1
$\begingroup$

A separable state is defined as follows:

$\rho_{AB}$ = $\sum_{i} p_{i} \rho_{A}\otimes\rho_{B}$, where $\rho_{A,B}$ are pure states.

Essentially it is a classical mixture of unentangled states. Such a state is guaranteed to have zero entanglement.

My question is, would this still hold if $p_{i}$ were to be a continuous probability distribution, and I were to have a state like:

$\rho_{AB}(x,x';y,y')$ = $\int da p(a) \rho_{A}(x,x';a)\otimes\rho_{B}(y,y';a)$

Is this still a separable state with zero entanglement? My concern is if the integration process can lead to changes in the density matrix's functional form, which might lead to entanglement.

$\endgroup$
1
  • 1
    $\begingroup$ that is the definition of an unentangled state. Or in other words, entanglement is defined as not being able to write the state that way $\endgroup$
    – glS
    Jul 22, 2022 at 12:23

2 Answers 2

3
$\begingroup$

The key point here is Caratheodory's theorem: If a state can be written as a convex combination of tensor products of density matrices, such as your integral, then it can also be expressed as a convex combination of a small (in particular, finite) number of such tensor products of density matrices, which depends on the dimension of the space.

In brief: Whenever a state has a separable form with an integral, it can also be written in separable form with a finite sum (with a fixed maximum number of terms in the sum).

See e.g. this answer for more details: What is the minimum number of separable pure states needed to decompose arbitrary separable states?

$\endgroup$
0
$\begingroup$

Yes, provided that $$ \int p(a)\ \text{d}a = 1 $$ the mixed state defined in terms of such an integral represents a valid separable state.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.