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The Born rule, adding amplitudes for indistinguishable paths, which-path information and complementarity, counts and their spreads, and what quantum mechanics does not predict.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to predict counts from quantum amplitudes, explain how which-path information destroys interference, and state the limits of the model's predictions.
From the last lessons you know that photons and electrons interfere even when sent one at a time, that each is detected whole at one place, and that position and momentum cannot both be sharp. From statistics you know expected values and spreads. This lesson states the rules that connect waves to what detectors actually record, and marks where the quantum model's predictions stop.
| Term | What it means |
|---|---|
| Amplitude | A complex number $\psi$ whose squared magnitude gives a probability. |
| Born rule | The probability of an outcome is $\vert \psi\vert ^2$. |
| Superposition | A combination of states, whose amplitudes add. |
| Which-path information | Any record, read or not, of which route a particle took. |
| Complementarity | Bohr's principle that wave and particle aspects cannot be seen in the same experiment. |
| Decoherence | The loss of interference when a system becomes entangled with its surroundings. |
| Binomial spread | $\sqrt{Np(1 - p)}$, the typical scatter in a count of $N$ independent trials. |
Quantum mechanics predicts the odds of outcomes, not the outcomes themselves. Each possible way for something to happen has an amplitude $\psi$, and the probability is its squared magnitude, the Born rule:
$$P = |\psi|^2.$$
When an outcome can happen in two ways that nothing distinguishes, their amplitudes add before squaring:
$$P = |\psi_1 + \psi_2|^2 = |\psi_1|^2 + |\psi_2|^2 + 2|\psi_1||\psi_2|\cos\phi.$$
The last term is interference. If anything records which way happened, the ways become distinguishable, the probabilities add, $|\psi_1|^2 + |\psi_2|^2$, and the interference vanishes. Each run gives one definite result; over $N$ runs an outcome of probability $p$ occurs about $Np$ times, with a random spread of $\sqrt{Np(1 - p)}$.
Another way: picture
Picture flipping a coin that is somehow both heads and tails until it lands. The odds of each face are fixed, but no single toss can be predicted. In quantum mechanics the odds come from adding and squaring amplitudes, which can cancel, so some outcomes become impossible when two routes to them interfere.
Another way: steps
Classically, a polarizer at angle $\theta$ to a light wave's polarization passes the fraction $\cos^2\theta$ of its intensity, Malus's law. Send photons one at a time and each one either passes whole or is absorbed; none passes as a fraction. The fraction that pass, over many photons, is $\cos^2\theta$.
So for a single photon, $\cos^2\theta$ is a probability. Nothing about the photon beforehand decides whether it will pass; experiments on entangled photons, discussed below, rule out hidden instructions carried by each photon. The classical intensity law and the quantum probability law are the same law, read at different scales.
In the double-slit experiment, a particle can reach a point on the screen through either slit. If $N_1$ arrive per minute with only slit 1 open and $N_2$ with only slit 2, the amplitudes are proportional to $\sqrt{N_1}$ and $\sqrt{N_2}$. With both open, at a point where the paths are in phase, the count is $(\sqrt{N_1} + \sqrt{N_2})^2$, more than $N_1 + N_2$; where they are out of phase, $(\sqrt{N_1} - \sqrt{N_2})^2$, less.
Opening a second slit can therefore reduce the number of particles reaching a point, which no classical picture of particles can explain: more ways to arrive, fewer arrivals. Averaged across a fringe, the counts still add, so no particles are created or destroyed; they are redistributed.
Put a detector at the slits that records which one each particle uses, and the fringes vanish. It does not matter whether anyone reads the record, or whether the detector disturbs the particle's momentum much. What matters is that the information exists somewhere in the universe.
Quantum eraser experiments make the point sharply: record the path in a way that can later be erased, and the fringes return for exactly those particles whose path record was erased. Interference depends on indistinguishability, not on the presence of a human observer.
Niels Bohr summarized these results in the principle of complementarity: the wave and particle descriptions are both needed, but no single experiment shows both at once. An arrangement that reveals the path, a particle property, destroys interference, a wave property, and vice versa.
Intermediate cases trade one for the other smoothly. A detector that tells the path only partly, say with $80$ percent reliability, leaves faint fringes. Experiments have confirmed a precise inequality between how visible the fringes are and how well the path is known.
Einstein disliked the randomness, saying God does not play dice, and suspected that particles carry hidden instructions that decide their fates. In 1964 John Bell showed that any such local hidden instructions would limit the correlations between measurements on pairs of entangled particles, and that quantum mechanics predicts stronger correlations.
John Clauser at Berkeley in 1972, Alain Aspect in the 1980s, and later loophole-free experiments, including one at NIST in Boulder in 2015, found the quantum predictions correct. The randomness is genuine. Clauser, Aspect and Anton Zeilinger shared the 2022 Nobel Prize for these experiments.
Checking an answer. Probabilities must lie between 0 and 1 and sum to 1 over all outcomes. Bright fringes must exceed $N_1 + N_2$ and dark fringes fall below it, averaging to $N_1 + N_2$. Spreads must shrink relative to the mean as $1/\sqrt{N}$. And a which-path record must remove the cosine term.
Squaring amplitudes to get probabilities is a postulate of quantum mechanics, the Born rule, supported by every experiment. Adding amplitudes for indistinguishable alternatives is the superposition principle. Together they reproduce all the interference results of wave optics while explaining why each detection is localized.
The binomial spread assumes independent trials with the same probability, true for particles sent one at a time through a fixed apparatus. It is why quantum predictions are tested by counting: a single outcome says almost nothing, while ten thousand outcomes pin a probability to about one percent.
Quantum mechanics gives probabilities with extraordinary accuracy, but it says nothing about which outcome a single trial will produce. Interpretations disagree about what, if anything, happens during a measurement: whether the wave collapses, whether all outcomes occur in branching worlds, or whether the wave describes only our knowledge.
These interpretations make the same predictions for every experiment done so far, which is why physicists can use quantum mechanics with complete confidence while disagreeing about its meaning. A useful habit is to separate what the theory predicts, which is testable, from what it means, which so far is not.
If atoms can be in superpositions, why not cats or chairs? The answer is decoherence. A large object is constantly struck by air molecules and photons, each collision carrying away a little information about where it is. That information is a which-path record, and it destroys interference almost instantly.
For a dust grain in air, superpositions of positions a micrometer apart decohere in far less than a nanosecond. Quantum computers must therefore isolate their qubits from the environment extremely well; the machines built by IBM, Google and others cool their circuits to a hundredth of a degree above absolute zero to keep decoherence at bay for long enough to compute.
Two particles can share a single quantum state in which neither has definite properties of its own, but their measurement results are correlated. A pair of photons made together can be entangled so that, whatever polarizer angle is chosen, the two always give the same result when measured at the same angle, though each result on its own is perfectly random.
Entanglement does not let anyone send a message faster than light. Each experimenter alone sees only random outcomes; the correlation appears only when the two records are brought together and compared, which requires ordinary communication. What entanglement does is supply correlations stronger than any pair of objects carrying pre-set instructions could produce, which is precisely what Bell's theorem quantifies.
Entangled photons now carry encryption keys: quantum key distribution networks in several countries, including test links between research centers in the United States, use them so that any eavesdropper disturbs the correlations and reveals the intrusion. A satellite link, demonstrated in 2017, shared entangled photons between ground stations more than $1000$ km apart.
NIST's Randomness Beacon in Boulder publishes a fresh $512$-bit random number every minute for public use in lotteries, audits and cryptographic protocols. In 2018 NIST researchers demonstrated a source whose randomness is certified by a loophole-free Bell test: the bits could not have been predicted by any local hidden process.
Simpler quantum sources send photons at beam splitters and record which way each goes. A fair source gives $256$ ones in $512$ bits on average, with a spread of about $11$. Testing that the counts and their spreads match the binomial law, over billions of bits, is how engineers check the source has no bias.
The quantum computers built by companies such as IBM in New York and Google in California store information in qubits, which can be in superpositions of 0 and 1. A computation arranges for the amplitudes of wrong answers to cancel and those of right answers to reinforce, exactly as fringes form in a double slit.
Each run ends with a measurement that gives one definite string of bits, so the machines repeat a computation thousands of times and read the answer from the counts. Decoherence, the leak of which-path information to the surroundings, is the main obstacle, which is why the chips are cooled to within a hundredth of a degree of absolute zero.
It is natural to suppose that each photon secretly "knows" whether it will pass a polarizer, and that quantum probabilities only reflect our ignorance. Bell's theorem and the experiments that test it show that no such local hidden instructions can reproduce what is observed. The randomness is built into nature.
A related error is to think interference disappears only when a human looks. Any record that could distinguish the paths, in a detector, a stray photon or a molecule's recoil, destroys it, whether or not anyone ever reads the record.
Vertically polarized photons meet a polarizer at $60°$. Find the probability each passes.
$p = \cos^2 60° = 0.25$
Born rule for polarization.
Find the expected number of $400$ that pass.
$Np = 100$
Average.
Find the spread.
$\sqrt{400 \times 0.25 \times 0.75} = 8.7$
Binomial.
Those that pass now meet a horizontal polarizer, $30°$ further. Find the probability.
$\cos^2 30° = 0.75$
Measured from their new polarization.
Find the overall fraction through both.
$0.25 \times 0.75 = 0.19$
Though vertical and horizontal alone would pass none.
Slit 1 alone gives $64$ counts per minute at a point; slit 2 alone gives $16$. Find the amplitudes.
$\sqrt{64} = 8, \quad \sqrt{16} = 4$
Relative units.
Find the count at a bright fringe.
$(8 + 4)^2 = 144$
In phase.
Find the count at a dark fringe.
$(8 - 4)^2 = 16$
Out of phase.
Find the average of the two.
$\dfrac{144 + 16}{2} = 80 = 64 + 16$
Totals conserved.
Add a which-path detector.
$64 + 16 = 80\ \text{everywhere across the fringe}$
Interference gone.
Find the fringe visibility.
$V = \dfrac{144 - 16}{144 + 16} = 0.8$
Unequal amplitudes give incomplete dark fringes.
A process has probability $p = 0.3$. Find the spread in $100$ trials.
$\sqrt{100 \times 0.3 \times 0.7} = 4.6$
Binomial.
Find the relative spread.
$\dfrac{4.6}{30} = 15\%$
Spread over mean.
Repeat for $10{,}000$ trials.
$\sqrt{10{,}000 \times 0.21} = 46, \quad \dfrac{46}{3000} = 1.5\%$
Ten times better.
State the scaling.
$\text{relative spread} \propto \dfrac{1}{\sqrt{N}}$
A hundred times the trials for ten times the precision.
Find the trials to measure $p$ to $0.1\%$ relative.
$N \approx \dfrac{1 - p}{p}\cdot\dfrac{1}{(0.001)^2} = 2.3 \times 10^6$
From $\sqrt{(1-p)/Np} = 0.001$.
Explain why single runs prove little.
$\text{one outcome is consistent with almost any } p$
Quantum tests are statistical.
Relate to a real experiment.
$\text{Bell tests collect millions of pairs}$
To beat the statistical spread decisively.
Find the amplitudes.
$\sqrt{49} = 7, \quad \sqrt{9} = 3$
Square roots of counts.
Subtract them out of phase.
$(7 - 3)^2$
Destructive interference.
Evaluate the count.
Electrons sent one at a time through two slits build up an interference pattern of $5$ clear bright fringes. A detector is then placed at the slits that records which slit each electron passes. What pattern builds up?
Complete the worked solution: at a spot on a screen, $81$ particles per minute arrive through slit 1 alone and $4$ through slit 2 alone. Find the count per minute there with both slits open at a bright fringe, at a dark fringe, and with both open but a detector recording which slit each particle used.
Add the amplitudes in phase.
$N_{\text{bright}} = \left(\sqrt{N_1} + \sqrt{N_2}\right)^2 =$ r
Constructive interference.
Subtract them out of phase.
$N_{\text{dark}} = \left(\sqrt{N_1} - \sqrt{N_2}\right)^2 =$ d
Destructive interference.
Add the counts with a which-path record.
$N_{\text{record}} = N_1 + N_2 =$ m
No interference.
Check the average.
$\tfrac{1}{2}(N_{\text{bright}} + N_{\text{dark}}) = N_1 + N_2$
Interference moves particles around; it does not create them.
Match each idea of quantum mechanics to its statement.
| $P = |\psi|^2$ | a combination of states | one definite outcome each time | path knowledge excludes interference | |
|---|---|---|---|---|
| Born rule | ||||
| superposition | ||||
| measurement | ||||
| complementarity |
Vertically polarized photons meet a polarizer whose axis is $60°$ from vertical, where $\cos^2\theta = 0.25$. For $10{,}000$ photons sent one at a time, fill in the probability each passes, the expected number passing, and the spread $\sqrt{Np(1 - p)}$ in that number.
| value | |
|---|---|
| probability of passing | |
| expected number passing | |
| spread in the number |
A particle can reach a detector by two indistinguishable paths with real amplitudes $4$ and $3$ (in arbitrary units) and phase difference $p$ between them. Write the relative probability of detection as a formula in $p$.
Answer:
At a point on a screen, a detector counts $64$ electrons per minute with only slit 1 open and $16$ per minute with only slit 2 open. The point is at a bright fringe when both are open. How many per minute does it count then?
Answer: per minute
A quantum random-number source like those studied at NIST sends single photons at a beam splitter; each photon's detection at one output records a 1, at the other a 0, with equal probability. For a string of $512$ bits, what is the standard spread in the number of 1s?
Answer: spread in the count of 1s
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A particle can reach a detector by two indistinguishable paths with real amplitudes $3$ and $1$ (in arbitrary units) and phase difference $p$ between them. Write the relative probability of detection as a formula in $p$.
Answer:
You can reason with quantum probabilities. Explain to someone why opening a second slit can reduce the number of particles reaching a point.
21. Your turn: slit 1 alone gives $49$ counts per minute at a point and slit 2 alone gives $9$. Find the count at a dark fringe., step 3
$16\ \text{per minute}$
Fewer than either slit alone gives.