What does position entangled mean?

San K
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what does position entangled mean (physically)?

The below is an attempt to understand position entanglement. It may require a lot of corrections and modifications. Look forward to your responses.

A photon (A) would have a probability distribution, so would it's entangled twin/photon (B).

The probabilities distribution would actually be spherical in a 3D space (or 4D time-space, if you will).

To simplyfy...in the below analysis... let's look at just one dimension...say x-axis

So if we calculate that the 90% probability for Photon A lies between say x = 1 to 5

and the 90% probability for Photon B lies between say x = 100 to 105

i.e. the photon are separated by 99 units.

Now if we were to find:

Photon A at say x=1 then we would find Photon B at 100
Photon A at say x=2.5 then we would find Photon B at 102.5

[Assuming there are (symmetric or anti-symmetric, depending upon which is the correct one) entangled etc]
 
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San K said:
A photon (A) would have a probability distribution, so would it's entangled twin/photon (B).

The probabilities distribution would actually be spherical in a 3D space (or 4D time-space if you will).

Not necessarily. Position entanglement just means that the two-photon (position) state is described by some wavefunction that doesn't factorise, i.e. \psi(\mathbf{x}_{\mathrm{A}}, \mathbf{x}_{\mathrm{B}}) \neq \phi(\mathbf{x}_{\mathrm{A}}) \chi(\mathbf{x}_{\mathrm{B}}). Almost every possible two-photon wavefunction is entangled.
To simplyfy...in the below analysis... let's look at just one dimension...say x-axis

So if we calculate that the 90% probability for Photon A lies between say x = 1 to 5

and the 90% probability for Photon B lies between say x = 100 to 105

i.e. the photon are separated by 99 units.

Now if we were to find:

Photon A at say x=1 then we would find Photon B at 100
Photon A at say x=2.5 then we would find Photon B at 102.5

[Assuming there are (symmetric or anti-symmetric, depending upon which is the correct one) entangled etc]


If you're just looking at correlations in position, then not necessarily (entanglement implies correlation, but it's possible to have correlation without entanglement). You can certainly imagine an entangled state in which the positions of the two photons are individually unknown, but they're known to be a fixed distance apart. But if they're entangled then it means a lot more than just that their positions are correlated. For an entangled state with the type of behaviour your describing, the momenta of the photons would also be highly correlated for instance.
 
Thanks wle, and welcome to the physics forum

wle said:
it's possible to have correlation without entanglement).

interesting...can you give an example?

you don't mean - correlation (w/o entanglement) that works instantaneously across space-time?


wle said:
You can certainly imagine an entangled state in which the positions of the two photons are individually unknown, but they're known to be a fixed distance apart. But if they're entangled then it means a lot more than just that their positions are correlated. For an entangled state with the type of behavior your describing, the momenta of the photons would also be highly correlated for instance.

ok thanks
 
San K said:
interesting...can you give an example?

you don't mean - correlation (w/o entanglement) that works instantaneously across space-time?

I mean just plain, old, boring classical correlation that doesn't need a wavefunction to describe (and actually shouldn't be described by a wavefunction). For instance if I hide a photon at some location x and a second photon at location x + a for you to find, and I pick x randomly (and for argument's sake we imagine that these particular photons, once I drop them somewhere, aren't going to start dispersing like mad).

If you happen to know what a density operator is, then a correlated but non-entangled state would mean anything of the form
\rho_{\mathrm{AB}} = \sum_{i} p_{i} \rho^{(i)}_{\mathrm{A}} \otimes \rho^{(i)}_{\mathrm{B}} \,.
 
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