Measurement probabilities and what a superposition is not
Energy odds come from |cₙ|²; relative phase is still physical.
Students leave the expansion lesson able to compute every cₙ and still believing that a superposition is a particle sitting in one level nobody has looked at yet. Here the |cₙ|² become measurement probabilities, the projection rule says what to carry forward, and one measurable number shows that the phases matter.
Build the model
The |cₙ|² fix the energy statistics; the phases they leave out are just as physical.
1The big idea
Expand a state of the well as Ψ = Σ cₙψₙ and the measurement postulate reads off everything an energy measurement can say: the only possible results are the eigenvalues Eₙ, the chance of each is |cₙ|², and a result Eₙ leaves the system in ψₙ, so an immediate repeat gives Eₙ again. The rule is built to be consistent: Σ |cₙ|² = 1 because Ψ is normalised, and the average Σ |cₙ|²Eₙ equals ∫Ψ*ĤΨ dx. What it does not license is the reading that the particle secretly had one Eₙ all along. The |cₙ|² are only moduli; each cₙ also carries a phase, and relative phases reshape |Ψ|² through cross terms such as 2|c₁||c₂| cos φ ψ₁ψ₂. So a superposition and a coin-toss mixture with the same weights share every energy statistic yet give different position statistics, and one number, the chance of finding the particle in the left half of the well, tells them apart. Only a phase common to every cₙ is unobservable. The rule has edges. It describes an ideal, sharp, non-destructive measurement; a detector that absorbs the particle, or cannot resolve neighbouring levels, needs more. It assumes each Eₙ belongs to one ψₙ, true in this well and false once levels are degenerate. And it says which state to carry forward, not how the change happens: that question divides the interpretations of quantum mechanics and stays open here.
2In one sentence
In a state Ψ = Σ cₙψₙ, an energy measurement returns one eigenvalue Eₙ with probability |cₙ|² and leaves the system in ψₙ, so an immediate repeat returns Eₙ again.
Example. An electron in a 1.00 nm well prepared in (ψ₁ + ψ₂)/√2 gives 0.376 eV or 1.504 eV, each in about half of many runs, and never the mean 0.940 eV. After a 1.504 eV result, an immediate repeat of the energy measurement returns 1.504 eV again.
3An energy measurement returns a level, never an average
Take a state already expanded on the well's eigenfunctions, Ψ = Σ cₙψₙ. The measurement postulate for energy says two things. The only possible results are the eigenvalues Eₙ = n²π²ħ²/(2mL²). The probability of Eₙ is |cₙ|², which is why Σ |cₙ|² = 1 matters: it makes the list of probabilities complete. It follows that a level whose coefficient is zero never appears; a state symmetric about the centre of the well has cₙ = 0 for every even n, so E₂, E₄, … are never found. For an electron in a 1.00 nm well prepared in (ψ₁ + ψ₂)/√2, every run returns 0.376 eV or 1.504 eV. The mean over many runs, Σ |cₙ|²Eₙ = 0.940 eV, equals ∫Ψ*ĤΨ dx, which is the consistency check on the rule, but no single run ever returns it. Limiting case: if Ψ = ψ₃, then c₃ = 1, every other coefficient is zero, and the measurement is certain to give E₃.
4A probability is a statement about a preparation
One electron, measured once, gives one number. The |cₙ|² describe the procedure that prepared it: repeat that procedure N times, measure each copy once, and the count of Eₙ approaches N|cₙ|². Each run is an independent trial, so the count scatters binomially, by √(Np(1 − p)) with p = |cₙ|². For 100 runs of (ψ₁ + ψ₂)/√2 that is 50 ± 5 results of E₁; for 10 000 runs it is 5000 ± 50, so the fraction tightens as 1/√N while each individual outcome stays as unpredictable as ever. Two things follow. Measuring one electron twice is not a second sample of the distribution, because the first measurement has changed the state. And a single result, E₂ say, does not show that |c₂|² was large. Only the histogram over many identical preparations tests the numbers, which is how every prediction in this unit is checked.
5The projection rule, and what it leaves unexplained
Experiment adds a second fact: measure the energy again straight away and the first result comes back every time. The projection postulate encodes it. After the result Eₙ we carry the state forward as ψₙ, whose expansion has the coefficient 1 on ψₙ and zero elsewhere, so the repeat is certain. In the isolated well it stays certain later too, because ψₙ is stationary. For position the word immediately matters: a particle just found in a narrow region is left in a narrow state, a superposition of many ψₙ, which spreads at once. The rule idealises a sharp, non-destructive measurement; a detector that absorbs the particle leaves nothing to repeat on, and one that cannot resolve adjacent levels cannot select a single ψₙ. What the rule does not say is how or when Ψ becomes ψₙ. A physical collapse, a branching of the world, a hidden trajectory, an update of what we assign to the preparation: the interpretations in common use agree on every probability this course computes and disagree about that question, which we leave open.
6A superposition is not a coin you have not looked at
Compare two sources. The first prepares (ψ₁ + ψ₂)/√2; the second tosses a coin and prepares ψ₁ or ψ₂. Their energy histograms are identical: half E₁, half E₂. Now measure position the moment each particle is prepared, and ask only whether it is in the left half of the well. For the coin-toss source, classical probability applies: P(left) = ½P(left | ψ₁) + ½P(left | ψ₂) = ½, since each ψₙ² is symmetric about L/2. For the superposition, |Ψ|² = ½ψ₁² + ½ψ₂² + ψ₁ψ₂, and the cross term is positive in the left half and negative in the right, because sin(2πx/L) changes sign at the centre. With u = πx/L, ∫₀(L/2) ψ₁ψ₂ dx = (4/π)∫₀(π/2) sin²u cos u du = 4/(3π), so P(left) = ½ + 4/(3π) = 0.924. That is a difference of 0.42 in one directly measurable probability. Weighting conditional probabilities is exactly the rule a classical ensemble obeys; the superposition breaks it, and the cross term is the size of the break.
7Relative phase is physical, global phase is not
Write c₂/c₁ = (|c₂|/|c₁|)e(iφ). The probabilities |cₙ|² ignore φ, but the density does not: for real ψₙ the cross term is 2|c₁||c₂| cos φ ψ₁ψ₂. Change c₂ from +1/√2 to −1/√2 and P(left) falls from 0.924 to 0.076: the particle is now found in the right half in 92.4% of runs. Change it to i/√2 and cos φ = 0, so P(left) = 0.500, the coin-toss value; yet this state is still not the mixture, since its mean momentum is not zero and its density starts to move, as UQMI-06.08 shows. Multiply every coefficient by the same e(iα), on the other hand, and every |cₙ|², every cross term and every prediction is unchanged: a global phase is not physics. Counting makes the point exact. N complex coefficients carry 2N real numbers; normalisation and the global phase remove two, leaving 2N − 2. The moduli supply N − 1 of those, so the other N − 1, the relative phases, are information that no energy histogram can hold.
8Measuring the energy erases the phase
Run the superposition source again, but measure energy first and position straight after. Each particle is now in ψ₁ or ψ₂, and each ψₙ² is symmetric about L/2, so P(left) = ½ for every φ: the 0.924 has gone, and the ensemble is indistinguishable from the coin-toss source. So an energy measurement is not a passive reading of a value already present: a passive reading could not change the position statistics that follow, and this one moves them from 0.924 to 0.500. The standing checks all pass. P(left) + P(right) = 1 for every φ. |Ψ|² carries m⁻¹ and every probability is dimensionless. As |c₂| → 0 the left-half probability returns to ½, the value for ψ₁ alone. And once the phases are scrambled, by a measurement or by anything else that records the energy, only the |cₙ|² survive: the ensemble behaves as the mixture and ordinary probability takes over.
Key equations
What each relation says, and the units it is written in.
Possible results of an energy measurement
P(Eₙ) = |cₙ|², for Ψ = Σ cₙψₙ
The squared modulus of each coefficient is the chance of that level. A level with cₙ = 0 never appears, whatever the other coefficients are.
cₙ dimensionless, ψₙ in m⁻¹⁄², Eₙ in J or eV. Only eigenvalues occur, and Σ |cₙ|² = 1 makes the list complete.
Projection rule
result Eₙ found ⇒ Ψ → ψₙ
An immediate repeat then finds the coefficient of ψₙ equal to 1 and returns Eₙ with certainty. The rule says which state to use next, not how the change happens.
Up to an overall phase. Each level of this well has one eigenfunction, so the result fixes the state.
Mean of many energy results
⟨E⟩ = Σ |cₙ|² Eₙ
An average over identically prepared systems. For (ψ₁ + ψ₂)/√2 it is 2.5E₁, which sits between the levels and is never a single result.
⟨E⟩ in J or eV. In this well Eₙ = n²E₁, so ⟨E⟩ = E₁ Σ n²|cₙ|².
Position density of a two-level superposition
|Ψ|² = |c₁|²ψ₁² + |c₂|²ψ₂² + 2|c₁||c₂| cos φ ψ₁ψ₂
The cross term depends on the relative phase and moves probability across the well while |c₁|² and |c₂|² stay fixed.
Real ψₙ in m⁻¹⁄², so |Ψ|² in m⁻¹; φ = arg(c₂/c₁) in rad. A phase shared by c₁ and c₂ cancels.
Coin-toss mixture with the same weights
ρ(x) = |c₁|²ψ₁² + |c₂|²ψ₂²
Identical energy statistics to the superposition. Any difference in position statistics is the cross term, and so the relative phase, at work.
ρ in m⁻¹. Each run is prepared in one eigenstate, so the density is a weighted average with no cross term.
Left-half probability in the infinite well
P(0 ≤ x ≤ L/2) = ½ + (8/(3π)) |c₁||c₂| cos φ
One measurable number that reads cos φ: 0.924 for (ψ₁ + ψ₂)/√2, 0.076 for (ψ₁ − ψ₂)/√2, 0.500 for the mixture.
Dimensionless. Uses ∫₀(L/2) ψ₁ψ₂ dx = 4/(3π) = 0.424, and ½ for each ψₙ² over half the well.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Turn φ from 0° to 180°: the solid curve swings from the left half to the right while the bars and the dashed mixture never move. At 90° the curves coincide, yet that density still moves in time and the mixture does not. Then set the switch to 1: every φ gives the dashed curve.
Live results
- P(E₂) = |c₂|²
- 0.50
- MEAN ENERGY ⟨E⟩
- 2.50 E₁
- P(LEFT HALF)
- 0.924
- MEAN POSITION ⟨x⟩
- 0.320 L
Catch the common trap
The mistake this lesson is built to prevent.
The tempting reading of Ψ = (ψ₁ + ψ₂)/√2 is that each particle really is in ψ₁ or really in ψ₂, and the ½ records our ignorance, like a coin under a cup. That picture makes a definite, testable prediction: every probability is the weighted average of what the two eigenstates give, so the chance of finding the particle in the left half of the well is ½ × ½ + ½ × ½ = 0.50. The superposition gives 0.92 at preparation, and 0.08 if c₂ changes sign, because |Ψ|² carries the cross term ±ψ₁ψ₂, which no average of eigenstates contains. The ½ is real, but it is the probability of what an energy measurement will return, not a report of an energy the particle already has. Nor does the particle hold the average energy 2.5E₁ instead: no run ever returns it.
Test it: what changes?
- P(E₁), P(E₂) and ⟨E⟩ unchanged
- P(left half) at preparation falls 0.924 → 0.076
- Equals the mixture's 0.500 at φ = 90°
- Coin-toss mixture stays at 0.500
- P(left half) becomes 0.500 for every φ
- Cross term drops out of the position histogram
- Energy probabilities |cₙ|² unchanged
- Ensemble matches the coin-toss mixture
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron in an infinite well of width 0.500 nm is prepared, run after run, in Ψ = (2ψ₁ − iψ₃)/√5. Which energies can a measurement return, with what probabilities, and what is the mean over many runs? In 1000 runs, how many results of E₃ should you expect, and how far will that count scatter? One run returns E₃: what does an immediate second energy measurement give?
View solution →
- Check normalisation first: |2/√5|² + |−i/√5|² = 4/5 + 1/5 = 1, so the |cₙ|² can be read as probabilities. The factor −i has modulus 1 and changes no probability.
- Only eigenvalues can occur. With ħ²/(2m) = 0.0381 eV nm² for an electron, E₁ = π² × 0.0381 eV nm² ÷ (0.500 nm)² = 1.504 eV, and E₃ = 9E₁ = 13.54 eV. So P(E₁) = 0.80 and P(E₃) = 0.20, while P(E₂) = 0 because Ψ has no ψ₂ component.
- Mean: ⟨E⟩ = 0.80 × 1.504 eV + 0.20 × 13.54 eV = 3.91 eV, which is 2.6E₁. No level has n² = 2.6, so no single run ever returns 3.91 eV: it describes the list of results, not any one of them.
- Counts: each run is an independent trial with p = 0.20, so 1000 runs give about 200 results of E₃, with binomial scatter √(1000 × 0.20 × 0.80) = 12.6. Batches between about 187 and 213 are typical.
- Repeat: the result E₃ leaves the electron in ψ₃, whose only coefficient is 1 on ψ₃ itself, so an immediate second measurement returns 13.54 eV with probability 1. The first result was random; the second is not.
Answer. E₁ = 1.50 eV with probability 0.80 and E₃ = 13.5 eV with probability 0.20, never E₂; ⟨E⟩ = 3.91 eV is not itself a possible result; about 200 ± 13 of 1000 runs give E₃; after an E₃ result a repeat gives 13.5 eV with certainty.
MediumExample 2
In an infinite well of width L, compare three preparations: Ψ = (√3ψ₁ + ψ₂)/2, Ψ′ = (√3ψ₁ − ψ₂)/2, and a source that emits ψ₁ in three runs out of four and ψ₂ otherwise. For each, find P(E₁), P(E₂), ⟨E⟩ and the probability of finding the particle in the left half, 0 ≤ x ≤ L/2, at the moment of preparation. How many of 2000 runs land in the left half?
View solution →
- Energy: all three give P(E₁) = (√3/2)² = 3/4 and P(E₂) = (1/2)² = 1/4, so ⟨E⟩ = (3/4)E₁ + (1/4)(4E₁) = 1.75E₁ = 7π²ħ²/(8mL²). The minus sign in Ψ′ and the coin in the mixture leave no trace on an energy histogram.
- Densities: with real ψₙ, |Ψ|² = (3/4)ψ₁² + (1/4)ψ₂² ± (√3/2)ψ₁ψ₂, the cross term 2c₁c₂ψ₁ψ₂ taking the sign of c₂. The mixture has no cross term, because every run is in one eigenstate: its density is (3/4)ψ₁² + (1/4)ψ₂².
- Integrals over 0 ≤ x ≤ L/2: each ψₙ² is symmetric about L/2 and contributes ½. With u = πx/L, ∫ψ₁ψ₂ dx = (4/π)∫₀(π/2) sin²u cos u du = 4/(3π) = 0.4244.
- So P(left) = ½ ± (√3/2)(0.4244) = 0.5 ± 0.3676: 0.868 for Ψ, 0.132 for Ψ′ and 0.500 for the mixture. Checks: P(left) + P(right) = 1 in each case, and letting c₂ → 0 returns ½, the value for ψ₁ alone.
- Counts in 2000 runs: about 1735, 265 and 1000. The binomial scatter is at most √(2000 × 0.25) = 22, so the three sources sit more than 30 standard deviations apart: easily told apart by position, never by energy.
Answer. P(E₁) = 3/4, P(E₂) = 1/4 and ⟨E⟩ = 1.75E₁ for all three; P(left) = 0.868 for Ψ, 0.132 for Ψ′ and 0.500 for the mixture, about 1735, 265 and 1000 of 2000 runs.
HardExample 3
Electrons in an infinite well of width L = 1.00 nm are prepared in Ψ = (ψ₁ + e(iφ)ψ₂)/√2 with φ unknown. In batch 1, position is measured at once in 4000 runs, and 2960 electrons are found in the left half. In batch 2, also of 4000 runs, energy is measured first and position straight after. Find φ with its uncertainty, say what the data leave open, predict batch 2, and test the claim that each electron already had E₁ or E₂.
View solution →
- Energy data cannot help: for every φ, P(E₁) = P(E₂) = ½, so each run returns 0.376 eV or 1.504 eV, each with probability ½. The phase has to be read from something that feels the cross term.
- At preparation |Ψ|² = ½ψ₁² + ½ψ₂² + cos φ ψ₁ψ₂. Over the left half each ψₙ² gives ½ and ψ₁ψ₂ gives 4/(3π), so P(left) = ½ + 0.4244 cos φ.
- Data: P(left) = 2960/4000 = 0.7400, with binomial uncertainty √(0.74 × 0.26/4000) = 0.0069. Then cos φ = (0.7400 − 0.5)/0.4244 = 0.565 ± 0.016, so φ = ±55.6°, and the uncertainty in φ is 0.0163/sin 55.6° = 0.0198 rad = 1.1°.
- The sign stays open. Replacing φ by −φ turns Ψ into its complex conjugate, since ψ₁ and ψ₂ are real, and |Ψ*|² = |Ψ|² at every x. The two are still different states: their mean momenta, ⟨p⟩ = −(8ħ/(3L)) sin φ, are ∓2.32 × 10⁻²⁵ kg m s⁻¹, so momentum measurements on many runs would tell them apart.
- Batch 2: the energy measurement leaves each electron in ψ₁ or ψ₂, each with a density ψₙ² symmetric about L/2, so P(left) = ½ whatever φ was. Expect 2000 ± 32 left-half counts, and 2000 ± 32 results of 1.504 eV. The first measurement has erased the phase.
- The claim that each electron already had E₁ or E₂ makes batch 1 a coin-toss mixture, predicting 2000 ± 31.6 left-half counts. The observed 2960 sits (2960 − 2000)/31.6 = 30 standard deviations away, so that reading is ruled out.
Answer. cos φ = 0.565 ± 0.016, so φ = ±(55.6 ± 1.1)°, the sign left open by position data at preparation; batch 2 gives about 2000 ± 32 left-half counts and 2000 ± 32 results of 1.504 eV; the already-had-an-energy claim misses batch 1 by 30 standard deviations.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Each run returns one eigenvalue Eₙ, with probability |cₙ|².
An ideal energy measurement returns Eₙ with probability |cₙ|² and leaves ψₙ, so an immediate repeat gives Eₙ again. ⟨E⟩ = Σ|cₙ|²Eₙ describes many runs, not one. Weights alone do not fix the state: relative phases change position statistics, which is how a superposition differs from a mixture.
Next: Time evolution: each eigenstate carries its own phase →