Lean, informalized

Nielsen & Chuang · Chapter 12, Quantum information theory · p. 530

Exercise 12.1 - One unitary on two qubits copies either of two orthogonal pure states onto a blank qubit

Two orthogonal states of a qubit can be cloned.

Hover over a blue or green term in the text, or a node in the graph, to find it in the other. Select it to read its declaration both in Lean and in words.

The Exercise

A cloning machine for a qubit takes the qubit in some pure state |ψ⟩|\psi\rangle together with a second, blank qubit prepared in a fixed state, and leaves both qubits in |ψ⟩|\psi\rangle. By the no cloning theorem no unitary does this for every |ψ⟩|\psi\rangle. The exercise restricts the input to one of two orthogonal pure states |ψ⟩|\psi\rangle and |φ⟩|\varphi\rangle, known in advance, and asks for a unitary on the pair of qubits that copies whichever of the two it is given onto a blank qubit in |0⟩|0\rangle.

Qubits and Pure States

A quantum system SS is a physical object whose state space is a finite dimensional complex Hilbert space.✦ The qubit is the two level system whose state space is ℂ2\mathbb{C}^2. The composite system S⊗TS \otimes T of two quantum systems SS and TT is the quantum system whose state space is the tensor product of the state spaces of SS and TT; for two qubits it is ℂ2⊗ℂ2\mathbb{C}^2 \otimes \mathbb{C}^2.✦

A pure state |ψ⟩|\psi\rangle of SS is a unit vector in the state space of SS, ⟨ψ|ψ⟩=1\langle\psi|\psi\rangle = 1.✦ For a bit ii, the basis state |i⟩|i\rangle of the qubit is the pure state whose vector is the standard basis vector eie_i of ℂ2\mathbb{C}^2; these are the computational basis states |0⟩|0\rangle and |1⟩|1\rangle.✦ For pure states |ψ⟩|\psi\rangle of SS and |φ⟩|\varphi\rangle of TT, the product |ψ⟩⊗|φ⟩|\psi\rangle \otimes |\varphi\rangle, written |ψ⟩|φ⟩|\psi\rangle|\varphi\rangle, is the pure state of S⊗TS \otimes T whose vector is the tensor product of the two vectors.✦

A unitary UU of SS, a closed evolution of SS, is an operator on the state space of SS with U†U=UU†=I.U^\dagger U = U U^\dagger = I .(1)✦ Applied to a pure state |ψ⟩|\psi\rangle of SS, UU gives the pure state U|ψ⟩U|\psi\rangle, the vector of |ψ⟩|\psi\rangle acted on by UU.✦

Fidelity

A state ρ\rho of SS is a positive operator of unit trace on the state space of SS.✦ A pure state |ψ⟩|\psi\rangle of SS determines the state |ψ⟩⟨ψ||\psi\rangle\langle\psi| of SS, the rank one projector onto its vector.✦ The fidelity of two states ρ\rho and σ\sigma of SS is the nonnegative real number F(ρ,σ)=tr⁡ρσρ,F(\rho,\sigma) = \operatorname{tr}\sqrt{\sqrt{\rho}\,\sigma\sqrt{\rho}} ,(2)✦ where the square root of a positive operator is taken eigenvalue by eigenvalue.

The Claim

Theorem 1. Let |ψ⟩|\psi\rangle and |φ⟩|\varphi\rangle be pure states of the qubit such that the states |ψ⟩⟨ψ||\psi\rangle\langle\psi| and |φ⟩⟨φ||\varphi\rangle\langle\varphi| have fidelity zero. Then there is a unitary UU of the composite system of two qubits with U(|ψ⟩|0⟩)=|ψ⟩|ψ⟩andU(|φ⟩|0⟩)=|φ⟩|φ⟩.U\bigl(|\psi\rangle|0\rangle\bigr) = |\psi\rangle|\psi\rangle \quad\text{and}\quad U\bigl(|\varphi\rangle|0\rangle\bigr) = |\varphi\rangle|\varphi\rangle .(3)✦

What Is Checked

The claim is stated over pure states of the qubit and of the pair of qubits, the basis state |0⟩|0\rangle, the product of pure states, the fidelity (2) of the states |ψ⟩⟨ψ||\psi\rangle\langle\psi| and |φ⟩⟨φ||\varphi\rangle\langle\varphi|, and unitaries (1) of the two qubit system acting on its pure states.

The library proves that each of these constructions produces a state. The standard basis vectors of ℂ2\mathbb{C}^2 have unit norm; the norm of a product vector is the product of the norms, so |ψ⟩|φ⟩|\psi\rangle|\varphi\rangle is a unit vector; a unitary preserves norms, so U|ψ⟩U|\psi\rangle is one as well.