LOCC Preview

LOCC stands for local operations and classical communication. It is the standard operational setting for asking which transformations are possible when separated parties can manipulate their own subsystems, exchange ordinary classical messages, but cannot send quantum systems to each other.

Separable states can be prepared by local devices coordinated through classical randomness; entangled states cannot be created that way. LOCC nevertheless can concentrate already shared entanglement into a successful branch, at the expense of failure probability.

Helpful background. Conditional states gives branch probabilities and updates; Schmidt decomposition supplies the pure-state data used in conversion tests.

Local Operations

Fix a bipartite decomposition

$$\mathcal H_{AB} = \mathcal H_A\otimes\mathcal H_B.$$

A local operation acts on one laboratory’s subsystem and any local apparatus. Added ancillas are initially unentangled across the Alice–Bob split; supplying a shared entangled ancilla would itself supply a resource. The finite-dimensional formulas below allow input and output spaces to differ.

For example, a nonselective local operation on $A$ can be represented in finite dimensions by operators $\{K_\alpha\}$ from $\mathcal H_A$ to an output space satisfying

$$\sum_\alpha K_\alpha^\dagger K_\alpha = I_A.$$

It acts on the joint state as

$$\rho_{AB} \longmapsto \rho'_{AB} = \sum_\alpha (K_\alpha\otimes I_B) \rho_{AB} (K_\alpha^\dagger\otimes I_B).$$

Local unitaries are the reversible special case with one Kraus operator $K=U_A$. Local measurements, local noise, adding and discarding local ancillas, and resetting a local system are also local operations.

The word local is relative to the specified subsystem split. An operation may be local for one decomposition and nonlocal for another. This is the same convention used in Local Unitary Equivalence and Entanglement Depends on a Decomposition.

Classical Communication

Classical communication means that the parties can send classical data, such as measurement outcomes, basis choices, or instructions. A message may influence which local operation is performed later.

A simple one-way adaptive protocol has this form:

1. Alice applies a measurement operation with outcome $a$.
2. Alice sends the classical label $a$ to Bob.
3. Bob chooses a local operation depending on $a$.

In a simplified single-Kraus-per-outcome notation, an unnormalized branch state has the form

$$\widetilde\rho_{ab} = (K_a\otimes L_{b\vert a}) \rho_{AB} (K_a^\dagger\otimes L_{b\vert a}^\dagger),$$

where $K_a$ acts on $A$ and $L_{b\vert a}$ acts on $B$. The branch probability is

$$p(a,b) = \operatorname{Tr}(\widetilde\rho_{ab}),$$

and, when $p(a,b)>0$, the normalized branch state is

$$\rho_{AB\vert ab} = \frac{\widetilde\rho_{ab}}{p(a,b)}.$$

Realistic instruments may have several Kraus operators for one recorded outcome. The essential point is unchanged: the operations act locally, while the choice of later operations may depend on classical records. The Conditional States page explains how to calculate outcome probabilities and conditional states.

An LOCC protocol may have many rounds and may be adaptive in both directions. The precise mathematical closure of all possible LOCC protocols has subtleties, so this page uses LOCC as an operational class rather than as a formal taxonomy.

Why LOCC Cannot Create Entanglement

The key structural fact is:

$$\text{separable input} \quad \xrightarrow{\ \operatorname{LOCC}\ } \quad \text{separable output}.$$

First consider a product local channel $\mathcal E_A\otimes\mathcal F_B$. If

$$\rho_{AB} = \sum_r p_r\, \sigma_A^{(r)} \otimes \tau_B^{(r)}$$

is separable, then

$$(\mathcal E_A\otimes\mathcal F_B)(\rho_{AB}) = \sum_r p_r\, \mathcal E_A(\sigma_A^{(r)}) \otimes \mathcal F_B(\tau_B^{(r)}),$$

which is still a convex mixture of product states.

Measurements and classical messages only refine this argument. Suppose an LOCC branch labeled by $m$ has nonzero probability. Starting from a separable input, the unnormalized branch can always be written as a sum of product positive operators:

$$\widetilde\rho_{AB\vert m} = \sum_r q_{r,m}\, \widetilde\sigma_{A,m}^{(r)} \otimes \widetilde\tau_{B,m}^{(r)}, \qquad q_{r,m}\ge0.$$

After division by the branch probability, the conditioned branch remains separable:

$$\rho_{AB\vert m} = \frac{\widetilde\rho_{AB\vert m}} {\operatorname{Tr}\widetilde\rho_{AB\vert m}}.$$

Thus even postselection cannot turn a genuinely separable input into an entangled output by LOCC. Postselection can increase the entanglement of some branches when the input was already entangled, but the probabilities and the other branches must be included in any monotonicity statement.

This is the operational content behind the definition of Separable Mixed States: they are exactly the bipartite states that can be assembled from local quantum states with classical randomness.

Reversible and irreversible local changes

Local unitaries preserve entanglement because both the operation and its inverse are LOCC. Measurements and discarding systems are irreversible and can reduce entanglement. Classical communication can create shared classical correlations, but the branch argument above shows why it cannot turn a separable state into an entangled one.

The division into parties is essential. A laboratory may entangle two qubits that it controls locally; this does not create entanglement across the Alice–Bob split.

A Bell-Pair Measurement Example

Consider the Bell state

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr).$$

If Alice measures her qubit in the computational basis and sends the outcome to Bob, the conditioned branch states are product:

$$\rho_{AB\vert 0} = \lvert00\rangle\langle00\rvert, \qquad \rho_{AB\vert 1} = \lvert11\rangle\langle11\rvert.$$

If the outcome is ignored but the measurement has occurred, the joint state becomes

$$\rho'_{AB} = \frac12 \lvert00\rangle\langle00\rvert + \frac12 \lvert11\rangle\langle11\rvert.$$

This state is separable but classically correlated. Alice’s classical message lets Bob sort the data into correlated branches, but it does not recreate the lost Bell entanglement. The example shows how LOCC can convert entanglement into ordinary shared classical information.

Entanglement Conversion Preview

For bipartite pure states, the Schmidt coefficients provide a sharp first glimpse of LOCC convertibility. Write

$$\lvert\psi\rangle = \sum_i \sqrt{\lambda_i^\psi}\, \lvert i_A i_B\rangle, \qquad \lvert\phi\rangle = \sum_i \sqrt{\lambda_i^\phi}\, \lvert i'_A i'_B\rangle,$$

where the Schmidt probability vectors are sorted in decreasing order:

$$\lambda_1\ge\lambda_2\ge\cdots, \qquad \sum_i\lambda_i=1.$$

For finite-dimensional bipartite pure states, with probability vectors padded with zeros to the same length, Nielsen’s theorem characterizes exact deterministic conversion by LOCC without an additional entangled catalyst. The conversion

$$\lvert\psi\rangle \longrightarrow \lvert\phi\rangle$$

is possible exactly when

$$\boldsymbol\lambda^\psi \prec \boldsymbol\lambda^\phi,$$

meaning

$$\sum_{i=1}^k \lambda_i^\psi \le \sum_{i=1}^k \lambda_i^\phi \quad \text{for every }k, \qquad \sum_i\lambda_i^\psi = \sum_i\lambda_i^\phi =1.$$

The direction is important. More entangled pure states have more uniform Schmidt vectors. A uniform vector is majorized by a more peaked vector, so deterministic LOCC can degrade pure-state entanglement but cannot make it more uniform.

For two qubits,

$$\left( \frac12,\frac12 \right) \prec (1,0),$$

so a Bell pair can be converted to a product state by LOCC. The reverse conversion is impossible by LOCC because

$$(1,0) \not\prec \left( \frac12,\frac12 \right).$$

Similarly,

$$(0.6,0.4) \prec (0.8,0.2),$$

so a partially entangled two-qubit pure state with Schmidt vector $(0.6,0.4)$ can be deterministically converted into a less entangled one with Schmidt vector $(0.8,0.2)$.

Here is a protocol for that last conversion. Alice measures with

$$K_0=\operatorname{diag}(\sqrt{8/9},\sqrt{1/3}), \qquad K_1=\operatorname{diag}(1/3,\sqrt{2/3}).$$

Their squared operators sum to $I$. On the input $\sqrt{0.6}|00\rangle+\sqrt{0.4}|11\rangle$, outcome zero has probability $2/3$ and yields $\sqrt{0.8}|00\rangle+\sqrt{0.2}|11\rangle$. Outcome one has probability $1/3$ and yields $\sqrt{0.2}|00\rangle+\sqrt{0.8}|11\rangle$. Alice announces that outcome, and both apply $X$ in the second branch. Every branch then has the same desired final state: randomness in the measurement is compatible with deterministic final conversion.

For dimensions above two, Schmidt vectors need not be comparable. Entropy alone does not decide exact conversion; all partial-sum inequalities must hold.

LOCC Monotonicity

An entanglement monotone is a quantity that does not increase under LOCC. For a selective protocol with outcomes $m$, probabilities $p_m$, and output states $\rho_m$, the usual average monotonicity condition is

$$E(\rho) \ge \sum_m p_m E(\rho_m).$$

For pure bipartite states, the Entanglement Entropy is the central example:

$$E(\psi) = S(\rho_A) = S(\rho_B).$$

Two-qubit concurrence and negativity also satisfy average LOCC monotonicity. In contrast, not every useful spectral diagnostic does: Rényi entropies explains the distinction for orders greater than one.

Average monotonicity does not say that every selected branch must have no more entanglement than the input. A rare successful branch may be more entangled than the initial state, provided the success probability and the other branches keep the average from increasing. This is why postselection must be handled carefully.

What LOCC Is Not

LOCC is not the same as applying only local unitaries. Local unitaries are reversible LOCC operations; LOCC also includes irreversible operations such as measurements, noise, discarding systems, and conditioning on classical messages.

LOCC is distinct from no signaling. A joint channel that ignores its input and prepares a Bell pair has output marginals independent of the other party’s input, hence is no-signaling in that sense, but it cannot be performed by LOCC on a separable input. Conversely, LOCC permits ordinary classical signals between laboratories, with the time needed to send them.

LOCC is not the same as all separable operations. A separable operation has Kraus operators that can be written in product form, but not every separable operation is achievable by LOCC; allowing more rounds does not remove the general distinction. The distinction matters in quantum information, though it is usually beyond what is needed for a first entanglement course.

LOCC is also not permission to send a quantum system. If Alice physically sends a qubit to Bob through a quantum channel, the operation is no longer LOCC. Shared entanglement plus classical communication can simulate some communication tasks when the right extra resources are present, but those protocols are not the same as creating entanglement from nothing.

Continuing to entanglement protocols

Entanglement distillation applies repeated local measurements and communication to noisy pairs, keeping selected outcomes. Resource theories compares this setting with other choices of allowed operations. Those extensions distinguish finite-copy success probabilities from asymptotic rates and entanglement cost from distillable entanglement.

Exercises

1. Let

$$\rho_{AB} = \sum_r p_r\, \sigma_A^{(r)} \otimes \tau_B^{(r)}$$

be separable. Show that a local unitary $U_A\otimes U_B$ maps it to another separable state.

Solution

The transformed state is

$$(U_A\otimes U_B) \rho_{AB} (U_A^\dagger\otimes U_B^\dagger) = \sum_r p_r\, U_A\sigma_A^{(r)}U_A^\dagger \otimes U_B\tau_B^{(r)}U_B^\dagger.$$

Each term is still a product density operator, and the coefficients are still nonnegative and sum to one. Therefore the state remains separable.

2. Alice and Bob share $\lvert\Phi^+\rangle$. Alice measures in the computational basis and Bob learns the outcome. Compute the unconditioned post-measurement state and decide whether it is entangled.

Solution

The two outcomes occur with probability $1/2$. The conditioned states are $\lvert00\rangle\langle00\rvert$ and $\lvert11\rangle\langle11\rvert$. If the outcome is ignored, the state is

$$\rho'_{AB} = \frac12 \lvert00\rangle\langle00\rvert + \frac12 \lvert11\rangle\langle11\rvert.$$

This is a convex mixture of product states, so it is separable. It contains classical correlation but no entanglement.

3. Decide whether either exact deterministic LOCC conversion between Schmidt vectors $(0.5,0.4,0.1)$ and $(0.6,0.2,0.2)$ is possible, without a catalyst.

Solution

The first vector is not majorized by the second: its first partial sum $0.5$ is smaller than $0.6$, but its two-term sum $0.9$ exceeds $0.8$. The reverse fails already at the first sum, since $0.6>0.5$. Neither conversion is possible. Pure-state entanglement forms a partial order, not a single list ranked by entropy.

4. A protocol sometimes produces a maximally entangled pair from a partially entangled input, but succeeds only with probability $p<1$. Why does this not contradict LOCC monotonicity?

Solution

LOCC monotonicity is normally an average statement:

$$E(\rho) \ge \sum_m p_m E(\rho_m).$$

A successful branch may have more entanglement than the input, but the failed branches and the success probability must be included. The protocol is not creating free entanglement; it is probabilistically concentrating entanglement already present in the input.

5. Explain why sending a qubit from Alice to Bob is not an LOCC operation, even if Alice also sends classical messages.

Solution

LOCC allows local quantum operations and classical communication only. A transmitted qubit is quantum communication: it carries an unknown quantum state and can transmit entanglement or coherent superposition. That physical resource is outside LOCC. Classical messages can coordinate Bob’s local operations, but they cannot replace the act of sending a quantum system unless some additional resource, such as shared entanglement, is explicitly supplied in a protocol.

References

• C. H. Bennett, H. J. Bernstein, S. Popescu, and B. Schumacher, “Concentrating partial entanglement by local operations”, Physical Review A 53, 2046–2052 (1996).
• R. Horodecki, P. Horodecki, M. Horodecki, and K. Horodecki, “Quantum entanglement”, Reviews of Modern Physics 81, 865–942 (2009).
• M. A. Nielsen, “Conditions for a class of entanglement transformations”, Physical Review Letters 83, 436–439 (1999).
• G. Vidal, “Entanglement monotones”, Journal of Modern Optics 47, 355–376 (2000).
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018), Ch. 6.