Day 3 — Quantum Operations

Math Prep

Solve this BEFORE lesson.md. Day 3 is qiskit.quantum_info — the module that is the math. Everything here maps 1:1 onto a class you'll use: Pauli, SparsePauliOp, Statevector, Operator, DensityMatrix, partial_trace, state_fidelity.

Requires: Day 1–2 math prep. Video back-up: Watrous Lessons 01–02 and (deeper, optional) Lesson 09 Density Matrices; notes in resource-artifactory/notes/video-notes/.

1. Pauli algebra — the multiplication table

X^2 = Y^2 = Z^2 = I,\qquad XY = iZ,\quad YZ = iX,\quad ZX = iY

Reversed order flips the sign: $YX = -iZ$ , etc. So distinct Paulis anticommute: $XY = -YX$ .

Every product of Paulis is again a Pauli times a phase from $\{\pm1, \pm i\}$ — that's why Pauli objects in Qiskit carry a phase and print things like -iY. Multi-qubit Paulis are tensor strings like $Z\otimes I\otimes X$ = "ZIX" (little-endian trap: in the string "ZIX", the RIGHTMOST letter is qubit 0 — so X acts on q0, I on q1, Z on q2).

Commutation of Pauli strings: count the positions where the two strings differ and neither is I. Odd count ⇒ anticommute; even ⇒ commute. (Pauli.commutes does exactly this.)

2. Hermitian vs unitary — the two big families

Property	Definition	Role	Qiskit
Hermitian	$A^\dagger = A$	observables; real eigenvalues	`SparsePauliOp`, measurement operators
Unitary	$U^\dagger U = I$	evolution/gates; eigenvalues $e^{i\varphi}$ on unit circle	`Operator`, gates

Paulis are both (Hermitian AND unitary) — eigenvalues $\pm1$ . H is also both. S and T are unitary but NOT Hermitian ( $S^\dagger = S^{-1} \ne S$ ).

Eigen-facts to memorize (basis of measurement-in-a-basis questions):

Operator	+1 eigenvector	−1 eigenvector
$Z$	$\\|0\rangle$	$\\|1\rangle$
$X$	$\\|+\rangle$	$\\|-\rangle$
$Y$	$\\|{+i}\rangle = \tfrac{1}{\sqrt2}(\\|0\rangle{+}i\\|1\rangle)$	$\\|{-i}\rangle$

3. Outer products and projectors

$|\psi\rangle\langle\phi|$ = column × row = a matrix. Key example:

|0\rangle\langle0| = \begin{pmatrix}1&0\\0&0\end{pmatrix},\quad |1\rangle\langle1| = \begin{pmatrix}0&0\\0&1\end{pmatrix},\quad Z = |0\rangle\langle0| - |1\rangle\langle1|

A projector satisfies $\Pi^2 = \Pi = \Pi^\dagger$ . Measurement probability via projectors: $P(x) = \langle\psi|\Pi_x|\psi\rangle$ — same Born rule, new clothes. Spectral theorem: every Hermitian $A = \sum_j \lambda_j \Pi_j$ (eigenvalues × projectors), e.g. $Z = (+1)|0\rangle\langle0| + (-1)|1\rangle\langle1|$ .

4. Expectation values — the number Estimator computes

\langle A \rangle_\psi = \langle\psi|A|\psi\rangle = \sum_j \lambda_j \,P(\text{outcome } j)

Worked examples you should be able to do cold:

$\langle Z\rangle_{|0\rangle} = 1$ , $\langle Z\rangle_{|1\rangle} = -1$ , $\langle Z\rangle_{|+\rangle} = 0$ , $\langle X\rangle_{|+\rangle} = 1$ .
General 1-qubit state: $\langle Z\rangle = |\alpha|^2 - |\beta|^2 = P(0) - P(1)$ . ⟨Z⟩ is just "probability difference."
Bell state $|\Phi^+\rangle$ : $\langle Z\otimes Z\rangle = 1$ (outcomes always agree), $\langle Z \otimes I\rangle = 0$ (each side alone is random), $\langle X\otimes X\rangle = 1$ .

Linearity: $\langle 2ZZ + 3XI\rangle = 2\langle ZZ\rangle + 3\langle XI\rangle$ — this is why SparsePauliOp Hamiltonians are evaluated term by term.

5. Density matrices — states as matrices

\rho = |\psi\rangle\langle\psi| \;\text{(pure)},\qquad \rho = \sum_k p_k |\psi_k\rangle\langle\psi_k| \;\text{(mixed)}

Properties: $\rho^\dagger = \rho$ , $\rho \succeq 0$ , $\mathrm{Tr}(\rho) = 1$ . Purity $= \mathrm{Tr}(\rho^2)$ : equals 1 iff pure; minimum $1/2^n$ for the maximally mixed state $I/2^n$ .

$|+\rangle\langle+| = \tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix}$ — pure superposition: off-diagonals present (coherence).
$\tfrac12 I$ — 50/50 classical mixture: no off-diagonals, purity $\tfrac12$ . Both give 50/50 Z-measurements; only $|+\rangle$ gives deterministic X-measurements. That's the entire pure-vs-mixed distinction.

Born rule and expectations in ρ-form: $P(x) = \mathrm{Tr}(\Pi_x \rho)$ , $\langle A\rangle = \mathrm{Tr}(A\rho)$ .

6. Partial trace — the state of a subsystem

For $\rho_{AB}$ , "trace out B" produces A's reduced state $\rho_A = \mathrm{Tr}_B(\rho_{AB})$ : keep the terms whose B-index matches in bra and ket, drop the rest.

The flagship fact: each qubit of a Bell state is maximally mixed:

\mathrm{Tr}_B\big(|\Phi^+\rangle\langle\Phi^+|\big) = \tfrac12 I

Entanglement of the pure whole ⇔ mixedness of the parts. (partial_trace(Statevector(bell), [0]) → DensityMatrix of $\tfrac12 I$ ; purity 0.5.)

7. Fidelity — how close are two states

F(|\psi\rangle, |\phi\rangle) = |\langle\psi|\phi\rangle|^2 \in [0,1]

(Qiskit's state_fidelity uses the squared-overlap convention: 1 = same state, 0 = orthogonal. Watrous's book uses the square root of this — beware when cross-reading.) Global phase drops out: $F(|\psi\rangle, e^{i\gamma}|\psi\rangle) = 1$ . Example: $F(|0\rangle, |+\rangle) = |\langle0|+\rangle|^2 = \tfrac12$ .

PROBLEMS

P1. Compute $XZ$ and express it as a phase times a Pauli. Do $X$ and $Z$ commute or anticommute?

P2. Do the Pauli strings "XX" and "ZZ" commute? "XI" and "ZZ"? (Use the differ-and-not-I counting rule.)

P3. Is $T$ Hermitian? Is it unitary? What are its eigenvalues?

P4. Write the projector onto $|-\rangle$ as a 2×2 matrix and verify $\Pi^2 = \Pi$ .

P5. Compute $\langle X\rangle$ for $|\psi\rangle = \cos\frac{\theta}{2}|0\rangle + \sin\frac{\theta}{2}|1\rangle$ (real amplitudes).

P6. For the Bell state $|\Phi^+\rangle$ , compute $\langle Z\otimes Z\rangle$ and $\langle Z\otimes I\rangle$ from the outcome probabilities.

P7. Write $\rho$ for the classical mixture "70% $|0\rangle$ , 30% $|1\rangle$ " and for the pure state $\sqrt{0.7}|0\rangle + \sqrt{0.3}|1\rangle$ . Which matrix entries differ? Compute both purities.

P8. Show $\mathrm{Tr}_{q_0}(|\Phi^+\rangle\langle\Phi^+|) = \tfrac12 I$ by expanding $|\Phi^+\rangle\langle\Phi^+|$ and applying the keep-matching-B-indices rule.

P9. Compute state_fidelity between $|+\rangle$ and $|{+i}\rangle = \tfrac{1}{\sqrt2}(|0\rangle + i|1\rangle)$ .

P10. The exam asks: "Which is true of $\rho = I/2$ ?" — (a) it's a pure superposition, (b) measuring in ANY basis gives 50/50, (c) it equals $|+\rangle\langle+|$ . Pick and justify.

ANSWERS

A1. $XZ = \begin{pmatrix}0&-1\\1&0\end{pmatrix} = -iY$ . And $ZX = iY \ne XZ$ ⇒ they anticommute ( $XZ = -ZX$ ).

A2. "XX" vs "ZZ": differ at 2 positions ⇒ even ⇒ commute (both are symmetries of the Bell basis). "XI" vs "ZZ": differ at 1 non-identity position ⇒ odd ⇒ anticommute.

A3. $T^\dagger = \mathrm{diag}(1, e^{-i\pi/4}) \ne T$ ⇒ not Hermitian. $T^\dagger T = I$ ⇒ unitary. Eigenvalues: $1$ and $e^{i\pi/4}$ (it's diagonal).

A4. $|-\rangle\langle-| = \tfrac12\begin{pmatrix}1&-1\\-1&1\end{pmatrix}$ . Squaring gives $\tfrac12\begin{pmatrix}1&-1\\-1&1\end{pmatrix}$ again. ✓

A5. $\langle\psi|X|\psi\rangle = 2\cos\frac{\theta}{2}\sin\frac{\theta}{2} = \sin\theta$ . Sanity check at $\theta=\pi/2$ : the state is $|+\rangle$ ... no — at $\theta = \pi/2$ the state is $\cos\frac{\pi}{4}|0\rangle + \sin\frac{\pi}{4}|1\rangle = |+\rangle$ and $\sin(\pi/2) = 1 = \langle X\rangle_{|+\rangle}$ . ✓

A6. $ZZ$ outcomes: $|00\rangle, |11\rangle$ each prob $\tfrac12$ , bits always equal ⇒ eigenvalue $+1$ always ⇒ $\langle ZZ\rangle = 1$ . Single qubit is uniformly random ⇒ $\langle Z\otimes I\rangle = \tfrac12(+1)+\tfrac12(-1) = 0$ .

A7. $\rho_{\text{mix}} = \begin{pmatrix}0.7&0\\0&0.3\end{pmatrix}$ ; $\rho_{\text{pure}} = \begin{pmatrix}0.7&\sqrt{0.21}\\\sqrt{0.21}&0.3\end{pmatrix}$ . Difference: the off-diagonal coherences ( $\sqrt{0.21}\approx0.458$ ). Purities: mix $0.49+0.09 = 0.58$ ; pure $= 1$ (check: $0.49+0.09+2\cdot0.21 = 1$ ✓).

A9. $\langle+|{+i}\rangle = \tfrac12(1 + i)$ , so $F = \left|\tfrac{1+i}{2}\right|^2 = \tfrac{2}{4} = \tfrac12$ .

A10. (b). $P(x) = \mathrm{Tr}(\Pi_x \tfrac{I}{2}) = \tfrac12$ for any rank-1 projector, in every basis. (a),(c) false: $I/2$ has purity $\tfrac12$ (mixed); $|+\rangle\langle+|$ has off-diagonals.

Self-check: you can multiply any two Paulis with the right phase, compute ⟨Pauli⟩ for the standard states instantly, tell pure from mixed by looking at a density matrix, and state why a Bell state's reduced state is $I/2$ .