Can a Local Operation Disentangle Two Parties?

The claim that “you cannot disentangle two parties by operating on only one” is true for local unitary operations, but false if “operation” includes measurements or other irreversible processes. A local measurement on one half of a Bell pair can leave the two halves in a product state.

The distinction is between changing a system reversibly and discarding information about it. Here, “local” means that the operation acts on Alice’s subsystem only; it does not act directly on Bob’s subsystem.

Local unitaries preserve entanglement

For a pure bipartite state, the Schmidt decomposition has the form

$$|\psi\rangle_{AB}=\sum_i \sqrt{\lambda_i}\,|i\rangle_A|i\rangle_B,$$

where the nonzero coefficients satisfy $\sum_i\lambda_i=1$. The coefficients $\lambda_i$ determine the state’s entanglement: the reduced density operator of either party has eigenvalues $\lambda_i$, and its von Neumann entropy is the entanglement entropy.

If Alice applies a unitary $U_A$, the state becomes

$$(U_A\otimes I_B)|\psi\rangle_{AB} =\sum_i\sqrt{\lambda_i}\,(U_A|i\rangle_A)|i\rangle_B.$$

The vectors $U_A|i\rangle_A$ remain orthonormal, so the Schmidt coefficients have not changed. The entanglement is unchanged. The same reasoning applies to a unitary on Bob’s side. For mixed states, Schmidt coefficients generally are not defined, but local unitaries still preserve entanglement because they are reversible local transformations.

This argument does not apply to every operation that acts locally. A measurement is generally not unitary, and it can remove entanglement.

Worked example: measure one half of a Bell pair

Consider the Bell state

$$|\Phi^+\rangle=\frac{|00\rangle+|11\rangle}{\sqrt{2}}.$$

Suppose Alice measures her qubit in the computational basis, with outcomes $0$ and $1$. If she gets outcome $0$, the joint state after the measurement is $|00\rangle$. If she gets outcome $1$, it is $|11\rangle$. In either case, the joint state is a product state, so it is not entangled.

If you do not condition on, or retain, Alice’s measurement result, the joint density operator is

$$\rho_{AB}=\frac{1}{2}|00\rangle\langle00|+\frac{1}{2}|11\rangle\langle11|.$$

This is still separable: it is a probabilistic mixture of product states. The original Bell state was entangled, but the measurement destroyed that entanglement even though Alice measured only her own qubit.

The conditional and unconditioned descriptions answer slightly different questions. Given Alice’s known result, the state is a particular product state. If the result is ignored, the state is a mixture of two product states. Both are unentangled.

General local operations can reduce entanglement

A general deterministic local quantum operation is described by a completely positive, trace-preserving map. For example, Alice can discard her system and prepare a fixed state $|0\rangle$. Applied to a joint state $\rho_{AB}$, this gives

$$|0\rangle\langle0|_A\otimes\operatorname{Tr}_A(\rho_{AB}),$$

which is a product state. This is an extreme example: the original state of Alice’s subsystem has been thrown away. It shows why the blanket claim that no operation on one party can disentangle the pair is false.

More generally, local operations and classical communication cannot increase entanglement on average. They may preserve it, reduce it, or destroy it. A selective operation—one where you keep only a particular measurement outcome—can change the entanglement of the selected subensemble. The average over all outcomes obeys the non-increase rule.

Why this does not allow faster-than-light signalling

A local trace-preserving operation on Alice’s side does not change Bob’s reduced state when Alice’s outcome or operation result is unavailable to him. If Alice’s operation has Kraus operators $K_i$, then $\sum_i K_i^\dagger K_i=I_A$, and Bob’s reduced state after the operation is

$$\begin{aligned} \rho'_B &=\operatorname{Tr}_A\left(\sum_i(K_i\otimes I_B)\rho_{AB}(K_i^\dagger\otimes I_B)\right)\\ &=\operatorname{Tr}_A(\rho_{AB})=\rho_B. \end{aligned}$$

In the Bell-pair example, Bob’s state before Alice measures is the maximally mixed state $I/2$. Afterward, if he does not know her result, it is still $I/2$. He cannot determine from his qubit alone whether she measured it. To use the conditional outcome, Alice must communicate it through an ordinary classical channel.

Common mistakes

• Treating every local operation as a unitary. Measurements, discarding a subsystem, and noisy channels are local too, but they need not preserve entanglement.
• Confusing a mixture with an entangled state. A mixture can be entangled, but the Bell-pair measurement mixture above is explicitly a mixture of product states and is separable.
• Assuming collapse lets Bob detect Alice’s action. Bob’s reduced density operator is unchanged by Alice’s unconditioned local operation. The conditional state description does not provide a signalling channel.
• Claiming local operations can create entanglement from nothing. Local operations cannot create entanglement between initially unentangled parties. The relevant rule is non-increase on average, not invariance under every local operation.