Lean, informalized

Nielsen & Chuang · Chapter 12, Quantum information theory · p. 546

Exercise 12.6 - Overlap and fidelity of a block of the source with a product of rotated basis states

Block fidelity depends only on the Hamming weight.

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The Exercise

A source emits qubits one at a time, each in one of two pure states: |0⟩|0\rangle, or the state obtained from |0⟩|0\rangle by a rotation through π/2\pi/2 about the yˆ\hat y axis.✦ A block of nn emissions is a product state of nn qubits, labelled by the string kk of nn bits that records which of the two states each qubit received. Rotating |0⟩|0\rangle and |1⟩|1\rangle through π/4\pi/4 about the same axis gives a second basis of the qubit, and its products, labelled by strings xx of nn bits, are product states of the nn qubits as well. The exercise asks for the overlap of a block of the source with one of these products, and for the fidelity between the two. The library states both for every length nn and every pair of labels.

States of a Quantum System

A quantum system SS is a physical system whose state space is a finite dimensional complex Hilbert space.✦ The qubit is the system whose state space is ℂ2\mathbb{C}^2 itself✦, and its computational basis states |0⟩|0\rangle and |1⟩|1\rangle are the standard basis vectors of ℂ2\mathbb{C}^2.✦

A pure state |ψ⟩|\psi\rangle of SS is a unit vector in the state space of SS, fixed physically up to a global phase.✦ A state ρ\rho of SS is a positive operator of unit trace on the state space of SS.✦ Each pure state |ψ⟩|\psi\rangle gives the state |ψ⟩⟨ψ||\psi\rangle\langle\psi|, the projector onto its vector.✦

The overlap ⟨φ|ψ⟩\langle\varphi|\psi\rangle of two pure states |φ⟩|\varphi\rangle and |ψ⟩|\psi\rangle of SS is the complex inner product of their vectors, conjugate linear in |φ⟩|\varphi\rangle.✦ The fidelity of two states ρ\rho and σ\sigma of SS is the nonnegative number F(ρ,σ)=tr⁡ρσρ,F(\rho,\sigma) = \operatorname{tr}\sqrt{\sqrt{\rho}\,\sigma\sqrt{\rho}},(1)✦ where the square root of a positive operator takes the square root of each eigenvalue.

The composite ⨂iSi\bigotimes_i S_i of a finite family of systems SiS_i is the system whose state space is the tensor product of the state spaces of the SiS_i.✦ Given a pure state |χi⟩|\chi_i\rangle of each SiS_i, their product is the pure state of the composite with vector ⨂i|χi⟩\bigotimes_i |\chi_i\rangle.✦

Rotations of a Qubit

A unitary UU of SS is an operator on the state space of SS with U†U=UU†=1U^\dagger U = U U^\dagger = 1.✦ It carries a pure state |ψ⟩|\psi\rangle of SS to the pure state U|ψ⟩U|\psi\rangle.✦

On the qubit, the Pauli matrices are 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) For a real vector n→=(nx,ny,nz)\vec n = (n_x, n_y, n_z) the matrix n→⋅σ→\vec n \cdot \vec\sigma is nxX+nyY+nzZn_x X + n_y Y + n_z Z✦, and the rotation about n→\vec n through a real angle θ\theta is Rn→(θ)=exp⁡(−iθ2n→⋅σ→),R_{\vec n}(\theta) = \exp\Bigl(-\frac{i\theta}{2}\,\vec n \cdot \vec\sigma\Bigr),(3)✦ written Rnˆ(θ)R_{\hat n}(\theta) when the axis is a unit vector nˆ\hat n. About the axis yˆ=(0,1,0)\hat y = (0, 1, 0) this is Ry(θ)=Ryˆ(θ)=e−iθY/2=(cos⁡θ2−sin⁡θ2sin⁡θ2cos⁡θ2).R_y(\theta) = R_{\hat y}(\theta) = e^{-i\theta Y/2} = \begin{pmatrix} \cos\frac{\theta}{2} & -\sin\frac{\theta}{2} \\[2pt] \sin\frac{\theta}{2} & \cos\frac{\theta}{2} \end{pmatrix}.(4)✦ For a real angle θ\theta, the yy rotation gate Ry(θ)R_y(\theta) of the qubit is the unitary of the qubit whose matrix is the one displayed in (4).

The Source and the Rotated Basis

The two states of the source are |ψ0⟩=|0⟩,|ψ1⟩=Ry(π/2)|0⟩=|0⟩+|1⟩2.|\psi_0\rangle = |0\rangle, \qquad |\psi_1\rangle = R_y(\pi/2)\,|0\rangle = \frac{|0\rangle + |1\rangle}{\sqrt 2}.(5)✦ The rotated basis has one state for each bit bb, |ϕb⟩=Ry(π/4)|b⟩,|ϕ0⟩=cos⁡π8|0⟩+sin⁡π8|1⟩,|ϕ1⟩=−sin⁡π8|0⟩+cos⁡π8|1⟩.|\phi_b\rangle = R_y(\pi/4)\,|b\rangle, \qquad |\phi_0\rangle = \cos\tfrac{\pi}{8}\,|0\rangle + \sin\tfrac{\pi}{8}\,|1\rangle, \quad |\phi_1\rangle = -\sin\tfrac{\pi}{8}\,|0\rangle + \cos\tfrac{\pi}{8}\,|1\rangle.(6)✦

For strings of bits x=x1⋯xnx = x_1 \cdots x_n and k=k1⋯knk = k_1 \cdots k_n, the blocks are the product states of the composite of nn qubits, with state space (ℂ2)⊗n(\mathbb{C}^2)^{\otimes n}, |Φx⟩=|ϕx1⟩⊗⋯⊗|ϕxn⟩,|Ψk⟩=|ψk1⟩⊗⋯⊗|ψkn⟩.|\Phi_x\rangle = |\phi_{x_1}\rangle \otimes \cdots \otimes |\phi_{x_n}\rangle, \qquad |\Psi_k\rangle = |\psi_{k_1}\rangle \otimes \cdots \otimes |\psi_{k_n}\rangle.(7)

The Claim

Theorem 1. Let xx and kk be strings of nn bits, and let |x||x|, the Hamming weight of xx, be the number of ones in xx. The overlap of the blocks |Φx⟩|\Phi_x\rangle and |Ψk⟩|\Psi_k\rangle is the product of the overlaps of their qubits, ⟨Φx|Ψk⟩=∏i=1n⟨ϕxi|ψki⟩,\langle \Phi_x | \Psi_k \rangle = \prod_{i=1}^{n} \langle \phi_{x_i} | \psi_{k_i} \rangle,(8)✦ and the fidelity of the states they give is F(|Φx⟩⟨Φx|,|Ψk⟩⟨Ψk|)=(cos⁡π8)n−|x|(sin⁡π8)|x|.F\bigl(|\Phi_x\rangle\langle\Phi_x|,\ |\Psi_k\rangle\langle\Psi_k|\bigr) = \Bigl(\cos\frac{\pi}{8}\Bigr)^{n-|x|} \Bigl(\sin\frac{\pi}{8}\Bigr)^{|x|}.(9)✦

What Is Checked

The claim is stated over the blocks (7), products of the source states (5) and of the rotated basis states (6) as pure states of the composite of nn qubits.

The library proves that each of these is a state of its system. A tensor product of unit vectors is a unit vector, so each block is a pure state of the composite, whose state space is finite dimensional because the products of basis vectors of the factors form a basis of it.