Measurement, Protocols, Circuits

1. Quantum Measurement

Quantum measurement is the process of extracting classical information from a quantum state.

An orthonormal basis is a set of vectors ${| φ_{i} ⟩}$ such that their inner product satisfies:

⟨ φ_{i} | φ_{j} ⟩ = δ_{i j} = {\begin{cases} 1 & i = j \\ 0 & i \neq j \end{cases}

Examples for a 1-qubit system:

Computational Basis: ${| 0 ⟩, | 1 ⟩}$
Hadamard Basis: ${| + ⟩, | - ⟩}$ , where:
- $| + ⟩ = \frac{1}{\sqrt{2}} (| 0 ⟩ + | 1 ⟩)$
- $| - ⟩ = \frac{1}{\sqrt{2}} (| 0 ⟩ - | 1 ⟩)$

For a 2-qubit system, the basis expands to ${| 00 ⟩, | 01 ⟩, | 10 ⟩, | 11 ⟩}$ .

Let $B = {| φ_{i} ⟩}$ be an orthonormal basis. If a system is in state $| ψ ⟩ = \sum_{i} a_{i} | φ_{i} ⟩$ :

Important Note: The global phase does not matter, as $| a_{i} |^{2} = | e^{i θ} a_{i} |^{2}$ .

Example: Measuring $| ψ ⟩ = \frac{1}{2} | 0 ⟩ + \frac{\sqrt{3}}{2} | 1 ⟩$ in the ${| + ⟩, | - ⟩}$ basis:

$⟨ + | ψ ⟩ = \frac{1 + \sqrt{3}}{2 \sqrt{2}}$ , so $P r (+) = | ⟨ + | ψ ⟩ |^{2} = \frac{2 + \sqrt{3}}{4}$ .
$⟨ - | ψ ⟩ = \frac{1 - \sqrt{3}}{2 \sqrt{2}}$ , so $P r (-) = \frac{2 - \sqrt{3}}{4}$ .

When measuring only the first qubit of a multi-qubit system:

Given $| ψ ⟩ = \sqrt{\frac{1}{10}} | 00 ⟩ + \sqrt{\frac{2}{10}} | 01 ⟩ + \sqrt{\frac{3}{10}} | 10 ⟩ + \sqrt{\frac{4}{10}} | 11 ⟩$ :

Grouping by the first qubit: $| ψ ⟩ = \sqrt{\frac{3}{10}} | 0 ⟩ \otimes (\sqrt{\frac{1}{3}} | 0 ⟩ + \sqrt{\frac{2}{3}} | 1 ⟩) + \sqrt{\frac{7}{10}} | 1 ⟩ \otimes (\sqrt{\frac{3}{7}} | 0 ⟩ + \sqrt{\frac{4}{7}} | 1 ⟩)$ .
$P r (0) = \frac{3}{10}$ ; State becomes $| 0 ⟩ \otimes (\sqrt{\frac{1}{3}} | 0 ⟩ + \sqrt{\frac{2}{3}} | 1 ⟩)$ .
$P r (1) = \frac{7}{10}$ ; State becomes $| 1 ⟩ \otimes (\sqrt{\frac{3}{7}} | 0 ⟩ + \sqrt{\frac{4}{7}} | 1 ⟩)$ .

Superdense coding allows Alice to send two classical bits to Bob by sending only one qubit, provided they share an entangled pair (ebit).

The Procedure:

Initial State: Alice and Bob share $| β_{00} ⟩ = \frac{1}{\sqrt{2}} (| 00 ⟩ + | 11 ⟩)$ .
Alice's Operation: Depending on the bits $(x y)$ she wants to send, Alice applies a gate to her qubit:
- 00: Apply $I \to | β_{00} ⟩ = \frac{1}{\sqrt{2}} (| 00 ⟩ + | 11 ⟩)$ .
- 01: Apply $X \to | β_{01} ⟩ = \frac{1}{\sqrt{2}} (| 10 ⟩ + | 01 ⟩)$ .
- 10: Apply $Z \to | β_{10} ⟩ = \frac{1}{\sqrt{2}} (| 00 ⟩ - | 11 ⟩)$ .
- 11: Apply $i Y$ (or $Z X$ ) $\to | β_{11} ⟩ = \frac{1}{\sqrt{2}} (| 01 ⟩ - | 10 ⟩)$ .
Transmission: Alice sends her qubit to Bob.
Decoding: Bob measures both qubits in the Bell basis to retrieve $x y$ .

Teleportation is the reverse: moving a quantum state $| ψ ⟩$ using two classical bits and one ebit.

The Procedure:

Setup: Alice has a qubit in state $| ψ ⟩ = a_{0} | 0 ⟩ + a_{1} | 1 ⟩$ and shares $| β_{00} ⟩$ with Bob.
Alice's Measurement: Alice performs a Bell measurement on her two qubits (the unknown state and her half of the ebit).
The State Shift: The total state can be rewritten such that Bob’s qubit is in one of four states: $| ψ ⟩$ , $X | ψ ⟩$ , $Z | ψ ⟩$ , or $X Z | ψ ⟩$ , depending on Alice's outcome.
Correction: Alice sends her classical outcomes (2 bits) to Bob. Bob applies the corresponding Pauli gates ( $Z^{a} X^{b}$ ) to recover exactly $| ψ ⟩$ .

Any single-qubit state can be represented as a point on the Bloch Sphere:

| ψ ⟩ = \cos \frac{θ}{2} | 0 ⟩ + e^{i ϕ} \sin \frac{θ}{2} | 1 ⟩

Gates rotate the state vector around axes on the Bloch Sphere:

$R_{x} (θ) = e^{- i θ X / 2} = (\begin{matrix} \cos \frac{θ}{2} & - i \sin \frac{θ}{2} \\ - i \sin \frac{θ}{2} & \cos \frac{θ}{2} \end{matrix})$
$R_{y} (θ) = (\begin{matrix} \cos \frac{θ}{2} & - \sin \frac{θ}{2} \\ \sin \frac{θ}{2} & \cos \frac{θ}{2} \end{matrix})$
$R_{z} (θ) = (\begin{matrix} e^{- i θ / 2} & 0 \\ 0 & e^{i θ / 2} \end{matrix})$

CNOT Gate: Flips the target qubit if the control qubit is $| 1 ⟩$ .
- $| 0 ⟩ | ψ ⟩ \to | 0 ⟩ | ψ ⟩$
- $| 1 ⟩ | ψ ⟩ \to | 1 ⟩ X | ψ ⟩$
Creating Entanglement: A Hadamard gate (H) followed by a CNOT gate converts $| 00 ⟩$ into the Bell state $| β_{00} ⟩$ .