The Exercise

A qubit is prepared in the state $|+\rangle = (|0\rangle + |1\rangle)/\sqrt{2}$, the equal superposition of the two computational basis states. Two such qubits side by side are in the joint state $|+\rangle^{\otimes 2}$, and three are in $|+\rangle^{\otimes 3}$. The exercise asks for both joint states written out explicitly in two ways: as sums of products of basis states such as $|0\rangle|1\rangle$, and as vectors of amplitudes computed with the Kronecker product.

Basis Kets and the Kronecker Product

For $i$ in $\{0, 1\}$, the computational basis ket $|i\rangle$ is the vector of $\mathbb{C}^2$ with entry $1$ in position $i$ and $0$ in the other, so that $|0\rangle = (1, 0)$ and $|1\rangle = (0, 1)$.^✦ The state $|+\rangle$ is the vector of $\mathbb{C}^2$ whose two entries are both the number $1/\sqrt{2}$: $$|+\rangle = \left( \tfrac{1}{\sqrt{2}}, \, \tfrac{1}{\sqrt{2}} \right).$$(1)^✦

For a vector $|a\rangle$ with complex entries $a_i$ indexed by a set $I$ and a vector $|b\rangle$ with complex entries $b_j$ indexed by a set $J$, the Kronecker product $|a\rangle \otimes |b\rangle$ is the vector indexed by the pairs $(i, j)$ in $I \times J$ with entries $$\bigl( |a\rangle \otimes |b\rangle \bigr)_{(i,j)} = a_i \, b_j .$$(2)^✦

The twofold power of $|+\rangle$ is the vector $|+\rangle^{\otimes 2} = |+\rangle \otimes |+\rangle$, whose entries are indexed by pairs of bits. For bits $i$ and $j$, the basis product $|ij\rangle$ is the vector $|i\rangle \otimes |j\rangle$.

The threefold power of $|+\rangle$ is the vector $|+\rangle^{\otimes 3} = |+\rangle \otimes (|+\rangle \otimes |+\rangle)$, whose entries are indexed by a bit and a pair of bits. For bits $i$, $j$ and $k$, the basis product $|ijk\rangle$ is the vector $|i\rangle \otimes (|j\rangle \otimes |k\rangle)$.

The Claim

What Is Checked