The Exercise

Alice holds a quantum system $A$ and Bob a system $B$, and the two share a pure state $|\psi\rangle$ of the joint system. The exercise asks to show that $|\psi\rangle$ is entangled exactly when the conditional entropy $S(A|B)$ of $A$ given $B$ is negative.^✦

A classical conditional entropy is never negative.

Pure States of Two Systems

A quantum system $A$ is a physical object whose state space 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 of the state spaces of $A$ and $B$.^✦

A pure state $|\psi\rangle$ of a system $A$ is a unit vector in the state space of $A$, $\lVert |\psi\rangle \rVert = 1$.^✦ For pure states $|a\rangle$ of $A$ and $|b\rangle$ of $B$, the product $|a\rangle \otimes |b\rangle$ 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 $$|\psi\rangle = |a\rangle \otimes |b\rangle ,$$(1)^✦ and it is entangled when there are none.

States and Entropy

A state $\rho$ of a system $A$ is a positive operator of unit trace on the state space of $A$; pure and mixed states alike are states in this sense.^✦ A pure state $|\psi\rangle$ of $A$ determines the state $|\psi\rangle\langle\psi|$ of $A$, the rank one projector onto its vector.^✦

The von Neumann entropy of a state $\rho$ is $$S(\rho) = -\operatorname{tr}(\rho \ln \rho) ,$$(2)^✦ the Shannon entropy of the eigenvalues of $\rho$, in nats.

The reduced state $\rho_B$ of a state $\rho$ of $A \otimes B$ is the state of $B$ obtained by tracing out $A$.^✦ In an orthonormal basis $\{|i\rangle\}$ of the state space of $A$, $$\rho_B = \operatorname{tr}_A \rho = \sum_i \bigl(\langle i| \otimes I\bigr)\, \rho \, \bigl(|i\rangle \otimes I\bigr) .$$(3)^✦

The conditional entropy $S(A|B)$ of a state $\rho$ of $A \otimes B$ is the von Neumann entropy of $\rho$ less that of its reduced state on $B$, $$S(A|B) = S(\rho) - S(\rho_B) .$$(4)^✦

The Claim

What Is Checked

The claim is stated over the pure states of the composite system $A \otimes B$, the product states (1) among them, and the conditional entropy (4), built from the state $|\psi\rangle\langle\psi|$, its reduced state (3) on $B$ and the von Neumann entropy (2).

The norm of a product vector is the product of the norms, so $|a\rangle \otimes |b\rangle$ is a unit vector.