Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

It seems as though I've come across a rather unusual conclusion that could either simply be a misinterpretation or a contradictory discovery. I seem to have found a way to utilize the Heisenberg Uncertainty Principle (HUP) to our benefit to communicate faster than the speed of light (FTL). I am aware of many proposals that try to utilize entangled spin particles and try to communicate, but this schemes fail because the measurement outcome can NOT be controlled. Therefore, even if two people share an entangled pair of particles, and Alice measures spin up, although she knows Bob has a spin down particle, Bob didn't measure yet and has no way of knowing that Alice has measured her particle. Hence, FTL is impossible based on the reason that the measurement outcome can not be controlled.

This leads to my proposal of using position and momentum entangled pairs. Consider Alice and Bob hold an ensemble of entangled particles in position and momenta, where each particle is trapped in a separate harmonic potential. At some specified time agreed upon in the future, Alice measures all of her particles' position to high precision. Alice and Bob have synchronized clocks and Bob measures all of his entangled particle's momentum to whatever accuracy he chooses. Alice can calculate the average value and also the standard deviation of position. The standard deviation of position would be extremely narrow, i.e, the spread of her measurements would be very small.

From the entanglement relation, x1 = x2 and p1 = -p2, we know that when Bob measures his particle's momentum, the spread of momentum will be very large. This must be true because position and momentum cannot be measured to arbitrary accuracy at the same time. This is very similar to Einstein's proposal to violate the HUP, however I am exploiting HUP.

What this all means is that the if Bob measures a very large momentum spread, it must mean that Alice has made her measurements. If Bob measures a relatively moderate momentum spread, then he knows Alice did not measure her particles. Since the position measurement can be made to arbitrary accuracy, we are "controlling the spread or the standard deviation as the means to communicate." (MAIN RATIONALE)

Say Alice and Bob have multiple ensembles of entangled particles. Alice can relay a message by simultaneously measuring her first ensemble, meaning its a "1" and not touch her second ensemble meaning its a "0", and perhaps she chose to measure the third ensemble, "1", etc. Hence generating the series 101..., where each ensemble of entangled particles represents one bit of information.

What is flawed in this proposal? Entanglement in position and momenta is well established. We can also choose to measure the position of a particle to arbitrary precision. The HUP must hold for Bob and everyone involved.

share|improve this question
1  
You can't determine the spread of a distribution by a single measurement. You need an ensemble of many repeated measurements, but Bob has no way of knowing that Alice has done the same thing each time. Bob still has no way of knowing whether Alice made a measurement or not unless they communicate classically. –  Michael Brown Apr 4 '13 at 5:55
    
I did mention that Alice makes a measurement on an ensemble of her particles with each of her particles entangled with Bob's particles. Each ensemble represents one bit. Moreover, Alice and Bob agree ahead of time that Alice only measures position and Bob only measures momentum. They also have synchronized clocks and measure at the same time or very close in time interval. –  QEntanglement Apr 4 '13 at 6:00
    
Could you give a few formulas? The argument is rather fuzzy now. E.g., what is the state they share, and what observable(s) do they measure? –  Norbert Schuch Apr 4 '13 at 8:01
    
The only equation needed is the standard deviation equation, which everyone knows and the entanglement relation in terms of position and momentum. This is x1 = x2 and p1 = -p2, where x1 is position measurement of particle 1 in Lab A and x2 is position measurement of particle 2 in Lab B. The observables measured is the position and momentum measurement of an atom. There really isn't any other equation needed. –  QEntanglement Apr 4 '13 at 8:31
1  
The spread of position and momentum in the entangled particles would both be large. The entanglements would only show up when you go to compare notes. The proof is local commmutativity of all operators. –  Ron Maimon Apr 4 '13 at 9:58
show 1 more comment

1 Answer 1

This is a general fact independent of any particular realization of the system: if Alice cannot control the outcome of her measurements, and does not communicate the results of her measurements classically to Bob, entanglement cannot be used to communicate between the two.

Here's why. Let Alice and Bob be in control of two systems $A$ and $B$ which can be arbitrarily entangled with each other. The state is given by a density matrix $\rho_{AB}$ on the combined Hilbert space $\mathcal{H}_A \otimes \mathcal{H}_B$. It is well known that if Bob measures an observable $\mathcal{\hat{O}}_B$ ($ = \mathbb{I}_A \otimes \mathcal{\hat{O}}_B$) the expectation value is given by the trace:

$$ \langle \mathcal{O}_B \rangle = \mathrm{Tr}\left[ \mathcal{\hat{O}}_B \rho_{AB} \right] = \mathrm{Tr_B}\left[ \mathcal{\hat{O}}_B \rho_{B} \right], $$

where $\rho_B = \mathrm{Tr_A}\left[ \rho_{AB} \right]$ is the reduced density matrix of system $B$. Any measurement Bob can do can be described by this reduced density matrix.

Now suppose Alice performs a projective measurement with the outcome $a$ (could be any measured property - position, momentum, spin, whatever). This causes a "collapse of the wavefunction" which is implemented by a projection operator $\Pi_a$ which acts on $\rho_{AB}$ by

$$ \rho_{AB} \to \Pi_a \rho_{AB} \Pi_a. $$

The resulting reduced density matrix for Bob is

$$ \rho_B = \mathrm{Tr_A}\left[ \Pi_a \rho_{AB} \Pi_a \right] = \mathrm{Tr_A}\left[ \Pi_a \rho_{AB} \right], $$

using the cyclicity of the trace and the properties of a projection operator.

Now Alice can neither control the outcome $a$. Further, she does not communicate classically the result to Bob (so he can do post-selection). So the effective density matrix for Bob is the sum of all possible reduced density matrices, since there is no way he can distinguish between them without classical information from Alice. Thus

$$ \rho^{\text{eff}}_B = \sum_a \mathrm{Tr_A}\left[ \Pi_a \rho_{AB} \right] = \mathrm{Tr_A}\left[ (\sum_a \Pi_a) \rho_{AB} \right] = \mathrm{Tr_A}\left[ \rho_{AB} \right] = \rho_B, $$

exactly the same as if Alice had not done any measurements! There is no communication.

share|improve this answer
    
I am an undergraduate physics student and have not taken graduate level quantum mechanics that uses the trace and notation you've shown here. I do not follow completely, but I get the idea. Are you talking about a single position measurement or multiple measurements in an ensemble of entangled pairs? I believe that the single measurement case, you can't control the outcome. However, I'm not concerned with the single measurement outcome. I'm looking at the standard deviation of multiple position measurements. Does your answer address this and I've missed it completely? –  QEntanglement Apr 4 '13 at 8:33
1  
@QEntanglement "have not taken graduate level quantum mechanics that uses the trace and notation you've shown here" Fair enough. Density matrices are not that complicated. They tend to be more convenient than wavefunctions for entanglement problems like this, though the two formalisms are ultimately equivalent. "Does your answer address this" Yes. I didn't actually specify what systems $A$ and $B$ are, so they could just as well be ensembles. Then you could get a variance from something like $\mathcal{\hat{O}}_B \sim \sum_i \hat{p}_i^2$. –  Michael Brown Apr 4 '13 at 9:27
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.