4
$\begingroup$

From my limited understanding of quantum entanglement, it seems like qubits act the same way as pseudo-random-number-generators (except as far as we can tell, these ones really are random). When you entangle two qubits, you give them both the same seed.

Then a measurement is like running an (otherwise deterministic) program. As long as Alice and Bob both run the same program on their two (identically seeded) PRNG's, they'll both get the same results. There's no teleportation here, they just both have access to the same information.

Why is this referred to with the fantastic name "Quantum Teleportation"? Since Alice and Bob can't pass information to each other through their PRNG's (they're just copies after all, they're not really linked).

$\endgroup$

1 Answer 1

5
$\begingroup$

You have pinpointed an important nuance of quantum information theory.

A perfectly entangled state is, in some sense, like a single bit in a one time pad: just two copies of a shared random bit. In fact, the teleportation protocol is perfectly analogous — not the same, but certainly analogous — to transmitting a message securely using a one-time pad: Alice performs a Bell-basis measurement (which corresponds to taking the parity of a data bit and a bit of the one-time pad); she communicates the result of the measurement to Bob (corresponding to transmitting the outcome of the parity computation); and Bob performs a Pauli transformation on his qubit (corresponding to a single-bit operation depending on the parity bit that he recieves). The correspondance between the protocols of teleportation and encoding with one-time pads is perfect.

The reason why teleportation is not the same as encoding with a one time pad is the same why entanglement is not the same as shared randomness: a perfectly entangled state cannot be described by random bits which describe the as-yet unmeasured outcome of any measurement. That is to say, there is no local hidden variable model for the entangled state — it is in this sense totally unlike a shared random bit. Indeed, no number of shared random bits can represent the entangled state.

Note that this lack of hidden variable models (at least, models in which some random variable is simply revealed by measurement) is not just restricted to entangled state. By the Kochen-Specker theorem, this holds even for single-spin states, at least for spin-1 particles. Furthermore, the difference between an entangled state and a merely shared random bit is expressible precisely by the correlations in the multiple degrees of freedom for measuring a single spin. So the mystery behind entanglement and teleportation is basically the same mystery as there is behind the existence of more than one incompatible basis of measurement for a single spin. It is in some sense a happy coincidence that protocols such as sending secret messages securely, as if by a one-time pad, carry through for quantum states.

$\endgroup$
4
  • $\begingroup$ When you say, "a perfectly entangled state cannot be described by random bits", that means that the qubits provide actual randomness as opposed to pseudo-randomness, right? In the sense that there's no underlying algorithm determining what's the next bit that might be measured. But how can we possibly know that? What measurement could be made that would tell whether the universe is random or just pseudo-random (following some unknown unobservable background pseudo-randomness generator). Isn't it likely the state is described by some property of the qubit that we have no way of measuring? $\endgroup$
    – dspyz
    Dec 5, 2013 at 0:50
  • 1
    $\begingroup$ No: I mean that there is no way to model the entangled state as involving any randomly generated bits (using 'genuine' randomness or otherwise) for which the measurement outcome is a function of those random bits. Not only does entanglement produce 'genuine' randomness, it produces randomness by a means which does not have any analogue in classical random variables, and the reason for this is ultimately because it would also produce a uniformly random outcome for any choice of measurement. $\endgroup$ Dec 5, 2013 at 1:31
  • $\begingroup$ In short: "Isn't it likely the state is described by some property of the qubit that we have no way of measuring?" No --- and that is precisely what is meant by saying that it cannot be described by local hidden variables. $\endgroup$ Dec 5, 2013 at 1:32
  • 1
    $\begingroup$ You really need to steep yourself in quantum mechanics in order to understand the distinction, Bell's theorem might help. $\endgroup$ Dec 5, 2013 at 1:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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