4
$\begingroup$

Consider we have two atoms $a$ and $b$. They are entangled with each other in position and momentum, with some wavefuction describing them in position space that is $\Psi(x_a, x_b)$. This initialization of the entangled state is achieved as described in this paper: http://arxiv.org/abs/quant-ph/9907049

We can trap both atoms by putting them in a harmonic oscillator potential with lasers and get them to both entangle with methods described by the paper I referenced above and by using a nondegenerate optical parametric amplifier (NOPA) as the entanglement source.

From what I understand, we can think of each particle having its own local axis and separated by some arbitrary global distance. Let us also consider only one dimension, the $x$-dimension, to keep things simple.

Say we apply the position operator, $\hat x_a$ to particle $a$ and given that it was in a harmonic oscillator potential, $V$, we will obtain a single measurement $x_a$. If we had identically, or as close as we can, prepared systems and we applied the same operator, or measurement, multiple times for an ensemble of entangled pairs, and we put all of these measurements together, then we'd get the probability distribution or the "quantum state of position" of atom $a$. Essentially, we do quantum state tomography. Referring the figure I've attached, I've made up a scenario where the atom happens to be between $x$ = -4 and 4 with arbitrary units with respect to their local axis.

enter image description here

Atoms $a$ and $b$ should roughly have the same distribution ... because each single measurement of position for each entangled pair in the ensemble should yeild $x_a = x_b$ with respect to their local axes ... because this is what it means to be entangled. (Need a check here)

Now, let us consider an two ensembles of atoms called $A$ and $B$. They are separated in Lab A and Lab B. They are entangled such that we will get the same distribution if position is measured at either laboratory.

We know that given a certain potential, we will obtain a characteristic distribution when we measure the ensemble. Basically, applying operator, $\hat x$, and potential, $V_0$, will produce a different distribution than $\hat x$, and potential, $V_1$, whatever those potentials maybe. (Need a check here)

We also know that we if we apply $\hat x$, and potential, $V_0 + V_1 = V_2$, it too will produce a different distribution, different than $\hat x$, and potential, $V_0$, or $\hat x$, and potential, $V_1$, alone. (Need a check here)

Scientist at Lab A has ensembles $A_1, A_2, A_3, ... $ and a scientist at Lab B has ensembles $B_1, B_2, B_3, ... $. Ensemble $A_1$ is entangled with $B_1$, $A_2$ with $B_2$, $A_3$ with $B_3$, and so on.

Scientist A chooses to apply $V_0$ to $A_1$ in their "local" area, Note this is actually applying to $\Psi(x_a, x_b)$, because the ensemble is entangled. Scientist A skips ensemble $A_2$ and chooses to apply $V_0$ to $A_3$, etc. Scientist B applies $V_1$ blindly to all the ensembles $B_1, B_2, B_3, ...$ at the same time agreed ahead of time with synchronized clocks

The distribution scientist B should get is the lower right hand side of the following figure: (Note, I made up the distribution shapes. They are just shown to be different to make the case or explain the thought experiment.)

enter image description here

The logical deduction based on my previous understanding and setup, if its all consistent is that one could send a message using ensembles of entangled atoms, synchronized clocks, and simultaneous measurements at a specified time agreed ahead of time.

If scientist A wants to send a message to scientist B, then he or she would choose which ensembles to apply $V_0$ to and then scientist B would blindly apply $V_1$ regardless of what scientist A did. Scientist B would know what scientist A sent because they both were applying their respective potentials at the same time to the same shared or entangled system, $\Psi(x_a, x_b)$ resulting in characteristic distributions that can be assigned to "1" "0" "1".

Does this proposed thought experiment have a major flaw? If so where? (First violation is obviously FTL information exchange. It seems to violate it, but where?)

$\endgroup$
  • $\begingroup$ It is not clear if you consider the $(A_i, B_i)$ as several distinct entangled states (one for each $i$), of if you think to one global entangled state with total correlation between the states $A_i$ and $B_i$, for each $i$ $\endgroup$ – Trimok Jul 15 '14 at 11:45
  • $\begingroup$ You've completely misunderstood what entanglement is in the first place, and of the impact a potential has on an entangled state. $\endgroup$ – ticster Jul 17 '14 at 2:03
3
+25
$\begingroup$

You've completely misunderstood the impact of adding a potential to an entangled state.

Atoms a and b should roughly have the same distribution ... because each single measurement of position for each entangled pair in the ensemble should yeild $x_a=x_b$ with respect to their local axes ... because this is what it means to be entangled. (Need a check here)

I just want to point out here that entanglement doesn't necessarily mean that the 2 distributions are the same. You could indeed have 2 perfectly identical states that are entangled, just as you could have 2 perfectly identical states that aren't entangled. Also, you could also have 2 completely different states and have them be entangled. What entanglement means is that you will find that there is a correlation between the 2 measurements, not that the 2 have a similar distribution. Let's take a simple example that doesn't even require the slightest bit of math to understand. Imagine a spin 0 particle decaying into 2 spin 1/2 particles. The states describing these 2 particles are now entangled, because whatever spin one is, the other must be the opposite so that the sum total is 0. Note that they both have the same distribution. In other words, if I did quantum tomography on both, I would find that they both are spin up half the time, and spin down half the time. But that's not why I say they're entangled. They're entangled because for each measurement, the states of both are correlated. In other words, they're entangled because measuring the spin of one particle implies the spin of another. This is due to the physical mechanism that produced these 2 particles such that the sum total of their spins must be 0. In other words, if one is spin 1/2, the other is spin -1/2, and vice versa. It is important to keep this mechanism in mind to understand the next mistake you've made.

Scientist A chooses to apply $V_0$ to $A_1$ in their "local" area, Note this is actually applying to $\psi(x_a,x_b)$, because the ensemble is entangled.

This is technically true, but it doesn't mean what you think it means. Sure applying $V_0$ in lab $A$ affects $\psi(x_a,x_b)$, but only because particle A is part of that system. In fact it's also part of $\psi(x_a,x_c)$ where C is any other particle in the universe, but again only because particle A is part of that system. That doesn't mean that it necessarily affects particle B, even if they're entangled. In other words, if $\psi(x_a)$ changes, then of course $\psi(x_a,x_b)$ changes because it's part of it. But that doesn't mean $\psi(x_b)$ changes. In fact it doesn't. Why would it ? Remember, the two particles are entangled because a physical mechanism entangled them. To understand the significance of this, let's go back to our spin example.

First recall that because the 2 particles being entangled implies that if we measure A to be spin 1/2, then B must be spin -1/2, and vice versa, so that the total spin is 0. Now, say we applied a potential to particle A that added 1 to its spin (whether or not this is easily feasible is irrelevant here, just entertain the hypothetical). Why on earth would that affect particle B ? You might think "well because it would shift the spin of particle B as well since they're entangled !". It wouldn't, and here's why. Again, remember, A and B are entangled because there's a physical reason for the total spin to be 0. If some other physical mechanism shifted the spin of particle A, then it would have to be somehow transferring this spin from somewhere (where that is would just depend on the physical mechanism we use to change the spin), and so we lose the condition that spin of A + spin of B is 0. Instead, it is now spin of A + spin of B is 1. The two are still correlated, and the only thing my physical mechanism did (the equivalent of $V_1$ in your example) was shift the correlation. Now it's such that if I measured A to be spin 3/2, then B must be spin -1/2, and if A is spin 1/2, then B must be spin 1/2. And so, like I said, we've just shifted the correlation. Remember, entanglement is nothing more or less than that correlation. And now, the usual reasoning applies for why this does not allow FTL communication. B doesn't know if A applied a potential or not until he uses some traditional (speed of light) communication to look at the measurements of A. The distribution of spin for B remains the exact same either way, likewise for any other potential that is applied to A. Such a potential only affects the relative value of the 2 distributions, but it doesn't actually change distribution B. The same reasoning would apply to a potential that would shift the position of a spatially entangled state.

Hope that helps, feel free to ask if you have any questions.

$\endgroup$
1
$\begingroup$

Your thought experiment does have a major flaw. According to quantum mechanics in any measurement of two spatially separated atoms a and b what happens to b has absolutely no effect at all on the probabilities of measurement outcomes on b. I'm not going to work out exactly what the flaw is in your proposed experiment, but just indicate why quantum theory rules out FTL communication.

Before I do that I will give a more precise description of entanglement than you gave. What sort of thing do you have to explain to understand what's going on in an entanglement experiment? The problems often look somewhat like this.

(1) There are observables on A and B, call them Acorr, Bcorr such that if you compare the results of the measurements after they have been completed and the information about the measurement results has been transmitted to the same location, you will find that they are correlated. So it might be that if you're measuring electron spin the spins will be found to be opposite with probability 1, or with some probability that differs from 1/2 when they are compared.

(2) There are observables on A and B, call them Anoncorr, Bnoncorr such that if you compare the results of the measurements after they have been completed and the information about the measurement results has been transmitted to the same location, you will find that they are not correlated. So it might be that if you're measuring electron spin the spins will be found to be opposite with probability 1/2 and the same with probability 1/2.

(3) There are intermediate cases. And overall the correlations are such that they do not match what you would get if you had two systems represented by local classical stochastic variables.

There might be complications due to measurement error or whatever in assessing the results of a particular experiment. You can also come up with formulae for precisely how the probabilities of matching differ depending on what observables you pick. The basic problem is that the probability of seeing correlations when you compare the results, differs from what you would expect from a local theory using classical stochastic variables. Experiments find such correlations but it does not follow from those correlations that quantum mechanics is non-local. No matter what measurement you do to on system b you cannot tell what has been done to system a.

What happens when b is measured? The measurement apparatus differentiates into multiple versions, each of which has recorded one of the possible outcomes. The records of measurements on a become correlated with records of measurement on b when the measurement results are compared because the decoherent systems that carry the measurement outcomes also carry locally inaccessible information. Their observables depend on what Bob has measured but the expectation values of those observables do not depend on the measurement, see

http://xxx.lanl.gov/abs/quant-ph/9906007

http://arxiv.org/abs/1109.6223.

These information transfers take place entirely locally and cannot take place in any other way because the underlying equations of motion are local. Quantum mechanics is not non-local, it rules out non-locality. People sometimes get confused about this because they believe in something called collapse of the wave function. They think that when they do a measurement only one outcome happens, but in reality all of the outcomes happen and each version of the observer sees only one of them as a result of decoherence. Your experiment will not produce FTL communication because quantum systems are governed by local equations of motion.

$\endgroup$
-3
$\begingroup$

You describe a communication protocol exploiting EPR/Bell-type setups. If I understand your protocol correctly, you envisage a 'stream' of entangled ensembles, and preagreed measurements being performed on each block of the stream to yield a 'bit'. There are various ways we could do this: you use harmonic oscillator potential. Fine. We could do the classic $x$ vs $z$ spin, or various other measurements.

You ask where FTL information exchange takes place. I'll assume, though you don't seem to specify, that your two parties are at spacelike separation (a stronger and more specific constraint than merely being 'separated'), and that the speed of the 'transmission' is such that it implies non-locality. Ultimately, the analysis of your protocol breaks down to a standard analysis of Bell experiments.

This will come down to your chosen interpretation of quantum mechanics. This is not a settled debate, but clearly some positions are more tenable than others. In a 'naive' Copenhagen interpretation, there is FTL information exchange. This is the whole point of the Einstein-Podolsky-Rosen paradox (their argument was that QM couldn't be complete, because the violations of local realism were physically unacceptable). Bell built on this by showing that it couldn't be explained away by local hidden variables; we can probably disregard Bohm at this stage. Alanf gave a good -- and strident -- analysis from a position inherited from Everett (+1). I'll look at the relational interpretation.

Decoherence poses problems for your protocol, much as it does for many quantum information type ideas. The struggle is to get some way of isolating your systems so that they do not decohere. Let us assume you have done this (a non trivial task).

We then have the problem of verifying that the information exchange has maintained integrity: ie that what B receives actually corresponds to what A sent. How do we do this? Well, they'd have to 'meet': not necessarily in person, but their light cones would have to intersect in some form; email will suffice. Thus it is impossible to state, with certainty, that your information exchange has been conducted with integrity until local constraints again apply. And, as if by magic, we find it does; A and B will agree. We could argue that we could have certainty, if we accept that our theory gives an adequate account of the world; but the point is that we can't actually verify the theory until we've confirmed, via an alternative channel, that our results are what the theory predicts (or they falsify the theory, or corroborate, or whatever your philosophy of science says is the best way to approach it).

This is not to say that we should doubt the validity of the underlying theory, or that we have reason to believe that the measurements would, for some reason, not correspond (on the contrary, multiple Bell type experiments suggest they do, and Zeilinger's work extended that to three-way GHZ entanglements). Rather, it strongly implies that the underlying theory is telling us something about local realism...ie that ascribing objectively definite (non-relational) states to quantum systems at spacelike separation is unphysical; they have to be given 'in relation to' another system. For brevity, this paper goes into much more detail on the relational analysis of EPR. It gives a very hypothetical, thought-experiment style explanation, but one which is (in my opinion) analytically valid, with no 'spooky action at a distance'. This is not exclusive of the idea of decoherence, but rather a different interpretation of it at the limit; most decoherence doesn't even approach these kinds of scenarios, but is rather why the moon is there when I'm not looking; quantum systems leak information into the world. The difference between this and Everett-based interpretations is that where they say 'all states, single description', the relational interpretation says 'single state relative to single observer, and ultimately all will be coherent'. Choose your poison -- the issue is not settled, and this probably isn't a forum for arguing which is better (philosophy, really, not physics, as empirically they couldn't be distinguished).

Thus the FTL issue isn't clear-cut. It may be that it is happening, but most modern interpretations would argue that it isn't. But EPR/Bell phenomena have certainly been shown experimentally to happen.

In summary: does your proposed experiment correspond with EPR/Bell type scenarios for transmitting information: yes, in principle, if you could construct measurements that satisfy your $V_0 + V_1 = V_2$ criteria. There are other measurements anyway that would enable essentially the same protocol, ie B performs a constant measurement the results of which appear to be affected by a chosen measurement by A (eg, B always measures spin along $z$, and A measures spin at a different angle for only those ensembles $A_1, A_2, A_3 \dots$ he wants to be '1' bits, so B's distribution for each ensemble differs depending on whether A measured or not, because the total spin of each singlet pair is always 0). Does it mean FTL communication? Depends on your interpretation of quantum mechanics, but there is no reason why it should.

$\endgroup$
  • 1
    $\begingroup$ His mistake is a mistake regardless of any interpretation of QM that you choose. $\endgroup$ – ticster Jul 17 '14 at 2:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.