The Exercise

Three pure states of two qubits are given: the Bell state $|\Phi^+\rangle$, the product state $|+\rangle|+\rangle$, and $(|00\rangle + |01\rangle + |10\rangle)/\sqrt{3}$. For each, the exercise asks for its Schmidt coefficients and for orthonormal vectors of the two qubits that write it as a sum of products, one product for each coefficient.

Two Qubits

A quantum system $S$ is given by its state space, a finite dimensional complex Hilbert space.^✦ A pure state $|\psi\rangle$ of $S$ is a unit vector in the state space of $S$.^✦ 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 $T$.^✦

The qubit is the quantum system whose state space is $\mathbb{C}^2$.^✦ The computational basis states $|0\rangle$ and $|1\rangle$ of the qubit are the pure states whose vectors are the two standard basis vectors of $\mathbb{C}^2$, and for two qubits $|jk\rangle$ stands for $|j\rangle \otimes |k\rangle$.^✦ The state $|+\rangle$ of the qubit, the eigenvector of the Pauli operator $X$ with eigenvalue $+1$, is the pure state whose vector is the column $(1/\sqrt{2}, 1/\sqrt{2})$: $$|+\rangle = \frac{1}{\sqrt{2}}\bigl(|0\rangle + |1\rangle\bigr).$$(1)^✦

The Schmidt Decomposition

A Schmidt decomposition of a pure state $|\psi\rangle$ of a composite system $S \otimes T$ consists of $n$ positive numbers $\lambda_1, \dots, \lambda_n$ whose squares sum to one, orthonormal vectors $|a_1\rangle, \dots, |a_n\rangle$ in the state space of $S$, and orthonormal vectors $|b_1\rangle, \dots, |b_n\rangle$ in the state space of $T$, such that $$|\psi\rangle = \sum_{i=1}^{n} \lambda_i \, |a_i\rangle \otimes |b_i\rangle .$$(2)^✦ The positive numbers $\lambda_i$ of a Schmidt decomposition of $|\psi\rangle$ are the Schmidt coefficients of $|\psi\rangle$. The number $n$ of terms in a Schmidt decomposition of $|\psi\rangle$ is the Schmidt number of $|\psi\rangle$.

The Three States

The Bell state $|\Phi^+\rangle$ is the pure state of two qubits whose vector is $|00\rangle + |11\rangle$ divided by its norm $\sqrt{2}$: $$|\Phi^+\rangle = \frac{1}{\sqrt{2}}\bigl(|00\rangle + |11\rangle\bigr).$$(3)^✦

For pure states $|\psi\rangle$ of $S$ and $|\varphi\rangle$ of $T$, the product state $|\psi\rangle \otimes |\varphi\rangle$ is the pure state of $S \otimes T$ whose vector is the tensor product of the two vectors.^✦ The second state is the product of two copies of (1): $$|+\rangle|+\rangle = |+\rangle \otimes |+\rangle = \frac{1}{2}\bigl(|00\rangle + |01\rangle + |10\rangle + |11\rangle\bigr).$$(4)^✦

For a nonzero vector $|v\rangle$ in the state space of $S$, its normalization is the pure state of $S$ with vector $|v\rangle / \sqrt{\langle v|v\rangle}$.^✦ The third state $|\chi\rangle$ is the normalization of $|00\rangle + |01\rangle + |10\rangle$, a nonzero vector of norm $\sqrt{3}$.^✦

For a real number $s$, the vector $|e_s\rangle$ of the qubit is $s|0\rangle + |1\rangle$ divided by its norm, $$|e_s\rangle = \frac{s|0\rangle + |1\rangle}{\sqrt{1 + s^2}} ,$$(5)^✦ The two values of $s$ used below are the golden ratio $\phi = (1+\sqrt{5})/2$ and its conjugate $\phi' = (1-\sqrt{5})/2$, the two roots of $x^2 = x + 1$.

The Claim

What Is Checked

The claim is stated over the Schmidt decomposition (2) of a pure state of two qubits, and over the three states as constructed: $|\Phi^+\rangle$ from the vector $|00\rangle + |11\rangle$, $|+\rangle|+\rangle$ as a product of pure states, and $|\chi\rangle$ as the normalization of $|00\rangle + |01\rangle + |10\rangle$, with the vectors $|e_\phi\rangle$ and $|e_{\phi'}\rangle$ of (5).

About these constructions the library proves that each gives a state of unit norm: $|+\rangle$ and $|\Phi^+\rangle$; the product of any two pure states, since the norm of a product vector is the product of the norms; and $|\chi\rangle$, since $|00\rangle + |01\rangle + |10\rangle$ is nonzero and a nonzero vector divided by its norm has norm one.