> 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 :-)
> 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").
> 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.
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.
> 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.
> 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.