# Particle/antiparticle annihilation and entanglement

+ 7 like - 0 dislike
44 views

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
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 like - 0 dislike

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 (1,710 points)
+ 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!

This post imported from StackExchange Physics at 2014-03-31 16:05 (UCT), posted by SE-user Heterotic
answered Mar 22, 2014 by (515 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverfl$\varnothing$wThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.