The Exercise

Two qubits $A$ and $B$ are in one of the four Bell states $|\beta_{xy}\rangle$, labelled by two bits $x$ and $y$. The exercise asks for the reduced density operator of each qubit: the state of $A$ alone, with $B$ traced out, and the state of $B$ alone, with $A$ traced out, for each of the four Bell states. The library states the answer as two theorems, one for each qubit, each covering the four Bell states at once.

A Quantum System

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 $S$, $\langle\psi|\psi\rangle = 1$.^✦ A state $\rho$ of $S$ is a density operator on the state space of $S$, an operator with $$\rho \geq 0, \qquad \operatorname{tr}\rho = 1.$$(1)^✦ A pure state $|\psi\rangle$ of $S$ determines the state $|\psi\rangle\langle\psi|$ of $S$, the rank one projector onto its vector.^✦

The qubit is the quantum system whose state space is $\mathbb{C}^2$, the two dimensional complex Euclidean space.^✦ The basis state $|i\rangle$ of the qubit, for a bit $i$, is the pure state whose vector is the standard basis vector $e_i$ of $\mathbb{C}^2$; $i = 0$ and $i = 1$ give the computational basis states $|0\rangle$ and $|1\rangle$.^✦

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 $B$.^✦ For pure states $|\psi\rangle$ of $A$ and $|\varphi\rangle$ of $B$, the product $|\psi\rangle \otimes |\varphi\rangle$, written $|\psi\rangle|\varphi\rangle$, is the pure state of $A \otimes B$ whose vector is the tensor product of the two vectors.

The Bell States

For bits $x$ and $y$, the Bell state $|\beta_{xy}\rangle$ is the pure state of the composite system of two qubits whose vector is the Bell vector $$|\beta_{xy}\rangle = \frac{1}{\sqrt{2}}\bigl(|0\rangle|y\rangle + (-1)^x\,|1\rangle|\bar{y}\rangle\bigr),$$(2) where $\bar{y} = y + 1 \bmod 2$ is the flip of $y$. Writing $|ab\rangle$ for $|a\rangle|b\rangle$, the four Bell states are $|\beta_{00}\rangle = (|00\rangle + |11\rangle)/\sqrt{2}$, $|\beta_{01}\rangle = (|01\rangle + |10\rangle)/\sqrt{2}$, $|\beta_{10}\rangle = (|00\rangle - |11\rangle)/\sqrt{2}$ and $|\beta_{11}\rangle = (|01\rangle - |10\rangle)/\sqrt{2}$.

Reduced States

The reduced state $\rho_A$ of a state $\rho$ of a composite system $A \otimes B$ is the state of $A$ given by the partial trace over $B$, taken against an orthonormal basis $\{|j\rangle\}$ of the state space of $B$: $$\rho_A = \operatorname{Tr}_B \rho = \sum_j (I \otimes \langle j|)\,\rho\,(I \otimes |j\rangle).$$(3)^✦ The reduced state $\rho_B$ of a state $\rho$ of a composite system $A \otimes B$ is the state of $B$ given by the partial trace over $A$, taken against an orthonormal basis $\{|i\rangle\}$ of the state space of $A$: $$\rho_B = \operatorname{Tr}_A \rho = \sum_i (\langle i| \otimes I)\,\rho\,(|i\rangle \otimes I).$$(4)

The maximally mixed state of a quantum system $S$ whose state space is nonzero, of dimension $d$, is the state of $S$ whose operator is the identity divided by $d$, $$\rho = \frac{I}{d}.$$(5)^✦ The qubit has $d = 2$, and its maximally mixed state is $I/2$.

The Claim

What Is Checked

The claim is stated over the Bell states (2) of two qubits, the projector $|\beta_{xy}\rangle\langle\beta_{xy}|$ that each of them determines, the two reduced states (3) and (4) of a state of a composite system, and the maximally mixed state of the qubit.

About these constructions the library proves that each Bell vector has unit norm, and that the product of two pure states is a unit vector, since the norm of a product vector is the product of the norms. It also proves that the projector onto a unit vector, either partial trace of a density operator, and the operator $I/d$ are density operators.