The Exercise

Two qubits have a state space of dimension four, with the computational basis $|00\rangle$, $|01\rangle$, $|10\rangle$, $|11\rangle$. The Bell states $|\beta_{xy}\rangle$, one for each pair of bits $x$ and $y$, are four other vectors of that space, each a superposition of two computational basis states with amplitudes $\pm 1/\sqrt{2}$. The exercise is to show that they too form an orthonormal basis: each has unit norm, two different ones are orthogonal, and four orthonormal vectors in a space of dimension four span it.

Two Qubits

A quantum system $S$ is a physical system whose state space is a finite dimensional complex Hilbert space.^✦ The qubit $Q$ is the two level system, with state space $\mathbb{C}^2$.^✦ The composite system $S \otimes T$ of two quantum systems $S$ and $T$ is the quantum system whose state space is the tensor product of the state spaces of $S$ and $T$; two qubits form the composite $Q \otimes Q$, with state space $\mathbb{C}^2 \otimes \mathbb{C}^2$.^✦

A pure state $|\psi\rangle$ of a quantum system $S$ is a unit vector in the state space of $S$.^✦ For a bit $i$, the basis state $|i\rangle$ of the qubit is the pure state whose vector is the standard basis vector $e_i$ of $\mathbb{C}^2$, so that $|0\rangle$ and $|1\rangle$ are the computational basis states.^✦ The product $|\psi\rangle \otimes |\varphi\rangle$ of a pure state $|\psi\rangle$ of $S$ and a pure state $|\varphi\rangle$ of $T$ is the pure state of $S \otimes T$ whose vector is the tensor product of the two vectors; its norm is the product of the two norms, which is one.^✦ For bits $x$ and $y$ we write $|x\,y\rangle$ for $|x\rangle \otimes |y\rangle$.

The Bell States

For a bit $y$ let $\bar y = y + 1 \bmod 2$ be the other bit. The Bell vector $|\beta_{xy}\rangle$, for bits $x$ and $y$, is the vector of $\mathbb{C}^2 \otimes \mathbb{C}^2$ $$|\beta_{xy}\rangle = \frac{1}{\sqrt{2}} \bigl( |0\,y\rangle + (-1)^{x}\, |1\,\bar y\rangle \bigr) .$$(1)^✦

The Bell state $|\beta_{xy}\rangle$ is the pure state of two qubits whose vector is the Bell vector (1): $$|\beta_{xy}\rangle \in \mathbb{C}^2 \otimes \mathbb{C}^2 , \qquad \bigl\lVert\, |\beta_{xy}\rangle \,\bigr\rVert = 1 .$$(2)^✦

The four Bell states are orthonormal, $$\langle \beta_{xy} | \beta_{x'y'} \rangle = \delta_{xx'}\, \delta_{yy'} ,$$(3) and there are as many of them as the dimension of the two qubit state space: $$\#\{0,1\}^2 = 4 = \dim \bigl( \mathbb{C}^2 \otimes \mathbb{C}^2 \bigr) .$$(4)

The Bell basis $\mathcal{B}$ of two qubits is the orthonormal basis of $\mathbb{C}^2 \otimes \mathbb{C}^2$ obtained from the Bell states by (3) and (4), its vectors $|b_{xy}\rangle$ labelled by pairs of bits: $$\mathcal{B} = \bigl( |b_{00}\rangle,\, |b_{01}\rangle,\, |b_{10}\rangle,\, |b_{11}\rangle \bigr) .$$(5)^✦

The Claim

What Is Checked

The claim (6) is stated over the Bell basis (5) and the Bell states (2).

About these constructions the library proves the unit norm of each Bell vector, the orthonormality (3) of the Bell states and the count (4).