
Deterministic quantum teleportation through fiber channels - lainon
http://advances.sciencemag.org/content/4/10/eaas9401
======
abdullahkhalids
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

~~~
krastanov
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).

------
dsr_
The number one problem with quantum teleportation is the name. It should have
been named quantum telegraphy. There would be fewer cranks.

~~~
abdullahkhalids
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.

[1]
[https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.70...](https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.70.1895)

~~~
buboard
but quantum state is all that can ever be measured about the system so seems a
redundant word

~~~
nine_k
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.

~~~
FeepingCreature
Quantum copypaste.

------
lisper
For anyone wanting a layman's description of QT:

[https://blog.rongarret.info/2014/12/quantum-teleportation-
de...](https://blog.rongarret.info/2014/12/quantum-teleportation-
demystified.html)

~~~
kgwgk
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.

~~~
lisper
> 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:

[http://www.flownet.com/ron/QM.pdf](http://www.flownet.com/ron/QM.pdf)

~~~
kgwgk
> 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.

~~~
lisper
> These statements are not equivalent.

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.

~~~
kgwgk
>> 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?

~~~
lisper
> 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.

~~~
kgwgk
In case anyone else is reading this, I want to make clear that I'm giving the
standard description of quantum teleportation.

[https://en.wikipedia.org/wiki/Quantum_teleportation#Formal_p...](https://en.wikipedia.org/wiki/Quantum_teleportation#Formal_presentation)
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?

~~~
lisper
> 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 :-)

[https://patents.google.com/patent/US7126691B2/en](https://patents.google.com/patent/US7126691B2/en)

~~~
kgwgk
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).

~~~
lisper
> Do you really think your idea works?

Of course not. That's the whole point.

~~~
kgwgk
I'm lost. It doesn't work according to QM. And you agree that it doesn't work.
How does it pose a problem for QM again?

~~~
lisper
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.

~~~
kgwgk
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).

~~~
lisper
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.

~~~
kgwgk
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.

~~~
lisper
> 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.

~~~
kgwgk
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).

~~~
lisper
> this approach leads nowhere.

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...](http://blog.rongarret.info/2018/05/a-quantum-mechanics-puzzle.html)

[https://blog.rongarret.info/2018/05/a-quantum-mechanics-
puzz...](https://blog.rongarret.info/2018/05/a-quantum-mechanics-puzzle-part-
deux.html)

[https://blog.rongarret.info/2018/05/a-quantum-mechanics-
puzz...](https://blog.rongarret.info/2018/05/a-quantum-mechanics-puzzle-part-
drei.html)

> 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.

~~~
kgwgk
> how B was transformed from its initial state into its final state, and more
> to the point, when this transition happened.

There is a full description here:
[https://en.wikipedia.org/wiki/Quantum_teleportation#Formal_p...](https://en.wikipedia.org/wiki/Quantum_teleportation#Formal_presentation)

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>
    

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 :-)

~~~
lisper
> 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...

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.

~~~
kgwgk
> The problem with that description is that the word "now" is not well defined
> when Alice and Bob are far apart.

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").

~~~
lisper
> Do you have an alternative description to work around this "problem"?

I like this approach:

[https://arxiv.org/abs/quant-ph/9605002](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.

~~~
kgwgk
I thought you were referring to the relativity of simultaneity. I'm glad you
insist on the FTL communication issue, maybe now I will be able to convince
you that there is no issue at all...

> 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)
    

You say that, before any message from Alice can reach Bob, he can determine
wheter or not she has performed the measurement.

Can you explain how? (Hint: no, you cannot.)

~~~
lisper
> I thought you were referring to the relativity of simultaneity.

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.

~~~
kgwgk
> I'm not sure you're clear on what we're actually disagreeing about.

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.

~~~
lisper
> your claim that the QM description is not correct

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.

~~~
kgwgk
>> your claim that the QM description is not correct

> 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](https://arxiv.org/abs/1804.03267)

~~~
lisper
Oh, Hi! Sorry, I didn't recognize you.

> 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.

~~~
kgwgk
There goes my HN anonymity! (Not a big deal, but if you can edit it out I
prefer it, just in case...)

I have some follow up comments, I'll send you an email when I have time to
write something detailed.

------
thomastjeffery
Papers like this are so terse.

I understand the need to use correct verbiage; but to the layman, this is
incomprehensible.

~~~
abdullahkhalids
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.

~~~
copperx
I think OP was questioning how is this paper fits in an generalist IT/business
forum.

~~~
abdullahkhalids
I totally agree with them that it should not be posted here.

------
rightbyte
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?

~~~
kgwgk
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.

------
modzu
it begins

------
jcwilde
Is this not the holy grail?

~~~
adrianN
The holy grail of what?

------
copperx
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.

~~~
abdullahkhalids
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.

~~~
TomMarius
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.

