The Exercise

Two quantum systems $A$ and $B$ share a pure state $|\psi\rangle$. The exercise asks for two tests that tell a product state from an entangled one. The first counts: $|\psi\rangle$ is a product state exactly when its Schmidt number, the number of nonzero Schmidt coefficients, is one.^✦ The second looks at the parts: $|\psi\rangle$ is a product state exactly when the reduced state $\rho^A$, the state of $A$ alone, is pure, and exactly when the reduced state $\rho^B$, the state of $B$ alone, is pure. The library states the three equivalences separately, for any two quantum systems.

A Quantum System

A quantum system $S$ is a physical system whose state space $\mathcal{H}_S$ is a finite dimensional complex Hilbert space.^✦ The composite system $A \otimes B$ of two quantum systems $A$ and $B$ is the quantum system whose state space is the tensor product $\mathcal{H}_A \otimes \mathcal{H}_B$.^✦

A state $\rho$ of $S$ is a positive operator of unit trace on $\mathcal{H}_S$.^✦ A pure state $|\psi\rangle$ of $S$ is a unit vector of $\mathcal{H}_S$, with $\lVert \psi \rVert = 1$, and every pure state $|\psi\rangle$ of $S$ determines the state $|\psi\rangle\langle\psi|$ of $S$, the rank one projector onto $|\psi\rangle$.

Among the states, the pure ones are these projectors: a state $\rho$ of $S$ is pure when there is a pure state $|\varphi\rangle$ of $S$ with $$\rho = |\varphi\rangle\langle\varphi| .$$(1)^✦

Products, Reduced States and the Schmidt Number

For pure states $|a\rangle$ of $A$ and $|b\rangle$ of $B$, the product $|a\rangle|b\rangle = |a\rangle \otimes |b\rangle$ is the pure state of $A \otimes B$ whose vector is the tensor product of the two vectors.^✦ A pure state $|\psi\rangle$ of $A \otimes B$ is a product state when there are pure states $|a\rangle$ of $A$ and $|b\rangle$ of $B$ with $$|\psi\rangle = |a\rangle|b\rangle .$$(2)^✦

The reduced state $\rho^A$ of a state $\rho$ of $A \otimes B$ is the state of $A$ obtained by tracing out $B$: the partial trace of $\rho$ over $B$, computed in an orthonormal basis $\{|j\rangle\}$ of $\mathcal{H}_B$, $$\rho^A = \operatorname{tr}_B \rho = \sum_j \bigl(I \otimes \langle j|\bigr)\, \rho \,\bigl(I \otimes |j\rangle\bigr) .$$(3)^✦ The reduced state $\rho^B$ of a state $\rho$ of $A \otimes B$ is the state $\operatorname{tr}_A \rho$ of $B$, the partial trace of $\rho$ over $A$ in an orthonormal basis of $\mathcal{H}_A$.^✦

Fix the standard orthonormal basis $\{|k\rangle\}$ of $\mathcal{H}_B$. Any pure state $|\psi\rangle$ of $A \otimes B$ expands as $|\psi\rangle = \sum_k |v_k\rangle|k\rangle$, with each $|v_k\rangle$ a vector of $\mathcal{H}_A$. The coefficient operator $M_\psi$ of $|\psi\rangle$ is the linear operator from $\mathcal{H}_B$ to $\mathcal{H}_A$ that sends each $|k\rangle$ to $|v_k\rangle$, $$M_\psi = \sum_k |v_k\rangle\langle k| .$$(4)^✦ The Schmidt number $\mathrm{Sch}(\psi)$ of a pure state $|\psi\rangle$ of $A \otimes B$ is the number of its nonzero Schmidt coefficients, which are the singular values of $M_\psi$.^✦

The Claim

What Is Checked

The three equivalences are stated over a pure state $|\psi\rangle$ of the composite system $A \otimes B$, the product states (2), the reduced states $\rho^A$ and $\rho^B$ of $|\psi\rangle\langle\psi|$ with the purity of a state (1), and the Schmidt number of $|\psi\rangle$, counted from the singular values of its coefficient operator (4).

The library proves that the product $|a\rangle|b\rangle$ of two pure states is a unit vector, since the norm of a product vector is the product of the norms. It proves that the projector $|\psi\rangle\langle\psi|$ onto a unit vector is a state, and that the partial trace of a state over either factor is a state, so that $\rho^A$ and $\rho^B$ are states of $A$ and of $B$.