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.

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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.

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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. 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 and 1.3.1 Projective Measurement; the uncertainty relation is derived properly in 1.3.2 Expectation & Uncertainty; the analyzer-as-black-box generalizes to POVMs in 1.3.3 POVMs & Generalized Measurement; 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.

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Hands-on (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

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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.

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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.

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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 and 1.3.2 Expectation & Uncertainty — the formal versions of this lesson's postulate and inequality.
• 1.4.3 Nonlocality & CHSH — the road to Bell's theorem, which closes the loophole this lesson deliberately leaves open.

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