Lean, informalized

Nielsen & Chuang · Chapter 7, Quantum computers: physical realization · p. 314

Exercise 7.26 - The matrices of J2J^2 and jzj_z for two qubits, and their four common eigenvectors

Two spins combine into a triplet and a singlet.

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The Exercise

Two particles of spin 12\tfrac12, here two qubits, have a total angular momentum, the sum of their two spins. Its components jxj_x, jyj_y, jzj_z and its square J2J^2 act on the state space of the pair, which has dimension four. The exercise is to write J2J^2 and jzj_z as matrices in the computational basis, and to show that the three triplet states and the singlet are eigenvectors of both: of J2J^2 with eigenvalue j(j+1)j(j+1), where the total spin is j=1j = 1 on the triplet and j=0j = 0 on the singlet, and of jzj_z with eigenvalue mm, the zz component of the total spin.

Two Spins

An operator on two qubits is written as a 4×44 \times 4 complex matrix in the computational basis, with rows and columns in the order |00⟩|00\rangle, |01⟩|01\rangle, |10⟩|10\rangle, |11⟩|11\rangle: the basis state |ab⟩|ab\rangle stands in position 2a+b2a + b, counting from 00.✦

With ℏ=1\hbar = 1, the spin of a single qubit is half its Pauli matrices XX, YY, ZZ. The components jxj_x, jyj_y, jzj_z of the total angular momentum of two qubits are the operators jx=12(X⊗I+I⊗X),j_x = \tfrac{1}{2} \left( X \otimes I + I \otimes X \right),(1) jy=12(Y⊗I+I⊗Y),j_y = \tfrac{1}{2} \left( Y \otimes I + I \otimes Y \right),(2) jz=12(Z⊗I+I⊗Z),j_z = \tfrac{1}{2} \left( Z \otimes I + I \otimes Z \right),(3) on the pair, where II is the 2×22 \times 2 identity. The square J2J^2 of the total angular momentum is the operator J2=jx2+jy2+jz2.J^2 = j_x^2 + j_y^2 + j_z^2 .(4)✦

Four states of the pair, the coupled states, carry the labels |j,m⟩J|j, m\rangle_J of a total spin jj and its zz component mm; the subscript JJ marks them as eigenstates of the total angular momentum, to tell them from the product states |ab⟩|ab\rangle. The triplet states, with j=1j = 1, are |1,1⟩J=|00⟩,|1, 1\rangle_J = |00\rangle,(5) |1,0⟩J=12(|01⟩+|10⟩),|1, 0\rangle_J = \frac{1}{\sqrt{2}} \left( |01\rangle + |10\rangle \right),(6) |1,−1⟩J=|11⟩,|1, -1\rangle_J = |11\rangle,(7) and the singlet, with j=0j = 0, is |0,0⟩J=12(|01⟩−|10⟩).|0, 0\rangle_J = \frac{1}{\sqrt{2}} \left( |01\rangle - |10\rangle \right).(8)

The Claim

Theorem 1. Let jxj_x, jyj_y, jzj_z be the components (1) to (3) of the total angular momentum of two qubits, and J2J^2 its square (4). In the computational basis, J2=(2000011001100002)J^2 = \begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix}(9)✦ and jz=(100000000000000−1).j_z = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} .(10)✦ The triplet states and the singlet are eigenvectors of J2J^2, J2|1,1⟩J=2|1,1⟩J,J^2 \, |1, 1\rangle_J = 2 \, |1, 1\rangle_J ,(11)✦ J2|1,0⟩J=2|1,0⟩J,J^2 \, |1, 0\rangle_J = 2 \, |1, 0\rangle_J ,(12)✦ J2|1,−1⟩J=2|1,−1⟩J,J^2 \, |1, -1\rangle_J = 2 \, |1, -1\rangle_J ,(13)✦ J2|0,0⟩J=0,J^2 \, |0, 0\rangle_J = 0 ,(14)✦ and of jzj_z, jz|1,1⟩J=|1,1⟩J,j_z \, |1, 1\rangle_J = |1, 1\rangle_J ,(15)✦ jz|1,0⟩J=0,j_z \, |1, 0\rangle_J = 0 ,(16)✦ jz|1,−1⟩J=−|1,−1⟩J,j_z \, |1, -1\rangle_J = - |1, -1\rangle_J ,(17)✦ jz|0,0⟩J=0.j_z \, |0, 0\rangle_J = 0 .(18)✦

What Is Checked

The claim is stated over the operators jzj_z and J2J^2 of (3) and (4) and over the coupled states (5) to (8). For these operators and states the library shows that J2J^2 and jzj_z have the matrices (9) and (10) in the computational basis, and that each of the triplet states and the singlet is an eigenvector of both, with the eigenvalues (11) to (18).