In this work, continuous-variable quantum states are being teleported. The reason they almost-incorrectly call it deterministic is that continuous-variable [1](think analog) quantum states even when degraded by signal transmission errors will always transmit some information to the destination. Qubit (or discrete state) teleportation on the other hand will usually have some probabilistic errors that completely destroy the signal.
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
I work in this field and I can not really make sense of your statement. "Fidelity" the way they define it has nothing to do with whether they have a qubit or any other system (infinite dimensional or not).
The original paper [1], written by some of the best scientists of the era, correctly called it "Teleporting an Unknown Quantum State", and nowhere did it use the term "quantum teleportation". Later, in interest of conciseness, the ambiguous term "quantum teleportation" got introduced. It would have been better if "quantum state teleportation" had been used.
No mass is teleported, that's the key. A state of one mass is copied to another mass, but no mass is moved in the sense that would invoke General Relativity, as far as I understand.
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.
Names have the power to guide perception about what they identify.
"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.
"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 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.
I agree that the cartoon that you criticize is confusing. Starting with the first panel, because it doesn't make clear that Alice doesn't know the color/state of her photon. If she knows it's yellow she only has to tell Bob to prepare a yellow photon and there is no need for an entangled pair to make it possible. The whole point of the setup is to be able to "teleport" an arbitrary unknown quantum state.
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.
> sometimes the state of his photon is already the state of Alice's original photon [but] this is true only after the measurement performed by Alice
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:
> 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.
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.
> 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.
>> C was a pure state and now B is described by the same density matrix [after teleportation, i.e. after the whole procedure including the two-bit message reception and corresponding transformation]
> 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?
> The density matrix describing C at the beginning and B at the end is the same
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.
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
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 :-)
People get patents for perpetual motion machines as well... Do you really think your idea works?
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 horizontal polarization, what you get in the "receiver" is a 50/50 mixture of pure states |H> and |V> which is described by the same density matrix.
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).
I never said it did. I said that if you can transform a mixed state into a pure state then that poses a problem for QM because it would lead to FTL signaling. But you can't, so it doesn't.
Every time you do your measurement in the “transmiter” you will have a pure state in the receiver and it doesn’t lead to FTL signaling.
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).
Are you distinguishing between a "mixed state" and an "ensemble" and a "mixture of pure states"? Because AFAIK these are all the same thing. If you think they are different, what distinguishes them?
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.
A mixed state can be produced as a subset of a entangled system. It can also be produced as a mixture of pure independent states. They need not correspond to the same physical situation: you can prepare the same mixture state in different ways. But these ensembles can not be distinguished later (unless you “remember” the lost information about the precise state for each element!).
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.
> 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.
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.
Fine, but this approach leads nowhere. Before you said that “Before teleportation C is in a pure state”. But that’s surely impossible, because wherever C is coming from it’s entangled with a gazillion things already... If the original state C can be “pure”, then the final state B can be “pure” as well (and identical to C).
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.
> 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.
Alice knows which one of the four is the pure state of B after the measurement and tells Bob how to rotate the state (if needed) to recover alpha|0> + beta|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 :-)
> 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.
> The full-size coin that Alice inserted into Vending Machine A is no more (more precisely, its original state is lost).
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.
The presumption is that the original state is unknown. If the state was known, Bob would just need to prepare it in the same way. There is no need for an entangled pair and a destructive procedure to create a copy in that case.
I still don't understand what do you mean by "the color will be the same as that on the corresponding side of the full-size coin that Alice [had]". The original one will be lost; you can recreate a new one in the same state but this doesn't mean that "the color will be the same". If they have each a coin prepared as "heads = blue", looking at the other side they will find "tails = red" or "tails = green" without any correlation.
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". :-)
With all due respect, these papers are not written for the layman. They are written for the few thousand people who work in the field or near enough to understand this work. Single results in physics (and science) in general are almost never individually relevant to the public. The public should read either review or meta papers, or they should read the articles/books written by scientists for public consumption. As a physicist, I rarely upvote physics papers on HN, unless there is insightful commentary in the comments.
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.
This paper is really not meant for the layman. Even PhD physicists would find it terse unless they are working in the field of quantum information or AMO physics.
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.
To those saying the paper isn't for the "layperson", or average reader - papers like this exist to fluff the ego of the author more than they exist to convey information. It's very possible to write in a way that is both accessible to everyone and still informative to experts.
I find this statement very offensive. Papers like this exist to update people on the state of the art in their field (a target audience of rarely more than 100k people in the whole world). As others have mentioned already, there are great review articles, specifically meant for people just entering the field, and even they require quite a bit of dedication for a layperson to understand.
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.
What point are you trying to make with the comics? Of course a textbook or a review paper should be written clearly and simply, we already said that, and it has nothing to do with the original point of writing efficient letter when talking to your colleagues (as opposed to the public).
Do I get this straight:
"Quantum teleportation" is like receiving a letter with four pages and then telling another receiver of the same letter which page to read?
With the caveat that the four pages in Bob's letter are not written until Alice does her thing with her copy of the letter and the photon she wants to "teleport". Note also that she doesn't know the state of the photon, which is destroyed as a result of the procedure.
It's not surprising that the article is in English even though it's an all-Chinese team, because it was published on an English journal. But I can't stop wondering how many years are left before most scientific output is in Chinese. The dominance of English might be over soon.
It took centuries of British colonization to make British English the lingua franca of the world, and decades of American cultural and economic imperialism to switch the default to American English. It is unlikely that a complete switch to Chinese will happen on any shorter scale, if ever.
Decades aren't really a long scale, though. 3 decades ago almost nobody in the Czech Republic spoke any English, today it's a common day-to-day language in Prague as many foreigners live there, even the homeless speak good English.
Scientific results are primary communication. It can happen in English because lots and lots of non-English scientists worldwide understand English, just as in earlier times it could happen in Latin, French or German. It's pointless to produce articles which don't reach the audience that matters, so your language is irrelevant compared to the language of the audience.
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.
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