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.

  • 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


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?


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.


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