Math Prep

Solve this BEFORE lesson.md. Goal: when the lesson says "x, y, z, h, s, sdg, t, tdg, p, rx, ry, rz, u, sx, sxdg, id are methods on the circuit," you know exactly what 2×2 matrix each name is and what it does to a state.

Video back-up: freeCodeCamp Quantum Computing – Math and Theory for Beginners (resource-artifactory/videos/freecodecamp-qc-math/) and Watrous Lesson 01 Single Systems, Lesson 02 Multiple Systems (resource-artifactory/videos/watrous-uqic/). Notes: resource-artifactory/notes/video-notes/.

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1. Complex numbers in 10 minutes

A complex number: $z = a + bi$, with $i^2 = -1$.

• Modulus (length): $|z| = \sqrt{a^2 + b^2}$. Note $|z|^2 = z \cdot z^*$ where $z^* = a - bi$ is the conjugate.
• Euler form: $z = r e^{i\varphi} = r(\cos\varphi + i\sin\varphi)$.
• The unit circle values you will use constantly:

Multiplying by $e^{i\varphi}$ rotates a complex number by angle $\varphi$ and never changes its length. That's why phases don't change measurement probabilities on their own.

2. States are column vectors (Dirac notation)

A qubit state is a length-1 column vector in $\mathbb{C}^2$:

• |·⟩ ("ket") = column vector. ⟨·| ("bra") = its conjugate transpose (a row vector): $\langle\psi| = (|\psi\rangle)^\dagger$.
• General state: $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle = \begin{pmatrix}\alpha\\\beta\end{pmatrix}$ with the normalization rule $|\alpha|^2 + |\beta|^2 = 1$.
• $\alpha, \beta$ are amplitudes (complex). $|\alpha|^2$ = probability of measuring 0, $|\beta|^2$ = probability of measuring 1 (Born rule).
• Inner product: $\langle\phi|\psi\rangle$ = row·column = a complex number. $\langle\psi|\psi\rangle = 1$ for normalized states. Two states are orthogonal when $\langle\phi|\psi\rangle=0$.

The four named single-qubit states you must recognize on sight:

All four give 50/50 outcomes in the standard basis — they differ only in relative phase, which other gates can later convert into probability differences. That's the whole trick of quantum interference.

3. Gates are unitary matrices

A gate is a $2^n \times 2^n$ matrix $U$ applied by multiplication: $|\psi_{\text{out}}\rangle = U|\psi_{\text{in}}\rangle$. It must be unitary: $U^\dagger U = I$ (where $U^\dagger$ = conjugate-transpose, called the adjoint or dagger). Unitary ⇔ preserves lengths ⇔ probabilities always sum to 1 ⇔ reversible (inverse is $U^\dagger$).

Every gate in the Day-1 list, decoded

Pauli gates — the "classical-ish" ones:

Hadamard — the superposition-maker:

$H$ is its own inverse ($H^2 = I$) and converts between X-basis and Z-basis: $HXH = Z$, $HZH = X$.

Phase-family gates — all diagonal, all only touch the $|1\rangle$ amplitude:

So: $T^2 = S$, $S^2 = Z$, $T^4=Z$, $S\cdot S^\dagger = I$. "dg" = dagger = adjoint = inverse.

Rotation gates — continuous versions, angle $\theta$ around a Bloch-sphere axis (half-angle $\theta/2$ appears in the matrix — memorize that):

$R_z(\theta)$ vs $P(\theta)$: same up to the global phase $e^{-i\theta/2}$ — physically identical alone, but the difference matters once controlled. (This is exactly the lesson's p(pi) == z exact vs rz(pi) = -iZ trap.)

Square-root-of-X:

The universal one-qubit gate u: every single-qubit unitary (up to global phase) is

Check: $U(\pi/2, 0, \pi) = H$, $\;U(0,0,\lambda) = P(\lambda)$, $\;U(\pi, 0, \pi) = X$.

Global vs relative phase — the exam's favorite trap

• $|\psi\rangle$ and $e^{i\gamma}|\psi\rangle$ are the same physical state (global phase, unobservable).
• $\tfrac{1}{\sqrt2}(|0\rangle + |1\rangle)$ vs $\tfrac{1}{\sqrt2}(|0\rangle - |1\rangle)$ differ by a relative phase — different states ($|+\rangle$ vs $|-\rangle$), distinguishable after an H.

4. Multiple qubits: the tensor product ⊗

Two qubits live in $\mathbb{C}^4$; states of independent qubits combine with the Kronecker/tensor product:

Basis order: $|00\rangle, |01\rangle, |10\rangle, |11\rangle$ = indices 0,1,2,3 (binary!). Gates combine the same way: applying $A$ to one qubit and $B$ to the other is $A \otimes B$, a 4×4 matrix. $n$ qubits ⇒ $2^n$ amplitudes — that's the exponential state space.

Qiskit's little-endian twist: qubit 0 is the LEAST significant bit. A state written $|q_1 q_0\rangle$ means the left symbol is qubit 1. x(0) on 2 qubits gives bitstring '01'. Matrix-wise Qiskit's 2-qubit operator for "A on q1, B on q0" is $A \otimes B$ — matching a.tensor(b) putting b on the lower wires.

Controlled gates. CNOT (cx) flips the target iff the control is $|1\rangle$. With control = q0 (Qiskit order, control on the low qubit):

General recipe: $C\text{-}U = |0\rangle\langle0|_{\text{ctl}} \otimes I + |1\rangle\langle1|_{\text{ctl}} \otimes U$. Because $CZ$'s action ($-1$ only on $|11\rangle$) doesn't care which qubit is "control," CZ/CP/SWAP-type gates are symmetric; CX/CY/CH are not — exactly the lesson's symmetric/asymmetric table.

Entanglement. A state is entangled iff it cannot be written as $|a\rangle\otimes|b\rangle$. The Bell state

has no such factorization (try it: you'd need $a_0b_0 = a_1b_1 = \tfrac{1}{\sqrt2}$ but $a_0b_1 = 0$ — impossible).

5. Measurement (just enough for Day 1)

Measuring in the computational basis: outcome $x$ with probability $|\langle x|\psi\rangle|^2$ = |amplitude|², then the state collapses to $|x\rangle$. Measurement is NOT unitary and NOT reversible — that's why inverse() refuses circuits containing measure.

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PROBLEMS — do on paper, answers below

P1. Normalize $\begin{pmatrix}3\\4i\end{pmatrix}$ and give the probability of measuring 1.

P2. Compute $H|1\rangle$ by matrix multiplication. Which named state is it?

P3. Show $S \cdot S = Z$ and $T^\dagger T = I$ by multiplying matrices.

P4. Compute $HZH$ and identify the resulting gate.

P5. What does $X$ do to $|-\rangle$? Give the resulting state and explain why measuring it in the standard basis can't tell the difference.

P6. Verify $R_z(\pi) = e^{-i\pi/2} Z = -iZ$. Why is $R_z(\pi)$ alone physically identical to $Z$?

P7. Write out $|+\rangle \otimes |1\rangle$ as a 4-vector. If this is "q1 = |+⟩, q0 = |1⟩" in Qiskit convention, which bitstrings can be measured and with what probabilities?

P8. Prove $\tfrac{1}{\sqrt2}(|00\rangle + |11\rangle)$ is entangled (no product decomposition exists).

P9. Apply CNOT (control q0, target q1) to $|+\rangle$ on q0 and $|0\rangle$ on q1, i.e. the state $\tfrac{1}{\sqrt2}(|00\rangle+|01\rangle)$ written as $|q_1q_0\rangle$. What comes out?

P10. Using $U(\theta,\phi,\lambda)$, verify $U(\pi/2,0,\pi) = H$.

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ANSWERS

A1. Norm $= \sqrt{|3|^2 + |4i|^2} = \sqrt{9+16} = 5$. State: $\tfrac{1}{5}\begin{pmatrix}3\\4i\end{pmatrix}$. $P(1) = |4i/5|^2 = 16/25$.

A2. $H|1\rangle = \tfrac{1}{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix} = \tfrac{1}{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix} = |-\rangle$.

A3. $S^2 = \begin{pmatrix}1&0\\0&i\end{pmatrix}^2 = \begin{pmatrix}1&0\\0&i^2\end{pmatrix} = \begin{pmatrix}1&0\\0&-1\end{pmatrix} = Z$. $T^\dagger T = \begin{pmatrix}1&0\\0&e^{-i\pi/4}\end{pmatrix}\begin{pmatrix}1&0\\0&e^{i\pi/4}\end{pmatrix} = I$.

A4. $HZH = X$. (Compute: $ZH = \tfrac{1}{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix}$, then $H(ZH) = \begin{pmatrix}0&1\\1&0\end{pmatrix}$.) H swaps the X and Z bases.

A5. $X|-\rangle = \tfrac{1}{\sqrt2}(X|0\rangle - X|1\rangle) = \tfrac{1}{\sqrt2}(|1\rangle - |0\rangle) = -|-\rangle$. Only a global phase changed; standard-basis probabilities are 50/50 before AND after, and global phase is unobservable.

A6. $R_z(\pi) = \begin{pmatrix}e^{-i\pi/2}&0\\0&e^{i\pi/2}\end{pmatrix} = \begin{pmatrix}-i&0\\0&i\end{pmatrix} = -i\begin{pmatrix}1&0\\0&-1\end{pmatrix} = -iZ$. The factor $-i$ is a global phase ⇒ same physical action as $Z$ (but Operator(rz(π)) == Operator(z) is False while .equiv is True).

A7. $|+\rangle\otimes|1\rangle = \tfrac{1}{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}\otimes\begin{pmatrix}0\\1\end{pmatrix} = \tfrac{1}{\sqrt2}(0,1,0,1)^T$, i.e. $\tfrac{1}{\sqrt2}(|01\rangle + |11\rangle)$ reading $|q_1q_0\rangle$. Measured bitstrings: '01' and '11', 50% each — q0 is always 1 (rightmost bit), q1 is random. In code: plus.tensor(one) since a.tensor(b) puts b (here $|1\rangle$) on qubit 0.

A8. Suppose $(a_0|0\rangle + a_1|1\rangle)\otimes(b_0|0\rangle + b_1|1\rangle)$ has amplitudes $a_0b_0 = \tfrac{1}{\sqrt2}$, $a_0b_1 = 0$, $a_1b_0 = 0$, $a_1b_1 = \tfrac{1}{\sqrt2}$. From $a_0b_1=0$: either $a_0=0$ (contradicts $a_0b_0\ne0$) or $b_1=0$ (contradicts $a_1b_1\ne0$). No factorization ⇒ entangled. ∎

A9. CNOT maps $|00\rangle \mapsto |00\rangle$ and (control q0 = 1) $|01\rangle \mapsto |11\rangle$. Result: $\tfrac{1}{\sqrt2}(|00\rangle + |11\rangle) = |\Phi^+\rangle$ — the Bell state. This is exactly h(0); cx(0,1) from the lesson.

A10. $U(\pi/2,0,\pi)$: $\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \tfrac{1}{\sqrt2}$, so the matrix is $\begin{pmatrix}\tfrac{1}{\sqrt2} & -e^{i\pi}\tfrac{1}{\sqrt2}\\ \tfrac{1}{\sqrt2} & e^{i\pi}\tfrac{1}{\sqrt2}\end{pmatrix} = \tfrac{1}{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix} = H$. ✓

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Self-check before starting lesson.md: you should be able to write from memory the matrices for X, Y, Z, H, S, T, P(λ), Rx, Ry, Rz, and CNOT, and state the Born rule and the tensor-product recipe. If not, redo P2–P4 and re-read §3.