# 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.
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.
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.
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."
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.
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.