In short

A projective measurement is a set of orthogonal projectors \{P_m\} that sum to the identity, \sum_m P_m = I. Given an input state |\psi\rangle, the probability of outcome m is the Born rule, p(m) = \langle \psi | P_m | \psi\rangle. After the measurement, the state collapses to the projected, renormalised component: |\psi\rangle \mapsto P_m |\psi\rangle / \sqrt{p(m)}. Measurement is the only non-unitary operation in the entire quantum postulate system — it is irreversible, it destroys phase information, and it is where quantum becomes classical. One measurement yields one bit per qubit; everything else about |\psi\rangle is gone the instant you read the outcome.

You have just spent several chapters building a state. Complex amplitudes, normalisation, inner products, gates that rotate the state into elaborate superpositions — every piece of it governed by the crisp, reversible rule U^\dagger U = I. Then you measure the qubit.

You get a single bit. A 0 or a 1. And the state you so carefully prepared is gone.

This is not a bug in the formalism. It is the rule. Every quantum computation you will ever do ends in a measurement, because a quantum computer's job is to produce classical output — numbers, answers, bits you can email. The step that converts a quantum superposition into a classical bit is projective measurement, and it is the only move in the entire quantum postulate system that is not unitary. It is the one event that deletes information, the one event that is irreversible, the one event where "quantum" becomes "classical."

This chapter tells you exactly what happens in that step. What the probabilities are. What the post-measurement state is. Why you cannot undo it. And why, despite its destructiveness, measurement is the machine that actually extracts answers from quantum algorithms.

Setup — what a projective measurement needs

The ingredients of a projective measurement are embarrassingly simple.

Pick an orthonormal basis of your state space — call it \{|m\rangle\}, where m runs over some index set (for a qubit, m \in \{0, 1\}). From each basis vector, build a projector:

P_m = |m\rangle\langle m|.

The object |m\rangle\langle m| is the outer product — the column |m\rangle times the row \langle m| — which gives back a matrix. For the computational basis on a qubit,

P_0 = |0\rangle\langle 0| = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \qquad P_1 = |1\rangle\langle 1| = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}.

Why these are called "projectors": P_m has the defining property P_m^2 = P_m. Applied twice, it does the same thing as applied once — like the shadow operation in 3D space, where "project onto the floor" applied twice is the same as "project onto the floor" applied once. The shadow is already on the floor; projecting it again doesn't move it.

The projectors of a projective measurement have two structural properties you should write on your palm:

  1. Orthogonality: P_m P_{m'} = 0 when m \ne m'. Different outcomes correspond to different, non-overlapping subspaces.
  2. Completeness: \sum_m P_m = I. The projectors together account for the entire state space — every state is "somewhere" in the decomposition.
Projector P_m in picturesA 2D sketch showing a state vector |psi⟩ as an arrow, with its projection onto the |m⟩ axis drawn as a vertical drop. The projection has length ⟨m|psi⟩, and this length squared is the probability of outcome m. The remaining perpendicular component has length ⟨m-perp|psi⟩, whose squared length is the probability of the other outcome.|m⊥⟩ axis|m⟩ axis|ψ⟩P_m |ψ⟩ = ⟨m|ψ⟩ |m⟩perpendicularcomponentlength = |⟨m|ψ⟩|p(m) = |⟨m|ψ⟩|²
The projector $P_m = |m\rangle\langle m|$ drops $|\psi\rangle$ onto the $|m\rangle$ axis. The shadow's length (squared) is the probability of getting outcome $m$. Whatever falls onto the perpendicular direction is the part of $|\psi\rangle$ that "would have given the other outcome."

The Born rule — probabilities from amplitudes

Given |\psi\rangle and the projectors \{P_m\}, the probability of measurement outcome m is

\boxed{p(m) = \langle \psi | P_m | \psi\rangle.}

This is the Born rule, first stated by Max Born in 1926, and it is the bridge between the abstract quantum formalism and the classical statistics you actually observe in the lab.

For the computational basis on a qubit, write |\psi\rangle = \alpha|0\rangle + \beta|1\rangle and compute:

p(0) = \langle \psi | P_0 | \psi \rangle = \langle \psi | 0 \rangle \langle 0 | \psi \rangle = |\langle 0 | \psi \rangle|^2 = |\alpha|^2,
p(1) = \langle \psi | P_1 | \psi \rangle = |\langle 1 | \psi \rangle|^2 = |\beta|^2.

Why \langle \psi | 0 \rangle \langle 0 | \psi \rangle = |\langle 0 | \psi \rangle|^2: the two factors are \langle \psi | 0 \rangle and \langle 0 | \psi \rangle, which are complex conjugates of each other (because \langle a|b\rangle = \overline{\langle b|a\rangle}). A complex number times its conjugate is the squared modulus.

So the probability of getting outcome m is the squared modulus of the m-th amplitude. This is "the squared amplitudes are the probabilities" in one formal line.

And crucially: the probabilities always sum to 1.

\sum_m p(m) = \sum_m \langle \psi | P_m | \psi \rangle = \langle \psi | \left(\sum_m P_m\right) | \psi \rangle = \langle \psi | I | \psi \rangle = \langle \psi | \psi \rangle = 1,

where the last step uses the fact that |\psi\rangle is a unit vector. This is not a lucky coincidence — it is why the completeness condition \sum_m P_m = I was imposed in the first place. Completeness is what makes the Born rule produce a valid probability distribution.

Bar chart of probabilities before and after measurementTwo bar charts side by side. Left chart: the state alpha|0⟩ + beta|1⟩ shows two bars with heights |alpha|² and |beta|² representing probabilities of outcomes 0 and 1. Right chart: after measurement, only one bar survives at height 1 (either the |0⟩ bar or the |1⟩ bar, depending on outcome).before measurement|0⟩|1⟩|α|²|β|²10after measurement (outcome = 0)|0⟩|1⟩1010
Before measurement, the probability of each outcome is governed by the squared amplitudes $|\alpha|^2$ and $|\beta|^2$. After measurement, the distribution has collapsed to a single certain outcome — the post-measurement state gives outcome $0$ or outcome $1$ with probability $1$ if re-measured in the same basis.

Born's 1926 paper was the first time anyone wrote down p = |\psi|^2 as an interpretation of the quantum amplitude, rather than just a mathematical object. Before him, Schrödinger had introduced the wave function but was not sure what it meant physically; Born's insight that |\psi|^2 is a probability density is the interpretation every quantum physicist has used since. It also earned him the 1954 Nobel Prize in Physics — almost three decades after the paper, because the interpretation took that long for the community to accept over competing ideas.

Collapse — the post-measurement state

Given that outcome m happened, what is the state now?

\boxed{|\psi\rangle \longmapsto \frac{P_m |\psi\rangle}{\sqrt{p(m)}}.}

Three things are happening in that formula. Apply the projector P_m to |\psi\rangle to pick out the component in the |m\rangle direction. Divide by \sqrt{p(m)} to renormalise, because the projected vector has shrunk and you need to restore the unit-vector condition. The result is what the qubit's state is the instant after the measurement.

For the computational basis, this simplifies dramatically. If outcome 0 occurred:

\frac{P_0 |\psi\rangle}{\sqrt{p(0)}} = \frac{|0\rangle\langle 0|(\alpha|0\rangle + \beta|1\rangle)}{\sqrt{|\alpha|^2}} = \frac{\alpha |0\rangle}{|\alpha|} = e^{i\theta}|0\rangle,

where e^{i\theta} is a global phase (since \alpha / |\alpha| has modulus 1 but in general a phase). Physically the state is |0\rangle — the global phase is irrelevant.

Why the global phase drops out: \alpha / |\alpha| is a unit complex number, a pure phase. Multiplying a state by a pure phase does not change any measurement probability or any expectation value, so the physical state is the same as |0\rangle. You can safely write "the post-measurement state is |0\rangle" without loss.

Likewise, if outcome 1 occurred, the post-measurement state is |1\rangle (up to global phase). So for computational-basis measurement, the post-measurement state is just the basis vector corresponding to the outcome you observed. The superposition is gone. The complex amplitudes \alpha and \beta are gone. All that is left is one definite eigenstate, and one bit of classical information that says which one.

What the collapse means operationally

"Collapse" sounds dramatic. The operational content is quieter: the formula for the state has been updated. Every subsequent gate, every subsequent measurement, every subsequent calculation now treats the qubit as if it is |m\rangle — because that is what it is, as far as any future experiment can tell.

In particular:

Whether "collapse" is a real physical event (Copenhagen interpretation) or merely an update of your knowledge (the many-worlds and relational-QM interpretations handle this differently) is a topic philosophers still argue about. The mathematics is the same either way, and for the purposes of quantum computing — where all that matters is predicting measurement outcomes — the operational rule |\psi\rangle \mapsto P_m |\psi\rangle / \sqrt{p(m)} is all you need.

Why measurement is not unitary

Look at a projector: P_m^2 = P_m. Applying it twice is the same as applying it once. But for a unitary, U^\dagger U = I — applying a unitary and then its dagger gives back the identity. Does P_m satisfy P_m^\dagger P_m = I? For a projector, P_m^\dagger = P_m (projectors are Hermitian), so P_m^\dagger P_m = P_m^2 = P_m — and P_m \ne I unless P_m is the identity (which would make the measurement trivial). So projectors are not unitary.

Two deeper consequences of this algebraic fact:

Non-injectivity. A unitary is a bijection — every input has a unique output and every output can be traced back to a unique input. A projector squashes an entire subspace (every state of the form |\psi\rangle = \alpha|0\rangle + \beta|1\rangle with different (\alpha, \beta)) down to a single state (|0\rangle, if outcome 0 happened) scaled by different amplitudes. Many different input states produce the same post-measurement state. From the post-measurement state alone, you cannot reconstruct what came before. Information has been destroyed.

Irreversibility. There is no "reverse measurement" operation V with V \cdot \text{measure} = \text{identity}. In the formalism of unitary evolution, you could always apply U^\dagger to undo U. There is no P_m^\dagger whose action is "un-project" — because the projected-away part is genuinely gone and cannot be retrieved by any further operation on the qubit alone.

Connection to no-cloning. The no-cloning theorem (Wootters & Zurek, 1982) says you cannot duplicate an unknown quantum state. If you could (a) clone |\psi\rangle to produce two copies, (b) measure one of them, and (c) somehow un-measure it to recover |\psi\rangle, you would have both the classical outcome and the original state. This is impossible in quantum mechanics. Two pieces of the impossibility argument are at work: cloning is forbidden, and measurement is irreversible. Remove either one and the prohibition collapses. Both are structural facts about the formalism.

Measurement in other bases

The Born rule is not specific to the computational basis. Any orthonormal basis of the state space defines a valid projective measurement. Some bases you will meet constantly:

For an X-basis measurement, the projectors are P_+ = |+\rangle\langle +| and P_- = |-\rangle\langle -|. The probability of outcome + on a state |\psi\rangle is

p(+) = |\langle + | \psi\rangle|^2,

and the post-measurement state (if + occurred) is |+\rangle, up to global phase.

Example. Apply X-basis measurement to |0\rangle. Since |0\rangle = (|+\rangle + |-\rangle)/\sqrt{2}, the amplitudes in the X basis are both 1/\sqrt{2}. The probabilities are |1/\sqrt{2}|^2 = 1/2 for each outcome. Measuring |0\rangle in the X basis is a fair coin flip — even though |0\rangle gave outcome 0 deterministically in the computational basis. The same state gives very different statistics in different bases. This is one of the defining features of quantum mechanics.

Implementing arbitrary-basis measurement via gates

Real quantum hardware typically only has the physical infrastructure for computational-basis measurement — it can detect whether a superconducting qubit is in its ground state or its excited state, and that is the Z-basis, period. How do you measure in a different basis on such hardware?

Rotate the state first, then measure. To measure in the X basis, apply a Hadamard to rotate the X basis onto the Z basis, then do a computational-basis measurement. Concretely: "measuring |\psi\rangle in the X basis" is implemented as "measuring H|\psi\rangle in the computational basis." The Hadamard pre-rotation is the software side; the hardware only does one kind of measurement. This pattern — apply a unitary, then measure in the computational basis — covers every projective measurement you will ever need.

Same state, different basesTwo Bloch spheres side by side, each showing the same state |psi⟩ drawn at the |+⟩ equator point. The left sphere shows a measurement in the Z (computational) basis: a projection to either the north or south pole with equal probability. The right sphere shows a measurement in the X basis: a deterministic projection to |+⟩, no randomness.|0⟩|1⟩|ψ⟩=|+⟩measure in Z basisp(0) = p(1) = ½ (random)|0⟩|1⟩|ψ⟩=|+⟩measure in X basisp(+) = 1 (certain)
The same state $|+\rangle$ gives a random coin flip if measured in the Z basis, but a deterministic outcome if measured in the X basis. Which basis you choose is part of the experiment. The state has no "real" underlying value that is revealed by measurement — the outcomes depend on both the state and the basis.

Repeated measurements and statistics

One measurement gives you one bit. To learn anything more about |\psi\rangle than a single bit, you need many runs — prepare the state, measure, record the outcome, repeat. This is called "shot-based" measurement on quantum hardware. Every number a quantum computer reports, from the output of Shor's algorithm to a simple sampling experiment, ultimately comes from histograms over many shots.

Say |\psi\rangle = \alpha|0\rangle + \beta|1\rangle with |\alpha|^2 = 0.7 and |\beta|^2 = 0.3. Run the circuit 1000 times. On average you will see 0 about 700 times and 1 about 300 times. The statistical uncertainty is \sqrt{1000 \cdot 0.7 \cdot 0.3} \approx 14 in the count of zeros — so you would expect 700 \pm 14 zeros, giving you the probability to about \pm 0.014. To pin down p(0) to three decimal places, you need around 100{,}000 shots.

This is a real engineering constraint. Quantum hardware today runs circuits at typical repetition rates of maybe 10^4 shots per second. A million shots is a couple of minutes. Estimating probabilities finer than \sim 10^{-3} starts to take meaningful wall time — and this is one reason quantum-classical hybrid algorithms (like variational quantum eigensolvers) care so much about sample efficiency.

The Indian National Quantum Mission (2023, ₹6000 crore) funds hardware at IITs and TIFR, and the shot-statistics bottleneck applies equally to those machines. Sampling every quantum result takes time; the cleverness of an algorithm is often measured in how few shots it needs per useful bit.

The no-communication theorem — an important aside

You may have heard that entangled qubits can "communicate faster than light" because measuring one instantly affects the other. The phrase is wrong, and it is worth stating this clearly here, because projective measurement is the tool used in Einstein's original "spooky action at a distance" argument.

When two qubits are entangled (say, in a Bell state \tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle)) and you measure one of them, the measurement outcome on the other is correlated with your outcome — if yours is 0, theirs will also be 0 when they measure. But here is the crucial point: you cannot use this to send information. Your outcome is random (50-50), and the other party's outcome is also random (50-50), with the same statistics regardless of whether or when you measure. The correlation shows up only when you compare outcomes afterwards, which requires a classical (light-speed or slower) communication channel.

This is the no-communication theorem (Ghirardi, Rimini, Weber and independently others, around 1980). It is a consequence of the linearity of quantum mechanics plus the Born rule, and it is what makes quantum mechanics consistent with special relativity. Entanglement is surprising; it is not a telegraph.

The proper treatment belongs in later chapters on entanglement and teleportation. For this chapter, just record that measurement outcomes on one qubit do not signal information to the holder of another qubit, however entangled they are.

Worked examples

Example 1: Computational-basis measurement on |+⟩

Measure |+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle) in the computational basis. Give the probabilities and the post-measurement state for each outcome.

Step 1. Write the state with explicit amplitudes: |+\rangle = \alpha|0\rangle + \beta|1\rangle with \alpha = \beta = 1/\sqrt{2}. Why: the definition of |+\rangle gives the amplitudes directly — both are 1/\sqrt{2}.

Step 2. Apply the Born rule.

p(0) = |\alpha|^2 = (1/\sqrt{2})^2 = 1/2, \qquad p(1) = |\beta|^2 = 1/2.

Why: in the computational basis, P_0 = |0\rangle\langle 0| gives p(0) = |\langle 0|\psi\rangle|^2 = |\alpha|^2, and similarly for P_1.

Step 3. Compute the post-measurement states.

If outcome 0 was obtained:

\frac{P_0 |+\rangle}{\sqrt{p(0)}} = \frac{(1/\sqrt{2})|0\rangle}{\sqrt{1/2}} = \frac{1/\sqrt{2}}{1/\sqrt{2}}|0\rangle = |0\rangle.

Why: P_0 |+\rangle picks out the |0\rangle component — amplitude 1/\sqrt{2} times |0\rangle. Dividing by \sqrt{p(0)} = 1/\sqrt{2} normalises the result to a unit vector, which gives exactly |0\rangle.

If outcome 1 was obtained:

\frac{P_1 |+\rangle}{\sqrt{p(1)}} = \frac{(1/\sqrt{2})|1\rangle}{\sqrt{1/2}} = |1\rangle.

Result. Measuring |+\rangle in the computational basis gives 0 or 1 with equal probability 1/2. The post-measurement state is |0\rangle or |1\rangle respectively — a complete basis vector, no superposition, no phase information. The original |+\rangle is gone.

Remark. Run this circuit 1000 times: expect about 500 zeros and 500 ones, with statistical fluctuation \pm\sqrt{1000 \cdot 0.5 \cdot 0.5} \approx \pm 16. This is a quantum coin flip — fair by construction, statistically indistinguishable from a very well-made classical coin.

Example 2: X-basis measurement on the same |+⟩

Measure the same state |+\rangle in the X basis \{|+\rangle, |-\rangle\}. Give the probabilities and the post-measurement state.

Step 1. Identify the projectors.

P_+ = |+\rangle\langle +|, \qquad P_- = |-\rangle\langle -|.

Why: the X-basis measurement uses projectors built from the X-basis vectors. This is the same construction as the computational-basis measurement, just with different basis vectors.

Step 2. Apply the Born rule. The probability of outcome + is p(+) = |\langle + | + \rangle|^2.

Compute \langle + | + \rangle. This is the inner product of |+\rangle with itself, which is 1 (because |+\rangle is a unit vector).

p(+) = |1|^2 = 1.

And the probability of outcome - is p(-) = |\langle - | + \rangle|^2 = |0|^2 = 0 (because |+\rangle and |-\rangle are orthogonal). Why \langle - | + \rangle = 0: the X basis is orthonormal — two different basis vectors have inner product zero. You built |+\rangle and |-\rangle exactly to be orthogonal so that they could serve as a basis.

Step 3. Compute the post-measurement state. Since p(+) = 1, the outcome is always + (deterministically). The post-measurement state:

\frac{P_+ |+\rangle}{\sqrt{p(+)}} = \frac{|+\rangle\langle + | + \rangle}{1} = |+\rangle.

Result. Measuring |+\rangle in the X basis gives outcome + with probability 1. The state is unchanged.

Contrast with Example 1. The exact same state gave a random coin flip when measured in the Z basis but a deterministic outcome when measured in the X basis. The state |+\rangle is an eigenstate of the X-basis measurement — it has a definite value in that basis — but it is a superposition in the Z basis. "A qubit's value" is not a property of the qubit alone; it is a property of the qubit plus the basis you choose to measure in.

This is the lesson that trips up every first-time student of quantum mechanics, and it is the reason "measure the qubit" is ambiguous. Always say in what basis.

Common confusions

Going deeper

Everything you need to do quantum computing is in the first half of this chapter: projectors, the Born rule, collapse to the projected component, irreversibility. The rest of this section goes deeper: the general measurement formalism beyond projective (POVMs), a sketch of why the Born rule is the only consistent probability rule, the quantum Zeno effect, and the interpretive zoo.

The general measurement formalism — POVMs

Projective measurement is a special case of a more general structure. A generalised measurement on a quantum system is a set of operators \{M_m\} (one per outcome) satisfying \sum_m M_m^\dagger M_m = I. The probability of outcome m is p(m) = \langle \psi | M_m^\dagger M_m | \psi\rangle, and the post-measurement state is M_m |\psi\rangle / \sqrt{p(m)}.

When each M_m is a projector (and the projectors are orthogonal), this reduces to projective measurement. When the M_m are not projectors, you get a more general family — POVMs (positive operator-valued measures). POVMs let you do measurements that are not achievable by any projective measurement, including:

Projective measurement is the version you will use in nearly every quantum algorithm. POVMs are the framework you need when characterising quantum detectors, analysing protocols like BB84, and optimising measurement strategies.

The Born rule from Gleason's theorem

Why p(m) = \langle \psi | P_m | \psi\rangle, and not some other function of the state and the projector?

Gleason's theorem (1957) says: for a Hilbert space of dimension \ge 3, the only probability assignment on projectors that is consistent with the algebra (additivity over orthogonal projectors, probability 1 on the identity) is the trace rule p(P) = \text{Tr}(\rho P) — which, for pure states, is exactly the Born rule \langle \psi | P | \psi\rangle.

The theorem is deep. It says the Born rule is not an extra postulate bolted on to make the theory predictive — it is the unique probability assignment consistent with the rest of the structure. Assume states live in Hilbert space, assume measurements are projectors, assume probabilities are additive over orthogonal outcomes; Gleason's theorem forces |\langle m | \psi \rangle|^2 out of those three inputs and nothing else.

(The theorem fails in 2 dimensions, which is an amusing technicality — a single qubit has 2-dimensional Hilbert space. But extensions and alternative derivations, including Deutsch's many-worlds argument and Zurek's envariance derivation, recover the Born rule in 2D as well.)

The quantum Zeno effect

If you measure a quantum system frequently enough, you can freeze its evolution. This is the quantum Zeno effect (Misra and Sudarshan, 1977).

Concretely: suppose a qubit is about to evolve continuously from |0\rangle to |1\rangle under some unitary, on a timescale T. If you measure it in the computational basis at time \delta t \ll T, the probability of finding it still in |0\rangle is roughly 1 - O(\delta t^2 / T^2) — a tiny deviation. The measurement collapses it back to |0\rangle. Do this N times in rapid succession with \delta t = T/N: the probability of still being in |0\rangle at time T is (1 - O(1/N^2))^N \to 1 as N \to \infty.

So: if you watch the pot often enough, it never boils. The effect has been experimentally demonstrated (Itano et al., 1990; Fischer et al., 2001). It is a direct consequence of the collapse rule plus the quadratic-in-time initial deviation of unitary evolution (the so-called "quadratic short-time behaviour" that comes from \cos(Ht/\hbar) \approx 1 - (Ht/\hbar)^2/2).

The Zeno effect is not just a curiosity. In quantum error correction, frequent syndrome measurements freeze the state against slow-drifting errors — a deliberately engineered application of the same physics.

Interpretations of collapse

The mathematics of measurement is settled. The interpretation — what is actually happening when the state "collapses" — is not. A few of the major options, with no endorsement:

For the purposes of this track — quantum computing, where the job is to predict measurement outcomes — all of these give the same answers. The interpretive choice does not affect any calculation you will do. But being aware that there is a choice matters for honest writing about the field.

Where this leads next

References

  1. Max Born, Zur Quantenmechanik der Stoßvorgänge (1926) — the paper where |\psi|^2 first becomes a probability. Wikipedia: Born rule has a summary with the original citation [1].
  2. Nielsen and Chuang, Quantum Computation and Quantum Information (2010), §2.2.3 and §2.2.5 — the definitive reference on projective and general measurement. Cambridge University Press.
  3. John Preskill, Lecture Notes on Quantum Computation, Ch. 2 (measurement, POVMs, Gleason's theorem) — theory.caltech.edu/~preskill/ph229.
  4. Wikipedia, Measurement in quantum mechanics — projectors, the Born rule, and collapse in encyclopaedic form.
  5. Wikipedia, Wave function collapse — the collapse postulate and its interpretive status.
  6. Qiskit Textbook, Measurement — hands-on measurement in the Qiskit framework, with live histograms.