University Physics IV · Foundations of Quantum Mechanics · 5.1
Probability Amplitudes & Superposition
Everything strange about quantum mechanics that you can actually calculate starts here. Learn when to add amplitudes and when to add probabilities, get the cross term right, and you can predict any two-path interference pattern in this course — including the ones that switch off the moment something keeps a record.
Build the model
Connect the measurement to the mechanism.
Classical probability says that if an event can happen two ways, add the two probabilities. Quantum mechanics keeps the arithmetic but replaces the objects: each way of reaching an outcome carries a complex number, the amplitude, those are added, and only the sum is squared. The order matters. |ψ₁ + ψ₂|² is not |ψ₁|² + |ψ₂|², and the difference is a cross term 2Re(ψ₁*ψ₂) = 2√(P₁P₂) cos φ that swings from +2√(P₁P₂) to −2√(P₁P₂) as the relative phase runs through a cycle.
That single term is the whole of interference: it is why electrons fired one at a time still build fringes, and why blocking a path can make a detector click more often. The rule comes from experiment rather than argument — Young with light, Davisson–Germer with electrons, then neutrons, atoms and molecules — and it arrives with a condition attached. Amplitudes may be added only across alternatives left indistinguishable; if anything anywhere records which path was taken, the cross term dies and probabilities add after all.
What it costs is an explanation. The rule says the record matters but not why, nor when a record counts as made. That debt is the measurement problem, and this topic hands it forward rather than paying it.
- Simple definition
- A probability amplitude is a complex number attached to one way an outcome can happen; amplitudes for alternatives that stay indistinguishable are added first, and the probability is the squared modulus of the sum.
- Example
- In a balanced interferometer each path alone puts probability 0.250 on detector D₁, so adding probabilities predicts 0.500; adding amplitudes gives |0.500 + 0.500e(iφ)|², which is 1.000 in phase and exactly 0 antiphase.
The order of the two operations is the entire content of the rule — square first and every interference effect in this course disappears.
ψⱼ is the dimensionless amplitude for alternative j; P is a probability, so both sides are pure numbers.
Names the interference explicitly. It is bounded by ±2√(P₁P₂), so it is largest for balanced paths and vanishes the moment either path is blocked.
Pⱼ = |ψⱼ|² is the probability with alternative j alone available; φ is a relative phase in radians.
Compresses a whole pattern into one measurable number, and shows that imbalance alone dims fringes with no which-path record anywhere.
Dimensionless, with 0 ≤ V ≤ 1; V = 1 requires exactly balanced paths, P₁ = P₂.
Turns geometry into phase: 50 keV electrons have λdB = 5.4 pm, so a path difference of only 2.7 pm flips a maximum into a minimum.
ΔS in J s, ħ = 1.055 × 10⁻³⁴ J s, Δℓ the path-length difference in m, λdB = h/p.
Tells you which phases you may discard when normalising, and which one carries the entire pattern.
α is real and dimensionless; only phase differences between amplitudes being added are observable.
Makes 'indistinguishable' quantitative: a marker of 50% overlap leaves half-depth fringes and still lets you name the path 93% of the time.
D is the distinguishability of the path record, 0 ≤ D ≤ 1; equality holds when particle and marker are both pure.
Square last: the one change from classical probability
Classical probability has a rule for alternatives: if an outcome can be reached two ways, add the two probabilities. Quantum mechanics keeps that arithmetic but changes the objects. Each way of reaching the outcome carries a complex amplitude, the amplitudes are added, and only the sum is squared. Take a balanced two-path interferometer in which each path delivers an amplitude of modulus 0.500 to detector D₁. Block either path and D₁ fires with probability 0.250, so probability-adding predicts 0.500 no matter what else you do. Amplitude-adding predicts |0.500 + 0.500e(iφ)|² = 0.500(1 + cos φ): 1.000 when the two arrive in phase, 0.500 in quadrature, and exactly 0 when they arrive antiphase. At antiphase, blocking one path raises the count at D₁ from 0 to 0.250 — removing a way to arrive has made arrivals more likely. No assignment of non-negative probabilities to the two paths can produce that, which is why the change of order is not a bookkeeping preference.
Where the cross term comes from
Multiply out the square and the extra term appears on its own. Writing each amplitude in polar form, ψⱼ = √Pⱼ e(iθⱼ), we get |ψ₁ + ψ₂|² = (ψ₁ + ψ₂)*(ψ₁ + ψ₂) = |ψ₁|² + |ψ₂|² + ψ₁*ψ₂ + ψ₂*ψ₁ = P₁ + P₂ + 2Re(ψ₁*ψ₂) = P₁ + P₂ + 2√(P₁P₂) cos φ, with φ = θ₂ − θ₁. Three consequences follow. The factor of 2 is not decoration: it counts both cross products in the expansion, and dropping it is the commonest slip in this topic. The term is bounded by ±2√(P₁P₂), so it is largest for balanced paths and dies when either path is blocked. And it depends only on the difference of the phases, so multiplying the whole state by e(iα) changes nothing physical. With P₁ = 0.09 and P₂ = 0.01 the cross term ranges over ±0.06, giving a maximum of 0.16, a minimum of 0.04 and a visibility of 0.12/0.20 = 0.60. Integrated over the whole screen the cross term sums to zero: it moves probability from the dark fringes into the bright ones, and never manufactures any.
What sets the relative phase
The phase is not a free parameter to be fitted; it is accumulated along the path. For a free particle the action over a path length ℓ is pℓ, so φ = ΔS/ħ = pΔℓ/ħ = 2πΔℓ/λdB, with λdB = h/p. Every geometrical feature of an interference experiment enters at this point. In a two-slit geometry with slit separation d, the path difference to a screen point at angle θ is d sin θ, so maxima sit where d sin θ = mλ. Put numbers on it: 50 keV electrons have λdB = 5.4 pm, so slits 100 nm apart throw fringes λL/d = 27 µm apart on a screen 0.50 m away — a pattern you can record, generated by path differences of a few picometres. Phase also accumulates in time, as e(−iEt/ħ), so a path that spends time in a region of different potential energy arrives shifted. That is the mechanism behind the Aharonov–Bohm phase, and behind neutron interferometers that detect a gravitational phase shift across a height difference of a few centimetres.
Add amplitudes only across indistinguishable alternatives
The rule carries a condition, and the condition is about the rest of the world. Add amplitudes when the two alternatives leave everything else — apparatus, environment, any stray photon — in the same final state. Add probabilities when they leave it in perfectly distinguishable states. Nothing in the condition mentions a person: it is the existence of the record that kills the cross term, not anyone reading it, and a record already buried in a detector's thermal noise counts in full. This is testable, not philosophical. Fullerene molecules interfering in a Talbot–Lau interferometer lose their fringes as the molecules are heated, because a hot C₇₀ emits thermal photons whose wavelength eventually becomes short enough to localise which path it came from; visibility falls as the internal temperature rises, with no mechanical disturbance applied at all. The same condition run in reverse explains identical-particle statistics: exchange paths for two identical particles are indistinguishable in principle, so their amplitudes add, and that is the origin of bosonic and fermionic behaviour.
Partial records grade the interference
Between 'no record' and 'perfect record' lies a continuum, and it is quantitative. Let the path tag end in states |d₁⟩ and |d₂⟩. For balanced paths the cross term is multiplied by the overlap, so the visibility is V = |⟨d₁|d₂⟩|, while the best distinguishability the tag can support is D = √(1 − |⟨d₁|d₂⟩|²), and the two obey D² + V² ≤ 1, with equality when both states are pure. Overlap 1 gives V = 1 and D = 0: full fringes, no path information. Overlap 0.6 gives V = 0.6 and D = 0.8, so you can name the path correctly ½(1 + D) = 90% of the time and still watch fringes at 60% depth. Overlap 0 gives V = 0 and D = 1. Notice the second, quieter route to poor fringes: unequal path amplitudes. With P₁ ≠ P₂ the predictability |P₁ − P₂|/(P₁ + P₂) is already non-zero and V is already below 1, with no marker anywhere in the apparatus. Dim fringes are evidence of path knowledge available in principle, not evidence that a measurement was performed.
What the rule does not do
This is a calculational postulate with a hole in it. It tells you to add amplitudes when the alternatives are indistinguishable and probabilities when they are not, and it does not say why the existence of a record anywhere should change the arithmetic, nor when a record counts as made. Decoherence fills part of the gap — it shows how fast an environment records path information, and so why fringes are invisible for anything much larger than a molecule — but the step from a spread-out state to one definite click is still postulated separately, and topic 5.6 states it as exactly that. Two further cautions. The rule does not license the sentence 'the particle went through both slits': no amplitude is attached to that claim and no measurement tests it. And the amplitude belongs to a single particle's state, not to a beam. Electrons fired one per second at a double slit, hours apart, still build the complete fringe pattern, so the cross term is never two particles interfering with each other.
Change one variable at a time
Make the relationship visible.
Set amplitude ratio and marker overlap to 1 and sweep φ: P swings from 1 to 0, even though blocking either path alone pins it at the dashed 0.5. Then pull the overlap down to 0.5 — the fringes halve in depth without switching off, and D climbs to 0.87.
VISIBILITY V1.00
DISTINGUISHABILITY D0.00
P AT THIS PHASE1.000
CROSS TERM0.500
Live interpretationVISIBILITY V: 1.00. DISTINGUISHABILITY D: 0.00. P AT THIS PHASE: 1.000. CROSS TERM: 0.500
Catch the common trap
Explain before calculating.
A point P on the screen of a two-slit electron experiment receives detection probability 0.20 with slit 1 alone open, and 0.20 with slit 2 alone open. With both slits open and nothing anywhere recording which slit was used, the two amplitudes arrive at P with a relative phase of 60°. What is the detection probability at P?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyIn a balanced two-path interferometer, each path alone delivers an amplitude of modulus 0.500 to detector D₁, so blocking either path leaves probability 0.250 there. Find the probability at D₁ with both paths open at relative phases of 0, 90° and 180°, and say what happens to the count when you block one path at 180°.
- Each path contributes amplitude of modulus 0.500, so P₁ = P₂ = 0.500² = 0.250, and the classical probability-adding prediction is 0.250 + 0.250 = 0.500 at every phase.
- Add the amplitudes instead: ψ = 0.500 + 0.500e(iφ), so P = |ψ|² = 0.250 + 0.250 + 2(0.500)(0.500) cos φ = 0.500(1 + cos φ).
- φ = 0: P = 0.500(1 + 1) = 1.000. φ = 90°: P = 0.500(1 + 0) = 0.500, which is the classical value — the cross term is zero here, not absent. φ = 180°: P = 0.500(1 − 1) = 0.
- At φ = 180° both paths open gives 0, while blocking one path gives 0.250. Removing an alternative has made the detector click more often, which no assignment of non-negative probabilities to the two paths can reproduce: taking away a way to arrive can never add arrivals.
AnswerP = 1.000, 0.500 and 0 at φ = 0, 90° and 180°. Blocking one path at 180° raises D₁ from 0 to 0.250 — the fingerprint of a cross term, not of extra particles.
MediumA double-slit electron beam gives a count rate at point P of 900 s⁻¹ with slit 1 alone open and 100 s⁻¹ with slit 2 alone open. With both slits open and no which-path record anywhere, find the maximum and minimum rates as the relative phase is scanned, the fringe visibility, the rate at a relative phase of 120°, and check the answer against the duality relation.
- Rates are proportional to probabilities, so the same rule applies: R = R₁ + R₂ + 2√(R₁R₂) cos φ = 1000 + 2√(900 × 100) cos φ = 1000 + 600 cos φ, in s⁻¹.
- Maximum at cos φ = +1: Rₘₐₓ = 1600 s⁻¹. Minimum at cos φ = −1: Rₘᵢₙ = 400 s⁻¹. Equivalently (√900 ± √100)² = (30 ± 10)², which is 1600 and 400.
- Visibility: V = (1600 − 400)/(1600 + 400) = 1200/2000 = 0.600. The shortcut 2√(R₁R₂)/(R₁ + R₂) = 600/1000 gives the same 0.600.
- At φ = 120°, cos φ = −0.500, so R = 1000 + 600(−0.500) = 700 s⁻¹ — below the classical 1000 s⁻¹ but well clear of the 400 s⁻¹ minimum.
- The imbalance by itself already carries path information: predictability D = |R₁ − R₂|/(R₁ + R₂) = 800/1000 = 0.800, and D² + V² = 0.640 + 0.360 = 1.000. Nothing is watching the slits, yet the fringes are already down to 60% depth because any hit is nine times more likely to have come through slit 1.
AnswerRₘₐₓ = 1600 s⁻¹, Rₘᵢₙ = 400 s⁻¹, V = 0.600, and R(120°) = 700 s⁻¹. The path predictability is D = 0.800, with D² + V² = 1.000 — the imbalance alone saturates the trade.
HardAn atom interferometer with balanced paths tags each path with an internal state: |d₁⟩ on path 1 and |d₂⟩ on path 2, with a real overlap ⟨d₁|d₂⟩ = 0.500. The atom flux is 2000 s⁻¹ and the internal state is never measured. Find the fringe visibility and the rates at relative phase 0 and 180°, the distinguishability and the best possible single-shot guess of the path, and the overlap that would be needed to restore V = 0.900.
- After recombination the joint state is |ψ⟩ = (1/√2)(|1⟩|d₁⟩ + e(iφ)|2⟩|d₂⟩). The tag is never read, so it is summed over, and the cross term picks up the overlap factor: it becomes Re(⟨d₁|d₂⟩e(iφ)) = 0.500 cos φ.
- So P(φ) = ½[1 + 0.500 cos φ], and the visibility is V = |⟨d₁|d₂⟩| = 0.500 — half-depth fringes, not no fringes.
- Rates: R(0) = 2000 × ½(1 + 0.500) = 1500 s⁻¹, and R(180°) = 2000 × ½(1 − 0.500) = 500 s⁻¹. Check: (1500 − 500)/(1500 + 500) = 0.500, as required.
- Distinguishability of the tag: D = √(1 − |⟨d₁|d₂⟩|²) = √(1 − 0.250) = 0.866. The best possible single-shot guess of the path then succeeds with probability ½(1 + D) = ½(1.866) = 0.933.
- Check the trade: D² + V² = 0.750 + 0.250 = 1.000, saturated because atom and tag are both in pure states. You can name the path 93.3% of the time and still see fringes at half depth — the record grades the interference rather than switching it off.
- To reach V = 0.900 the overlap must rise to 0.900, which drops D to √(1 − 0.810) = 0.436 and the guessing rate to ½(1.436) = 0.718.
AnswerV = 0.500, with 1500 s⁻¹ and 500 s⁻¹ at φ = 0 and 180°; D = 0.866, so the path is guessed correctly 93.3% of the time, and D² + V² = 1.000. Reaching V = 0.900 needs overlap 0.900, cutting the guessing rate to 71.8%.