Lets consider there is channel between Alice and Bob, sequered with quantum cryptography. To put it simple, Alice is left, Bob is right and the quantum states are 'up' ~ U and 'down' ~ D. Alice and Bob generates pair of bounded particles ,always one u one d but you dont know wtigth one flyes left or right , what fly always in both direction. Also they can capture particles. If captured, a particle is destroyed.

Alice captures the particle what flyes left and sends free what flyes to the right, Bob does the mirrored. If anyone captures the particle, he would know that other particle of the pair is.

Lets consider the MiM attacker Eve, what can do like Alice and Bob do.

Eve captures the next particle send by Alice, let it be U. So she knows, that the particle what Alice captured locally was D. But the particle U is destroyed, Bob will know that way, that there is something wrong with the channel.

Important : Eve then generates own particles, while capturing the left one. If the captured is D, Eve knows that the pair is U and it might be send further to Bob. In the model that there are only 2 states, the probability of producing the right pair is 50% , therefore Eve in average need to be capable to generate and send particles twice faster that Alice and Bob do, that generally fair possible practicly.

Alice can not distinguish between cases whether a sent particle was captured by Bob or Eve, neither Bob can distinguish, because the quantum features of particles match.

Later Eve can, as usually in MiM take over a channel and start to send false data to them.

Is there something wrong it this szenario or generaly quantum cryptography is a big talk about nothing?

  • $\begingroup$ See also crypto.stackexchange.com/a/51314 for a lot more details. Yes, QKD is a lot of hype for not a lot of value—not that it doesn't work in the sense that it fails to function, but it works only hop-by-hop over dedicated fibre links with exorbitantly expensive hardware and doesn't provide much security that we can't already get with classical cryptography end-to-end on conventional CPUs and any communication medium from copper wire to avian carrier. $\endgroup$ May 22 '19 at 4:03

Is there something wrong it this scenario?

Obviously, yes. Any such QKD system would require an authenticated channel back from Bob to Alice; that's to make sure that Bob is the one that's actually receiving the quantum signal, and not a man-in-the-middle.

More fundamentally, your system doesn't actually take advantage of any Quantum Effects; by measuring a particle, you can deduce its entire quantum state (or, at least, as much of the state that the protocol cares about); as such, it can be entirely described with classical physics.

There are actually several proposed QC systems; the simplest is probably the original BB84 system. This system has the photon being in one of four quantum states, and takes advantage of the fact that a measurement will give only 1 bit (and this is the crucial quantum effect); using the authenticated channel, Alice and Bob can coordinate which of Bob's measurements gave answers consistent with what Alice sent; because the attacker cannot measure things without losing information, any attempt by him to measure and retransmit will introduce a measurable error rate.

[Or] quantum cryptography is a big talk about nothing?

A number of people (including myself) are quite skeptical about quantum cryptography; however that doesn't have with us not thinking it doesn't work. Instead, we look at the limitations (need for special quantum channel, cost of equipment, low data rate, low range, vulnerability to side channels), and ask "what advantage does it have compared with classical cryptography (either symmetric or asymmetric) to justify dealing with these limitations".

  • $\begingroup$ 1) Eve captures particle from Alice, she knows what pair is. 2) Eve makes pair, capturing one of them in attemt to create the second one like Alice sent and send the right one to Bob, he can not distinguish is it sent from Alice or Eve cause it is identical.3) Eve does the same to channel back from Bob to Alice, 4) The normal channel could be asymetrical plus chaining : the encoding key is private key + hash of previous message, with high speed like 1000 messages per second to ensure channel consistency, plus passive sensor on wire, makes imposible to physicly do MiM without any quants $\endgroup$ May 22 '19 at 16:44
  • $\begingroup$ @user8426627: "Eve captures particle from Alice, she knows what pair is"; "...from Alice or Eve cause it is identical", well, perhaps in your system, but not in a real QC system. In QM, you cannot measure the full quantum state of a particle from a measurement, and you cannot recreate a quantum state after measuring it. You can measure it on one basis, but that irretrievably destroys any possibility of measuring it on another basis (or otherwise recreating an identical state). QC relies on this... $\endgroup$
    – poncho
    May 22 '19 at 18:09
  • $\begingroup$ why? lets see a spin: it can be only like in up and down measure by some basis, and if you capture in one spin state, you know the other's particle in pair will be opposite. Thinking Eve have the info about how Alice and Bob measure(what is up and donw for them), she can also measure that way and recreate pairs... or i dont understand something fundamentally in this stuff... $\endgroup$ May 22 '19 at 20:39
  • $\begingroup$ @user8426627: "Thinking Eve have the info about how Alice and Bob measure"; actually, what you have isn't that far from BB84; however (in that protocol) Bob doesn't announce what basis he used to measure (up/down vs right/left) until after he took his measurement. So, if the attacker tries to take a measurement, he has to guess which basis Bob will use (and get it wrong some fraction of the time); if he guesses wrong, he won't be able to regenerate a particle that Bob will measure correctly (and hence introducing errors). I suggest you review the BB84 protocol; it's about the simplest QC $\endgroup$
    – poncho
    May 22 '19 at 21:12
  • $\begingroup$ ohh, Bob announce AFTER he took his measuremen, thats the trick, finaly, thanks !! But will that be still consistent, if results of Bob are random because of different basis? $\endgroup$ May 22 '19 at 23:47

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