Lean, informalized

Nielsen & Chuang · Chapter 11, Entropy and information · p. 515

Exercise 11.15 - A generalized measurement on a qubit, its outcome unread, that lowers the von Neumann entropy of some state

Resetting a qubit by measurement can lower its entropy.

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The Exercise

A qubit undergoes a generalized measurement with the two measurement operators |0⟩⟨0||0\rangle\langle 0| and |0⟩⟨1||0\rangle\langle 1|, and nobody reads the outcome. The exercise is to show that this measurement can lower the von Neumann entropy of the qubit.

A Quantum System

A quantum system AA is a physical object whose state space, its mathematical realisation, is a finite dimensional complex Hilbert space.✦ Each finite dimensional complex Hilbert space HH gives a system, the system whose state space is HH itself.✦ The qubit is the two level system given in this way by ℂ2\mathbb{C}^2, the standard two dimensional complex space.✦

A pure state ψ\psi of AA is a vector of unit norm, ‖ψ‖=1\lVert \psi \rVert = 1, in the state space of AA, determined physically up to a global phase.✦ The basis states |0⟩|0\rangle and |1⟩|1\rangle of the qubit are the pure states whose vectors are the standard basis vectors e0e_0 and e1e_1 of ℂ2\mathbb{C}^2.✦

A state ρ\rho of AA is a positive operator of unit trace on the state space of AA; pure and mixed states are states alike.✦ The von Neumann entropy of a state ρ\rho of AA is the real number S(ρ)=−tr⁡(ρln⁡ρ),S(\rho) = -\operatorname{tr}(\rho \ln \rho),(1)✦ the Shannon entropy of the eigenvalues of ρ\rho, measured in nats.

Measurement Without Reading the Outcome

A measurement on AA with a finite set of outcomes is a family of measurement operators MiM_i on the state space of AA, one for each outcome ii, that satisfies the completeness relation ∑iMi†Mi=I.\sum_i M_i^\dagger M_i = I.(2)✦

For a measurement with operators MiM_i and a state ρ\rho of AA, the unread postmeasurement state is the state ρ′\rho' of AA with density operator ρ′=∑iMiρMi†.\rho' = \sum_i M_i \rho M_i^\dagger.(3)✦ It describes the system after the measurement when the outcome is not learned.

The reset measurement is the measurement of the exercise, on the qubit, with outcomes i=0,1i = 0, 1 and operators Mi=|0⟩⟨i|.M_i = |0\rangle\langle i|.(4)✦ Both operators map into |0⟩|0\rangle. ρ′=|0⟩⟨0|ρ|0⟩⟨0|+|0⟩⟨1|ρ|1⟩⟨0|=(⟨0|ρ|0⟩+⟨1|ρ|1⟩)|0⟩⟨0|=|0⟩⟨0|.\rho' = |0\rangle\langle 0|\rho|0\rangle\langle 0| + |0\rangle\langle 1|\rho|1\rangle\langle 0| = \bigl(\langle 0|\rho|0\rangle + \langle 1|\rho|1\rangle\bigr)\,|0\rangle\langle 0| = |0\rangle\langle 0|.(5)

The Claim

Theorem 1. Let M0M_0 and M1M_1 be the operators (4) of the reset measurement on the qubit. There is a state ρ\rho of the qubit whose unread postmeasurement state has strictly lower von Neumann entropy than ρ\rho: S(∑iMiρMi†)<S(ρ).S\Bigl(\sum_{i} M_i \rho M_i^\dagger\Bigr) < S(\rho).(6)✦

What Is Checked

The claim is stated over the qubit and its states, the reset measurement (4) with its unread postmeasurement state (3), and the von Neumann entropy (1).

The library proves that the reset operators satisfy the completeness relation (2), M0†M0+M1†M1=|0⟩⟨0|+|1⟩⟨1|=IM_0^\dagger M_0 + M_1^\dagger M_1 = |0\rangle\langle 0| + |1\rangle\langle 1| = I, and that the basis vectors e0e_0 and e1e_1 have unit norm. For any measurement and any state ρ\rho, it proves that ∑iMiρMi†\sum_i M_i \rho M_i^\dagger is again a state: the operator is positive because ρ\rho is, and its trace is tr⁡(∑iMi†Miρ)=tr⁡ρ=1\operatorname{tr}\bigl(\sum_i M_i^\dagger M_i \rho\bigr) = \operatorname{tr} \rho = 1 by completeness.