Separable Quantum State

quantum-computing

Definition

Separable Quantum State

Let ${\mathcal{H}}_{AB}$ be the composite state space of two quantum systems $A$ and $B$. A pure quantum state $|\psi ⟩\in {\mathcal{H}}_{AB}$ is separable across the cut $A\mid B$ iff there exist normalised states $|a⟩\in {\mathcal{H}}_A$ and $|b⟩\in {\mathcal{H}}_B$ such that

$$|\psi ⟩=|a⟩\otimes |b⟩.$$

A state that is not separable is an entangled quantum state.

Equivalently, the amplitude matrix factors as one outer product:

$$M_{\psi }=\left(\begin{array}{cc}{\alpha }_{00}& {\alpha }_{01}\\ {\alpha }_{10}& {\alpha }_{11}\end{array}\right)=\left(\begin{array}{c}{a}_0\\ a_1\end{array}\right)\left(\begin{array}{cc}{b}_0& b_1\end{array}\right)$$

Examples

Product state

The state

$$|\psi ⟩=\frac{1}{\sqrt{2}}(|0⟩+|1⟩)\otimes |0⟩$$

is separable because it is explicitly a tensor product.

Expanding gives

$$|\psi ⟩=\frac{1}{\sqrt{2}}|00⟩+\frac{1}{\sqrt{2}}|10⟩.$$

Bell state

The Bell state

$$|{\Phi }^{+}⟩=\frac{1}{\sqrt{2}}(|00⟩+|11⟩)$$

has coefficient matrix

$$C=\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& 0\\ 0& \frac{1}{\sqrt{2}}\end{array}\right).$$

Since

$$\det (C)=\frac{1}{2}\ne 0,$$

the state is not separable.

Separable mixed state with correlation

The state

$$\rho =\frac{1}{2}|00⟩⟨00|+\frac{1}{2}|11⟩⟨11|$$

is separable because

$$\rho =\frac{1}{2}(|0⟩⟨0|\otimes |0⟩⟨0|)+\frac{1}{2}(|1⟩⟨1|\otimes |1⟩⟨1|).$$

It is correlated, but the correlation is classical rather than entangled.