Wave–Particle Duality

Beginner

For centuries scientists argued about light. Sent through two narrow slits, it paints bright and dark stripes, something only waves can do. Yet when light frees electrons from a metal (the photoelectric effect), it behaves like a hail of tiny energy packets called photons. Each packet carries the energy $E=hf$, where $f$ is the frequency, which sets the colour, and $h$ is a tiny fixed number, Planck's constant; so a violet photon carries more energy than a red one. Light is neither just a wave nor just a stream of particles.

Electrons, which we picture as tiny balls, share this double nature. Follow the animation step by step.

Step 1. The gun fires electrons one by one, so slowly that only one is ever on its way. Each hits the screen at one spot and leaves one dot, like a marble striking a target. So far: particles.

Step 2. Where the next dot lands is pure chance. But after a few hundred dots a pattern shows up: bright stripes separated by almost empty gaps. That is the mark of waves from the two slits adding in some places and cancelling in others, like ripples on a pond.

Step 3. The electrons travel alone, so none has a partner to interfere with. Each electron's own wave must pass through both slits at once.

Step 4. Put a detector by the slits to see which one each electron uses: the stripes vanish and the dots pile up into one smooth patch.

Check yourself: the experiment is stopped after just five electrons. Do you see stripes? Answer: no, only five scattered dots. Stripes need many dots.

Remember: quantum objects travel like waves and arrive like particles.

Intermediate

One electron, one dot

In the animation an electron gun fires electrons at a wall with two narrow slits, and a screen behind it records each arrival. Only one electron is on its way at a time: the first six are drawn in flight, then the animation speeds up while the counter climbs. Every arrival is a single dot, the whole electron with all its charge detected at one point. Yet after a few hundred arrivals the dots crowd into bright stripes separated by nearly empty gaps, the interference pattern that light makes behind two slits.

The electron has a momentum $p=mv$ and a kinetic energy $E$, quoted in electronvolts: $1\mathrm{eV}=1.602\times 10^{-19}\mathrm{J}$ is the energy an electron gains across 1 V. The slits are a distance $d$ apart, the screen is a distance $L$ beyond them, and the fringe spacing $\Delta y$ separates neighbouring bright stripes.

The de Broglie wavelength

Light is a wave, but the lesson Photoelectric Effect showed that it also arrives in packets, photons of energy $E=hf$, each carrying a momentum $p=h/\lambda$. In 1924 Louis de Broglie turned the idea around and proposed that every particle with momentum $p$ has a wavelength
$\lambda =\frac{h}{p}$

Here $h=6.626\times 10^{-34}\mathrm{J}\cdot \mathrm{s}$ is the Planck constant. Because $h$ is so tiny, a wavelength is noticeable only for particles of very small mass at modest speeds: electrons, neutrons, atoms. The blue wave before the slits stands for this matter wave: not a ripple of electron substance but a wave of probability.

A slow particle has kinetic energy $E=\frac{1}{2}m{v}^2$, and since $p=mv$ this is $E=p^2/(2m)$, so
$p=\sqrt{2mE}$

An electron accelerated from rest through a voltage $V$ gains $E=eV$, where $e=1.602\times 10^{-19}\mathrm{C}$ is the size of its charge. Four times the energy gives twice the momentum and half the wavelength.

From energy to fringe spacing

Beyond the slits the two parts of the matter wave overlap and interfere as light does in Diffraction and the Double Slit: where the paths differ by a whole number of wavelengths many electrons land, and where they differ by an odd number of half wavelengths almost none do. For small angles this gives a chain of three equations that solves most problems:

• $\lambda =\frac{h}{p}$
• $p=\sqrt{2mE}$
• $\Delta y=\frac{\lambda L}{d}$

Energy gives momentum, momentum gives wavelength, and wavelength gives the fringe spacing. Faster electrons have a shorter wavelength and narrower fringes; closer slits or a more distant screen spread the fringes out. The last equation is the double-slit formula for light, with the de Broglie wavelength.

Worked example: the electrons in the animation

The electrons in the animation have $E=50\mathrm{eV}$. Convert the energy to joules, find the momentum with $m=9.11\times 10^{-31}\mathrm{kg}$, then the wavelength:
$E=50\cdot 1.602\times 10^{-19}=8.01\times 10^{-18}\mathrm{J}$
$p=\sqrt{2\cdot 9.11\times 10^{-31}\cdot 8.01\times 10^{-18}}=3.82\times 10^{-24}\mathrm{kg}\cdot m/s$
$\lambda =\frac{6.626\times 10^{-34}}{3.82\times 10^{-24}}=1.73\times 10^{-10}\mathrm{m}=0.173\mathrm{nm}$

This is the 0.17 nm on the blue readout, about the spacing of atoms in a crystal. The speed, $v=p/m=4.2\times 10^{6}m/s$, is only 1.4 % of the speed of light, so ignoring relativity in $p=\sqrt{2mE}$ is safe.

The drawing is labelled not to scale. Suppose the real slits are $d=0.50\mu \mathrm{m}$ apart and the screen is $L=2.0\mathrm{m}$ away. Then
$\Delta y=\frac{1.73\times 10^{-10}\cdot 2.0}{0.50\times 10^{-6}}=6.9\times 10^{-4}\mathrm{m}=0.69\mathrm{mm}$

The purple $\Delta y$ mark beside the screen stands for this spacing. Seen from the slits, neighbouring stripes are only $\lambda /d=3.5\times 10^{-4}\mathrm{rad}$, or 0.02°, apart, which is why electron interference needs slits finer than a micrometre, a long flight path or electron lenses that magnify the pattern.

Worked example: one wavelength, three particles

To see how atoms are arranged in a crystal you need a wavelength close to their spacing. What energy must a photon, an electron and a neutron have for $\lambda =0.173\mathrm{nm}$, the value in the animation?

All three need the same momentum, $p=h/\lambda =3.82\times 10^{-24}\mathrm{kg}\cdot m/s$, but momentum and energy are related differently. For the photon, $E=pc$:
$E=\frac{hc}{\lambda }=\frac{6.626\times 10^{-34}\cdot 3.00\times 10^8}{1.73\times 10^{-10}}=1.15\times 10^{-15}\mathrm{J}=7.2\mathrm{keV}$
which is an X-ray photon. The electron needs 50 eV, as above. For the neutron, with $m=1.675\times 10^{-27}\mathrm{kg}$,
$E=\frac{p^2}{2m}=\frac{(3.82\times 10^{-24}{)}^2}{2\cdot 1.675\times 10^{-27}}=4.36\times 10^{-21}\mathrm{J}=0.027\mathrm{eV}$

The same wavelength costs 7.2 keV, 50 eV or 0.027 eV, a factor of 260,000 in energy, because the wavelength is fixed by momentum and a heavier particle reaches a given momentum with far less kinetic energy. The neutron, at only $p/m=2280m/s$, has about the thermal energy of room temperature, which is why neutrons slowed in a reactor suit crystal studies. X-rays, electrons and neutrons are used this way.

Reading the pattern on the screen

No rule says where the next dot will land: the wave fixes only the probability of each place. With 10 or 20 dots the screen looks random; around 200 the stripes are clear, and after all 560 the dark gaps are almost empty, with the stripe centres one $\Delta y$ apart.

In a dark gap the waves from the two slits arrive half a wavelength out of step and cancel, as in Interference and Superposition, so fewer electrons land there with both slits open than with one: a second route makes arriving less likely, which no stream of ordinary particles could do.

The stripes fade towards the edges because each slit, a quarter of $d$ wide, spreads its wave through only a limited angle. That fading envelope takes the values 1, 0.81, 0.41 and 0.09 at the bright stripes, and the next would sit on a zero of it, just beyond the screen's edge.

Common mistakes

• The electrons do not interfere with one another: only one is in flight at a time, yet 560 electrons sent singly build the fringes. Each electron's own wave passes through both slits.
• An electron is never split at the slits: every detection is one whole electron, with the full charge $1.602\times 10^{-19}\mathrm{C}$, at a single point.
• The wavelength is set by momentum, not by energy: a 50 eV electron and a 0.027 eV neutron share the same 0.17 nm. And $p=\sqrt{2mE}$ does not apply to photons, for which $p=E/c$.
• Large objects have a wavelength too, just an unobservable one: a 0.15 kg ball at 30 m/s has $p=4.5\mathrm{kg}\cdot m/s$ and $\lambda =1.5\times 10^{-34}\mathrm{m}$, some $10^{19}$ times smaller than a proton.
• Faster electrons give narrower fringes, not wider ones: doubling the speed halves both $\lambda$ and $\Delta y$, from 0.69 mm to 0.35 mm.

Where you meet it

The first proof came in 1927, when Davisson and Germer scattered 54 eV electrons off a nickel crystal and found a strong beam at 50°, just where waves of $\lambda =h/p=0.167\mathrm{nm}$ reflected from rows of atoms 0.215 nm apart reinforce, since $0.215\sin 50^{\circ }=0.165\mathrm{nm}$. Electron microscopes run at tens of kilovolts, giving wavelengths of a few picometres, some $10^5$ times shorter than visible light, and can image single atoms. Molecules of 60 carbon atoms have produced fringes too.

Atomic Models: the Bohr Atom introduced its allowed orbits as a rule; here the rule gets a reason, because each orbit holds a whole number of electron wavelengths. Schrödinger's Equation and Tunnelling gives the matter wave its own equation of motion and lets it leak through barriers, and the expert level explains why watching the electrons destroys the fringes.

Advanced

Lumps that arrive like a wave

This section sets up the puzzle that the rest of the lesson resolves. Electrons are fired one at a time at a plate with two slits, and a screen behind it records where each lands. Pick a small stretch of the screen and call $P_1$ the probability that an electron lands there with only slit 1 open, $P_2$ the same with only slit 2 open, and $P_{12}$ the probability with both open.

If electrons were pellets, each would pass through one slit or the other. With both slits open, the pellets through slit 1 would land just as they did with slit 2 closed, and likewise for slit 2, so the counts would simply add: $P_{12}=P_1+P_2$. A classical wave behaves differently. It delivers its energy continuously, not in lumps, and what adds is the height of the wave, not its energy. Because the intensity goes as the square of the height, the intensity with both slits open is not the sum of the two single-slit intensities, and fringes appear.

Electrons do what neither picture allows. The detector clicks with one whole electron at a time, never a fraction, as for pellets. Yet the clicks build up a fringe pattern, as for a wave, and $P_{12}\ne P_1+P_2$: at a dark fringe, opening the second slit makes arrivals rarer. The pellet argument shows why this matters. Any account in which each electron goes through one definite slit, and behaves there as it would with the other slit closed, leads straight to $P_{12}=P_1+P_2$. The experiment therefore rules out every such account, however detailed. The next section gives the rule that replaces it.

Amplitudes add, probabilities do not

Quantum mechanics keeps the arithmetic of waves but applies it to probabilities. Every way an event can happen (here, the electron reaching the chosen point through slit 1, or through slit 2) is given a complex number, its probability amplitude: ${\psi }_1$ for the first way, ${\psi }_2$ for the second. The rule has two cases.

When nothing in the world distinguishes the ways, their amplitudes add, and the probability is the squared magnitude of the total:
$\psi ={\psi }_1+{\psi }_2$
$|\psi {|}^2=|{\psi }_1{|}^2+|{\psi }_2{|}^2+2\mathrm{Re}({\psi }_1^{\ast }{\psi }_2)$
The first two terms are the single-slit probabilities. The last one, the cross term (or interference term), can be positive or negative, and it is the whole difference from classical counting. When some record shows which way was taken, the probabilities of the separate ways add instead, and the cross term is absent.

To use the rule we need the amplitudes themselves. A free particle of momentum $p$ has the de Broglie wavelength $\lambda =h/p$. Along a path of length $r$ the phase of its amplitude advances by $kr$, where $k=p/\hslash =2\pi /\lambda$ is the wave number and $\hslash =h/(2\pi )$. For two equal slits the amplitudes are ${\psi }_j=A{e}^{ik{r}_j}$, where $r_1$ and $r_2$ are the lengths of the two routes to the chosen point. They have the same size $A$ and differ only in phase, by
$\delta =k(r_2-r_1)$
Everything that follows comes from adding two such numbers.

Two arrows and their sum

Legend

${\psi }_1$	Amplitude for arriving through slit 1: an arrow of length $A$ whose direction is its phase.
${\psi }_2$	Amplitude for arriving through slit 2: also of length $A$, turned from ${\psi }_1$ by $\delta$.
$\psi$	Total amplitude with both slits open: the diagonal ${\psi }_1+{\psi }_2$.
$A$	Length of each of the two arrows (equal slits).
$\delta$	Phase difference of the two amplitudes: the angle between the arrows.	$\mathrm{rad}$

Adding two complex numbers is adding two arrows, and Figure 1 uses this to make the cross term visible. Each amplitude is drawn as an arrow in the complex plane: its length is its magnitude $A$ and its direction is its phase. Only the angle $\delta$ between ${\psi }_1$ and ${\psi }_2$ matters, because turning both arrows together by the same angle changes no length, and so no probability.

The sum $\psi$ is a diagonal of the rhombus built on the two arrows (a rhombus, because both sides have the same length $A$). Now look at the other diagonal, dashed. The diagonals of a rhombus cut each other in half at a right angle, marked in the figure. So half of $\psi$ is one leg of a right triangle whose hypotenuse is ${\psi }_1$, of length $A$, and whose angle at the origin is $\delta /2$. That leg is $A\cos (\delta /2)$, and doubling it gives
$|\psi |=2A\cos \frac{\delta }{2}$
for $\delta$ between 0 and $\pi$; for larger angles take the magnitude of the cosine.

Squaring, and using $2{\cos }^{2}x=1+\cos 2x$, splits the result into the two parts of the rule:
$|\psi {|}^2=4A^2{\cos }^2\frac{\delta }{2}=2A^2+2A^2\cos \delta$
The first term, $2A^2$, is the which-path sum $|{\psi }_1{|}^2+|{\psi }_2{|}^2$; the second, $2A^2\cos \delta$, is the cross term. Two cases fix the picture. When $\delta =0$ the arrows line up, so $|\psi |=2A$ and $|\psi {|}^2=4A^2$, twice the which-path value. When $\delta =\pi$ they point in opposite directions and cancel, so $|\psi {|}^2=0$: no electron ever arrives there, although either slit alone would send some.

Why must the amplitudes be complex numbers rather than real waves? Each amplitude also carries a time factor $e^{-iEt/\hslash }$, where $E$ is the electron's energy. Energy is conserved along both routes, so the factor is the same for both: it turns both arrows of Figure 1 together and drops out of $|\psi {|}^2$. A steady beam therefore gives a steady count rate. A real wave such as $\cos (kr-\omega t)$ would instead make the probability itself oscillate in time, which is never seen.

From path difference to fringes

Legend

$d$	Distance between the two slits.	$\mathrm{m}$
$L$	Distance from the slit plate to the screen.	$\mathrm{m}$
$y$	Height of $P$ above the axis through the middle of the slits.	$\mathrm{m}$
$P$	Point of the screen where the two amplitudes are added.
$\theta$	Angle between the axis and the line from the middle of the slits to $P$.	$\mathrm{rad}$
$r_1$	Length of the route from slit 1 to $P$.	$\mathrm{m}$
$r_2$	Length of the route from slit 2 to $P$.	$\mathrm{m}$
$r_2-r_1$	Extra length of route 2, left over once the arc has marked off $r_1$; about $d\sin \theta$ for a distant screen.	$\mathrm{m}$

The phase difference $\delta$ depends on where the electron lands. Figure 2 gives the geometry that turns a position on the screen into a path difference, and so into a phase.

Two slits, labelled 1 and 2, are a distance $d$ apart, and the screen stands a distance $L$ behind the plate. A point $P$ of the screen lies at height $y$ above the axis through the middle of the slits; seen from that middle point, $P$ is at an angle $\theta$ from the axis. Two routes reach $P$: one of length $r_1$ from slit 1 and one of length $r_2$ from slit 2. Because $P$ is above the axis, route 2 is the longer one.

To find how much longer, follow the arc centred on $P$ through slit 1. Every point of the arc is a distance $r_1$ from $P$; it is a piece of a wavefront converging on $P$, whose points all arrive in step. Where the arc meets route 2 it has therefore marked off a length $r_1$, measured back from $P$. What is left of route 2 at the slit end, the thick segment, is the extra length $r_2-r_1$.

The real screen is millions of times farther away than $d$, so the two routes are almost parallel and the short arc is almost a straight line square to both. The arc, the thick segment and the plate between the slits then form a small right triangle with hypotenuse $d$. Its angle at slit 1 is $\theta$, because the arc is square to the routes and the plate is square to the axis. The side opposite that angle is the extra length:
$r_2-r_1\approx d\sin \theta$
and the phase difference of Figure 1 is
$\delta =k(r_2-r_1)=\frac{2\pi d\sin \theta }{\lambda }$

Bright fringes are where the arrows of Figure 1 line up, $\delta =2\pi m$ with $m$ a whole number, that is $d\sin \theta =m\lambda$: the extra length is a whole number of wavelengths. The angles are tiny, so $\sin \theta \approx y/L$, and the bright fringes sit at $y_m=m\lambda L/d$. They are equally spaced, by
$\Delta y=\frac{\lambda L}{d}$
and the dark fringes lie halfway between them, where $\delta$ is an odd multiple of $\pi$.

Reading the histogram

Legend

$n$	Number of electrons that landed in a bin of width $\Delta y/3$.
$y$	Position along the screen, measured from the centre of the pattern.	$\mathrm{m}$
$|\psi {|}^2$	Probability density with both slits open (solid curve), scaled to counts per bin.
$|{\psi }_1{|}^2+|{\psi }_2{|}^2$	Which-path sum: the pattern when every electron's slit is recorded (dashed curve), with nearly the same area.
$\Delta y$	Distance between neighbouring bright fringes, $\lambda L/d$; the single-slit factor first vanishes at $4\Delta y$.	$\mathrm{m}$

Figure 3 puts the theory against counted electrons. It is what the expert layer of the animation builds up: 560 electrons, sorted by where they landed along the screen coordinate $y$ into bins a third of a fringe wide. The height $n$ of each gold bar is the number of electrons in its bin. Two curves are drawn over the bars, both scaled to counts per bin: the solid curve is the quantum prediction $|\psi {|}^2$, and the dashed curve is the which-path sum $|{\psi }_1{|}^2+|{\psi }_2{|}^2$, the pattern expected if every electron's slit were recorded.

First compare the two curves. For equal slits $|\psi {|}^2=4A^2{\cos }^2(\delta /2)$, which is twice the dashed value $2A^2$ at each bright fringe and zero at each dark one. The cross term $2A^2\cos \delta$ is positive over half of each fringe and negative over the other half, so it averages to zero across a fringe. The two curves therefore enclose nearly the same area: interference moves arrivals from the dark bands into the bright ones, but creates and destroys none.

Next, the slow fall-off of both curves away from the centre. This is the single-slit factor. Each slit has a width of its own, here $a=d/4$, and waves from different parts of one slit also interfere with each other. As a result $A$ is not constant but depends on $y$ through the envelope $\sin \beta /\beta$, with $\beta =\pi ay/(\lambda L)$. The envelope first vanishes at $\beta =\pi$, that is at $y=\lambda L/a=4\Delta y$, the right-hand mark under the axis. At the bright fringes of orders 0 to 3 the intensity envelope $(\sin \beta /\beta {)}^2$ is 1, 0.81, 0.41 and 0.09 of its central value, and order 4 is missing because it falls exactly on the zero. The screen ends at about $3.2\Delta y$, where the bars stop.

Finally, the bars against the solid curve. In the animation the central bin ends with 94 of the 560 electrons and each first-order bin with 77. The curve peaks near 104 per bin, yet even a perfect run would leave the central bar a little short of it. A bar counts everything across its bin, so it averages the curve over a third of a fringe, and the mean of ${\cos }^2$ over that width, centred on a peak, is 0.91.

In a real run each count also fluctuates: a bin expecting $n$ electrons receives about $n\pm \sqrt{n}$. For the central bin that is a relative scatter of 10 %, and the scatter falls to 1 % only at 10,000 counts per bin. This is why the first few dozen dots look like noise. The wave predicts the distribution of many arrivals; it never says where the next single electron will land.

Watching the electrons

Why not simply look at which slit each electron uses? Light up the region just behind the slits, so that each passing electron scatters a photon and a flash shows where it went. Then every electron is seen at one slit, never at both. Sort the electrons by their flash: those seen at slit 1 land with the pattern $|{\psi }_1{|}^2$, those seen at slit 2 with $|{\psi }_2{|}^2$. Together they make the dashed curve of Figure 3. Once the path is recorded, probabilities add, as the rule says, and the fringes are gone.

Could the light be made gentle enough not to disturb the electrons? Dimmer light does not help, because light comes in photons: dimmer means rarer photons, not weaker ones. An electron either scatters a whole photon, is located and joins the smooth pattern, or scatters none and joins the fringes.

The other option is softer photons. A photon of wavelength ${\lambda }_{\gamma }$ carries momentum $h/{\lambda }_{\gamma }$, so a longer wavelength gives a gentler kick. But a flash can be located only to within about its own wavelength. Telling apart slits a distance $d$ apart therefore needs ${\lambda }_{\gamma }$ below about $d$, and so a photon momentum of at least about $h/d$. The next section shows that a kick of this size is just enough to wash out the fringes.

A kick the size of a fringe

Legend

$\vec{p}$	Momentum of the electron before the photon scatters.	$\mathrm{kg}\cdot m/s$
${\vec{p}}^{\prime }$	Momentum after the kick: the same length, turned through a small angle.	$\mathrm{kg}\cdot m/s$
$h/d$	Largest sideways kick from a photon short enough in wavelength to tell the slits apart.	$\mathrm{kg}\cdot m/s$
$\lambda /d$	Largest angle through which the kick turns the electron's path; also the angle between neighbouring bright fringes.	$\mathrm{rad}$
$\Delta y$	Distance on the screen from the centre of the pattern to the next bright fringe, $\lambda L/d$.	$\mathrm{m}$

Figure 4 compares the kick from one such photon with the spacing of the fringes. On the left, the electron arrives with momentum $\vec{p}$. The photon scatters off it in an unknown direction and hands it a sideways momentum of up to about $h/d$: the arrow from the tip of $\vec{p}$ to the tip of the new momentum ${\vec{p}}^{\prime }$. The kick is small and sideways, so ${\vec{p}}^{\prime }$ keeps the length of $\vec{p}$ but is turned through an angle of up to about $(h/d)/p$. With $p=h/\lambda$ that angle is $\lambda /d$.

On the right, the same angle appears in the apparatus. Seen from the slits, the middle of the pattern and the next bright fringe, a distance $\Delta y$ away on a screen at distance $L$, are an angle $\Delta y/L$ apart, and $\Delta y/L=\lambda /d$ because $\Delta y=\lambda L/d$. So a kick strong enough to find the slit turns the electron by up to the angle between neighbouring fringes, in a random direction. Each landing point is shifted at random by up to about a whole fringe, and the bright and dark bands smear into each other.

In numbers, for the 50 eV electrons of the worked problem below and slits 0.50 µm apart, a 0.50 µm photon carries only 1/2900 of the electron's momentum, yet on a screen 2.0 m away it can shift the landing point by up to 0.69 mm, a whole fringe. Once ${\lambda }_{\gamma }$ exceeds $d$ the fringes begin to return, because the flash no longer shows which slit was used.

This trade-off is not a weakness of using light. It is the uncertainty principle, which limits the spreads $\Delta x$ and $\Delta p$ of a position and a momentum along the same direction:
$\Delta x\Delta p\ge \frac{\hslash }{2}$
Locating the electron at one slit pins its sideways position to better than about $d$, so its sideways momentum must spread by at least the order of $\hslash /d$, whatever does the locating. The estimate above is good to factors of order 1; the worked problem below does the exact sum for a different detector.

The limit has to hold for every apparatus. An electron that was located and yet undisturbed would have to make fringes, because nothing disturbed it, and make none, because its path is known and probabilities add: the theory would contradict itself. The limit belongs to nature, not to clumsy instruments.

Partial which-path information

Real detectors can lie anywhere between recording nothing and recording the path perfectly, and the fringes fade by just the right amount in between. To see how, treat the detector itself as a quantum system. If the electron passes slit 1, the detector ends in the state written $|D_1⟩$; if it passes slit 2, in $|D_2⟩$. Their overlap $⟨D_1|D_2⟩$ is a complex number, of size between 0 and 1, that measures how alike the two states are. Electron and detector end together in the state ${\psi }_1|D_1⟩+{\psi }_2|D_2⟩$, and when only the electron's landing point is observed, the squared magnitude of this sum carries the overlap into the cross term:
$|\psi {|}^2=|{\psi }_1{|}^2+|{\psi }_2{|}^2+2\mathrm{Re}({\psi }_1^{\ast }{\psi }_2⟨D_1|D_2⟩)$

Identical states (overlap 1) record nothing and leave full fringes; orthogonal states (overlap 0) record the path perfectly and leave none. In between, two numbers describe the result. The fringes are measured by their visibility
$V=\frac{P_{\max }-P_{\min }}{P_{\max }+P_{\min }}$
where $P_{\max }$ and $P_{\min }$ are the probabilities at a bright fringe and at the neighbouring dark one, so that $V=1$ for perfect fringes and $V=0$ for none. For equal amplitudes $V=|⟨D_1|D_2⟩|$. The path information is measured by the distinguishability $D$, defined so that the best possible guess of the slit, made from the detector, is right with probability $(1+D)/2$: $D=0$ is a coin toss and $D=1$ is certainty. Then
$V^2+D^2\le 1$
with equality when the detector starts in a definite quantum state. Partial knowledge of the path buys partial fringes; full knowledge and full fringes never come together.

Nobody has to read the record; it only has to exist. A gas molecule that bounces off the electron is such a detector, and so, for a large molecule sent through the slits, is a photon of heat radiation that the warm molecule emits in flight. Each such event multiplies the overlap by another factor smaller than 1, and after many of them the cross term is gone. This loss of interference to the surroundings is called decoherence, and it is why matter-wave experiments need a high vacuum.

Worked problem: a recoiling slit plate

In the late-1920s debates between Einstein and Bohr, one proposal was to read the electron's path from the recoil of the slit plate. Take the animation's 50 eV electrons at the real scale of the intermediate example ($\lambda =0.173\mathrm{nm}$, $d=0.50\mu \mathrm{m}$, $L=2.0\mathrm{m}$, so $\Delta y=0.69\mathrm{mm}$), with a plate free to slide sideways. How well must the plate's momentum be known, and what happens to the fringes?

Step 1, the two recoils. An electron that reaches the centre of the screen through slit 1, a distance $d/2$ above the axis, is deflected towards the axis through an angle $d/(2L)$; through slit 2 it is deflected the other way by the same angle. Its sideways momentum changes by $pd/(2L)$ one way or the other, and by conservation of momentum the plate takes the opposite kick. The two possible kicks of the plate differ by
$\delta p=\frac{pd}{L}=\frac{h}{\Delta y}=\frac{6.626\times 10^{-34}}{6.94\times 10^{-4}}=9.5\times 10^{-31}\mathrm{kg}\cdot m/s$
where the middle step uses $p=h/\lambda$ and $\Delta y=\lambda L/d$. (Here $\delta p$ is a momentum difference, not the phase $\delta$.)

Step 2, the plate's position. To tell the two kicks apart reliably, the spread of the plate's momentum must be well below $\delta p$; take a standard deviation ${\sigma }_p=\delta p/4=2.4\times 10^{-31}\mathrm{kg}\cdot m/s$. Standard deviations are the spreads of the uncertainty principle, so the plate's position must then be spread by at least
${\sigma }_x=\frac{\hslash }{2{\sigma }_p}=\frac{1.055\times 10^{-34}}{2\cdot 2.4\times 10^{-31}}=2.2\times 10^{-4}\mathrm{m}$

Step 3, the fringes. That is $\Delta y/\pi$, about a third of a fringe. The pattern is fixed to the slits, so it is as uncertain in position as the plate. Averaging the ${\cos }^2$ fringes over a Gaussian spread ${\sigma }_x$ multiplies their contrast by $e^{-2{\pi }^2{\sigma }_x^2/\Delta y^2}=e^{-2}=0.14$. The detector rule of the previous section gives the same answer from the other side. The plate's two possible final states are spreads of width ${\sigma }_p$ whose centres differ by $\delta p$, and they overlap by $e^{-\delta p^2/(8{\sigma }_p^2)}=e^{-2}$. So $V=0.14$ and $D=\sqrt{1-e^{-4}}=0.99$: the path is known almost for certain, and the fringes are nearly gone.

Step 4, a looser plate. If its momentum is known only to ${\sigma }_p=\delta p$, the fringes keep 88 % visibility, but the path is revealed only with $D=0.47$. Both cases give $V^2+D^2=1$: a plate in a definite quantum state sits on the boundary, and no plate does better.

Limits of validity

The relation $p=\sqrt{2mE}$, with $E$ the kinetic energy, ignores relativity. That is harmless at 50 eV, a ten-thousandth of the electron's rest energy $m{c}^2=511\mathrm{keV}$. In a 100 kV electron microscope, though, the exact momentum, from $pc=\sqrt{E^2+2Em{c}^2}$, gives $\lambda =3.70\mathrm{pm}$, while the simple formula gives 3.88 pm, about 5 % too long. The relation $\lambda =h/p$ itself stays exact; only the link between $p$ and $E$ changes.

The fringe formula needs small angles and a screen in the far field, $L\gg d^2/\lambda$, which is 1.4 mm in the example against $L=2\mathrm{m}$. Fringes also need a coherent beam: a small source, so that both slits receive the same wave, and nearly a single energy, so that all the electrons share one wavelength and one fringe spacing.

For large objects the wave is real but invisible, for two reasons. The first is the wavelength. A 0.15 kg ball at 30 m/s has $\lambda =1.5\times 10^{-34}\mathrm{m}$: behind slits 10 cm apart, a screen 10 m away would carry fringes $1.5\times 10^{-32}\mathrm{m}$ apart, and a detector the size of a proton would average some $10^{17}$ of them into a smooth curve. The second is decoherence: in air the ball is struck by about $5\times 10^{25}$ molecules a second, each one a which-path record.

Yet no sharp boundary between quantum and classical objects is known. Molecules of 60 carbon atoms, with $\lambda =2.5\mathrm{pm}$ at 220 m/s, have made fringes behind a grating of 100 nm period, and so, later, have molecules of about 2000 atoms.

Where this leads

Schrödinger's Equation and Tunnelling turns the amplitude into a wave function obeying its own equation, with $E=\hslash \omega$ and $p=\hslash k$, and lets it leak through barriers that a classical particle could never cross. The orbits of Atomic Models: the Bohr Atom are those that fit a whole number of wavelengths, $2\pi r=n\lambda$, which is equivalent to an angular momentum $n\hslash$. In Semiconductors, Diodes and LEDs, electron waves in a periodic crystal form energy bands. Particles and Forces: the Standard Model treats every particle as a quantum of a field and computes each process by adding the amplitudes of its alternative histories, drawn as interaction diagrams: the rule of this lesson in its most general form.