When combining two quantum states is there any rules that say its going to be in a mixed or pure state? I have two quantum systems, a probe and a target that have been entangled. The probe is prepared in a mixed state and so has the target. Is the combined system therefore a mixed state?
If I had prepared my target in a pure state and probe in a mixed is the resulting combined system still a mixed state?
 A: So, if I understand you correctly, you have a system with two parts, the probe and the target, i.e. the density matrices $\rho_{probe}$ and $\rho_{target}$ are the reduced density matrices of a state $\rho$ describing the whole system.
Let us first consider the case that you don't have entanglement. If the two states are not correlated in any way, it'll just be the product of the two density matrices. Hence, given $\rho_{probe}$ and $\rho_{target}$, the combined system will be $\rho_{probe}\otimes \rho_{target}$. Then we obtain:


*

*two pure states add up to a pure state

*a mixed state and a pure state add up to a product state.

*two mixed states add up to a mixed state


The reason is that the pure states are (let's stick to finite dimensional systems for simplicity, everything should also be true in infinite dimensions) rank-one density matrices and the rank is multiplicative. Basically the same holds true for classical correlations (i.e. the combined system is separable).
Now, if the combined system has entanglement, hence the state is not a product, you need to be more careful. Let's first consider the case that one state is pure (wlog $\rho_{probe}$ is pure) and the other is mixed. In this case, the state $\rho$ of the two systems will always be mixed. The reason is that if $\rho$ were pure, $S(\rho_{target})=S(\rho_{mixed})$, i.e. the two von Neumann entropies would be equal. Since $S(\rho_{target})=0$ as it is pure and $S(\rho_{mixed})\neq 0$ as it is mixed, this cannot happen. In fact, when $\rho_{probe}$, which is the reduced density matrix of $\rho$ is pure, then the whole system must be in a product state, i.e. there cannot be any entanglement at all!
So if you have genuine entanglement in $\rho$, then both reduced density matrices must be mixed. In this case, the above reasoning tells you that if both target and probe are mixed, the state of the whole system may well be pure - this can happen, when $S(\rho_{target})=S(\rho_{probe})$. The easiest example is given by a bipartite maximally entanlged pure state, since then each of the two parts (i.e. probe and target) is in a maximally mixed state.
