What does it mean to be momentum entangled?

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Discussion Overview

The discussion revolves around the concept of momentum entanglement, particularly in the context of photons, and its implications for improving the resolution of optical microscopes. Participants explore the nature of momentum entanglement compared to polarization entanglement and raise questions about the distribution of phase and wavelength in this context.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant asks for clarification on what it means to be momentum entangled and how this relates to the resolution of optical microscopes.
  • Another participant explains that when photons emerge from a PDC crystal, they are entangled in terms of frequency and wavelength, which is analogous to momentum entanglement.
  • It is noted that knowing one photon's frequency allows for the deduction of the other's frequency, leading to implications for their positions being indeterminate.
  • A participant questions whether the frequency of the photons remains indeterminate until measured, suggesting this could create an illusion of frequency being transferred between the entangled photons at the time of measurement.
  • There is a query regarding the quantization of frequency and whether a photon with a specific quanta of frequency could lead to the emergence of multiple photons from the PDC.
  • Participants discuss the commutation of position and frequency, with one correcting a statement about their relationship.

Areas of Agreement / Disagreement

Participants express differing views on the implications of frequency entanglement and its relationship to position indeterminacy. There is no consensus on the quantization of frequency or the specifics of how momentum entanglement operates compared to other forms of entanglement.

Contextual Notes

Participants highlight the complexity of the relationship between frequency, position, and momentum entanglement, with some assumptions about measurement and indeterminacy remaining unresolved.

San K
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what does it mean to be momentum entangled? and how can it improve resolution?

what does it mean to be momentum entangled? - say for photons
and
how can it improve the resolution of an optical microscope?

Understanding polarization entanglement is easier/familiar - i.e. - the twins have opposite spins.

in momentum entanglement how is the phase and/or wave-length divided/distributed?
 
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When photons come out of a PDC crystal, they are entangled as to frequency and wavelength. This is an analog (more or less) to momentum since their velocity is fixed as c.

The entanglement is such that frequency(Alice) + frequency(Bob) = fixed amount. Wavelength is essentially the reciprocal of the frequency (times a constant).

If you know Alice's frequency, you can deduce Bob's. And vice versa. So they are entangled. This may seem obvious, but remember that position commutes with this observable. Once you know Alice's frequency, you make Bob's position indeterminate.
 
Great response DrChinese. You covered all the questions well.
DrChinese said:
The entanglement is such that frequency(Alice) + frequency(Bob) = fixed amount.

frequency(Alice) + frequency(Bob) = fixed amount = frequency(photon that struck the PDC) - frequency(that is lost due to conversion/interaction?)

also is frequency (Alice or Bob) indeterminate till measured?

this could give the illusion that -- frequency is being transferred/distributed/balanced between Alice and Bob at the time of measurement/decoherence

DrChinese said:
Once you know Alice's frequency, you make Bob's position indeterminate.

this strengthens the QE hypothesis (and makes the LHV hypothesis weaker) I guess.

on a separate note: is frequency quantized?

so if struck the PDC with a photon having 1 quanta of frequency, we could never have two photons emerging out of the PDC even after trillions of strikes by such photons?
 
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DrChinese said:
If you know Alice's frequency, you can deduce Bob's. And vice versa. So they are entangled. This may seem obvious, but remember that position commutes with this observable. Once you know Alice's frequency, you make Bob's position indeterminate.

I suspect you meant doesn't commute.
 
Nabeshin said:
I suspect you meant doesn't commute.

Ah yes, thanks for that. I misspoke myself. :smile:

They are conjugates so they DON'T commute.
 

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