B Quantum entanglement and relativistic simultaneity

[Chris Miller]

Wasn't sure whether to post this in SR or QP here, but chose the latter.

Assuming:

1. The results of a quantum measurement are random, and that Alice and Bob (performing simultaneous measurements on widely separated, entangled particles) end up with measurements that are perfectly correlated.

2. This would be a way of exchanging (i.e., agreeing on) encryption keys, in that both Alice and Bob would measure the same bit stream. (If so, would one pair of entangled particles be sufficient, or would they need a unique pair per key bit?)

3. Alice and Bob travel away from each other at a relative velocity of .866c for a Lorentz factor of 2, each taking one measurement per minute by their own clock (two by the other's).

Question: Are their measurements still correlated? In other words, do they arrive at the same key?

 

[DrChinese]

Wasn't sure whether to post this in SR or QP here, but chose the latter.

Assuming:

1. The results of a quantum measurement are random, and that Alice and Bob (performing simultaneous measurements on widely separated, entangled particles) end up with measurements that are perfectly correlated.

2. This would be a way of exchanging (i.e., agreeing on) encryption keys, in that both Alice and Bob would measure the same bit stream. (If so, would one pair of entangled particles be sufficient, or would they need a unique pair per key bit?)

3. Alice and Bob travel away from each other at a relative velocity of .866c for a Lorentz factor of 2, each taking one measurement per minute by their own clock (two by the other's).

Question: Are their measurements still correlated? In other words, do they arrive at the same key?

There is nothing in particular about Alice and Bob's relative motion that changes the results of measurements on entangled particles.

 

[Chris Miller]

There is nothing in particular about Alice and Bob's relative motion that changes the results of measurements on entangled particles.

So they'd arrive at the same key?

 

[DrChinese]

So they'd arrive at the same key?

Yes, assuming you mean reading a common set of random values.

 

[Chris Miller]

Yes, assuming you mean reading a common set of random values.

The particles seem to be in a frame of reference for which simultaneity is different than in either Alice's or Bob's. E.g., the particles' "now" would be when Alice and Bob each measure their 60th bit, even though Bob's 60th minute is, in his frame, simultaneous with Alice's 30th, and vice versa.

 

[Nugatory]

2. This would be a way of exchanging (i.e., agreeing on) encryption keys, in that both Alice and Bob would measure the same bit stream. (If so, would one pair of entangled particles be sufficient, or would they need a unique pair per key bit?)

One pair per key bit, because Alice and Bob only get one measurement and hence only one bit from each pair. Practical quantum key distribution algorithms actually need more than one pair per useful key bit, both to protect against man-in-the-middle attacks and to handle detector noise. Googling for "quantum key distribution" will find some really fascinating stuff.

Alice and Bob travel away from each other at a relative velocity of .866c for a Lorentz factor of 2, each taking one measurement per minute by their own clock (two by the other's).

Question: Are their measurements still correlated? In other words, do they arrive at the same key?

As long as can you arrange to get a steady stream of new entangled particles from the common source to them, yes. It's worth noting that which one measures any given pair first would be frame-dependent even if they were at rest relative to one another; the relative velocity is something of a red herring here.

 

[Nugatory]

The particles seem to be in a frame of reference for which simultaneity is different than in either Alice's or Bob's.

It feels natural to say "<something> is in a frame of reference...", but you will save yourself much confusion if you break yourself of that bad habit. Everything is in all frames of reference, always (at least in special relativity) and forgetting this is an invitation to misunderstanding.

E.g., the particles' "now" would be when Alice and Bob each measure their 60th bit, even though Bob's 60th minute is, in his frame, simultaneous with Alice's 30th, and vice versa.

No, because the particles are not at rest relative to one another.

You are starting to see why collapse interpretations ("A measurement instaneously collapses the wave function everywhere") don't work very well with spacelike-separated measurements; the cure is to not use these interpretations when you're trying to understand these situations.

And do remember that the quantum mechanics you learn in your first year or so of college physics is non-relativistic QM - it's not supposed to work when relativistic effects are not negligible.

 

[Khashishi]

The measurements by Alice and Bob do not have to be simultaneous to show a correlation.

 

[SlowThinker]

But... if you have a pair of photons with equal but opposite spin, you can produce a long stream of bits by measuring the spin's projection to x-axis, then y-axis, then x-axis, then y-axis, and so on? Or am I wrong?
Would the results only be correlated for the first measurement?

 

[Denis]

But... if you have a pair of photons with equal but opposite spin, you can produce a long stream of bits by measuring the spin's projection to x-axis, then y-axis, then x-axis, then y-axis, and so on? Or am I wrong?
Would the results only be correlated for the first measurement?

The results would be correlated for the first measurement, but not for later measurements. After the first measurement, it would not be an entangled state anymore. You would need other yet untouched pairs for other bits.

 

[Chris Miller]

The results would be correlated for the first measurement, but not for later measurements. After the first measurement, it would not be an entangled state anymore. You would need other yet untouched pairs for other bits.

I had the same question as answered here. I'm a little confused by what I thought I read, though, which was that passive measurements didn't disentangle, only state manipulations, which is why using quantum entanglement to communicate isn't possible (although here I'd "argue" that the two entangled particles are communicating, just not in any way that's useful to us [except in encryption key agreement]).

 

[Chris Miller]

It feels natural to say "<something> is in a frame of reference...", but you will save yourself much confusion if you break yourself of that bad habit. Everything is in all frames of reference, always (at least in special relativity) and forgetting this is an invitation to misunderstanding.

No, because the particles are not at rest relative to one another.

You are starting to see why collapse interpretations ("A measurement instaneously collapses the wave function everywhere") don't work very well with spacelike-separated measurements; the cure is to not use these interpretations when you're trying to understand these situations.

And do remember that the quantum mechanics you learn in your first year or so of college physics is non-relativistic QM - it's not supposed to work when relativistic effects are not negligible.

Thanks, Nugatory, for correcting my vernacular. Not quite sure what the right terminology would be now. Possessive? E.g., A's frame of reference vs. B's frame of reference? Do you agree that QE key agreement is possible despite relativistic effects, as DrChinese seems to have said?

 

[Nugatory]

Thanks, Nugatory, for correcting my vernacular. Not quite sure what the right terminology would be now. Possessive? E.g., A's frame of reference vs. B's frame of reference?

"The frame of reference of <something>", "<something>'s frame", and similar constructions are a convenient shortcuts for "doing our calculations using coordinates in which <something> is at rest". Given how awkward that latter phrase is, it's not surprising that people get sloppy and don't use it.

Do you agree that QE key agreement is possible despite relativistic effects, as DrChinese seems to have said?

Yes, of course. The measurements performed by Alice and Bob are events. You can use any coordinate system you wish to assign numbers to these events (no law requires you to use coordinates in which you yourself are at rest, although that is often a convenient choice) but that doesn't affect the event itself, any more than a pinhole in a sheet of paper changes its nature if you choose to use polar coordinates instead of cartesian coordinates to label points on the surface of the paper. Likewise, the existence of an infinite number of different coordinate systems in which Alice, Bob, and the particles are moving at different speeds relative to the origin of the coordinate system has no bearing on what happens when Alice and Bob have measured.