Math Prep

Solve this BEFORE lesson.md. SamplerV2 is a probability machine: it samples bitstrings from $P(x) = |\alpha_x|^2$. Day 5's API (counts, BitArray, get_counts, analysis) is bookkeeping on top of the probability theory below.

Requires: Day 1 + Day 4 math prep. Video back-up: Watrous Lesson 01–02 (measurement sections); notes in resource-artifactory/notes/video-notes/.

---

1. From statevector to distribution

An $n$-qubit state defines a distribution over $2^n$ bitstrings: $P(x) = |\alpha_x|^2$, $\sum_x P(x) = 1$. The Sampler never returns amplitudes — only samples. Two different states can induce the SAME distribution ($|+\rangle$ and $|-\rangle$, or any state vs its phase-tweaked cousin): sampling in one basis loses all phase information.

Bitstring convention (drill it again): key 'q_{n-1} … q_1 q_0' — qubit 0 rightmost. The integer value of the bitstring = the statevector index.

2. Counts, frequencies, probabilities

$N$ shots ⇒ counts $\{x: n_x\}$. Estimated probability $\hat p_x = n_x / N$.

• Expected count: $\mathbb{E}[n_x] = N p_x$; fluctuation $\sim \sqrt{N p_x(1-p_x)}$ (binomial).
• Bell state, 1000 shots: expect {'00': ~500, '11': ~500} with typical wobble $\pm\sqrt{1000\cdot\tfrac12\cdot\tfrac12} \approx \pm16$. Seeing {'00': 503, '11': 497} is normal; seeing '01': 200 means your circuit (or noise) is wrong, not statistics.
• A never-appearing bitstring is evidence its probability is $\lesssim 1/N$, not proof it's zero.

3. Marginal and joint distributions

Marginalizing = ignoring some bits and summing probabilities over them. If $P(q_1 q_0)$ has $P(00) = P(01) = P(11) = \tfrac13$:

Independence check: $P(q_1 q_0) \ne P(q_1)P(q_0)$ here ⇒ correlated bits. For the Bell distribution $\{00: \tfrac12, 11: \tfrac12\}$: each marginal is uniform, but the joint is maximally correlated — knowing one bit determines the other. (API hook: BitArray slicing / marginal_counts compute exactly these sums.)

Conditional probability: $P(q_1{=}1 \mid q_0{=}1) = \frac{P(11)}{P(q_0{=}1)}$. For Bell: $= \frac{1/2}{1/2} = 1$.

4. Classical registers and multiple measurements

With several classical registers, the joint outcome is a tuple of bitstrings — mathematically still one joint distribution, keyed per register (result[0].data.<regname>). Mid-circuit measurement gives a sequential sample: first outcome ~ its marginal, state collapses, second outcome ~ the conditional distribution given the first. Deferred measurement (Day 2 §6) guarantees the joint distribution matches the measure-at-the-end circuit.

5. Comparing distributions

Two tools you'll meet in analysis questions:

• Total variation / trace distance between distributions: $d(P,Q) = \tfrac12\sum_x |P(x) - Q(x)| \in [0,1]$. 0 = identical, 1 = disjoint support.
• Hellinger fidelity (used by hellinger_fidelity): $F(P,Q) = \big(\sum_x \sqrt{P(x)Q(x)}\big)^2$. 1 = identical. For ideal-vs-noisy Bell counts this is the standard "how good was my run" number.

6. Sampling error vs shot budget (the recurring exam numbers)

Halving error costs 4× shots (error $\propto 1/\sqrt N$). Default shots in most V2 primitives: 4096-class values — enough for ~1% resolution.

---

PROBLEMS

P1. State: $|\psi\rangle = \tfrac{1}{\sqrt2}|00\rangle + \tfrac12|10\rangle - \tfrac12|11\rangle$ (labels $|q_1q_0\rangle$). Give the full outcome distribution, then both marginals $P(q_0)$, $P(q_1)$. Are the bits independent?

P2. For P1's state: 4000 shots are run. Expected counts per outcome, and typical fluctuation size for the '00' count?

P3. GHZ state $\tfrac{1}{\sqrt2}(|000\rangle + |111\rangle)$: what distribution do you get if you sample only qubits 0 and 1 (marginalize q2)? Are q0, q1 independent in that marginal?

P4. Distribution A = {'00': 0.5, '11': 0.5}; distribution B = {'00': 0.25, '01': 0.25, '10': 0.25, '11': 0.25}. Compute the total variation distance and the Hellinger fidelity.

P5. After h(0) and measuring q0 into c0, the circuit applies x(1) only if c0 = 1, then measures q1 into c1. Give the joint distribution of (c1, c0).

P6. A sampler run of a supposed Bell circuit returns {'00': 480, '11': 470, '01': 30, '10': 20} at 1000 shots. Statistical fluke or systematic error? Justify with a quick σ calculation.

P7. You need to distinguish $p = 0.50$ from $p = 0.51$ reliably (3σ separation). Roughly how many shots?

---

ANSWERS

A1. $P(00) = \tfrac12$, $P(10) = \tfrac14$, $P(11) = \tfrac14$ (the minus sign dies in $|\cdot|^2$ — phases are invisible to sampling). Marginals: $P(q_0{=}0) = \tfrac12 + \tfrac14 = \tfrac34$, $P(q_0{=}1) = \tfrac14$; $P(q_1{=}0) = \tfrac12$, $P(q_1{=}1) = \tfrac12$. Independence would need $P(00) = P(q_1{=}0)P(q_0{=}0) = \tfrac38 \ne \tfrac12$ ⇒ correlated.

A2. Expected: '00': 2000, '10': 1000, '11': 1000. Fluctuation for '00': $\sqrt{4000 \cdot \tfrac12 \cdot \tfrac12} \approx 32$.

A3. Marginal over q2: outcomes '00' and '11' (of $q_1q_0$) each with prob $\tfrac12$ — a Bell-like distribution. NOT independent: the marginal is still perfectly correlated (knowing q0 fixes q1). GHZ hides its 3-way correlation in every pairwise marginal.

A4. TV distance: $\tfrac12(|0.5{-}0.25| + |0{-}0.25| + |0{-}0.25| + |0.5{-}0.25|) = 0.5$. Hellinger: $\big(\sqrt{0.5\cdot0.25} + 0 + 0 + \sqrt{0.5\cdot0.25}\big)^2 = (2\cdot\tfrac{1}{\sqrt8})^2 = \tfrac{4}{8} = 0.5$.

A5. c0 is 0 or 1 with prob ½ each. If c0=0: q1 stays $|0\rangle$ ⇒ c1=0. If c0=1: X fires ⇒ c1=1. Joint: $P(c_1c_0{=}00) = \tfrac12$, $P(11) = \tfrac12$ — a perfectly correlated distribution created by feed-forward (this is the deferred-measurement picture of CNOT).

A6. Systematic. Wrong-parity outcomes total 50/1000 = 5%. Under the ideal distribution their probability is 0, and a true-0 event has σ ≈ 0 — 50 counts can't be a fluctuation. This is hardware noise (readout/gate error), the Day-6 topic.

A7. Need $3\sigma \le 0.01$ with $\sigma \approx \tfrac{1}{2\sqrt N}$ ⇒ $\tfrac{3}{2\sqrt N} \le 0.01$ ⇒ $\sqrt N \ge 150$ ⇒ $N \gtrsim 22\,500$ shots. (Order of magnitude is what matters: ~$10^4$–$10^5$.)

---

Self-check: you can turn any small statevector into its distribution instantly, marginalize and condition it, size up statistical fluctuations, and explain why sampling can't see phases.