I Superposition of particle identities

bsaucer
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Suppose a neutral meson decays into an electron and a positron. Are the two particles entangled as they fly apart? Before any measurement takes place, are the particles in a mixed superposition as to which one is the electron, and which one is the positron? Is there a way to test for such superposition?
 
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Technically yes, but there is no way to use this superposition in any way. For all practical purposes you just have particles flying in random but fixed directions. The first electromagnetic interaction will destroy the superposition.
 
mfb said:
The first electromagnetic interaction will destroy the superposition.

And that will happen as soon as the x = t/c wavefront hits a charged particle or piece of material.
 
Vanadium 50 said:
And that will happen as soon as the x = t/c wavefront hits a charged particle or piece of material.
Wouldn't the Earth's magnetic field be sufficient to do the job before even that happens?
 
kuruman said:
Wouldn't the Earth's magnetic field be sufficient to do the job before even that happens?

Yup.
 
kuruman said:
Wouldn't the Earth's magnetic field be sufficient to do the job before even that happens?
Not exactly. The destruction of the superposition is a kind of wavefunction collapse. Collapse causes endless arguments and can be interpreted in many ways. However collapse models that have been updated in the last 30 years all agree with non-collapse models - that collapse, or the appearence of collapse, begins with an interaction that leaves a record of the value followed by decoherence. A very feeble interaction may not leave a record that's big enough to be recovered under the quantum uncertainty. So the entanglement may not be broken, just reduced a little.

The connection between a measurement-interaction and the entanglement is quite simple to understand. Interaction always creates an entanglement, that is, after all, what an observation means (to set up a correlation between the observing system and the observed system). Since this second entanglement appears to collapse to a definite eigenstate, it destroys the original superposition of the correlated states. A feeble interaction doesn't create a strong enough second entanglement to create a record, there is no full collapse and the superposition is not broken.
 
Derek P said:
A feeble interaction doesn't create a strong enough second entanglement to create a record, there is no full collapse and the superposition is not broken.
OK, then at what point does the interaction become "strong enough" to lead to decoherence? When the interaction energy is greater than ... what energy?
 
You have an electric dipole with a time rate of change, rotating in a magnetic field. That will radiate, and the pattern of radiation depends on the sign of the rotation: i.e. the identities of the particles. As far as I can tell, the wavefunction collapses immediately because the charge distribution imprints itself on the radiation field. This is mfb's point "The first electromagnetic interaction will destroy the superposition" Because this experiment is in an applied magnetic field, this interaction happens effectively instantaneously.

Rather than quibble about how instantaneous is instantaneous, wouldn't it be better to pick a better example? Neutral B and anti-B mesons produced in e+e- collisions do not have an identity - more specifically, are not in a flavor eigenstate - until one decays, which determines its eigenstate as well as its partners.
 
The model is the Stern-Gerlach apparatus. I am thinking of the spin singlet state of the electron-positron pair in the Earth's non-uniform field. My question is simple and straightforward. In that field and assuming no shielding, would there be decoherence? I assume "yes", but if the answer is "no", I would like to know why because I don't know why.
 

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