Entangled States: A First Definition

An entangled pure state describes a whole whose parts cannot each be assigned a separate pure-state vector. Entanglement is a property of the joint preparation relative to a specified subsystem split; it is not a force between the parts. The full treatment of entangled states includes broader criteria and the mixed-state qualifications.

Helpful background. Product states gives the factorization test used here.

Pure-State Entanglement

For a specified bipartition $A|B$, a pure state is entangled when no local vectors satisfy

$$|\Psi\rangle_{AB}=|\psi\rangle_A\otimes|\phi\rangle_B.$$

The Bell state

$$|\Phi^+\rangle=\frac{|00\rangle+|11\rangle}{\sqrt2}$$

has coefficient matrix $C=I/\sqrt2$, of rank two, so it cannot factor. In contrast, $|+\rangle_A|-\rangle_B$ is product despite containing several computational-basis terms.

For an entangled bipartite pure state, both reduced states are mixed. For this Bell state they are $I/2$: either subsystem gives equal probabilities in every orthonormal qubit basis. The joint state still contains correlations that the local density operators alone cannot specify.

Correlation Is Not Enough

The mixed state

$$\rho_{\mathrm{mix}}=\frac12|00\rangle\langle00| +\frac12|11\rangle\langle11|$$

also gives perfectly correlated computational-basis outcomes, but it is a classical mixture of product preparations and is therefore separable. For mixed states, “not a product” is weaker than “entangled”: entanglement means that no convex mixture of product states represents the density operator.

Neither entanglement nor conditioning on a remote outcome permits signaling by an unread local measurement. The complete local instrument leaves the other subsystem’s marginal unchanged; local observables explains this calculation.

Quick Check

Compare the expectation of $X\otimes X$ in $|\Phi^+\rangle$ and $\rho_{\mathrm{mix}}$, where $X$ interchanges $|0\rangle$ and $|1\rangle$. Why does this comparison require both subsystems?

Solution

$X\otimes X$ interchanges $|00\rangle$ and $|11\rangle$, leaving $|\Phi^+\rangle$ unchanged, so its expectation is $1$. Each product term in $\rho_{\mathrm{mix}}$ has expectation zero, so the mixture gives $0$. Both preparations have the same reduced states $I/2$, and therefore every single-subsystem measurement agrees. This joint observable distinguishes these two preparations; its value alone is not a universal entanglement test, since $|++\rangle$ also has expectation $1$.

References

• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, ch. 2.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, ch. 6. Author’s manuscript.