Quantum computing hardware basics: Worked examples

6.5.13 · D3 · Hardware › Advanced & Emerging Architectures › Quantum computing hardware basics

Hum sirf parent note aur Boltzmann Distribution mein bane tools reuse karte hain. Kuch naya assume nahin kiya gaya hai.

Scenario matrix

Neeche har worked example ko us cell ke saath tag kiya gaya hai jo wo fill karta hai. Columns hain quantity jo hum compute karte hain; rows hain input ka type.

Input class	Thermal ratio $P_{1} / P_{0}$	Coherence $T_{1}, T_{2}, T_{ϕ}$	Gate budget / readout
Cold / large gap (nice regime)	Ex 1	Ex 4	Ex 6
Hot / small gap (formula theek hai par scary)	Ex 2	—	Ex 7 (word problem)
Zero / degenerate input ( $T \to 0$ , $T_{ϕ} \to \infty$ )	Ex 3	Ex 5	—
Limiting / edge ( $T \to \infty$ , $T_{2} = 2 T_{1}$ ceiling)	Ex 2, Ex 3	Ex 5	Ex 8 (exam twist)

Ek conversion trick jo neeche har jagah use hoti hai. $Δ E$ ko joules mein compute karne ki bajaye, hum dimensionless exponent banate hain: $x = \frac{Δ E}{k _{B} T} = \frac{h f}{k _{B} T} .$

Yeh step kyun? Dono $h f$ aur $k_{B} T$ energies hain (joules); unka ratio ek pure number hai. Ek pure number ko $e^{- x}$ mein feed karna safe hai — koi units nahin bachi jो mismatch ho sakein. Agar $x$ bada hai to qubit cold-and-happy hai; agar $x$ zero ke paas hai to woh hot-and-scrambled hai. Toh $x$ hi poori story hai.

Thermal scenarios (Boltzmann machine)

Worked example

Ex 1 — Cold / large gap (nice regime). Ek transmon jiska $f = 5 GHz$ hai, $T = 15 mK$ par rakha hai. $P_{1} / P_{0}$ kya hai, aur initialization kitni achi hai?

Forecast: padhne se pehle andaaza lagao — wrong-state probability $1 0^{- 2}$ , $1 0^{- 7}$ , ya $0.5$ ke kareeb hai?

Exponent banao $x = h f / (k_{B} T)$ . $h f = (6.626 \times 1 0^{- 34}) (5 \times 1 0^{9}) = 3.313 \times 1 0^{- 24} J$ . Yeh step kyun? Hume energy gap joules mein chahiye taaki use thermal energy se compare kar sakein.
$k_{B} T = (1.381 \times 1 0^{- 23}) (0.015) = 2.072 \times 1 0^{- 25} J$ . Yeh step kyun? Same units (J) toh ratio clean hai.
$x = 3.313 \times 1 0^{- 24} /2.072 \times 1 0^{- 25} \approx 15.99$ . Yeh step kyun? Yahi pure number hai jo sab kuch decide karta hai.
$P_{1} / P_{0} = e^{- 15.99} \approx 1.1 \times 1 0^{- 7}$ . Yeh step kyun? Exponent ko Boltzmann mein plug karo.

Verify: parent note ke " $\approx 1 0^{- 7}$ " claim aur uske $x \approx 16$ se match karta hai. Sanity: $x > 15$ matlab "bahut cold" toh near-zero excitation bilkul expected hai. Initialization criterion (DiVincenzo #2) comfortably meet ho raha hai.

Worked example

Ex 2 — Hot / small gap + $T \to \infty$ edge. Same qubit ( $f = 5 GHz$ ) lekin room temperature $T = 300 K$ par chhoda gaya. $P_{1} / P_{0}$ compute karo. Phir true high-temperature limit ke baare mein argue karo.

Forecast: kya coin mostly heads hogi, mostly tails, ya 50/50 mush?

$x = h f / (k_{B} T)$ . Ex 1 se numerator ( $3.313 \times 1 0^{- 24}$ J) ke saath aur $k_{B} T = (1.381 \times 1 0^{- 23}) (300) = 4.143 \times 1 0^{- 21}$ J: $x = 3.313 \times 1 0^{- 24} /4.143 \times 1 0^{- 21} \approx 7.996 \times 1 0^{- 4}$ . Yeh step kyun? Bada $T$ matlab bada $k_{B} T$ , toh $x$ zero ki taraf collapse karta hai.
$P_{1} / P_{0} = e^{- 0.0008} \approx 0.9992$ . Yeh step kyun? Plug in karo — almost exactly 1 matlab equal populations.
The limit: jab $T \to \infty$ , $x \to 0$ , toh $P_{1} / P_{0} = e^{- x} \to e^{0} = 1$ . Yeh step kyun? Yeh "totally scrambled" ka mathematical statement hai: dono levels equally likely, qubit ke paas koi reliable starting state nahin.

Verify: $0.9992 \approx 1$ parent note se match karta hai. Units: $x$ dimensionless ✓. Limiting behaviour: $lim_{T \to \infty} e^{- h f / k_{B} T} = 1$ , yeh "coin slapped flat" case hai — isi liye hum refrigerate karte hain, yeh baat consistent hai.

Worked example

Ex 3 — Zero / degenerate inputs ( $T \to 0$ aur $Δ E \to 0$ ). Do degenerate cases: (a) $T \to 0^{+}$ fixed gap ke saath; (b) ek degenerate qubit jiska gap $Δ E \to 0$ fixed $T$ par. Dono mein $P_{1} / P_{0}$ ka kya hota hai?

Forecast: ek $0$ ki taraf jaata hai, doosra $1$ ki taraf — kaun sa kaun sa hai?

(a) $T \to 0^{+}$ : $x = h f / (k_{B} T) \to + \infty$ , toh $P_{1} / P_{0} = e^{- \infty} = 0$ . Yeh step kyun? Absolute zero saari thermal jiggle hataa deta hai; qubit excited ho hi nahin sakta — woh $∣0 ⟩$ mein frozen hai. Yeh ideal (lekin unreachable) limit hai.
(b) $Δ E \to 0$ : $x = Δ E / (k_{B} T) \to 0$ , toh $P_{1} / P_{0} = e^{0} = 1$ . Yeh step kyun? Agar dono levels ki same energy hai toh $∣0 ⟩$ prefer karne ki koi wajah nahin; thermal noise dono ko equally fill karta hai. Isi liye hum gap engineer karte hain (Josephson Junction anharmonicity ke zariye) — zero gap ek useless qubit hai.

Verify: limits $e^{- x}$ ki monotone shape se agree karte hain: $e^{- \infty} = 0$ , $e^{0} = 1$ . Dono Ex 1–2 ke degenerate endpoints hain, toh koi naya physics nahin, sirf boundary. $T \to 0$ literally tak pahunchne ke liye Cryogenics and Dilution Refrigerators chahiye.

Coherence scenarios ( $T_{1}, T_{2}, T_{ϕ}$ )

Figure — Quantum computing hardware basics

Figure ceiling relationship dikhata hai: $1/ T_{2}$ (chalk-blue bar) $1/ (2 T_{1})$ (yellow) relaxation share aur $1/ T_{ϕ}$ (pink) pure-dephasing share ka sum hai. Kyunki rates add hoti hain aur $1/ T_{ϕ} \geq 0$ , blue bar yellow bar se kabhi chhoti nahin ho sakti — yeh bilkul wahi statement hai $T_{2} \leq 2 T_{1}$ .

Worked example

Ex 4 — Cold / nice regime: $T_{ϕ}$ solve karo. Ek qubit $T_{1} = 80 μ s$ aur $T_{2} = 100 μ s$ report karta hai. Pure-dephasing time $T_{ϕ}$ nikalo.

Forecast: kya $T_{ϕ}$ , $T_{2}$ se bada hoga ya chhota?

$\frac{1}{T _{2}} = \frac{1}{2 T _{1}} + \frac{1}{T _{ϕ}}$ ko rearrange karo: $\frac{1}{T _{ϕ}} = \frac{1}{T _{2}} - \frac{1}{2 T _{1}}$ . Yeh step kyun? $T_{ϕ}$ ek hi unknown hai; known relaxation share subtract karke isolate karo.
$\frac{1}{T _{2}} = \frac{1}{100} = 0.01 μ s^{- 1}$ ; $\frac{1}{2 T _{1}} = \frac{1}{160} = 0.00625 μ s^{- 1}$ . Yeh step kyun? Dono known times ko rates (per-microsecond) mein convert karo taaki subtract ho sakein.
$\frac{1}{T _{ϕ}} = 0.01 - 0.00625 = 0.00375 μ s^{- 1} \Rightarrow T_{ϕ} = 266.7 μ s$ . Yeh step kyun? Rate ko wapas time mein invert karo.

Verify: $\frac{1}{2 ( 80 )} + \frac{1}{266.7} = 0.00625 + 0.00375 = 0.01 = \frac{1}{100}$ ✓. $T_{ϕ} = 266.7 > T_{2}$ , sensible: yahan dephasing slow hai, toh relaxation bada limiter hai.

Worked example

Ex 5 — Degenerate / ceiling edge ( $T_{ϕ} \to \infty$ , phir $T_{ϕ} \to 0$ ). (a) Ek qubit jisme koi pure dephasing nahin, $T_{ϕ} \to \infty$ , ka $T_{1} = 50 μ s$ hai. $T_{2}$ nikalo. (b) Agar dephasing catastrophic ho, $T_{ϕ} \to 0$ ?

Forecast: case (a) famous ceiling hit karta hai — kya value?

(a) $\frac{1}{T _{2}} = \frac{1}{2 T _{1}} + \frac{1}{\infty} = \frac{1}{2 T _{1}} + 0$ , toh $T_{2} = 2 T_{1} = 100 μ s$ . Yeh step kyun? Pure dephasing hatane par sirf relaxation-induced phase loss bachta hai — yeh ek given $T_{1}$ ke liye best possible hai, ceiling $T_{2} \leq 2 T_{1}$ .
(b) $\frac{1}{T _{2}} = \frac{1}{2 T _{1}} + \frac{1}{0 ^{+}} \to \infty$ , toh $T_{2} \to 0$ . Yeh step kyun? Infinitely fast phase noise coherence ko instantly khatam kar deti hai $T_{1}$ ki parwah kiye bina — coin energetically ghoomti rehti hai lekin uska facing track karna turant impossible ho jaata hai.

Verify: (a) $T_{2} = 2 T_{1}$ inequality saturate karta hai — parent ke " $T_{2}$ is Two-limited" se match karta hai. (b) $lim_{T_{ϕ} \to 0} (\frac{1}{2 T _{1}} + \frac{1}{T _{ϕ}})^{- 1} = 0$ ✓. Yeh do endpoints har real device ko bracket karte hain; koi bhi measured value $0 < T_{2} \leq 2 T_{1}$ mein hi hogi. $T_{ϕ}$ se ladna Quantum Error Correction aur better shielding ka roz ka kaam hai.

Gate-budget & readout scenarios

Worked example

Ex 6 — Cold / nice regime: gate depth. $T_{2} = 100 μ s$ aur gate time $t_{g} = 20 ns$ ke saath, (a) error per gate aur (b) usable circuit depth nikalo.

Forecast: roughly kitne gates ke baad decoherence dominate karta hai — dozens, hundreds, ya thousands?

Ek unit mein convert karo: $T_{2} = 100 μ s = 100000 ns$ . Yeh step kyun? Error ek ratio $t_{g} / T_{2}$ hai; dono nanoseconds mein hone chahiye warna ratio meaningless hai.
(a) error per gate $\approx t_{g} / T_{2} = 20/100000 = 2 \times 1 0^{- 4}$ . Yeh step kyun? Short-time expansion $e^{- t / T_{2}} \approx 1 - t / T_{2}$ , toh per gate loss $t / T_{2}$ hai.
(b) depth $\approx T_{2} / t_{g} = 100000/20 = 5000$ gates. Yeh step kyun? Yeh reciprocal hai — woh gates ki count jahan accumulated error order 1 tak pahunche.

Verify: parent ke $2 \times 1 0^{- 4}$ aur $5000$ se match karta hai. Cross-check: $(error per gate) \times (depth) = 2 \times 1 0^{- 4} \times 5000 = 1$ ✓ — construction se, depth wahan hai jahan total error $\approx 1$ ho.

Worked example

Ex 7 — Hot / real-world word problem. Ek engineer ka fridge warm ho jaata hai aur qubit frequency lower ho ke $f = 4 GHz$ (softer junction) ho jaati hai jabki $T = 100 mK$ . Kya initialization abhi bhi acceptable hai (maan lo hum $P_{1} / P_{0} < 1 0^{- 3}$ demand karte hain)?

Forecast: chhota gap aur warm fridge ke saath, kya hum pass karte hain?

$h f = (6.626 \times 1 0^{- 34}) (4 \times 1 0^{9}) = 2.650 \times 1 0^{- 24}$ J. Yeh step kyun? Chhoti frequency ⇒ chhota gap ⇒ chhota $x$ ⇒ worse initialization; quantify karo.
$k_{B} T = (1.381 \times 1 0^{- 23}) (0.100) = 1.381 \times 1 0^{- 24}$ J. Yeh step kyun? Warm fridge ⇒ badi thermal energy ⇒ yeh bhi $x$ ko shrink karta hai.
$x = 2.650 \times 1 0^{- 24} /1.381 \times 1 0^{- 24} \approx 1.919$ . Yeh step kyun? Dono effects ne $x$ ko $\sim 16$ (Ex 1) se $\sim 2$ tak push kiya — danger sign.
$P_{1} / P_{0} = e^{- 1.919} \approx 0.147$ . Yeh step kyun? Requirement se compare karo.

Verify: $0.147 > 1 0^{- 3}$ ⇒ badly fail — wrong state mein start hone ki lagbhag 15% chance. Decision: gap badhao ya $T$ ghataao. Sanity: $x \approx 2$ na cold ( $x \approx 16$ ) hai na scrambled ( $x \approx 0$ ) — yeh "marginal" middle mein baitha hai, exactly jahan ek real fault rehti.

Worked example

Ex 8 — Exam-style twist (limiting logic, no calculator). "Ek vendor claim karta hai $T_{1} = 40 μ s$ aur $T_{2} = 120 μ s$ . $T_{ϕ}$ compute kiye bina, kya yeh physically possible hai?"

Forecast: vendor par trust karo ya jhooth pakdo?

Ceiling yaad karo: $T_{2} \leq 2 T_{1}$ . Yeh step kyun? Yeh ek aisi inequality hai jiske liye $T_{ϕ}$ ki koi knowledge nahin chahiye — fastest possible test.
Ceiling compute karo: $2 T_{1} = 2 \times 40 = 80 μ s$ . Yeh step kyun? Yeh given $T_{1}$ ke liye maximum allowed $T_{2}$ hai.
Compare karo: claimed $T_{2} = 120 μ s > 80 μ s$ . Yeh step kyun? Agar $T_{2}$ , $2 T_{1}$ se zyaada hai toh coherence relation ko ek negative $1/ T_{ϕ}$ chahiye hogi, matlab dephasing jo coherence create kare — impossible.

Verify: $\frac{1}{T _{ϕ}} = \frac{1}{120} - \frac{1}{80} = 0.00833 - 0.0125 = - 0.00417 < 0$ ⇒ negative rate ⇒ spec unphysical hai. Confirm karta hai ki ceiling ne isse instantly pakad liya. Yahi reasoning parent ke mnemonic " $T_{2}$ is Two-limited" ke peechhe hai.

Recall Har cell ka one-line recap

Cold thermal :: $x \approx 16$ , $P_{1} / P_{0} \approx 1 0^{- 7}$ (Ex 1). Hot thermal :: $x \to 0$ , ratio $\to 1$ , scrambled (Ex 2). Zero-input thermal :: $T \to 0 \Rightarrow 0$ ; gap $\to 0 \Rightarrow 1$ (Ex 3). Coherence solve :: $1/ T_{ϕ} = 1/ T_{2} - 1/ (2 T_{1})$ (Ex 4). Coherence edges :: $T_{ϕ} \to \infty \Rightarrow T_{2} = 2 T_{1}$ ; $T_{ϕ} \to 0 \Rightarrow T_{2} \to 0$ (Ex 5). Gate budget :: error $= t_{g} / T_{2}$ , depth $= T_{2} / t_{g}$ (Ex 6). Word problem :: warm+soft gap initialization fail kar sakta hai (Ex 7). Exam twist :: $T_{2} > 2 T_{1}$ impossible hai (Ex 8).

Related building blocks: Boltzmann Distribution, Bloch Sphere, Superposition and Entanglement, Classical Bits vs Qubits, Unitary Operators and Reversible Computing.

Scenario matrix

Thermal scenarios (Boltzmann machine)

Coherence scenarios (T1​,T2​,Tϕ​)

Gate-budget & readout scenarios

Coherence scenarios ( $T_{1}, T_{2}, T_{ϕ}$ )