# Mixed separable state with a continous probability distribution

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.

• 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
– glS
Jul 22, 2022 at 12:23

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.