Quantum computing hardware basics: Worked examples

We only reuse tools built in the parent note plus the Boltzmann Distribution. Nothing new is assumed.

The scenario matrix

Every worked example below is tagged with the cell it fills. The columns are the quantity we compute; the rows are the kind of input.

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 still fine but 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)

A conversion trick used everywhere below. Instead of computing $Δ E$ in joules, we form the dimensionless exponent $x = \frac{Δ E}{k _{B} T} = \frac{h f}{k _{B} T} .$

Why this step? Both $h f$ and $k_{B} T$ are energies (joules); their ratio is a pure number. Feeding a pure number into $e^{- x}$ is safe — no units left to mismatch. If $x$ is large the qubit is cold-and-happy; if $x$ is near zero it is hot-and-scrambled. So $x$ is the whole story.

Thermal scenarios (the Boltzmann machine)

Worked example

Ex 1 — Cold / large gap (nice regime). A transmon with $f = 5 GHz$ sits at $T = 15 mK$ . What is $P_{1} / P_{0}$ , and how good is the initialization?

Forecast: guess before reading — is the wrong-state probability closer to $1 0^{- 2}$ , $1 0^{- 7}$ , or $0.5$ ?

Form the exponent $x = h f / (k_{B} T)$ . $h f = (6.626 \times 1 0^{- 34}) (5 \times 1 0^{9}) = 3.313 \times 1 0^{- 24} J$ . Why this step? We need the energy gap in joules to compare it to thermal energy.
$k_{B} T = (1.381 \times 1 0^{- 23}) (0.015) = 2.072 \times 1 0^{- 25} J$ . Why this step? Same units (J) so the ratio is clean.
$x = 3.313 \times 1 0^{- 24} /2.072 \times 1 0^{- 25} \approx 15.99$ . Why this step? This is the pure number that decides everything.
$P_{1} / P_{0} = e^{- 15.99} \approx 1.1 \times 1 0^{- 7}$ . Why this step? Plug the exponent into Boltzmann.

Verify: matches the parent note's " $\approx 1 0^{- 7}$ " claim and its $x \approx 16$ . Sanity: $x > 15$ means "very cold" so a near-zero excitation is exactly what we expect. Initialization criterion (DiVincenzo #2) is comfortably met.

Worked example

Ex 2 — Hot / small gap + the $T \to \infty$ edge. Same qubit ( $f = 5 GHz$ ) but left at room temperature $T = 300 K$ . Compute $P_{1} / P_{0}$ . Then argue the true high-temperature limit.

Forecast: will the coin be mostly heads, mostly tails, or a 50/50 mush?

$x = h f / (k_{B} T)$ . With the numerator from Ex 1 ( $3.313 \times 1 0^{- 24}$ J) and $k_{B} T = (1.381 \times 1 0^{- 23}) (300) = 4.143 \times 1 0^{- 21}$ J: $x = 3.313 \times 1 0^{- 24} /4.143 \times 1 0^{- 21} \approx 7.996 \times 1 0^{- 4}$ . Why this step? Huge $T$ makes $k_{B} T$ huge, so $x$ collapses toward zero.
$P_{1} / P_{0} = e^{- 0.0008} \approx 0.9992$ . Why this step? Plug in — almost exactly 1 means equal populations.
The limit: as $T \to \infty$ , $x \to 0$ , so $P_{1} / P_{0} = e^{- x} \to e^{0} = 1$ . Why this step? This is the mathematical statement of "totally scrambled": both levels equally likely, the qubit carries no reliable starting state.

Verify: $0.9992 \approx 1$ matches parent note. Units: $x$ dimensionless ✓. Limiting behaviour: $lim_{T \to \infty} e^{- h f / k_{B} T} = 1$ , the "coin slapped flat" case — consistent with why we refrigerate.

Worked example

Ex 3 — Zero / degenerate inputs ( $T \to 0$ and $Δ E \to 0$ ). Two degenerate cases: (a) $T \to 0^{+}$ with fixed gap; (b) a degenerate qubit whose gap $Δ E \to 0$ at fixed $T$ . What happens to $P_{1} / P_{0}$ in each?

Forecast: one goes to $0$ , the other goes to $1$ — which is which?

(a) $T \to 0^{+}$ : $x = h f / (k_{B} T) \to + \infty$ , so $P_{1} / P_{0} = e^{- \infty} = 0$ . Why this step? Absolute zero removes all thermal jiggle; the qubit cannot be excited — it is frozen in $∣0 ⟩$ . This is the ideal (but unreachable) limit.
(b) $Δ E \to 0$ : $x = Δ E / (k_{B} T) \to 0$ , so $P_{1} / P_{0} = e^{0} = 1$ . Why this step? If the two levels have the same energy there is no reason to prefer $∣0 ⟩$ ; thermal noise fills both equally. This is why we engineer a gap (via the Josephson Junction anharmonicity) — a zero gap is a useless qubit.

Verify: limits agree with the monotone shape of $e^{- x}$ : $e^{- \infty} = 0$ , $e^{0} = 1$ . Both are degenerate endpoints of Ex 1–2, so no new physics, just the boundary. Reaching $T \to 0$ literally needs Cryogenics and Dilution Refrigerators.

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

Figure — Quantum computing hardware basics

The figure shows the ceiling relationship: $1/ T_{2}$ (chalk-blue bar) is the sum of the relaxation share $1/ (2 T_{1})$ (yellow) and the pure-dephasing share $1/ T_{ϕ}$ (pink). Because rates add and $1/ T_{ϕ} \geq 0$ , the blue bar can never be shorter than the yellow one — which is exactly the statement $T_{2} \leq 2 T_{1}$ .

Worked example

Ex 4 — Cold / nice regime: solve for $T_{ϕ}$ . A qubit reports $T_{1} = 80 μ s$ and $T_{2} = 100 μ s$ . Find the pure-dephasing time $T_{ϕ}$ .

Forecast: will $T_{ϕ}$ be bigger or smaller than $T_{2}$ ?

Rearrange $\frac{1}{T _{2}} = \frac{1}{2 T _{1}} + \frac{1}{T _{ϕ}}$ to $\frac{1}{T _{ϕ}} = \frac{1}{T _{2}} - \frac{1}{2 T _{1}}$ . Why this step? $T_{ϕ}$ is the only unknown; isolate it by subtracting the known relaxation share.
$\frac{1}{T _{2}} = \frac{1}{100} = 0.01 μ s^{- 1}$ ; $\frac{1}{2 T _{1}} = \frac{1}{160} = 0.00625 μ s^{- 1}$ . Why this step? Convert both known times to rates (per-microsecond) so they can be subtracted.
$\frac{1}{T _{ϕ}} = 0.01 - 0.00625 = 0.00375 μ s^{- 1} \Rightarrow T_{ϕ} = 266.7 μ s$ . Why this step? Invert the rate back to a time.

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: dephasing here is slow, so relaxation is the bigger limiter.

Worked example

Ex 5 — Degenerate / ceiling edge ( $T_{ϕ} \to \infty$ , then $T_{ϕ} \to 0$ ). (a) A qubit with no pure dephasing, $T_{ϕ} \to \infty$ , has $T_{1} = 50 μ s$ . Find $T_{2}$ . (b) What if dephasing is catastrophic, $T_{ϕ} \to 0$ ?

Forecast: case (a) hits the famous ceiling — what value?

(a) $\frac{1}{T _{2}} = \frac{1}{2 T _{1}} + \frac{1}{\infty} = \frac{1}{2 T _{1}} + 0$ , so $T_{2} = 2 T_{1} = 100 μ s$ . Why this step? Removing pure dephasing leaves only relaxation-induced phase loss — this is the best a given $T_{1}$ allows, the ceiling $T_{2} \leq 2 T_{1}$ .
(b) $\frac{1}{T _{2}} = \frac{1}{2 T _{1}} + \frac{1}{0 ^{+}} \to \infty$ , so $T_{2} \to 0$ . Why this step? Infinitely fast phase noise kills coherence instantly regardless of $T_{1}$ — the coin still spins energetically but you lose all track of its facing immediately.

Verify: (a) $T_{2} = 2 T_{1}$ saturates the inequality — matches parent's " $T_{2}$ is Two-limited". (b) $lim_{T_{ϕ} \to 0} (\frac{1}{2 T _{1}} + \frac{1}{T _{ϕ}})^{- 1} = 0$ ✓. These two endpoints bracket every real device; anything measured must land in $0 < T_{2} \leq 2 T_{1}$ . Fighting $T_{ϕ}$ is the daily job of Quantum Error Correction and better shielding.

Gate-budget & readout scenarios

Worked example

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

Forecast: roughly how many gates before decoherence dominates — dozens, hundreds, or thousands?

Convert to one unit: $T_{2} = 100 μ s = 100000 ns$ . Why this step? Error is a ratio $t_{g} / T_{2}$ ; both must be nanoseconds or the ratio is meaningless.
(a) error per gate $\approx t_{g} / T_{2} = 20/100000 = 2 \times 1 0^{- 4}$ . Why this step? Short-time expansion $e^{- t / T_{2}} \approx 1 - t / T_{2}$ , so the loss per gate is $t / T_{2}$ .
(b) depth $\approx T_{2} / t_{g} = 100000/20 = 5000$ gates. Why this step? It's the reciprocal — the count of gates whose accumulated error reaches order 1.

Verify: matches parent's $2 \times 1 0^{- 4}$ and $5000$ . Cross-check: $(error per gate) \times (depth) = 2 \times 1 0^{- 4} \times 5000 = 1$ ✓ — by construction, depth is where total error hits $\approx 1$ .

Worked example

Ex 7 — Hot / real-world word problem. An engineer's fridge drifts warmer and the qubit frequency is lowered to $f = 4 GHz$ (softer junction) while $T = 100 mK$ . Is initialization still acceptable (say, we demand $P_{1} / P_{0} < 1 0^{- 3}$ )?

Forecast: with a smaller gap and a warmer fridge, do we still pass?

$h f = (6.626 \times 1 0^{- 34}) (4 \times 1 0^{9}) = 2.650 \times 1 0^{- 24}$ J. Why this step? Smaller frequency ⇒ smaller gap ⇒ smaller $x$ ⇒ worse initialization; quantify it.
$k_{B} T = (1.381 \times 1 0^{- 23}) (0.100) = 1.381 \times 1 0^{- 24}$ J. Why this step? Warmer fridge ⇒ bigger thermal energy ⇒ also shrinks $x$ .
$x = 2.650 \times 1 0^{- 24} /1.381 \times 1 0^{- 24} \approx 1.919$ . Why this step? Both effects pushed $x$ from $\sim 16$ (Ex 1) down toward $\sim 2$ — danger sign.
$P_{1} / P_{0} = e^{- 1.919} \approx 0.147$ . Why this step? Compare to the requirement.

Verify: $0.147 > 1 0^{- 3}$ ⇒ fails badly — about a 15% chance of starting in the wrong state. Decision: raise the gap or lower $T$ . Sanity: $x \approx 2$ is neither the cold ( $x \approx 16$ ) nor scrambled ( $x \approx 0$ ) case — it sits in the "marginal" middle, exactly where a real fault would live.

Worked example

Ex 8 — Exam-style twist (limiting logic, no calculator). "A vendor claims $T_{1} = 40 μ s$ and $T_{2} = 120 μ s$ . Without computing $T_{ϕ}$ , is this physically possible?"

Forecast: trust the vendor or catch the lie?

Recall the ceiling: $T_{2} \leq 2 T_{1}$ . Why this step? It's the one inequality that needs no knowledge of $T_{ϕ}$ — the fastest possible test.
Compute the ceiling: $2 T_{1} = 2 \times 40 = 80 μ s$ . Why this step? This is the maximum $T_{2}$ the given $T_{1}$ allows.
Compare: claimed $T_{2} = 120 μ s > 80 μ s$ . Why this step? If $T_{2}$ exceeds $2 T_{1}$ the coherence relation would need a negative $1/ T_{ϕ}$ , i.e. dephasing that creates coherence — impossible.

Verify: $\frac{1}{T _{ϕ}} = \frac{1}{120} - \frac{1}{80} = 0.00833 - 0.0125 = - 0.00417 < 0$ ⇒ negative rate ⇒ the spec is unphysical. Confirms the ceiling caught it instantly. This is the reasoning behind the parent's mnemonic " $T_{2}$ is Two-limited".

Recall One-line recap of each cell

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 can fail initialization (Ex 7). Exam twist :: $T_{2} > 2 T_{1}$ is impossible (Ex 8).

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

The scenario matrix

Thermal scenarios (the Boltzmann machine)

Coherence scenarios (T1​,T2​,Tϕ​)

Gate-budget & readout scenarios

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