Lean, informalized

Nielsen & Chuang · Chapter 11, Entropy and information · p. 511

Exercise 11.12 - The von Neumann entropy of a qubit mixture, computed and compared with the Shannon entropy of its weights

The entropy of a mixture of two nonorthogonal states.

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The Exercise

A qubit is prepared in the state |0⟩|0\rangle with probability pp and in the state |+⟩=(|0⟩+|1⟩)/2|+\rangle = (|0\rangle + |1\rangle)/\sqrt{2} with probability 1−p1-p. The two pure states are not orthogonal, ⟨0|+⟩=1/2\langle 0|+\rangle = 1/\sqrt{2}, and the density operator of the preparation is ρ=p|0⟩⟨0|+(1−p)|+⟩⟨+|.\rho = p\,|0\rangle\langle 0| + (1-p)\,|+\rangle\langle +| .(1)✦ The exercise asks for the von Neumann entropy S(ρ)S(\rho) of this state and for its comparison with the Shannon entropy H(p,1−p)H(p, 1-p) of the weights.

A Quantum System

A quantum system SS is a physical system whose state space is a finite dimensional complex Hilbert space.✦ Any such space HH is itself the state space of a system✦, and the qubit is the two level system obtained in this way from ℂ2\mathbb{C}^2✦. A state ρ\rho of SS is a positive operator of unit trace on the state space of SS.✦

The Pauli matrices are the complex 2×22 \times 2 matrices X=(0110),Y=(0−ii0),Z=(100−1).X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.(2) The Pauli vector is σ→=(X,Y,Z)\vec\sigma = (X, Y, Z), and for a real vector r→=(r1,r2,r3)\vec r = (r_1, r_2, r_3) the matrix r→⋅σ→\vec r \cdot \vec\sigma is r1X+r2Y+r3Zr_1 X + r_2 Y + r_3 Z✦. Write II for the 2×22 \times 2 identity matrix. The Bloch matrix of r→\vec r is 12(I+r→⋅σ→).\tfrac12 \left( I + \vec r \cdot \vec\sigma \right).(3)✦ The Bloch state of a real vector r→\vec r with ∥r→∥≤1\|\vec r\| \le 1 is the state of the qubit whose density operator is the Bloch matrix (3).✦

The mixture of two states ρ\rho and σ\sigma of SS with weight pp, where 0≤p≤10 \le p \le 1, is the state of SS with density operator pρ+(1−p)σp\rho + (1-p)\sigma: the state prepared by choosing ρ\rho with probability pp and σ\sigma otherwise.✦

The State of the Exercise

The state of the exercise is the mixture, with weight pp, of the Bloch states of the unit vectors zˆ=(0,0,1)\hat z = (0, 0, 1) and xˆ=(1,0,0)\hat x = (1, 0, 0).✦

For probabilities q1q_1 and q2q_2 the Shannon entropy is H(q1,q2)H(q_1, q_2), and the von Neumann entropy of a state ρ\rho is the Shannon entropy of the eigenvalues of its density operator, measured in nats: H(q1,q2)=−q1ln⁡q1−q2ln⁡q2,S(ρ)=−tr⁡(ρln⁡ρ),H(q_1, q_2) = -q_1 \ln q_1 - q_2 \ln q_2 , \qquad S(\rho) = -\operatorname{tr}(\rho \ln \rho),(4)✦ with 0ln⁡0=00 \ln 0 = 0.

The Claim

The Bloch matrix depends linearly on r→\vec r, so the density operator of ρ\rho is the Bloch matrix of r→=pzˆ+(1−p)xˆ=(1−p,0,p)\vec r = p\,\hat z + (1-p)\,\hat x = (1-p, 0, p). Its eigenvalues are (1±∥r→∥)/2(1 \pm \|\vec r\|)/2, with ∥r→∥=(1−p)2+p2\|\vec r\| = \sqrt{(1-p)^2 + p^2}.

Theorem 1. Let 0≤p≤10 \le p \le 1 and let ρ\rho be the state (1). Its von Neumann entropy is S(ρ)=H(1+(1−p)2+p22,1−(1−p)2+p22),S(\rho) = H\!\left( \frac{1 + \sqrt{(1-p)^2 + p^2}}{2},\ \frac{1 - \sqrt{(1-p)^2 + p^2}}{2} \right),(5)✦ and it is at most the Shannon entropy of the weights: S(ρ)≤H(p,1−p).S(\rho) \le H(p, 1-p).(6)✦ If 0<p<10 < p < 1, the inequality is strict: S(ρ)<H(p,1−p).S(\rho) < H(p, 1-p).(7)✦

What Is Checked

The three statements are made over the qubit and the state ρ\rho, built as the mixture with weight pp of the Bloch states of zˆ\hat z and xˆ\hat x; they compare its von Neumann entropy with the Shannon entropy of two probabilities.

About these constructions the library proves that the Bloch matrix (3) is positive exactly when ∥r→∥≤1\|\vec r\| \le 1 and that its trace is one, so the Bloch state is a state of the qubit, and that the convex combination pρ+(1−p)σp\rho + (1-p)\sigma of two density operators is again a density operator, so the mixture is a state.