What does it mean to be momentum entangled?

[San K]

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?

 

[DrChinese]

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.

 

[San K]

Great response DrChinese. You covered all the questions well.

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

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?

 

[Nabeshin]

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.

 

[DrChinese]

I suspect you meant doesn't commute.

Ah yes, thanks for that. I misspoke myself.

They are conjugates so they DON'T commute.