Unit 3 · The Mathematical MachineryChapter 15 of 25

Superposition & Measurement

Both, neither, and the moment the dice are thrown

  • 25 min read
  • 36 min listen
  • 26 videos
01Video briefing

Start with the preview

Start here — today's preview. Watch “Quantum Superposition, Explained Without Woo Woo” from The Science Asylum before (or after) this session. The lab will wait. Mostly.

Today's preview

Quantum Superposition, Explained Without Woo Woo

The Science Asylum13 min

Nick Lucid of The Science Asylum takes superposition back to its roots in ordinary wave physics and strips away the mystical talk that usually surrounds it. As you watch, test what you hear against this session's central claim: a superposition is one perfectly definite state written as a sum, and whether a state counts as a superposition depends on which measurement you have in mind.

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More from the video vault

Deep dive

Lecture 1: Introduction to Superposition

MIT OpenCourseWare76 min

The opening lecture of MIT's 8.04 Quantum Physics I (Spring 2013), in which Allan Adams uses thought experiments with measuring boxes to build superposition and uncertainty from scratch. Skip the first ten minutes of course logistics, then watch how a few simple box experiments force exactly the rules of this session: repeat a measurement and the answer stays put; measure an incompatible property and it is scrambled.

Watch on YouTube
Bonus

The Problem with Quantum Measurement

Sabine Hossenfelder7 min

A compact statement of the measurement problem from the theoretical physicist Sabine Hossenfelder, who explains why the measurement update cannot simply be derived from the Schrödinger equation. Her conclusion, that quantum mechanics therefore needs to be completed, is one position among several; chapter 22 lays out the others.

Watch on YouTube
  • Quantum Superposition, Explained Without Woo Woo — The Science Asylum — https://www.youtube.com/watch?v=ZUipVyVOm-Y
  • Lecture 1: Introduction to Superposition — MIT OpenCourseWare — https://www.youtube.com/watch?v=lZ3bPUKo5zc
  • The Problem with Quantum Measurement — Sabine Hossenfelder — https://www.youtube.com/watch?v=Be3HlA_9968
02Cold open

Today's mission

Dr. Q · cold open

Psigor, put down that coin! No, wait, keep it, and fetch a teacup. Tonight we gamble with the universe itself. Last session you learned the grammar of quantum mechanics: kets, operators, eigenvalues. Now we put that grammar to work on the two most argued-over words in physics, superposition and measurement. I shall prepare a single qubit that is neither secretly 0 nor secretly 1, prove that it is not a coin hiding under a cup, freeze it in place merely by checking on it, and then confess the deepest unsolved puzzle in the whole theory: why a world run by smooth, reversible equations ever produces one definite answer. Place your bets, students. Observe... and COLLAPSE!

By the end of this session you will be able to:

  1. Write a qubit state as α|0⟩ + β|1⟩, check its normalization, and calculate measurement probabilities and expected counts from its amplitudes.
  2. Distinguish a superposition from a classical mixture by the interference it produces, and choose a measurement that tells the two apart.
  3. State the three rules of projective measurement (outcome, probability and post-measurement state) and use them to predict repeated measurements in the same basis and in a different one.
  4. Explain the quantum Zeno effect and calculate how frequent measurements suppress a transition.
  5. State the measurement problem precisely as a conflict between linear, unitary evolution and collapse, and explain why consciousness plays no role in the physics of measurement.
  6. Describe how decoherence hides the superpositions of large objects, and what it does and does not explain.
03The lesson

Read it — or listen along

7 parts. Press play on any part and the paragraph being read lights up as Dr. Q speaks.

Part 1

One State, Many Descriptions

Today we apply the grammar of chapter 14 to the simplest system there is, a qubit, short for quantum bit: anything with exactly two perfectly distinguishable states, labelled |0⟩ and |1⟩, such as a photon polarized horizontally or vertically, or an electron whose spin along a chosen axis is up or down. Because the Schrödinger equation is linear, as chapter 10 showed, any weighted sum of allowed states is also allowed. That sum is a superposition, |ψ⟩ = α|0⟩ + β|1⟩, and the weights α and β are complex numbers called probability amplitudes.

The amplitudes carry the odds. By the Born rule of chapter 9, asking 'zero or one?' returns 0 with probability |α|² and 1 with probability |β|², so |α|² + |β|² = 1, the normalization condition; the state 0.6|0⟩ + 0.8|1⟩ qualifies, since 0.36 + 0.64 = 1. Amplitudes also carry a phase. An overall phase shared by both terms changes nothing measurable, but the relative phase between α and β is physical. A general qubit needs just two angles: the amplitude of |0⟩ is the cosine of θ/2, and the amplitude of |1⟩ is the sine of θ/2, turned in the complex plane by a phase angle φ. The angle θ sets the odds; φ decides how the state interferes.

Now the crucial subtlety. A superposition is not 'both 0 and 1 at once', a slogan that sounds profound and explains nothing. It is one perfectly definite state, and what makes it a superposition is its relation to a particular measurement. Take |+⟩ = (1/√2)(|0⟩ + |1⟩). Asked 'zero or one?', it answers at random. But it is an eigenstate of the Pauli matrix σₓ of chapter 14, whose eigenstates |+⟩ and |−⟩ = (1/√2)(|0⟩ − |1⟩) form the X basis, so asked 'plus or minus?', it answers plus every time. For a photon, |+⟩ is light polarized at 45 degrees, which a polarizer turned to 45 degrees passes with certainty. Superposition is a relationship between a state and a basis.

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Part 2

The Coin Under the Cup

Here is the most tempting mistake in quantum mechanics. I flip a coin, slam a teacup over it, and ask you the odds. Fifty-fifty, you say, and you are right, but only about your knowledge: the coin is already heads or tails. A probabilistic blend of definite states like this is called a mixture, and its probabilities measure ignorance. Is |+⟩ just a quantum coin under a cup? Measure only in Z and you cannot tell, because both give 0 half the time. To catch the difference, ask a question that makes the possibilities interfere.

Ask the X question. The amplitude for plus is the sum α + β divided by √2; for |+⟩, with α = β = 1/√2, it is exactly 1, so plus comes up every time, while the amplitude for minus, α − β divided by √2, cancels to zero. That cancellation is destructive interference, and only amplitudes, which can be negative or complex, can do it. Each hidden coin is really |0⟩ or really |1⟩, and each of those gives plus with even odds, so any blend gives plus exactly half the time. Probabilities add; amplitudes add and then get squared. That rule from the double slit of chapter 7 is the whole difference between superposition and ignorance.

Experiment settles it. In 1986 Philippe Grangier, Gérard Roger and Alain Aspect at the Institut d'Optique in Orsay sent single photons to a beam splitter and showed that each was registered on one side or the other, with joint detections on both sides far rarer than any classical wave allows. Then, with the same source and detectors, they sent the photons through a Mach-Zehnder interferometer, which splits and then recombines the two possible paths, and recorded fringes with a visibility above 98 percent. Had every photon secretly taken one arm, each exit would have received half the photons whatever the arm lengths. Instead, at the right setting, nearly every photon left by one exit, steered by a relative phase no hidden-coin story contains.

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Part 3

Throwing the Dice: Projective Measurement

Now make measurement precise. In chapter 14, each observable became a Hermitian operator whose eigenvalues are the possible results and whose eigenstates |k⟩ form a basis. A projective measurement, the textbook ideal, obeys three rules. First, the outcome is always an eigenvalue. Second, the probability of result k is the squared magnitude of ⟨k|ψ⟩, the overlap between |k⟩ and the state: |α|² for 0 and |β|² for 1 in a Z measurement. Third, immediately afterwards the system is in the matching eigenstate |k⟩. This third rule is the projection postulate, better known as collapse: α|0⟩ + β|1⟩ jumps to whichever of |0⟩ or |1⟩ was found.

Max Born proposed the probability rule in 1926; in 1927 Werner Heisenberg wrote that every determination of position reduces the wave packet; Paul Dirac's textbook of 1930 said a measurement makes the system jump into an eigenstate of the quantity measured. In 1932 John von Neumann made the jump an axiom, distinguishing two utterly different kinds of change: process 1, the abrupt, unpredictable change caused by measurement, and process 2, smooth, deterministic Schrödinger evolution. Keep that pair in mind. It returns to haunt us.

This randomness is not that of a sloppy instrument. Prepare a million qubits in exactly the state 0.6|0⟩ + 0.8|1⟩ and measure each perfectly: nothing in the theory says which will give 1, yet very nearly 64 percent will. Single outcomes are random; the statistics are exact. Whether collapse is a physical process, an update of information, or never happens at all is interpretation-dependent, the quarrel of chapter 22. But every physicist uses the rule the same way: after a result, predict the future from the eigenstate found.

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Part 4

Ask Twice, Same Answer; Ask Sideways, Start Over

The aftermath rule has a testable consequence. Measure Z and get 1, and the state is now |1⟩; measure Z again at once and, since the overlap of |1⟩ with itself is 1, you get 1 with certainty. This repeatability is what makes a measurement a measurement: a reading that changed every time you glanced at it would record nothing. It also motivated the collapse rule, since the second answer is certain even though the first was not.

Now ask sideways. Prepare |0⟩ and measure X: plus and minus come up fifty-fifty, because |0⟩ = (1/√2)(|+⟩ + |−⟩). Suppose you get plus. Measure Z again and you find 0 or 1 at random: the X measurement erased the Z information. Z and X are incompatible observables. Their matrices do not commute, as you found in chapter 14, so no state is an eigenstate of both, and a sharp answer to one question forces a maximally uncertain answer to the other. In chapter 17, silver atoms will do exactly this in chains of Stern-Gerlach magnets.

You can watch the same logic with three polarizing filters. Two crossed at 90 degrees, vertical then horizontal, block essentially all light. Slide a third, at 45 degrees, between them, and one-eighth of the original unpolarized light reappears. Wave optics gets the intensities from Malus's law; the quantum account follows each indivisible photon. The diagonal filter measures it in the diagonal basis: it passes with probability equal to the cosine squared of 45°, one half, emerging diagonally polarized with its vertical history wiped out, and gets another fifty-fifty chance at the last filter. If filters merely read fixed labels, adding one could only remove light. Here it lets more through, because measurement changes the state.

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Part 5

A Watched Quantum Pot: The Zeno Effect

Repeatability has a startling corollary. Suppose a resonant radio wave steadily rotates a qubit from |0⟩ toward |1⟩, which it would reach after a time T. At short times the amplitude of |1⟩ grows in proportion to the elapsed time, so the probability grows only as the time squared. Check early and you almost certainly find 0, and collapse resets the state to |0⟩, restarting the slow creep. With N checks spread over T, each finds a flip with probability of only about (π/2N)², so the total risk is roughly π²/(4N), shrinking toward zero as N grows. In 1977 Baidyanath Misra and George Sudarshan named this the quantum Zeno effect, after Zeno of Elea, whose paradox argued that a flying arrow, motionless at every instant, never moves.

It was tested in 1990 at the National Institute of Standards and Technology in Boulder by Wayne Itano, Daniel Heinzen, John Bollinger and David Wineland. They held about 5,000 beryllium ions in a Penning trap and applied a radio-frequency pulse, lasting 256 ms, that would carry every ion between two sublevels of its ground state. During it they fired 1 to 64 brief ultraviolet laser pulses: an ion in one sublevel scattered light, an ion in the other stayed dark, so each pulse was a Z measurement. With one pulse, essentially every ion made the transition; with two, half did; with 64, fewer than 8 percent did, in agreement with the calculations. Wineland later shared the 2012 Nobel Prize in Physics with Serge Haroche.

One detail matters enormously. Each ion in the scattering sublevel threw off about 72 photons per pulse, but the collecting optics could catch only about 2 in every 10,000 of them, far fewer than one per ion. The authors stressed that what counts is the number of photons scattered, not detected: the measurement was complete once photons flew off carrying the information, and no eye, screen or mind was needed. Nor does interruption always slow things down. In 2001 Mark Raizen's group at the University of Texas at Austin saw both the Zeno effect and the anti-Zeno effect, in which frequent measurement speeds decay up, in sodium atoms tunnelling out of a trap made of laser light.

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Part 6

The Measurement Problem, Stated Precisely

Now the confession. Quantum mechanics, as usually taught, runs on two rules for change. Rule A is the Schrödinger equation: deterministic, continuous, reversible and linear, so a superposition going in comes out as the same superposition of the separately evolved parts. Such evolution is called unitary, because it preserves the length of every state and every overlap between states. Rule B is collapse: random, abrupt, irreversible and not linear at all, since it discards every branch but one. Rule B is invoked for measurements, but a measuring device is made of atoms obeying Rule A like everything else. Which rule applies, and when?

Apply Rule A to the measurement itself. A perfect detector that shows '0' for |0⟩ and '1' for |1⟩ produces |0⟩|A₀⟩ and |1⟩|A₁⟩, where |A₀⟩ and |A₁⟩ are pointer readings. Linearity then forces a qubit that starts as α|0⟩ + β|1⟩ to end as α|0⟩|A₀⟩ + β|1⟩|A₁⟩: not one reading but a superposition of two, with the pointer entangled with the qubit, a word chapter 20 makes precise. Yet every notebook records one result, with frequencies |α|² and |β|². In 1995 the philosopher Tim Maudlin put the clash crisply: the wavefunction is a complete description; it always evolves by the linear Schrödinger equation; measurements have single, definite outcomes. All three cannot be true. Collapse theories modify the equation, pilot-wave theory adds particle positions, and many-worlds denies the single outcome.

Because the chain from qubit to detector to eye to brain can be extended indefinitely, some physicists once wondered whether it ends only at a conscious mind; Eugene Wigner floated the idea in 1961 and later abandoned it, and no experiment supports it. The mainstream view is plain: a measurement is a physical interaction that records information in an apparatus or its environment, and, as the beryllium ions showed, quantum systems respond to that record whether or not anyone reads it. Interpretations disagree about whether collapse is real, but none of the mainstream ones lets a mind cause it. The genuine puzzle is Rule A against Rule B.

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Part 7

Decoherence: Why Cats Never Look Superposed

In 1935 Erwin Schrödinger imagined a cat sealed in a box with a radioactive atom, a Geiger counter and a vial of poison: Rule A, applied without limit, would leave the cat in a superposition of alive and dead. He meant it as a criticism of taking the formalism too literally. So why do we never see dust or cats in superpositions? The leading answer is decoherence, developed by Dieter Zeh from 1970 and by Wojciech Zurek, Erich Joos and others in the 1980s. A large object is never alone: air molecules strike it and light bounces off it, and every encounter leaves a faint record of where it is, just as the detector recorded the qubit.

Once the surroundings hold a record, the object is entangled with them, and for any measurement on the object alone, the interference terms are multiplied by the overlap between the environment's two records. When the records differ, that overlap plunges toward zero and the object behaves exactly like a mixture. The speed is staggering. For a dust grain a hundredth of a millimetre across, superposed across its own width, standard estimates give a decoherence time of about 10⁻³¹ seconds in air, 10⁻¹⁴ seconds in the best laboratory vacuum, and one second in deep space, lit only by the cosmic microwave background. A warm, breathing cat never stands a chance.

Decoherence has been switched on and off in the laboratory. In 1996 Michel Brune, Serge Haroche and colleagues in Paris watched a superposition of two microwave fields of a few photons, held in a superconducting cavity, lose its coherence, faster the more different its components were. In 2003 Klaus Hornberger, Markus Arndt, Anton Zeilinger and colleagues in Vienna leaked gas into an interferometer for C₇₀ fullerene molecules: the higher the pressure, the more collisions recorded the molecules' paths, and the fainter the fringes, as predicted.

Decoherence explains why interference vanishes for big things, why the everyday world looks definite in position rather than in some exotic basis, and why records are stable. What it does not do, by itself, is pick one outcome: the total state of object plus environment still holds both branches, which merely can no longer interfere. Whether that ends the story is where the interpretations of chapter 22 part company.

Watch this part

04Video shelf

Go deeper on video

23 more videos for this chapter, 17 of them from universities, research institutes and Nobel lectures. The clips under “Watch this part” in the lesson start at the moment that matters.

23 videos · 9 h 32 min

  • Lecture · MIT

    The nature of superposition. Mach-Zehnder interferometer

    MIT OpenCourseWare · Barton Zwiebach

    Zwiebach contrasts classical superposition of fields with the quantum kind using single photons in a Mach-Zehnder interferometer, then states the rule this section builds on: measure a property on αA + βB and you never get anything in between, only a or b, with odds set by |α|² and |β|², after which the state is the one you found. When he says a photon is 'in both beams at the same time', translate it into this section's language (one definite state written as a sum of two path states), and catch his point at about 7:30 that the same vector can be written as a sum in many different ways.

  • Lecture · IBM Quantum

    Superposition: The Quantum Principle That Changes Everything

    Qiskit

    An IBM Quantum classroom lesson that tries to understand superposition by comparing measurements on a qubit in superposition with a classical coin flip, and explores how the 'quantum coin' and the classical coin behave differently. That is this section's coin-under-the-cup question; keep its test in mind as you watch: asked only 'zero or one?', |+⟩ and a fair coin give identical statistics, so the difference has to come from asking a different question.

  • Lecture · MIT

    Completeness of eigenvectors and measurement postulate

    MIT OpenCourseWare · Barton Zwiebach

    Starts at 7:55, where Zwiebach states the measurement postulate as this section does: the result is always an eigenvalue, its probability is |αᵢ|², the squared overlap with the eigenstate, and the state collapses to that eigenstate, so an immediate repeat gives the same value; rewind to the beginning to see how any state is expanded in eigenfunctions and why the |αᵢ|² add up to 1. Stay for 15:10, where he calls it 'a very strange postulate' that splits quantum mechanics into the realm of the Schrödinger equation and the realm of measurement, a preview of section 6.

  • Demo · YouTube

    Three polarizing filters: a simple demo of a creepy quantum effect

    Orion Lawlor

    A 90-second tabletop demo: two crossed polarizers block the light, and a third at a diagonal angle lets light through again, but only when it sits between them, not before or after. Watch it right after this section's three-filter paragraph and narrate each filter as a measurement in a new basis; the uploader's notes also point to Dirac, who opened his 1930 textbook with the polarization of single photons.

  • Demo · YouTube

    A quantum measurement you can do at home

    Looking Glass Universe

    Looking Glass Universe uses a laser pointer, a calcite crystal and polarizing film to make a polarization measurement at home, then works through rotating the measurement (4:15), orthogonal states (5:47), equivalent measurements (6:41) and the quantum versus classical view (7:47). It makes 'ask sideways' tangible: watch how rotating the measurement changes which question is being put to the light.

  • Explainer · YouTube

    Zeno's Paradox & The Quantum Zeno Effect

    PBS Space Time

    PBS Space Time starts from Zeno's arrow paradox and builds the quantum Zeno effect out of superposition and collapse (2:24), with chapters on stopping the quantum arrow (3:33), its experimental verification (5:03) and the debate over what counts as a measurement (8:16). Compare its account with the 1990 NIST beryllium-ion experiment in this section, and with the point that scattered photons, not detected ones, do the freezing.

  • Short · Washington University in St. Louis

    Zeno Effect

    Washington University in St. Louis

    A three-minute university video on Kater Murch's group, who probed the role of measurement in the Zeno effects with an artificial-atom qubit, including a 'quasimeasurement' interaction that only disturbs it; Murch calls the results the first observation of the two Zeno effects on a single quantum system. It sharpens this section's lesson that what freezes a quantum state is a physical interaction, not anybody looking.

  • Explainer · YouTube

    "Standard" quantum mechanics is obviously wrong

    Looking Glass Universe

    Starts at 2:37, just after the sponsor segment; YouTube's chapter markers then run from quantum determinism to the measurement contradiction (5:23), testing the measurement rule (8:02), the Heisenberg cut (10:41) and alternatives to Copenhagen (13:10), the same arc as this section. Set its Schrödinger equation against Rule A and its measurement rule against Rule B, and pair the Heisenberg-cut chapter with this section's 'sliding cut' callout.

  • Lecture · Uppsala University

    Nobel Laureate in physics Serge Haroche – Nobel Lectures in Uppsala 2012

    Uppsala universitet · Serge Haroche

    Haroche, whose group ran the 1996 Paris experiment in this section, gives a public version of his Nobel talk 'Controlling photons in a box and exploring the quantum to classical boundary' (the Stockholm lecture itself is in the Lecture Hall). It starts at 40:10, where a single atom prepares a Schrödinger-cat state of light, followed at 42:37 by 'Fifty milliseconds in the life of a Schrödinger cat': the laboratory version of this section's question of how fast a superposition of distinguishable states fades.

  • Explainer · YouTube

    How Quantum Mechanics produces REALITY & perhaps ARROW of TIME | wave collapse & Decoherence

    Arvin Ash

    A beginner-level account of how the quantum world turns into the definite classical one, starting at 'Switch from quantum to classical' (5:03) and 'Quantum decoherence explained' (6:06), then asking whether decoherence is really irreversible (8:52) and whether there is a point of no return (11:26). Read any talk of 'looking' as interacting and leaving a record, keep this section's caveat in mind (decoherence explains why interference vanishes, not by itself which outcome occurs), and treat the closing link to the arrow of time (12:18) as speculation.

  • Lecture · Yale

    22. Quantum mechanics IV: Measurement theory, states of definite energy

    YaleCourses · Ramamurti Shankar · 2010

    The lecture from Yale University PHYS 201 Fundamentals of Physics II (Ramamurti Shankar, 2010) that goes with Chapter 15. The whole course is in the Lecture Hall.

  • Lecture · IBM Quantum

    Single Systems | Understanding Quantum Information & Computation | Lesson 01

    Qiskit · John Watrous · 2022–25

    The lecture from IBM Quantum Understanding Quantum Information & Computation (John Watrous, 2022–25) that goes with Chapter 15. The whole course is in the Lecture Hall.

  • Lecture · Caltech

    Ph CS 219A Lecture 3A Measurement

    John Preskill · 2021

    The lecture from Caltech Ph/CS 219A Quantum Computation (John Preskill, 2021) that goes with Chapter 15. The whole course is in the Lecture Hall.

  • Lecture · Nobel Prize

    2012 Nobel Lectures in Physics

    Nobel Prize · Serge Haroche & David J. Wineland · 2012

    Serge Haroche on trapping photons in a box to watch the quantum-to-classical boundary — decoherence — in action, then David Wineland on superposition and entanglement with trapped ions. The experiments that turned thought experiments into laboratory routine.

  • Panel · World Science Festival

    Measure for Measure: Quantum Physics and Reality

    World Science Festival · Brian Greene with David Z. Albert, Sean Carroll, Sheldon Goldstein and Ruediger Schack · 2014

    Brian Greene moderates a 2014 panel in which Sean Carroll argues for many worlds, Sheldon Goldstein for Bohmian mechanics and Rüdiger Schack for QBism, with the philosopher of physics David Albert pressing all three. Watch it after Part 6: every move in the debate maps onto today's trilemma.

  • Series · Perimeter

    Quantum 101

    Perimeter Institute for Theoretical Physics · Katie Mack · 2023

    Ten five-minute episodes hosted by the cosmologist Katie Mack: duality, quantization, uncertainty, superposition, entanglement, probability, tunnelling and quantum computing. The friendliest possible on-ramp, and a quick refresher before any chapter.

  • Course · Yale

    Fundamentals of Physics II

    YaleCourses · Ramamurti Shankar · 2010

    Ramamurti Shankar, author of a classic graduate textbook, gives one of the gentlest rigorous introductions anywhere: seven lectures from the double-slit experiment through measurement, the particle in a box, the time-dependent Schrödinger equation and the postulates.

  • Course · MIT

    Quantum Physics II

    MIT OpenCourseWare · Barton Zwiebach · 2013

    The sequel: vector spaces and Dirac notation, the uncertainty principle done properly, the Heisenberg picture, two-state systems, tensor products and the addition of angular momentum. The natural next step once Unit 3 of this course feels comfortable.

  • Course · Stanford

    The Theoretical Minimum: Quantum Mechanics

    Stanford · Leonard Susskind · 2012

    Leonard Susskind's course for curious adults, the basis of his book with Art Friedman. It starts from a single spin, builds states, operators and measurement from scratch, then reaches entanglement — one of the cleanest routes from zero to the real formalism.

  • Course · Stanford

    Quantum Entanglements, Part 1

    Stanford · Leonard Susskind · 2006

    Nine evenings on the strangest resource in physics — spin, density matrices, entangled states and Bell's theorem — taught by Susskind to a room of non-specialists. Pairs naturally with Unit 5.

  • Course · IBM Quantum

    Understanding Quantum Information & Computation

    Qiskit · John Watrous · 2022–25

    The clearest modern course on quantum information, from IBM Quantum's John Watrous: single and multiple systems, circuits, teleportation and superdense coding, Shor's and Grover's algorithms, density matrices and error correction. Mathematically honest without being heavy.

  • Course · Caltech

    Quantum Computation

    John Preskill · 2021

    The graduate course from the physicist who coined the terms "quantum supremacy" and "NISQ": density operators, measurement, channels, Bell inequalities, teleportation, circuits, Shor's factoring, Grover's search and quantum simulation. Advanced, and worth every minute.

Every video for Chapter 15 in the Video Library

05To the blackboard

Key equations

Equation 1

A qubit superposition

∣ψ⟩=α ∣0⟩+β ∣1⟩,∣α∣2+∣β∣2=1|\psi\rangle = \alpha\,|0\rangle + \beta\,|1\rangle, \qquad |\alpha|^2 + |\beta|^2 = 1

Any combination of the two basis states is a legitimate state, provided the squared magnitudes of the amplitudes add up to one, because they are the probabilities of the two outcomes of a Z measurement.

What each symbol means
SymbolMeaningUnit
|ψ⟩the state of the qubitdimensionless
|0⟩, |1⟩the two Z-basis statesdimensionless
α, βcomplex probability amplitudesdimensionless
|α|², |β|²probabilities of measuring 0 and 1dimensionless

Say it: “ket psi equals alpha times ket zero plus beta times ket one, where the magnitude of alpha squared plus the magnitude of beta squared equals one”

Equation 2

Projective measurement: the Born rule and collapse

P(k)=∣⟨k∣ψ⟩∣2,∣ψ⟩  → result k   ∣k⟩P(k) = \left|\langle k|\psi\rangle\right|^2, \qquad |\psi\rangle \;\xrightarrow{\ \text{result } k\ }\; |k\rangle

Measuring an observable gives one of its eigenvalues, with probability equal to the squared overlap between the state and the matching eigenstate, and leaves the system in that eigenstate. An immediate repeat therefore gives the same result.

What each symbol means
SymbolMeaningUnit
|k⟩eigenstate of the measured observable for result kdimensionless
⟨k|ψ⟩overlap (inner product) between that eigenstate and the statedimensionless
P(k)probability of result kdimensionless

Say it: “the probability of result k equals the squared magnitude of the overlap of k with psi, and after the result k, the state becomes ket k”

Equation 3

The general qubit and the X basis

∣ψ⟩=cos⁡θ2 ∣0⟩+eiφsin⁡θ2 ∣1⟩,∣±⟩=12(∣0⟩±∣1⟩)|\psi\rangle = \cos\tfrac{\theta}{2}\,|0\rangle + e^{i\varphi}\sin\tfrac{\theta}{2}\,|1\rangle, \qquad |\pm\rangle = \tfrac{1}{\sqrt{2}}\big(|0\rangle \pm |1\rangle\big)

Every qubit state, up to an unobservable overall phase, is fixed by two angles: theta sets the odds of 0 and 1, and phi is the relative phase. The states plus and minus are the eigenstates of the X observable, an equal superposition in the Z basis but definite outcomes in their own.

What each symbol means
SymbolMeaningUnit
θangle setting the split of probability between |0⟩ and |1⟩, from 0 to 180 degreesrad or degrees
φrelative phase between the two amplitudesrad or degrees
|+⟩, |−⟩eigenstates of σₓ, the X basisdimensionless

Say it: “ket psi equals cosine of theta over two times ket zero, plus e to the i phi times sine of theta over two times ket one. The X-basis states, ket plus and ket minus, equal one over root two times ket zero plus or minus ket one”

Equation 4

Odds in the Z and X bases

P(0)=cos⁡2θ2,P(+)=12(1+sin⁡θ cos⁡φ)P(0) = \cos^2\tfrac{\theta}{2}, \qquad P(+) = \tfrac{1}{2}\left(1 + \sin\theta\,\cos\varphi\right)

The Z odds depend only on theta, but the X odds also depend on the relative phase phi. A mixture of 0 and 1 with any odds gives plus exactly half the time, so an X measurement can unmask a superposition whenever cosine phi is not zero.

What each symbol means
SymbolMeaningUnit
P(0)probability of result 0 in a Z measurementdimensionless
P(+)probability of result plus in an X measurementdimensionless
θ, φthe preparation angles of equation e3rad or degrees

Say it: “the probability of zero equals cosine squared of theta over two, and the probability of plus equals one half times one plus sine theta times cosine phi”

Equation 5

The quantum Zeno effect

Pstay(N)=[cos⁡2 ⁣(π2N)]N≈1−π24N  ⟶  1(N→∞)P_{\text{stay}}(N) = \left[\cos^2\!\left(\frac{\pi}{2N}\right)\right]^{N} \approx 1 - \frac{\pi^2}{4N} \;\longrightarrow\; 1 \quad (N \to \infty)

A drive that would carry a qubit fully from 0 to 1 is interrupted by N equally spaced Z measurements. Each check catches a flip with probability of only about pi over two N, squared, so the total leakage falls like one over N and frequent enough checking freezes the qubit.

What each symbol means
SymbolMeaningUnit
Nnumber of equally spaced measurements during the would-be flipdimensionless
P_stayprobability the qubit is found in |0⟩ at every checkdimensionless

Say it: “the probability of staying in zero at every one of N checks equals cosine squared of pi over two N, raised to the power N, which is approximately one minus pi squared over four N, and tends to one as N grows without limit”

Equation 6

The measurement problem in one line

(α∣0⟩+β∣1⟩) ∣Aready⟩  → U^   α ∣0⟩∣A0⟩+β ∣1⟩∣A1⟩\big(\alpha|0\rangle + \beta|1\rangle\big)\,|A_{\text{ready}}\rangle \;\xrightarrow{\ \hat{U}\ }\; \alpha\,|0\rangle|A_0\rangle + \beta\,|1\rangle|A_1\rangle

Linear, unitary evolution turns a superposed system and a ready detector into a superposition of two pointer readings, entangled with the system. Experiments always show one reading, so something more, collapse or an interpretation, must account for the single outcome.

What each symbol means
SymbolMeaningUnit
|A_ready⟩the detector before the measurementdimensionless
|A₀⟩, |A₁⟩pointer states showing 0 and 1dimensionless
Ûunitary operator describing the measurement interactiondimensionless

Say it: “alpha ket zero plus beta ket one, times the ready state of the apparatus, evolves under the unitary U into alpha times ket zero with pointer A zero, plus beta times ket one with pointer A one”

06Worked examples

Real numbers, step by step

Cover the steps, try the problem yourself, then reveal the solution one step at a time.

Example 1

Reading the odds off a state

Problem

A qubit is prepared in |ψ⟩ = 0.6|0⟩ + 0.8|1⟩. Check that it is normalized, find the Z-measurement probabilities, predict the count of 0 results in 1,000 runs with its typical scatter, say what a second Z measurement gives after a result of 1, and find the X-measurement probabilities of the original state.

  1. Step 1: Normalization: |α|² + |β|² = 0.6² + 0.8² = 0.36 + 0.64 = 1. The state is legitimate.

  2. Step 2: Z probabilities: 0 with probability 0.36 and 1 with probability 0.64.

  3. Step 3: Counts: in 1,000 runs expect 1,000 × 0.36 = 360 zeros. The binomial standard deviation is √(1,000 × 0.36 × 0.64) = √230.4 ≈ 15, so a real run lands near 360 ± 15.

  4. Step 4: Repeat: a result of 1 leaves the state |1⟩, so an immediate second Z measurement gives 1 with probability 1.

  5. Step 5: X basis: the amplitude for plus is (0.6 + 0.8) ÷ √2, so the probability is one half of 1.4², which is 0.98. The amplitude for minus is (0.6 − 0.8) ÷ √2, so the probability is one half of 0.2², which is 0.02.

Answer

Normalized; 0 with probability 0.36 and 1 with 0.64; about 360 ± 15 zeros in 1,000 runs; a repeat after 1 gives 1 with certainty; in the X basis, plus 98 percent and minus 2 percent

Why it matters: A state that looks like a lopsided coin in Z is almost a sure thing in X. Change the sign of the second amplitude and the X odds swap to 2 percent and 98 percent, while the Z odds do not budge: the relative phase is invisible in one basis and decisive in another.

Try it yourself first — then check each step.
Example 2

Superposition versus the coin under the cup

Problem

Three boxes each give 0 in 75 percent of Z measurements. Box A holds the superposition with θ = 60° and φ = 0; box B holds θ = 60° and φ = 180°; box C is a mixture, a qubit secretly prepared as |0⟩ with probability 0.75 and |1⟩ with probability 0.25. What fraction of X measurements gives plus for each box?

  1. Step 1: Z check: for A and B, the probability of 0 is the cosine squared of θ/2, which is the cosine squared of 30°, or 0.75. For C it is 0.75 by construction. The Z statistics are identical.

  2. Step 2: Box A: the probability of plus is one half of (1 + sine 60° × cosine 0°) = one half of (1 + 0.866) = 0.933.

  3. Step 3: Box B: the cosine of 180° is −1, so the probability of plus is one half of (1 − 0.866) = 0.067.

  4. Step 4: Box C: each hidden |0⟩ gives plus with probability 0.5, and so does each hidden |1⟩, so the total is 0.75 × 0.5 + 0.25 × 0.5 = 0.5.

Answer

Plus comes up 93.3 percent of the time for A, 6.7 percent for B, and exactly 50 percent for the mixture C

Why it matters: Only a superposition can push the X odds away from fifty-fifty, because only amplitudes interfere. Beware one trap: with φ = 90° a superposition also gives 50 percent in X, and you would need a third basis, Y, to unmask it.

Try it yourself first — then check each step.
Example 3

Freezing a qubit by checking on it

Problem

A radio-frequency pulse lasting T = 256 ms would rotate a qubit completely from |0⟩ to |1⟩. During the pulse you make N equally spaced Z measurements. Find the probability that the qubit is found in |0⟩ at every check for N = 1, 4, 10 and 64, and compare N = 64 with the approximation 1 − π²/(4N).

  1. Step 1: Per interval the drive turns the state through one N-th of the full flip, so a check finds 1 with probability equal to the sine squared of π/2N, and finds 0, resetting the state, with probability equal to the cosine squared of π/2N.

  2. Step 2: Surviving all N checks: the cosine squared of π/2N, raised to the power N.

  3. Step 3: N = 1: the cosine squared of 90° is 0, so the qubit always flips. N = 4: the cosine squared of 22.5° is 0.8536, and 0.8536 to the fourth power is 0.531. N = 10: the cosine squared of 9° is 0.9755, and to the tenth power that is 0.781.

  4. Step 4: N = 64, one check every 4 ms: the cosine squared of π/128 is 0.99940, and to the 64th power that is 0.962. The approximation gives 1 − 9.870 ÷ 256 = 0.961.

Answer

The qubit stays in |0⟩ with probability 0 for N = 1, 0.53 for N = 4, 0.78 for N = 10 and 0.96 for N = 64

Why it matters: Each check leaks only about (π/2N)², so N checks leak about π²/(4N) in total. The NIST team's matching prediction for 64 pulses was a 3.7 percent transition probability, slightly below our 3.8 percent chance of leaving at least once, because an ion that flips can later flip back.

Try it yourself first — then check each step.
Example 4

Three polarizers and a resurrected beam

Problem

A 2.00 milliwatt beam of unpolarized red light, wavelength 633 nm, passes a vertical polarizer and then a horizontal one. What power gets through? A third polarizer at 45 degrees is then slipped between them. What power emerges now, and how many photons per second is that?

  1. Step 1: First filter: unpolarized light loses half its power, leaving 1.00 milliwatt, now vertically polarized.

  2. Step 2: Horizontal filter straight after: the cosine squared of 90° is 0, so essentially nothing gets through.

  3. Step 3: With the 45-degree filter inserted: it passes the cosine squared of 45°, one half, so 1.00 milliwatt becomes 0.500 milliwatt, now diagonally polarized; then the horizontal filter passes half again, leaving 0.250 milliwatt.

  4. Step 4: Photon energy: E = hc/λ = (6.626×10⁻³⁴ J·s × 2.998×10⁸ m/s) ÷ (633×10⁻⁹ metres) = 3.138×10⁻¹⁹ joules. Photon rate: 0.250×10⁻³ watts ÷ 3.138×10⁻¹⁹ joules ≈ 7.97×10¹⁴ photons per second.

Answer

Essentially zero with two crossed filters; 0.250 milliwatt, about 8.0×10¹⁴ photons per second, after inserting the 45-degree filter

Why it matters: Each photon meets each filter as a yes-or-no measurement. The middle filter re-prepares every survivor in the diagonal basis, erasing its vertical history, which is why adding an obstacle lets light through.

Try it yourself first — then check each step.
07Myth busters

What everyone gets wrong

Read the myth, decide whether you believed it, then bust it.

  • Myth #1

    A superposition is secretly one thing or the other; we just don't know which, like a coin hidden under a cup.

    Truth

    A hidden coin is a mixture, and mixtures cannot interfere: measured in the X basis, any blend of |0⟩ and |1⟩ gives plus exactly half the time. The superposition |+⟩ gives plus every time, because its amplitudes add before squaring. Single-photon interferometers, such as Grangier, Roger and Aspect's of 1986 with fringe visibility above 98 percent, show the difference directly, and Bell's theorem in chapter 21 rules out a whole class of hidden-value stories.

  • Myth #2

    Collapse is caused by consciousness: the wavefunction waits for a mind to look.

    Truth

    In physics a measurement is a physical interaction that leaves a record in an apparatus or environment. In the 1990 NIST Zeno experiment, photons scattered by the ions froze the transition even though almost none were ever detected. Eugene Wigner, the best-known advocate of the consciousness idea, later abandoned it, and no experiment supports it. Whether collapse is a real process at all depends on the interpretation, but none of the mainstream interpretations makes it depend on a mind.

  • Myth #3

    A watched pot never boils: staring at anything slows it down, thanks to the quantum Zeno effect.

    Truth

    The Zeno effect needs genuine measurements repeated faster than the system's coherent evolution, such as the 64 laser pulses in 256 ms that froze the NIST ions. A pot of water is already being 'measured' by its environment countless times every second, and your gaze adds nothing. Frequent interruption can even speed up some decays, the anti-Zeno effect observed in 2001.

08Interactive lab

Your turn at the controls

Welcome to the Measurement Casino, where the house always wins and the house is quantum mechanics. You prepare a qubit with two dials, θ for the odds and φ for the relative phase, then send it to a table that asks either the Z question or the X question. Every draw throws genuine quantum dice, shows the collapse, and adds to a running tally, so you can watch single random outcomes build into exact statistics.

Your tasks

0/4

How it connects to the theory

Your dials set the amplitudes cos(θ/2) and sin(θ/2), with relative phase φ. The Z table returns 0 with probability cos²(θ/2), the X table returns plus with probability one half of (1 + sin θ · cos φ), and every result replaces the state with the matching eigenstate, which is why repeats agree and switching bases re-randomizes. The mixture box gives plus exactly half the time in X however its odds are set, and that is how you unmask it. The Zeno meter applies the survival probability cos²(π/2N) per check from the worked example, so its frozen qubit is the same physics as the NIST beryllium ions.

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09Memory aids

Make it stick

Memory trick

Walk through the doors of the quantum CASINOS. C: Coefficients squared give the odds, and they must sum to 1. A: Amplitudes Add before squaring, so superpositions interfere and coins under cups do not. S: Snap, the state jumps to the eigenstate matching the result. I: Immediate repeats give Identical answers. N: New basis, New dice, because an incompatible measurement re-randomizes. O: Often-checked qubits freeze, the Zeno effect. S: Smooth versus Snap, the measurement problem, with the environment's records hiding the other branches. To use it, run the seven letters in order on any measurement question: compute the odds, look for interference, snap the state, repeat, switch basis, watch the clock, then ask which rule you just used.

Analogy — and where it breaks

A superposition is like a direction on a map. Northeast is not secretly north or secretly east; it is one perfectly definite direction that happens to have components along both. Ask the 'north or east?' question and you must get one of two answers, with odds set by the components; ask 'northeast or northwest?' and northeast answers with certainty. Where it breaks: you can read both components of a real arrow at once without disturbing it, but a quantum measurement answers only one question, randomly, and resets the arrow to the answer it gave. And quantum components are complex numbers whose phases interfere, which no map arrow can do.

The whole chapter in one line

A superposition is one definite state that can interfere; measurement picks one answer at random and snaps the state to match.

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10Games

Play it into long-term memory

Match the vocabulary, then take the Quiz Show. Score 70% or better to tick it off your checklist.

Vocab match · timed

Vocab Match

Match each term to its definition, against the clock.

  • 12 pairs in 2 rounds
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  • Finish with 3 mistakes or fewer to win
Live from the lab

Dr. Q's Quantum Quiz Show

Dr. Q, your host

Welcome, contestants! 8 questions, 30 seconds each, and one mad scientist judging you, lovingly. Answer fast for a speed bonus; answer right in a row and your points multiply.

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11Worksheet

Prove it on paper

Answer every question, check your work, and read the explanations — the stretch challenge is for the truly mad.

14 questions12 terms1 challenge

Work through every question with the lesson closed, then check yourself. In the X basis, |+⟩ = (1/√2)(|0⟩ + |1⟩) and |−⟩ = (1/√2)(|0⟩ − |1⟩). For a qubit cos(θ/2)|0⟩ + sin(θ/2)|1⟩ with relative phase φ, the probability of 0 in Z is cos²(θ/2) and the probability of plus in X is one half of (1 + sin θ · cos φ). Unpolarized light loses half its intensity at the first polarizer, and polarized light passes a polarizer at angle θ to its polarization with probability cos²θ. Show every step in the calculation questions.

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Part A

Vocabulary

Superposition
A state written as a weighted sum of other states, such as α|0⟩ + β|1⟩; a single definite state whose outcomes are uncertain only for measurements in a basis that does not contain it.
Probability amplitude
A complex coefficient in a superposition; the squared magnitude of an amplitude gives the probability of the matching outcome.
Normalization
The requirement that the squared magnitudes of all amplitudes add to one, so that the outcome probabilities sum to 1.
Basis
A complete set of perfectly distinguishable states, such as |0⟩ and |1⟩ or |+⟩ and |−⟩, in terms of which any state can be written.
Relative phase
The phase difference between the amplitudes in a superposition; invisible to some measurements but decisive for interference.
Mixture
A probabilistic blend of definite states that reflects ignorance, like a coin under a cup; unlike a superposition, it produces no interference.
Projective measurement
An ideal measurement whose results are the eigenvalues of an observable, with probabilities given by the Born rule, leaving the system in the matching eigenstate.
Collapse
The projection postulate: the update of the state to the eigenstate matching a measurement result; whether it is a physical process is interpretation-dependent.
Incompatible observables
Observables whose operators do not commute, such as Z and X, so that no state has definite values of both and measuring one randomizes the other.
Quantum Zeno effect
The suppression of a quantum transition by frequent measurements, which repeatedly return the system to its initial state.
Measurement problem
The conflict between linear, unitary evolution, which predicts superpositions of pointer readings, and the single definite outcomes every experiment records.
Decoherence
The loss of observable interference when a system becomes entangled with its environment, which records information about it.

Part B

Questions

Auto-checked except short answers, which you mark yourself.

  1. Question 1.Multiple choice

    Which of these is a properly normalized qubit state?

  2. Question 2.Multiple choice

    Box P holds qubits in the state |+⟩; box Q holds qubits secretly set to |0⟩ or |1⟩ with even odds. Which single type of measurement, repeated on many qubits from each box, tells you which box is which?

  3. Question 3.Multiple choice

    A qubit is measured in Z and gives 0. It is then measured in X and gives minus. What is the probability that a third measurement, in Z, gives 0?

  4. Question 4.True or false

    Two qubit states with the same values of |α|² and |β|² give the same statistics for every possible measurement.

  5. Question 5.True or false

    An ideal projective Z measurement repeated immediately on the same qubit can give 0 the first time and 1 the second.

  6. Question 6.True or false

    Decoherence explains why interference between macroscopically different states becomes unobservable, but by itself it does not select which single outcome occurs.

  7. Question 7.Fill in the blank

    A probabilistic blend of definite states, whose probabilities reflect only our ignorance, is called a .

  8. Question 8.Fill in the blank

    The rule that leaves a system in the eigenstate matching the measured result is called the projection postulate, or .

  9. Question 9.Short answer

    State the measurement problem in your own words. Your answer should mention linearity.

  10. Question 10.Short answer

    A friend says the quantum Zeno effect proves that a conscious observer controls reality. Use the NIST experiment to answer them.

  11. Question 11.Calculation

    A qubit is prepared in |ψ⟩ = (1/√3)|0⟩ + √(2/3)|1⟩. Show that it is normalized, find the probability of each Z result, and predict the number of 1 results in 600 runs, with the binomial standard deviation √(N·p·(1 − p)).

    This one asks for more than one value — give each, with its units.

  12. Question 12.Calculation

    A qubit is prepared with θ = 90° and φ = 60°. Find the probability of 0 in a Z measurement and of plus in an X measurement. What would a fifty-fifty mixture of |0⟩ and |1⟩ give in X?

    This one asks for more than one value — give each, with its units.

  13. Question 13.Calculation

    A drive would rotate a qubit fully from |0⟩ to |1⟩ in a time T, and you make N equally spaced Z measurements during it. (a) Find the probability the qubit is found in |0⟩ at every check for N = 20. (b) Use the approximation 1 − π²/(4N) to estimate how many checks keep the survival probability at 99 percent or more, then check your estimate with the exact formula.

    This one asks for more than one value — give each, with its units.

  14. Question 14.Calculation

    An 8.0 mW unpolarized beam passes four polarizers in a row, with axes at 0°, 30°, 60° and 90°. What power emerges? Compare with two crossed polarizers at 0° and 90° alone.

    Give a number and units. 6.63×10⁻³⁴, 6.63e-34 and 6.63 x 10^-34 all work.

Part C · optional, dangerous, extremely rewarding

Mad scientist challenge

The measurement problem meets decoherence. A qubit starts in α|0⟩ + β|1⟩ with real amplitudes, and a detector records it, so that the combined state becomes α|0⟩|A₀⟩ + β|1⟩|A₁⟩, where the pointer states are normalized and have real overlap c = ⟨A₀|A₁⟩. (a) Rewrite the combined state using |0⟩ = (1/√2)(|+⟩ + |−⟩) and |1⟩ = (1/√2)(|+⟩ − |−⟩), and show that the probability of finding the qubit in |+⟩ is one half of (1 + 2αβc). (b) Evaluate it for α = β = 1/√2 when the detector records nothing (c = 1), records partially (c = 0.6), and records perfectly (c = 0). (c) Evaluate it for α = 0.6, β = 0.8 and c = 0.5. (d) Explain what this has to do with why a dust grain in air never shows interference.

12Wrap-up

Recap & sign-off

Key takeaways

  • A superposition α|0⟩ + β|1⟩ is one definite state; the odds are |α|² and |β|², which must sum to 1.
  • Superposition is relative to a basis: |+⟩ is random in Z but certain in X.
  • Superposition is not ignorance: amplitudes add before squaring and can interfere, while a hidden-coin mixture always gives fifty-fifty in X.
  • A projective measurement returns an eigenvalue with Born-rule probability and leaves the matching eigenstate, so immediate repeats agree and an incompatible measurement re-randomizes.
  • Frequent measurement freezes a slowly driven qubit, the quantum Zeno effect, demonstrated with trapped beryllium ions at NIST in 1990.
  • The measurement problem is the clash between linear, unitary evolution, which predicts superposed pointer readings, and the single outcomes every experiment records.
  • Measurement is a physical interaction that leaves a record; consciousness plays no role in the physics.
  • Decoherence hides the superpositions of large objects within tiny fractions of a second, but by itself it does not pick an outcome.
Dr. Q signs off

Mwahaha, measurably! Tonight you prepared superpositions, unmasked a coin hiding under a cup, collapsed states on demand, froze a qubit just by pestering it, and stared into the measurement problem without blinking. Next session we leave our two-level toy for a real atom. We solve Schrödinger's equation for hydrogen, the atom Bohr could describe only with hand-made rules, and discover that its electron does not orbit at all: it occupies orbitals, standing-wave clouds labelled by quantum numbers. Bring your sense of shape.

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