Quantum computing hardware basics: Exercises — Hinglish deep dive

6.5.13 · D4 · Hardware › Advanced & Emerging Architectures › Quantum computing hardware basics

Constants jo tumhe chahiye ho sakte hain (ek baar likho, har jagah use karo): $h = 6.626 \times 1 0^{- 34} J\cdots, k_{B} = 1.381 \times 1 0^{- 23} J/K$

Level 1 — Recognition

Recall Solution 1.1

$T_{1}$ ::: relaxation time — excited-state population $1/ e$ tak decay karta hai; energy bahar nikal jaati hai (coin flat gir jaata hai).
$T_{2}$ ::: dephasing time — phase coherence (off-diagonal amplitude) $1/ e$ tak decay karta hai (coin abhi bhi spin kar raha hai lekin tum uski direction track nahi kar sakte).
$α$ ::: $∣ ψ ⟩ = α ∣0 ⟩ + β ∣1 ⟩$ mein $∣0 ⟩$ ka complex amplitude.
$∣ α ∣^{2}$ ::: $∣0 ⟩$ measure karne ki probability.
$U$ ::: ek unitary operator = ek quantum gate ( $U^{†} U = I$ , total probability preserve karta hai).

Recall Solution 1.2

Criterion 2 — Initialization ek known state mein. $P_{1} / P_{0} = e^{- Δ E / k_{B} T}$ se, warmth $∣1 ⟩$ ko re-populate karti hai; agar qubit reliably $∣0 ⟩$ se start nahi karta, toh tum initialize nahi kar sakte. (Cold, criterion 3 yaani coherence mein bhi help karta hai, lekin initialization direct wala hai — dekho Boltzmann Distribution.)

Level 2 — Application

Recall Solution 2.1

WHAT: $Δ E / k_{B}$ tumhe woh temperature batata hai jis par thermal energy gap ke barabar hoti hai. $\frac{Δ E}{k _{B}} = \frac{h \cdot ( 5 \times 1 0 ^{9} )}{k _{B}} = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 5 \times 1 0 ^{9} )}{1.381 \times 1 0 ^{- 23}} \approx 0.240 K$ Toh gap thermal energy ke hisaab se lagbhag 0.24 K "worth" hai. Isse kaafi neeche (15 mK tak) cool karne par thermal excitation negligible ho jaata hai.

Recall Solution 2.2

Temperature form use karo: $\frac{Δ E}{k _{B} T} = \frac{0.240}{0.015} = 16.0$ . $\frac{P _{1}}{P _{0}} = e^{- 16} \approx 1.1 \times 1 0^{- 7}$ Har dस million mein se lagbhag ek qubit thermally excited hai — essentially hamesha $∣0 ⟩$ mein. WHY it matters: numerically "reliably initialized" ka matlab yahi hai.

Recall Solution 2.3

WHY the approximation: $t ≪ T_{2}$ ke liye, coherence $e^{- t / T_{2}} \approx 1 - t / T_{2}$ , toh lost fraction $\approx t / T_{2}$ hai. (a) $ε \approx \frac{t _{g}}{T _{2}} = \frac{20 \times 1 0 ^{- 9}}{100 \times 1 0 ^{- 6}} = 2 \times 1 0^{- 4}$ . (b) $N \approx \frac{T _{2}}{t _{g}} = \frac{100 \times 1 0 ^{- 6}}{20 \times 1 0 ^{- 9}} = 5000$ gates. Yeh ratio, raw $T_{2}$ nahi, algorithm depth set karta hai.

Level 3 — Analysis

Recall Solution 3.1

WHY the formula: relaxation ( $1/ T_{1}$ ) aur pure phase noise ( $1/ T_{ϕ}$ ) dono phase destroy karte hain; independent rates add hote hain, aur relaxation sirf $\frac{1}{2 T _{1}}$ contribute karta hai kyunki populations midpoint tak relax hote hain. $\frac{1}{T _{2}} = \frac{1}{2 ( 80 )} + \frac{1}{120} = \frac{1}{160} + \frac{1}{120} (μ s^{- 1})$ $\frac{1}{T _{2}} = 0.006250 + 0.008333 = 0.014583 μ s^{- 1} \Rightarrow T_{2} \approx 68.6 μ s$ Check: $T_{2} = 68.6 < 2 T_{1} = 160 μ s$ . ✓ (Dekho Quantum Error Correction kyun $T_{2}$ raise karna matter karta hai.)

Recall Solution 3.2

Kyunki $T_{ϕ} > 0$ hai, term $1/ T_{ϕ} \geq 0$ hai. Therefore $\frac{1}{T _{2}} = \frac{1}{2 T _{1}} + \frac{1}{T _{ϕ}} \geq \frac{1}{2 T _{1}} ⟹ T_{2} \leq 2 T_{1} .$ Equality sirf limit $T_{ϕ} \to \infty$ mein hold hoti hai (koi pure dephasing nahi) — tab $T_{2} = 2 T_{1}$ . Yeh degenerate case hai: energy relaxation single coherence killer hai.

Recall Solution 3.3

WHAT: $∣0 ⟩$ probability hai $cos^{2} \frac{θ}{2} = 250/1000 = 0.25$ . $cos \frac{θ}{2} = 0.5 \Rightarrow \frac{θ}{2} = 60° \Rightarrow θ = 120°.$ Check: $sin^{2} 60° = 0.75$ = $P_{1}$ . ✓ Bloch sphere par yeh north pole se $120°$ neeche hai — southern hemisphere mein, $∣1 ⟩$ ki taraf tila hua. Note karo $ϕ$ is measurement ko invisible hai (figure dekho).

Level 4 — Synthesis

Recall Solution 4.1

Depth budget $N = T_{2} / t_{g}$ compare karo:

A: $N_{A} = 100 μ s /50 ns = 2000$ gates.
B: $N_{B} = 1 s /10 μ s = 100, 000$ gates. Algorithm ko 3000 gates chahiye. A fail karta hai ( $3000 > 2000$ ); B succeed karta hai ( $3000 ≪ 100000$ ). Total-error estimate $\approx N t_{g} / T_{2} = N / N_{m a x}$ : A deta hai $3000/2000 = 1.5$ (coherence gone), B deta hai $3000/100000 = 0.03$ . Raw $T_{2}$ kyun mislead karta hai: ek trapped-ion ka slow lekin ultra-long-lived qubit (B) ek fast superconducting qubit (A) se better ho sakta hai bhalee A "faster" feel hota ho. Ratio $T_{2} / t_{g}$ , akela $T_{2}$ nahi, true figure of merit hai.

Recall Solution 4.2

(a) Hadamard gate: $H ∣0 ⟩ = \frac{1}{2} (∣0 ⟩ + ∣1 ⟩)$ . Yeh "coin spinning shuru karo" wala move hai. (b) $H = \frac{1}{2} (11 1 - 1)$ . Tab $H^{†} H = \frac{1}{2} (11 1 - 1) (11 1 - 1) = \frac{1}{2} (2002) = I .$ ✓ Unitary → probability preserved. (c) $P_{0} = P_{1} = \frac{1}{2} \Rightarrow cos^{2} \frac{θ}{2} = \frac{1}{2} \Rightarrow θ = 90°$ : equator par, $+ x$ axis ke along (kyunki $ϕ = 0$ ). Dekho Superposition and Entanglement.

Level 5 — Mastery

Recall Solution 5.1

WHAT: demand karo $ε \approx t_{g} / T_{2} \leq 1 0^{- 3}$ . $t_{g} \leq 1 0^{- 3} \times T_{2} = 1 0^{- 3} \times 50 μ s = 0.05 μ s = 50 ns .$ Toh gates ≤ 50 ns hone chahiye. Physical gates per $T_{2}$ : $T_{2} / t_{g} = 50 μ s /50 ns = 1000$ . Kyunki error correction har logical operation ke liye bahut saari physical operations consume karta hai, usable logical depth kaafi chhota hota hai — exactly yahi wajah hai ki "1000 physical qubits ≈ 1 logical qubit" aur isliye fidelity, qubit count nahi, bottleneck hai.

Recall Solution 5.2

Grain of truth: 300 qubits ki state genuinely $2^{300}$ amplitudes ka vector hai; computer ki internal evolution sach mein unhe sab par ek saath act karti hai. Refutation: measurement catch hai. 300 qubits read karne par sirf 300 classical bits milte hain, ek collapsed outcome — tum kabhi bhi $2^{300}$ amplitudes extract nahi kar sakte. Advantage storage nahi hai; yeh interference hai: gates arrange karna taaki wrong-answer amplitudes cancel ho jayein (waves ki tarah) aur sahi answer ki amplitude final measurement mein dominate kare. Quantum speedup cancellation ki choreography hai, koi giant hard drive nahi.

Recall Solution 5.3

Q1: $\frac{1}{T _{2}} = \frac{1}{2 ( 200 )} + \frac{1}{40} = 0.0025 + 0.025 = 0.0275 \Rightarrow T_{2} \approx 36.4 μ s$ . Q2: $\frac{1}{T _{2}} = \frac{1}{2 ( 60 )} + \frac{1}{300} = 0.008333 + 0.003333 = 0.011667 \Rightarrow T_{2} \approx 85.7 μ s$ . Depth budgets: $N_{Q 1} = 36.4 μ s /30 ns \approx 1213$ ; $N_{Q 2} = 85.7 μ s /30 ns \approx 2857$ . Winner: Q2. Physics: Q1 ka ek great $T_{1}$ hai lekin uski coherence dephasing-limited hai (chhota $T_{ϕ}$ sum dominate karta hai) — coin spin karta hai lekin uska phase jaldi scramble ho jaata hai. Q2 ka phase quiet hai (bada $T_{ϕ}$ ), toh weaker $T_{1}$ ke baad bhi uska $T_{2}$ double se zyada hai. Lesson: apna worst rate fix karo; $\frac{1}{T _{2}}$ mein sabse bada term woh hai jis par attack karna hai. (Iska Josephson-junction flux noise, Josephson Junction, often $T_{ϕ}$ set karta hai.)

Recall Self-test checklist

Energy gap ko temperature mein convert karna? ::: $k_{B}$ se divide karo (Ex 2.1).
Error per gate? ::: $\approx t_{g} / T_{2}$ (Ex 2.3).
$T_{1}, T_{ϕ}$ ko $T_{2}$ mein combine karna? ::: $\frac{1}{T _{2}} = \frac{1}{2 T _{1}} + \frac{1}{T _{ϕ}}$ (Ex 3.1).
Probability se latitude? ::: $P_{0} = cos^{2} \frac{θ}{2}$ , root lo, angle double karo (Ex 3.3).
Kaun sa qubit jeetta hai? ::: highest $T_{2} / t_{g}$ , aur worst rate fix karo (Ex 5.3).