The Exercise

Three particles of spin one half, or three qubits, have a state space of dimension eight. Their total angular momentum $\vec{J}$ is the sum of the three spins. In units where $\hbar = 1$, a state of total spin $j$ and spin component $m$ along $z$ is an eigenstate of $J^2 = \vec{J} \cdot \vec{J}$ with eigenvalue $j(j+1)$ and of the component $J_z$ with eigenvalue $m$. By the rule for adding angular momenta, the eight dimensions split into a quartet with $j = \tfrac{3}{2}$ and $m = \tfrac{3}{2}, \tfrac{1}{2}, -\tfrac{1}{2}, -\tfrac{3}{2}$, and two doublets with $j = \tfrac{1}{2}$ and $m = \pm\tfrac{1}{2}$.

The exercise takes eight explicit states of the three qubits, one for each place in this decomposition, and asks to show that each is an eigenstate of $J^2$ and of $J_z$ with the $j$ and $m$ of its place.

Three Qubits

The register of three qubits has eight configurations, the strings $abc$ of three binary digits, $$abc \in \{0,1\} \times \{0,1\} \times \{0,1\},$$(1)^✦ the $k$th digit belonging to qubit $k$. Its state space is $\mathbb{C}^2 \otimes \mathbb{C}^2 \otimes \mathbb{C}^2 = \mathbb{C}^8$: a vector assigns a complex amplitude to each string, and an operator is a complex $8 \times 8$ matrix with rows and columns labelled by the strings. The basis vectors $|abc\rangle$ are the strings’ own vectors and satisfy $$\langle a'b'c' | abc \rangle = \delta_{aa'} \, \delta_{bb'} \, \delta_{cc'}.$$(2)^✦

On a single qubit the Pauli matrices are $$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},$$(3) so that $|0\rangle$ is spin up along $z$ and $|1\rangle$ spin down. For a complex $2 \times 2$ matrix $M$, the operator $M_k$ on the register is $M$ acting on qubit $k$: the tensor product of $M$ in the $k$th factor with the identity $I$ in the other two, $$M_1 = M \otimes I \otimes I,$$(4) $$M_2 = I \otimes M \otimes I,$$(5) $$M_3 = I \otimes I \otimes M.$$(6)

The spin of qubit $k$ is $\tfrac{1}{2} \vec{\sigma}_k$, with Pauli vector $\vec{\sigma}_k = (X_k, Y_k, Z_k)$, so the components of $\vec{J}$ are $$J_x = \tfrac{1}{2} (X_1 + X_2 + X_3),$$(7) $$J_y = \tfrac{1}{2} (Y_1 + Y_2 + Y_3),$$(8) $$J_z = \tfrac{1}{2} (Z_1 + Z_2 + Z_3),$$(9) and its square is the operator $$J^2 = J_x^2 + J_y^2 + J_z^2.$$(10)

The Eight States

Four of the states are symmetric under every exchange of two qubits, each the normalized equal superposition of the strings with a given number of ones: $$|\tfrac{3}{2}, \tfrac{3}{2}\rangle = |000\rangle,$$(11) $$|\tfrac{3}{2}, \tfrac{1}{2}\rangle = \tfrac{1}{\sqrt{3}} \bigl( |100\rangle + |010\rangle + |001\rangle \bigr),$$(12) $$|\tfrac{3}{2}, -\tfrac{1}{2}\rangle = \tfrac{1}{\sqrt{3}} \bigl( |011\rangle + |101\rangle + |110\rangle \bigr),$$(13) $$|\tfrac{3}{2}, -\tfrac{3}{2}\rangle = |111\rangle.$$(14)

The other four form two pairs, marked $a$ and $b$. In pair $a$, qubits 1 and 3 are in the singlet $(|10\rangle - |01\rangle)/\sqrt{2}$, and qubit 2 is up in one state and down in the other: $$|\tfrac{1}{2}, \tfrac{1}{2}\rangle_a = \tfrac{1}{\sqrt{2}} \bigl( |100\rangle - |001\rangle \bigr),$$(15) $$|\tfrac{1}{2}, -\tfrac{1}{2}\rangle_a = \tfrac{1}{\sqrt{2}} \bigl( |110\rangle - |011\rangle \bigr).$$(16) The two states of pair $b$ are $$|\tfrac{1}{2}, \tfrac{1}{2}\rangle_b = \tfrac{1}{\sqrt{6}} \bigl( |001\rangle - 2\,|010\rangle + |100\rangle \bigr),$$(17) $$|\tfrac{1}{2}, -\tfrac{1}{2}\rangle_b = \tfrac{1}{\sqrt{6}} \bigl( 2\,|101\rangle - |110\rangle - |011\rangle \bigr).$$(18)

The Claim

What Is Checked

The claim is stated over two operators on the register of three qubits (1), the square of the total angular momentum (10) and its component along $z$ (9), both built from the Pauli matrices (3) placed on single qubits by (4) to (6), and over the eight states (11) to (18), each a combination of the basis vectors (2) with explicit coefficients.