Visual walkthrough — Quantum computing hardware basics

6.5.13 · D2 · Hardware › Advanced & Emerging Architectures › Quantum computing hardware basics

Hum Classical Bits vs Qubits, Superposition and Entanglement par build karte hain, aur Bloch Sphere par land karte hain. Yeh parent note ka visual companion hai.

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Step 1 — Raw state se shuru karo: do complex numbers

KYA HAI. Ek qubit ki state likhi jaati hai $|\psi ⟩=\alpha |0⟩+\beta |1⟩.$

Har symbol ko use karne se pehle mein zor se padhunga:

• $|0⟩$ aur $|1⟩$ — do "pure" answers, jaise heads aur tails. Bracket-jaisi notation $|\cdot ⟩$ (ek "ket") bas ek fancy box hai jo kehta hai "yeh ek quantum state hai". Kuch nahi zyada.
• $\alpha$ (alpha) — ek number jo batata hai ki mix mein kitna heads hai.
• $\beta$ (beta) — ek number jo batata hai ki kitna tails hai.
• $|\psi ⟩$ (psi) — poori mixed state; coin jabki woh abhi bhi spin kar raha hai.

Complex numbers kyun, plain fractions kyun nahi? Kyunki $\alpha ,\beta$ complex numbers hain — har ek ek flat plane mein ek arrow ki tarah size aur direction (ek angle) carry karta hai. Direction ki zaroorat hai kyunki quantum waves ko ek doosre ko cancel karne ki zaroorat padti hai (interference). Plain positive fractions sirf add ho sakte hain; woh kabhi subtract hokar zero nahi ban sakte.

PICTURE. $\alpha$ aur $\beta$ mein se har ek apne chhote flat plane mein ek arrow hai. Do arrows = char real dials (har arrow ko ek horizontal amount aur ek vertical amount chahiye).

Toh abhi hamare state ko 4 real numbers chahiye hain. Yeh bahut zyada hain — sphere ko sirf 2 chahiye. Agle steps ek demolition job hain: hum woh numbers hatate hain jo aap kabhi observe nahi kar sakte.

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Step 2 — Dial #1 hatao: normalization

KYA HAI. Hum yeh rule lagate hain $|\alpha {|}^2+|\beta {|}^2=1.$

Term by term:

• $|\alpha {|}^2$ — $\alpha$ arrow ki length, squared. Yeh woh probability hai ki coin heads aayega.
• $|\beta {|}^2$ — tails ke liye wahi cheez.
• $=1$ — dono probabilities ko certainty mein add hona chahiye. Coin kahin na kahin land karta hai.

KYUN. Probabilities ko $1$ mein add hona chahiye; ek coin $70\%$ heads aur $70\%$ tails ke saath land nahi kar sakta. Yeh akela equation dono arrow-lengths ko baandhta hai: ek baar jab aap ek length choose karte hain, doosri force ho jaati hai.

PICTURE. $|\alpha |$ aur $|\beta |$ lengths ko ek right triangle ki do sides samjho jiska hypotenuse $1$ par fixed hai. Pythagoras kehta hai $|\alpha {|}^2+|\beta {|}^2=1$. Ek side upar slide karo aur doosri zaroor choti ho jaayegi — woh free nahi hain.

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Step 3 — Dial #2 hatao: global phase invisible hai

KYA HAI. Poori state ko ek spin $e^{i\gamma }$ se multiply karo: $|\psi ⟩⟶e^{i\gamma }|\psi ⟩.$

• $e^{i\gamma }$ — exactly $1$ length ka arrow jo angle $\gamma$ par point karta hai. Isse multiply karna dono $\alpha$ aur $\beta$ arrows ko ek hi angle $\gamma$ se rotate karta hai, bina koi length badlaaye.

KYUN yeh kuch measure nahi badalta. Measurement sirf kabhi squared lengths $|\alpha {|}^2,|\beta {|}^2$ maangti hai. Ek arrow ko rotate karne se uski length nahi badlati, isliye $|e^{i\gamma }\alpha {|}^2=|\alpha {|}^2$. Har physical prediction identical hai. Ek dial jo kuch bhi observable nahi badlata woh dial hai jise hum freeze karne ki permission rakhte hain.

PICTURE. Dono amplitude-arrows ek saath aise swing karte hain jaise clock hands ek shaft par glued hon. Un ke beech ka shape untouched hai — sirf jo maayane rakhta hai woh hai $\alpha$ aur $\beta$ ke beech ka relative angle, unka shared starting angle nahi.

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Step 4 — Do bachne waale, aur angle $\theta$

KYA HAI. Steps 2–3 ke baad, sirf do real numbers bachte hain. Ek heads aur tails ke beech ke split ko control karta hai; split-controller ko $\theta$ kaho. Hum dono lengths ko yun define karte hain $|\alpha |=\cos \frac{\theta }{2},|\beta |=\sin \frac{\theta }{2}.$

Yeh legal kyun hai? Kyunki Step 2 ne $|\alpha {|}^2+|\beta {|}^2=1$ force kiya, aur identity ${\cos }^2\frac{\theta }{2}+{\sin }^2\frac{\theta }{2}=1$ kisi bhi $\theta$ ke liye automatically true hai. Toh lengths ko cosine aur sine ke roop mein likhna koi extra assumption nahi hai — yeh Pythagoras ko ek single knob $\theta$ se satisfy karne ka sabse natural tarika hai.

• $\cos \frac{\theta }{2}$ — heads-length; sabse bada ($=1$) jab $\theta =0$.
• $\sin \frac{\theta }{2}$ — tails-length; sabse bada ($=1$) jab $\theta =\pi$.

Cosine/sine bilkul kyun? Kyunki hum ek akela number chahte hain jo smoothly saari probability ko "all heads" se "all tails" aur wapas slide kare — exactly wahi karta hai jo ek point quarter-circle ke around move karte waqt karta hai. Cosine aur sine ek circle par ek point ke coordinates hain hi. Woh woh tool hai jo is sawaal ka jawab deta hai: "Main do Pythagoras-linked lengths ko ek angle se kaise trade karun?"

PICTURE. Jaise $\theta$ sweep karta hai, point ek circle par ride karta hai; uska horizontal shadow $\cos \frac{\theta }{2}$ (heads) hai, uska vertical shadow $\sin \frac{\theta }{2}$ (tails) hai.

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Step 5 — Doosra bachne wala: relative phase $\varphi$

KYA HAI. Poori bachne wali state hai $|\psi ⟩=\cos \frac{\theta }{2}|0⟩+e^{i\varphi }\sin \frac{\theta }{2}|1⟩.$

• $\varphi$ (phi) — relative phase: woh angle jitna tails-arrow ko (ab real) heads-arrow ke relative mein rotate kiya gaya hai. Yeh woh dial hai jo Step 3 se bacha.
• $e^{i\varphi }$ — sirf tails part ko spin karta hai.

Yeh doosra real dial kyun hai. Hum ne global-phase freedom ka use karke $\alpha$ ko real banaya (zero angle). Lekin $\beta$ ka angle $\alpha$ ke relative mein nahi hataya ja sakta — yeh physical hai. Toh do survivors hain:

1. $\theta$ — probability kaise split hoti hai (ek tilt),
2. $\varphi$ — relative twist (ek turn around).

PICTURE. $\varphi$ koi length nahi badlata, isliye yeh abhi heads-vs-tails odds nahi badlata. Balki yeh state ko ek axis ke around rotate karta hai. Do dials — ek tilt aur ek turn — exactly wahi hai jo aapko ek globe par koi bhi spot name karne ke liye chahiye: $\theta$ = latitude-jaisa, $\varphi$ = longitude.

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Step 6 — SPHERE kyun, aur half-angle kyun

KYA HAI. $(\theta ,\varphi )$ ko ordinary spherical coordinates use karke unit ball par ek point par map karo: $\vec{r}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta ).$

• $\theta$ polar angle hai jo north pole se neeche measure kiya jaata hai: yeh $0\to \pi$ range karta hai.
• $\varphi$ azimuth hai (equator ke around spin): yeh $0\to 2\pi$ range karta hai.

State ke andar $\theta /2$ kyun lekin sphere par $\theta$? Endpoints dekho:

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

Yahan khoobsoorat part hai. State mein, $|0⟩$ aur $|1⟩$ orthogonal hain (maths mein $90°$ ka relationship). Lekin ball par woh opposite poles par baithte hain — $180°$ apart. $\frac{1}{2}$ ka factor exactly woh gear-ratio hai jo state ke $90°$ ko globe ke $180°$ mein turn karta hai. Andar angle ko halva karna baahr spread ko double karta hai, isliye orthogonal states physical opposites ban jaati hain.

PICTURE. North pole $=|0⟩$, south pole $=|1⟩$, equator = even 50/50 superpositions (jaise parent note mein Hadamard ka output), aur equator ke around jaana $\varphi$ badlata hai — pure relative phase.

Recall

$\theta /2$ kyun aur $\theta$ kyun nahi? ::: Taaki dono poles ($\theta =0$ aur $\theta =\pi$) exactly $|0⟩$ aur $|1⟩$ dein, aur orthogonal states ball par $180°$ apart land karein. Ek qubit mein actually kitne real dials hote hain? ::: Do — $\theta$ aur $\varphi$ — normalization aur global phase ko khatam karne ke baad.

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Step 7 — Edge aur degenerate cases (koi gap mat chhodna)

Formula ko apne corners par survive karna chahiye. Har ek check karte hain.

Case A — Poles ($\theta =0$ ya $\theta =\pi$). Ek pole par, $\sin \frac{\theta }{2}$ ya $\cos \frac{\theta }{2}$ $0$ hai, isliye $e^{i\varphi }$ term zero se multiply ho jaata hai. Twist $\varphi$ kuch nahi karta — har longitude ek hi pole par milta hai. Yeh geography se match karta hai: North Pole ka koi meaningful longitude nahi hota. Toh $\varphi$ exactly $|0⟩$ aur $|1⟩$ par undefined hai, aur yeh theek hai — twist karne ke liye kuch nahi hai.

Case B — Equator ($\theta =\pi /2$). Tab $\cos \frac{\pi }{4}=\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$: ek perfect 50/50 coin. Ab $\varphi$ sabse zyada matter karta hai — yeh aapko saari "equally heads-and-tails but different phase" states ke around slide karta hai, jaise $\varphi =0$ par $\frac{|0⟩+|1⟩}{\sqrt{2}}$ versus $\varphi =\pi$ par $\frac{|0⟩-|1⟩}{\sqrt{2}}$. Yeh physically alag hain (opposite equator points) phir bhi heads/tails basis mein identical measurement odds hain.

Case C — Agar hum ne angle half NAHI kiya hota? $\theta$ directly use karte hue, $\theta$ aur $\theta +2\pi$ same state dete, lekin aap sirf $\theta \in [0,\pi ]$ se northern hemisphere tak pahunchte — poles $|0⟩,|1⟩$ nahi hote aur orthogonal states $90°$ apart hoti, sab kuch half sphere par cramming ho jaata. Half-angle hi woh cheez hai jo map ko pure poore sphere par, exactly ek baar banati hai.

Case D — Degenerate "classical" limit. Agar phase forbidden hota ($e^{i\varphi }$ hatao, real $\ge 0$ amplitudes force karo), aap ek meridian par stuck ho jaate — pole to pole tak ek semicircle. Woh semicircle woh closest hai jo ek "phaseless" theory pahunch sakti hai: woh probabilities mix kar sakti hai lekin kabhi interfere nahi kar sakti. Poora sphere woh extra room hai jo quantum mechanics phase ke saath khareedta hai.

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Ek-picture summary

Yahan ek page par poora demolition hai: 4 real dials → (−1 normalization) → (−1 global phase) → 2 dials → ek sphere.

Recall Poore walkthrough ki Feynman retelling

Hum ek spinning coin describe karna chahte the jo part heads, part tails hai. Humne ise do arrows ke roop mein likha: ek "kitna heads" ke liye ($\alpha$), ek "kitna tails" ke liye ($\beta$). Har arrow ko ek length aur ek direction chahiye, isliye woh char numbers hain ek coin describe karne ke liye — bahut zyada.

Phir humne ghar saaf kiya. Pehla: dono lengths dono bade nahi ho sakte, kyunki heads ki chance plus tails ki chance $1$ ke barabar honi chahiye — woh $1$ hypotenuse ke saath Pythagoras hai, aur yeh do numbers ko ek mein fuse karta hai. Doosra: agar aap dono arrows ko ek saath spin karo, kuch bhi jo aap kabhi measure kar sako woh nahi badlata — isliye humne us shared spin ko freeze kar diya. Isse ek aur number mar gaya.

Do numbers bache. Ek kehta hai coin heads aur tails ke beech kitna tila hua hai (ise $\theta$ kaho); doosra kehta hai tails-arrow heads ke relative kitna twisted hai (ise $\varphi$ kaho). Ek tilt aur ek twist — woh exactly latitude aur longitude hai. Toh har possible qubit ek globe par ek dot hai: $|0⟩$ upar, $|1⟩$ neeche, aur fifty-fifty spins equator ke around. Sirf ek quirk hai ki hum formula ke andar half angle se tilt karte hain, jo woh gear hai jo top aur bottom ko exactly opposite rakhta hai. Woh globe Bloch sphere hai.