Math Prep

Solve this BEFORE lesson.md. Day 7's tools (plot_bloch_multivector, plot_state_qsphere, plot_state_city, histograms) each draw a specific mathematical object. Know the object, and every plot question becomes a computation.

Requires: Day 1 + Day 3 math prep. Video back-up: freeCodeCamp course (Bloch sphere section); Watrous Lesson 09 (Bloch ball); notes in resource-artifactory/notes/video-notes/.

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1. The Bloch sphere — every 1-qubit state is a point on a ball

Any 1-qubit pure state can be written (global phase fixed away) as

$(\theta, \phi)$ are spherical coordinates: θ from the north pole, φ around the equator. Equivalently, the Bloch vector

Landmarks (memorize as pictures):

Orthogonal states are antipodal (NOT perpendicular) on the sphere — Hilbert-space 90° becomes 180° on the sphere because of the half-angle.

2. Gates are rotations of the sphere

• $X, Y, Z$: 180° rotations about their own axes. ($X$ swaps poles — hence NOT.)
• $R_x(\theta), R_y(\theta), R_z(\theta)$: rotations by exactly θ about x/y/z. (Matrix half-angle ↔ full sphere angle — Day 2.)
• $H$: 180° rotation about the diagonal axis $(x{+}z)/\sqrt2$ — swaps x-axis ↔ z-axis, so $|0\rangle \leftrightarrow |+\rangle$, $|1\rangle \leftrightarrow |-\rangle$.
• $S$: 90° rotation about z: $|+\rangle \to |{+i}\rangle \to |-\rangle \to |{-i}\rangle \to |+\rangle$. $T$: 45° about z.
• Phase gates never move the poles (only rotate about z) — that's "Z-family gates don't change measurement probabilities" drawn geometrically.

Mixed states live INSIDE the ball: $\rho = \tfrac12(I + \vec r\cdot\vec\sigma)$ with $|\vec r| \le 1$; pure ⇔ surface, maximally mixed $I/2$ ⇔ center ($\vec r = 0$). Each qubit of a Bell pair plots at the center — that's how plot_bloch_multivector betrays entanglement (it shows per-qubit reduced states, and can never show the correlations themselves).

3. The Q-sphere — one ball for the WHOLE register

plot_state_qsphere is a different object: one node per basis state $|x\rangle$ of all $n$ qubits.

• Node position: latitude by Hamming weight of $x$ ($|0\cdots0\rangle$ north pole, $|1\cdots1\rangle$ south pole).
• Node size ∝ $|\alpha_x|^2$ (probability).
• Node color = phase $\arg(\alpha_x)$: 0 and π sit on opposite sides of the color wheel.

So Bell $|\Phi^+\rangle$: two equal nodes ($|00\rangle$ north, $|11\rangle$ south), same color. $\tfrac{1}{\sqrt2}(|00\rangle - |11\rangle)$: same two nodes, opposite colors — the plot that distinguishes what histograms can't.

4. City / matrix plots — drawing ρ itself

plot_state_city draws the density matrix entries $\rho_{jk}$ as bars (real and imaginary parts separately).

• Diagonal bars $\rho_{xx}$ = probabilities.
• Off-diagonal bars = coherences: present for superpositions, ZERO for classical mixtures — the visual pure-vs-mixed test (Day 3 §5).
• Bell state: 4 corner bars of height 0.5 at positions (00,00), (00,11), (11,00), (11,11).

5. Histograms — drawing the sampled distribution

plot_histogram(counts) is Day 5's $\hat p_x$ as bars; plot_distribution normalizes to probabilities. Phases are invisible here (Day 5 §1). GHZ and Bell histograms look identical in shape (two bars) — only the labels ('000'/'111' vs '00'/'11') differ.

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PROBLEMS

P1. Give $(\theta, \phi)$ and the Bloch vector for $|\psi\rangle = \cos\frac{\pi}{8}|0\rangle + e^{i\pi/2}\sin\frac{\pi}{8}|1\rangle$. Where does it sit on the sphere?

P2. Start at $|0\rangle$. Apply H, then S, then H. Track the Bloch point through each step and name the final state.

P3. Is ry(π/2) on $|0\rangle$ the same state as h on $|0\rangle$? Is ry(π/2) the same gate as h? Check on both $|0\rangle$ and $|1\rangle$.

P4. A Bloch-multivector plot of a 2-qubit state shows BOTH qubit arrows with zero length. Give two different global states consistent with the plot — one entangled pure state, one unentangled (mixed) state. What does zero-length force if the global state is known to be pure?

P5. Q-sphere shows nodes at $|00\rangle$ and $|11\rangle$, equal size, colors 180° apart on the wheel. Write the state. What would its counts histogram look like vs $|\Phi^+\rangle$'s?

P6. For $|\psi\rangle = \tfrac{\sqrt3}{2}|0\rangle + \tfrac12|1\rangle$: compute the Bloch vector and the density-matrix (city-plot) bars.

P7. $T$ is applied 8 times to $|+\rangle$. Describe the trajectory on the Bloch sphere and the final state.

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ANSWERS

A1. $\theta = \pi/4$, $\phi = \pi/2$. Bloch vector $(\sin\frac{\pi}{4}\cos\frac{\pi}{2},\ \sin\frac{\pi}{4}\sin\frac{\pi}{2},\ \cos\frac{\pi}{4}) = (0, \tfrac{1}{\sqrt2}, \tfrac{1}{\sqrt2})$ — in the y–z plane, halfway between +y and +z (45° down from the north pole toward +y).

A2. $|0\rangle \xrightarrow{H} |+\rangle$ (+x) $\xrightarrow{S} |{+i}\rangle$ (+y) $\xrightarrow{H}$: H maps $(x,y,z)\to(z,-y,x)$, so $(0,1,0) \to (0,-1,0)$ = $|{-i}\rangle = \tfrac{1}{\sqrt2}(|0\rangle - i|1\rangle)$ (up to global phase).

A3. Same state on $|0\rangle$: $R_y(\pi/2)|0\rangle = \cos\frac{\pi}{4}|0\rangle + \sin\frac{\pi}{4}|1\rangle = |+\rangle = H|0\rangle$. Different gates: on $|1\rangle$, $R_y(\pi/2)|1\rangle = -\tfrac{1}{\sqrt2}|0\rangle + \tfrac{1}{\sqrt2}|1\rangle = -|-\rangle$ while $H|1\rangle = +|-\rangle$. Same ray but opposite sign — and that sign is relative once these gates act inside superpositions or controlled contexts, so the operators are genuinely different ($H = H^\dagger$ but $R_y(\pi/2)^\dagger = R_y(-\pi/2) \ne R_y(\pi/2)$).

A4. Zero arrows = both reduced states are $I/2$. Entangled pure option: any Bell state. Unentangled option: the product of two maximally-mixed qubits $\tfrac{I}{2}\otimes\tfrac{I}{2}$ (two independently randomized qubits). If the global state is pure, zero-length arrows force entanglement — a pure product state always shows two unit-length arrows.

A5. $\tfrac{1}{\sqrt2}(|00\rangle - |11\rangle) = |\Phi^-\rangle$ (phase π on the $|11\rangle$ node). Histogram: identical to $|\Phi^+\rangle$'s — two equal bars at '00' and '11'. Q-sphere sees phase; histograms never do.

A6. $\cos\frac{\theta}{2} = \tfrac{\sqrt3}{2} \Rightarrow \theta = \tfrac{\pi}{3}$, $\phi = 0$. Bloch vector $(\sin\frac{\pi}{3}, 0, \cos\frac{\pi}{3}) = (\tfrac{\sqrt3}{2}, 0, \tfrac12)$. $\rho = \begin{pmatrix}3/4 & \sqrt3/4\\ \sqrt3/4 & 1/4\end{pmatrix}$: city bars 0.75 and 0.25 on the diagonal, $\sqrt3/4 \approx 0.433$ on both off-diagonals, all real (imaginary city empty).

A7. Each T rotates 45° about z along the equator: $|+\rangle$ visits 8 equally-spaced equatorial points and returns to $|+\rangle$, since $T^8 = (T^4)^2 = Z^2 = I$.

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Self-check: you can place any $\alpha|0\rangle + e^{i\phi}\beta|1\rangle$ on the sphere, rotate it by any named gate mentally, decode q-sphere size/color, and explain why per-qubit Bloch arrows shrink for entangled states.