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Particle/antiparticle annihilation and entanglement

+ 7 like - 0 dislike
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This is a basic question. Suppose that A and B are completely entangled particles and so are C and D. If B and C are antiparticles that annihilate each other will A and D be entangled as a result. It seems they should be in order to satisfy conservation laws. Also if the initial pairs were the results of pair creation events then the Feynman diagram of the entire can also be interpreted as a single particle starting as A emitting and absorbing photons and emerging as D.

More generally if the state of the 4 particle system just before annihilation was represented as $$\alpha_{ij}\, |A_i\rangle\, |B_j\rangle\; \beta_{kl}\, |C_k\rangle\, |D_l\rangle$$ would the state of A and D just the after the B-C annihilation be $$\sum_j\alpha_{ij}\beta_{jk}\,|A_i\rangle\, |D_k\rangle$$ in a product state with the state of the emitted photons.

Is this correct?

This post imported from StackExchange Physics at 2014-03-31 16:05 (UCT), posted by SE-user Daniel Mahler
asked Mar 9, 2014 in Theoretical Physics by Daniel Mahler (250 points) [ no revision ]
What conservation law are you thinking about when you say A and D will now be entangled?

This post imported from StackExchange Physics at 2014-03-31 16:05 (UCT), posted by SE-user BMS
If A and B are entangled in a way that the sum of say their momenta has a definite value and so are C and D, then afterwards A and D (and the emitted photon?) should be entangled such that the total sum of momenta has a definite value equal to the sum of the definite momenta of the initial pairs.

This post imported from StackExchange Physics at 2014-03-31 16:05 (UCT), posted by SE-user Daniel Mahler

2 Answers

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An experimentalist's view:

Let us take a concrete example: antihydrogen scatters on hydrogen.

"Entanglement", an unnecessarily confusing term to describe that one wave function describes the system coherently, i.e. all phases are also known, will depend on the way we set up the experiment, i.e. the initial conditions.

To assume we can approximate the four particle system with one wavefunction we should know all the variables concerning them, including polarization, not only energy and momentum balance. If we can manage this, then the output particles from the experiment will be correlated ( entangled) with explicit functions and the variables of the unmeasured outgoing particle will be predictable from the measured ones. There will be probabilities for the various outgoing channels, but that is another story.

If we do not have all the incoming variable information but can guess at a reasonable initial wavefunction where the unknown variables are averaged over, the loss of correlation at the output stage will not be absolute, but there will exist angular or other type correlations that will allow rigorous conclusions for the initial state. After all the Higgs was found by correlations in the invariant mass spectrum of the outgoing particles ( resonance bump).

This post imported from StackExchange Physics at 2014-03-31 16:05 (UCT), posted by SE-user anna v
answered Mar 17, 2014 by anna v (1,710 points) [ no revision ]
+ 1 like - 0 dislike

I would just like to emphasize that the word "completely" in the question makes a big difference. If a state A is maximally entangled with a state B, then A cannot be maximally entangled with a third state C as well. This is known as the "monogamy of entanglement". In general, the more states something is entangled with, the less the entanglement will be. Some more information and references can be found at [1]. Hope this helps!

[1]http://www.quantiki.org/wiki/Monogamy_of_entanglement

This post imported from StackExchange Physics at 2014-03-31 16:05 (UCT), posted by SE-user Heterotic
answered Mar 22, 2014 by Heterotic (515 points) [ no revision ]

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