Visual walkthrough — Quantum computing hardware basics

We build on Classical Bits vs Qubits, Superposition and Entanglement, and land on the Bloch Sphere. This is the visual companion to the parent note.

---

Step 1 — Start with the raw state: two complex numbers

WHAT. A qubit's state is written $|\psi ⟩=\alpha |0⟩+\beta |1⟩.$

Let me read every symbol out loud before we use it:

• $|0⟩$ and $|1⟩$ — the two "pure" answers, like heads and tails. The bracket-ish notation $|\cdot ⟩$ (a "ket") is just a fancy box that says "this is a quantum state". Nothing more.
• $\alpha$ (alpha) — a number that says how much of heads is in the mix.
• $\beta$ (beta) — a number that says how much of tails is in the mix.
• $|\psi ⟩$ (psi) — the whole mixed state; the coin while it is still spinning.

WHY complex numbers and not plain fractions? Because $\alpha ,\beta$ are complex numbers — each carries a size and a direction (an angle), like an arrow in a flat plane. We need the direction because quantum waves must be able to cancel each other (interference). Plain positive fractions can only add up; they can never subtract to zero.

PICTURE. Each of $\alpha$ and $\beta$ is an arrow in its own little flat plane. Two arrows = four real dials to turn (each arrow needs a horizontal amount and a vertical amount).

So right now our state costs us 4 real numbers. That's too many — a sphere needs only 2. The next steps are a demolition job: we knock out the numbers you can never observe.

---

Step 2 — Knock out dial #1: normalization

WHAT. We impose the rule $|\alpha {|}^2+|\beta {|}^2=1.$

Term by term:

• $|\alpha {|}^2$ — the length of the $\alpha$ arrow, squared. This is the probability the coin lands heads.
• $|\beta {|}^2$ — same for tails.
• $=1$ — the two probabilities must sum to certainty. The coin lands somewhere.

WHY. Probabilities have to add to $1$; a coin cannot land with $70\%$ heads and $70\%$ tails. This single equation ties the two arrow-lengths together: once you pick one length, the other is forced.

PICTURE. Think of the two lengths $|\alpha |$ and $|\beta |$ as the two sides of a right triangle whose hypotenuse is fixed at $1$. Pythagoras says $|\alpha {|}^2+|\beta {|}^2=1$. Slide one side up and the other must shrink — they are not free.

---

Step 3 — Knock out dial #2: global phase is invisible

WHAT. Multiply the whole state by a spin $e^{i\gamma }$: $|\psi ⟩⟶e^{i\gamma }|\psi ⟩.$

• $e^{i\gamma }$ — an arrow of length exactly $1$ pointing at angle $\gamma$. Multiplying by it rotates both $\alpha$ and $\beta$ arrows by the same angle $\gamma$, without changing any length.

WHY it changes nothing you can measure. Measurement only ever asks for the squared lengths $|\alpha {|}^2,|\beta {|}^2$. Rotating an arrow does not change its length, so $|e^{i\gamma }\alpha {|}^2=|\alpha {|}^2$. Every physical prediction is identical. A dial that changes nothing observable is a dial we are allowed to freeze.

PICTURE. Both amplitude-arrows swing together like clock hands glued to one shaft. The shape between them is untouched — only what matters is the relative angle between $\alpha$ and $\beta$, not their shared starting angle.

---

Step 4 — The two survivors, and the angle $\theta$

WHAT. After Steps 2–3, only two real numbers remain. One controls the split between heads and tails; call the split-controller $\theta$. We define the two lengths as $|\alpha |=\cos \frac{\theta }{2},|\beta |=\sin \frac{\theta }{2}.$

Why is this legal? Because Step 2 forced $|\alpha {|}^2+|\beta {|}^2=1$, and the identity ${\cos }^2\frac{\theta }{2}+{\sin }^2\frac{\theta }{2}=1$ is automatically true for any $\theta$. So writing the lengths as a cosine and sine is not an extra assumption — it is the most natural way to satisfy Pythagoras with a single knob $\theta$.

• $\cos \frac{\theta }{2}$ — the heads-length; biggest ($=1$) when $\theta =0$.
• $\sin \frac{\theta }{2}$ — the tails-length; biggest ($=1$) when $\theta =\pi$.

WHY cosine/sine at all? Because we want one number that smoothly slides all the probability from "all heads" to "all tails" and back — exactly what a point moving around a quarter-circle does. Cosine and sine are the coordinates of a point on a circle. That is the tool that answers "how do I trade two Pythagoras-linked lengths using a single angle?"

PICTURE. As $\theta$ sweeps, the point rides a circle; its horizontal shadow is $\cos \frac{\theta }{2}$ (heads), its vertical shadow is $\sin \frac{\theta }{2}$ (tails).

---

Step 5 — The other survivor: the relative phase $\varphi$

WHAT. The full surviving state is $|\psi ⟩=\cos \frac{\theta }{2}|0⟩+e^{i\varphi }\sin \frac{\theta }{2}|1⟩.$

• $\varphi$ (phi) — the relative phase: the angle by which the tails-arrow is rotated relative to the (now real) heads-arrow. This is the dial that survived from Step 3.
• $e^{i\varphi }$ — spins only the tails part.

WHY it's the second real dial. We used the global-phase freedom to make $\alpha$ real (zero angle). But $\beta$'s angle relative to $\alpha$ cannot be removed — it is physical. So the two survivors are:

1. $\theta$ — how the probability splits (a tilt),
2. $\varphi$ — the relative twist (a turn around).

PICTURE. $\varphi$ does not change any length, so it does not change heads-vs-tails odds right now. Instead it rotates the state around an axis. Two dials — a tilt and a turn — is exactly what you need to name any spot on a globe: $\theta$ = latitude-like, $\varphi$ = longitude.

---

Step 6 — Why a SPHERE, and why the half-angle

WHAT. Map $(\theta ,\varphi )$ to a point on a unit ball using ordinary spherical coordinates: $\vec{r}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta ).$

• $\theta$ is the polar angle measured down from the north pole: it ranges $0\to \pi$.
• $\varphi$ is the azimuth (spin around the equator): it ranges $0\to 2\pi$.

WHY $\theta /2$ inside the state but $\theta$ on the sphere? Watch the endpoints:

• $\theta =0$: state $=\cos 0|0⟩+\dots =|0⟩$. On the ball, $\vec{r}=(0,0,1)$ = north pole. ✓
• $\theta =\pi$: state $=\cos \frac{\pi }{2}|0⟩+\sin \frac{\pi }{2}|1⟩=|1⟩$. On the ball, $\vec{r}=(0,0,-1)$ = south pole. ✓

Here is the beautiful part. In the state, $|0⟩$ and $|1⟩$ are orthogonal (a $90°$ relationship in the maths). But on the ball they sit at opposite poles — $180°$ apart. The factor of $\frac{1}{2}$ is exactly the gear-ratio that turns the state's $90°$ into the globe's $180°$. Halving the angle inside doubles the spread outside, so orthogonal states become physical opposites.

PICTURE. North pole $=|0⟩$, south pole $=|1⟩$, equator = the even 50/50 superpositions (like Hadamard's output from the parent note), and going around the equator changes $\varphi$ — pure relative phase.

Recall

Why $\theta /2$ and not $\theta$? ::: So that the two poles ($\theta =0$ and $\theta =\pi$) give $|0⟩$ and $|1⟩$ exactly, and orthogonal states land $180°$ apart on the ball. How many real dials does a qubit truly have? ::: Two — $\theta$ and $\varphi$ — after killing normalization and global phase.

---

Step 7 — Edge and degenerate cases (never leave a gap)

The formula must survive its corners. Let's check each.

Case A — The poles ($\theta =0$ or $\theta =\pi$). At a pole, $\sin \frac{\theta }{2}$ or $\cos \frac{\theta }{2}$ is $0$, so the $e^{i\varphi }$ term is multiplied by zero. The twist $\varphi$ does nothing — every longitude meets at the same pole. This matches geography: the North Pole has no meaningful longitude. So $\varphi$ is undefined exactly at $|0⟩$ and $|1⟩$, and that's fine — there's nothing to twist.

Case B — The equator ($\theta =\pi /2$). Then $\cos \frac{\pi }{4}=\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$: a perfect 50/50 coin. Now $\varphi$ matters most — it slides you around all the "equally heads-and-tails but different phase" states, e.g. $\frac{|0⟩+|1⟩}{\sqrt{2}}$ at $\varphi =0$ versus $\frac{|0⟩-|1⟩}{\sqrt{2}}$ at $\varphi =\pi$. These are physically different (opposite equator points) yet have identical measurement odds in the heads/tails basis.

Case C — What if we had NOT halved the angle? Using $\theta$ directly, $\theta$ and $\theta +2\pi$ would give the same state, but you'd only reach the northern hemisphere by $\theta \in [0,\pi ]$ — the poles wouldn't be $|0⟩,|1⟩$ and orthogonal states would be $90°$ apart, cramming everything onto half a sphere. The half-angle is what makes the map onto the whole sphere, exactly once.

Case D — Degenerate "classical" limit. If phase were forbidden (drop $e^{i\varphi }$, force real $\ge 0$ amplitudes), you'd be stuck on one meridian — a semicircle from pole to pole. That semicircle is the closest a "phaseless" theory gets: it can mix probabilities but never interfere. The full sphere is the extra room quantum mechanics buys with phase.

---

The one-picture summary

Here is the whole demolition on one page: 4 real dials → (−1 normalization) → (−1 global phase) → 2 dials → a sphere.

Recall Feynman retelling of the whole walkthrough

We wanted to describe a spinning coin that's part heads, part tails. We wrote it as two arrows: one for "how much heads" ($\alpha$), one for "how much tails" ($\beta$). Each arrow needs a length and a direction, so that's four numbers to describe one coin — way too many.

Then we cleaned house. First: the two lengths can't both be big, because the chance of heads plus the chance of tails must equal 1 — that's Pythagoras with a hypotenuse of 1, and it fuses two numbers into one. Second: if you spin both arrows together, nothing you can ever measure changes — so we froze that shared spin. That killed a second number.

Two numbers left. One says how the coin is tilted between heads and tails (call it $\theta$); the other says how the tails-arrow is twisted relative to heads (call it $\varphi$). A tilt and a twist — that's exactly latitude and longitude. So every possible qubit is one dot on a globe: $|0⟩$ at the top, $|1⟩$ at the bottom, and the fifty-fifty spins around the equator. The only quirk is we tilt by half the angle inside the formula, which is the gear that puts the top and bottom exactly opposite each other. That globe is the Bloch sphere.