Wave-particle duality and the double slit
Indistinguishable routes add amplitudes; distinguishable routes add probabilities.
This is the experiment the rest of the course exists to explain, so we take it slowly: first what the detector actually records, then the one piece of arithmetic that reproduces it, then what it costs to ask which slit the electron used. By the end, adding amplitudes before squaring should feel forced on you by data, not handed down.
Build the model
Every electron lands as a dot; the dots are distributed like a wave.
1The big idea
The double slit delivers two facts that classical physics cannot hold together. Each electron is detected whole, at one point, like a particle; and the points from many electrons, sent so sparsely that two are almost never in flight together, pile up into fringes with the spacing λL/d that a wave of λ = h/p would make. The fringes include places where opening a second slit lowers the count, so whatever combines the two routes must be able to cancel. The model that fits gives each route a complex amplitude, adds the amplitudes of routes that nothing distinguishes, and squares the modulus of the sum to get the detection probability; the cross term 2|ψ₁||ψ₂| cos δ is the interference. The model stops at two boundaries. The moment the routes become distinguishable, through any record of the slit whether or not anyone reads it, probabilities add again, and partial records give partial fringes. And in the classical limit λ is so small that the fringes are finer than any detector, so averaging leaves P₁ + P₂. Wave and particle are not two modes the electron switches between: they are two questions an experiment can ask, and complementarity names the observed fact that no arrangement answers both at once.
2In one sentence
Wave-particle duality is the observed fact that a quantum object is always detected whole at a single point, while the statistics of many detections show interference, reproduced by adding the amplitudes of indistinguishable routes and then squaring.
Example. If slit 1 alone gives 100 counts per minute at some point and slit 2 alone gives 100, opening both gives anywhere from 0 to 400 per minute there, set by the phase difference; only the average over a whole fringe is 200.
3What the detector records: one dot at a time
Put a position-sensitive detector behind a double slit and turn the source down until electrons arrive one at a time. Each electron registers as a single localised event, one pixel, never a faint smear across the screen and never a fraction of an electron's charge. One at a time is a claim you can check with numbers. A 2.0 keV electron travels at 2.6 × 10⁷ m s⁻¹, so at 5000 electrons per second successive electrons are 5.3 km apart along the beam; in an apparatus 0.8 m long an electron therefore has company only about 1.5 × 10⁻⁴ of the time. Yet after 10⁴ to 10⁵ detections the dots have piled into bright and dark bands with exactly the spacing a wave of λ = h/p predicts. The fringes therefore belong to the statistics of single electrons, not to electrons interacting with one another. Electron biprism experiments in Bologna in the 1970s and at Hitachi in 1989 recorded this build-up frame by frame.
4Opening a second slit can lower the count
Count at one detector position y three ways: slit 1 open alone gives a rate R₁, slit 2 alone gives R₂, both open gives R₁₂. If each electron passed through one slit or the other, and neither slit cared whether the other was open, every electron reaching y would have come by exactly one route and R₁₂ = R₁ + R₂ everywhere. The data refuse this. At the centre of a dark fringe R₁₂ is close to zero while R₁ and R₂ are each healthy: opening the second slit has removed electrons from a place the first slit alone was sending them. No rule that adds non-negative numbers can produce a decrease. At a bright fringe with R₁ = R₂ the count is 4R₁, twice the sum. Averaged across a whole fringe, R₁₂ comes back to R₁ + R₂, so nothing is created or destroyed; electrons are moved from dark bands into bright ones. Whatever combines the two routes must be able to cancel, which means it carries a sign or a phase.
5Add the amplitudes, then square
The rule that fits is the one waves obey. Give each route a complex amplitude, ψ₁ = A₁e(iφ₁) and ψ₂ = A₂e(iφ₂), chosen so that A₁² and A₂² are the single-slit densities P₁ and P₂. With both slits open and nothing recording the route, add them, then multiply the sum by its complex conjugate: |ψ₁ + ψ₂|² = A₁² + A₂² + 2A₁A₂ cos(φ₂ − φ₁). The first two terms are the particle answer; the third is new, and it is the whole of the interference. Each route advances its phase by 2π per de Broglie wavelength, so δ = φ₂ − φ₁ = 2π(r₂ − r₁)/λ, which becomes 2πd sin θ/λ on a far screen. Now check the limits. Close slit 2 and A₂ = 0 returns the single-slit pattern. Equal amplitudes send the minima exactly to zero. With P₂ = P₁/9, A₂ = A₁/3, so the minima sit at (2/3)²P₁ = 0.44P₁ and the maxima at (4/3)²P₁ = 1.78P₁, giving V = 0.60. Complex amplitudes are the bookkeeping for phase that makes cancellation possible.
6Where the fringes fall, and why a ball shows none
On a screen a distance L ≫ d away, sin θ ≈ y/L, so δ = 2πdy/(λL) and bright fringes repeat every Δy = λL/d. Electrons with kinetic energy K = 150 eV have λ = h/√(2mₑK) = 100 pm; slits 250 nm apart and a screen 1.5 m away give Δy = 0.60 mm, easy to resolve. Now take the classical limit. A detector bin of width w averages cos δ across its width, which multiplies the interference term by sin(πw/Δy)/(πw/Δy). A bin half a fringe wide keeps 64% of the contrast, one exactly a fringe wide keeps none, and once w ≫ Δy only P₁ + P₂ survives. A 0.15 kg ball thrown at 30 m s⁻¹ has λ = 1.5 × 10⁻³⁴ m, and slits 1 m apart with a screen 10 m away put its fringes 1.5 × 10⁻³³ m apart, some eighteen orders of magnitude finer than a proton. The fringes are present in the arithmetic and invisible to every possible detector, which is why adding probabilities works for baseballs: classical physics is the limit in which the cross term averages away.
7Asking which slit costs the fringes
Shine light just behind the slits and look for the photon each electron scatters. When the light can tell the slits apart, every electron is seen at one slit or the other, never both and never half, and the detections sorted by that result reproduce P₁ and P₂; their sum has no fringes. The price can be counted. To resolve slits d apart the light needs λᵧ ≲ d, so each scattered photon carries momentum h/λᵧ ≳ h/d and hands the electron a transverse kick Δpy of up to that size. Divided by the electron's momentum p = h/λ, the kick is an angle of order λ/d, the angular period of the fringes. For the 150 eV electrons above, 250 nm light turns each electron through up to 4.0 × 10⁻⁴ rad, moving its landing point by up to 0.60 mm, one full fringe. The kicks are random, so the fringes smear into P₁ + P₂; lengthen λᵧ to spare them and the image blurs until it no longer says which slit. The kick is sufficient, not necessary: an atom interferometer in 1998 stored the path in the atom's internal state, with a recoil far too small to smear anything, and the fringes vanished all the same.
8Complementarity is read off the experiments
Collect the results and a rule of experience emerges, not a theorem. Any arrangement that could in principle reveal the route gives P₁ + P₂; any arrangement that could not gives |ψ₁ + ψ₂|²; partial information gives partial fringes. If a fraction f of the electrons leave a record of their slit and the rest leave none, the pattern is (1 − f)P₁₂ + f(P₁ + P₂) and the visibility falls to 1 − f of its full value, whether or not anyone ever reads the records. Bohr called the trade-off complementarity: an interference pattern and a which-way record answer mutually exclusive experimental questions, and no arrangement delivers both in full. This course takes it from the data rather than deriving it. The working rule is short enough to carry into every later problem: for routes that nothing distinguishes, add amplitudes; for routes that something distinguishes, add probabilities.
Key equations
What each relation says, and the units it is written in.
Amplitudes add for indistinguishable routes
ψ(y) = ψ₁(y) + ψ₂(y)
Holds only while nothing records the slit. The detection density is then |ψ₁ + ψ₂|², never |ψ₁|² + |ψ₂|².
ψ₁, ψ₂: complex amplitudes at screen position y with only slit 1 or only slit 2 open, in m⁻¹⁄² on a 1D screen
Two-path detection density
P₁₂ = P₁ + P₂ + 2√(P₁P₂) cos δ
The cross term is all of the interference. It swings between ±2√(P₁P₂) and averages to zero over a fringe, so electrons are redistributed, never created or lost.
P₁ = |ψ₁|² and P₂ = |ψ₂|², in m⁻¹; δ is the phase of ψ₂ relative to ψ₁, in rad
Phase difference on a distant screen
δ = 2π d sin θ / λ
Each route gains 2π of phase per wavelength travelled. Bright fringes sit where d sin θ = nλ, dark ones halfway between.
d: slit separation (m); θ: angle from the slit midpoint (rad); λ = h/p (m); valid for L ≫ d
Fringe spacing
Δy = λL/d = hL/(pd)
Higher momentum, finer fringes: doubling an electron's accelerating voltage shrinks Δy by √2, and a thrown ball's fringes lie far below any detector's resolution.
L: slit-to-screen distance (m); p: electron momentum (kg m s⁻¹); small angles and L ≫ d
Fringe visibility
V = (Pₘₐₓ − Pₘᵢₙ)/(Pₘₐₓ + Pₘᵢₙ)
Unequal slits dim the fringes without erasing them: P₂ = P₁/4 still gives V = 0.80. Only path information or averaging over fringes drives V to zero.
dimensionless, 0 ≤ V ≤ 1; for two coherent routes V = 2√(P₁P₂)/(P₁ + P₂)
Partial which-path record
P = (1 − f) P₁₂ + f (P₁ + P₂)
Visibility falls by the factor 1 − f. A record nobody reads counts in full: its existence, not anyone looking at it, removes the cross term.
f: fraction of electrons whose slit is recorded (dimensionless); P₁₂, P₁, P₂ in m⁻¹
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Start at N = 100, where the jagged trace is noise, and raise N to 10⁵ to watch fringes condense out of single dots. Then push f to 1: they sink onto the dashed sum. Lowering P₂/P₁ lifts the dark fringes off zero without removing them.
Live results
- ELECTRONS DETECTED N
- 1000
- FRINGE VISIBILITY V
- 1.00
- DARK FRINGE ÷ (P₁ + P₂)
- 0.00
- EXPECTED COUNTS, CENTRAL BIN
- 20
Catch the common trap
The mistake this lesson is built to prevent.
The tempting reading is that each electron really goes through one slit or the other and we simply do not know which. Follow it through and it predicts the wrong pattern. If every electron had a definite slit and landed as that slit alone would send it, the runs would split into two groups: those through slit 1, landing with density P₁, and those through slit 2, landing with density P₂. The total would be P₁ + P₂, a smooth hump with no zeros inside it. The measured pattern has zeros where P₁ + P₂ is large, so that picture fails, and for an unobserved electron the question of which slit it used has no answer in quantum mechanics, not merely an unknown one. The opposite picture, that the electron splits and half goes each way, fails too: a detector at the slits always fires at one slit with a whole electron's charge, and then the fringes are gone. Quantum mechanics claims neither; it assigns an amplitude to each route and adds them.
Test it: what changes?
- Visibility drops to (1 − f) of its full value
- Dark fringes rise, bright fringes fall
- Fringe-averaged count stays P₁ + P₂
- At f = 1 only P₁ + P₂ remains
- Dark fringes lift off zero
- Visibility 2√(P₁P₂)/(P₁ + P₂) falls
- Fringe spacing λL/d unchanged
- Fringes survive until P₂ = 0
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
A feeble-light double-slit experiment uses a 633 nm laser attenuated to 1.0 pW, and the light travels 1.2 m from the attenuator to the camera. How many photons arrive per second, how far apart are successive photons on average, and how often does a photon share the apparatus with another?
View solution →
- Photon energy: E = hc/λ = (6.626 × 10⁻³⁴ J s)(2.998 × 10⁸ m s⁻¹)/(633 × 10⁻⁹ m) = 3.14 × 10⁻¹⁹ J.
- Arrival rate: R = P/E = (1.0 × 10⁻¹² W)/(3.14 × 10⁻¹⁹ J) = 3.2 × 10⁶ photons per second.
- Mean spacing along the beam: c/R = (2.998 × 10⁸ m s⁻¹)/(3.19 × 10⁶ s⁻¹) = 94 m, nearly eighty times the length of the apparatus.
- Mean number in flight: μ = RL/c = 1.2 m/94 m = 0.013. Laser photons arrive at random, so given one photon in the apparatus, the chance that another is there too is 1 − e(−0.013) ≈ 1.3%.
- Limiting case: raise the power a millionfold to 1.0 μW and μ = 1.3 × 10⁴, some thirteen thousand photons in flight at once, an ordinary classical beam. The fringes look the same at both powers; only the exposure needed to see them changes.
Answer. 3.2 × 10⁶ photons per second, 94 m apart on average; a photon has company only about 1.3% of the time, so fringes built up under these conditions are made one photon at a time.
MediumExample 2
Electrons accelerated from rest through 600 V pass two slits 400 nm apart and land on a detector plane 0.60 m away. Near the centre a small counter records 250 per minute with only slit 1 open and 90 per minute with only slit 2 open, slit 2 being partly masked by an absorber that adds no phase. Treat those rates as constant over the central 100 μm. With both slits open, what does the counter record at the centre, and 30.0 μm from it?
View solution →
- Momentum and wavelength: p = √(2mₑeV) = √(2 × 9.109 × 10⁻³¹ kg × 1.602 × 10⁻¹⁹ C × 600 V) = 1.323 × 10⁻²³ kg m s⁻¹, so λ = h/p = 5.007 × 10⁻¹¹ m. The relativistic correction at 600 eV is about 0.03% and can be dropped.
- Fringe spacing: Δy = λL/d = (5.007 × 10⁻¹¹ m)(0.60 m)/(4.00 × 10⁻⁷ m) = 7.51 × 10⁻⁵ m = 75.1 μm, so the phase difference at position y is δ = 2πy/Δy.
- Add amplitudes, then square: R(y) = R₁ + R₂ + 2√(R₁R₂) cos δ = 250 + 90 + 2√22500 cos δ = 340 + 300 cos δ per minute.
- At the centre δ = 0, so R = 340 + 300 = 640 per minute, nearly twice the 340 that adding probabilities would predict.
- At y = 30.0 μm: δ = 2π × 30.0/75.1 = 2.510 rad, cos δ = −0.807, and R = 340 − 242 = 98 per minute.
- Checks: R runs between (√250 − √90)² = 40 and (√250 + √90)² = 640 per minute, averages 340 = R₁ + R₂ over a fringe, and has V = 300/340 = 0.88. Unmask slit 2 so that R₂ = 250 and the dark fringes would reach zero.
Answer. 640 per minute at the centre and about 98 per minute at 30.0 μm, where adding probabilities predicts 340 at both; fringes 75.1 μm apart swing between 40 and 640 per minute, V = 0.88.
HardExample 3
Each 600 eV electron of the medium problem (slits 400 nm apart, screen 0.60 m away) scatters one photon just behind the slits. The light arrives perpendicular to the line joining the slits and scatters isotropically, so the photon's final momentum along that line is uniform between −h/λᵧ and +h/λᵧ. Find the factor F by which the fringe visibility falls, and evaluate it for λᵧ = 2.0 μm and λᵧ = 300 nm.
View solution →
- The electron takes the equal and opposite kick, Δpy uniform on ±h/λᵧ; the incoming light carries no momentum along the slit line. Dividing by p = h/λ gives a deflection uniform on ±λ/λᵧ, which shifts the landing point by s, uniform on ±s₀ with s₀ = Lλ/λᵧ.
- In fringe units s₀/Δy = (Lλ/λᵧ)/(λL/d) = d/λᵧ: the screen distance, the electron's wavelength and its energy all drop out.
- Average the cross term over the shift: the mean of cos(2π(y − s)/Δy) is cos(2πy/Δy) × sin(2πs₀/Δy)/(2πs₀/Δy). So F = sin(2πd/λᵧ)/(2πd/λᵧ). Take the slits narrow enough that the single-slit envelope is broad beside these shifts; then P₁ + P₂ is unchanged.
- λᵧ = 2.0 μm: d/λᵧ = 0.20 and F = sin(0.4π)/(0.4π) = 0.951/1.257 = 0.76; landing points move by at most 15 μm, a fifth of a fringe. But light gathered from every direction resolves only about λᵧ/2 = 1.0 μm, wider than d, so it cannot say which slit.
- λᵧ = 300 nm: d/λᵧ = 1.33 and F = sin(2.67π)/(2.67π) = 0.866/8.38 = 0.10. The resolution λᵧ/2 = 150 nm is well below d, so the photon carries most of the path information, and the fringes keep only a tenth of their contrast.
- Limits: as λᵧ → ∞, F → 1, no information and full fringes. F first reaches zero at λᵧ = 2d = 800 nm, exactly where the resolution λᵧ/2 equals d. For λᵧ between d and 2d, F is negative and the surviving fringes are shifted by half a period; at all wavelengths |F| ≤ λᵧ/(2πd), a bound that keeps falling as λᵧ shrinks.
Answer. F = sin(2πd/λᵧ)/(2πd/λᵧ): 0.76 at 2.0 μm, where the light cannot resolve the slits, and 0.10 at 300 nm, where it can. Path information and fringe contrast trade against each other.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Add amplitudes, then square, unless something records the route.
With nothing recording the slit, P₁₂ = P₁ + P₂ + 2√(P₁P₂) cos δ, with bright fringes every λL/d. The cross term averages to zero over a fringe. Any which-path record, read or not, removes it; a recorded fraction f scales visibility by 1 − f.