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: $\frac{1}{\sqrt{2}}( |+e\rangle \pm |-e\rangle)$?

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    $\begingroup$ You are correct. By conservation of angular momentum (which splits into non-relativistic orbital and spin components), and linear momentum, you find quite a complex entangled system. Notice however, that there is nothing exotic about such entanglement, and far more trivial things like collisions between two hard spheres can be similar. $\endgroup$
    – genneth
    Oct 12, 2012 at 20:35
  • $\begingroup$ @genneth that could be an answer $\endgroup$
    – David Z
    Oct 13, 2012 at 0:08
  • $\begingroup$ @genneth I don't quite get it. Is it possible to design some kind of experiment to prove such wavefunction really exists? $\endgroup$
    – pipsi
    Oct 15, 2012 at 7:44
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    $\begingroup$ @Pipsi: the wave function isn't a "real object" that can be measured or seen by one experiment. It is a collection of numbers storing all the probabilities that any outcome of a measurement will actually materialize, and all these probabilities following from the wave function not only can be tested but have been tested in more than a quadrillion of collisions at the LHC and at many other places. So yes, there is absolutely no doubt that the final states predicted from QFT, with all their counterintuitive properties (for the laymen), are right within a tiny error margin. $\endgroup$ Oct 16, 2012 at 8:33

4 Answers 4


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.

  • $\begingroup$ It's important to note, though, that there's an extra step to go from correlations to entanglement. From a brief look at the references it looks like they measure correlations that are well described by an entangled state, but they do not make efforts to prove they do not admit classical-correlations models. This is fine, but it is spooky how many "quantum" effects admit hidden-variable explanations. $\endgroup$ Oct 15, 2012 at 15:50
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    $\begingroup$ Dear @Emilio, I think that your whole "spooky" comment nicely shows what Kostya called "QC people are in awe of entanglement". All your propositions are spun for an extra effect. First, entanglement is just a quantum description of the most general correlation in the quantum world. Second, the correlated quantum numbers of particles (with nonzero information) coming from collisions are entanglement that never admits a classical hidden-variable description. Also, correlations/entanglement situations admitting a classical description are clearly of measure zero in the Hilbert space. $\endgroup$ Oct 16, 2012 at 6:12
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    $\begingroup$ I deleted parts of the comment discussion which were degenerating into personal attacks. $\endgroup$
    – David Z
    Oct 23, 2012 at 5:46
  • $\begingroup$ Allow me to apologize for using argumentative language. I simply wanted to discuss the differences between Motl's continuous description of entanglement as a mere correlation and what Bell demonstrated. A complete explanation of why you can't retain locality is given in this document pages 10-12. bslps.be/meaningWF.pdf $\endgroup$
    – user7348
    Oct 23, 2012 at 16:24

Yes, electron and positron are entangled. But entangled wave function should be of products:

$\frac{1}{\sqrt{2}}( |+e_1\rangle |-e_2\rangle + |-e_1\rangle |+e_2\rangle)$?


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.


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.


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