# The Conundrum of Projections > Three lessons of experimental facts have been accumulating an unpaid debt: no set of little arrows, > however cleverly arranged, can give two equal and opposite projections along *every* axis. This > lesson pays the debt — first by proving the classical picture impossible, then by replacing it with > the postulate that measurement is a projection. ## Learning Objectives After this lesson you will be able to: 1. State the **conundrum of projections** precisely and prove that no fixed assignment of a classical moment vector to each atom reproduces the observations. 2. State the **projection postulate**, express outcomes as orthogonal **projectors**, and derive repeatability from idempotence. 3. Compute the probability of a *sequence* of outcomes with the Wigner formula $\lVert P_c P_b P_a \lvert{\psi}\rangle\rVert^2$, and show why inserting a measurement changes later results. 4. Compute commutators of spin observables and explain **incompatibility** as the absence of a common eigenbasis. 5. Apply the Robertson **uncertainty relation** to spin, and explain why a vanishing bound does *not* mean two observables are compatible. 6. Analyze the $N$-analyzer staircase, and state precisely what the conundrum does and does not rule out. --- ## Intuition Everything so far has been "here is a strange fact." Now we prove that the strangeness is unavoidable — that it is not a failure of imagination on our part but a genuine impossibility. The classical hypothesis is as reasonable as hypotheses get: each atom leaves the oven carrying a definite magnetic moment $\boldsymbol{\mu}$, pointing somewhere, and the analyzer merely reveals the projection $\mu_z = \mu\cos\alpha$ that was already there. Every deflection is a fact about the atom; the apparatus is a passive reader. That hypothesis makes a prediction we can check with a protractor. If an atom carries a vector, then for any two directions the vector makes some angle with each, and the projections must be *related* in the way projections of a fixed vector are related. In particular, a vector of length $\mu$ can have projection $\pm\mu$ — the extreme values — only on the one axis it is parallel to. Along any other axis its projection is strictly smaller in magnitude; along the perpendicular axis it is zero. But the experiment gives the same full-size splitting for *every* orientation of the magnet. Turn the apparatus anywhere you like, and you get $+\mu$ and $-\mu$, never a smaller value and never a zero. A vector cannot do that. Nothing that has a direction can do that. Whatever the atom carries, it is not an arrow, and the number the analyzer reports cannot have been sitting there waiting to be read. --- ## Theory ### The conundrum, stated precisely Let $\hat{n}$ be the analyzer axis. The experimental facts from Lesson 2 are: - **(F1)** For every $\hat{n}$, exactly two outcomes occur, with deflections corresponding to $\mu_n = +\mu$ and $\mu_n = -\mu$, the *same* $\mu$ for every $\hat{n}$. - **(F2)** No intermediate value is ever observed, for any $\hat{n}$. - **(F3)** Measurements repeated along the same axis agree; measurements along a perpendicular axis are 50–50; along a general relative angle $\theta$ the probabilities are $\cos^2(\theta/2)$ and $\sin^2(\theta/2)$. The **conundrum of projections** is the observation that (F1) and (F2) together are incompatible with the atom's carrying any pre-existing vector at all [Fre, §1.4]. It is worth killing the classical hypothesis in three escalating versions, because each patch someone might propose fails for its own reason. **Version 1 — the source emits two fixed directions.** Suppose the oven emitted atoms with moments along only two directions, $\hat{a}$ and $\hat{b}$, chosen so that measuring along $\hat{z}$ gives one positive and one negative projection. Then consider the axis $\hat{c} \propto \hat{a}\times\hat{b}$. By construction $\hat{c}$ is perpendicular to both, so $\mu_c = 0$ for *every* atom in the beam: an analyzer oriented along $\hat{c}$ would show a single undeflected spot. This never happens. (If $\hat a$ and $\hat b$ are parallel or antiparallel the cross product vanishes, but then any axis perpendicular to them serves the same purpose.) **Version 2 — each atom carries its own vector.** Drop the two-direction assumption and let each atom carry an arbitrary $\boldsymbol{\mu}$ of length $\mu$. Fact (F1) requires $|\mu_n| = \mu$ for the observed outcomes along every axis $\hat{n}$. But $$ \mu_n = \boldsymbol{\mu}\cdot\hat{n} = \mu\cos\alpha_n, \qquad |\mu_n| = \mu \iff \cos\alpha_n = \pm 1 \iff \boldsymbol{\mu} \parallel \hat{n} . $$ A single vector cannot be parallel to every axis. So no atom, carrying any fixed vector whatsoever, can produce full-magnitude projections along all axes. **Version 3 — a probability distribution over vectors.** The most general classical patch: let the source draw $\boldsymbol{\mu}$ from *any* distribution over the sphere. This changes nothing, because Version 2's argument applies atom by atom. For any distribution, the set of atoms whose moment is within a few degrees of $\hat{n}$ has some probability; all the rest have $|\mu_n| < \mu$ strictly, and a continuum of intermediate deflections must appear on the plate. Fact (F2) says they never do. The only escape is to let the drawn vector depend on the analyzer setting $\hat{n}$ — but then the value is not a property the atom carried in, it is manufactured by the measurement. Which is the conclusion we were trying to avoid. **Therefore:** the projection reported by the analyzer is not a pre-existing property of the atom. It is **indeterminate** until measured, and the measurement is what makes it definite. > **Scientific honesty about the scope.** What the conundrum rules out is that the atom carries a > classical *vector* whose components are read off. It does not, by itself, rule out every conceivable > hidden-variable theory — a sufficiently contrived one could carry extra machinery keyed to the > apparatus. Closing that door requires an experiment whose statistics no local hidden-variable model > can reproduce at all: Bell's theorem, which Freericks takes up in his Chapter 2 and which this > program covers in [1.4.3 Nonlocality & CHSH](../../term-1-quantum-mechanics/c04-entanglement/03-nonlocality-chsh.md). > Do not overclaim this argument; it is decisive against the naive picture, and that is already a lot. ### Measurement as projection Replace the classical picture with a rule that reproduces (F1)–(F3) exactly. For each analyzer axis $\hat{n}$, the two outcomes correspond to two orthogonal states $\lvert{\pm\hat{n}}\rangle$, and to each we associate a **projector** $$ P_{\pm}^{\hat n} = \lvert{\pm\hat{n}}\rangle\langle{\pm\hat{n}}\rvert . $$ The **projection postulate** then says: measuring along $\hat n$ on a system in state $\lvert{\psi}\rangle$ yields outcome $\pm$ with probability $\lVert P_{\pm}^{\hat n}\lvert{\psi}\rangle\rVert^2 = \langle{\psi}\rvert P_{\pm}^{\hat n}\lvert{\psi}\rangle$, and leaves the system in the normalized projected state $$ \lvert{\psi'}\rangle = \frac{P_{\pm}^{\hat n}\lvert{\psi}\rangle}{\bigl\lVert P_{\pm}^{\hat n}\lvert{\psi}\rangle\bigr\rVert} . $$ Explicitly, in the $z$ basis, $$ P_+^{z} = \begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}, \qquad P_+^{x} = \frac12\begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}, \qquad P_-^{x} = \frac12\begin{pmatrix}1 & -1\\ -1 & 1\end{pmatrix} . $$ Three properties do all the work, and each corresponds to an experimental fact. *Hermiticity*, $P^\dagger = P$, makes the probabilities real. *Completeness*, $P_+^{\hat n} + P_-^{\hat n} = \mathbb{1}$, makes them sum to one — every atom leaves by exactly one port. And **idempotence**, $P^2 = P$, is repeatability: measuring twice in a row along the same axis gives the same answer with certainty, because the state after the first measurement is already in the range of $P$, so the second projection does nothing. Experiment 1 of Lesson 2 is the statement $P^2 = P$, made of brass and vacuum. ### Sequences of measurements: the Wigner formula For a chain of measurements with outcomes $a$, then $b$, then $c$, the joint probability is $$ P(a, b, c) = \bigl\lVert P_c\,P_b\,P_a \lvert{\psi}\rangle \bigr\rVert^2 , $$ with the projectors applied in time order, innermost first. Apply it to the three-analyzer experiment starting from $\lvert{+z}\rangle$: $$ P_+^{x}\lvert{+z}\rangle = \frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \frac12\begin{pmatrix}1\\1\end{pmatrix}, \qquad \left\lVert\cdot\right\rVert^2 = \frac12 , $$ then $$ P_+^{z}\left[\frac12\begin{pmatrix}1\\1\end{pmatrix}\right] = \frac12\begin{pmatrix}1\\0\end{pmatrix}, \qquad \left\lVert\cdot\right\rVert^2 = \frac14 . $$ So one quarter of the atoms entering the middle analyzer survive both stages — matching the counts of Lesson 2. Notice how the norm bookkeeping automatically produces the *conditional* probability $\tfrac14 / \tfrac12 = \tfrac12$ for the last stage: the $z$ information really is gone. The same algebra states the erasure as an operator identity. If inserting the $x$ measurement were harmless, the composite would act like the plain $z$ projection. Instead $$ P_+^{z}\,P_+^{x}\,P_+^{z} = \frac{1}{2}\,P_+^{z} \;\ne\; P_+^{z} , $$ whereas $P_+^{z}P_+^{z}P_+^{z} = P_+^{z}$ exactly. The factor $\tfrac12$ is the attenuation caused by looking at $x$ in between. Measurement is not a passive reading; it is an operation on the state. ### Incompatible observables Package the two outcomes of an axis into a single observable — the Hermitian operator whose eigenvalues are the measured values and whose eigenvectors are the outcome states: $$ S_z = \frac{\hbar}{2}\bigl(P_+^{z} - P_-^{z}\bigr) = \frac{\hbar}{2}\sigma_z, \qquad S_x = \frac{\hbar}{2}\sigma_x, \qquad S_y = \frac{\hbar}{2}\sigma_y , $$ with the Pauli matrices $\sigma_z = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$, $\sigma_x = \begin{pmatrix}0&1\\1&0\end{pmatrix}$, $\sigma_y = \begin{pmatrix}0&-i\\i&0\end{pmatrix}$. Their commutators are $$ [S_x, S_y] = i\hbar S_z, \qquad [S_y, S_z] = i\hbar S_x, \qquad [S_z, S_x] = i\hbar S_y . $$ Two observables are **compatible** when they commute, and compatibility is exactly the condition for a common eigenbasis to exist — a set of states in which *both* have definite values. Since the spin components do not commute, no such basis exists, and therefore **no state assigns sharp values to two different spin axes at once.** That single algebraic fact explains the entire pattern of Lesson 2: the atom "forgets" the previous axis not because of clumsy apparatus, but because there was never a state in which both values were defined. Correspondingly the projectors fail to commute, $[P_+^z, P_+^x] \ne 0$, which is why the order of measurements matters. ### The uncertainty relation, with a caution The Robertson relation quantifies incompatibility for a given state: $$ \Delta A\,\Delta B \;\ge\; \frac{1}{2}\bigl|\langle[A, B]\rangle\bigr| , $$ so for spin, $\Delta S_z\,\Delta S_x \ge \tfrac{\hbar}{2}\lvert\langle S_y\rangle\rvert$. Now a subtlety that repays attention. Evaluate the relation in $\lvert{+z}\rangle$: there $\langle S_y\rangle = 0$, so the bound reads $\Delta S_z\,\Delta S_x \ge 0$ — completely vacuous. And indeed $\Delta S_z = 0$ while $\Delta S_x = \hbar/2$, so the product is zero and the inequality is satisfied trivially. **A vanishing Robertson bound does not mean the observables are compatible.** The bound is state-dependent and can collapse; the incompatibility, $[S_z, S_x] = i\hbar S_y \ne 0$ as an operator statement, does not. For contrast, evaluate it in $\lvert{+y}\rangle = (\lvert{+z}\rangle + i\lvert{-z}\rangle)/\sqrt2$. Here $\langle S_y\rangle = \hbar/2$, so the bound is $\Delta S_z\,\Delta S_x \ge \hbar^2/4$; and since both $z$ and $x$ measurements are 50–50 in this state, $\Delta S_z = \Delta S_x = \hbar/2$ and the product is exactly $\hbar^2/4$. The relation is *saturated* — this state is as close to sharp on both axes as any state can be, which is to say not close at all. ### The staircase: dragging a spin by measuring it One last consequence, both surprising and useful. Lesson 2 showed that a $+z$ atom facing a $180°$ analyzer never emerges from the $+$ port: $\cos^2 90° = 0$. Now interpose $N-1$ analyzers, so the axis turns in $N$ equal steps of $\pi/N$, and keep only the atoms that come out $+$ at every stage. Each step succeeds with probability $\cos^2(\pi/2N)$, and the steps are independent given survival, so $$ P_N = \left[\cos^2\!\left(\frac{\pi}{2N}\right)\right]^{N} . $$ For $N = 1$ this is 0, as it must be. For $N = 2$, $(\tfrac12)^2 = \tfrac14$. For $N = 10$, about $0.78$; for $N = 100$, about $0.976$. And in the limit, using $\cos^2 x \approx 1 - x^2$, $$ P_N \approx \left(1 - \frac{\pi^2}{4N^2}\right)^{N} \longrightarrow 1 \quad \text{as } N \to \infty . $$ A sequence of gentle measurements can turn a spin completely around, from $+z$ to $-z$, with probability approaching certainty — even though a single measurement at $180°$ would achieve it with probability zero. Repeated observation drags the state along with the apparatus. This is the mechanism behind the quantum Zeno effect and its cousins, and it is a striking demonstration that measurement in quantum mechanics is a *dynamical* act with consequences, not a passive glance. ### Where this goes Everything in this course is now a special case of machinery the main program develops formally. The projection postulate becomes [1.1.2 Observables & the Measurement Postulate](../../term-1-quantum-mechanics/c01-postulates/02-observables-measurement-postulate.md) and [1.3.1 Projective Measurement](../../term-1-quantum-mechanics/c03-measurement/01-projective-measurement.md); the uncertainty relation is derived properly in [1.3.2 Expectation & Uncertainty](../../term-1-quantum-mechanics/c03-measurement/02-expectation-uncertainty.md); the analyzer-as-black-box generalizes to POVMs in [1.3.3 POVMs & Generalized Measurement](../../term-1-quantum-mechanics/c03-measurement/03-povm-generalized.md); and the question the conundrum leaves open — whether *any* local pre-existing values could work — is settled by Bell in Term 1.4. What the Stern-Gerlach cascades give you that the axioms cannot is the memory that all of it was forced by two spots on a glass plate. --- ## Hands-on (Python) ```python import numpy as np ket_pz, ket_mz = np.array([1, 0], complex), np.array([0, 1], complex) def proj(theta, sign=+1): """Projector onto the +/- state of an analyzer at angle theta (radians).""" t = theta if sign > 0 else theta + np.pi v = np.array([np.cos(t / 2), np.sin(t / 2)], complex) return np.outer(v, v.conj()) Pz, Px = proj(0.0), proj(np.pi / 2) # --- Projectors: Hermitian, idempotent, complete ------------------------- print("P^2 = P :", np.allclose(Pz @ Pz, Pz)) print("P = P^dagger :", np.allclose(Pz, Pz.conj().T)) print("complete :", np.allclose(proj(0.0, +1) + proj(0.0, -1), np.eye(2))) # --- Sequential measurement: the Wigner formula -------------------------- psi = ket_pz after_x = Px @ psi after_zx = Pz @ after_x print(f"\nP(+x) = {np.linalg.norm(after_x)**2:.3f}") # 0.500 print(f"P(+x then +z) = {np.linalg.norm(after_zx)**2:.3f}") # 0.250 # The erasure, as an operator identity: Pz Px Pz = (1/2) Pz, not Pz. print("Pz Px Pz == 0.5 * Pz :", np.allclose(Pz @ Px @ Pz, 0.5 * Pz)) print("[Pz, Px] == 0 :", np.allclose(Pz @ Px - Px @ Pz, 0)) # --- Commutators of the spin operators (hbar = 1) ------------------------ sx = np.array([[0, 1], [1, 0]], complex) sy = np.array([[0, -1j], [1j, 0]], complex) sz = np.array([[1, 0], [0, -1]], complex) Sx, Sy, Sz = sx / 2, sy / 2, sz / 2 print("[Sx, Sy] == i Sz :", np.allclose(Sx @ Sy - Sy @ Sx, 1j * Sz)) # --- The staircase: turning a spin around by measuring it ---------------- print("\n N P_N") for N in (1, 2, 3, 10, 100, 1000): print(f"{N:5d} {np.cos(np.pi / (2 * N))**(2 * N):.4f}") # 0.0000, 0.2500, 0.4219, 0.7805, 0.9756, 0.9975 -> 1 ``` --- ## Exercises **E1 (easy).** Verify that $P_+^{x} = \tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix}$ is a legitimate projector, compute $P_+^{x} + P_-^{x}$, and state which experimental fact each property corresponds to.

Solution

*Hermitian:* the matrix is real and symmetric, so $P^\dagger = P$ — this makes $\langle\psi\rvert P\lvert\psi\rangle$ real, as a probability must be. *Idempotent:* $P^2 = \tfrac14\begin{pmatrix}1&1\\1&1\end{pmatrix}^2 = \tfrac14\begin{pmatrix}2&2\\2&2\end{pmatrix} = \tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix} = P$ — this is repeatability (Experiment 1): measuring twice along the same axis gives the same answer. *Complete:* $P_+^{x} + P_-^{x} = \tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix} + \tfrac12\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}$ — every atom leaves by exactly one of the two ports, so the probabilities sum to 1.

**E2 (easy).** Atoms from the $-$ exit of a vertical analyzer are fed into an analyzer oriented at $45°$ [Fre, §1.4.1]. What is the probability of exiting its $+$ port? Compute it two ways: from the relative angle, and by explicit inner product.

Solution

*Relative angle.* The incoming atoms are $\lvert{-z}\rangle$, i.e. prepared along the axis at $180°$. The analyzer sits at $45°$, so the relative angle is $135°$ and $P(+) = \cos^2(135°/2) = \cos^2 67.5° = 0.1464$. *Inner product.* $\lvert{+\hat n(45°)}\rangle = \cos 22.5°\lvert{+z}\rangle + \sin 22.5°\lvert{-z}\rangle$, so $\langle{+\hat n(45°)}|{-z}\rangle = \sin 22.5° = 0.3827$ and $P(+) = 0.3827^2 = 0.1464$. ✓ Equivalently $P(+) = \sin^2(\theta/2)$ with $\theta = 45°$ measured from the *original* $z$ axis, which is the complement rule of Lesson 2 in action. About 15% get through — a $-z$ atom is unlikely, but not forbidden, to register $+$ on a nearby axis.

**E3 (medium).** Compute $P_+^{z}P_+^{x}P_+^{z}$ explicitly and interpret the result. Then compute $P_+^{x}P_+^{z}P_+^{x}$ and comment on the symmetry.

Solution

$$ P_+^{x}P_+^{z} = \frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix} = \frac12\begin{pmatrix}1&0\\1&0\end{pmatrix}, \qquad P_+^{z}\left[\frac12\begin{pmatrix}1&0\\1&0\end{pmatrix}\right] = \frac12\begin{pmatrix}1&0\\0&0\end{pmatrix} = \frac12 P_+^{z} . $$ So $P_+^{z}P_+^{x}P_+^{z} = \tfrac12 P_+^{z}$. The direction of the surviving state is unchanged (it is still $\lvert{+z}\rangle$), but the amplitude is halved, i.e. the intensity quartered — the same $\tfrac14$ the Wigner formula gave. Had the $x$ measurement been harmless we would have found $P_+^{z}$ itself. By the identical computation with the roles swapped, $P_+^{x}P_+^{z}P_+^{x} = \tfrac12 P_+^{x}$. The symmetry reflects that $z$ and $x$ are at $90°$ from each other in either order — neither axis is privileged, and the disturbance is mutual. It is *not* generally true that $P_aP_bP_a = |\langle a|b\rangle|^2 P_a$ gives the same constant when $a$ and $b$ swap for non-orthogonal, non-symmetric pairs — but here $|\langle{+z}|{+x}\rangle|^2 = \tfrac12$ both ways.

**E4 (medium).** Evaluate the Robertson bound for $S_z$ and $S_x$ in the states $\lvert{+z}\rangle$ and $\lvert{+y}\rangle = (\lvert{+z}\rangle + i\lvert{-z}\rangle)/\sqrt2$. In one case the bound is vacuous. Explain carefully why that does *not* mean $S_z$ and $S_x$ are compatible in that state.

Solution

*In $\lvert{+z}\rangle$:* $\langle S_y\rangle = 0$, so the bound is $\Delta S_z\Delta S_x \ge 0$. Directly: $S_z$ is sharp, $\Delta S_z = 0$; $S_x$ is 50–50, $\Delta S_x = \hbar/2$; the product is 0. The inequality holds with room to spare and says nothing. *In $\lvert{+y}\rangle$:* $\langle S_y\rangle = \hbar/2$, so $\Delta S_z\Delta S_x \ge \tfrac{\hbar}{2}\cdot\tfrac{\hbar}{2} = \hbar^2/4$. Both $z$ and $x$ measurements are 50–50 here ($|\langle{\pm z}|{+y}\rangle|^2 = |\langle{\pm x}|{+y}\rangle|^2 = \tfrac12$), so $\Delta S_z = \Delta S_x = \hbar/2$ and the product equals $\hbar^2/4$ — the bound is saturated. *Why a vacuous bound is not compatibility.* Robertson bounds the product of spreads *in a given state*, and its right-hand side is itself a state-dependent expectation that can vanish. Compatibility is a property of the *operators*: $A$ and $B$ are compatible iff $[A,B] = 0$, which is what guarantees a common eigenbasis and hence states where both are sharp. Here $[S_z,S_x] = i\hbar S_y \ne 0$ as an operator, so no such state exists — and indeed even in $\lvert{+z}\rangle$, where the product of spreads is zero, $S_x$ is maximally uncertain. A zero bound means "this inequality has nothing to say here," not "both can be sharp."

**E5 (hard).** Derive $P_N = [\cos^2(\pi/2N)]^N$ for the $N$-step staircase from $+z$ to $-z$, evaluate it for $N = 2, 10, 100$, and find its large-$N$ behaviour including the leading correction. Then explain the apparent paradox: a single measurement at $180°$ succeeds with probability 0, yet the limit of many small measurements succeeds with probability 1.

Solution

*Derivation.* The axis turns by $\pi$ in $N$ equal steps, so consecutive analyzers differ by $\theta = \pi/N$. Keeping only the $+$ port at each stage, each step succeeds with probability $\cos^2(\theta/2) = \cos^2(\pi/2N)$, and after a successful step the state is exactly the $+$ eigenstate of that analyzer — so the next step is statistically identical. Multiplying $N$ independent factors gives $P_N = [\cos^2(\pi/2N)]^N$. *Values.* $P_2 = (\cos^2 45°)^2 = \tfrac14$; $P_{10} = (\cos^2 9°)^{10} = 0.7805$; $P_{100} = (\cos^2 0.9°)^{100} = 0.9756$. *Asymptotics.* With $x = \pi/2N$ and $\cos^2 x = 1 - x^2 + \tfrac{x^4}{3} + \cdots$, $$ \ln P_N = N\ln\left(1 - \frac{\pi^2}{4N^2} + \cdots\right) = -\frac{\pi^2}{4N} + O(N^{-3}), \qquad P_N \approx e^{-\pi^2/4N} \approx 1 - \frac{\pi^2}{4N} . $$ Check: $N = 100$ gives $1 - 0.0247 = 0.9753$, against the exact $0.9756$. ✓ The approach to certainty is like $1/N$. *The paradox resolved.* There is no contradiction, because the two experiments are not the same experiment. The failure probability per step scales as $\theta^2/4$ — *quadratically* in the step angle — while the number of steps grows only linearly, so the total failure probability $\sim N(\pi/2N)^2 = \pi^2/4N$ vanishes. Physically: each measurement projects the state onto the new axis, and a projection onto a nearby direction costs almost nothing but *carries the state with it*. The state does not survive the rotation passively; it is repeatedly re-prepared. This is the quantum Zeno mechanism, and its practical descendants include adiabatic state transfer and measurement-based control.

--- ## Checkpoint 1. State the conundrum of projections, and give the cross-product argument against a source emitting two fixed directions. 2. Why can no distribution over classical moment vectors — however contrived — reproduce fact (F2)? 3. Which property of a projector corresponds to the repeatability of measurement, and why? 4. Compute $[S_z, S_x]$ and explain what its non-vanishing implies about states in which both are sharp. 5. In $\lvert{+z}\rangle$ the Robertson bound for $\Delta S_z\Delta S_x$ is zero. Does that mean $S_z$ and $S_x$ are compatible? Explain.

Answers

1. The experiment shows two outcomes of *equal and full magnitude* along every analyzer axis, which no assignment of pre-existing moment vectors can produce. If the source emitted only $\hat a$ and $\hat b$, then an analyzer along $\hat a\times\hat b$ would find zero projection for every atom — one undeflected spot, never observed. 2. Version 2's argument applies atom by atom: $|\boldsymbol\mu\cdot\hat n| = \mu$ forces $\boldsymbol\mu \parallel \hat n$, so for any fixed vector, almost every axis yields an intermediate projection. Averaging over a distribution cannot remove intermediate values from individual atoms; it can only reweight them, and (F2) says they never appear at all. 3. Idempotence, $P^2 = P$. After the first measurement the state lies in the range of $P$, so applying $P$ again leaves it unchanged with probability 1 — exactly Experiment 1. 4. $[S_z, S_x] = i\hbar S_y \ne 0$. Non-commuting Hermitian operators have no common eigenbasis, so there is no state in which both $S_z$ and $S_x$ have definite values — the "memory loss" of the cascades is this fact, not an apparatus defect. 5. No. The Robertson bound is state-dependent and its right-hand side $\tfrac\hbar2|\langle S_y\rangle|$ happens to vanish in $\lvert{+z}\rangle$; the inequality then simply carries no information. Compatibility is the operator statement $[S_z,S_x] = 0$, which is false. Indeed in that very state $\Delta S_x = \hbar/2$ is maximal.

--- ## Further Reading - **[Fre]** J. K. Freericks, *Quantum Mechanics Done Right*, §1.4 — the conundrum of projections and the rotating-analyzer problems, following Styer's formulation. - **[Sak]** Sakurai & Napolitano, §§1.3–1.4 — kets, bras, operators, and the measurement/compatibility formalism this lesson sketches. - **[NC]** Nielsen & Chuang, §§2.2.3–2.2.5 — projective measurement, the measurement postulate, and the generalization to POVMs. - **[Mer]** Mermin, *Quantum Computer Science*, Ch. 1 — a compact and unusually careful account of what measurement does to a qubit. - [1.3.1 Projective Measurement](../../term-1-quantum-mechanics/c03-measurement/01-projective-measurement.md) and [1.3.2 Expectation & Uncertainty](../../term-1-quantum-mechanics/c03-measurement/02-expectation-uncertainty.md) — the formal versions of this lesson's postulate and inequality. - [1.4.3 Nonlocality & CHSH](../../term-1-quantum-mechanics/c04-entanglement/03-nonlocality-chsh.md) — the road to Bell's theorem, which closes the loophole this lesson deliberately leaves open. --- ← Prev: [Probability in the Quantum World](03-probability-quantum-world.md) · Up: [Resources](../README.md) · Next: [The Analyzer Loop](../c02-freericks-advanced-spin/01-analyzer-loop.md) →