Past exam of the short options paper of the physics Prelims of the University of Oxford / 2022 / S03 / 3

A pure bipartite quantum state is entangled if it cannot be written as a tensor product $|a⟩\otimes |b⟩$. More generally a mixed quantum state is entangled if its density matrix cannot be expressed as a convex mixture of product density matrices. Two applications are quantum teleportation, which transfers an unknown qubit using a shared entangled pair plus classical communication, and superdense coding, which uses a shared entangled pair and transmission of one qubit to communicate two classical bits. Neither permits faster-than-light communication.

Measuring the first qubit applies $P_0=|0⟩⟨0|\otimes I$ or $P_1=|1⟩⟨1|\otimes I$. The Born rule givesThe normalized conditional quantum states areEach expression applies only when its denominator is nonzero; an impossible outcome has no defined conditional state.

When $p_0,p_1>0$, the observed relative frequencies converge, for large sample size, to the conditional probabilitiesEquality says that the second binary outcome has the same distribution whichever first outcome occurred; this is independence of these two particular measurement results. Cross-multiplication givesFor finite samples the two empirical fractions fluctuate; they are estimates, not exact deterministic equalities. If $p_0=0$ or $p_1=0$, the missing conditional fraction is undefined, but one outcome is constant, so the joint measurement results are independent and the displayed modulus identity still holds.

This criterion only addresses the computational-basis statistics. For examplehas all four joint probabilities equal to $1/4$, yet is maximally entangled: its first-qubit density matrix is $I/2$ and $\alpha \delta =-1/4\ne 1/4=\beta \gamma$. Moreover classical correlations of mixed separable states also exist, for example $\frac{1}{2}|00⟩⟨00|+\frac{1}{2}|11⟩⟨11|$. Correlations in a single basis alone neither certify nor exclude quantum entanglement.

Expand a normalized product state asThus $\alpha =ac$, $\beta =ad$, $\gamma =bc$, $\delta =bd$, givingConversely the nonzero coefficient matrix $C=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)$ has determinant zero precisely when it has rank one, so the condition also suffices for a pure two-qubit state to factor.

For any fixed pair of local measurement settings with operators $E_i,F_j$, the Born rule factorizes:Hence a product state has independent local results in every pair of local bases. This assumes independent settings and local measurements; shared classical randomization of preparations or a subsequent joint interaction can generate correlations. A separable mixture is not necessarily a single product state.

Alice's outcome is random. Conditional on knowing it, Bob may assign a different quantum state, but without that classical record he sees the same density matrix. For the displayed pure two-qubit state, his reduced density matrix before Alice's measurement isAfter Alice's computational-basis measurement with the outcome unannounced, the weighted mixture $p_0{\rho }_{B|0}+p_1{\rho }_{B|1}$ is exactly this same matrix. More generally, for any local trace-preserving operation with measurement operators $K_a$,The equality follows from the cyclic property of the partial trace for operators acting on Alice's subsystem. Bob's every local outcome probability therefore remains unchanged. This is the no-communication theorem: $\boxed{\text{entanglement alone cannot send information faster than light}}$. Conditional state changes become usable correlations only when the classical outcomes are compared.