The Exercise

For a state of two systems $A$ and $B$, the inequality of Araki and Lieb bounds the entropy $S(A,B)$ of the joint state from below by the entropies $S(A)$ and $S(B)$ of its two reduced states, $S(A,B) \geq |S(A) - S(B)|$. A pure joint state attains the bound, with $S(A,B) = 0$ and $S(A) = S(B)$. The exercise asks for a mixed state that attains it too, with the larger entropy on $B$.

Systems and States

A quantum system $Q$ is given by its state space, a finite dimensional complex Hilbert space.^✦ The composite system $A \otimes B$ of two systems $A$ and $B$ is the system whose state space is the tensor product of their state spaces.^✦ The qubit is the system whose state space is $\mathbb{C}^2$^✦, with computational basis kets $|0\rangle = (1,0)$ and $|1\rangle = (0,1)$^✦.

A state $\rho$ of $Q$ is a density operator on the state space of $Q$: a positive operator of unit trace.^✦ A pure state $|\psi\rangle$ of $Q$ is a unit vector in the state space of $Q$, fixed physically up to a global phase^✦; it determines the state $|\psi\rangle\langle\psi|$ of $Q$, the projector onto $|\psi\rangle$^✦. A state $\rho$ of $Q$ is pure when $\rho = |\psi\rangle\langle\psi|$ for some pure state $|\psi\rangle$ of $Q$, and mixed otherwise.^✦

The product state $\rho \otimes \sigma$ of a state $\rho$ of $A$ and a state $\sigma$ of $B$ is the state of $A \otimes B$ whose density operator is the tensor product of the two, acting as $\rho$ on the first factor and as $\sigma$ on the second.^✦ The maximally mixed state of a system $Q$ whose state space has dimension $d \geq 1$ is the state $I/d$, the identity divided by $d$; on the qubit it is $I/2$.^✦

The reduced state $\rho_A$ of a state $\rho$ of $A \otimes B$ is the state of $A$ obtained by tracing out $B$^✦, and the reduced state $\rho_B$ is the state of $B$ obtained by tracing out $A$: $$\rho_A = \operatorname{Tr}_B \rho, \qquad \rho_B = \operatorname{Tr}_A \rho.$$(1)^✦

Entropy

The von Neumann entropy $S(\rho)$ of a state $\rho$ of $Q$ is the Shannon entropy of the eigenvalues of $\rho$, measured in nats: $$S(\rho) = -\operatorname{tr}(\rho \ln \rho).$$(2)^✦ For a state $\rho$ of $A \otimes B$, the three entropies of the exercise are $S(A,B) = S(\rho)$, $S(A) = S(\rho_A)$ and $S(B) = S(\rho_B)$.

The Claim

The example is the product of the pure state $|0\rangle$ of a qubit $A$ with the maximally mixed state of a second qubit $B$: $$\rho = |0\rangle\langle 0| \otimes \frac{I}{2}.$$(3)^✦

What Is Checked

The claim is stated over the product state $\rho$ of (3), its reduced states (1), the von Neumann entropy (2) of each, and purity as equality with a projector $|\psi\rangle\langle\psi|$ onto a pure state of $A \otimes B$. The projector onto a unit vector is a density operator, the tensor product of two density operators is again positive of unit trace, and so is the identity divided by the dimension, so the product state (3) is a state of $A \otimes B$. Tracing out either factor of a state again gives a state, so the reduced states (1) are states of $A$ and of $B$.