2
$\begingroup$

suppose that quantum teleportation of a macroscopic composite object, like say, a grain of rock, was physically and technologically possible

The traditional teleportation scheme would have to prepare a pair of entangled 'blank' systems ahead of time (A and B), and then make a complete set of 'comparison measurements' between the grain to be teleported and A, such that the measurements do not measure the individual states of the grain, but only how they compare to a "corresponding" (whatever that might mean) state of the blank system A.

So, the information about comparisons is sent back to the carrier of blank system B via a classical information channel, which then is used somehow to complete the teleportation of the grain object at B

This part is what I don't understand: how does the carrier of B uses the classical information to 'restore' the grain into its blank system B?

Does the carrier have to actively interact with their blank system B in some specific way prescribed by the classical information?

$\endgroup$
2
$\begingroup$

Even without sending classical infotmation - the state was completley teleported.

Unfortunatley, by teleportation identity - after teleporting state $ |\psi\rangle$ at A you get a state $U^{\mu \nu}|\psi\rangle$ at B (where $U^{\mu \nu} $ is a unitary matrix which depend on mesurement results at A), i.e. the state was teleported to B and transformed to another basis (via $U^{\mu \nu} $). Without sending the classical information - you can't determine at B which matrix $(U^{\mu \nu})^\dagger $ to apply on the state to recover it. You can think of it as the teleported state is incripted, and trying to measure it (before applying $(U^{\mu \nu})^\dagger $) probably will ruin it and won't get the original teleported state.

In practical use, the classical information tell you what basis to use (which measurements to use) to recover the teleported state.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ does the number of nondiagonal basis entries of $U^{\mu \nu}$ grow in linear proportion with the degrees of freedom of the teleported grain? or a mol worth of teleported matter will require less than a mol worth of classical basis information? $\endgroup$ – diffeomorphism Sep 28 '15 at 21:14
  • $\begingroup$ I'm not sure what you're asking. $U^{\mu \nu}$ are known matrices, calculated via the teleportation identity. The difficulty is to know which one to use to recover the teleported states (there may be more than 2 indices to them).Theoretically, with infinite precision instruments, you can encrypt infinite amount of information (in amplitudes and phases) even in a single qubit and recover the information by very low number of classical bits. See dense coding. $\endgroup$ – Alexander Sep 28 '15 at 21:28
  • $\begingroup$ I'm asking how does grow the amount of information required to encode $U^{\mu \nu}$, when the number of degrees of freedom on the teleported object grows? $\endgroup$ – diffeomorphism Sep 28 '15 at 21:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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