Entangled electron-positron pair Usually when we talk about entanglement, we mean entangled spin states (of electrons) or polarizations (of photons). My questions are: 
Does pair production guarantee the product electron and positron entangled?
If there's no observer measuring either particle, can we say the types, or charge, of the particles are also entangled, with a wavefunction like: $\frac{1}{\sqrt{2}}( |+e\rangle \pm |-e\rangle)$?
 A: I've already quite a long time ago noticed that in particle physics we usually do stuff that quantum-computing people will call an "entaglement". We just don't phrase it like that, because we are used to it and we aren't much "in awe" about it.  
So the "entanglement" you are talking about is long known in particle physics.
The earliest reference I know is this:
“Pion-Pion Correlations in Antiproton Annihilation Events”, Phys. Rev. Lett. 3 (1959), no. 4, 181–183.
As you see, it is for pions (charged, actually).
The more "modern" review is this:
“Bose–Einstein and Fermi–Dirac interferometry in particle physics”, Rep.
Prog. Phys 66 (2003) 481.
A: Yes, electron and positron are entangled. But entangled wave function should be of products:
$\frac{1}{\sqrt{2}}( |+e_1\rangle |-e_2\rangle + |-e_1\rangle |+e_2\rangle)$?
A: If you know the total linear and angular momentum of the system before pair production it would be entangled. The entangled state could be something like 
$$
\begin{aligned}
  |\psi\rangle 
    &= \frac{1}{\sqrt{2}}\left(|+e,p,m\rangle|-e,p_{tot}-p,m_{tot}-m\rangle  
  \right.  \\  &  \qquad\quad  \left.
    +  |-e,p,m\rangle|+e,p_{tot}-p,m_{tot}-m\rangle\right)
\end{aligned}
$$ 
although it could be far more complicated than that.
However, as Genneth rightly points out such entanglement can arise from much more accessible systems (basically any collision). We could perform an experiment to violate Bell inequalities with particle anti-particle pairs but it would be a lot of effort to check physics that we already have a good understanding of.
A: Pondering on the meaning of pair production and entanglement I am inclined to think that if it occurs at all, then there must be an equal probability of an electron being found in either state. 
For an entanglement calculation I think it should be the inner product:
$$ \left( \frac{1}{\sqrt{2}}|+\rangle + \frac{1}{\sqrt{2}} |-\rangle \right) \times \left( \frac{1}{\sqrt{2}}|+\rangle - \frac{1}{\sqrt{2}} |-\rangle \right)$$
$$= \left( \frac{1}{2}|++\rangle + \frac{1}{2} |--\rangle \right) $$
Notation wise, I don't think you need the electron symbol to be there to signify the state it is in.
I am not sure about my answer so if I am wrong please let me know.
