When a particle and its antiparticle annihilate can we talk about their previous entanglement? Since antiparticles like an electron and a positron can annihilate each other, and since this seems to be possible only because they are in completely opposite phase of some sort, involving quantum numbers too, I was wondering if this implied that mathematically the two particles were actually entangled before they encountered each other.
I am writing this because of the mostly hostile reception given to my question about entanglement. It was made clear that it was pure nonsense to want to examine what a world without entanglement would be as QM foundations was entirely dependent on it.
As I presume the gamma rays resulting from particles annihilation are correlated, I assume the two particles were correlated too in order to meet each other.
Or is it that this correlation happened "just in the moment", particles reorienting each other because of each other's fields or something.
 A: This is an interesting question, with interesting implications.
Usually, when people compute the probability amplitude for the annihilation of two particles in the context of quantum field theory, they assume a tensor product of two states for the two initial particles. As such, the initial state is separable. However, the pair of gamma rays that are produced is in an entangled state, because they have to maintain energy and momentum conservation. But that still leaves some degrees of freedom unspecified. Therefore, the final state is in some superposition of all the possible combinations that would satisfy the conservation requirements.
The question is, what would be the case if the initial state was entangled? Well each term in the expansion of the initial state would produce a specific superposition of photon pairs. Added together, these different terms would cause a modified spectrum/superposition for the final states.
With a suitably entangled initial state, one may be able to produce a separable final state where the states of the two photons are fixed. What would such an entangled initial state look like? Well, one can consider the reverse process where two particles are created from two photons. If the two initial photons are given as a tensor product of initial states, then the two particles that are produced would be in an entangled state. Inverting this entangled state, one would obtain the ideal initial state in an annihilation process that would produce a separable final state for the two photons.
To make it clearer, let me add some mathematics. In particle physics, you would typically get a statement (apart from notational differences) such as
$$ \langle \mathbf{k}_1,\mathbf{k}_2|\hat{U}|\mathbf{p}_+,\mathbf{p}_-\rangle = i\mathcal{M}(\mathbf{k}_1,\mathbf{k}_2,\mathbf{p}_+,\mathbf{p}_-) . $$
Here it represents two charged fermions with momenta $\mathbf{p}_+$ and $\mathbf{p}_-$ that annihilate to produce two photons with momenta $\mathbf{k}_1$ and $\mathbf{k}_2$. In words, it says that the probability amplitude for the initial particles with these fixed momenta to annihilate into two photons with their fixed momenta is given by $\mathcal{M}(\mathbf{k}_1,\mathbf{k}_2,\mathbf{p}_+,\mathbf{p}_-)$. In practice, it would mean that one prepares the two initial particles in their fixed states and then make projective measurements of the photon states after the annihilation process.
Instead of doing it this way, let's ask what is the complete state$^{\dagger}$ that is produced by the process. It would mean that we imply a state given by
$$ |\psi\rangle = \hat{U}|\mathbf{p}_+,\mathbf{p}_-\rangle , $$
such that
$$ \langle \mathbf{k}_1,\mathbf{k}_2|\psi\rangle = i\mathcal{M}(\mathbf{k}_1,\mathbf{k}_2,\mathbf{p}_+,\mathbf{p}_-) . $$
Well it turns out that, due to the orthogonality of the photon states, we can write that state as
$$ |\psi\rangle = i\int|\mathbf{k}_a,\mathbf{k}_b\rangle \mathcal{M}(\mathbf{k}_a,\mathbf{k}_b,\mathbf{p}_+,\mathbf{p}_-)\ \text{d}\mathbf{k}_a\ \text{d}\mathbf{k}_b . $$
Note that $\mathcal{M}(\mathbf{k}_1,\mathbf{k}_2,\mathbf{p}_+,\mathbf{p}_-)$ must maintain the conservation of momentum-energy. It implies that for a fixed combination of input momenta the value of $\mathbf{k}_2$ would depend on the value of $\mathbf{k}_1$. Therefore, it follows that $|\psi\rangle$ is an entangled state. But $|\mathbf{p}_+,\mathbf{p}_-\rangle$ is not an entangled state. So, it then follows that the separable state produces an entangled state via the unitary process for the pair annihilation.
One can follow the exact same argument for the reverse process to show that for a special entangled initial state, the final state would be separable.
$\dagger$ Strictly speaking, we consider only the two photon part of the final state. So, we should include a projection operator to select this part and then normalize the resulting state. However, these details don't affect the essence of what is being demonstrated.
