Lean, informalized

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

Exercise 11.11 - Two pure states of a qubit have zero entropy, a mixed one the entropy of its two eigenvalues

The entropy of two pure qubit states and a mixed one.

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 qubit is prepared in one of three states, with density matrices in the computational basis {|0⟩,|1⟩}\{|0\rangle, |1\rangle\} (1000),12(1111),13(2111).\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \qquad \frac12 \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, \qquad \frac13 \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}.(1) The first two are pure, the projectors |0⟩⟨0||0\rangle\langle 0| and |+⟩⟨+||+\rangle\langle +| with |+⟩=(|0⟩+|1⟩)/2|+\rangle = (|0\rangle + |1\rangle)/\sqrt2; the third is mixed. The exercise asks for the von Neumann entropy of each.

States of a Qubit

A quantum system QQ 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.✦ A state ρ\rho of QQ is a positive operator of unit trace on the state space of QQ.✦

States of the qubit are written with the Pauli 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) Let σ→=(X,Y,Z)\vec\sigma = (X, Y, Z) be the Pauli vector and II the identity matrix. 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✦, and the Bloch matrix of r→\vec r is 12(I+r→⋅σ→)=12(1+r3r1−ir2r1+ir21−r3).\frac12 \left( I + \vec r \cdot \vec\sigma \right) = \frac12 \begin{pmatrix} 1 + r_3 & r_1 - i r_2 \\ r_1 + i r_2 & 1 - r_3 \end{pmatrix}.(3)✦ The Bloch state of a real vector r→\vec r with r12+r22+r32≤1r_1^2 + r_2^2 + r_3^2 \le 1 is the state of the qubit whose density operator is its Bloch matrix (3).✦

The three states of (1) are the Bloch states of (0,0,1)(0, 0, 1), (1,0,0)(1, 0, 0) and (23,0,13)(\tfrac23, 0, \tfrac13).

The Entropy

The von Neumann entropy of a state ρ\rho of a system QQ is S(ρ)=−tr⁡(ρln⁡ρ),S(\rho) = -\operatorname{tr}(\rho \ln \rho),(4)✦ the sum of −λln⁡λ-\lambda \ln \lambda over the eigenvalues λ\lambda of ρ\rho, taken with the natural logarithm and so measured in nats. The Shannon entropy of two probabilities p1p_1 and p2p_2 is H(p1,p2)=−p1ln⁡p1−p2ln⁡p2H(p_1, p_2) = -p_1 \ln p_1 - p_2 \ln p_2, with 0ln⁡0=00 \ln 0 = 0; for a state with two eigenvalues p1p_1 and p2p_2, S(ρ)=H(p1,p2)S(\rho) = H(p_1, p_2).

The eigenvalues of the Bloch matrix (3) are (1±|r→|)/2(1 \pm |\vec r|)/2, with |r→||\vec r| the length of r→\vec r.

The Claim

Theorem 1. The von Neumann entropies of the three states (1) of the qubit are S(|0⟩⟨0|)=0,S(|0\rangle\langle 0|) = 0,(5)✦ S(|+⟩⟨+|)=0,S(|+\rangle\langle +|) = 0,(6)✦ and S(13(2111))=H(3+56,3−56).S\left( \frac13 \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \right) = H\left( \frac{3 + \sqrt5}{6}, \frac{3 - \sqrt5}{6} \right).(7)✦

What Is Checked

The claim is stated over the von Neumann entropy (4) of three Bloch states of the qubit, those of the vectors (0,0,1)(0, 0, 1), (1,0,0)(1, 0, 0) and (23,0,13)(\tfrac23, 0, \tfrac13), and over the Shannon entropy HH of two numbers.

The Bloch matrix of a real vector r→\vec r is positive exactly when |r→|≤1|\vec r| \le 1, and its trace is one. The Bloch state of a vector in the closed unit ball is therefore a density operator, a state of the qubit.