Quantifying Entanglement

A two-qubit state can be "a little" entangled or "maximally" entangled, and we want a number that says how much. But "how much entanglement" is not one question — it is a family of questions, each answered by a different measure. This lesson explains what an entanglement measure is supposed to capture, why no single scalar suffices, and which axioms a sensible measure must obey.

Entanglement as a resource

The modern viewpoint treats entanglement as a physical resource, like energy or fuel. Two distant parties, Alice and Bob, can freely perform local operations and classical communication (LOCC): each manipulates their own subsystem, and they may phone each other. The one thing LOCC cannot do is create entanglement out of a product state. Therefore entanglement is exactly the part of a bipartite state's correlation that is not free under LOCC — it must be supplied, distributed, and spent.

This resource picture fixes the central requirement: an entanglement measure $E(\rho)$ must be a monotone, never increasing under LOCC,

E\big(\Lambda_{\mathrm{LOCC}}(\rho)\big) \leq E(\rho).

If a quantity could be pumped up by local actions alone, it would not be measuring a resource that local actions cannot manufacture.

The pure-state case is settled

For a pure bipartite state $|\psi\rangle_{AB}$ the answer is essentially unique. The Schmidt decomposition writes

|\psi\rangle_{AB} = \sum_i \sqrt{\lambda_i}\,|i\rangle_A |i\rangle_B, \qquad \sum_i \lambda_i = 1,

and all the entanglement is encoded in the Schmidt coefficients $\{\lambda_i\}$ . The canonical measure is the entanglement entropy — the von Neumann entropy of either reduced density operator $\rho_A = \operatorname{Tr}_B |\psi\rangle\langle\psi|$ :

E(|\psi\rangle) = S(\rho_A) = -\sum_i \lambda_i \log_2 \lambda_i.

It is $0$ for a product state ( $\rho_A$ pure, one nonzero $\lambda_i$ ) and $\log_2 d$ for a maximally entangled state of two $d$ -level systems (flat $\lambda_i = 1/d$ ). For two qubits the maximum is $1$ , achieved by any Bell state.

Mixed states break the uniqueness

For mixed states the single clean number splinters. The reduced-state entropy $S(\rho_A)$ is no longer a valid measure: a classically correlated separable state such as $\tfrac12|00\rangle\langle00| + \tfrac12|11\rangle\langle11|$ has $S(\rho_A) = 1$ bit yet contains zero entanglement. The entropy now mixes genuine entanglement with mere classical correlation.

Worse, the distillable entanglement and the entanglement cost split apart: there exist states from which no entanglement can be distilled even though entanglement was required to create them — bound entanglement, treated later in this module. With one operational rate no longer pinning down the value, several inequivalent measures arise, each capturing a different operational task:

Entanglement of formation $E_F$ — the asymptotic cost in Bell pairs to prepare the state.
Distillable entanglement $E_D$ — the rate of Bell pairs extractable by LOCC.
Concurrence $C$ — an algebraically computable two-qubit quantity, tied to $E_F$ .
Negativity $\mathcal{N}$ — built from the partial transpose, easy to compute and an LOCC monotone, but blind to bound entanglement.

Axioms a measure should satisfy

A function $E(\rho)$ deserving the name "entanglement measure" is generally asked to satisfy:

Vanishing on separable states: $E(\rho) = 0$ if $\rho$ is separable.
LOCC monotonicity: $E$ does not increase under LOCC (often strengthened to non-increase on average even under selective measurements).
Reduction on pure states: $E(|\psi\rangle)$ equals the entanglement entropy $S(\rho_A)$ .
Convexity: mixing states cannot increase entanglement, $E\!\left(\sum_i p_i \rho_i\right) \leq \sum_i p_i E(\rho_i)$ — losing the label of which state you hold can only destroy a resource, never create it.

Different measures meet different subsets of an extended axiom list (additivity, asymptotic continuity, normalization), and that is precisely why a whole zoo of measures exists. The rest of this module builds the most useful members of that zoo and shows how to compute them.