# Projective Measurement > Measurement is where the deterministic, unitary world of quantum mechanics hands you a *random* > classical number — and then violently rewrites the state to match. This lesson makes that precise. > The **Born rule** assigns the probabilities, **Hermitian observables** package the question you ask, > and **projection** describes the collapse. Get this exactly right and the rest of quantum information > — uncertainty, POVMs, entanglement, every algorithm's read-out step — follows without mysticism. ## Learning Objectives After this lesson you will be able to: 1. State the **Born rule** and compute outcome probabilities $p_k = \langle\psi|P_k|\psi\rangle$. 2. Decompose an **observable** $A=A^\dagger$ into its spectral projectors $A=\sum_k\lambda_k P_k$ and read off measurement outcomes and probabilities. 3. Apply the **projection postulate**: the post-measurement state $|\psi\rangle\to P_k|\psi\rangle/\sqrt{p_k}$. 4. Prove **repeatability** of projective measurement and explain measuring in **different bases** (computational $Z$ vs. $X$). 5. Show that **orthogonal states are perfectly distinguishable** by a single projective measurement. --- ## Intuition A measurement asks the state a *question* with a fixed list of allowed answers. The question is an observable; the allowed answers are its eigenvalues. Quantum mechanics adds two twists that classical physics does not have: 1. **The answer is random** unless the state was already an eigenstate of the question. The amount of "overlap" the state has with each eigenstate sets the probability — that's the Born rule. 2. **Asking changes the state.** Once you get answer $\lambda_k$, the state snaps onto the subspace of states that *would have given* $\lambda_k$ with certainty. This is **collapse**, and it is described by a **projector**. Geometrically: a normalized state $|\psi\rangle$ is a unit vector. Measuring in an orthonormal basis $\{|k\rangle\}$ drops a perpendicular onto each axis; the squared length of the shadow on axis $k$ is the probability of outcome $k$, and after the measurement the vector *is* that axis. Pythagoras ($\sum_k |\langle k|\psi\rangle|^2 = 1$) is exactly why the probabilities sum to one. --- ## Theory ### The Born rule > **Born rule (basis form).** If a system is in normalized state $|\psi\rangle$ and we measure in the > orthonormal basis $\{|k\rangle\}$, the probability of outcome $k$ is > $$ p_k = |\langle k|\psi\rangle|^2 = \langle\psi|k\rangle\langle k|\psi\rangle. $$ These are genuine probabilities: each $p_k = |\langle k|\psi\rangle|^2 \ge 0$, and using completeness $\sum_k |k\rangle\langle k| = I$ ([0.1.3](../../term-0-foundations/c01-linear-algebra/03-dirac-notation.md)), $$ \sum_k p_k = \sum_k \langle\psi|k\rangle\langle k|\psi\rangle = \langle\psi|\Big(\sum_k |k\rangle\langle k|\Big)|\psi\rangle = \langle\psi|I|\psi\rangle = \langle\psi|\psi\rangle = 1. $$ So **normalization is conservation of probability**. This is also why states are only defined up to a **global phase**: $|\psi\rangle$ and $e^{i\gamma}|\psi\rangle$ give $|\langle k|\psi\rangle|^2$ identically (the $|e^{i\gamma}|^2 = 1$ cancels), so no measurement can ever distinguish them. ### Observables and spectral decomposition A measurable quantity is an **observable**: a Hermitian operator $A = A^\dagger$. By the spectral theorem for Hermitian operators ([0.1.5](../../term-0-foundations/c01-linear-algebra/05-eigen-spectral-theorem.md), [0.1.6](../../term-0-foundations/c01-linear-algebra/06-special-operators.md)) it has **real** eigenvalues and an orthonormal eigenbasis. Grouping eigenvectors that share an eigenvalue into **eigenspaces**, write the **spectral decomposition** $$ \boxed{\;A = \sum_k \lambda_k\, P_k,\qquad P_k = \sum_{j:\,\lambda_j=\lambda_k} |v_j\rangle\langle v_j|\;} $$ where the $\lambda_k$ are the *distinct* eigenvalues and $P_k$ is the **orthogonal projector** onto the $\lambda_k$-eigenspace. These projectors form a **complete set of orthogonal projectors**: $$ P_k = P_k^\dagger,\qquad P_j P_k = \delta_{jk} P_k,\qquad \sum_k P_k = I. $$ The first two say each $P_k$ is a projector ($P_k^2=P_k$) and distinct eigenspaces are orthogonal; the last is completeness (the eigenvectors span $\mathcal H$). The eigenvalues $\lambda_k$ are the **possible measured values**; the projectors $P_k$ encode the question's answer-slots. ### Projective (von Neumann) measurement > **Projective measurement postulate.** A projective measurement is specified by a complete set of > orthogonal projectors $\{P_k\}$ ($P_jP_k=\delta_{jk}P_k$, $\sum_k P_k=I$), one per outcome. Measuring > state $|\psi\rangle$ yields outcome $k$ with probability > $$ p_k = \langle\psi|P_k|\psi\rangle = \|P_k|\psi\rangle\|^2, $$ > and the state immediately after, *given* outcome $k$, is > $$ |\psi'\rangle = \frac{P_k|\psi\rangle}{\sqrt{p_k}}\qquad (p_k>0). $$ The basis-form Born rule is the special case where every $P_k = |k\rangle\langle k|$ is rank one: then $p_k = \langle\psi|k\rangle\langle k|\psi\rangle = |\langle k|\psi\rangle|^2$, recovering the boxed rule above. The projector form is more general because it allows **degenerate** outcomes (rank $> 1$): a measurement need not resolve which vector *inside* an eigenspace you landed on. **Why these are probabilities, again.** Positivity: $p_k = \langle\psi|P_k|\psi\rangle = \langle\psi|P_k^\dagger P_k|\psi\rangle = \|P_k|\psi\rangle\|^2 \ge 0$, using $P_k=P_k^\dagger=P_k^2$. Normalization: $\sum_k p_k = \langle\psi|(\sum_k P_k)|\psi\rangle = \langle\psi|\psi\rangle = 1$. **Renormalization of the collapsed state.** The raw post-measurement vector $P_k|\psi\rangle$ has norm $\|P_k|\psi\rangle\| = \sqrt{p_k}$, so dividing by $\sqrt{p_k}$ restores a unit vector — collapse keeps us on the unit sphere. Note this is a **non-linear, non-unitary, irreversible** update: information about the components in the other eigenspaces is gone. > **Expectation preview.** Once you have $p_k$, the average measured value is > $\langle A\rangle = \sum_k \lambda_k p_k = \langle\psi|A|\psi\rangle$. We make this — and its spread — > the whole subject of [the next lesson](02-expectation-uncertainty.md). ### Repeatability Projective measurement is **repeatable**: measure the same observable twice in immediate succession and the second result equals the first, with certainty. **Proof.** Suppose the first measurement returned $k$, collapsing to $|\psi'\rangle = P_k|\psi\rangle/\sqrt{p_k}$. The probability that an immediate second measurement of the same $\{P_j\}$ returns outcome $j$ is $$ p_{j\mid k} = \langle\psi'|P_j|\psi'\rangle = \frac{\langle\psi|P_k\,P_j\,P_k|\psi\rangle}{p_k} = \frac{\langle\psi|P_k\,(\delta_{jk}P_k)|\psi\rangle}{p_k} = \delta_{jk}\,\frac{\langle\psi|P_k|\psi\rangle}{p_k} = \delta_{jk}, $$ using $P_jP_k = \delta_{jk}P_k$ and $P_k^2=P_k$. So $p_{j\mid k}=1$ if $j=k$ and $0$ otherwise. ∎ Repeatability is what makes the eigenvalue a *property* of the post-measurement state: after measuring, the system is in an eigenstate of $A$ with definite value $\lambda_k$. (This idealization assumes no evolution between the two measurements; dynamics — [Term 1.6](../c06-dynamics/01-time-evolution.md) — will generally move the state off the eigenstate.) ### Measuring in different bases The *same* state gives *different* statistics depending on which observable you measure — i.e. which orthonormal basis you project onto. - **Computational ($Z$) basis.** $Z = (+1)|0\rangle\langle0| + (-1)|1\rangle\langle1|$, so the projectors are $P_0 = |0\rangle\langle0|$, $P_1 = |1\rangle\langle1|$. Outcome "$0$" ($\lambda=+1$) with probability $|\langle 0|\psi\rangle|^2$, outcome "$1$" ($\lambda=-1$) with $|\langle1|\psi\rangle|^2$. - **Hadamard ($X$) basis.** $X = (+1)|+\rangle\langle+| + (-1)|-\rangle\langle-|$ with $|\pm\rangle = \tfrac1{\sqrt2}(|0\rangle\pm|1\rangle)$. The projectors are $P_+ = |+\rangle\langle+|$, $P_- = |-\rangle\langle-|$, and outcome probabilities are $|\langle\pm|\psi\rangle|^2$. Hardware almost always measures in the **computational basis only**. To measure in another basis you **rotate that basis onto the computational basis first**, then measure $Z$. Since $H|+\rangle=|0\rangle$ and $H|-\rangle=|1\rangle$ (the Hadamard swaps the $X$- and $Z$-bases, $HXH=Z$; [Appendix E](../../appendix/E-math-identities.md)), measuring $X$ on $|\psi\rangle$ is identical to applying $H$ and measuring $Z$: $$ |\langle\pm|\psi\rangle|^2 = |\langle 0/1|H|\psi\rangle|^2 . $$ This "rotate, then read $Z$" pattern is how *all* basis changes (and expectation-value estimation) are implemented on real devices — we use it from [Term 1.2.3](../c02-qubits-bloch/03-bases-and-expectation.md) onward and in the Braket code below. ### Perfect distinguishability of orthogonal states > **Claim.** Two orthogonal states $|\psi_0\rangle\perp|\psi_1\rangle$ can be distinguished with > **certainty** by a single projective measurement. Non-orthogonal states cannot > (proved in [Lesson 3](03-povm-generalized.md)). **Proof.** Extend $\{|\psi_0\rangle,|\psi_1\rangle\}$ to an orthonormal basis and take the projective measurement $\{P_0,P_1,P_{\text{rest}}\}$ with $P_i = |\psi_i\rangle\langle\psi_i|$ and $P_{\text{rest}} = I - P_0 - P_1$. If the system is prepared in $|\psi_0\rangle$, then $p_0 = \langle\psi_0|P_0|\psi_0\rangle = |\langle\psi_0|\psi_0\rangle|^2 = 1$, while $p_1 = |\langle\psi_1|\psi_0\rangle|^2 = 0$. Symmetrically for $|\psi_1\rangle$. So the outcome reveals the prepared state with zero error. ∎ This is the bedrock fact behind quantum communication and read-out: an orthonormal basis is a set of *perfectly reliable codewords*. The moment states are non-orthogonal, perfect discrimination is impossible — the seed of the uncertainty principle and of quantum cryptography. --- ## Worked Examples ### Example 1 — Measuring $|+\rangle$ in the $Z$ basis Let $|\psi\rangle = |+\rangle = \tfrac1{\sqrt2}(|0\rangle + |1\rangle)$ and measure the observable $Z$, i.e. project with $P_0 = |0\rangle\langle0|$, $P_1=|1\rangle\langle1|$. **Probabilities.** $$ p_0 = \langle+|P_0|+\rangle = |\langle 0|+\rangle|^2 = \Big|\tfrac1{\sqrt2}\Big|^2 = \tfrac12, \qquad p_1 = |\langle 1|+\rangle|^2 = \tfrac12 . $$ A coin flip, as expected for an equatorial Bloch state. The associated eigenvalues are $\lambda_0=+1$ and $\lambda_1=-1$, so $\langle Z\rangle = (+1)\tfrac12 + (-1)\tfrac12 = 0$. **Collapse.** If we get outcome "$0$", $$ |\psi'\rangle = \frac{P_0|+\rangle}{\sqrt{p_0}} = \frac{\tfrac1{\sqrt2}|0\rangle}{\sqrt{1/2}} = |0\rangle. $$ The superposition is destroyed; a re-measurement of $Z$ now yields "$0$" with certainty (repeatability). ### Example 2 — Measuring $|+\rangle$ in the $X$ basis Same state $|\psi\rangle = |+\rangle$, now measure the observable $X$ with projectors $P_+ = |+\rangle\langle+|$, $P_- = |-\rangle\langle-|$. **Probabilities.** Because $|+\rangle$ is itself an eigenstate of $X$ (with eigenvalue $+1$), $$ p_+ = \langle+|P_+|+\rangle = |\langle+|+\rangle|^2 = 1, \qquad p_- = |\langle-|+\rangle|^2 = 0 . $$ So in the $X$ basis the outcome is **deterministic**: always "$+$". The state did not need to collapse — $P_+|+\rangle/\sqrt{1} = |+\rangle$ leaves it unchanged. **The lesson.** A single state $|+\rangle$ is *maximally uncertain* in $Z$ ($50/50$) yet *perfectly certain* in $X$. "Definiteness" is a relationship between a state and a chosen observable, not an intrinsic property of the state. This complementarity is quantified by the uncertainty relation next lesson, and the "rotate then read $Z$" recipe ($X$-measurement $=$ apply $H$, measure $Z$) is exactly the Braket experiment below. --- ## Hands-on (Python) A self-contained NumPy simulator for projective measurement, then a brief Braket sanity check. We do all quantitative work in NumPy; Braket here is only an illustration of sampling on the (free) local simulator. ```python import numpy as np # ---- States and observables (big-endian; |0> is index 0) ----------------- ket0 = np.array([1, 0], dtype=complex) ket1 = np.array([0, 1], dtype=complex) plus = (ket0 + ket1) / np.sqrt(2) # |+> minus = (ket0 - ket1) / np.sqrt(2) # |-> Z = np.array([[1, 0], [0, -1]], dtype=complex) X = np.array([[0, 1], [1, 0]], dtype=complex) def projectors_from_observable(A): """Spectral decomposition A = sum_k lambda_k P_k. Returns (distinct eigenvalues, projectors) with eigenvalues grouped so that degenerate eigenspaces yield a single (possibly rank>1) projector. """ A = (A + A.conj().T) / 2 # symmetrize for numerical safety eigvals, eigvecs = np.linalg.eigh(A) # eigh: Hermitian, orthonormal columns distinct = [] projectors = [] for lam, vec in zip(eigvals, eigvecs.T): P = np.outer(vec, vec.conj()) # |v> for each projector.""" return np.array([np.real(psi.conj() @ (P @ psi)) for P in projectors]) def collapse(psi, P, p): """Post-measurement state P|psi>/sqrt(p).""" return (P @ psi) / np.sqrt(p) def measure_once(psi, projectors, eigvals, rng): """Sample one projective-measurement outcome; return (eigenvalue, new_state).""" p = born_probabilities(psi, projectors) k = rng.choice(len(projectors), p=p) return eigvals[k], collapse(psi, projectors[k], p[k]) # ---- Measuring |+> in the Z basis vs the X basis ------------------------- valsZ, projZ = projectors_from_observable(Z) valsX, projX = projectors_from_observable(X) print("Z-basis probs on |+>:", np.round(born_probabilities(plus, projZ), 6)) # [0.5 0.5] print("X-basis probs on |+>:", np.round(born_probabilities(plus, projX), 6)) # deterministic # Sanity checks against the postulate for P in projZ: assert np.allclose(P @ P, P) and np.allclose(P, P.conj().T) # projector assert np.allclose(sum(projZ), np.eye(2)) # completeness assert np.isclose(born_probabilities(plus, projZ).sum(), 1.0) # normalization ``` ```python # ---- Empirical sampling: 50/50 for Z, and repeatability ------------------ rng = np.random.default_rng(0) shots = 10_000 counts = {+1: 0, -1: 0} for _ in range(shots): val, _ = measure_once(plus, projZ, valsZ, rng) counts[int(round(val.real))] += 1 print("Z-measurement of |+> :", counts) # ~5000 / ~5000 # Repeatability: after collapse, remeasure -> same outcome, every time val1, psi_after = measure_once(plus, projZ, valsZ, rng) val2, _ = measure_once(psi_after, projZ, valsZ, rng) assert val1 == val2 # always equal print("Collapsed eigenvalue", int(val1.real), "is repeatable:", val1 == val2) ``` A brief look on **Braket's local simulator** (free — no AWS charges, no account needed). Measuring $|+\rangle$ in $Z$ is `H` on $|0\rangle$ then a default (computational-basis) measurement; we expect a roughly even split. To measure $X$ instead, you would apply a second `H` before measuring (rotate the $X$-basis onto $Z$). ```python # Braket LocalSimulator: prepare |+> = H|0>, sample in the computational (Z) basis. from braket.circuits import Circuit from braket.devices import LocalSimulator device = LocalSimulator() # free local state-vector simulator circ = Circuit().h(0) # |0> -> |+> counts = device.run(circ, shots=1000).result().measurement_counts print(counts) # e.g. Counter({'0': ~500, '1': ~500}) # Big-endian: the single bit is qubit 0. Outcome '0' <-> Z eigenvalue +1. ``` > The NumPy `measure_once` is the *reference semantics*; Braket samples the same Born distribution on a > high-performance simulator. We keep expectation/probability computations in NumPy (exact), and use > Braket only for sampling — the deep Braket introduction is [Term 2.1](../../term-2-qc-core/c01-gates-circuits/04-braket-first-circuits.md). --- ## Exercises **E1 (easy).** A qubit is in $|\psi\rangle = \tfrac35|0\rangle + \tfrac45|1\rangle$. Compute the probabilities of outcomes "$0$" and "$1$" in a computational-basis measurement, and the post-measurement state for each.

Solution

$p_0 = |\tfrac35|^2 = \tfrac9{25}$, $p_1 = |\tfrac45|^2 = \tfrac{16}{25}$; they sum to $1$. ✓ Collapse: outcome "$0$" gives $\dfrac{P_0|\psi\rangle}{\sqrt{p_0}} = \dfrac{\frac35|0\rangle}{3/5} = |0\rangle$; outcome "$1$" gives $\dfrac{\frac45|1\rangle}{4/5} = |1\rangle$. (Computational-basis projectors are rank one, so collapse always lands exactly on $|0\rangle$ or $|1\rangle$.)

**E2 (easy).** Measure the state $|0\rangle$ in the $X$ basis. What are the probabilities, and what is the collapsed state for each outcome?

Solution

$\langle+|0\rangle = \langle-|0\rangle = \tfrac1{\sqrt2}$, so $p_+ = p_- = \tfrac12$. Outcome "$+$" collapses to $|+\rangle$, outcome "$-$" to $|-\rangle$. ($|0\rangle$ is unbiased in the $X$ basis, mirroring Example 2 with the roles of $X$ and $Z$ swapped.)

**E3 (medium).** Let $A = \begin{psmallmatrix}2&0\\0&5\end{psmallmatrix}$ and $|\psi\rangle = \tfrac1{\sqrt5}(|0\rangle + 2|1\rangle)$. Find the spectral projectors of $A$, the probability of each measured value, and the expectation $\langle A\rangle$ two ways (via $\sum_k\lambda_k p_k$ and via $\langle\psi|A|\psi\rangle$).

Solution

$A$ is already diagonal: eigenvalues $\lambda=2$ (eigenvector $|0\rangle$) and $\lambda=5$ ($|1\rangle$), so $P_2 = |0\rangle\langle0|$, $P_5 = |1\rangle\langle1|$. Probabilities: $p_2 = |\tfrac1{\sqrt5}|^2 = \tfrac15$, $p_5 = |\tfrac2{\sqrt5}|^2 = \tfrac45$. Then $\langle A\rangle = 2\cdot\tfrac15 + 5\cdot\tfrac45 = \tfrac{2+20}{5} = \tfrac{22}{5}$. Directly, $\langle\psi|A|\psi\rangle = \tfrac15(1\cdot2 + 4\cdot5) = \tfrac{22}{5}$. ✓ Agreement is the identity $\langle A\rangle = \sum_k\lambda_k p_k$.

**E4 (medium).** Consider the degenerate observable on two qubits $A = Z\otimes I$ acting on $|\psi\rangle = \tfrac1{\sqrt2}(|00\rangle + |11\rangle)$. Find the spectral projectors of $A$ (note the eigenvalues are degenerate, rank-2), the outcome probabilities, and the collapsed states.

Solution

$Z\otimes I$ has eigenvalue $+1$ on the span of $\{|00\rangle,|01\rangle\}$ and $-1$ on $\{|10\rangle,|11\rangle\}$, so $P_{+1} = |0\rangle\langle0|\otimes I$ and $P_{-1} = |1\rangle\langle1|\otimes I$ (each rank 2). On $|\psi\rangle$: $P_{+1}|\psi\rangle = \tfrac1{\sqrt2}|00\rangle$, giving $p_{+1} = \tfrac12$ and collapse $|00\rangle$; likewise $p_{-1}=\tfrac12$ with collapse $|11\rangle$. The degenerate measurement learns only qubit 0's value, but here that already pins the full state because of the entanglement — a preview of [Term 1.4](../c04-entanglement/01-multi-qubit-states.md).

**E5 (hard).** Prove that the **average post-measurement state** (averaging over outcomes, weighted by their probabilities, before you learn the outcome) is the "decohered" operator $\sum_k P_k\,\rho\,P_k$ where $\rho = |\psi\rangle\langle\psi|$. Then show its diagonal in the measurement basis matches the Born probabilities.

Solution

With probability $p_k$ the state becomes $\rho_k = \dfrac{P_k|\psi\rangle\langle\psi|P_k}{p_k}$ (a pure state). Averaging over outcomes: $\bar\rho = \sum_k p_k\,\rho_k = \sum_k P_k|\psi\rangle\langle\psi|P_k = \sum_k P_k\,\rho\,P_k$ (the $p_k$ cancels). For rank-1 projectors $P_k = |k\rangle\langle k|$, $\bar\rho = \sum_k |\langle k|\psi\rangle|^2\,|k\rangle\langle k| = \sum_k p_k|k\rangle\langle k|$, whose $(k,k)$ entry is $p_k$ and whose off-diagonal entries vanish: measurement (unconditioned) *destroys coherences*. This map $\rho\mapsto\sum_k P_k\rho P_k$ is the prototype of a quantum channel — [Term 4.1](../../term-4-noise-hardware/c01-open-systems/02-kraus-operators-channels.md). ∎

**E6 (hard).** Prove the perfect-distinguishability claim **in reverse**: if a projective measurement $\{P_0, P_1\}$ distinguishes states $|\psi_0\rangle, |\psi_1\rangle$ with zero error (outcome $0\Rightarrow$ prepared $\psi_0$, outcome $1\Rightarrow\psi_1$, each with certainty), then $\langle\psi_0|\psi_1\rangle = 0$.

Solution

--- ## Checkpoint 1. State the Born rule and prove that the outcome probabilities sum to $1$. 2. What three algebraic conditions define a complete set of orthogonal projectors $\{P_k\}$? 3. Write the post-measurement state after outcome $k$, and explain the role of the $\sqrt{p_k}$. 4. Prove that projective measurement is repeatable. 5. How do you measure the observable $X$ on hardware that can only measure $Z$? 6. Why are orthogonal states perfectly distinguishable but $|0\rangle$ and $|+\rangle$ are not?

Answers

1. $p_k = \langle\psi|P_k|\psi\rangle$. $\sum_k p_k = \langle\psi|\sum_k P_k|\psi\rangle = \langle\psi|I|\psi\rangle = 1$ by completeness and normalization. 2. $P_k=P_k^\dagger$ (Hermitian), $P_jP_k=\delta_{jk}P_k$ (orthogonal idempotents), and $\sum_k P_k=I$ (complete). 3. $|\psi'\rangle = P_k|\psi\rangle/\sqrt{p_k}$. The raw vector $P_k|\psi\rangle$ has norm $\sqrt{p_k}$; dividing renormalizes it to a unit vector so it is a valid state. 4. After collapse to $P_k|\psi\rangle/\sqrt{p_k}$, a second measurement gives outcome $j$ with probability $\langle\psi|P_kP_jP_k|\psi\rangle/p_k = \delta_{jk}$ using $P_jP_k=\delta_{jk}P_k$. 5. Rotate the $X$-eigenbasis onto the computational basis (apply $H$, since $H|\pm\rangle = |0/1\rangle$), then measure $Z$; $|\langle\pm|\psi\rangle|^2 = |\langle 0/1|H\psi\rangle|^2$. 6. A projective measurement assigns $|\psi_0\rangle\mapsto$ outcome 0 with certainty only if $P_0|\psi_0\rangle=|\psi_0\rangle$; orthogonality lets $P_0,P_1$ be the corresponding rank-1 projectors with no overlap. $|0\rangle,|+\rangle$ have $\langle0|+\rangle=\tfrac1{\sqrt2}\ne0$, so no projective measurement (indeed no measurement) separates them with zero error.

--- ## Further Reading - **[NC]** Nielsen & Chuang, §2.2.3–2.2.5 — the measurement postulate, projective measurements, and the global-phase / observable discussion. - **[Sak]** Sakurai & Napolitano, §1.4 — measurements, observables, and the collapse postulate, with the Stern–Gerlach intuition. - **[Pre]** Preskill, *Ph219*, Ch. 2 — the axioms of measurement and the orthogonal-projector formalism. --- ← Prev: [Bases & Expectation Values](../c02-qubits-bloch/03-bases-and-expectation.md) · Up: [Term 1](../README.md) · Next: [Expectation & Uncertainty](02-expectation-uncertainty.md) →