How alternatives interfere
Fire electrons, one at a time, at a wall with two narrow slits in it, and record where each one lands on a screen behind. Every electron arrives as a single dot at a single place, as a particle should. Yet as the dots accumulate they form bright and dark stripes, the signature of a wave passing through both slits. If an electron is a particle, why are there fringes?
This experiment has been done, with the electrons far enough apart in time that only one is in the apparatus at once, and the stripes still build up dot by dot. So the pattern is not electrons jostling one another. Something about a single electron's passage involves both slits. This essay finds the rule that predicts the pattern, computes the stripe spacing for a real electron, and then asks what happens to the pattern when the apparatus can tell which slit was used. The answer to that last question is the most important thing in the essay.
Three experiments with the same wall
The clearest way into the puzzle, due to Richard Feynman, is to run the same two-slit experiment with three different things. In each case the source is the same and is left unchanged; only the slits are opened or blocked. At the screen we place a small detector of fixed width, and slide it to different positions $x$.
Bullets. They arrive whole, one at a time. Block slit 2, fire many bullets, and record the fraction of them that land in the detector at $x$; call that probability $P_1(x)$. Block slit 1 instead and get $P_2(x)$. With both slits open, each bullet went through one slit or the other, so $P_{12} = P_1 + P_2$. Probabilities of alternatives add.
Water waves. Nothing arrives in lumps; a detector measures a continuous intensity. The wave passes through both slits, and at the screen the two heights add before the intensity, which goes as the square of the height, is taken. Where crest meets crest the intensity is large; where crest meets trough it is zero. So $I_{12} \neq I_1 + I_2$: there are fringes.
Electrons. They arrive whole, like bullets: the detector clicks, or it does not, and a click is always one full electron. But the distribution of clicks has fringes, like waves. In particular there are places where electrons arrive when either slit alone is open, and stop arriving when both are open. Opening a second door reduces the traffic. No account in which each electron goes through one slit, unaffected by the other, can produce that, since it would give $P_1 + P_2$, which is never less than $P_1$.
The rule: add arrows, then square
The rule that works borrows the arithmetic of waves but applies it to probability. To each way an event can happen we attach a complex number, called a probability amplitude, or just an amplitude. Three statements then cover everything in this essay.
- The probability of an event is the squared length of its amplitude: $P = |a|^2$.
- If the event can happen in several alternative ways, and nothing anywhere records which way it happened, the amplitudes add: $a = a_1 + a_2$.
- If a way consists of successive stages, its amplitude is the product of the amplitudes of the stages.
Let $a_1$ be the amplitude for an electron to land in the detector at $x$ by way of slit 1, and $a_2$ by way of slit 2. Throughout, $P(x)$ means the probability of landing in that small, fixed-width detector; since the width never changes, $P(x)$ plotted against $x$ has the shape of the arrival pattern. With only slit 1 open, $P_1 = |a_1|^2$. With both open, rule 2 applies, and
$$\begin{aligned} P_{12} &= |a_1 + a_2|^2\\[4pt] &= (a_1 + a_2)^*(a_1 + a_2)\\[4pt] &= |a_1|^2 + |a_2|^2 + 2\,\mathrm{Re}\,(a_1^*a_2). \end{aligned}$$The first two terms are the bullet answer, $P_1 + P_2$. The third is new. It is called the cross term or interference term, and it can have either sign. To see what it does, write each amplitude as an arrow. Take the two slits to be equally well illuminated, so both arrows have the same length $A$, and let their phases be $\varphi_1$ and $\varphi_2$:
$$a_1 = A\,e^{i\varphi_1},\qquad a_2 = A\,e^{i\varphi_2}.$$Then $a_1^*a_2 = A^2 e^{i(\varphi_2-\varphi_1)}$, whose real part is $A^2\cos(\varphi_2-\varphi_1)$, and
$$P_{12} = 2A^2\,\bigl[\,1 + \cos(\varphi_2 - \varphi_1)\,\bigr].$$Only the difference of the phases matters. When the arrows point the same way, $P_{12} = 4A^2$: twice what bullets would give. When they point opposite ways, $P_{12} = 0$: that is the place where opening the second slit stops the electrons arriving. At a phase difference of a quarter turn the cosine vanishes and the bullet answer $2A^2$ comes back.
Where the phases come from
An electron with momentum $p$ has wavelength $\lambda = h/p$, and its amplitude turns through one full revolution, $2\pi$, for every wavelength of distance travelled. A route of length $r$ therefore contributes a phase $\varphi = 2\pi r/\lambda$, and the phase difference between the two routes is $2\pi(r_2 - r_1)/\lambda$: the number of wavelengths by which one route is longer, counted in turns.
For a point at height $x$ on a distant screen, the geometry in the figure gives a path difference $r_2 - r_1 \approx d\,x/L$, where $d$ is the separation of the slits and $L$ the distance to the screen. (The approximation holds when $x$ and $d$ are both much smaller than $L$.) So
$$P_{12}(x) = 2A^2\left[\,1 + \cos\!\left(\frac{2\pi\,d\,x}{\lambda L}\right)\right].$$The pattern repeats each time the cosine's argument grows by $2\pi$, that is, each time $x$ increases by the fringe spacing
$$s = \frac{\lambda L}{d}.$$Longer wavelength or a more distant screen spreads the fringes; slits further apart crowd them together.
Worked example: a 100 eV electron
The wavelength. Take electrons accelerated through 100 volts, so each has kinetic energy $E = 100$ eV $= 1.602\times10^{-17}$ J. This is tiny compared with the electron's rest energy of 511,000 eV, so the ordinary relation $E = p^2/2m$ applies, with $m = 9.109\times10^{-31}$ kg:
$$\begin{aligned} p &= \sqrt{2mE} = 5.40\times10^{-24}\ \text{kg m/s},\\[4pt] \lambda &= \frac{h}{p}\\[4pt] &= \frac{6.626\times10^{-34}\ \text{J s}}{5.40\times10^{-24}\ \text{kg m/s}}\\[4pt] &= 1.23\times10^{-10}\ \text{m}\\[4pt] &= 0.123\ \text{nm}. \end{aligned}$$That is about the size of an atom, and thousands of times shorter than visible light, which is why electron interference needs very fine slits.
The fringes. Slits of this kind have been cut in thin membranes with separations of a few hundred nanometres. Take $d = 200$ nm and a screen at $L = 1$ m:
$$\begin{aligned} s &= \frac{\lambda L}{d}\\[4pt] &= \frac{1.23\times10^{-10}\ \text{m}\times 1\ \text{m}}{2.00\times10^{-7}\ \text{m}}\\[4pt] &= 0.61\ \text{mm}. \end{aligned}$$Stripes two-thirds of a millimetre apart: large enough to see once magnified a little, and produced by a wavelength the size of an atom.
Check: nothing is created or lost. Averaged over one full fringe the cosine is zero, so the average of $P_{12}$ is $2A^2 = P_1 + P_2$. Interference does not change how many electrons arrive; it moves them from the dark stripes to the bright ones.
Check: why bullets show no fringes. A 10 g bullet at 300 m/s has $p = 3$ kg m/s and $\lambda = h/p = 2.2\times10^{-34}$ m. With slits a centimetre apart and a target 100 m away, $s = 2.2\times10^{-30}$ m, some $10^{14}$ times smaller than an atomic nucleus. Any real detector averages over an astronomical number of fringes, the cosine averages to zero, and what is left is $P_1 + P_2$. The bullet rule is what the electron rule looks like when the wavelength is far too small to resolve.
Finding out which slit
Rule 2 carried a condition: the amplitudes add if nothing records which way it happened. To see why the condition is there, put a marker next to the slits: anything at all, a single atom will do, that can end up in one of two conditions, which we will call flag up and flag down. Arrange that an electron passing slit 1 leaves the flag up, and one passing slit 2 leaves it down.
Now ask for the probability that an electron arrives at $x$. There are two distinct final outcomes to consider: "electron at $x$, flag up" and "electron at $x$, flag down". The first can happen only by way of slit 1, so its amplitude is $a_1$; the second only by way of slit 2, with amplitude $a_2$. Neither outcome has two alternative routes, so rule 2 has nothing to add. Each outcome gets its own probability by rule 1, and since they are different outcomes their probabilities add in the everyday way:
$$P(x) = |a_1|^2 + |a_2|^2 .$$The cross term is gone and so are the fringes: the dashed curve in the figure. We did not assume that the marker jolted the electron, and nobody needs to read the flag. What matters is that the two routes now lead to different final states of the world. Amplitudes add only for routes to the same final state.
Records need not be perfect. Suppose the marker is unreliable: an electron through slit 1 always leaves the flag up, but one through slit 2 sometimes leaves it up too. Then the flag's final state no longer settles the question. Part of the time, "electron at $x$, flag up" could have happened by either route, and for that part the amplitudes add and interfere. The result is fringes of reduced contrast: the dark stripes are no longer completely dark.
What sets the contrast is how alike the two possible final states of the marker are: the one slit 1 leaves and the one slit 2 leaves. If they are the same, the marker records nothing and the fringes are complete. If they are completely different, the record is perfect and the fringes vanish. In between, the more distinguishable the two marker states, the fainter the fringes. Nothing in this depends on how hard the marker pushes the electron.
Optional: the partial record, calculated
Suppose slit 1 always leaves the flag up, while slit 2 leaves it up with amplitude $\gamma$ and down with amplitude $\delta$, where $|\gamma|^2 + |\delta|^2 = 1$. By rule 3, the amplitude for "via slit 2, flag up" is $\gamma a_2$. The outcome "electron at $x$, flag up" now has two routes, and "flag down" has one:
$$\begin{aligned} P(x) &= |a_1 + \gamma a_2|^2 + |\delta a_2|^2\\[4pt] &= |a_1|^2 + |a_2|^2 + 2\,\mathrm{Re}\,(\gamma\,a_1^*a_2). \end{aligned}$$The cross term survives, multiplied by $\gamma$, which measures how much the flag state left by slit 2 overlaps the one left by slit 1. With equal arrows as before, the pattern swings between $2A^2(1+|\gamma|)$ and $2A^2(1-|\gamma|)$. If $\gamma = 1$ the flag tells us nothing and the fringes are complete. If $\gamma = 0$ the record is perfect and they vanish. If $\gamma = \tfrac12$ the bright stripes are three times as bright as the dark ones, instead of infinitely brighter.
Optional: the same bookkeeping in bracket notation
Later essays write amplitudes in a notation invented by Paul Dirac. The amplitude for something that starts in condition $s$ to end in condition $x$ is written $\langle x\,|\,s\rangle$, read from right to left: "start at $s$, end at $x$".
Rule 3 says the route through slit 1 has amplitude $\langle x\,|\,1\rangle\langle 1\,|\,s\rangle$: from the source to slit 1, then from slit 1 to $x$. Rule 2 then gives the whole two-slit amplitude as
$$\begin{aligned} \langle x\,|\,s\rangle = {}&\langle x\,|\,1\rangle\langle 1\,|\,s\rangle\\[4pt] &+ \langle x\,|\,2\rangle\langle 2\,|\,s\rangle . \end{aligned}$$Our $a_1$ and $a_2$ are the two terms on the right. The same notation will express how alike two marker states are, which returns in the final essay of the core sequence, on measurement, where the "flag" is a detector.
What an amplitude is not
It is tempting to picture the amplitude as a physical ripple spreading through the apparatus, with the electron somehow smeared along it. Two facts resist that picture. The electron is never found smeared: it always arrives whole, at one place. And for two electrons the amplitude is not a pair of ripples in ordinary space; it is a single complex number for each pair of positions, a function of six coordinates. The amplitude is better read as the quantity from which probabilities are calculated, obeying the three rules above. What it "really is" remains a fair question, taken up in the essay on measurement. The rules themselves have never failed.
Two questions
1. Turn up the voltage. The accelerating voltage in the worked example is raised from 100 V to 400 V, with the same slits and screen. What is the new fringe spacing?
Worked answer
The kinetic energy is four times larger, and since $p = \sqrt{2mE}$ the momentum is only twice as large. So $\lambda = h/p$ halves, from 0.123 nm to 0.061 nm, and the spacing $\lambda L/d$ halves with it, from 0.61 mm to 0.31 mm.
The easy slip is to halve the wavelength once for each doubling of energy and arrive at a quarter. Wavelength follows momentum, not energy, and for a slow particle momentum grows as the square root of the energy.
2. A gentler detector. A colleague argues: "The fringes disappear because detecting the electron at a slit gives it a kick. Build a detector gentle enough, and we could know the slit and keep the fringes." Using the section on finding out which slit, say what is wrong with this.
Worked answer
The argument puts the cause in the wrong place. In the calculation, the fringes disappeared without the electron's amplitudes $a_1$ and $a_2$ changing at all: no kick was assumed. They disappeared because the two routes ended in different final states of the marker, and amplitudes for different final states never add. So what controls the fringes is how distinguishable the two marker states are, not how hard the marker pushes.
Fringes come back when the marker states left by the two routes become less distinguishable, which is the same as the detector telling us less about the slit. Making the kick smaller does not, by itself, bring them back: a detector whose two records are fully distinct wipes out the fringes however gently it works. Knowing the slit and keeping the fringes are incompatible, whatever the detector is made of.
The tempting wrong turn is to picture loss of interference as mechanical damage to the electron. Disturbance can happen in real detectors, but it is not what the rules say removes the fringes. The deciding question is always whether the alternatives lead to the same final state of the world.
Further reading
- Feynman, Lectures on Physics, Vol. III, Ch. 1 "Quantum Behavior" (bullets, waves, electrons, and watching the electrons, §1-2 to §1-6) and Ch. 3 "Probability Amplitudes" (§3-1 for the three rules, §3-2 for the two-slit pattern with a which-slit record). The source of this essay's approach, and still the best extended account.
- Dirac, Principles of Quantum Mechanics (4th ed.), Ch. I "The Principle of Superposition", §4–§5: the same principle stated abstractly, using the polarisation of a photon instead of slits.
- Landau & Lifshitz, Quantum Mechanics (2nd ed.), §2 "The principle of superposition": a compact statement, with the consequence that the theory's equations must be linear.
Where this leaves us
The opening question has an answer of a definite kind. An electron arrives as a particle, but the probability of where it arrives is computed by adding one arrow for each route it could have taken and squaring the total. Fringes are the cross term of that square. They appear when the routes lead to the same final state of the world and fade, in proportion, as the routes become distinguishable.
Two routes needed two complex numbers. But an electron leaving the slits could be detected at any of a continuum of places, and other systems have their own lists of possibilities. We need a way to describe all the amplitudes belonging to a system at once, and to understand what changes, and what does not, when we choose a different list of alternatives to describe the same situation. That is the idea of a quantum state, and the next essay approaches it through something familiar: a vector drawn on a page, described along two different pairs of axes.