Compatible and incompatible observables
Commuting observables never spoil each other's answers.
Unit 3 proved that commuting operators share eigenfunctions. This lesson turns that theorem into predictions at the bench: for any two quantities you will say whether both can be pinned down at once, whether the order of two measurements changes the odds, and what a measurement in the middle does to the answer you already had.
Build the model
Whether a second measurement can spoil the first is decided by one commutator.
1The big idea
Two observables are compatible when measuring one never spoils an answer already obtained for the other: measure A, then B, then A again, and the third result matches the first in every run, from every preparation. Unit 3 supplies the test in one line: Hermitian  and B̂ share a complete set of eigenfunctions exactly when [Â, B̂] = 0 (UQMI-03.09). Read physically, a complete shared set means every state is a superposition of functions sharp in both, so the pair has one joint distribution whichever is measured first, and a B measurement, even one nobody reads, leaves the odds for A untouched. When the commutator is not zero there is no such set, so some preparation loses its A value to a B measurement, and the order of two measurements changes the odds. That verdict is about the set, not every state. A non-commuting pair can still share the functions its commutator sends to zero, like the ring state with zero momentum and even parity at once; only a commutator that annihilates nothing, as iħ does for x̂ and p̂, leaves no state sharp in both. The picture assumes ideal measurements in quick succession: a delay lets the Schrödinger equation move the state unless  also commutes with Ĥ, and a real detector reports a band rather than an eigenvalue, so in practice a repeat is confirmed only to within its resolution.
2In one sentence
Two observables are compatible when their operators commute: a complete set of states is then sharp in both, and measuring one never changes a value already found for the other.
Example. On a ring of circumference 1.00 nm, Ĥ = p̂²/2m commutes with p̂: an electron found at 1.50 eV and then at p = +6.63 × 10⁻²⁵ kg m s⁻¹ returns 1.50 eV on every repeat energy measurement.
3Compatible is a promise about sequences of results
Define the word by experiment. Prepare many copies of a state and measure A, then B, then A again, each measurement ideal and each following at once. A and B are compatible if the third result always equals the first, from every preparation and whatever B gave. Unit 3 has already said when that happens, and it is quoted here, not re-proved: Hermitian  and B̂ commute exactly when they have a complete set of shared eigenfunctions ψₙₘ, each sharp in both (UQMI-03.09). Read it as physics. The first result aₙ leaves a state built only from shared functions carrying aₙ (UQMI-04.04). The B measurement keeps those with one value bₘ, and every one of them still carries aₙ, so the repeat cannot fail. Completeness adds two more testable promises. Any state expands as ψ = Σ cₙₘψₙₘ, so the pair has one joint distribution, P(aₙ, bₘ) = |cₙₘ|², whichever is measured first. And summing over bₘ returns P(aₙ): a B measurement placed in the path, even one nobody reads, leaves the odds for A exactly as they were.
4One function per level: the partner comes free
Put an electron in an infinite well from −a/2 to a/2. The potential is symmetric, V(−x) = V(x), so parity, Π̂ψ(x) = ψ(−x), commutes with Ĥ. Each level Eₙ = n²h²/(8ma²) holds a single function, so a corollary from Unit 3 applies: on a one-function level, an eigenfunction of Ĥ is also an eigenfunction of anything that commutes with Ĥ. Check it: √(2/a) cos(πx/a) and √(2/a) cos(3πx/a) are even, √(2/a) sin(2πx/a) is odd. In a 1.20 nm well, an electron found at E₃ = 2.35 eV then reads even with certainty, and a parity measurement placed between two energy measurements cannot change the second. Parity here acts as a coarse energy measurement that reports only whether n is odd or even: from 0.6ψ₁ + 0.8ψ₂, the result “even” occurs in 36% of runs and leaves ψ₁, exactly what the result E₁ would have left. This is why every bound energy eigenfunction of a symmetric one-dimensional well, the harmonic oscillator's included, is even or odd: its levels hold one function each, and compatibility does the rest.
5A shared level: the value survives, the state moves
Compatibility does not promise that the second measurement leaves the state alone. Take an electron on a ring of circumference L = 1.00 nm, with ψ(s + L) = ψ(s) and Ĥ = p̂²/2m, so [Ĥ, p̂] = 0. Level n has Eₙ = n²h²/(2mL²) = n² × 1.50 eV, and for n ≥ 1 it holds two functions, e(2πins/L)/√L and e(−2πins/L)/√L. Prepare √(2/L) cos(2πs/L). Its energy is sharp at 1.50 eV, but it is an equal sum of the two travelling waves, so p = +h/L or −h/L, ±6.63 × 10⁻²⁵ kg m s⁻¹, each with probability ½. Afterwards the electron is in a travelling wave, a different function, yet a repeat energy measurement returns 1.50 eV every time, because both travelling waves sit on the same level. The theorem promises that some basis is shared, not that the function you prepared belongs to it. It also says what a complete measurement is: energy alone leaves the level's two functions unresolved, while energy with momentum names every ring state uniquely, and so does energy with parity.
6Incompatible: order changes the odds
Parity on the ring, Π̂ψ(s) = ψ(−s) with s measured from a mirror line, also commutes with Ĥ, but not with p̂: differentiating ψ(−s) brings out a minus sign, so p̂Π̂ = −Π̂p̂. Run the sequence. An electron found with p = +h/L is left in e(2πis/L)/√L. Its even part, cos(2πs/L)/√L, carries probability ½ and its odd part the other ½. Either parity eigenfunction is an equal mix of the two travelling waves, so a repeat momentum reading gives +h/L only half the time: the middle measurement has erased the sign of the momentum and left the energy intact. There is no joint distribution either. From the same start, “+h/L, then even” has probability ½, while “even, then +h/L” has ¼. Within one level the two read different things: momentum sees only the sizes of the two travelling-wave amplitudes, parity depends on their relative phase as well, and each measurement wipes out what the other would read. Compatibility does not chain, either: Ĥ commutes with both, yet they do not commute with each other.
7A shared state is not a shared basis
Incompatibility forbids a complete shared set, not every shared state. If ψ is sharp in both, [Â, B̂]ψ = (ab − ba)ψ = 0, so shared eigenfunctions live among the states the commutator annihilates. On the ring, [p̂, Π̂]ψ = 2p̂Π̂ψ vanishes only if Π̂ψ is constant, so the one survivor is the n = 0 state 1/√L, with p = 0, even parity and E = 0 at once; a momentum–parity–momentum sequence started there repeats every time. It is also why every other ring level is doubled: a level of one function would have to be sharp in both, and only the constant is. Two mirror lines at angle α show how much can survive. Their parities commute on level n only when nα is a multiple of 90°, so at α = 22.5° levels 0, 4, 8, … each have a basis sharp in both mirrors, while no state on levels 1 to 3 is. Angular momentum will add a three-dimensional case: its components do not commute, yet the state with zero angular momentum is sharp in all three. Whether a given state can be sharp in both starts with the commutator acting on that state; whether a complete set can is settled by the commutator itself.
8Position and momentum: nothing escapes
For x̂ and p̂ on a line the commutator is the constant iħ (UQMI-04.05), and the shared-state test demands iħψ = 0, which only ψ = 0 passes. There is no counterpart of the ring's constant state, and idealised states do not help: the position eigenfunction δ(x − x₀) and the momentum eigenfunction e(ikx) are different functions, and neither can be normalised (UQMI-04.02). So no preparation, however refined, leaves a particle whose position and momentum results are both certain, and a position–momentum–position sequence need not repeat: a narrow position band leaves a narrow packet, the momentum result that follows leaves a long wavetrain, and the second position reading can land far outside the first band. The scale of the conflict is ħ, which is why classical mechanics never noticed it. Localise a 1.0 g bead to 1.0 μm and the smallest velocity spread UQMI-04.08 will allow is 5.3 × 10⁻²⁶ m s⁻¹, far below any instrument, so for everyday bodies x and p behave as if compatible.
Key equations
What each relation says, and the units it is written in.
Compatibility, quoted from UQMI-03.09
[Â, B̂] = 0 ⇔ a complete set ψₙₘ with Âψₙₘ = aₙψₙₘ and B̂ψₙₘ = bₘψₙₘ
Every state expands on functions sharp in both, so pinning down one value never forces the other to spread: the two can be sharp together.
 and B̂ Hermitian; aₙ and bₘ real, in the SI units of each observable; ψₙₘ normalised, in m⁻¹⁄².
Joint probability, the same in either order
P(aₙ, bₘ) = |∫ ψₙₘ* ψ dx|²
A then B or B then A give the same numbers, and Σₘ P(aₙ, bₘ) = P(aₙ): a compatible measurement, even one never read, leaves the other's odds alone.
ψ the prepared state, ψₙₘ the shared eigenfunction for that pair (if several carry the pair, add their squared overlaps); both in m⁻¹⁄², P a pure number.
A one-function level fixes the partner
[Â, B̂] = 0 and aₙ non-degenerate ⇒ B̂ψₙ = bψₙ
Finding aₙ already fixes b, so a B measurement next is certain and harmless. In a symmetric well it hands every energy level a definite parity.
ψₙ the only eigenfunction of  with eigenvalue aₙ; b real, in the units of B. Quoted from UQMI-03.09.
A–B–A repeat probability
P(aₙ again) = Σₘ |∫ χₘ* ψₙ dx|⁴
It is 1 only if ψₙ is itself an eigenfunction of B̂. Two outcomes at overlaps cos θ and sin θ give cos⁴θ + sin⁴θ, which never falls below ½.
aₙ non-degenerate, so ψₙ is the one state its result leaves; χₘ the eigenfunctions of B̂, one per outcome bₘ; every overlap is a pure number.
Where a shared eigenfunction can live
[Â, B̂]ψ = (ab − ba)ψ = 0 if Âψ = aψ and B̂ψ = bψ
Necessary, not sufficient. A non-commuting pair can share only states its commutator annihilates: on a ring p̂Π̂ = −Π̂p̂, which spares just the constant.
a and b are numbers, so ab = ba. The commutator carries the units of A × B.
Position and momentum share nothing
[x̂, p̂]ψ = iħψ ≠ 0 for every ψ ≠ 0
No function, normalisable or not, is an eigenfunction of both, so no preparation makes x and p sharp together. UQMI-04.08 says how wide they must be.
ħ = 1.055 × 10⁻³⁴ J s. x̂ in m and p̂ in kg m s⁻¹, so the commutator carries J s, the units of x × p.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Hold n = 1 and drag α from 0° to 45°: the repeat falls from 1 to its floor of ½. Then set α = 22.5° and step n to 2 and on to 4: the mirrors still do not commute, yet on level 4 the curves are orthogonal and both parities are sharp. On n = 0 the constant state is shared at every angle.
Live results
- P(EVEN ABOUT M₂)
- 0.854
- P(M₁ REPEATS)
- 0.750
- |[Π̂₁, Π̂₂]| ON LEVEL n
- 1.41
- SHARP ENERGY, L = 1.00 nm
- 1.50 eV
Catch the common trap
The mistake this lesson is built to prevent.
“They do not commute, so they can never both be sharp” and its mirror, “they commute, so every eigenfunction of one is an eigenfunction of the other”, both read the commutator as a verdict on each state. It is a verdict on the set. Momentum and parity on a ring anticommute, yet the constant state has p = 0 and even parity exactly, because a nonzero commutator can still annihilate particular functions, and those are free to be sharp in both. And Ĥ commutes with p̂, yet √(2/L) cos(2πs/L) has a sharp energy and a momentum that comes out +h/L or −h/L at random, because on a level holding two functions the theorem guarantees that some basis is shared, not that yours is. The state-by-state reading is safe in two cases only: a level holding one function, where a commuting partner is automatically sharp, and a commutator such as iħ that annihilates nothing, where no state is sharp in both.
Test it: what changes?
- P(even about M₂) falls from 1 to ½
- P(M₁ repeats) falls from 1 to ½
- Commutator size grows from 0 to 2
- Energy stays sharp at 1.50 eV
- Commutator on the level vanishes
- M₂ reads odd with certainty
- M₁ repeat becomes certain
- Sharp energy rises to 24.1 eV
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron in an infinite well from x = −0.60 nm to x = +0.60 nm is prepared in ψ = (2ψ₁ + ψ₂ + 2ψ₃)/3, where ψₙ are the energy eigenfunctions. Find the probability of the pair of results “E₁ and even parity” when energy is measured first and when parity is measured first, and give the state after each step.
View solution →
- Parity of each level. Centred on x = 0, ψ₁ = √(2/a) cos(πx/a) and ψ₃ = √(2/a) cos(3πx/a) are even and ψ₂ = √(2/a) sin(2πx/a) is odd. The well is symmetric, so [Ĥ, Π̂] = 0, and each level holds one function, so every ψₙ is sharp in both. Normalisation: 4/9 + 1/9 + 4/9 = 1.
- Levels, with a = 1.20 nm: E₁ = h²/(8ma²) = (6.626 × 10⁻³⁴ J s)² ÷ (8 × 9.109 × 10⁻³¹ kg × (1.20 × 10⁻⁹ m)²) = 4.18 × 10⁻²⁰ J = 0.261 eV, so E₂ = 4E₁ = 1.04 eV and E₃ = 9E₁ = 2.35 eV.
- Energy first: P(E₁) = (2/3)² = 4/9, leaving ψ₁. That function is even, so parity then reads even with certainty. Joint probability: 4/9 × 1 = 0.444.
- Parity first: P(even) = 4/9 + 4/9 = 8/9, leaving (2ψ₁ + 2ψ₃)/3 ÷ √(8/9) = (ψ₁ + ψ₃)/√2. Energy then gives E₁ with probability ½, leaving ψ₁. Joint probability: 8/9 × ½ = 4/9 = 0.444, the same number and the same final state.
- Checks: P(odd) = 1/9 = P(E₂), because parity only asks whether n is odd or even. Summed over both parity results P(E₁) is still 4/9, so a parity measurement nobody reads leaves ⟨E⟩ = (4 × 1 + 1 × 4 + 4 × 9)E₁/9 = 44E₁/9 = 1.28 eV unchanged.
Answer. P(E₁ and even) = 4/9 ≈ 0.444 in either order, ending in ψ₁ both ways; after “even” alone the state is (ψ₁ + ψ₃)/√2. The levels are E₁ = 0.261 eV, E₂ = 1.04 eV and E₃ = 2.35 eV.
MediumExample 2
An electron on a ring of circumference L = 1.00 nm is prepared in ψ = [e(2πis/L) + 2e(−2πis/L)]/√(5L), with s the distance round the ring from a mirror line. Find P(even parity) when parity is measured directly, when an unread momentum measurement comes first, and when an unread energy measurement comes first. What does the comparison show?
View solution →
- Check the state: (1 + 4)/5 = 1, so ψ is normalised, and both terms lie on level n = 1, so E = h²/(2mL²) = 1.50 eV with certainty. Momentum is ±h/L = ±6.63 × 10⁻²⁵ kg m s⁻¹, with P(+) = 1/5 = 0.20 and P(−) = 4/5 = 0.80.
- Parity directly. The even part [ψ(s) + ψ(−s)]/2 = 3cos(2πs/L)/√(5L), and ∫ 9cos²(2πs/L)/(5L) ds round the ring = 9/10. The odd part, −i sin(2πs/L)/√(5L), carries 1/10. So P(even) = 0.90, and 0.90 + 0.10 = 1.
- Momentum first. Either result leaves a single travelling wave, e(±2πis/L)/√L, whose even part carries exactly ½. Summed over the two unread results: P(even) = 0.20 × 0.5 + 0.80 × 0.5 = 0.50.
- Energy first. The result is 1.50 eV with certainty and ψ lies wholly on level 1, so the state is left as it was and P(even) stays 0.90.
- Meaning: [Ĥ, Π̂] = 0, so an energy measurement, read or not, leaves the parity odds alone; [p̂, Π̂] ≠ 0, so a momentum measurement moves them from 0.90 to 0.50. Parity depends on the relative phase of the two travelling waves, which a momentum result discards: with 2i in place of 2 the momentum odds are unchanged, but P(even) = |1 + 2i|²/10 = 0.50 directly.
Answer. P(even) = 0.90 measured directly, 0.50 after an unread momentum measurement, and 0.90 after an unread energy measurement. Only the incompatible measurement in between changes the odds.
HardExample 3
Two mirror lines through the centre of a ring of circumference 1.00 nm cross at α = 22.5°; Π̂₁ and Π̂₂ reflect ψ in each. An electron found even about M₁ is tested for M₂ parity, then M₁ again: on level n = 1 in one run, n = 2 in another. For each, find the M₂ odds, the chance M₁ repeats, and the commutator's size on the level; then find the lowest level above n = 0 sharp in both mirrors, and its energy.
View solution →
- Set up. Let φ = 2πs/L be the angle round the ring, with M₁ at φ = 0 and M₂ at φ = α; reflection in M₂ sends φ to 2α − φ. On level n, “even about M₁” leaves √(2/L) cos nφ, and M₂'s eigenfunctions there are √(2/L) cos n(φ − α), even, and √(2/L) sin n(φ − α), odd. Ĥ commutes with both mirrors, so Eₙ = n² × 1.50 eV stays sharp throughout.
- Overlaps round the ring: (2/L)∫ cos nφ cos n(φ − α) ds = cos nα and (2/L)∫ cos nφ sin n(φ − α) ds = −sin nα. So P(M₂ even) = cos²nα: 0.854 on n = 1, where nα = 22.5°, and 0.500 on n = 2, where nα = 45°. P(M₂ odd) is 0.146 and 0.500; each pair sums to 1.
- M₁ again. Each M₂ branch returns to “even about M₁” with the same squared overlap it was reached by, so P(repeat) = cos⁴nα + sin⁴nα = 1 − ½sin²(2nα): 0.750 on n = 1 and 0.500 on n = 2, the two-outcome floor.
- Commutator. Π̂₁Π̂₂ψ(φ) = ψ(φ + 2α) and Π̂₂Π̂₁ψ(φ) = ψ(φ − 2α), so on cos nφ it gives cos(nφ + 2nα) − cos(nφ − 2nα) = −2 sin(2nα) sin nφ, of size 2|sin(2nα)|: √2 = 1.41 on n = 1 and 2.00 on n = 2. Here P(repeat) = 1 − ⅛[2 sin(2nα)]², so the larger commutator gives the worse repeat.
- Shared level. Both parities are sharp when nα is a multiple of 90°, first at n = 4: reflecting cos 4φ in M₂ gives cos(4(45° − φ)) = cos(180° − 4φ) = −cos 4φ, even about M₁ and odd about M₂, and 2|sin 180°| = 0. Its energy is E₄ = 16h²/(2mL²) = 16 × 1.504 eV = 24.1 eV.
- Checks: at α = 0 every overlap is 1 and every repeat certain; at α = 90° the rotations by +180° and −180° coincide, so perpendicular mirrors commute outright. Overlaps and reflections are pure numbers, and each pair of probabilities sums to 1.
Answer. n = 1: M₂ even 0.854, odd 0.146, M₁ repeats 0.750, commutator size 1.41. n = 2: 0.500 each, repeat 0.500, size 2.00. The lowest shared level is n = 4, at 24.1 eV; no state on levels 1 to 3 is sharp in both.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Commuting means a complete shared basis, not every state shared.
If [Â, B̂] = 0, every state expands on functions sharp in both, the joint odds are the same in either order, and an A–B–A sequence always repeats. A nonzero commutator permits shared states only among those it annihilates; iħ annihilates none. All of this assumes ideal measurements in quick succession.
Next: The general uncertainty relation from the Schwarz inequality →