Excitation of two-atom state by single photon : FTL hope?

[Swamp Thing]

This concept is rather interesting : "Excitation of two atoms by a propagating single photon pulse" -- http://arxiv.org/abs/1411.3445 .

They say that, in principle, one can tailor an optical pulse so that it will excite two atoms from the ground state (i.e., the |gg> state) so it ends up in any state that we want, as long as it is a linear combination of the following two states:
|s> = 1/√2 ( |eg> + |ge> )
and
|a> = 1/√2 (|eg> - |ge> )

and we can do so with 100% certainty of hitting the desired state, in theory.

As a special case of this, we can predictably select between |eg> and |ge>.

Here's what is interesting: the control over the final state is perfect even for an arbitrarily large distance between the two atoms. They "cooperatively" end up in states that are anti-correlated, AND you have control over the state of each atom, which you manipulate by changing the phase relation between the electromagnetic fields near the respective atoms.

Assuming that all this is correct, does this in any way lead towards FTL communication?

 

[bhobba]

Assuming that all this is correct, does this in any way lead towards FTL communication?

Why would it? Its just a correlation.

Thanks
Bill

 

[Swamp Thing]

Why would it? Its just a correlation.

Ah, but Alice gets to decide whether :

her atom will be |g> and Bob's will be |e> ,
OR
Alice's atom will be in |e> and Bob's in |g>.

She does this by controlling the phase of the field arriving at her atom.
When Bob looks at his atom, he knows what setting Alice selected.

(Alternately, Bob can decide the atoms' fates and thus send a bit of information to Alice).

 

[bhobba]

But Alice gets to decide whether :.

So?

You can't use it to send information. All it is, is a correlation.

Thanks
Bill

 

[Swamp Thing]

Is it not true that if Bob can find out the state of his atom, he knows what decision Alice took?
To me, this experiment differs from all the other ones (ho-hum, just a correlation) in that, in this case anyone experimenter can control the outcome of the measurement and yet the remote measurement is correlated with the sender's measurement.

Here I am assuming that Bob can do a local measurement whose bases are |bob ground> and |bob excited>. If so, then they don't have to wait for a classical signal to travel across before they can compare notes retrospectively and then conclude that their outcomes were random but correlated. The result(s) are not random.

 

[bhobba]

Is it not true that if Bob can determine the state of his atom, he knows what decision Alice took?

That's not the point - the point is how to use it to send information.

Describe exactly how you would use it to do that.

If you can do it an instant Nobel Prize awaits - so its worth trying. I personally wouldn't bother because I understand the premises QM is based on and it doesn't allow that - but if you doubt it try your best.

Oh - and BTW you would have proven QM incorrect.

Thanks
Bill

 

[Swamp Thing]

I'm not preparing my Nobel Lecture yet, because there could be something wrong in the paper I quoted. (Anyway, the prize would probably go to those authors).

If Alice wants to send a "1", she sets the phase shifter so that the atoms end up as "Bob's atom excited, Alice ground"
If Alice wants to send a "0", she sets the phase so that the atoms end up as "Bob's atom ground state, Alice's excited".

Bob does a measurement where the basis (eigen??) states are "bob's atom ground" and "bob's atom excited".
If he sees "excited", he considers that as "1"
If he sees "ground", he considers that as "0".

There may be some wrinkles in actually measuring the state.. for example, it might involve a waiting time to let the atom decay and emit a photon; in that case, the waiting time might just be the very gotcha that enables the no-signalling rule to operate. But if Bob can measure his state faster than a classical channel then...

 

[bhobba]

I'm not preparing my Nobel Lecture yet, because there could be something wrong in the paper I quoted. (Anyway, the prize would probably go to those authors).

And this works if the atoms are spatially separated?

Note what it said -
'To make a comparison with more experimentally accessible options, we consider how well one can excite two atoms using coherent state pulses with an average photon number of 1'

How do you excite the second atom if its at the other side of the universe?

Thanks
Bill

 

[fzero]

How do you excite the second atom if its at the other side of the universe?

In their model, presumably they would seek to smear the photon wavefunction as much as necessary. I think fig. 1 had the setup.

Besides that, I'm also fairly suspicious of the methods in the paper since they seem to be claiming that they can achieve an absorption probability of 1 with their process. I would have expected the probability that neither atom absorbs the photon to be fairly large no matter what profile the photon pulse had if their model remotely resembled a real system. Perhaps I am missing something about these toy two-state models.

 

[Swamp Thing]

How do you excite the second atom if its at the other side of the universe?

Use a beam splitter and send the two output beams to Alice and Bob? You could perhaps even use fibers in the two branches.

 

[bhobba]

Use a beam splitter and send the two output beams to Alice and Bob? You could perhaps even use fibers in the two branches.

And that would allow FTL precisely how? Remember what they are doing is exiting two atoms. You can smear the wave-function but exactly how would you excite different atoms where each atom is at different sides of the universe? And in exiting it exactly how does it do it FTL?

Thanks
Bill

 

[Swamp Thing]

how would you excite different atoms where each atom is at different sides of the universe? And in exiting it exactly how does it do it FTL?

The source (located exactly midway between Alice and Bob) would transmit a series of pulses, and we assume that when Alice and Bob's conversation is due to begin, the first pulses have already started arriving at the two stations. (I believe this requirement is taken as given in most schemes of this type. Even if it works across the campus, it proves FTL in principle).

Bob grabs the first pulse and focuses it onto his atom, and Alice does the same. Only, Alice chooses to set up a "1" or a "0" while Bob just waits to see what comes across. Since both atoms are at the same distance from the source, Bob's atom's fate is sealed instantaneously when the pulses go past the two atoms. In contrast, if Alice were to send the "1" or "0" over a newly generated light pulse, then Bob would have to wait until that pulse made its way across to him.

 

[bhobba]

The source (located exactly midway between Alice and Bob) would transmit a series of pulses, and we assume that when Alice and Bob's conversation is due to begin

At what speed would the pulses travel? And how would either determine what state - remember it depends crucially on the shape of the pulse.

Thanks
Bill

 

[Swamp Thing]

At what speed would the pulses travel?

At the speed of light.

-----------------

And how would either determine what state?

The phase relationship of the two delocalized dipoles and the bandwidth of the rising exponential will determine the state in which the excitation is created.

 

[bhobba]

At the speed of light.

Yes. So?

A specifically shaped pulse, with the shape determining what state the two atoms go into arrives at the speed of light and the atoms go into the state determined by the rise time. There is no way for either atom to determine what state it goes into so information can be sent. And the phase relationship - how does either atom decide on that to send information when they are on the other side of the universe?

Thanks
Bill

 

[Swamp Thing]

It is instructive to see the case of exciting one atom perfectly without creating any excitation in the other by trying to excite the bi-atomic system into |eg>. As expected, when r is large, this reduces correctly to a dipole pattern around the first atom alone.

Well, it seems that to get a |eg> or |ge> state, the beam has to be configured in a way that Alice or Bob cannot really achieve : it's not only a matter of the phase relationship but also the amplitude. If the beam is symmetric, then Alice can choose between |eg>+|ge> and |eg>-|ge> but in this case the measurements are of course going to be totally random!

Bang goes my Nobel...

 

[bhobba]

Bang goes my Nobel...

Don't worry about it.

Its only after some acquaintance with QM you realize its impossible. It really hard to use that acquaintance at the start, when part of getting it in the first is what you did.

I hope it makes sense.

Thanks
Bill