Quantum Key Distribution confusion.

Click For Summary
Quantum Key Distribution (QKD) using the E91 protocol allows Alice and Bob to share a secret key through the measurement of photon polarizations. They send pairs of photons with the same polarization, and by randomly changing their measurement orientations, they determine when their results align to establish shared bits. However, simply knowing the binary values does not constitute a secure message; the key must be used with a strong encryption algorithm. Using encryption methods like Blowfish or AES with the derived key enhances security, preventing potential eavesdropping. Thus, the combination of QKD and robust encryption ensures safe communication.
Silversonic
Messages
121
Reaction score
1
Hey, I'm slightly confused on a part of the process of Quantum Key Distribution (E91 protocol).

I have something which tells me how Alice would send a message to Bob. An atomic transition from an outside source would produce 2 photons both with the same circular polarisation. One of these is sent to Alice, the other to Bob. Alice and Bob have analysers to measure the vertical (V) or horizontal (H) polarization of the photons and they get the same answer with 100% probability (supposing there are no eaves droppers).

They could also measure diagonal (D) and antidiagonal (A) polarisations using a different orientation of their analysers. Alice and Bob change the orientation of their analysers randomly for each photon received.

These are the binary keys;

V (vertical): Binary 1
H: Binary 0

D: Binary 1
A: Binary 0.

So after the outside source sends all the photons, Alice and Bob exchange information on the the orientation of their analysers for each photon pair received to deduce when both of them had the same orientation, but NOT the results of the measurements. So for each photon pair where the orientation of the analysers was aligned the same, Alice and Bob have a shared secret bit (which is the value of whatever polarisation state of the photon they got).

But that's all I'm left with.

So exactly, how has Alice sent a secret message to Bob? All that's happened is they've both been sent a continuous stream of photon pairs, and now they know when BOTH of them had binary 1 and when BOTH of them had binary 0. So after Alice is sent Bob's orientation information for each photon pair, she can deduce the Binary keys for each pair. For example;

Pair 1: 1
Pair 2: 0
Pair 3: -
Pair 4: -
Pair 5: 1
Pair 6: 1
Pair 7: 0
Pair 8: 1
Pair 9: -

"-" corresponds to when Alice and Bob had different orientations and therefore there is no secret key between them. What would happen next? Would Alice then send a subsequent message to Bob saying "Look at Pair 2, 5 and then 8" and then Bob would deduce the secret binary key "011"?Edit: After thought I realize I'm probably right in this assumption. If there was a delete topic button I'd take this down.
 
Last edited:
Physics news on Phys.org
The bits where orientations were the same give you a key. How you use the key to encode information is entirely up to you. Keep in mind, however, that different encryption has different security. What you suggest is not terribly strong. Because if you say that bit N of the message is bit K of the key, and later use the same bit K of the key to send bit M of the message, Eve knows that M and N are the same bits without knowing the key. With enough correlations, she can figure out the key and crack the message entirely by cryptography attack that has nothing to do with your key exchange.

What you need is an encryption algorithm that is secure in its own right. Use something like Blowfish or AES with the key you got, and send the encrypted message that way. This way, the complexity of getting the cipher cracked is complexity of corresponding algorithm with key being completely secret. Both of these are considered secure, so combined with quantum key distribution, it gives you a safe communication protocol.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

Similar threads

  • · Replies 2 ·
Replies
2
Views
476
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
17
Views
630
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 24 ·
Replies
24
Views
4K