Quantum computing hardware basics: Exercises — Hardware deep dive

Constants you may need (write them once, use them everywhere): $h = 6.626 \times 1 0^{- 34} J\cdots, k_{B} = 1.381 \times 1 0^{- 23} J/K$

Level 1 — Recognition

Recall Solution 1.1

$T_{1}$ ::: relaxation time — the excited-state population decays to $1/ e$ ; energy leaks out (coin falls flat).
$T_{2}$ ::: dephasing time — phase coherence (the off-diagonal amplitude) decays to $1/ e$ (coin still spins but you lose track of its direction).
$α$ ::: the complex amplitude of $∣0 ⟩$ in $∣ ψ ⟩ = α ∣0 ⟩ + β ∣1 ⟩$ .
$∣ α ∣^{2}$ ::: the probability of measuring $∣0 ⟩$ .
$U$ ::: a unitary operator = a quantum gate ( $U^{†} U = I$ , preserves total probability).

Recall Solution 1.2

Criterion 2 — Initialization to a known state. From $P_{1} / P_{0} = e^{- Δ E / k_{B} T}$ , warmth re-populates $∣1 ⟩$ ; if the qubit doesn't reliably start in $∣0 ⟩$ , you cannot initialize. (Cold also helps criterion 3, coherence, but initialization is the direct one — see Boltzmann Distribution.)

Level 2 — Application

Recall Solution 2.1

WHAT: $Δ E / k_{B}$ tells you the temperature at which thermal energy equals the gap. $\frac{Δ E}{k _{B}} = \frac{h \cdot ( 5 \times 1 0 ^{9} )}{k _{B}} = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 5 \times 1 0 ^{9} )}{1.381 \times 1 0 ^{- 23}} \approx 0.240 K$ So the gap is "worth" about 0.24 K of thermal energy. Cooling well below this (to 15 mK) makes thermal excitation negligible.

Recall Solution 2.2

Use the temperature form: $\frac{Δ E}{k _{B} T} = \frac{0.240}{0.015} = 16.0$ . $\frac{P _{1}}{P _{0}} = e^{- 16} \approx 1.1 \times 1 0^{- 7}$ About one qubit in ten million is thermally excited — essentially always in $∣0 ⟩$ . WHY it matters: this is what "reliably initialized" means numerically.

Recall Solution 2.3

WHY the approximation: for $t ≪ T_{2}$ , coherence $e^{- t / T_{2}} \approx 1 - t / T_{2}$ , so the lost fraction is $\approx t / T_{2}$ . (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. This ratio, not raw $T_{2}$ , sets algorithm depth.

Level 3 — Analysis

Recall Solution 3.1

WHY the formula: relaxation ( $1/ T_{1}$ ) and pure phase noise ( $1/ T_{ϕ}$ ) both destroy phase; independent rates add, and relaxation contributes only $\frac{1}{2 T _{1}}$ because populations relax to the midpoint. $\frac{1}{T _{2}} = \frac{1}{2 ( 80 )} + \frac{1}{120} = \frac{1}{160} + \frac{1}{120} (μ s^{- 1})$ $\frac{1}{T _{2}} = 0.006250 + 0.008333 = 0.014583 μ s^{- 1} \Rightarrow T_{2} \approx 68.6 μ s$ Check: $T_{2} = 68.6 < 2 T_{1} = 160 μ s$ . ✓ (See Quantum Error Correction for why raising $T_{2}$ matters.)

Recall Solution 3.2

Since $T_{ϕ} > 0$ , the term $1/ T_{ϕ} \geq 0$ . 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 holds only in the limit $T_{ϕ} \to \infty$ (no pure dephasing) — then $T_{2} = 2 T_{1}$ . This is the degenerate case: energy relaxation is the sole coherence killer.

Recall Solution 3.3

WHAT: $∣0 ⟩$ probability is $cos^{2} \frac{θ}{2} = 250/1000 = 0.25$ . $cos \frac{θ}{2} = 0.5 \Rightarrow \frac{θ}{2} = 60° \Rightarrow θ = 120°.$ Check: $sin^{2} 60° = 0.75$ = $P_{1}$ . ✓ On the Bloch sphere this is $120°$ down from the north pole — in the southern hemisphere, tilted toward $∣1 ⟩$ . Note $ϕ$ is invisible to this measurement (see figure).

Level 4 — Synthesis

Recall Solution 4.1

Compare the depth budget $N = T_{2} / t_{g}$ :

A: $N_{A} = 100 μ s /50 ns = 2000$ gates.
B: $N_{B} = 1 s /10 μ s = 100, 000$ gates. Algorithm needs 3000 gates. A fails ( $3000 > 2000$ ); B succeeds ( $3000 ≪ 100000$ ). Total-error estimate $\approx N t_{g} / T_{2} = N / N_{m a x}$ : A gives $3000/2000 = 1.5$ (coherence gone), B gives $3000/100000 = 0.03$ . Why raw $T_{2}$ misleads: A trapped-ion's slow but ultra-long-lived qubit (B) can be better than a fast superconducting one (A) despite A "feeling" faster. The ratio $T_{2} / t_{g}$ , not $T_{2}$ alone, is the true figure of merit.

Recall Solution 4.2

(a) The Hadamard gate: $H ∣0 ⟩ = \frac{1}{2} (∣0 ⟩ + ∣1 ⟩)$ . This is the "start the coin spinning" move. (b) $H = \frac{1}{2} (11 1 - 1)$ . Then $H^{†} H = \frac{1}{2} (11 1 - 1) (11 1 - 1) = \frac{1}{2} (2002) = I .$ ✓ Unitary → probability preserved. (c) $P_{0} = P_{1} = \frac{1}{2} \Rightarrow cos^{2} \frac{θ}{2} = \frac{1}{2} \Rightarrow θ = 90°$ : on the equator, along the $+ x$ axis (since $ϕ = 0$ ). See Superposition and Entanglement.

Level 5 — Mastery

Recall Solution 5.1

WHAT: demand $ε \approx t_{g} / T_{2} \leq 1 0^{- 3}$ . $t_{g} \leq 1 0^{- 3} \times T_{2} = 1 0^{- 3} \times 50 μ s = 0.05 μ s = 50 ns .$ So gates must be ≤ 50 ns. Number of physical gates per $T_{2}$ : $T_{2} / t_{g} = 50 μ s /50 ns = 1000$ . Since error correction consumes many physical operations per logical operation, the usable logical depth is far smaller — this is exactly why "1000 physical qubits ≈ 1 logical qubit" and why fidelity, not qubit count, is the bottleneck.

Recall Solution 5.2

The grain of truth: the state of 300 qubits genuinely is a vector of $2^{300}$ amplitudes; the computer's internal evolution really does act on all of them at once. The refutation: measurement is the catch. Reading 300 qubits yields only 300 classical bits, one collapsed outcome — you can never extract the $2^{300}$ amplitudes. The advantage is not storage; it is interference: arranging gates so wrong-answer amplitudes cancel (like waves) and the right answer's amplitude dominates the final measurement. Quantum speedup is a choreography of cancellation, not a giant hard drive.

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 has a great $T_{1}$ but its coherence is dephasing-limited (small $T_{ϕ}$ dominates the sum) — the coin spins but its phase scrambles fast. Q2's phase is quiet ( $T_{ϕ}$ large), so even with weaker $T_{1}$ its $T_{2}$ is more than double. Lesson: fix your worst rate; the largest term in $\frac{1}{T _{2}}$ is the one to attack. (Its Josephson-junction flux noise, Josephson Junction, often sets $T_{ϕ}$ .)

Recall Self-test checklist

Convert an energy gap to a temperature? ::: divide by $k_{B}$ (Ex 2.1).
Error per gate? ::: $\approx t_{g} / T_{2}$ (Ex 2.3).
Combine $T_{1}, T_{ϕ}$ into $T_{2}$ ? ::: $\frac{1}{T _{2}} = \frac{1}{2 T _{1}} + \frac{1}{T _{ϕ}}$ (Ex 3.1).
Latitude from probability? ::: $P_{0} = cos^{2} \frac{θ}{2}$ , take root, double the angle (Ex 3.3).
Which qubit wins? ::: highest $T_{2} / t_{g}$ , and fix the worst rate (Ex 5.3).