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.

The paper I am trying to understand is here: http://pra.aps.org/abstract/PRA/v49/i2/p1473_1

The paper describes the quantum teleportation protocol in a general case with continuous dynamical variables (position and momenta) in contrast to what is always discussed here, the discrete case.

I am very confused as to what is being teleported in this this scheme. I understand that in the discrete case, its its polarization state of a photon or the spin of an electron.

What I understand is that you start out with an entangled pair of particles in position and momentum in the original EPR State: http://prola.aps.org/abstract/PR/v47/i10/p777_1

So if you are to try to teleport the position and momenta information of a particle, what is this exactly? Don't we need an origin point?

Going on with the teleportation procedure, you interact the input particle with one of the entangled pairs, now somehow an analogous bell measurement is made on the position and momentum of the combined pair?? Now you communicate that result to the other entangled particle and you do appropriate "shifts" so that the particle is now the input particle in terms of position and momentum.

Does this mean that the position and momentum distribution of one particle is teleported to another? What happens if the input particle was moving? Does this mean that after teleportation, the other particle keeps on moving too? This is extremely confusing.

share|improve this question
1  
A free arXiv version of the Lev Vaidman paper is available here. –  Qmechanic Oct 16 '12 at 20:27
add comment

2 Answers 2

up vote 2 down vote accepted

You have several questions. Let's take them one by one :

What is teleported ?

In continuous variable (theoretical) papers, position and momentum are just convenient denominations for any pair of conjugate continuous variables, in the same way than a qubit can be a 2-level atom, a polarized single photon or a spin-1/2 particle. If a quantum object has continuous unbounded degree of freedom, it behaves like the position (along a given coordinate axis), and there exists a complementary observable which behave like momentum. Current experimental realizations include the quadrature of the light, the polarization of a bright beam, collective spin of atomic clouds, etc. Therefore, I'm not sure that focussing on the position of the particle itself helps in understanding these papers.

However, since, as told above, everything is equivalent to a position, the paper should apply to the position of a particle. So in the following, I will assume we are dealing with the position $x$ along the $x$ axis, and $p$ will be the momentum along the same axis.

What origin ?

The origin does not matter, as long as it is fixed. Suppose Alice and Bob have two different labs, and Alice wants to teleport the position and momentum of a particle from her lab to Bob's lab. That means that the initial position of particle A, relative to Alice's apparatus is the same as the final position of particle B relative to Bob's apparatus. In that sense, they can have the "same" position, even if they are in different rooms.

Of course, if one want to add details one should add that :

  • Since we are in the quantum world, "the same position" means that every position measurement would lead to the same measurement statistics.
  • Thank to Galilean relativity, if Alice and Bob are moving in respect to each other, the "same momentum" is also to be understood as relative to the movement of each lab.
  • Actually, teleporting both position and momentum allows to 'completely' teleport this degree of freedom, and the statistics of any measurement (e.g Fock-state projection) will be the same.

Teleportation procedure

I tend to think that continuous variable quantum information is often easier to understand in the Heisenberg picture, where the observable operators behave almost like in classical physics

Initial state

Alice's initial particle's position and momentum are described by the operators $\hat x_a$ and $\hat p_a$. The entangled pair is described by $\hat x'_a, \hat p'_a$ and $\hat x'_b,\hat p'_b$.

For convenience, I'll define $\hat x_\pm=\frac1{\sqrt2}(\hat x'_a\pm x'_b)$ and $\hat p_\pm=\frac1{\sqrt2}(\hat p'_a\pm p'_b)$. The entanglement of the pair means that, initially $\hat x_-$ and $\hat p_+$ are both small. It is allowed only if $\hat x_-$ and $\hat p_+$ both have big fluctuations.

The Bell measurement

The bell measurement is a simultaneous measure of $\hat x_m=\hat x_a-\hat x_a'$ and $\hat p_m=\hat p_a+\hat p_a'$. This measurement is possible because both observables commute. But after the measurement induces a back action on the complementary observables ($\hat x_a + \hat x_a'$ and $\hat p_a + \hat p_a'$), which become very noisy and cannot be measured later. In particular, the state of the particle A after the measurement does not contain information about the initial state anymore

The teleportation itself

Alice tells Bob the values of $x_m$ and $p_m$, and he moves/accelerates his particle accordingly particle. After this step, we have

$$ \hat x_b'^{\text{final}}=\hat x'_b + x_m = \hat x'_b + \hat x_a - \hat x'_a=\hat x_a-\sqrt2 \hat x_- $$

$$ \hat p_b'^{\text{final}}=\hat p'_b + \hat p_m = \hat p'_b + \hat p_a + \hat p'_a=p_a+\sqrt2 \hat p_+ $$

Since both $\hat x_-$ and $\hat p_+$ are small, the final observables $\hat x'_b$ and $\hat p'_b$ correspond to the initial observables $\hat x_a$ and $\hat p_a$. The position and momentum of the first particle have been teleported onto another particle.

So if the initial particle was moving ($\hat p_a\neq0$), then the "target" particle is indeed moving at the end of the process.

share|improve this answer
    
This is the first fantastic answer! This gets right to the point of what I'm asking. So from what I understand now is that the measurement statistic is teleported. I'm imagining we have 3 Hydrogen atoms, two of them entangled with one in Lab A and Lab B, then if we interact a third Hydrogen with Hydrogen at Lab A, then the measurement statistic of that input Hydrogen will manifest at Lab B where that Hydrogen is sitting? The "position" of the Hydrogen at Lab B before teleportation will remain the same but the position statistics of the input Hydrogen is teleported? –  QEntanglement Oct 18 '12 at 1:08
    
Yes. But of course, it is only true if the "postion" at Lab B is at a different scale than the "quantum position statistics" you're trying to reproduce. –  Frédéric Grosshans Oct 18 '12 at 12:22
    
Wait, I just got confused from your comment. Can you re-explain that? So testing my understanding by asking another question... Let's say we replace the input particle to be a Hydrogen molecule. There are position statistics regarding the two Hydrogen atoms making up the molecule. Say we have TWO entangled pairs, (4 particles). Can we simultaneously interact two entangled particles with the Hydrogen molecule, and then teleport the position statistics to the two corresponding entangled atoms? Thus, turn the two atoms into a molecule? Look at my other question I asked: –  QEntanglement Oct 19 '12 at 6:57
    
add comment

I don't know the particulars of this paper, but in general, when teleportation in continuous variables is performed in quantum optics, some coherent state (or superposition of coherent states) is teleported. You start with an EPR pair of modes (this is, of course, unphysical; in practice two-mode squeezed states are used) and you perform a joint measurement on one of the EPR modes and the mode to be teleported. This is achieved by mixing the two modes and measuring the $x$ and $p$ quadratures. The result of the measurement is then transmitted to Bob, who performs a displacement that depends on the result of the measurement and thus gets the teleported state.

For better understanding, I recommend reading some papers about experiments using CV quantum teleportation such as this one.

share|improve this answer
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.