No-Cloning and Quantum State Tomography

These notes continue from the Advanced Qubit Notes; everything covered there (complex numbers and phase, the Born rule, the Bloch sphere, the Pauli matrices, projective measurement) is assumed. The through-line is a detective case: someone hands you an unknown qubit, and your job is to find out what it is. Cracking the case takes three tools first, Euler’s formula (chapter 1), the tensor product (chapter 2), and the no-cloning theorem (chapter 3), before the investigation proper: three rounds of detection (chapter 4), one unifying formula (chapter 5), and the deeper morals (chapter 6).

1. Euler’s formula: why $e^{i\pi}$ equals $-1$

1.1 An imaginary power is not “multiplying that many times”

$e^{i\pi}$ is not “e multiplied by itself π times”; an imaginary power carries no such meaning in the first place. It is what mathematicians got by extending the exponential function to complex numbers, and the tool of extension is the Taylor series:

The right side uses only multiplication and addition, both of which complex numbers can do, so substitute $x = i\theta$ directly. Using the four-step cycle of the powers of $i$ ($i,\ i^2=-1,\ i^3=-i,\ i^4=1,\ldots$), the terms hop periodically between real and imaginary, and sorting them into two piles gives

Euler’s formula is not a decree; the series computes it on its own. Substituting $\theta=\pi$: $\cos\pi=-1,\ \sin\pi=0$, hence

1.2 The geometric meaning: $e^{i\theta}$ is “going around the circle”

$(\cos\theta,\sin\theta)$ is exactly the point on the unit circle at angle $\theta$, so $e^{i\theta}$ always lands on the unit circle with $\theta$ as the turning angle; multiplying by $e^{i\theta}$ = rotating counterclockwise by $\theta$. The four cardinal points:

The picture of $e^{i\pi}=-1$: start from $1$, walk half a turn counterclockwise along the unit circle, land precisely on $-1$. Conjugation is rotation in reverse: $(e^{i\theta})^*=e^{-i\theta}$.

1.3 Three identities these notes will keep reaching for

The computations of chapters 4 and 5 run entirely on them, so put them on file (each verifiable in one line by expanding with Euler’s formula):

Identity (i) will explain, in Round 1, why a Z measurement is blind to phase; (ii) and (iii) will fold $\langle X\rangle$ and $\langle Y\rangle$ into $\cos\phi$ and $\sin\phi$ in chapter 5.

2. The tensor product: how multi-qubit worlds are assembled

2.1 The rule itself

The joint state of two qubits is assembled with the tensor product $\otimes$, and the rule is “each component of the left vector multiplies the entire right vector”:

From this, the four basis vectors:

And the companion slot-by-slot inner product rule (chapter 3 will use it):

2.2 Why dimensions multiply: the classical world already works this way

One switch has 2 states; two switches have $2\times2=4$ combinations (off-off, off-on, on-off, on-on). The joint distribution of two dice needs $6\times6=36$ probability values, not $6+6=12$. “Combination counts multiply” is not a quantum novelty; it is simply what “joining two independent descriptions” means.

2.3 Why not the direct sum $\oplus$

The direct sum adds dimensions and describes “the system is either in A or in B” (as with superselection sectors of different charge). But two qubits are not an either-or; they exist simultaneously, each with its own freedom. Worse, the direct-sum structure has nowhere to store correlation: a state like $|00\rangle$, a relation between two qubits, simply does not exist inside $\oplus$.

2.4 Superposition pushes the Cartesian product up to the tensor product — entanglement is forced

Classically, the joint state of two switches can only be one of those 4 combinations. Quantum mechanics adds one clause: any superposition of distinguishable states is also a legal state. So the legal two-qubit states are arbitrary complex combinations of the four basis vectors,

and this set is precisely the 4-dimensional tensor-product space. In one line: Cartesian product (counts multiply) + superposition principle (closure under addition) = tensor product.

An inescapable by-product comes along. If only “independent” product states $|\psi\rangle\otimes|\varphi\rangle$ were allowed, that set would not be closed under addition. The fastest proof is by contradiction: suppose

This requires $ac=bd=\tfrac{1}{\sqrt2}$ and $ad=bc=0$. From $ad=0$: either $a=0$ (killing $ac$) or $d=0$ (killing $bd$), contradicting the first condition either way. So $|00\rangle$ and $|11\rangle$ are both product states, yet their equal-weight superposition is not any product state — an entangled state. The superposition principle demands closure under addition; completing all linear combinations of product states yields exactly the full tensor-product space, and the extra part is entanglement. Entanglement is not an added assumption; it is forced by superposition — you cannot build a quantum theory that has superposition and composite systems but no entanglement.

2.5 Physical evidence

• Parameter counting: if two qubits truly minded their own business, a joint state would need $2+2=4$ real parameters; the tensor product says 6 (4 complex amplitudes = 8 reals, minus normalization and global phase). The 2 extra degrees of freedom are correlation, directly measurable by state tomography.
• Bell inequality violations: any “local hidden variable” model has a hard ceiling on correlation strength (CHSH $\le 2$); the tensor-product structure predicts a ceiling of $2\sqrt2\approx2.83$, and experiments measure right around that value. This is the hardest evidence: the joint space really is bigger than “two independent systems”.
• Exponential blowup: $n$ qubits need $2^n$ amplitudes. The very difficulty classical computers have simulating quantum systems is empirical evidence of the tensor-product structure.

2.6 An honest footnote

The tensor product is, in the end, a postulate of quantum mechanics (the composite-system postulate). The “quantum reconstruction” program tries to derive it from plainer operational axioms; the key one is local tomography: the state of a composite system is fully determined by local measurements on the subsystems plus their correlations — which directly forces $d_{AB}=d_A\times d_B$ and, as a bonus, rules out “real-number quantum mechanics” (experimentally probed around 2021–2022).

3. The no-cloning theorem and fan-out 1

3.1 Fan-out: classical wires branch, quantum wires may not

In classical Boolean circuits, fan-out = how many downstream wires read one gate’s output. Classical circuits let wires branch freely: compute an intermediate value once, copy it endlessly, because “copying a voltage” is trivial.

In a quantum circuit, each horizontal line is one qubit’s time axis. Letting it branch is equivalent to demanding a machine

(feed in one original and one blank sheet $|0\rangle$, get two originals out). This chapter proves the machine does not exist, so quantum wires never branch — fan-out is identically 1.

3.2 Set the problem up straight: cloning ≠ preparation

A common reaction: “I just want to photocopy $|1\rangle$: input $|1\rangle\otimes|0\rangle$, output $|1\rangle\otimes|1\rangle$, isn’t that copying?” It is — but notice the premise: you already know it is $|1\rangle$. If you know, you need no copier; have the preparation device produce another one (that is preparation, not cloning). The strict setting of the cloning problem is: one sample in hand, identity unknown. “Measure it, then re-prepare” does not work (one measurement yields one bit and destroys the original; see chapter 4), so you would need a machine that copies blind, without looking. The theorem kills that machine.

3.3 A concrete-counterexample proof (no steps skipped)

The baseline: any allegedly universal $U$ must at least copy the two basis states (basis states are states too):

In passing: CNOT satisfies exactly these two lines. It is the “copier that only copies basis states”, and it is the essence of all classical copying (wire branching, cp, memory copies):

Verify the second line by row-by-row dot products, $\mathrm{CNOT}\cdot(0,0,1,0)^T$: row 1 $=0$; row 2 $=0$; row 3 $=0{\cdot}0+0{\cdot}0+0{\cdot}1+1{\cdot}0=0$; row 4 $=0{\cdot}0+0{\cdot}0+1{\cdot}1+0{\cdot}0=1$, giving $(0,0,0,1)^T=|11\rangle$ ✓.

Feed it a superposition. Input $|+\rangle\otimes|0\rangle$; compute the tensor product component by component and split over the basis:

Linearity locks the output. Matrix multiplication distributes, $U(av+bw)=aUv+bUw$, so any $U$ meeting the baseline has no choice:

which is exactly the Bell entangled state of §2.4. The copying contract, meanwhile, demands the output

The second component is $0$ on one side and $\tfrac12$ on the other — the two vectors are unequal; contradiction. The proof used only “linearity + works on basis states”, and the quantifier is $\forall U$: every machine fails either on the basis states or on $|+\rangle$. CNOT’s explicit matrix was never used in deriving the contradiction; it only made a cameo to show the baseline is satisfiable. $\blacksquare$

The mental picture: whatever copier you build, feed it a superposition and what comes out is not two copies but one entanglement — the copier turns into an entangler whether you like it or not.

3.4 The theorem’s true shape: clonable ⟺ pairwise orthogonal

The above only shot down the single sample $|+\rangle$. The general theorem is prettier: the states one machine can clone simultaneously must be pairwise orthogonal. The proof uses two tools: the slot-by-slot inner product rule of §2.1, and unitaries preserve inner products ($\langle Uu|Uv\rangle=\langle u|U^\dagger U|v\rangle=\langle u|v\rangle$).

Suppose some $U$ clones normalized states $|\psi\rangle,|\varphi\rangle$ simultaneously, and write $x=\langle\psi|\varphi\rangle$. The inner product of the two input vectors:

The inner product of the two output vectors:

Unitarity preserves inner products ⟹ the two ends are equal:

$x=1$: the two states were the same all along (nothing new copied); $x=0$: orthogonal. $\blacksquare$

Check case by case: $\{|0\rangle,|1\rangle\}$, $x=0$ ✓ — CNOT legally exists; $\{|0\rangle,|+\rangle\}$, $x=\tfrac{1}{\sqrt2}$ while $x^2=\tfrac12\ne x$ — no machine can copy both.

Conclusion: the set of states that fit inside one copier must be pairwise orthogonal — which is precisely the domain of “classical information”. The boundary of cloning = the boundary of orthogonality = the dividing line between classical and quantum information. Classical bits copy freely only because they live among orthogonal states; the contradiction detonates only on superpositions.

3.5 The consequence list

• Fan-out 1: quantum wires never branch. Together with “unitaries are square, $k$ in means $k$ out” (no merging either), the shape of quantum circuits is locked: $n$ parallel horizontal lines, wire count conserved.
• Move, but not copy: the SWAP gate and quantum teleportation are both legal — mv yes, cp no.
• Terminology alert: the “fanout gate” of circuit-complexity literature (multi-target CNOT, QAC circuits) still copies basis values only; no conflict with the theorem.
• Quantum money (Wiesner, 1970s): the anti-counterfeiting core is exactly “nobody can photocopy your banknote’s state”.
• BB84’s no-copy pillar: the eavesdropper cannot “keep a copy of the photon and measure at leisure once the basis is announced”. The protocol deliberately encodes bits in the two mutually non-orthogonal families $\{|0\rangle,|1\rangle\}$ and $\{|+\rangle,|-\rangle\}$ — standing, on purpose, on the side no copier can reach.
• Direct supply for chapter 4: the copies used in the investigation can only come from re-running the preparation procedure, never from photocopying the one in hand — which is why tomography must burn through a batch of copies.

4. The case: identify an unknown qubit

4.0 Notation: measuring Z, readings, and states

The everyday “measure in the computational basis (0 or 1)” is, formally, measuring the observable $Z$. The evidence is $Z$‘s spectral decomposition — eigenvalues times projectors, summed:

Projecting onto $|0\rangle$ or $|1\rangle$ is precisely that measurement. The raw readings the instrument physically outputs are the eigenvalues $\pm1$; converting to bits is pure relabeling: $+1\mapsto\text{"0"},\ -1\mapsto\text{"1"}$, formula $m=(-1)^{\text{bit}}$.

One distinction to nail down: $+1$ is not $|+\rangle$. A reading ($+1$, a number, an eigenvalue) and a state ($|+\rangle$, a vector, an eigenvector) are different kinds of thing. “Measuring X gave $+1$” is a reading, from which you infer the state is now $|+\rangle$; the same number $+1$ on the Z instrument corresponds to the state $|0\rangle$. Numbers are labels, states are the content — which label sticks to which state depends on which instrument you ran.

One typographic landmine while we are here: the Born probability is written $|\langle0|\psi\rangle|^2$, three vertical bars in total — the outer two take the modulus, the middle one separates bra from ket. When rendering breaks, the middle bar tends to vanish, producing illegal notation like $\vert\langle0\rangle\psi\vert^2$; recognize it as a typesetting accident, not a new symbol.

4.1 The case and the rules of the game

Someone hands you the unknown state $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$ and asks for $\alpha,\beta$ (equivalently, the Bloch angles $\theta,\phi$). The constraints:

• A single measurement is nearly useless: it emits one classical bit and the state collapses on the spot — you learn one bit and destroy all the evidence;
• No copying (chapter 3): you cannot “keep a copy and measure it slowly”;
• The only way out: re-run the preparation procedure as many times as needed (supply, not photocopy) and do statistics on a batch of identical copies.

The detective’s method: every kind of measurement is one clue, and every clue eliminates a crowd of suspects.

4.2 The landmark file (the whole case’s reference frame)

$\alpha=\cos\frac\theta2,\ \beta=e^{i\phi}\sin\frac\theta2$. Six landmark states:

(The coordinates column comes from chapter 5’s theorem; treat it as reference for now. Geometric shorthand: each orthogonal pair occupies the two ends of one axis; $\theta$ is latitude, $\phi$ is longitude.)

4.3 Round 1: measure Z — you only get the latitude

Measure Z on each copy: $P(0)=|\langle0|\psi\rangle|^2$. An inner product just “takes the component”: $\langle0|\psi\rangle=(1\;\;0)\binom{\alpha}{\beta}=\alpha$, $\langle1|\psi\rangle=\beta$. Suppose the statistics come out 50/50. Run the four equatorial suspects one by one:

Then settle the entire equator in one stroke: any equatorial state has $\beta=\tfrac{1}{\sqrt2}e^{i\phi}$, so

$\phi$ is annihilated exactly in the conjugate product (identity (i) of §1.3, cashed in). The geometric reading: a Z measurement reads the state arrow’s projection on the z axis — it sees latitude only and is blind to longitude. Conclusion: $|\alpha|^2=|\beta|^2=\tfrac12\Rightarrow\theta=\pi/2$; latitude in hand, the suspect pool shrinks from the whole sphere to the full equatorial circle — still infinitely many.

4.4 Round 2: change basis — “measuring X” ≡ prepend an H, then measure Z

You want to ask “are you $|+\rangle$ or $|-\rangle$” (measure X), but the lab instrument only measures Z. The trick: pad a unitary gate in front of the measurement. H is the two-way translator between the X world and the Z world; work out its four actions entry by entry ($H=\tfrac{1}{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$):

Protocol: $[H]-[\text{measure }Z]$, with the translation table: read 0 ⟹ “X gave $+1$, the state is now $|+\rangle$”; read 1 ⟹ “X gave $-1$, the state is now $|-\rangle$”.

Why reading 0 lets you conclude X = +1: H is a reversible, lossless renaming that renames $|+\rangle$ to $|0\rangle$. Statistical correctness in one line — first compute row vector times matrix:

which is exactly the Born probability a “direct X measurement” should have. The Z instrument reading 0 means the original state “won the draw” on its $|+\rangle$ component.

Run the suspects through the protocol:

• $|+\rangle\xrightarrow{H}|0\rangle$: reads 0 with certainty; $\quad|-\rangle\xrightarrow{H}|1\rangle$: reads 1 with certainty;
• $|{+i}\rangle$: $H|{+i}\rangle=\tfrac12\binom{1+i}{1-i}$, $P(0)=\big|\tfrac{1+i}{2}\big|^2=\tfrac{(1-i)(1+i)}{4}=\tfrac{1+1}{4}=\tfrac12$ — still 50/50 ($|{-i}\rangle$ likewise).

Case-closed branch one: the data reads 0 a hundred percent of the time ⟹ the only state that can do that is $|+\rangle$; the suspect circle collapses to a point, case closed. Branch two: X is also 50/50 ⟹ the only states unbiased toward both Z and X are the two ends of the y axis, $|{\pm i}\rangle$; proceed to Round 3.

4.5 Round 3: measure Y — pad $S^\dagger$, then H

We need a unitary $V$ that translates the y axis onto the z axis: $V|{+i}\rangle=|0\rangle,\ V|{-i}\rangle=|1\rangle$. Take $V=H\cdot S^\dagger$ with $S^\dagger=\begin{pmatrix}1&0\\0&-i\end{pmatrix}$, in two steps:

The circuit $[S^\dagger]-[H]-[\text{measure }Z]\equiv$ “measure Y”. Run it on $|{+i}\rangle$: reads 0 with certainty ⟹ Y gives $+1$ ⟹ case closed, $|\psi\rangle=|{+i}\rangle$.

4.6 Are the three instruments equally “precise”? Why exactly three rounds?

A natural illusion: “the Y instrument seems more powerful, more precise.” No. The three instruments are identical in precision; the only difference is orientation: Z reads the z-axis projection, X the x-axis, Y the y-axis — the same ruler mounted in three directions. Round 3 “settles it in one blow” only because the first two rounds had already squeezed the suspects onto the y axis, and it was simply the correctly oriented ruler’s turn.

The root of “three rounds, necessary and sufficient”: a Bloch arrow is fixed by exactly three Cartesian coordinates $(x,y,z)$ (chapter 5 proves the three expectation values are precisely them). One round fewer leaves a full circle of degeneracy (measure only Z and X, and you can never split the $|{\pm i}\rangle$ pair); one round more is redundant.

4.7 The universal recipe

“Measuring in any basis = prefix a suitable unitary + measure Z.” A lab needs only one physical measurement device plus a toolbox of unitary gates to own every measurement. For a polarization qubit, “padding a unitary” means inserting a wave plate before the beam splitter — the middle 45° sheet in the three-polarizer experiment is exactly one basis change.

5. Expectation values and Bloch coordinates: tomography’s unifying formula

5.1 The sandwich formula: a shortcut calculator for the average reading

For $\pm1$ readings, the expectation value (the average reading over many repetitions) is $\langle A\rangle=P(+1)-P(-1)$, estimated in the lab by counting frequencies. The sandwich $\langle\psi|A|\psi\rangle$ is its shortcut calculator, and the mechanism is “price-tagging”:

$Z$ multiplies each branch by its own reading (its eigenvalue). Close up with $\langle\psi|$ on the left:

Warning: $A|\psi\rangle$ is paper bookkeeping, not what physically happens during measurement (measurement = decompose + roll dice + collapse; see Advanced Qubit Notes §8). The shortcut and the direct method give the same number, spot-checkable on landmarks: $|0\rangle$ measuring X — direct method $P_X(\pm)=\tfrac12,\tfrac12$, difference 0; shortcut $X|0\rangle=|1\rangle,\ \langle0|1\rangle=0$ ✓. $|+\rangle$ measuring X — an eigenvector, both methods give 1 ✓.

5.2 All three, fully derived

Write $c=\cos\frac\theta2,\ s=\sin\frac\theta2$ (so $\alpha=c,\ \beta=e^{i\phi}s$), with the double-angle formulas $\sin\theta=2sc,\ \cos\theta=c^2-s^2$ at hand.

$\langle Z\rangle$: $Z|\psi\rangle=\binom{\alpha}{-\beta}$,

$\langle X\rangle$: $X|\psi\rangle=\binom{\beta}{\alpha}$,

$\langle Y\rangle$: $Y|\psi\rangle=\binom{-i\beta}{i\alpha}$,

(All three Euler identities filed in §1.3 are hereby cashed in.)

5.3 The theorem and the causal chain

The right side is exactly the Cartesian coordinates of the point on the unit sphere at polar angle $\theta$ and azimuth $\phi$ — every coordinate in the last column of the §4.2 table is a direct output of these three computations. Spot checks: $|+\rangle(\theta=\tfrac\pi2,\phi=0)\to(1,0,0)$ ✓; $|{+i}\rangle(\phi=\tfrac\pi2)\to(0,1,0)$ ✓; $|0\rangle(\theta=0)\to(0,0,1)$ ✓.

Keep the causal chain in order: definition — the three expectation values are the average readings of three kinds of measurement (each estimable from frequencies); theorem — after substituting the Bloch parameterization, they happen to equal the three coordinates. “Average reading = coordinate” is a beautiful thing that is computed, not decreed. Replay the three rounds: Round 1 estimates $\langle Z\rangle$ (latitude/z), Round 2 estimates $\langle X\rangle$, Round 3 estimates $\langle Y\rangle$ — three numbers pin the arrow to one point on the sphere. The procedure’s proper name: quantum state tomography (the name borrowed from medical CT — both reconstruct a whole from projections along several directions).

5.4 Statistical reality and optimal strategy

• The $1/\sqrt N$ law: frequencies only estimate probabilities; $N$ samples carry a standard error $\sim\tfrac{1}{2\sqrt N}$, so precision $\varepsilon$ costs $N\sim1/\varepsilon^2$ copies. The same mathematics as polling’s square-root statistics and the Chernoff bound of computational complexity ($\Pr[|\hat p-p|>\varepsilon]\le2e^{-2N\varepsilon^2}$, the engine of BPP error amplification).
• Finite data yields only a confidence region: a patch on the sphere, never a point; “knowing $|\psi\rangle$ exactly” requires the fiction of infinitely many copies.
• Adaptive tomography: choose the next basis from results so far (the end of §4.4, “if X is also 50/50, switch to Y”, is its prototype); provably better than a fixed schedule.
• Mutually unbiased bases (MUB): two orthonormal bases are mutually unbiased ⟺ every cross inner product satisfies $|\langle e_i|f_j\rangle|^2=\tfrac12$ — a state definite in one basis measures 50/50 in the other, zero informational overlap. The eigenbases of X, Y, Z are pairwise unbiased (verify one: $|\langle0|+\rangle|^2=\tfrac12$ ✓), and a two-dimensional space admits at most exactly 3 such bases — XYZ fills the quota, which is why it is the non-adaptive optimum. The two faces: the same property is, in tomography, the virtue that every measurement family squeezes out entirely fresh information; in BB84, the weapon that guarantees an eavesdropper who picks the wrong basis gets pure noise.

6. Deeper morals, and closing the single-qubit file

6.1 Amplitudes cannot be read

$\alpha,\beta$ are continuous complex parameters and physically real — they drive every gate; seen from the writing end, one qubit stores unlimited precision. But the reading end has one narrow door: any single qubit, walking out through measurement, surrenders at most one classical bit and dies as a superposition on the spot. Tomography does read out $\alpha,\beta$, but that is statistics bought by burning a whole batch of copies, not reading from a single one. (”$n$ qubits surrender at most $n$ classical bits” has a formal theorem behind it: the Holevo bound. Unlimited writing, restricted reading — a law, not an engineering defect.)

6.2 Quantum algorithms = the art of arranging interference

You cannot read intermediate results (reading = destroying), so the computation must proceed silently inside the superposition, using gates to make wrong answers’ amplitudes cancel and the right answer’s reinforce, so that the single final measurement emits the desired answer with high probability. The entire art of quantum algorithms: make the one bit you are allowed to take be exactly the bit you want. Deutsch, Grover, and Shor later replay the same pattern again and again: never try to read all the values; let interference float some global property up into that one readable bit.

6.3 The single-qubit pipeline, in one line

Every segment now holds its license: preparation ($|0\rangle$ plus a suitable $U$), the unitary stretch (the complete roster of gates), measurement (picking a basis = padding a unitary before measuring Z), output (the narrow door). A single qubit’s whole life — every legal move — fits on this line, and the line is also one full round of BB84 (Alice prepares, the channel, Bob picks a basis and measures, one bit comes out). Later in this series comes what happens when qubits are no longer alone — the exponential riches of $2^n$ dimensions and entanglement live there, and chapter 2 has already touched that door handle.

6.4 The eavesdropper = a tomographer with no supply

Assemble the three chapters and look at Eve’s predicament: she gets one photon per signal — no tomography (supply of one), no copying (chapter 3), and measuring in the wrong basis yields a random result while disturbing the state and leaving traces (MUB + collapse). Three pillars — cannot read straight (non-commutation), cannot copy (no-cloning), destroys the evidence (collapse) — compose the full picture: a quantum state carries continuous information, but a single copy surrenders only one bit, with self-destruction included. The security of QKD is this picture, engineered.

7. Misconceptions and traps

1. “CNOT copies, so the no-cloning theorem has a loophole” — the theorem targets a universal cloner; simultaneously clonable ⟺ pairwise orthogonal ($x=x^2$). CNOT is a legal “orthogonal-state copier”, and precisely the essence of classical copying (§3.4).
2. “I want to copy the $|1\rangle$ I know I have” — if you know it, re-prepare it; that is not cloning. The cloning problem = unknown + single copy (§3.2).
3. ”$\langle\psi|A|\psi\rangle$ is what happens during measurement” — no; it is a bookkeeping shortcut for the average reading. Measurement itself = decompose + roll dice + collapse (§5.1).
4. ”$+1$ is $|+\rangle$” — a reading (eigenvalue, a number) ≠ a state (eigenvector); the same $+1$ on the Z instrument labels $|0\rangle$ (§4.0).
5. “The Y instrument is more precise” — the three axes are equivalent, differing only in orientation; three rounds are necessary and sufficient because Bloch coordinates number exactly three (§4.6).
6. “Measuring Z should reveal some phase” — $\phi$ is annihilated exactly inside $\beta^*\beta$; Z is blind to longitude (§4.3).
7. “Measure enough times and you pin the state to a point” — finite data gives only a confidence region, shrinking as $1/\sqrt N$ (§5.4).
8. ”$e^{i\pi}$ is e multiplied by itself π times” — an imaginary power is a series extension, and its geometric meaning is a 180° turn (§1.1).
9. Notation trap: $|\langle0|\psi\rangle|^2$ has three vertical bars; broken rendering that drops the middle one produces the illegal $\vert\langle0\rangle\psi\vert^2$ (§4.0).
10. “Entanglement is an extra assumption” — product states are not closed under addition; superposition forces entanglement into existence (§2.4).

8. Self-test (answers included)

Problems

1. Verify $e^{i\phi}\cdot e^{-i\phi}=1$ two ways (exponent law; multiply the Euler expansions). What is $e^{i\pi/2}$?
2. Prove that $\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)$ is not a product state.
3. Using $x=x^2$, argue that no machine clones both $|0\rangle$ and $|+\rangle$.
4. Compute $\mathrm{CNOT}\big(|-\rangle\otimes|0\rangle\big)$ and interpret the result.
5. Write down the three expectation values of $|{-i}\rangle$ and check them against the landmark table.
6. To measure the observable Y, what do you pad before the Z instrument? Verify it.
7. If Round 1 statistics give $P(0)=0.8$, what is $\theta$? How far has the case progressed?
8. Why does BB84 deliberately choose two mutually non-orthogonal bases?

Answers

1. Exponent law: $e^{i\phi-i\phi}=e^0=1$; expansion: $(\cos\phi+i\sin\phi)(\cos\phi-i\sin\phi)=\cos^2\phi+\sin^2\phi=1$. $e^{i\pi/2}=\cos\tfrac\pi2+i\sin\tfrac\pi2=i$.
2. Suppose $(a,b)^T\otimes(c,d)^T=(ac,ad,bc,bd)^T$ equals it; then $ac=bd=\tfrac{1}{\sqrt2}$ and $ad=0$; $ad=0$ forces $a=0$ or $d=0$, contradicting $ac\ne0$ or $bd\ne0$ respectively.
3. $x=\langle0|+\rangle=\tfrac{1}{\sqrt2}$, while cloning requires $x=x^2=\tfrac12$ — unequal, so no such machine exists.
4. $|-\rangle\otimes|0\rangle=\tfrac{1}{\sqrt2}(1,0,-1,0)^T$; row-by-row through CNOT gives $\tfrac{1}{\sqrt2}(1,0,0,-1)^T=\tfrac{1}{\sqrt2}(|00\rangle-|11\rangle)$ — another Bell state: “copying” $|-\rangle$ fails and produces entanglement, consistent with the theorem.
5. $\theta=\tfrac\pi2,\phi=\tfrac{3\pi}2$: $\langle X\rangle=\cos\tfrac{3\pi}2=0$, $\langle Y\rangle=\sin\tfrac{3\pi}2=-1$, $\langle Z\rangle=0$ → $(0,-1,0)$ ✓, matching the table (negative end of the y axis).
6. Pad $S^\dagger$ then H: $S^\dagger|{+i}\rangle=|+\rangle\xrightarrow{H}|0\rangle$, $S^\dagger|{-i}\rangle=|-\rangle\xrightarrow{H}|1\rangle$ ✓.
7. $\cos^2\tfrac\theta2=0.8\Rightarrow\cos\theta=2(0.8)-1=0.6\Rightarrow\theta\approx53.1°$. Latitude fixed; the suspects are the entire circle at that latitude; Rounds 2/3 are still needed for the longitude.
8. A triple lock: non-orthogonal ⟹ out of any copier’s reach ($x=x^2$ fails); mutually unbiased ⟹ the wrong basis yields pure 50/50 noise; collapse ⟹ the disturbance leaves traces detectable by sample comparison.

9. Symbol quick reference (new in these notes)

The whole article in one sentence: Euler’s formula makes phase a turning angle, the tensor product multiplies dimensions and forces entanglement into existence, and $x=x^2$ locks clonability inside orthogonal states — so an unknown qubit can only be cracked by re-running its preparation and doing statistics across three rounds of Z, X, and Y; the three average readings turn out to be exactly the three coordinates of the Bloch arrow, and the three shackles, one copy surrenders one bit, copying yields entanglement, the wrong basis yields noise, are at once the reason tomography must burn copies and the lock that quantum cryptography closes on the eavesdropper.