Lean, informalized

Nielsen & Chuang · Chapter 12, Quantum information theory · p. 586

Proposition 12.18 - A unitary probe that leaves two nonorthogonal states unchanged cannot record which one it met

Telling nonorthogonal states apart disturbs them.

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The Exercise

A system SS is prepared in one of two pure states, |ψ⟩|\psi\rangle or |φ⟩|\varphi\rangle, which are not orthogonal, and an observer wants to learn which one without disturbing it. She brings in an ancilla AA prepared in a pure state |u⟩|u\rangle and lets the two evolve together under a unitary UU. The best she could hope for leaves SS exactly as it was and the ancilla in a state that depends on the preparation: |v⟩|v\rangle after |ψ⟩|\psi\rangle, a different |v′⟩|v'\rangle after |φ⟩|\varphi\rangle, so that AA carries information about which state was prepared. The proposition is that no unitary does this: any information gained about two nonorthogonal states disturbs at least one of them.✦

A Quantum System

A quantum system SS is a physical system whose state space is a finite dimensional complex Hilbert space.✦ The composite system S⊗AS \otimes A of two quantum systems SS and AA is the quantum system whose state space is the tensor product of the state spaces of SS and AA.✦

A pure state |ψ⟩|\psi\rangle of SS is a unit vector in the state space of SS.✦ The overlap ⟨ψ|φ⟩\langle\psi|\varphi\rangle of two pure states |ψ⟩|\psi\rangle and |φ⟩|\varphi\rangle of SS is the inner product of their vectors, a complex number, conjugate linear in the first vector and linear in the second; the two states are orthogonal when it vanishes.✦

For pure states |ψ⟩|\psi\rangle of SS and |u⟩|u\rangle of AA, the product state |ψ⟩⊗|u⟩|\psi\rangle \otimes |u\rangle is the pure state of S⊗AS \otimes A whose vector is the tensor product of the two vectors.✦

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

The Claim

Theorem 1. Let SS and AA be quantum systems, let |ψ⟩|\psi\rangle and |φ⟩|\varphi\rangle be pure states of SS that are not orthogonal, ⟨ψ|φ⟩≠0\langle\psi|\varphi\rangle \neq 0, and let |u⟩|u\rangle, |v⟩|v\rangle and |v′⟩|v'\rangle be pure states of AA with |v⟩≠|v′⟩|v\rangle \neq |v'\rangle. Then there is no unitary UU of S⊗AS \otimes A with U(|ψ⟩⊗|u⟩)=|ψ⟩⊗|v⟩andU(|φ⟩⊗|u⟩)=|φ⟩⊗|v′⟩.U\bigl(|\psi\rangle \otimes |u\rangle\bigr) = |\psi\rangle \otimes |v\rangle \quad\text{and}\quad U\bigl(|\varphi\rangle \otimes |u\rangle\bigr) = |\varphi\rangle \otimes |v'\rangle .(2)✦

What Is Checked

The claim (2) is stated over product states of the composite system S⊗AS \otimes A and a unitary of S⊗AS \otimes A acting on them, with the overlap ⟨ψ|φ⟩\langle\psi|\varphi\rangle as its hypothesis on the system.

The library proves that each side of those equations is a pure state of S⊗AS \otimes A. A product |ψ⟩⊗|u⟩|\psi\rangle \otimes |u\rangle is a unit vector because the norm of a tensor product of vectors is the product of their norms, and UU carries a unit vector to a unit vector because a unitary (1) preserves norms.