of UQMI-04.05 and UQMI-04.06. One kind of sharpness is permanent. For a time-independent Ĥ, Ψ = ψe(−iEt/ħ) remains an eigenfunction of Ĥ at every t, so a determinate energy never spreads, which is why stationary states are stationary (UQMI-02.05).
7A determinate state has to be a state
An eigenfunction describes a preparation only if it is normalisable and obeys the boundary conditions. On the open line the eigenfunctions of p̂ are e(ipx/ħ), with |ψ|² the same everywhere, and those of x̂ are δ(x − x₀), with ∫ δ² dx divergent (UQMI-03.04). Neither can be normalised, so no preparation has σₚ = 0 or σₓ = 0, and since every bounded eigenfunction of the free Ĥ is built from those same plane waves, a free particle on the line has no sharp energy either. Walls fail in another way: no e(ikx) vanishes at both walls of a box, so no box state is sharp in momentum, and ψₙ carries σₚ = nπħ/a. What a laboratory can make is a packet that closes in. The ratio test on ψ ∝ e(ik₀x − x²/(4σₓ²)) gives p̂ψ/ψ = ħk₀ + iħx/(2σₓ²), hence σₚ = ħ/(2σₓ): 5.3 × 10⁻²⁹ kg m s⁻¹ for σₓ = 1.0 μm, half that for 2.0 μm, and zero for no finite length. The limit is not a state, because the probability of finding the particle in any fixed metre falls to zero. Wrap the plane wave round a ring and it is normalisable, with p = nh/L exactly: sharpness depends on the space as well as the operator.
Key equations
What each relation says, and the units it is written in.
Substitute the eigenvalue equation
Q̂ψ = qψ ⇒ σQ² = ∫ ψ* (Q̂ − q)²ψ dx = 0
The eigenvalue equation read as a forecast: every ideal reading of Q on this preparation is q, a histogram with one bar, known before any run is made.
q is real, in the units of Q; ψ is normalised, in m⁻¹⁄². The first step is ⟨Q⟩ = ∫ ψ*qψ dx = q.
Only an eigenfunction has zero spread
σQ² = ∫ |(Q̂ − ⟨Q⟩)ψ|² dx = 0 ⇒ Q̂ψ = ⟨Q⟩ψ
Read in the laboratory: a sharp value can be prepared only as an eigenfunction, and the value it gives is an eigenvalue, never a number between two.
Q̂ Hermitian on a domain containing ψ; the identity is proved in UQMI-03.08. σQ carries the units of Q.
The local value and its scatter
σQ² = ∫ |Q̂ψ/ψ − ⟨Q⟩|² |ψ|² dx
A flat ratio means a sharp state, and the ratio's weighted scatter is σQ. Its values are not readings: every run still returns an eigenvalue.
Q̂ψ/ψ, the local value, is in the units of Q and defined where ψ ≠ 0; its |ψ|²-weighted mean is ⟨Q⟩.
Energy spread of a Gaussian in a harmonic well
σE = |½mω²s² − ħ²/(2ms²)|/√2, for ψ ∝ e(−x²/(2s²)) in V = ½mω²x²
The weighted scatter of the local energy ħ²/(2ms²) + (½mω² − ħ²/(2ms⁴))x² found in UQMI-03.04: any width but b leaves an x² term, and so a spread.
m in kg, ω in s⁻¹, s in m; σE in J. Zero only at s = b = √(ħ/mω), where every run reads ħω/2.
Sharpness passes to functions of Q
Q̂ψ = qψ ⇒ f(Q̂)ψ = f(q)ψ, but not conversely
Sharp momentum gives sharp free kinetic energy. Sharp p² does not give sharp p: one eigenvalue of f(Q̂) can come from several of Q̂, such as +p and −p.
f is any function given by a power series, such as Q² or p²/2m; f(q) carries the units of f.
Momentum sharpness is bought with length
σₚ = ħ/(2σₓ), for ψ = (2πσₓ²)⁻¹⁄⁴ e(ik₀x − x²/(4σₓ²))
On the open line σₚ → 0 only as σₓ → ∞, where the amplitude vanishes and ψ stops being a state: sharp as you like, never exactly sharp.
σₓ in m, k₀ in m⁻¹, σₚ in kg m s⁻¹; ħ = 1.055 × 10⁻³⁴ J s.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Slide s from 1.50b to 1.00b: the local-energy curve flattens onto ħω/2 and the band closes, the one setting where every reading agrees. Then set x₀ = 1.00b: the Gaussian keeps its shape, yet the curve tilts into a straight line and σE rises to 0.71ħω.
Live results
- MEAN ⟨E⟩
- 0.674 ħω
- SPREAD σE
- 0.638 ħω
- P(READING ħω/2)
- 0.923
- CLASSICAL ½mω²x₀²
- 0.000 ħω
Catch the common trap
The mistake this lesson is built to prevent.
Reading the eigenvalue equation as the act of measuring: “to measure Q, apply Q̂ to ψ, and Q̂ψ is the state afterwards.” For an eigenfunction the two happen to agree, because for q ≠ 0, qψ is the same state as ψ, and that coincidence is where the idea comes from. For any other state it fails at once. In a box take ψ = (ψ₁ + ψ₂)/√2. Then Ĥψ = (E₁ψ₁ + 4E₁ψ₂)/√2 is not normalised, carries the units of energy times ψ, and is not a multiple of ψ. Treated as the state after measurement, it would make a repeat energy reading give 4E₁ in 16 runs out of 17. In fact each run returns E₁ or 4E₁ with probability ½ (UQMI-04.03), and a repeat returns whatever the first run gave (UQMI-04.04). Applying an operator is a fixed, linear map; a measurement yields one random eigenvalue. The eigenvalue equation is a test a state passes or fails, not a picture of looking.
Test it: what changes?
- Local energy flattens onto ħω/2
- σE falls to zero
- ⟨E⟩ falls to ħω/2
- P(reading ħω/2) rises to 1
- Local energy tilts into a straight line
- σE rises from 0 to 0.71ħω
- ⟨E⟩ rises from ħω/2 to ħω
- Energy no longer determinate, though the shape is unchanged
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron moves freely on a ring of circumference L = 1.20 nm. Test ψA = e(4πix/L)/√L and ψB = √(2/L) cos(4πx/L) against p̂ = −iħ d/dx and Ĥ = p̂²/2m. For each, say whether momentum and energy are determinate, give the sharp values, and describe what momentum measurements on ψB return.
View solution →
- Both are states: |ψA|² = 1/L integrates to 1 round the ring, and (2/L) cos²(4πx/L) integrates to 1 because cos² averages to ½ round the ring. Both return to their starting value at x = L, as the ring demands.
- Momentum: p̂ψA = −iħ(4πi/L)ψA = (2h/L)ψA, a constant ratio, so every run on ψA reads p = 2h/L = 2 × 6.626 × 10⁻³⁴ J s ÷ 1.20 × 10⁻⁹ m = 1.10 × 10⁻²⁴ kg m s⁻¹. But p̂ψB is a multiple of sin(4πx/L), a different shape, so ψB is not determinate in p.
- Energy: two derivatives bring down −(4π/L)² for either function, so Ĥψ = [(2h/L)²/2m]ψ for both. E = (1.104 × 10⁻²⁴)² ÷ (2 × 9.109 × 10⁻³¹) J = 6.69 × 10⁻¹⁹ J = 4.18 eV, sharp in both states.
- Why ψB survives Ĥ: ψB = (ψA + ψC)/√2 with ψC = e(−4πix/L)/√L, whose momentum is −2h/L and whose energy is the same 4.18 eV. A combination within one degenerate eigenvalue stays determinate. Its momentum runs read +2h/L or −2h/L half the time each, so ⟨p⟩ = 0 and σₚ = 2h/L, and no run reads the mean, 0.
- Checks: E = p²/2m holds for either sign of p, as it does classically, and (kg m s⁻¹)² ÷ kg is J. As L → ∞ the normaliser 1/√L → 0, and the determinate momentum state dissolves into an unnormalisable plane wave.
Answer. ψA is sharp in both: p = 1.10 × 10⁻²⁴ kg m s⁻¹ and E = 4.18 eV. ψB is sharp in energy only, E = 4.18 eV; its momentum reads ±1.10 × 10⁻²⁴ kg m s⁻¹ half the time each, σₚ = 1.10 × 10⁻²⁴ kg m s⁻¹.
MediumExample 2
An electron beam is modelled on the open line by ψ = (2πσₓ²)⁻¹⁄⁴ e(ik₀x − x²/(4σₓ²)), with de Broglie wavelength λ = 2π/k₀ = 0.100 nm. Use the ratio p̂ψ/ψ to decide whether ψ is determinate in momentum, find ⟨p⟩ and σₚ, and find the packet length σₓ that makes σₚ/⟨p⟩ equal 1.0 × 10⁻³ and 1.0 × 10⁻⁶. What happens in the limit?
View solution →
- Act: p̂ψ = −iħ(ik₀ − x/(2σₓ²))ψ, so p̂ψ/ψ = ħk₀ + iħx/(2σₓ²). The ratio changes with x, so ψ is not an eigenfunction of p̂ and is not determinate in momentum for any finite σₓ.
- Mean: ⟨p⟩ = ∫ (p̂ψ/ψ)|ψ|² dx = ħk₀ + (iħ/2σₓ²)∫ x|ψ|² dx = ħk₀, because |ψ|² is even in x. With k₀ = 2π/λ, ⟨p⟩ = h/λ = 6.626 × 10⁻³⁴ J s ÷ 1.00 × 10⁻¹⁰ m = 6.63 × 10⁻²⁴ kg m s⁻¹.
- Spread: σₚ² = ∫ |p̂ψ/ψ − ħk₀|² |ψ|² dx = (ħ²/4σₓ⁴)⟨x²⟩ = ħ²/(4σₓ²), since |ψ|² is a normalised Gaussian of standard deviation σₓ. So σₚ = ħ/(2σₓ), in J s ÷ m = kg m s⁻¹.
- Required length: σₓ = ħ/(2σₚ) = λ/(4π σₚ/⟨p⟩). For σₚ/⟨p⟩ = 1.0 × 10⁻³, σₓ = 1.00 × 10⁻¹⁰ m ÷ (4π × 10⁻³) = 7.96 × 10⁻⁹ m, about 80 wavelengths. For 1.0 × 10⁻⁶, σₓ = 7.96 × 10⁻⁶ m.
- Limit: σₚ → 0 needs σₓ → ∞, where the peak density (2πσₓ²)⁻¹⁄² and the probability of any fixed interval both fall to zero. The limit is the plane wave e(ik₀x), which cannot be normalised (UQMI-03.04). Checks: ∫ |ψ|² dx = 1 for every σₓ, and setting k₀ = 0 moves the mean to zero while leaving σₚ unchanged, since the spread comes from the envelope alone.
Answer. Not determinate: p̂ψ/ψ = ħk₀ + iħx/(2σₓ²) varies with x. ⟨p⟩ = 6.63 × 10⁻²⁴ kg m s⁻¹ and σₚ = ħ/(2σₓ); σₓ = 8.0 nm gives 10⁻³ and 8.0 μm gives 10⁻⁶, while σₚ = 0 would need an unnormalisable infinite wave.
HardExample 3
An electron in a quantum dot is held in V = ½mω²x² with ħω = 40.0 meV and prepared in the trial state ψ = (πs²)⁻¹⁄⁴ e(−x²/(2s²)) with s = 2.00 nm. Starting from its local energy Ĥψ/ψ = ħ²/(2ms²) + (½mω² − ħ²/(2ms⁴))x², found in UQMI-03.04, find the width b at which ψ would be determinate and the energy it would then always return, and, for s = 2.00 nm, ⟨E⟩, σE and the probability that an energy measurement returns ħω/2.
View solution →
- Quote UQMI-03.04 rather than differentiate again: the local energy is A + Bx² with A = ħ²/(2ms²) and B = ½mω² − ħ²/(2ms⁴). Every term is in J: ħ²/(ms²) has units J² s² ÷ (kg m²) = J, and B carries J m⁻².
- Sharp only if B = 0: s⁴ = ħ²/(m²ω²), so s = b = √(ħ/mω) = ħ/√(m ħω). With ħω = 40.0 meV = 6.409 × 10⁻²¹ J, b = 1.055 × 10⁻³⁴ ÷ √(9.109 × 10⁻³¹ × 6.409 × 10⁻²¹) m = 1.38 nm, and then Ĥψ = (ħω/2)ψ: every run would read 20.0 meV.
- Mean at s = 2.00 nm: with ⟨x²⟩ = s²/2, ⟨E⟩ = ħ²/(2ms²) + (½mω² − ħ²/(2ms⁴))s²/2 = ħ²/(4ms²) + mω²s²/4. With r = s/b = 1.449 this is (ħω/4)(1/r² + r²) = 4.76 meV + 21.00 meV = 25.8 meV, above the determinate value.
- Spread: the local energy scatters only through Bx², so σE² = B²(⟨x⁴⟩ − ⟨x²⟩²) = B²s⁴/2, using ⟨x⁴⟩ = 3s⁴/4. Here Bs² = 2(21.00 − 4.76) meV = 32.5 meV, so σE = 32.5 meV ÷ √2 = 23.0 meV.
- Chance of the determinate value, anticipating UQMI-04.03: it is the squared overlap with the eigenfunction ψb, and two normalised Gaussians overlap by ∫ ψb ψ dx = √(2sb/(s² + b²)), so P(20.0 meV) = 2sb/(s² + b²) = 2(2.00)(1.380) ÷ (4.00 + 1.905) = 0.935. The other 6.5% of runs return higher eigenvalues, listed in Unit 9: a 93.5% majority at one value sits alongside σE = 23.0 meV.
- Checks: at s = b, B = 0, so σE = 0, P = 1 and ⟨E⟩ = ħω/2, the determinate case. For s much larger than b the kinetic term ħ²/(4ms²) fades and ⟨E⟩ → ½mω²⟨x²⟩, the classical potential energy of particles at rest spread over that width.
Answer. Ĥψ/ψ is flat only at s = b = 1.38 nm, where every run would read ħω/2 = 20.0 meV. At s = 2.00 nm, ⟨E⟩ = 25.8 meV, σE = 23.0 meV and P(20.0 meV) = 0.935.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Zero spread in Q means an allowed eigenfunction of Q̂.
A state is determinate for Q exactly when Q̂ψ = qψ, and every ideal run then reads the eigenvalue q. The eigenfunction must be normalisable and obey the boundary conditions, so on the open line nothing is sharp in x or p. Sharp Q gives sharp f(Q), not conversely.
Next: Eigenbasis expansion and the probability of an eigenvalue →