Deterministic quantum teleportation through fiber channels
advances.sciencemag.org
advances.sciencemag.org
Therefore, the correct metric to report for analog signals is the fidelity. They do report this as 69% - which is above a certain important threshold - but also fairly low for any practical application in the near future.
[1] as opposed to discrete-variable states eg. qubits
The most natural name would be "replication" (of the quantum state), but given than the original is lost and only the replica survives "teleportation" doesn't look so bad.
[1] https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.70...
I suppose teleporting a state in such a way that the target mass would change its motion in a significant way ("accelerated") without affecting the source mass is impossible, because it would break the conservation of impulse.
"Quantum teleportation" is as easily misused as that of "dark matter".
Is it actually matter? If the answer isn't a definite yes then the term doesn't help long term but maybe it's good enough for now. "Dark energy" is another term. Is it accurate? Maybe. It alse works for plenty of people for now.
Both terms assist as long as they help discussion but the lay person easily gets confused and experts can get mislead as well. If you think of it as matter then you design experiments in particular ways. You think about it in those terms. You don't think of it in other ways to the same extent.
Quote from Wikipedia: https://en.m.wikipedia.org/wiki/Dark_matter
"The primary evidence for dark matter is that calculations show that many galaxies would fly apart instead of rotating". Not exactly a ringing endorsement of matter as real. But as long as it helps the overall investigation then it's ok for now. Calculations can accurately describe reality and guide decisions so why not.
"Quantum teleportation" as a term is easily derailed no matter how helpful it describes a portion of reality. Maybe it simply needs a few more words nearby to put the brakes on people's imagination.
The map of the place is not the place.
https://blog.rongarret.info/2014/12/quantum-teleportation-de...
But the following doesn't seem correct:
> when Bob receives Alice’s bits, he uses that information to apply one of four transformations to his photon. It is Bob’s action that changes his photon’s state into the state of Alice’s original photon. And that is the only time that Bob’s photon changes state.
To recover the state being "teleported" by Alice, Bob has to apply one of the four pre-specified "rotations" (unitary transformations) according to the two-bit message he receives. In fact one of the operations is to do nothing, sometimes the state of his photon is already the state of Alice's original photon.
Of course this is true only after the measurement performed by Alice. Previously there was no relationship between the state of Bob's photon and the state of Alice's "source" photon: the probability of the state being the same (or related by one of the three other pre-specified rotations) is essentially zero.
No, that's wrong. Alice's measurement does not change the state of Bob's particle. There is no measurement Bob can perform on his particle which will tell him whether or not Alice has performed her measurement or not. Quantum teleportation is no different from regular old EPR in this regard.
> the probability of the state being the same (or related by one of the other three pre-specified rotations) is essentially zero.
Actually, it is precisely zero. The entire claim that the state of Alice's particle is teleported intact to Bob is actually a lie.
What is true is that no measurement can distinguish the state of Bob's particle (B) after teleportation from Alice's original (C). But this does not mean that the particles are in the same state. Before teleportation C is in a pure state, and B is in a mixed state (as is A). The Bell measurement that Alice performs on A and C entangles them, and the result is a three-particle mutually entangled system, plus two classical bits that tell you how to tweak B so that its measurement outcomes are the same as C's. But B is in the exact same (mixed) state after Alice's measurement as it was before.
BTW, there is a measurement Bob can perform, not on a single B, but on a collection of B's that will reveal that all the B's are in fact in a mixed state. If you want to go down that rabbit hole, read this:
> There is no measurement Bob can perform on his particle which will tell him whether or not Alice has performed her measurement or not.
These statements are not equivalent.
> Actually, it is precisely zero.
I meant that it's zero as in "the probability of a random real number in [0 1] to be 0.5 is zero", but not as in "the probability of a random real number in [0 1] to be 42 is zero".
> What is true is that no measurement can distinguish the state of Bob's particle (B) after teleportation from Alice's original (C). But this does not mean that the particles are in the same state.
C was a pure state and now B is described by the same density matrix so it is necessarily the same pure state.
> But B is in the exact same (mixed) state after Alice's measurement as it was before.
B is in a different mixed state afterwards and the two-bit message enables Bob to distinguish between the four potential pure states and recover Alice's original pure state C.
Why not?
> C was a pure state and now B is described by the same density matrix
No, it isn't. All of B's measurement outcomes are the same, but the density matrix is necessarily different because B is in an entangled state. If this were not the case then it would lead to FTL communication.
> B is in a different mixed state afterwards
That depends on what you mean. The only difference is that afterwards, B is entangled with C as well as A, so now the joint A-B system is in a mixed state whereas before the A-B system was pure. In that respect, yes, B is in a different state. But viewed in isolation, B's state is exactly the same after the A-C entanglement as before.
> No, it isn't. All of B's measurement outcomes are the same, but the density matrix is necessarily different because B is in an entangled state. If this were not the case then it would lead to FTL communication.
The density matrix describing C at the beginning and B at the end is the same (otherwise how could all measurement outcomes be the same?) and B is not in an entangled state. I don't see how this leads to FTL communication, the procedure requires Alice to send a (non-FTL) message to Bob.
>> B is in a different mixed state afterwards [after Alice's measurement]
> That depends on what you mean. The only difference is that afterwards, B is entangled with C as well as A, so now the joint A-B system is in a mixed state whereas before the A-B system was pure.
B is no longer entangled with the A/C system after Alice's measurement.
> In that respect, yes, B is in a different state. But viewed in isolation, B's state is exactly the same after the A-C entanglement as before.
For the sake of the argument, let's say that Alice knows that the original state C is |up> and that after performing her measurement she sends to Bob the message "you don't have to do any transformation, the state of B is now equal to C, i.e. |up>" (this case will happen eventually if we repeat the experience several times). If, according to you, the state of B is exactly the same after Alice's measurement as it was at the beginning of the procedure, how do you explain that the state of B is |up> at the end? Or what would you say is happening to B, if anything at all?
No, it isn't. C starts in a pure state. B is created and always remains in a mixed state.
> (otherwise how could all measurement outcomes be the same?)
Because you cannot distinguish a mixed state from a pure state when measuring only a single particle. You can distinguish them if you make measurements on multiple particles in the same state: pure states produce first-order interference. Mixed states don't. (This is just the old QM trope about "measurement destroys interference", which is not quite true. Entanglement "destroys" interference, except that it doesn't really, it just destroys first-order interference.)
> B is no longer entangled with the A/C system after Alice's measurement.
Yes, it is. The only way to "undo" an entanglement is to bring the originally entangled particles back together at the same location, i.e. it is essentially impossible. If you could cause an entangled particle to become unentangled (i.e. to transition from a mixed state to a pure state) by performing an operation on its partner then you could do FTL signaling by observing the presence or absence of first-order interference.
> how do you explain that the state of B is |up> at the end
It isn't. It's in a mixed state. It is no different than regular EPR with only two particles. If Alice measures A and gets the result "up" then Bob is guaranteed to get "down", but that does NOT mean that B is in the |down> state. It is still in a mixed state. |down> is a pure state.
It is important to remember that mixed states and pure states cannot be distinguished by making measurements on a single particle, so if there is only one particle this is a philosophical distinction, not a scientific one. But if the experiment can be repeated, then the fact that B is mixed can be experimentally demonstrated, with all the usual caveats about quantum weirdness. The point is that QT does not introduce any new weirdness that wasn't already there in EPR.
https://en.wikipedia.org/wiki/Quantum_teleportation#Formal_p... provides a clear description of the initial states of C and A/B, the operations performed by Alice and Bob and the final state of B (which is identical to the initial state of C).
Clearly you don't find this description correct but I would be surprised if you can write down an alternative description that makes sense.
You're free to have your own theory, but I don't understand why do you feel that you need one in the first place. What is the problem with measurement breaking entanglement?
> If you could cause an entangled particle to become unentangled (i.e. to transition from a mixed state to a pure state) by performing an operation on its partner then you could do FTL signaling by observing the presence or absence of first-order interference.
Let's say Alice and Bob have entangled spins in the pure state (leaving out the normalization term):
|up>_alice * |down>_bob - |down>_alice * |up>_bob
According to QM, if Alice measures the spin along the z-axis one of the following alternatives will happen:
a) she gets a positive measurement, her spin is in the pure state |up>, Bob's spin is in the pure state |down>, there is no entanglement between the spins
or
b) she gets a negative measurement, her spin is in the pure state |down>, Bob's spin is in the pure state |up>, there is no entanglement between the spins
How do you think this allows for FTL signaling?
I already told you: with a single particle, it's not possible. But with multiple particles you can tell the difference between a pure state and a mixed state by observing the presence or absence of first-order interference. So to communicate, you send a stream of EPR pairs to Alice and Bob. Alice transmits by either measuring batches of particles or not measuring them, and Bob receives by observing the presence or absence of first-order interference on his end.
Fun fact: I was able to get a patent on this idea :-)
This is what I think: the quantum state of your entangled pair is a pure state and if you look at the quantum state of the photon sent to the "receiver" it is described by the density matrix obtained by tracing out the other photon :
1/2 0
0 1/2
If you measure in the "transmiter" the 45-degrees polarization, what you get in the "receiver" is a 50/50 mixture of pure states |45> and |-45> which is again described by the same density matrix.
No matter what yo do or do not measure at the "transmiter" you will get the same measurements at the "receiver" because the density matrix describing the state is always the same (unpolarized light).
Of course not. That's the whole point.
What you have when you repeat the procedure is an ensemble of states: half of them the pure state |H> and half of them the pure state |V> (or half of them |H> and half of them |V>).
There is no entanglement, only (a mixture of) pure states. And it doesn’t lead to FTL signaling.
At the end of the “teleporting” procedure you have a pure state B which is identical to the original pure state C. And it doesn’t lead to FTL signaling.
Maybe your problem arises from the following misunderstanding:
> If you could cause an entangled particle to become unentangled (i.e. to transition from a mixed state to a pure state)
In the example I gave above, the density matrix for the photon arriving at the “detector” doesn’t change when the measurement is made at the “transmitter” and entanglement is broken. But the same density matrix describes different situations. When there is entanglement the photon is part of a pure state and the density matrix is obtained by tracing out the rest of the system. Later there is no entanglement and the density matrix describes a statistical mixture of pure states (50% |H> and 50% |V>, or whichever is the apropriate basis for the measurement that was made on the other photon destroying entanglement).
If you agree that they are all the same, then what you are saying is exactly the same as what I am saying, except for the "no entanglement" part. Whether or not you have entanglement or a mixed state depends on what point of view you choose to take. If you have an EPR state, then the joint system is in a pure state, but each individual particle is in a mixed state when viewed as an isolated system. That is what a mixed state is. A mixed state is nothing more or less than the state of a proper subset of an entangled system.
For example, you can get unpolarized light as a mixture of left- and right-polarized light, or as a mixture of horizontal- and vertical-polarized light, or getting one photon from entangled pairs (and discarding the other). I’m sure that you can find a way to describe that as an entanglement with something else, but there is no need to do so.
In particular, when Alice measures a photon in an entangled pair the entanglement is broken. Bob’s photon is no longer entangled with Alice’s photon. Hopefully you will agree that Alice’s photon is now in a pure state. Again, you may want to make things much more difficult but it’s not required to prevent FTL signaling.
That depends on your goals. If all you care about is making correct predictions of experimental results, then yes, one point of view is as good as another.
But if you care about actually understanding what is going on then explaining measurement in terms of entanglement is progress.
> when Alice measures a photon in an entangled pair the entanglement is broken
No. The entanglement remains, and is incorporated into a (much) larger system of mutually entangled particles, including Alice herself. As part of this process, decoherence happens, which makes the original entanglement impossible to detect as a practical matter. But it doesn't go away.
> Hopefully you will agree that Alice’s photon is now in a pure state.
No, I will not agree. When Alice measures the photon, the photon must be absorbed by some detector, and so can longer be said to exist at all in any meaningful sense.
I think you have a problem with the probabilistic nature of QM, if you only can accept a statistical mixture as being part of an entangled pure system. FTL signaling is just a excuse (and you don’t explain how standard QM would allow for it).
It has led to a great deal of mental clarity for me. YMMV.
> Before you said that “Before teleportation C is in a pure state”. But thats surely impossible, because wherever C is coming from it’s entangled with a gazillion things already...
That's right, and indeed figuring out how a laser actually works is quite challenging.
http://blog.rongarret.info/2018/05/a-quantum-mechanics-puzzl...
https://blog.rongarret.info/2018/05/a-quantum-mechanics-puzz...
https://blog.rongarret.info/2018/05/a-quantum-mechanics-puzz...
> If the original state C can be “pure”, then the final state B can be “pure” as well (and identical to C).
Only if you are willing to explain how B was transformed from its initial state into its final state, and more to the point, when this transition happened.
> I think you have a problem with the probabilistic nature of QM
Huh? What have I said that leads you to believe that? It's not true.
> if you only can accept a statistical mixture as being part of an entangled pure system
That has more to do with philosophy than physics. Does an apple really fall because gravity is pulling down on it? No, but it's a good enough approximation to the truth that it's usually not worth quibbling over.
There is a full description here: https://en.wikipedia.org/wiki/Quantum_teleportation#Formal_p...
It's difficult to copy formulas here but essentially A and B are in one Bell state at the beginning.
When Alice makes a measurement in the basis of the Bell states for C and A, the result is that B will now be in one of the states
alpha|0> + beta|1>
alpha|0> - beta|1>
beta|0> + alpha|1>
beta|0> - alpha|1>
So, again, when Alice does the measurement the state becomes one of the four listed states and when Bob rotates the state (if needed) B is the same state that C was originally. Maybe you agree that this is making correct predictions of experimental results, maybe not. I still don't know when and how does B change along the procedure according to your view.
> No, but it's a good enough approximation to the truth that it's usually not worth quibbling over.
Talking about "pure" states after a measurement is a good enough approximation to the truth, but you find it worth quibbling over :-)
The problem with that description is that the word "now" is not well defined when Alice and Bob are far apart.
> but you find it worth quibbling over
Well, yeah. Just because I'm willing to accept an approximation to the truth doesn't mean I stop caring about the actual truth.
Do you have an alternative description to work around this "problem"?
QM is a non-relativistic theory. In this case, the measurement on C/A is strictly before the rotation on B so there is not even the shadow of a problem. Relativistic extensions to QM are Lorentz-invariant and for space-separated observables there is a restriction for operators to be commuting (so the outcome is consistent with either of the observations being "first").
I like this approach:
https://arxiv.org/abs/quant-ph/9605002
> the measurement on C/A is strictly before the rotation on B so there is not even the shadow of a problem
Yes, obviously. That is not what I'm talking about.
Consider this variation on the theme: Alice and Bob share N>>1 entangled A-B pairs. Alice prepares N additional C particles, all in a known pure state P, and runs them all through the teleportation protocol. But instead of sending Bob rotation instructions, she sends him only the serial numbers of the particles that do not require Bob to perform a rotation. Bob keeps the particles corresponding to those serial numbers, and destroys the rest. I presume you will agree that Bob now has a collection of approximately N/4 particles in a pure state P. (I do not agree with this, BTW, but I accept it here for the sake of argument.)
Now imagine that Bob simply selects N/4 particles at random without any information from Alice. He has no idea what Alice has done on her end, or indeed if she has done anything at all. There are now three cases:
1. Alice has performed her measurements, and by pure chance Bob has selected the same set of particles that Alice would have told him to select. I presume that in this case you would agree that this set of particles is still in state P. (The odds of this happening are, of course, vanishingly small, but still >0, and this is a thought experiment.)
2. Alice has NOT performed any measurements. Are Bob's particles still in state P?
3. Alice has performed measurements, and Bob's random selection does not match Alice's list of particles that don't require rotation. (This is the overwhelmingly likely scenario, of course.) In this case, Bob's particles are not in state P (because 3/4 of them require a rotation). But are they in some other pure state Q?
The answer to 2 and 3 must be "yes" if you want to avoid FTL (because otherwise Bob could determine whether or not Alice has performed measurements). So the conclusion is: if you randomly select N/4 particles from a mixed population, the resulting particles are all in some pure state. Do you agree with this? If so, were those particles in that same pure state before they were selected?
The conclusion I'm driving towards is that purity of state has nothing to do with the actual physical situation, it's a matter of perspective. Any physical state can appear to be pure if you look at it in the right way, but finding the "right way" to look requires additional information, i.e. information from Alice. Nothing Alice does changes anything about the physical situation on Bob's side. It simply produces the information Bob needs in order to look at his particles in the "right way" to see a particular pure state.
> The answer to 2 and 3 must be "yes" if you want to avoid FTL (because otherwise Bob could determine whether or not Alice has performed measurements).
The answer to 2 is "no". If Alice has not performed any measurement, the N "B" particles that Bob has are not in a pure state. They are a subsystem of the N (indentical) entangled pairs formed by the N "A" particles that Alice has plus the corresponding N "B" particles that Bob has (let's call that state E).
So, depending on whether Alice has done the measurement or not, what we have is (I undertstand you don't really accept this description, it's for the sake of the argument where you will try to show how this description is unacceptable):
Measurement not done
Pairs remain entangled
The N particles held by Bob are not in a pure state
(The A/B pairs are in a pure state E)
Measurement done
Pairs are no longer entangled
The N particles held by Bob are each in one definite state P or P' or P'' or P'''
(Alice knows the precise state of Bob's particles, she knows how to rotate each particle to get the state P)
Can you explain how? (Hint: no, you cannot.)
Yes, I was.
> I'm glad you insist on the FTL communication issue,
It amounts to the same thing.
> maybe now I will be able to convince you that there is no issue at all...
Maybe, except that I'm not sure you're clear on what we're actually disagreeing about.
> You say that, before any message from Alice can reach Bob, he can determine wheter or not she has performed the measurement.
Only in a hypothetical case, not realizable in practice (probably -- see below), where Bob has chosen N/4 particles at random and those just happen to coincide with the particles that Alice measured and found to require no rotation. In that case, Bob will, on your view, observe first-order interference if Alice has measured, and not otherwise.
Maybe it would help if I describe it as a detailed thought experiment. The experiment takes place in four phases.
Phase 1: Alice and Bob choose a basis for a pure state P which they will use for the remaining two phases.
Phase 2: Alice teleports N copies of state P to Bob using the normal protocol, including the transmission of the results of her measurements. Bob takes the N/4 particles that require no rotation and tests them to see if they produce first-order interference relative to the basis of P. I trust you and I agree that he will observe interference in this case.
Phase 3:
This is the part that can't be realized in practice, and I have to number the steps here because this phase runs in a loop. Phase 3 is designed to answer the question: what would have happened in phase 2 if Bob had chosen the same N/4 particles by pure chance? So...
1. N EPR pairs are distributed.
2. Alice performs her part of the teleportation protocol, but does NOT tell Bob the results of her measurements.
3. Bob selects N/4 particles at random and tests for first-order interference relative to P. (The vast majority of the time the result will be negative.)
4. Bob and Alice compare Bob's random selections in step 3 with the results of Alice's measurements in step 2. If they happen to coincide, i.e. Bob just happened to pick the correct N/4 particles that did not require rotation, the experiment halts, with the result being whatever Bob observed on the last iteration of step 3.
I trust that you and I will agree that Bob will also observe interference in this case.
Phase 4 (this is the interesting one):
This is identical to phase 3, except that steps 2 and 3 are reversed, i.e. Bob makes his random selection and interference measurement BEFORE (in a fully relativistic sense) Alice makes her measurements. i.e.:
1. N EPR pairs are distributed.
2. Bob selects N/4 particles at random and tests for first-order interference relative to P.
3. Bob communicates to Alice that he is done.
4. AFTER Alice receives Bob's signal from step 3, she performs her side of the teleportation protocol for N copies of P.
5. They compare the results of Alice's measurements in step 4 with Bob's random choices made in step 2. If they coincide (i.e. Bob just happened to choose the correct N/4 particles that did not require rotation) the experiment concludes, with the result being whatever Bob observed on the final iteration of step 2. Otherwise, they go to step 1 and try again.
On your view, the predicted result of phase 4 will be negative. So if Alice cheats and performs her measurements before receiving Bob's signal, Bob can tell.
Postscript: now that I think about it, it might actually be possible to perform this experiment. A mach-zehnder interferometer can be pretty sensitive, so it might be possible to get N down low enough that the experiment could terminate in a reasonable amount of time.
To be clear, I disagree with your claim that the QM description is not correct because it leads to FTL communication.
If you stand by that, you're wrong. If that's not what you meant, then there is no disagreement.
Now, if I understand your latest comment, you say that it's "Only in a hypothetical case [...] where Bob has chosen N/4 particles at random and those just happen to coincide with the particles that Alice measured."
This is like saying that two standard 52-card decks can be used for FTL communication, in the hypothetical case where Bob picks the same card that Aliced picked. If you think that is FTL communication I cannot agree with that. This kind of "FTL communication" doesn't present any problem at all for the QM description.
Maybe I'm missing something, but I don't see in your description of the experiment any trace of potential FTL communication. You're not even trying! I would like to see something like the following (but I understand you cannot provide it, because it's impossible):
1) N EPR pairs are distributed
2) Bob does something ("selects N/4 particles at random and tests for first-order interference relative to P" or whatever) and concludes that "Alice has done something to her particles" or "Alice has not done anything to her particles" depending on the result of his measurement.
[The prediction is done at this point, the verification step follows.]
3) Alice tells Bob if she did or not measure her particles before he did. Or Bob tells Alice his prediction. Or Alice and Bob tell Charlie and he performs the check. In any case, this requires slower-than-light communication.
This experiment can be repeated several times to see if Bob's prediction is right every time or if at least there is some correlation between what Alice did and what Bob guessed.
Your description has additional steps involving communication between Alice and Bob before there is any result or prediction. It is not in any way incompatible with the standard QM formulation that you reject, as far as I can see.
It looks like if, in the card-picking example, Bob picks a card and then compares his card with Alice's card. If the card is the same they announce that FTL communication had been achieved. If the card is not the same, they try again. If you think this is not a fair characterisation, please make clear what prediction is Bob doing before communicating with Alice.
To make my point as clear as I can once again: according to QM, Bob cannot tell if Alice has performed or not a measurement on her side. For any measurement that Bob can do on his particle, the predicted distribution of outcomes conditional on "the entanglement is broken, the particle is now in an unknown pure state" is exactly the same as the predicted distribution of outcomes conditional on "the entanglement is intact, the particle is not in a pure state (the pair A/B is still in the pure state E)". Whatever the outcome he gets out of his measurement, it cannot be evidence for one or the other.
That's not my claim. My claim is that the change in the mathematical description of Bob's particles (from mixed states to pure states) when Alice measures her particles does not correspond to any actual physical change in Bob's particles.
We agree about the math, and we agree about the outcomes of all experiments performed by Alice and Bob.
There is one thing we disagree about, and that is the outcome of the thought experiment I described above (which may in fact be realizable in practice). I predict that you will observe "retrocausality" in phase 4, i.e. when the experiment halts, Bob's choices will correspond with Alice's measurements with (essentially) 100% certainty. (Of course, this isn't really retrocausality, but it appears that way.) AFAICT, on your view of the world, you would predict (essentially) a 0% chance of Bob's choices corresponding with Alice's measurements.
Just out of curiosity, are you a physicist? Because you obviously know what you're talking about.
BTW, just in case it wasn't obvious, I left out a detail in the description of the thought experiment: In phase 3, if the experiment does not halt at step 4 it loops back around to step 1, just as it does in phase 4.
> That's not my claim. My claim is that the change in the mathematical description of Bob's particles (from mixed states to pure states) when Alice measures her particles does not correspond to any actual physical change in Bob's particles.
Well, according to quantum mechanics there is a physical change when Alice measures her particles because:
before Alice's measurement => the pair "A/B" is entangled
after Alice's measurement => the pair "A/B" is not entangled
According to you alternative description, is the pair A/B entangled or not after Alice's measurement? Or is entanglement not an actual physical property of the pair?
I am (or used to be, for a brief time many years ago) a physicist. We had the opportunity to discuss in person last week. I enjoyed that (and this) discussion, thanks. By the way, I said I would send you a comment from David Mermin about the recent paper from Frauchiger and Renner. Actually I was thinking about this article from Jeffrey Bub: https://arxiv.org/abs/1804.03267
> Well, according to quantum mechanics there is a physical change when Alice measures her particles because: > before Alice's measurement => the pair "A/B" is entangled > after Alice's measurement => the pair "A/B" is not entangled
No. That's the whole point. The presence of absence of entanglement is a function of the perspective you choose to take when modeling the situation, not a function of the actual physical situation.
Thanks for the pointer to the Bub paper, that's a great reference.
Maybe we should move this discussion to email? It's getting pretty deeply nested and I don't think anyone else is paying attention.
I have some follow up comments, I'll send you an email when I have time to write something detailed.
> Whichever he chooses, the color will be the same as that on the corresponding side of the full-size coin that Alice inserted into Vending Machine A.
The full-size coin that Alice inserted into Vending Machine A is no more (more precisely, its original state is lost). I don't think it makes sense to say that the color would be the same as in a measurement that was never done and can never be done. But I guess you can say that the probability distribution of the colors he will see, no matter what side he chooses, will be the same as if he had looked at the original full-size coin held by Alice.
That's true, but the presumption is that the original was prepared in such a way that its state is known.
> the probability distribution of the colors he will see, no matter what side he chooses, will be the same as if he had looked at the original full-size coin held by Alice.
Exactly.
I guess I was thinking that if you wanted to experimentally verify that the state was being reproduced you could start with a known prepared state.
Edit: If you mean that the probability distribution of outcomes will be the same for any measurement then of course I do agree; this is why we can say that "the state of Alice's particle is teleported intact to Bob". :-)
I understand the need to use correct verbiage; but to the layman, this is incomprehensible.
This is not an insult to anyone who is not a scientist - just that jargon is necessary for rapid progress, and understanding jargon requires developing expertise by many years of work.
However, there are many exciting and great resources where anybody interested can learn more about advanced physics. "The Theoretical Minimum" or Feinman's Lecture series are a great long on boarding experience that starts slow but gets pretty serious towards the end.
And many researchers have more pop-sci type of blogs and media materials for those that want just a quick fix.
Or to phrase it in a different way: how often do emails on the dev mailing lists of open source project include a detailed description of the project that would be enough to bring laypeople up-to-speed.
I wouldn't expect much (not even most, but even a significant amount) of scientific output to happen in Chinese, even if produced by Chinese authors, until Chinese becomes a very popular second language worldwide which a critical mass of global researchers (e.g. most people who "matter" in a particular subfield) can use. Currently it still has a far way to go - while it's comparable in the total number of speakers (1.1 billion for both Chinese and English according to https://en.wikipedia.org/wiki/List_of_languages_by_total_num...), Chinese has just 200m secondary speakers compared to 740m of English, and that is the number that matters; not the total number of people but the "marketshare" of how many of worldwide 'educated class' know the language, and English has an extreme headstart in this regard. Any changes would require at least a full generation - new researchers growing up that had learned Chinese because it was widely taught for other reasons.