Day 2 — Parameterized & Dynamic Circuits, Transpilation

Math Prep

Solve this BEFORE lesson.md. Day 2 is about circuits whose gates carry angle parameters and about rewriting circuits into hardware basis gates. The math you need: rotation gates as matrix exponentials, phase bookkeeping, and universal decomposition.

Requires: Day 1 math prep. Video back-up: Watrous Lesson 03 Quantum Circuits; notes in resource-artifactory/notes/video-notes/.

1. Rotation gates are matrix exponentials

For any Pauli $P \in \{X, Y, Z\}$ (each satisfies $P^2 = I$ ):

R_P(\theta) = e^{-i\theta P/2} = \cos\tfrac{\theta}{2}\, I - i \sin\tfrac{\theta}{2}\, P

Plug in each Pauli and you get the Day-1 matrices. Two consequences you must internalize:

The half angle. $\theta$ is the rotation angle on the Bloch sphere, but the matrix contains $\theta/2$ . So a Bloch-sphere rotation by $\pi$ around X is $R_x(\pi) = -iX$ , i.e. X up to global phase.
Period $4\pi$ , not $2\pi$ : $R_P(\theta + 2\pi) = -R_P(\theta)$ (a global-phase flip), and only $R_P(\theta+4\pi) = R_P(\theta)$ exactly. Parameterized-circuit sweep questions use this.

Composition along the same axis is addition: $R_z(\alpha)R_z(\beta) = R_z(\alpha+\beta)$ . Rotations about different axes do NOT commute: $R_x R_z \ne R_z R_x$ in general — that's why gate order matters and why assign_parameters order never matters (it's just substituting numbers) but circuit order does.

2. Phase bookkeeping: p vs rz, one more time

P(\lambda) = \begin{pmatrix}1&0\\0&e^{i\lambda}\end{pmatrix}, \qquad R_z(\lambda) = e^{-i\lambda/2} P(\lambda)

Alone: identical physics (global phase). Controlled: different gates! Controlling a unitary promotes its global phase to a relative phase on the control:

C\text{-}(e^{i\gamma}U) = \begin{pmatrix}I & 0\\ 0 & e^{i\gamma}U\end{pmatrix} \ne C\text{-}U \text{ when } e^{i\gamma}\ne1

So cp(λ) ≠ controlled-rz(λ) — they differ by a p(λ/2) on the control. This is why Qiskit tracks qc.global_phase at all: the transpiler must keep it to make controlled/composed circuits exact.

3. Euler decomposition — why 3 parameters are enough

Any single-qubit unitary can be written (up to global phase) as three rotations:

U = e^{i\gamma}\, R_z(\phi)\, R_y(\theta)\, R_z(\lambda)

which is exactly Qiskit's $U(\theta,\phi,\lambda)$ from Day 1. Equivalently with the hardware-native set:

U(\theta,\phi,\lambda) \sim R_z(\phi+\pi)\; \sqrt{X}\; R_z(\theta+\pi)\; \sqrt{X}\; R_z(\lambda)

This identity is the whole reason IBM hardware only needs {rz, sx, x} plus one entangler (cx/cz/ecr): any 1q gate = at most 2 physical $\sqrt{X}$ pulses + virtual $R_z$ 's. And $R_z$ is "free" (virtual): it's implemented as a frame rotation in software, zero error, zero duration. That's why transpiled circuits are stuffed with rz and why optimization levels merge runs of 1q gates into at most rz-sx-rz-sx-rz.

4. Conjugation identities — the transpiler's rewrite rules

"Conjugating" $A$ by $B$ means $BAB^\dagger$ . Memorize this table ( $H$ and $S$ conjugations move rotation axes):

Identity	Meaning
$HXH = Z$ , $HZH = X$ , $HYH = -Y$	H swaps X↔Z axes
$H R_x(\theta) H = R_z(\theta)$	rewrite rx using rz
$S X S^\dagger = Y$	S rotates X-axis to Y-axis
$(H\otimes H)\,CX_{0\to1}\,(H\otimes H) = CX_{1\to0}$	CNOT direction flip
$CZ = (I\otimes H)\, CX\, (I\otimes H)$	CX ↔ CZ with 2 Hadamards on target
$SWAP = CX_{01}CX_{10}CX_{01}$	3 CNOTs (routing cost!)

The CNOT-direction flip and the SWAP identity explain transpiler behavior on real topologies: if the coupling map only has $0\!\to\!1$ , a cx(1,0) costs 4 extra H's; a gate between non-neighbors costs SWAPs = 3 CNOTs each.

5. Commutation — when the order doesn't matter

$[A,B] = AB - BA$ . Facts used in dynamic-circuit reasoning and optimization:

Diagonal gates (z, s, t, p, rz, cz, cp, rzz) all commute with each other.
$Z$ -family on the control commutes through CX; $X$ -family on the target commutes through CX.
Measurement in the Z basis commutes with Z-diagonal gates before it — that's why a z right before measure is dead code the optimizer deletes.

6. The deferred measurement principle (dynamic circuits)

Any circuit with a mid-circuit measurement + classically-controlled gate (if_test) is equivalent to one where the measurement is pushed to the end and the classical control becomes a quantum control:

measure q0 → if result: X on q1 ≡ CX(q0→q1) then measure q0.

That's the math behind Day 2's dynamic circuits and behind teleportation's "corrections."

PROBLEMS

P1. Expand $R_y(\theta) = \cos\frac{\theta}{2} I - i\sin\frac{\theta}{2} Y$ into a matrix and confirm it matches Day 1's $R_y$ .

P2. Show $R_x(2\pi) = -I$ . What measurement statistics distinguish the rx(2π) circuit from the identity circuit? What does this imply about Operator.equiv vs ==?

P3. Compute $R_z(\pi/2) R_z(\pi/2)$ and name the resulting gate (up to global phase).

P4. Using $R_z(\lambda) = e^{-i\lambda/2}P(\lambda)$ , find the matrix by which controlled- $R_z(\lambda)$ and controlled- $P(\lambda)$ differ, and express it as a gate on the control qubit.

P5. Verify $H R_x(\theta) H = R_z(\theta)$ using $R_x = \cos\frac{\theta}{2}I - i\sin\frac{\theta}{2}X$ and $HXH=Z$ .

P6. A device's coupling map has only the edge $0\!\to\!1$ (control 0, target 1). Count the total cx + h gates needed to implement cx(1,0).

P7. How many CNOTs does the transpiler insert to run a cx between qubits that are 2 hops apart (path 0–1–2, gate wanted between 0 and 2), if it SWAPs 0's state to 1 first and doesn't swap back?

P8. $\theta$ sweep: for ry(θ) acting on $|0\rangle$ , give $P(1)$ as a function of θ, and evaluate at $\theta = 0, \pi/2, \pi, 2\pi$ .

ANSWERS

A1. $-iY = \begin{pmatrix}0&-1\\1&0\end{pmatrix}$ , so $R_y(\theta) = \begin{pmatrix}\cos\frac{\theta}{2} & -\sin\frac{\theta}{2}\\ \sin\frac{\theta}{2} & \cos\frac{\theta}{2}\end{pmatrix}$ . ✓ (All real — that's why ry is the go-to for real-amplitude ansätze.)

A2. $R_x(2\pi) = \cos\pi\, I - i\sin\pi\, X = -I$ . None — no measurement can distinguish $-I$ from $I$ (global phase). Hence Operator(rx(2π)).equiv(Operator(id)) is True but == is False.

A3. $R_z(\pi/2)^2 = R_z(\pi) = -iZ$ — that is $Z$ (equivalently $S^2$ ) up to the global phase $-i$ .

A4. $C\text{-}R_z(\lambda) = C\text{-}(e^{-i\lambda/2}P(\lambda))$ : the control- $|1\rangle$ block gets an extra factor $e^{-i\lambda/2}$ . So $C\text{-}R_z(\lambda) = \big(P(-\lambda/2)\ \text{on the control}\big)\cdot C\text{-}P(\lambda)$ — the difference is a phase gate $P(-\lambda/2)$ on the control qubit.

A5. $H R_x(\theta) H = \cos\frac{\theta}{2}\, HIH - i\sin\frac{\theta}{2}\, HXH = \cos\frac{\theta}{2} I - i \sin\frac{\theta}{2} Z = R_z(\theta)$ . ∎

A6. cx(1,0) = H on both qubits, cx(0,1), H on both: 1 cx + 4 h.

A7. One SWAP (0↔1) = 3 CNOTs, then the desired cx on the now-adjacent pair = 1: 4 CNOTs total (and the logical→physical layout has changed — that's why the transpiled circuit carries a final layout, a concept in the lesson).

A8. $R_y(\theta)|0\rangle = \cos\frac{\theta}{2}|0\rangle + \sin\frac{\theta}{2}|1\rangle \Rightarrow P(1) = \sin^2\frac{\theta}{2}$ . Values: $\theta=0$ : 0; $\theta=\pi/2$ : $\sin^2(\pi/4) = 1/2$ ; $\theta=\pi$ : 1; $\theta=2\pi$ : 0. (Sweep questions: $P(1)$ oscillates with period $2\pi$ in θ even though the matrix period is $4\pi$ .)

Self-check: you can state $R_P(\theta) = \cos\frac{\theta}{2}I - i\sin\frac{\theta}{2}P$ , explain why rz is "free" on IBM hardware, and produce the CX↔CZ and SWAP identities from memory.