Entanglement as Resource

Entanglement becomes a resource when separated laboratories are restricted to local operations and classical communication (LOCC). A shared entangled state then enables transformations or tasks unavailable from separable inputs. This operational claim specifies a restriction and an advantage; it does not, by itself, select an interpretation of quantum mechanics.

Helpful background. LOCC defines the allowed operations, and Schmidt decomposition provides the pure-state description.

Free states and monotones

A bipartite state is separable when

$$\rho_{AB}=\sum_i p_i\,\rho_A^{(i)}\otimes\rho_B^{(i)}.$$

Shared classical randomness and local preparation can create every such state. LOCC cannot create entanglement from these inputs, even when parties exchange classical messages. Therefore an entangled input represents a capability absent from the free set.

An entanglement monotone quantifies an obstruction to free conversion. For pure bipartite states, the entropy of either reduced state gives

$$E(\psi)=S(\rho_A) =-\sum_i\lambda_i\log_2\lambda_i,$$

where $\lambda_i$ are Schmidt probabilities. It cannot increase on average under LOCC. Conditioning on a successful outcome can increase entanglement in that branch; ignoring the failure probability would misstate the monotonicity law.

A concentration example

Consider $\lvert\psi\rangle=\sqrt p\,\lvert00\rangle+ \sqrt{1-p}\,\lvert11\rangle$ with $1/2\leq p<1$. Alice applies a filter with success operator

$$K_s=\sqrt{\frac{1-p}{p}}\, \lvert0\rangle\langle0\rvert+ \lvert1\rangle\langle1\rvert.$$

On success the unnormalized output is $\sqrt{1-p}(\lvert00\rangle+\lvert11\rangle)$. The success probability is $2(1-p)$ and the normalized output is a Bell pair. A complementary failure operator makes the operation trace preserving and yields a product branch.

The branch gains entanglement, but the average yield is only $2(1-p)$ ebits, bounded by the input entropy $h_2(p)$. Quantum Information develops full entanglement distillation protocols; the example here explains why probabilistic resource conversion must count all branches.

Foundational distinctions

Entanglement, steering, and Bell nonlocality are different resources under different operational restrictions. Entanglement need not produce a Bell violation in a specified measurement scenario. A useful resource statement must identify the parties, allowed communication, available measurements, and target task.

The existence of an entanglement monotone also does not guarantee a total ordering of states: some transformations are incomparable, especially for mixed or multipartite states. The general framework belongs to resource theories.

Exercises

What happens to the filter at $p=1/2$ and as $p\to1$?

Solution

At $p=1/2$ it is the identity and succeeds with probability one because the input is already maximally entangled. As $p\to1$, success probability tends to zero. A product state cannot be converted into a Bell pair with positive probability by LOCC.

References

• Horodecki, Ryszard, Paweł Horodecki, Michał Horodecki, and Karol Horodecki, “Quantum entanglement,” Reviews of Modern Physics 81, 865–942 (2009). Review.
• Nielsen, Michael A., “Conditions for a class of entanglement transformations,” Physical Review Letters 83, 436–439 (1999). Paper.