Given a product state $\rho^{(1)} \otimes \rho^{(2)}$, can this state become non-product state under LOCC? Can LOCC create correlations between two systems?

  • $\begingroup$ I'd remember that non-product states can be correlated. Absence of entanglement doesn't mean absence of correlations. $\endgroup$
    – glS
    Nov 12, 2021 at 13:58

1 Answer 1


Sure, the classical communication allows you to generate the necessary correlations between two sites. For instance, suppose that Alice flips a coin and communicates the outcome to Bob. When the outcome is heads, they use their local operations to create the state $|0\rangle \langle 0| \otimes |0 \rangle \langle 0 |$. When the outcome is tails they use their local operations to create $|1\rangle \langle 1 | \otimes |1 \rangle \langle 1|$. Overall the state they generate (assuming the coin is fair) is $$ \frac12 |0\rangle \langle 0| \otimes |0 \rangle \langle 0 |+ \frac12 |1\rangle \langle 1| \otimes |1 \rangle \langle 1 | $$ which is not a product state. You can make this construction more general and create any separable state in this manner.

  • $\begingroup$ Does this mean LOCC can create entanglement from separable state? $\endgroup$
    – User101
    Nov 12, 2021 at 13:48
  • 1
    $\begingroup$ No. LOCC operations cannot create entanglement. Correlated systems are not necessarily entangled (non-separable). $\endgroup$
    – Condo
    Nov 12, 2021 at 13:57
  • $\begingroup$ Extending the above construction would allow you to create any probabilistic mixture of product states, i.e., any separable state. $\endgroup$
    – Rammus
    Nov 12, 2021 at 14:25

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