3
$\begingroup$

Just a random stupid thought.

Quantum entanglement cannot be used to transfer information faster then speed of light. But can it "embed" information until a key is received?

For exmample, assume Alice and Bob both have pairs of entangled particles:

  1. Alice & Bob measure their particles
  2. Alice send Bob, in classical channel, some information about her measurement results
  3. Bob use the received information to "decode" something from his measurement results.

Now if the information from Alice is less then information Bob extracted, then Shannon–Hartley theorem can be broken....?

(Since the information Bob received+extracted can then be greater than the limit of the classical channel)

$\endgroup$
4
$\begingroup$

A protocol of the form you have described does exist and is known as superdense coding. It allows Alice to transmit two classical bits to Bob by sending only one qubit through a quantum channel, providing that they already share an entangled state. However, this requires that they already share an entangled state. In other words, at some point in the preparation, one of them would need to create the entangled state and send half of it to the other using a quantum channel. After we take this into account we see that using this protocol we can send 2 classical bits with two uses of a quantum channel, but one of these uses can be done beforehand.

Can we do better than this (ie send more classical bits to Bob)? A result known as Holevo's bound says no. This result tells us that the maximum amount of classical information that can be extracted from a qubit is one classical bit. This is where the Bob's 'extra' bit of information came from.

How does this relate to the Shannon-Hartley theorem? I'm not too familiar with this theorem, but looking at wikipedia it seems to only deal with communication through classical channels. Naively, superdense coding would allow you to send 2 bits of information with one use of a quantum channel, 'breaking' the Shannon-Hartley theorem. However, as we have seen, this relies on using a quantum channel to exchange quantum information between Alice and Bob. Since Shannon-Hartley does not deal with quantum channels, it doesn't really apply in your situation.

Hope this helps!

$\endgroup$
  • $\begingroup$ Wow. I thought I am crazy. So they can prepare a large amount of entangled particles first and exploit it when needed to get a 2x speed boost ? $\endgroup$ – somebody4 Jul 10 '19 at 16:19
  • 1
    $\begingroup$ Yes, exactly! But they need a way of storing their entangled states for until they need them, which is possible in principle, but difficult in practice. Also, when they've measured their states, they have 'used them up', so will have to prepare more entangled states if they want to do it again. $\endgroup$ – asph Jul 10 '19 at 17:28
  • 1
    $\begingroup$ Superdense coding doesn't require the use of a classical channel. It uses a quantum channel once ahead of time to establish entanglement, and then once the 2-bit message is known it only needs to use the quantum channel once more (instead of twice). $\endgroup$ – Craig Gidney Jul 11 '19 at 4:50
  • $\begingroup$ Apologies, you're absolutely correct. I will edit the answer to reflect this later today. $\endgroup$ – asph Jul 11 '19 at 9:43
  • $\begingroup$ get a 2x speed boost?? No. One qubit is not one classical bit, nor is one qubit less than two classical bits. They simply cannot be compared like that. Sending 2 classical bits is much less demanding on hardware than sending one qubit in such a way as to take advantage of dense coding. $\endgroup$ – Andrew Steane Jul 17 '19 at 13:31

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.