# Alice/Bob Instant communication protocol. Does this work?

1. Apr 7, 2012

### al onestone

Here is an instant communication protocol for Alice and Bob. Alice and Bob are given one each of a pair of momentum entangled photons (or particles).

The two particles are prepared with entangled momentum states and the momentum distribution for both has uncertainty Δpi. The position distribution has the complementary uncertainty Δxi .
Alice modifies her particle by simply choosing to measure the position with accuracy Δxf where Δxf << Δxi. Because the position of Alice’s particle is now known with the much greater accuracy of Δxf, the accuracy of our knowledge of the position of Bob’s particle must also be Δxf. Since Bob’s particle now has positional uncertainty of Δxf, then Bob’s particle must now have a momentum uncertainty which is complementary to it, Δpf = Δxf/ħ >> Δpi. This being the case, all Bob has to do is prepare his particle at a greater distance from the source than Alice’s, by passing it through a band pass filter that only allows the momentum range Δpi , which should have the result that Bob’s particle will pass the filter with much less likelyhood due to the increased momentum distribution of Bob’s particle which was due to the measurement that Alice performed. Remember, if Alice doesn’t measure her particle then Bob’s particle should pass the filter with 100% reliability. (Rather than a bandpass filter you could use its inverse which would have opposite results). So Alice’s performance of a measurement on her particle should have measurable results on Bob’s particle by the above reasoning, results which are instantly projected from a distance. By modulating between the extremes(measurement and non-measurement) of Alice’s preparation we can use this thought experiment as an instant communication protocol.
Just to let you know, this is approximately my 100th instant communication protocol thought experiment. Usually I’ve gotten my reasoning wrong somewhere. So tell me, where have I gone wrong this time?

2. Apr 7, 2012

### San K

Quantum entanglement can, I gather, happen on a couple of properties of photons such as spin, etc

I don't know if two photons can be momentum entangled.

Let’s, for a moment, assume photons can be momentum entangled

In that case what will happen is that --

You cannot get/control what moment/position outcome you want ...just like when you measure the spin of a particle you cannot force it to be spin up (or down)......

this leads us to an interesting derivation:

only those properties can be Quantum entangled which are "inherently random"

Whenever we are trying for instant communication via QE, we need to check for, at the least, the following things:

1. can the properties be entangled
2. can you/alice/bob control the outcome of that property
3. do alice and bob need to compare notes (i.e. their individual outcomes) to arrive at a conclusion

Last edited: Apr 7, 2012
3. Apr 7, 2012

### lugita15

San K, in QM you can have entanglement with respect to ANY observable.

4. Apr 7, 2012

Staff Emeritus
Frankly, your general mistake is not learning QM first before trying to find a flaw in it. (QM says you cannot do what you are trying to do) This means all that boring stuff like scattering and bound states and separation of variables. But you need to start at the beginning and have it nailed before trying to find some loophole.

In this specific case, there is nothing in QM that says you cannot measure p and x of a single particle with arbitrary precision. What Heisenberg is telling us is something about the result of an ensemble of measurements of identically prepared states.

I'd encourage you to work on the basics of QM. You'll make more progress once you have it solidly under your belt.

5. Apr 7, 2012

### San K

ok that's an interesting piece of information

however the outcome of all such observable becomes "inherently random"

so i need to modify my statement to make it more precise, here goes (and you are welcome to modify it further):

outcome of all observable(s) that are quantum entangled becomes "inherently random"

6. Apr 9, 2012

### al onestone

To Vanadium 50: your quote "there is nothing in QM that says you cannot measure p and x of a single particle with arbitrary precision" is wrong. There is a clear and readily understandable principle of QM that states that the p and x of a particle cannot be known with arbitrary precision at an instant in time. However p and x can be known with arbitrary precision AFTER measurement (the unitary operation that transforms the system to a new state) but by then you have completely transformed the system to that of a new system. Clearly this does not apply to the above thought experiment where NO operation or measurement is being performed upon Bob's particle.
But in order for you to understand this with the necessary critical evaluation you might have to go back to pre-school quantum mechanics class where you can through around inappropriate terminology like scattering and bound states and ooooh "separation of variables". Hello, anybody in there Vanadium? This isn't a thread about solving a differential equation. Go back to kindergarten QM.

7. Apr 9, 2012

### al onestone

To San K: When I say momentum entangled I'm simply talking about the momentum conservation that occurs between all entangled pairs of particles. There is a momentum/wave vector conserving equation that determines the entanglement in all cases of "entangled particles".
With respect to your statement about inhierent randomness, I've recently had a discussion about this in a seperate thread, the conclusion is as follows. If the two particles are emmitted from a common origin at a source with a well defined position, then the uncertainty in position of the two particles is determined by the uncertainty in time of origin. When one's position is measured this knowledge indicates the exact time of origin of the particle, and so the exact position of the other particle(Bob's particle) can also be calculated in principle. For this reason we must conclude that the position distribution of Bob's particle is reduced. As a result of this the momentum uncertainty must be expanded to conform with Heisenberg uncertainty. This property should be measurable.

8. Apr 9, 2012

### DrChinese

Ouch, be careful before you blast away!

Vanadium 50 is quite correct, this is actually easy to do using entangled particle pairs. You can measure non-commuting observables to any degree of *accuracy*, spin/polarization is the easiest to discuss as the accuracy is high (but uncertainty remains complete at certain angle settings). Assume we are discussing photons, A and B, being measured by Alice and Bob at angle settings a=0 degrees and b=45 degrees and you get +1 for both.

There is nothing to stop us for *measuring* these accurately at the same time. The issue is whether A is actually in a state where it is +1 at angle settings a and b at the same time. As you would imagine, the answer is NO. If you measure A at setting b, your result will match Bob's outcome only 50% of the time (pure luck). Yet A is in a state where it is +1 at angle setting a and you can recheck that and verify as often as you desire without fear of altering A further. And similarly for B.

Last edited: Apr 9, 2012
9. Apr 9, 2012

### DrChinese

Yes, and a good try this is. But it won't actually work out as you anticipate.

You are assuming that the particle you are measuring the position on will be where you want it to be. Well, sometimes it won't be there! Once you correlate ALL of the results, you will simply have a random set of values that tells you nothing of use.

Despite what you might think, you CANNOT impart additional momentum to Bob's apparatus using this technique. If you could, you could also transfer energy using that technique (which would be nice). There was another thread in PF recently where this was discussed. See the following, which uses an idea very similar to yours:

Last edited: Apr 9, 2012
10. Apr 10, 2012

### al onestone

To DrChinese: OK, with regards to your comment "you CANNOT impart additional momentum to Bob's apparatus using this technique". I'm not trying to modify the momentum of Bob's particle by adding to it. I'm trying to modify the state of momentum of Bob's particle by expanding its distribution, its uncertainty, without changing its average momentum. The system I'm talking about is any entangled system of particles with a momentum/wave vector conserving process which determines that when you measure the position of one (Alice's) it will collapse the position of the other (Bob's) intantaneously. And these systems do exist and not just with photons. If you can predict with great accuracy the position of Bob's particle (WITHOUT measuring it) then you must admit that its positional uncertainty has reduced. Therefore we must conclude that its momentum uncertainty has increased. And this is measurable with an interferometer.

With regards to my comments to Vanadium 50, be sure you also discipline him on the matter for his harsh comments which were completely un-called for. And your polarization analysis of how one may measure commuting observables at precisely the same time, you're correct if we consider observables of separate particles, but it is impossible to measure p and x precisely for an individual particle at the same time, which is what he thought was possible("there is nothing in QM that says you cannot measure p and x of a single particle with arbitrary precision").

11. Apr 10, 2012

### DrChinese

Vanadium 50 is a mentor, so the chances of me "disciplining him are going to be vanishingly small. I personally thought his comment was fair, but hey, that's just me.

It might be possible to measure the non-commuting p and q at the same time, I'm not sure if that is possible really. But regardless, that is not what is being pointed out. What is being pointed out that if you had those values, you would still not know p and q to better than the HUP allows. You would just have a couple of numbers that would NOT accurately describe its state.

As to your scheme: it is the same thing really. What you are describing is sometimes called "quantum steering". I agree with: "If you can predict with great accuracy the position of Bob's particle (WITHOUT measuring it) then you must admit that its positional uncertainty has reduced. Therefore we must conclude that its momentum uncertainty has increased." I disagree with: "And this is measurable with an interferometer." I think there will be no discernible change for Bob to see. You can only see this when you correlate with Alice.

Although I am not familiar with the idea behind the observation at Bob. You mentioned using a band pass filter. Do you mean a filter which is tuned to a specific wavelength?

12. Apr 11, 2012

### al onestone

Your comment, "I disagree with: "And this is measurable with an interferometer." I think there will be no discernible change for Bob to see. You can only see this when you correlate with Alice".
If you admit that the position and momentum distribution of Bob's particle is changed then you have to also admit that these changes are measurable. The positional uncertainty determines the coherence length and the momentum uncertainty determines the bandwidth. Your only way out is to say that the two distributions are unchanged. You did say that "if you had those values, you would still not know p and q to better than the HUP allows. You would just have a couple of numbers that would NOT accurately describe its state." which means you've taken the position that neither distribution of Bob's particle is really changed at all. This is a more agreeable argument to me, but it assumes that measuring the position of Alice's particle(which drastically changes its position distribution) does not change the position distribution of Bob's particle. This might be correct. I don't know and this is what I'm trying to find out.
I've had similarly confusing discussions with other threads and the conclusion I've come to is that measuring Alice's particle with great certainty DOES affect the position distribution of Bob's particle. My difficulty is accepting that this change in the position uncertainty DOES NOT affect the momentum uncertainty? The discussion is not necessarily theoretical it is experimental. What would an interferometer measure with Bob's particle? What would the spectral distribution be and what would the coherence length be?
With regards to Vanadium 50 and his mentoring status I'll remind you of his mentoring quality "Frankly, your general mistake is not learning QM first before trying to find a flaw in it...I'd encourage you to work on the basics of QM. You'll make more progress once you have it solidly under your belt." I don't know about you but that does NOT sound like mentoring to me, it sounds more like an unnecessary outpouring of emotionally driven criticism.

13. Apr 11, 2012

### al onestone

Your other question "You mentioned using a band pass filter. Do you mean a filter which is tuned to a specific wavelength?" The band pass filter is a simple alternative to the interferometer. If the momentum distribution of Bob's particle changes then the interferometer will see this in the spectral distribution, and an alternative would be to use a band pass filter which only allows the original momentum distribution to pass. Assuming that the momentum distribution of Bob's particle is increased due to the position measurement of Alice's particle, if the Band pass filter only allows the original distribution to pass then the likelyhood of Bob's particle passing the filter decreases.

The whole thought experiment goes very simply;
Alice and Bob's particles are identically prepared with a very small momentum distribution and large position distribution.
Alice measures her particle's position.
The position distribution of Bob's particle is similarly reduced.
Therfore the momentum distribution of Bob's particle is increased.
This is measurable with an interferometer or a bandpass filter or a set of dichroic mirrors/beamsplitters etc.

14. Apr 11, 2012

### al onestone

Ah hah, the final resolution of the thought experiment.
DrChinese, I've got the answer that we can both agree on. Someone filled me in at another forum and the skinny goes like this;

The preparation of the entangled pair can either be of a complete momentum or a complete position entanglement. It cannot be both. The individual components do not have a finitely defined momentum when they are completely positionally entangled.
Therefore this thought experiment is invalid.

15. Apr 11, 2012

### Delta Kilo

al, it does not work

To see why, you can: (WARNING: actually involves doing some actual work!)

1) Define basis for the first particle |Ui>. second |Vj> (start with discrete sums, move to integrals later)
2) Define observable operator A with set of eigenvectors |Ai> for first particle,B and |Bj> for the second.
3) Define joint wavefunction of the two particles ψ = ƩƩ Cij |Ui>⊗|Vj>, and observables A⊗I and I⊗B.
4) Apply A⊗I to ψ, (measure first particle first) compute probabilities ρ(Ai) and corresponding ψ_i that the wavefunction will collapse to after the measurement.
5) Apply I⊗B to ψ (measure second particle first), compute probabilities ρ(Bj)
6) Apply I⊗B to each ψ_i (measure second particle after measurement of the first particle returned |Ai>), compute conditional probabilities ρ(Bj|Ai)
7) Compute apriori probability ρ'(Bj) = Ʃ ρ(Bj|Ai)ρ(Ai) of getting Bj when measuring second particle if you know that measurement of the first particle was done but the result of that measurement is not available.
8) Compare ρ(Bj) from 5 and ρ'(Bj) from 7
9) ..... ?
10) Profit!

16. Apr 11, 2012

### DrChinese

I think we are getting very close.

I believe the situation is more like this. You said: "Alice and Bob's particles are identically prepared with a very small momentum distribution and large position distribution." That means we essentially have a momentum eigenstate. You cannot have p/q entanglement when it is in a p eigenstate.

I don't believe it is technically correct that p and q cannot be simultaneously entangled. In fact, if they are p-entangled, I believe they MUST be entangled on both bases. I must admit I am not 100% certain about that. In fact, my uncertainty is greater than h/2$\pi$.

17. Apr 11, 2012

### lugita15

DrChinese, there shouldn't be any doubt in your mind about that. Entanglement occurs when the state of a two (or many) particle system cannot be written as a product state. But if you DO have a product state, i.e. a product of one-particle states, then obviously it will still be a product of one-particle states if you change the basis of some or all of the one-particle states.

18. Apr 11, 2012

### DrChinese

Ok, thanks for that.

PS Your post count is about to hit 4 digits. Better make #1000 another good one!

19. Apr 12, 2012

### al onestone

This thought experiment is invalid. For those who don't understand why yet, this is the reason.

When you have a combined state of complete momentum entanglement between two particles(1 & 2) the state description is as follows:

ψ = ∫f(p)*p>1p>2dp

where p> and x> are related by fourier relation p> = 1/√2∏∫eipxx>dx

so that

ψ = 1/2∏∫dx∫f'(x+x')eipxx>1x'>2dx'

where f'(x) is the fourier transform of f(p)

In this picture there is complete momentum entanglement but incomplete positional entanglement.

Alternatively, the state description when there is complete position entanglement is:

ψ = ∫f(x)*x>1x+a>2dx

so that

ψ = 1/2∏∫dp∫f'(p+p')eipap>1p'>2dp'

In this picture the positions are completely entangled but the momentum is not. In this state you simply CANNOT have a finitely defined uncertainty in momentum for either of the individual particles.
This is why I stated earlier "The preparation of the entangled pair can either be of a complete momentum or a complete position entanglement. It cannot be both. The individual components do not have a finitely defined momentum when they are completely positionally entangled."

Not to be alarmed though, I've failed in my attempts at superluminality thought experiments before, and my next one is sure to be perfect. You may all look forward to my next one which will be posted as a new thread within days.

20. Apr 12, 2012

### DrChinese

If you can send it superluminally, maybe it will get here a little faster.